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Students can Download Chapter 4 The Theory of The Firm Under Perfect Competition Questions and Answers, Plus Two Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Economics Chapter Wise Questions and Answers Chapter 4 The Theory of The Firm Under Perfect Competition

Plus Two Economics The Theory of The Firm Under Perfect Competition One Mark Questions and Answers

Question 1.
TR curve under perfect competition passes through the origin. Do you agree?
Answer:
Yes, I agree.

Question 2.
Pick the odd one out.

1. Homogeneous products, free entry, and exit, price maker.
2. Homogeneous product, freedom of entry and exit, large number of buyers and sellers, product differentiation.
3. MC = MR, MR = AFC, MC cuts MR from below, P = AC.

Answer:

1. Price maker. Others are features of perfect competition.
2. Product differentiation. Others are features of perfect competition.
3. MR = AFC. Others are conditions for firm’s equilibrium under perfect competition.

Question 3.
Pick out the odd one and justify it.
Free entry, profit maximization, perfect knowledge, price discrimination.
Answer:
Price discrimination. Others are features of perfect competition.

Question 4.
Which of the following represents normal profit?
(a) MR = MC
(b) AR = AC
(c) TR > TC
(d) AR < AC
Answer:
(b) AR = AC

Question 5.
Firm is price taker under:
(a) perfect competition
(b) monopoly
(c) monopolistic competition
(d) duopoly
Answer:
(a) perfect competition Question

6. Shut down point occurs at:
(a) rising part of
(b) following part of AVC
(c) minimum point AVC
(d) none of these
Answer:
(c) minimum point AVC

Question 7.
A firm is able to sell any quantity of the good at a given price. The firm’s marginal revenue will be
(a) Greater than AR
(b) Less than AR
(c) Equal to AR
(d) Zero
Answer:
(c) Equal to AR

Question 8.
The short run shut-down point of a firm in a perfectly competitive firm is.
(a) P = AVC
(b) P = AC
(c) P > AVC
(d) P < AVC
Answer:
(a) P = AVC

Question 9.
If P exceeds AVC but is smaller than AC at the best level of output, the firm in perfect competition is
(a) Making profit
(b) Minimizing losses in the short run
(c) Incurring a loss and should stop producing
(d) Breaking even
Answer:
(b) Minimizing losses in the short run

Plus Two Economics The Theory of The Firm Under Perfect Competition Two Mark Questions and Answers

Question 1.
A firm cannot make supernormal profit in the long run under perfect competition. Do you agree? Substantiate your answer.
Answer:
Yes, I do agree to the statement that a firm cannot make supernormal profit in the long run under perfect competition. This is because freedom of entry will prevent supernormal profit in the long run.

Question 2.
Define ‘Break even point’.
Answer:
The point on the supply curve at which a firm earns normal profit is called the break even point. The point of minimum average cost at which the supply curve cuts the LRAC curve is, therefore, the break-even point of a firm.

Question 3.
Examine the difference between the short run price and the long run price of a firm under perfect competition.
Answer:
There is a major difference between the short run and long run price under perfect competition. In the short run, price should be equal to or greater than the minimum AVC. If the price falls below this level, the firm will shut down production. On the other hand, the long run price should be equal to or greater than the minimum AC. Below this level, the firm will shut down production.

Question 4.
What is the supply curve the firm in the long run?
Answer:
In the long run, a firm will produce output only when its price is at least equal to the average cost of production. Therefore, average cost curve represents the supply curve of the firm.

Question 5.
How does technological progress affect the supply curve of the firm?
Answer:
The supply curve of the firm slopes upward from left to right indicating a direct relationship between price and quantity supplied. When there is technological progress, it will affect the supply. Now the firm will be in a position to produce and supply more output at the same price. Therefore the supply curve will shift towards the right side.

Question 6.
How does an increase in the number of firms in a market affect the market supply curve?
Answer:
As the number of firms changes the market supply curve shifts. When the number of firms increases the market supply curve shifts to the right. On the other hand, if the number of firms decreases, the market supply curve shifts to the left.

Question 7.
Define ‘shut down point’
Answer:
Shut down point refers to a situation where average revenue is equal to average variable cost. If the price fails to cover even average variable cost, firm will stop its production.

Question 8.
Make pairs:
Perfect competition, price marker, oligopoly, Cournot model, price taker, monopoly.
Answer:

• Perfect competition – price taker
• Monopoly – price maker
• Oligopoly – Cournot model

Question 9.
The diagram below shows two curves faced by a firm under perfect competition. Name them.

Answer:

1. Total revenue curve
2. Average/Marginal revenue curve

Question 10.
Choose the correct answer from the given multiple choices.
a. Identify the equilibrium condition of a firm under perfect competition.

1. AC=MR, & AC cuts MR from above.
2. MC=MR, & AC cuts MR from below.
3. AC=MR, & MC cuts MR from below.
4. MC=MR, &MC cuts MR from above.
5. MC=MR, & MC cuts MR from below.

b. The demand for the product of a firm is perfectly elastic in one of the following markets.
Identify the market.

1. monopoly
2. monopolistic competition
3. perfect competition
4. monopsony
5. oligopoly

Answer:
a. 5
b. 3

Plus Two Economics The Theory of The Firm Under Perfect Competition Three Mark Questions and Answers

Question 1.
Match the following.

Answer:

Question 2.
A firm maximizes profit when the difference between total revenue and total cost is the maximum. Profit is maximized when certain conditions are satisfied. Do you agree?
Answer:
Yes.
A firm maximizes profit when the following three conditions are satisfied.

1. The market price, p, is equal to the marginal cost.
2. The marginal cost is nondecreasing.
3. In the short run, the market price must be greater than or equal to the average variable cost. In the long run, the market price must be grater than or equal to the average cost.

Question 3.
Write the economic terms

1. Cost vary with output
2. Total Cost divided by quantity
3. TR_n – TR_n-1
4. TR = TC point
5. A case where an increase in all the inputs lead to a just proportionate increase in output.

Answer:

1. Variable cost
2. Average cost
3. Marginal revenue
4. Breakeven point
5. Constant returns

Question 4.
Perfect competition does not exist in the real world. Do you agree? Substantiate your view.
Answer:
Yes, I agree to the statement that in the real world perfect competition does not exist.

This is because, it is rare to find the features of perfect competition especially the features like perfect knowledge, perfect mobility of factors and product, etc. What we really find in the world is monopolistic competition which is a mix of perfect competition and monopoly.

Question 5.
A firm’s supply curve in the short run is the rising part of the SMC curve. Why?
Answer:
A firm under perfect competition in the short run will start supplying only if the price is equal to or greater than the short run average variable cost. Therefore, the rising part of the short run marginal cost which begins with the minimum SAVC is the supply curve of the firm in the short run.

Question 6.
Imagine that S° is the original supply curve of the firm. If a unit tax is imposed, what happens to the supply curve? Show the change in a diagram.
Answer:
If a unit tax is imposed, firm’s long run supply curve shifts to the left. It is shown in the following diagram.

Question 7.
Fill in the blanks.

1. Long run price under perfect competition will be equal to ……………….
2. Maximum price fixed for a product by the government is called ………..
3. The point denoted by the minimum of AVC is called …………

Answer:

1. Average cost
2. Price ceiling
3. Shutdown point

Question 8.
Choose the appropriate answer.
a. A fall in supply caused by a fall in price is known as:

1. Contraction of supply
2. Expansion of supply
3. Increase in supply ‘

b. When supply curve is a vertical straight line, supply is

1. Perfectly elastic
2. Perfectly inelastic
3. Unitary elastic

c. The price under perfect competition. short run should be at least equal to

1. Short run MC
2. Short run AC
3. Short run AVC

Answer:
a. Contraction of supply
b. Perfectly inelastic
c. Short run AVC

Question 9.
List the factors affecting elasticity of supply.
Answer:
The factors affecting elasticity of supply are

1. nature of the commodity
2. cost of production
3. time period
4. technique of production

Question 10.
Observe the following figures and answer the questions.

1. Point out the elasticities on the above supply curves.
2. Which method is applied here?

Answer:
1. supply curves:

• S₁ Elastic supply
• S₂ Unitary elastic supply
• S₃ Inelastic supply

2. Geometric method is applied here.

Question 11.

1. Identify the market structure that is represented by their curve in the diagram.
2. Explain why the AR curve is horizontal

Answer:
1. Perfect competition

2. It is assumed that under perfect competition compared to the industry the share of each firm is meager. No firm can influence the market supply. So even if a firm doubles the quantity supplied the market supply will not change. The price remains the same. So theARorthe demand curve is horizontal.

Question 12.
Information about a firm is given in the following

Find out the equilibrium level of output in terms of MC & MR. Give reasons for your answer.
Answer:

Equilibrium quantity is 4. At this level of output MC = MR & MC cuts MR from below.

Question 13.
A firm can sell any quantity of the output it produces at a given price. If so, what is the behaviour of marginal revenue and average revenue? Draw the two curves in a single diagram.
Answer:
The demand curve facing the firm is perfectly elastic. At this condition AR=MR=Price

Plus Two Economics The Theory of The Firm Under Perfect Competition Five Mark Questions and Answers

Question 1.
Consider the following table

1. At which production level there will be no profit or loss to the producer?
2. Comment on the profit and loss conditions as TC 1000 and TR750.

Answer:
1. At the production level of 100 units of output, the producer incurs 1500 TC and 1500 TR. So there will be no profit or loss to the producer

2. When TC =1000
TR = 750
Loss = TC-TR
= 1000-750 = 250

Question 2.
How would each of the following affect the market supply curve for wheat?

1. A new and improved technique is discovered.
2. The price of fertilizers falls
3. The government offers new tax concessions to farmers
4. Bad weather affects the crops.

Answer:

1. A new and improved technique is discovered: increase market supply
2. The price of fertilizers falls: increase market supply
3. The government offers new tax concessions to farmers: increase market supply
4. Bad weather affects the crops: decrease market supply

Question 3.
The following table shows the total cost schedule of a competitive firm. It is given that the price of the good is ₹10. Calculate the profit at each output level. Find the profit maximizing level of output.

Answer:

Question 4.
Explain briefly break even point with the help of an example.
Answer:
XUS Break even point refers to a situation when total revenue is equal to total cost assuming a given selling price per unit of output.
TR = TC
This can be explained with the help of an example.

If the firm produces less than 200 units, itsTR< TC. So it bears loss. If the firm produces more than 200 units, its TR > TC. So it earns profit.

If the firm produces 200 units, its TR = TC. This is the breakeven point. Break even point is a normal profit point as beyond it the firm earns super normal profits and below it, the firm incur losses.

Question 5.
Choose the correct answer
a. profit of a firm is the revenue earned:

1. zero of cost
2. net of cost
3. gross of cost
4. none of these

b. TMC curve cuts LAC curve :

1. at minimum point
2. at maximum point
3. below the LAC curve
4. none of these

c. under perfect competition, firm is :

1. price taker
2. price maker
3. both 1 and 2
4. none of the above

d. MR can be negative but AR is:

1. negative
2. positive
3. either positive or negative
4. none of the above

Answer:
a. net of cost
b. at minimum point
c. price taker
d. positive

Question 6.
Observe the following diagram.

Plot the break even point and shut down point.
Answer:

• Point ‘A’ represents shut down point
• Point ‘B’ represents the break even point.

Question 7.
Complete the following tables and identify the market structure


Answer:

This market represents a perfectly competitive market because in this market, P = AR = MR.

This market represents a monopoly market. In this market, AR and MR are different and AR > MR.

Question 8.
At the market price of ₹10, a firm supplies 4 units of orange. The market price rises to ₹30. The price elasticity of the firm’s supply is 1.25. What quantity will the firm supply at the new price?
Answer:
In the given example,
P = 10 Q = 4 P1 = 30 es = 1.25
ΔP = 30-10 = 20
Applying these values in the formula, we get, \(e_{s}=\frac{\Delta Q}{\Delta P} \times \frac{P}{Q}\)

ΔQ =1.25×8=10

Question 9.
Suppose that the market demand in a perfectly competitive industry is given by, Qd = 7000 – 500 p and the market supply function is given by, Qs = 4000 +250 P. Find the market equilibrium price.
Answer:
Equilibrium is determined by the condition,
Qd = Qs.
In this example,
7000-500 p = 4000 +250 p
7000-4000 = 250 p + 500 p
3000 = 750 P
∴ \(P=\frac{3000}{750}=4\)
Therefore, the equilibrium price in the market is ₹4.
7000 – 500 p = 4000 + 250 p
7000 – 4000 = 250 p + 500 p
3000 = 750 P
∴ \(P=\frac{3000}{750}=4\)

Question 10.
Point out the features of perfect competition.
Answer:
Features of perfect competion are pointed out below.

1. Large number of buyers and sellers
2. Homogeneous product
3. Freedom of entry and exit
4. Free movement of product and factors of production.
5. Profit motive.
6. Perfect knowledge of market conditions
7. Absence of transport cost.

Question 11.
Identify from the diagram below.

1. Shutdown point and breakeven point.
2. Distinguish between these two.

Answer:

1. c is the break even point. b is the shut even point

2. Break down point shows a situation where a firm earns no profit or no loss. It is the point where AR = AC. Shut down point shows a situation where a firm is compelled to stop the production since it is not able to cover its variable cost. This is the situation where the firm’s P > AVC.

Question 12.
Identify the profit maximizing level of output from the diagram below. List out the condition for profit maximization.

Answer:
b is the profit maximizing level of output.
MR = MC
At the profit maximizing level of output MC is non decreasing, that is the slope of MC is positive.
P > AVC.

Question 13.
Match the commodities given below with the diagram. Justify your answer.

Answer:
a. Mobile Phone

b. Coconut
The supply of mobile phones are relatively price elastic. This is because the purchase of mobile phones can easily react to a change in price. It is a manufactured good. All the raw materials needed to produce a mobile phone can easily be available. The producers can pile up stock and when price increases they can increase the supply.

The supply of coconut can be price inelastic. This is because the producers cannot easily react to a change in the price of coconut. Years needed for a coconut tree to get mature and start to produce coconuts. It is an agricultural product. The producers cannot hold the stock of coconut fora longer duration.

Question 14.
The following table gives you certain information about a firm.

1. Find the price at which output is sold and identify the form market,
2. Is this firm a price-taker or price-maker? Give reasons.
3. Find the firm’s equilibrium level of output in terms of MC &MR. Give reasons.
4. Also find profit of the firm at this level of output.

Answer:

1. Price = 8
2. Price-taker, a firm has no influence on price determination under perfect competition.
3. Equilibrium quantity is 3. At this level of output MC=MR.
4. Firm is at no profit-no loss condition, i. e., he is at break even point.

Plus Two Economics The Theory of The Firm Under Perfect Competition Eight Mark Questions and Answers

Question 1.
Prepare a seminar report on ‘perfect competitive markets’
Answer:
Respected teachers and dear friends,
The topic of my seminar paper is perfect competitive markets. The term market refers to all the places in which buyers and sellers are in contact with each other for the purchase and sale of any commodity or service. There are different kinds of markets based on their characteristics – say perfect competitive markets and noncompetitive markets.

Introduction:
Perfect competition is a market situation characterized by the existence of large number of buyers and sellers, homogeneous products, free mobility of factors of production, freedom of entry and exit, perfect knowledge of market conditions and absence of transportation costs.

Contents:

1. Profit Maximisation
2. Profit Maximisation in short-run: Diagrammatic representation
3. Long run profit maximisation

1. Profit Maximisation:
The main objective of the producer is to maximize the profit levels of his firm. The output level at which the firm maximizes the profit is called the equilibrium of the firm. The profit level of the firm is the difference between Total Revenue and Total Cost. Symbolically it is represented as X = TR – TC.

The firm under perfect competition maximizes its profit under three conditions. They are:

• The MC must be equal to MR (MC = MR)
• MC must be non-decreasing. It means that MC curve should cut the MR from below.
• Third condition has two parts, one for short-term and the other for long-term.

a. In the short run, price should be more than or equal to the minimum point of Average Variable Cost (AVC). It can be denoted as P> AVC.

b. In the long run, price should be more than or equal to the minimum point of Average Cost (AC). It can be denoted as P > AC.

2. Profit Maximisation in short run: Diagrammatic representation:
The profit maximizing condition of a firm in short-run can be understood from the figure. All the three profit maximizing conditions of a firm in short run are satisfied at point of output level q0.

• The price must be equal to MC (P = MC)
• MC must be non-decreasing.
• P > AVC

All the profit maximizing conditions are satisfied in the figure given above.

1. The prices become equal to MC. At point E, the firm reaches equilibrium at the output level q (MC = MR).
2. At q MC is non-decreasing at point E.
3. At point E, prices have become more than AVC (P > AVC at point E).

Therefore, three conditions for the equilibrium level of output are depicted in a single figure.

3. Long run profit maximisation:
The profit maximization level of the firm is reached when the long run supply curve of the firm is that portion of LRMC curve which lies over and above the minimum point of LRAC curve (P > minimum of LRAC). The supply curve of a firm, in the long run, is the rising portion of minimum point of LMRC. It can be explained with the help of the following figure.

Conclusion:
Thus it can be concluded that perfect competition is a market structure characterized by complete absence rivalry among individual sellers. Sellers in the market do not hold any freedom of influencing market prices. Therefore, they are known as price takers. However, it should be admitted that it is a rare form of market.

Question 2.
The diagram shows one of the short run equilibrium situations of a firm under perfect competition.

1. List out any four features of such market condition.
2. Identify the equilibrium situation of the firm profit, normal profit, loss diagrammatically. Show a situation in which the firm makes a profit.
3. With the help of a diagram explain the long run situation of a firm under perfect competition.

Answer:
1. Market Conditions:

• Large number of sellers
• Homegeneous product
• Firm is a price taker
• Free entry and exit

2. The firm is making a normal profit.
The firm produces ‘oq’ level of output and charges a price ‘op’. The shaded area in the diagram shows profit.

3. Firm under perfeert competition in the long run makes only normal profit. This is because if firms are making profit, new entrants will be attracted into the industry.

The price falls due to increase in supply and the extra profit will be taken away. If the firms are making loss some of them will leave the industry, price rises and the loss will be turned into profit. This is shown in the diagram below.

The firm under perfect competition in the long run will produce ‘oq’ level of output and charges a price ‘op’.

Question 3.
How are price and output determined under perfect competition in the short run? Compare the profit of a firm in the short run and long run and bring out the difference. Give suitable diagram.
Answer:
Under perfect competition, market determines the price – price taker and not price maker firms produce the output that maximises its profit – a firm may get abnormal profit or normal profit or loss in the short run – only normal profit will prevail in the long run – due to free entry and exit – Draws separate diagram for short run and long run-explains price and output determination as well as profit in short run and long run.

PROFIT MAXIMIZATION IN SHORT-RUN:
DIAGRAMMATIC REPRESENTATION:
The profit maximizing condition of a firm in short-rn can be understood from the figure. All the three profit maximizing conditions of a firm in short run are satisfied at point of output level q₀.

1. The price must be equal to MC (P = MC)
2. MC must be non-decreasing.
3. P > AVC

All the profit maximizing conditions are satisfied in the figure given above.

1. The prices become equal to MC. At point E, the firm reaches equilibrium at the output level q (MC = MR). 0
2. At q MC is non-decreasing at point E.
3. At point E, prices have become more than AVC (P > AVC at point E).

Therefore, three conditions for the equilibrium level of output are depicted in a single figure.

LONG RUN PROFIT MAXIMISATION:
The profit maximization level of the firm is reached when the long run supply curve of the firm is that portion of LRMC curve which lies over and above the minimum point of LRAC curve (P>minimum of LRAC). The supply curve of a firm, in the long run, is the rising portion of minimum point of LMRC. It can be explained with the help of the following figure.

Students can Download Chapter 7 Alternating Current Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 7 Alternating Current

Plus Two Physics Alternating Current NCERT Text Book Questions and Answers

Question 1.
A 100Ω resistor is connected to a 220V, 50 Hz ac supply.

1. What is the rms value of current in the circuit?
2. What is the net power consumed over a full cycle?

Answer:
Given R = 100Ω, E_ν = 220V, ν = 50 Hz.
1. Since l_ν = \(\frac{E_{ν}}{12}\)
So l_ν = \(\frac{220}{100}\) = 2.2 A

2. P_aν = E_ν I_ν = 220 × 2.2
or P_aν = 484 W.

Question 2.

1. The peak voltage of an ac supply is 300V. What is the rms voltage?
2. The rms value of current in an ac circuit is 10A. What is the peak current?

Answer:
1. Given E₀ = 300 V, E_aν = ?
Since E_ν = \(\frac{E_{0}}{\sqrt{2}}\) = 0.707 × 300
or E_ν = 212.1V.

2. Given I_ν = 10A, I₀=?
Since l₀ = \(\sqrt{2}\) E_ν = 1.414 × 10
or I = 14.14 A.

Question 3.
A 44 mH inductor is connected to 220V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Answer:
Given L = 44 mH = 44 × 10^-3H
E_ν = 220V, ν = 50Hz, I_ν = ?
Since I_ν = \(\frac{E_{ν}}{x_{L}}=\frac{220}{\omega L}\)

or Iν = 15.9A.

Question 4.
A 60µF capacitor is connected to a 11OV, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Answer:
Given C = 60µF = 60 × 10^-6F, E_ν = 110V, ν = 60 Hz.
Since I_ν = \(\frac{E_{ν}}{x_{c}}\)
∴ I_ν = ωCE_ν
= 2pνCE_ν = 2 × 3.142 × 60 × 60 × 10^-6 × 110
= 2.49A or
I_ν = 2.49V.

Question 5.
In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
Answer:
In both the cases the net power consumed is zero because in both the cases.
Net power consumed P = E_ν l_νcosΦ
and Φ =90°
∴ P = 0 (in each case).

Question 6.
Obtain the resonant frequency of a series LCR circuit with L=2.0H, C=32µV and R = 10?. What is the Q-value of this circuit?
Answer:
Given L = 2.0H, C = 32µF = 32 × 10^-6F
R = 10Ω, Q = ?, ω₀ = ?
Resonant frequency

Question 7.
A charged 30µF capacitor is connected to a 27mH inductor. What is the angular frequency of free oscillations of the circuit?
Answer:
Given C = 30µF=30 × 10^-6F,L = 27mH = 27 × 10^-3H
ω₀ = ?

or ω₀ = 1.1 × 10^-3S^-1.

Plus Two Physics Alternating Current One Mark Questions and Answers

Question 1.
Which type of transformer you use to operate the coffee maker at 220 V?
Answer:
Step down transformer.

Question 2.
In an A C. circuit, I_rms and I_o are related as.

Answer:
(d) I_rms = \(\frac{I_{0}}{\sqrt{2}}\)

Question 3.
A capacitor of capacitance C has reactance X. If capacitance and frequency become double then reactance will be
(a)  4X
(b) X/2
(c) X/4
(d) 2X
Answer:
(c) Explanation

Question 4.
Fill in the blanks

• Impedance: admittance
• ……………..: conductance

Answer:
Resistance

Question 5.
Why it is better to use an inductor rather than a resistor to limit the current through the fluorescent lamp?
Answer:
No power is developed across the inductor as heat.

Question 6.
In an a.c circuit with phase voltage V and current I, the power dissipated is
(a) V.l
(b) Depends on phase angle between V and I
(c) \(\frac{1}{2}\) × V.I
(d) \(\frac{1}{\sqrt{2}}\)
Answer:
(b) Depends on phase angle between V and I

Plus Two Physics Alternating Current Two Mark Questions and Answers

Question 1.
Fill in the blanks.

Answer:
(i) Current lags by π/2
(ii) X_c = 1/cω
(iii) R
(iv) Phase difference is zero.

Question 2.
A.C. adaptor converts household ac into low voltage dc. A stepdown transformer is a essential part of ac adapter.

1. What is the use of step down transformer?
2. What is the principle of a transformer? Explain.

Answer:

1. To decrease voltage
2. It works on the principle of mutual induction.

Plus Two Physics Alternating Current Three Mark Questions and Answers

Question 1.
Match the following

Answer:

Question 2.

The following figure is a part of a radio circuit.

1. Identify the circuit.
2. What happens to this circuit if X_L = X_C
3. lf X_C > X_L draw the phaser diagram.

Answer:
1. LCR circuit.

2. When X_L = X_C, the impedance of circuit becomes minimum and the current corresponding to that frequency is maximum.

3.

Plus Two Physics Alternating Current Four Mark Questions and Answers

Question 1.
Figure below shows a bulb connected in an electrical circuit.

1. When the key is switched ON the bulb obtains maximum glow only after a shorter interval of time which property of the solenoid is responsible for the delay?

• Self-induction
• Mutual Induction
• Inductive reactance
• None of the above

2. If the flux linked with the solenoid changes from 0 to 1 weber in 2 sec. Find the induced emf in the solenoid.

3. If the 3v battery is replaced by an AC source with the key closed, what will be observation? Justify your answer.
Answer:
1. Self-induction.

2. \(\frac{d \phi}{d t}=\frac{1}{2}\) = 0.5V.

3. When AC is connected the brightness of bulb will be decreased. This is due to the back emf in the circuit.

Question 2.
A graph connecting the voltage generated by an a.c. source and time is shown.

1. What is the maximum voltage generated by the source?
2. Write the relation connecting voltage and time
3. This a.c. source, when connected to a resistor, produces 40J of heat per second. Find the equivalent d.c. voltage which will produce the same heat in this resistor.

Answer:

1. 200v.
2. V = V₀sin ωt
3. \(V_{\max }=\frac{V_{0}}{\sqrt{2}}=\frac{200}{\sqrt{2}}\) = 141.8v.

Question 3.
An inductor, capacitor, and resister are connected in series to an a.c. source V = V₀sin ωt.

1. Draw a circuit diagram of L.C.R. series circuit with applied a.c. voltage.
2. Find an expression for impedance of L.C.R. series circuit using phasor diagram.
3. What is impedance of L.C.R. series circuit at resonance?

Answer:
1.

2.

For impedance of LCR circuit.
From the right angled triangle OAE,
Final voltage, V = \(\sqrt{v_{n}^{2}+\left(v_{L}-v_{c}\right)^{2}}\)

Where Z is called impedance of LCR circuit

3. Impedance, Z=R.

Question 4.
A voltage source is connected to an electrical component X as shown in figure.

1. Identify the device X.
2. Which of the following equations can represent the current through the circuit?

• i = i_m sin(ωt + π/2)
• i = i_m sin(ωt – π/2)
• i = i_m sin ωt
• i = i_m sin(ωt + π/4)

3. Draw the phasor diagram for the circuit. (2)
Answer:
1. Resistor
2. i = i_m sin ωt
3.

Question 5.
A friend from abroad presents you a coffee maker when she visited you. Unfortunately, it was designed to operate at 110V line to obtain 960W power that it needs.

1. Which type of transformer you use to operate the coffee maker at 220V? (1)
2. Assuming the transformer you use as ideal, calculate the primary and secondary currents. (2)
3. What is the resistance of the coffee maker? (1)

Answer:
1. Step down transformer.

2. Since the transformer is ideal
V_pI_p = V_s I_s = 960W, V_p = 220v, V_s = 110v
V_pI_p =960

3.

Plus Two Physics Alternating Current Five Mark Questions and Answers

Question 1.
The voltage-current values obtained from a transformer constructed by a student is shown in the following table.

1. Identify the transformer as step up or step down.
2. How much power is wasted by the transformer?
3. What are the possible energy losses in a transformer?
4. If the input voltage is 48v and input current is 1 A, is it possible to light 240v, 100w bulb using the above transformer. Justify.

Answer:
1. Step down transformer.

2. Power loss = 200w -10w = 190w

3. The possible energy losses in a transformer:

• Eddy current loss
• Copper loss
• Hysteries loss
• Flux leakage

4. In this case input power = VI = 48 × 1 = 48W.
If transformer does not waste energy, input power =out put power.
Hence maximum output power 48W. But bulb requires 100w. Hence the bulb does not glow with this low input power.

Question 2.
The current through fluorescent lights are usually limited using an inductor.

1. Obtain the relation i = i_m sin(ωt – π/2)for an inductor across which an alternating emf v = v_m sin ωt is applied. (2)
2. Why it is better to use an inductor rather than a resistor to limit the current through the fluorescent lamp? (1)
3. When 100 V DC source is connected across a coil a current of 1 A flows through it. When 100V, 50 Hz AC source is applied across the same coil only 0.5 A flows. Calculate the resistance and inductance of the coil. (2)

Answer:
1. AC voltage applied to an Inductor

Consider a circuit containing an inductor of inductance ‘L’ connected to an alternating voltage. Let the applied voltage be
V = V₀ sinωt…………(1)
Due to the flow of alternating current through coil, an emf, \(\mathrm{L} \frac{\mathrm{d} \mathrm{I}}{\mathrm{dt}}\) is produced in the coil. This induced emf is equal and opposite to the applied emf (in the case of ideal inductor).

2. No power is developed across the inductor as heat.

3. Resistance of the coil R = \(\frac{v}{I}=\frac{100}{1}\) = 100Ω. Current through the coil when ac source is applied.

R² + X²_L = 200²
X²_L = 200² – 100²
X_L = 173.2Ω
L_ω = 173.2

Question 3.
An alternating voltage is connected to a box with some unknown circuital arrangement. The applied voltage and current through the circuit are measured as v = 80 sinωt volt and i = 1.6 sin(ωt + 45°) ampere.

1. Does the current lead or lag the voltage?
2. Is the circuit in the box largely capacitive or inductive?
3. Is the circuit in the box at resonance?
4. What is the average power delivered by the box?

Answer:
1. Leads.

2. Capacitive

3. No Hint: Current and voltage are not in the same phase.

4. P = V_rmsl_rmsCosΦ, V_m = 80v, i_m = 1.6A

p = 45.25W

Students can Download Chapter 11 Alcohols, Phenols and Ethers Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 11 Alcohols, Phenols and Ethers

Plus Two Chemistry Alcohols, Phenols and Ethers One Mark Questions and Answers

Question 1.
Which of the following is most acidic?
(a) H₂0
(b) CH₃OH
(c) C₂H₅OH
(d) CH₃CH₂CH₂OH
Answer:
(a) H₂O

Question 2.
Propan-1 -ol on reaction with conc.H₂SO₄ at 413 K gives _____________
Answer:
1 -Propoxy propane

Question 3.
Which of the following alcohols can be obtained from HCHO?
(a) CH₃OH
(b) C₂H₅OH
(c) CH₃CH₂CH₂CH₂OH
(d) All of these
Answer:
(d) All of these

Question 4.
Phenol can be distinguished from ethyl alcohol by all reagents except
(a) NaOH
(b) Na
(c) Br₂/H₂O
(d) FeCl₃
Answer:
(b) Na

Question 5.
Anisole on reaction with HI forms
(a) phenol and methyl iodide
(b) iodobenzene and methanol
(c) benzene and methyl iodide
(d) phenol and methanal
Answer:
(a) phenol and methyl iodide

Question 6.
Arrange the following compounds in the increasing order of acidity: Phenol, Alcohol, and Water.
Answer:
Alcohol < Water < Phenol

Question 7.
Phenol is distinguished from ethanol by the following reagents except.
(a) Iron
(b) Sodium
(c) Bromin
(d) NaOH
Answer:
(b) Sodium

Question 8.
Phenol can be converted to o-hydroxy benzaldehyde by
Answer:
Reimer-Tiemann reaction.

Question 9.
4-methoxy acetophenone can be prepared from anisol by ______________
Answer:
Friedel crafts reaction.

Question 10.
Which of the following does not answer idoform test?
(a) 1-propanol
(b) Ethanol
(c) 2 propanol
(d) Ethanal
Answer:
(a) 1- propanol

Question 11.
Chlorination of toluene in presence of light and heat followed by treatment with aqeous KOH gives ________
Answer:
Benzyl alcohol.

Plus Two Chemistry Alcohols, Phenols and Ethers Two Mark Questions and Answers

Question 1.
Reaction of alcohol with metallic sodium is used as a test for alcohol. Substantiate this statement with the help of an equation.
Answer:
Alcohol reacts with metallic Na to form sodium alkoxide with the liberation of H₂.
ROH + Na → RONa + 1/2 H₂

Question 2.
Ethyl alcohol gives iodoform test. Methyl alcohol does not give iodoform test.

1. Do you agree with this?
2. If yes or no, substantiate your view.
3. How can you distinguish between 1 -butanol and 2-butanol?

Answer:
1. Yes.

2. Compounds containing CH₃CO- group and CH₃CH(OH)- group on reaction with iodine and alkali give yellow colour of iodoform. Ethanol contains CH₃CH (OH) – group.

3. 2-Butanol gives iodoform test as it contains CH₃CH(OH)-group, whereas 1-Butanol does not answer iodoform test.

Question 3.
When an organic compound is treated with neutral ferric chloride a violet colour is obtained. What will be the compound? Explain.
Answer:
Phenol. Phenol forms a violet-coloured water soluble complex with ferric chloride.

Question 4.
– O – is the functional group of ether. Classify the following into two groups with appropriate heading.
CH₃-O-CH₃, CH₃-O-C₂H₅, C₂H₅-O-C₂H₅, C₆H₅– O – CH₃.
Answer:

Question 5.
Explain how does the -OH group attached to a carbon of benzene ring activates it towards electrophilic substitution reaction?
Answer:
In phenol, the -OH group is directly attached to a carbon of benzene ring. The lone pair of oxygen participates into resonance with the benzene ring. As a result, electron density on benzene ring increases making it more easy to attack by an electrophile.

Question 6.
Write down the equations for the following conversions using Grignard reagent?
Methanal → Ethanol
Ethanal → 2-Propanol
Answer:

Question 7.
Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Answer:
Propanol has higher boiling point than butane because it has stronger interparticle forces. In propanol intermolecular hydrogen bonding is present whereas in butane intermolecular forces are weak van der Waals’ forces. A lot of heat is required to break intermolecular hydrogen bonding among propanol molecules.

Question 8.
While separating a mixture of ortho and para nitrophenols by stream distillation, name the isomer which will be steam volatile. Give reason.
Answer:
Ortho-Nitrophenol is steam volatile because in it there is intramolecular hydrogen bonding.

Due to intramolecular hydrogen bonding, the intermolecular forces in ortho-nitrophenol are weaker than that in para-nitrophenol (which has intermolecular hydrogen bonding) and hence it undergoes less association.

Question 9.
Give reason for higher boiling point of ethanol in comparison to methoxy methane.
Answer:
In ethanol the intermolecular forces are hydrogen bonds whereas in methoxymethane the intermolecular forces are dipole-diple forces. Since the intermolecular forces in ethanol are stronger than those in methoxymethane it has higher boiling point than methoxymethane.

Question 10.
While separating a mixture of ortho and para nitro phenols by steam distillation, name the isomer, which will be steam volatile. Give reason.
Answer:
Ortho-Nitro phenol exhibit intramolecular hydrogen bonding and para-Nitro phenol exhibit intermolecular hydrogen bonding. The ortho isomer is steam volatile because there is not intermolecular association.

Plus Two Chemistry Alcohols, Phenols and Ethers Three Mark Questions and Answers

Question 1.
Write a notes on:

1. Rectified Spirit
2. Power Alcohol
3. Denatured Spirit

Answer:
1. 95.6% ethyl alcohol is known as a rectified spirit.

2. A mixture of ethyl alcohol and gasoline can be used as a fuel in the internal combustion engine. It is known as power alcohol.

3. Commercial alcohol is made unfit for drinking by adding certain substances like pyridine, methanol etc. Spirit thus obtained is called denatured spirit.

Question 2.
Fill in the blanks:

Answer:

Question 3.
Ethyl alcohol can be prepared by fermentation of molasses.

1. What do you mean by fermentation?
2. Explain the process.
3. What do you mean by ‘wash’?

Answer:
1. Fermentation is the process of breaking up of large molecules into small molecule in the presence of biological catalyst called enzymes.

2. Molasses is the mother liquor left behind after the crystallisation of cane sugar from sugar cane juice. It is 40% sucrose solution. First it is diluted to 10% solution. Then yeast is added. Temperature of the system is kept at 305 K. The following reactions will take place.

3. 8-10% ethyl alcohol is known as ‘wash’.

Question 4.
Raju prepared soap in the chemistry lab. A liquid remained after the preparation. He argued that it was useless.

1. Do you agree with him? Why?
2. How can you prepare glycerol from spentlye?

Answer:

1. No. It can be used to prepare glycerol.
2. Spentlye contains unreacted alkali, glycerol, water, NaCI, and soluble soap. It is first treated with acid to remove alkali, then with aluminium sulphate to remove soluble soap. It is then evaporated under reduced pressure to remove NaCI.

The resulting solution is then decolorised using animal charcoal, Then the solution is distilled under reduced pressure to remove water. In this way we get glycerol.

Question 5.
When an old sample of ether was heated, it exploded.

1. What is the reason for this phenomenon?
2. How can you detect peroxide content in ether?
3. How can we remove peroxide from old sample of ether?

Answer:

1. Due to the formation of peroxide.
2. Presence of peroxide can be tested by adding ferrous salt solution followed by addition of KCNS solution. Formation of blood red colour indicates the presence of peroxide.
3. The peroxide can be removed by washing with ferrous salt solution.

Question 6.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:
Because of the presence of O-H group in them, alcohols are capable of forming hydrogen bonds with water molecules whereas hydrocarbons cannot form hydrogen bonds with water. As a result, alcohols are more soluble in water than the hydrocarbons of comparable molecular masses.

Question 7.
Give the structures and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.
Answer:

Question 8.
Explain why ortho-nitrophenol is more acidic than ortho-methoxyphenol?
Answer:
ortho-Nitrophenol is more acidic than ortho- methoxyphenol because nitro group by its electron withdrawing resonance effect stabilises the phenoxide ion whereas methoxy group by its electron releasing effect destabilises the phenoxide ion. Greater the stability of the phenoxide ion, greater is the dissociation of phenol and greater is its acid strength.

Question 9.
With the help of a mixture of con. HCI and ZnCI2 how can you distinguish between 1°, 2°, 3° alcohols.
Answer:

• 1° alcohol + Lucas reagent → no reaction
• 2° alcohol + Lucas reagent → turbidity within five minutes
• 3° alcohol + Lucas reagent → turbidity occurs immediately

Lucas reagent anhydrous ZnCI₂/HCI

Plus Two Chemistry Alcohols, Phenols and Ethers Four Mark Questions and Answers

Question 1.
Match the following:

Answer:

Question 2.

a) Alcohols are having high boiling points than corresponding alkyl halides and ethers. Why?
b) Phenol is more acidic than ethanol. Give the reason.
c) Predict the products :

Answer:
a) Due to the presence of polar hydroxyl group alcohols can associate through intermolecular hydrogen bonding.
b) Phenoxide ion is stabilized by resonance while alkoxide ion has no resonance stabilisation.
c)

Question 3

1. Phenol is acidic even though it has no carboxylic group, why?
2. Convert phenol to salicylic acid and name the reaction.

Answer:
1. By the removal of a proton from phenol a phenoxide ion is obtained. It is stabilised by resonance. Hence phenol acts as an acid.
2.

.

Question 4.
1. Compare the solubility of diethyl ether and n- butane in water.
2. Give the product(s) from the reaction of one mole of diethyl ether with

• one mole of conc. HI and
• excess of HI.

Answer:
1. Diethyl ether being weakly polar is capable of forming intermolecular hydrogen bonding with water. Hence, diethyl ether is soluble in water while n-butane is not.

2. The product(s) from the reaction of one mole of diethyl ether.

• C₂H₅OH and C₂H₅I
• 2C₂H₅I

Question 5.
Identify X and Y.

Answer:

Question 6.
1. Boiling point depends on the inter molecular hydrogen bonding.

• Ethanol and propane have comparable molecular masses but their boiling points differ widely. Why?

2. Williamson synthesis is an important method of ether synthesis. Which of the following reactions is better for ether synthesis? Justify.

Answer:
1. Ethanol molecules are associated by intermolecular hydrogen bonding.

• Ethanol molecules are associated by intermolecular hydrogen bonding. This is absent in propane. So Ethanol has higher boiling point.

2. (CH₃)₃C-ONa + CH₃-Br is better. The tertiary alkyl halide undergoes elimination in presence of base to form alkene.

Question 7.
Chloro methane reacts with aqueous sodium hydroxide to form methanol.

1. What happens when chlorobenzene reacts with aqueous sodium hydroxide? Justify.
2. Write the reaction by which chlorobenzene can be converted to phenol.

Answer:

1. No reaction. This is due to sp² state of carbon to which CI is attached, less polarity of C-X bond and resonance stabilisation.
2. Dow’s process.
   When chlorobenzene is heated with sodium hydroxide solution at 623 K under a pressure of 200 atm in the presence of copper catalyst, sodium phenoxide is obtained. This on hydrolysis gives phenol.

Question 8.
Phenol exhibit acidic character.

1. Why phenol exhibit acidic character?
2. Explain it with the help of resonating structure of phenoxide ion.

Answer:

1. Phenol can donate proton and phenoxide ion thus formed is stabilized by resonance.
2. The resonating structure of phenoxide ion is given below.

Question 9.
What is the action of phenol with

1. Aqueous Br₂?
2. Br₂ in CS₂?
3. Nitrating mixture?
4. dil. HNO₃?

Answer:
1. Phenol on action with aqueous Br₂ gives 2, 4, 6,- tribromophenol.

2. Phenol on action with Br₂ in CS₂ gives o- bromophenol and p-bromophenol.

3.

4.

Question 10.
Write notes on
a) Kolbe’s reaction
b) Reimer-Tiemann reaction
Answer:
a) Kolbe’s reaction: When Sodium phenoxide is heated with CO₂ at 400 K and under a pressure 6-7 atm, sodium salicylate is obtained. This on hydrolysis gives ortho-hydroxy benzoic acid or salicylic acid.

Reimer – Tiemman reaction: When phenol is heated with CHCI₃ at 340 K, o-hydroxy benzaldehyde or salicylaldehyde is obtained.

Question 11.

1. Fill in the blanks:
   CH₃CH₂OH + SOCI₂ → CH₃CH₂CI+ …?… +….?….
2. This method is used to prepare extra pure alkyl halide. Do you agree? Why?

Answer:

1. CH₃CH₂OH + SOCI₂ -> CH₃CH₂CI + SO₂ + HCI
2. Yes. The by products are escapable gases. Hence, the reaction gives pure alkyl halides.

Question 12.
1. Arrange the compounds in the increasing order of their strength.

• 4-Nitro phenol
• Phenol
• Propan-1-ol
• 4-Methyl phenol.

2. You are given benzene, conc.H₂SO_4, and NaOH. Prepare phenol using these compounds.
Answer:
1. 4-Nitrophenol > Phenol > 4-Methylphenol > 1- Propanol. Presence of electron with darwing groups at ortho and para positions increases the acidic strength of substituted phenols.

2. By the action of benzene with conc.H₂SO₄ & NaOH, sodium benzene sulphonate is formed. This on acidification gives phenol.

Plus Two Chemistry Alcohols, Phenols and Ethers NCERT Questions and Answers

Question 1.
Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Answer:
Propanol has higher boiling point than butane because it has stronger interparticle forces. In propanol intermolecular hydrogen bonding is present whereas in butane intermolecular forces are weak van der Waals’ forces. A lot of heat is required to break intermolecular hydrogen bonding among propanol molecules.

Question 2.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:
Because of the presence of O-H group in them, alcohols are capable of forming hydrogen bonds with water molecules whereas hydrocarbons cannot form hydrogen bonds with water. As a result, alcohols are more soluble in water than the hydrocarbons of comparable molecular masses.

Question 3.
Give the structures and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.
Answer:

Question 4.
Explain why ortho-nitrophenol is more acidic than ortho-methoxy phenol?
Answer:
ortho-Nitrophenol is more acidic than ortho methoxy phenol because nitro group by its electron with drawing resonance effect stabilises the phenoxide ion whereas methoxy group by its electron releasing effect destabilises the phenoxide ion. Greater the stability of the phenoxide ion, greater is the dissociation of phenol and greater is its acid strength.

Question 5.
Give IUPAC names of the following ethers:
i) C₂H₅OCH₂-CH(CH₃)CH₃
ii) CH₃OCH₂CH₂CI
iii) O₂N-C₆H₄-OCH₃(p)
iv) CH₃CH₂CH₂OCH₃

Answer:
i) 1-Ethoxy-2-methylpropane
ii) 2-Chloro-1-methoxyethane
iii) 4-Nitroanisole
iv) 1-Methoxypropane
v) 1-Ethoxy-4, 4-dimethyl cyclohexane
vi) Ethoxybenzene

Students can Download Chapter 6 Electro Magnetic Induction Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 6 Electro Magnetic Induction

Plus Two Physics Electro Magnetic Induction NCERT Text Book Questions and Answers

Question 1.
A horizontal straight wire 10m long extending from east to west is falling with a speed of 5.0ms^-1 at right angles to the horizontal component of the earth’s magnetic field 0.3 × 10^-4Wbm².

1. What is the instantaneous value of the e.m.f? induced in the wire?
2. What is the direction of the e.m.f.?
3. Which end of the wire is at the higher electrical potential?

Answer:
I – 10m, v = 5.0ms^-1, B = 0.30 × 10^-4Wbm²
1. Instantaneous value of e.m.f. induced
e = B lv
= 0.30 × 10^-4 × 10 × 5.0
or e = 1.5 × 10^-3V.

2. The direction of e.m.f. is from eastto west.

3. Since the current is flowing from east to west so the eastern end is at higher potential.

Question 2.
Current in a circuit falls from 5.0Ato 0.0A in 0.1s. If an average e.m.f. of 200V induced, give an estimate of the self-inductance of the circuit.
Answer:
dl = l₂ – l₁ = 0.0 – 5.0 = -5.0A,
dt = 0.1s, e = 200V, L=?
Since

or L = 4H

Question 3.
A pair of adjacent coils has a mutual inductance of 1 5H. If the current in one coil changes from 0 to 20A in 0.5s. What is the charge of flux linkage with the other coil?
Answer:
M= 1.5H, dl = l₂ – l₁ = 20 – 0 = 20A, dt = 0.5s, dΦ = ?
Since

And also

or dΦ = Mdl
or dΦ = 1.5 × 20 = 30 Wb.

Question 4.
A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing 25m long if the earth’s magnetic field at the location has a magnitude of 5.0 × 10^-4 and the dip angle is 30?
Answer:
v = 1800 km/h = 500ms^-1 I = 25m
Vertical component of B along horizontal
B = -Bsinθ = 5.0 × 10^-4 sin30° = 2.5 × 10^-4T
e = B/v sinθ
= 500 × 2.5 × 10^-4 × 25
= 31 V
The direction ofthe wing is immaterial.

Question 5.
Suppose the loop in Q.6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3T at the rate of 0.02TS^-1. If the cut is joined and the loop has a resistance of 1.6Ω, how much power is dissipated by the loops as heat? What is the source of this power?
Answer:
Induced e.m.f.

= 8 × 2 × 10^-4 × 0.02 = 3.2 × 10^-5V
Induced current,

= 2 × 10^-5A
Power Loss = I²r = (2 × 10^-5)² × 1.6 = 6.4 × 10^-10 W. Source of this power is the external agent responsible for changing the magnetic field with time.

Plus Two Physics Electro Magnetic Induction One Mark Questions and Answers

Question 1.
Eddy current are produced when.
(a) a metal is kept in varying magnetic field
(b) a metal is kept in steady magnetic field
(c) a circular coil is placed in a magnetic field
(d) through a circular coil, current is passed
Answer:
(a) a metal is kept in varying magnetic field

Question 2.
A100 milli henry coil carries a current of 1 A. energy stored in its magnetic field is
(a) 0.5 J
(b) 1 A
(c) 0.05 J
(d) 0.1 J
Answer:
(c) 0.05 J
Explanation : E = \(\frac{1}{2}\) LI² = \(\frac{1}{2}\) × (100 × 10^-3) × 1²
= 0.05 J.

Question 3.
A straight line conductor of length 0.4 m is moved with a speed of 7m/s perpendicular to a magnetic field of intensity 0.9 Wb/m². The induced e. m. f. across the conductor is.
(a) 5.04 V
(b) 25.2 V
(c) 1.26 V
(d) 2.52
Answer:
(d) 2.52
Induced e.m.f (s) = B/V
= 0.9 × 0.4 × 7 = 2.52V.

Question 4.
The core of a transformer is laminated because
(a) ratio of voltage in primary and secondary may be increased
(b) energy losses due to eddy currents may be minimized
(c) the weight of the transformer may be reduced
(d) rusting of the core may be prevented
Answer:
(b)

Plus Two Physics Electro Magnetic Induction Four Mark Questions and Answers

Question 1.

A conductor XY is moving through a uniform magnetic field of intensity B as shown in figure.

1. Name the emf.
2. The motion of XY towards right side causes an anticlockwise induced current. What will be the direction of magnetic induction in the region A?
3. The length of the conductor XY is 20cm. It is moving with a velocity 50m/s perpendicular to the magnetic fie Id. If the induced emf in the conductor is 2V find the magnitudes of magnetic field?

Answer:
1. Motional emf/induced emf.

2. Applying right hand grip rule at A direction of magnetic field is away from the paper.

3. ε = Blvsinθ, I = 20cm = 20 × 10^-2m, v
= 50m/s, ε = 2v

= 0.2 T.

Question 2.
1. When S₁ close bulb glows instantaneously when S₂ closes there is a delay in glowing the bulb.(1) This is due to………

• resistance of the coil
• back emf in the coil
• mutual induction
• none

2. Explain the phenomenon self-induction regarding above experiment (2)
3. Draw the graph with energy and current for a inductor. (1)
Answer:
1. Back emf in the coil.

2. When a.c. current is passed through the coil, a change in flux is produced. This change in flux produces a back e.m.f. in the coil. The phenomena of production of back emf is called self induction.

3. The graph with energy and current for a inductor:

Plus Two Physics Electro Magnetic Induction Five Mark Questions and Answers

Question 1.
When AC is switched on the thin metallic disc is found to thrown up in air.

1. Which makes the disc to thrown?
2. How will you explain the mechanism behind the movement of disc
3. Write the working principle of induction heater.

Answer:

1. Eddy current produced in the coil thrown disc into air.
2. Whenever the magnetic flux linked with a metal block changes, induced currents are produced due to this current, disc becomes a magnet. Hence disc thrown up in to air.
3. The change in flux produces eddy current in a metal. The heat produced by eddy current is used for cooking in induction heater.

Question 2.
AC generator is a device based on electromagnetic induction used to convert mechanical energy into electrical energy.

1. Draw a graph showing variation of induced emf over a cycle. Also indicate peak value of emf. (2)
2. How peak value of emf is related to its rms value? (1)
3. A rectangular wire loop of side 4cm × 6cm has 50 turns uniformly from 0.1 Tesla to 0.3 Tesla in 6 × 10^-2 second. Calculate induced emf in the coil. (2)

Answer:
1.

2. Erms = \(\frac{E_{0}}{\sqrt{2}}\)

3.

and Φ = BAN
Φ₁ = 0.1 × (24 × 10^-4) × 50
= 120 × 10^-4wb
Φ₂ = 0.3 × (24 × 10^-4) × 50
= 360 × 10^-4wb

= 40 × 10^-2V.

Students can Download Chapter 3 Production and Costs Questions and Answers, Plus Two Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Economics Chapter Wise Questions and Answers Chapter 3 Production and Costs

Plus Two Economics Production and Costs One Mark Questions and Answers

Question 1.
Which among the following cost curves is not ‘U’ shaped?
Answer:
AFC curve is rectangular hyperbola.

Question 2.
Identify the shape of the following AFC curve?
Answer:

The shape of AFC curve is rectangular hyperbola.

Question 3.
When a firm increased the number of labour from 10 to 11, keeping the capital fixed the total product increased from 120 to 130. Which of the following statement is correct in this regard?
(a) The total product fell
(b) This is a long run production
(c) The average product is rising
(d) 10 is the marginal product of an increased unit of labour.
Answer:
(d) 10 is the marginal product of an increased unit of labour.

Question 4.
Slope of an isoquant is:
(a) marginal cost
(b) DMRS
(c) DMRTS
(d) None of the above
Answer:
(c) DMRTS

Question 5.
When MP becomes zero, TP:
(a) increases
(b) decreases
(c) becomes maximum
(d) becomes negative
Answer:
(c) becomes maximum

Question 6.
Which among the following is not ‘u’ shape
(a) MC
(b) AC
(c) AVC
(d) AFC
Answer:
(d) AFC

Question 7.
When AC is minimum
(a) MC > AC
(b) MC = AC
(c) MC < AC
(d) None of these
Answer:
(b) MC = AC

Question 8.
All the following curves are U shaped except
(a) the AVC curve
(b) the AFC curve
(c) the AC curve
(d) the MC curve
Answer:
(b) the AFC curve

Plus Two Economics Production and Costs Two Mark Questions and Answers

Question 1.
Let the production function of a firm be Q = 5 L^1/2 K^1/2
Find the maximum output that the firm can produce with 100 units of L and 100 units of K.
Answer:
The production function is
Q = 5 L^1/2 K^1/2
Since the firm uses 100 units of L and 100 units of
K, we get the production function as
Q = 5 × 100^1/2 100^1/2
Q = 5 × 10x 10
Q = 5 × 100 = 500

Question 2.
Derive marginal product and average product from the total product schedule.

Answer:

Question 3.
Find the maximum possible output for a firm with zero units of labour and 10 units of capital when its production function is
Q = 5L + 2K
Answer:
The production function is Giving the values of L and K, we get
Q = 5 × 0 + 2 × 10
Q = 0 + 20 = 20

Question 4.
“In the long run, all costs are variable”. Do you agree? Justify your answer.
Answer:
Yes. In the long run, all costs are variable. This is because there is sufficient time available in the long run for any factor of production to get increased. As there are only variable factors, in the long run, all costs are variable.

Question 5.
“The AVC is U shaped” explain the reasons for the ‘U’ shape of AVC curve. Also, draw a diagram to clarify your points.
Answer:
The SMC and AVC curves start rising when production starts. As output increases, SMC falls. AVC being the average of marginal costs also falls but falls less than SMC. Then after a point, SMC start rising. AVC, however, continues to fall as long as value of SMC remains less than the prevailing value of AVC.

Once the SMC has risen sufficiently its value becomes greater than the value of AVC. The AVC starts rising. Therefore, the AVC curve is “U” shaped. It is given in the following diagram.

Question 6.
TFC of a firm is ₹2,000 and TVC is ₹3,000. It produces 20 units. Calculate the AVC and AC of the firm.
Answer:
TC = TFC + TVC
AC = 5000/20 = 250
AVC = 3000/20 = 150

Question 7.
Can there be some fixed cost in the long run?
Answer:
No, there cannot be some fixed cost in the long run. This is because, in the long run all factors of production can be adjusted and variable.

Question 8.
Let the production function of a firm be Q = 10 L^1/2 K^1/2 Find out the maximum possible output that the firm can produce with 100 units of L and 100 units of K.
Answer:
The production function is,
Q = 10 L^1/2 K^1/2
L =100 units
K =100 units
∴ Q = 10 × 1001/2 × 1001/2
= 10 × (10²)^1/2 × (10²)^1/2
= 10 × 10 × 10
= 1,000 units

Question 9.
The table given below shows the total product schedule of labour. Determine the AP and MP of labour?

Answer:

Question 10.
Prove that AC is the sum of AFC and AVC.
Answer:

Question 11.
A bus company started its operation with two buses, four labourers. It has paid a one-time road tax. As the passengers increased the company arranged more tripes, with the help of additional labour and for using more fuels. Classify the costs incurred by the bus company into fixed and variable.
Answer:

Question 12.
Both short run and long run average and marginal cost curves are U-shaped. But reasons for the U-shape are not the same. Bring out the differences.
Answer:
Behaviour of output in the short run as explained by the law of variable proportions Behaviour of output in the long run as explained by the laws of returns to scale.

Plus Two Economics Production and Costs Three Mark Questions and Answers

Question 1.
Match the following.

Answer:

Question 2.
Categorize the following into fixed cost
(a) wages of temporary workers
(b) cost of raw materials
(c) salary of permanent staff
(d) cost of transportation
(e) cost of plant
Answer:

Question 3.
Identify the following curves.

Answer:

1. TC
2. TVC
3. TFC

Question 4.
Identify the shapes of the following cost curves.

1. AFC
2. AC
3. TFC
4. TVC
5. TC

Answer:

1. AFC – Rectangular hyperbola
2. AC – ‘U’shape
3. TFC – Straight line parallel to X axis
4. TVC – Inverse‘S’ shape
5. TC – Inverse‘S’ shape

Question 5.
There exists vertical distance between 1. TVC and TC and (b) TC and TFC. What does this distance indicate?

1. TVC and TC
2. TC and TFC

Answer:

1. The vertical distance between TVC and TC represents the Total Fixed Costs (TFC).
2. The vertical distance between TC and TFC represents the Total Variable Costs (TVC)

Question 6.
Complete the following table

Answer:

Question 7.
Name the following curves. ,

Answer:

Question 8.
What do you mean by an ‘isoquant’?
Answer:
An isoquant is the set of all possible combinations of the two inputs that yield the same maximum possible level of output. Each isoquant represents a particular level of output and is labelled with that amount of output. The shape of isoquant is drawn below.

Question 9.
With the help of a diagram show the relation between average exists and average variable cost.

Answer:
Both AC and AVC are U shaped. As the output increases the gap between AC and AVC narrows.

Question 10.
In the short run, Total Variable Cost is zero when output is zero. When output rises, Total Cost also rises. Draw a suitable diagram and explain the relationship between Total Fixed Cost, Total Variable Cost and Total Cost.
Answer:

Question 11.
“Average Fixed Cost (AFC) curve is a continuously falling curve.”

1. Substantiate this statement by giving the reasons.
2. Graphically represent the AFC curve

Answer:
1. TFC is constant & hence AFC falls
2.

Plus Two Economics Production and Costs Five Mark Questions and Answers

Question 1.
The marginal product and total product of an input are related’. Prove this statement.’
Answer:
The marginal product (MP) and total product (TP) of an input are related. The points of relationship are given below.

1. When MP increases, TP also increases
2. When MP is zero, TP becomes maximum
3. When MP becomes negative, TP turns negative

Question 2.
Differentiate between returns to a factor and returns to scale.
Answer:
Returns to a factor refers to the effects on output of changes in one input with all other inputs are hold constant. On the other hand, returns to scale refers to the effect on total output of charges in some constant rate in all the inputs simultaneously.

Question 3.
The marginal cost and average cost are related to each other. Prove this using diagram and a numerical example?
Answer:

Marginal cost are addition made to the total costs by the production of an additional unit of the commodity.
MC = TC_n – TC_n-1

Average lost is per unit cost of production which is obtained by dividing the total cost by number of units of output produced.
\(A C=\frac{T C}{Q}\)

Since AC is obtained by dividing TC by all units, MC is the addition to TC by producing an additional unit so MC brings a change in AC.

From above:

1. When MC is less than AC, then MC will fall
2. When MC is equal to AC, AC remains costant.
3. When MC is more than AC, AC rises

Question 4.
Which of the following represent the long run production function?
1. MC is less than AC, then MC will fall

• Law of variable proportion
• Returns to scale

2.  Explain what may happen when all inputs vary simultaneously?
Answer:
Returns to scale.
When all inputs are varied simultaneously the output may change in three ways.
1. Constant returns to scale:
It is when a proportional increase in inputs result in an increase in output by the same proportion.

2. Increasing returns to scale:
It is when a proportional increase in all inputs result in an increase in output by more than proportion.

3. Decreasing returns to scale:
It is when a proportional increase in all inputs result in an increase in output by less than proportion

Question 5.
State whether the statement are true or false.

1. TC never becomes zero.
2. AC is sum total of AFC and AVC.
3. When AC rises, AC and MC are equal
4. Real cost is the cost in money terms
5. TFC curve is U shaped

Answer:

1. True
2. True
3. False
4. False
5. False

Question 6.
The following table shows the total cost schedule of a firm. What is the total fixed cost schedule of the firm? Calculate the JVC, AFC, AVC, SAC and SMC of the firm

Answer:

Question 7.
The diagram shows the total product curve of a factor in the law of variable proportion.

1. Explain the law
2. With the help of a diagram illustrate the average and marginal product curve and their relation.

Answer:
1. Law of variable proportion says that if one variable input is added with other fixed inputs the marginal product of a factor input initially rises but after reaching a certain level of output it starts falling.

2. The relation between the average and marginal product is given below. BothAPand MPare ‘n ’shaped. MPwill always pass through the maximum of AP.

Question 8.
Theory of production deals with producer’s behaviour, and according to this theory output produced by a firm passes through three stages in the short run.

1. Which are the three stages of production?
2. Analyse and bring out the salient features of each stage.
3. Which stage of production is very important as far as a firm is concerned? Why?
4. Give suitable diagram by drawing the Total Prod-uct, Average Product & Marginal Product curves.

Answer:

1. Increasing, diminishing and negative
2. First output increases at increasing rate, then at diminishing rate & finally starts falling.
3. Second stage, profit maximisation occurs in this Stage.
4.

Question 9.
If other things remaining same, graphically explain what happens to the supply curve for readymade shirts if there is

1. An increase in the wages paid to the tailors.
2. An increase in the price of ready-made shirts

Answer:

Plus Two Economics Production and Costs Eight Mark Questions and Answers

Question 1.
Prepare a seminar report on the topic “Production Function.”
Answer:
Respected teachers and dear friends:
The topic of my seminar paper is “Production Function”. The production function of a firm shows relationship between inputs and output. In this seminar paper, I would like to present the different production functions such as short run production function and the long run production function.

Introduction:
As we know the production function shows the transformation of inputs into output. The production function can be of two types – short run production function known as Law of Variable proportion and the long run production function known as returns to scale. Q =F(F₁ F₂… F_n)

Contents:
A. Law of Variable proportion

1. Increasing returns
2. Diminishing returns
3. Negative returns

A. The Law of Variable Proportions
When more and more units of a variable input are added with the fixed input, the marginal product would increase only upto a certain point. Therefore, the marginal product declines. This phenomenon is known as the Law of Variable Proportions. It is also known as returns to a factor.

The shape of TP, AP and MP suggests that they are specifically passing through three phases.
They are:
1. First phase:
In the first stage, both AP and MP increase. As a result TP also increases at an increasing rate. This stage is known as the stage of increasing return to a factor. AP reaches the maximum level in this stage.

2. Second phase:
Both AP and MP decrease at this stage. The TP increases at a decreasing rate. More importantly, TP reaches maximum and MP touches zero. This stage is also known as the stage of diminishing returns to a factor.

3. Third phase:
At this stage, the MP becomes negative. As a result, TP also starts declining. The decline of AP is continuous. In the graph, when TP reaches maximum and MP touches zero. When MP becomes negative, TP starts declining. This stage is known as the stage of negative returns to a factor.

B. Returns to scale

1. Increasing returns
2. Constant returns
3. Decreasing returns
   (for details, refer summary part)

B. Returns to scale
As stated earlier, all the factor inputs become variable in the long run and thereby no distinction can be made between fixed inputs and variable inputs. Return to scale is associated with long run production function.

Returns to scale refers to the change in output when all inputs are variable and the proportion between inputs remains constant. When all the inputs vary in the same proportion, the output (TP) behaves in different manner, which can be clubbed into three categories.

1. Increasing Returns to Scale (IRS):
When a proportionate change in all the inputs leads to more than proportionate change in output, it is known as the stage of increasing returns to scale. For instance, a 10% change in inputs results in more than 10% change in output.

2. Constant Returns to Scale (CRS):
When a proportionate change in all the inputs leads to change in output in the same proportion is known as the stage of constant returns to scale. It indicates that a 10% change in inputs leads to exactly 10% change in output.

3. Decreasing Returns to Scale (DRS):
When a proportionate change in all the inputs leads to less than proportionate change in output is known as the stage of decreasing returns to scale. It indicates that a 10% change in inputs leads to less than 10% change in output.

Conclusion:
Thus it can be concluded that there are two types of production function depending upon the use of inputs and the time period.

Question 2.
Prepare a seminar paper on Law of Returns to Scale.
Answer:
Returns to scale:
As stated earlier, all the factor inputs become variable in the long run and thereby no distinction can be made between fixed inputs and variable inputs. Return to scale is associated with long run production function.

Returns to scale refers to the change in output when all inputs are variable and the proportion between inputs remains constant. When all the inputs vary in the same proportion, the output (TP) behaves in different manner, which can be clubbed into three categories.

1. Increasing Returns to Scale (IRS):
When a proportionate change in all the inputs leads to more than proportionate change in output, it is known as the stage of increasing returns to scale. For instance, a 10 percentage change in inputs results in more than 10 percentage change in output.

2. Constant Returns to Scale (CRS):
When a proportionate change in all the inputs leads to change in output in the same proportion is known as the stage of constant returns to scale. It indicates that a 10 percentage change in inputs leads to exactly 10 percentage change in output.

3. Decreasing Returns to Scale (DRS):
When a proportionate change in all the inputs leads to less than proportionate change in output is known as the stage of decreasing returns to scale. It indicates that a 10 percentage change in inputs leads to less than 10 percentage change in output.

Question 3.
From the table given below

1. Fill the blank columns
2. Using the information in the table draw TC, TVC, TFC in one and AC, AVC, AFC, MC in other and prepare a short note on them.

Answer:
1.

2. Use the data to draw the diagram.
TC, TVC and TFC represent total costs. TC = TFC + TVC. TFC is the cost incurred on fixed factors. TVC is the cost incurred on variable costs. Even though the level of output is zero there will be some costs. This is known as fixed cost. When the level of output is zero total variable costs also will be zero.

It will increase as the level of output increases. AFC is the cost that always falls. It is a rectangular hyperbola. AVC, AC, MC are ‘U’ shaped. The MC will always pass through the minimum oftheAVCand AC.

Students can Download Chapter 12 Aldehydes, Ketones and Carboxylic Acids Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Plus Two Chemistry Aldehydes, Ketones and Carboxylic Acids One Mark Questions and Answers

Question 1.
The reaction is called
(a) Sandmeyer’s reaction
(b) Rosenmund’s reduction
(c) HVZ reaction
(d) Cannizaro’s reaction
Answer:
(b) Rosenmund’s reduction

Question 2.
Say TRUE or FALSE:
Aldol condensation is given by all aldehydes and ketones.
Answer:
False

Question 3.
A 40% solution of _________ is called formation
Answer:
formaldehyde (HCHO)

Question 4.
Benzamide on heating with bromine and caustic alkali gives
(a) benzene
(b) methylamine and benzene
(c) aniline
(d) m-Bromobenzaldehyde
Answer:
(c) aniline

Question 5.
In the reaction the product ‘B’ is
(a) acetanilide
(b) glycine
(c) ammonium acetate
(d) methane
Answer:
(b) glycine

Question 6.
Arrange the following in the decreasing order of acidity.
ClCH₂COOH, Cl₃CCOOH, CH₃COOH, Cl₂HCOOH
Answer:
Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH

Question 7.
Reaction of butanone with methyl magnesium bromiode followed by hydrolysis gives_________
Answer:
2 methyl -2- butanol

Question 8.
The major product of the addition of water molecule to propyne in the presence of mercuric sulphate and dil sulphuric acid is ________
Answer:
Propanone

Question 9.
One mole of propanone and one mole of formalde¬hyde are the products of ozonolysis of one mole of an alkene. The alkene is ________
Answer:
2 methyl propene

Question 10.
Which of the following is a better reducing agent for the following reduction.
RCOOH → RCH₂OH
(a) SnCI₂/HCI
(b) NaBH₄/ether
(c) H₂/pd
(d) N₂H₄/C₂H₅ONa
(e) B₂H₆/H₃O⁺
Answer:
(e) B₂H₆/H₃O⁺

Question 11.
The total number of acyclic structural isomers possible for compound with molecular formula C₄H₁₀O is ________
Answer:
7

Plus Two Chemistry Aldehydes, Ketones and Carboxylic Acids Two Mark Questions and Answers

Question 1.
Reactivity of ketone is less than that of aldehyde. Why?
Answer:
Due to steric hindrance and inductive effect of alkyl group.

Question 2.
Carboxylic acid decompose into carboxylate ion and H⁺ ion.

1. Explain this on the basis of resonating structure of carboxylic acid.
2. Arrange the following in the increasing order of acidity. HCOOH, CH₃COOH
3. Substantiate.

Answer:

1. Carboxylic acid decomposes to give proton and carboxylate ion and is stabilized by resonance. This explains the acidic character of carboxylic acid.
2. CH₃COOH < HCOOH
3. In acetic acid the electron donating effect (+l – effect) of -CH₃ group destabilises the carboxylate anion and decreases the acid strength. Whereas in formic acid the H atom has not electron withdrawing or electron donating effect.

Question 3.
What is Etard’s reaction?
Answer:
Etard’s reaction:
When toluene is oxidized using chromyl chloride, benzaldehyde is obtained.

Question 4.
What is HVZ reaction? Explain.
Answer:
HVZ reaction – When a carboxylic acid is treated with red P and halogen, the α-H atoms are successively replaced by halogen.

This reaction has great synthetic importance as the halogen atom can be replaced by a number of other groups giving useful products.

Question 5.
Predict the product and name the reaction:

1. HCHO + NaOH → B + C
2. CH₃COOH + CH₃OH → E

Answer:

1. CH₃OH + HCOONa – Cannizzaro reaction
2. CH₃COOCH₃ – Esterification

Question 6.
Write the name of any two tests to distinguish between acetaldehyde and acetone.
Answer:
Benedict’s test, Fehling’s test – Both tests are answered by acetaldehyde and not by aceotne.

Question 7.
An organic compound with molecular formula C₉H₁₀O forms 2, 4-DNP derivative, reduces Tollens’reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1, 2-benzenedicarboxylic acid. Identify the compound.
Answer:
From the given data it is clear that as the compound forms2, 4-DNP derivative it has /CO group. Since it reduces Tollens’ reagent therefore -CHO group is present. As it can also undergo Cannizzaro reaction therefore α -hydrogen is absent

The oxidation product suggests that the compound has a benzene ring. One of the – COOH groups have been obtained by the oxidation of – CHO group and the other from alkyl group. Hence on these basis, the structure of C₉H₁₀O is

Question 8.
Write a notes on

1. Reimer – Tiemann reaction
2. Rosenmund Reduction

Answer:
1. Reimer – Tiemman reaction: When phenol is heated with CHCI₃ at 340 K, o-hydroxy benzaldehyde or salicylaldehyde is obtained.

2. Rosenmund reduction: When an acid chloride is reduced by using hydrogen gas in presence of Pd and BaSO₄, an aldehyde is obtained.
OR

Question 9.
Distinguish the following compounds using any one test.
H₃C – CO – CH₃ and CH₃CH₂CHO
Answer:
CH₃COCH₃ give Iodoform test which CH₃CHO does not answer this test.

Question 10.
Aspirin is commonly used in medicine. How it is prepared? Give the equation.
Answer:
Aspirin is acetyl salicyclic acid. When salicyclic acid is treated with acetyl chloride, aspirin is obtained.

Salicylic acid + acetyl chloride → Aspirin

Question 11.
How will you prepare CH₃-CO-NH₂ and CH₃COOCH₃ from CH3COOH?
Answer:

Question 12.
Give the IUPAC name of the following compounds.
i) C₆H₅CH = CHCHO
ii) (CH₃)₃CCH₂COOH
Answer:

Question 13.
Give a test to distinguish between acetaldehyde and acetone.
Answer:
CH₃ – CO – CH₃ contains CH₃CO – group and hence it gives iodoform test.
CH₃ – CH₂ – CHO does not give iodoform test.

Question 14.
Predict the major product in the following reactions:

Answer:

Question 15.
Distinguish between formaldehyde & acetaldehyde.
Answer:

Question 16.
Which is more acidic, 2-chloropropanoic acid or 3- chloropropanoicacid? Why?
Answer:
2-chloropropanoic acid. Becasue the electron with drawing -Cl group is more closer to -COOH group in this compound.

Plus Two Chemistry Aldehydes, Ketones and Carboxylic Acids Three Mark Questions and Answers

Question 1.
Fill in the blanks:

Answer:

Question 2.
A student is given Tollens’ reagent for oxidation of aldehyde.

1. What is Tollens’ reagent?
2. Can you help him to do the experiment?
3. What is the result of the experiment?

Answer:

1. Tollens’ reagent is ammoniacal silver nitrate solution
2. Yes. To a little of the solution add Tollens’ reagent
3. A black precipitate of silver or silver mirror is obtained

Question 3.

1. What is the role of LiAIH₄?
2. Give one example of an oxidising agent?
3. What is the action of a carboxylic acid with alcohol?

Answer:

1. LiAIH₄ is a strong reducing agent. It reduces RCOOH to 1° alcohol (RCH₂OH)
2. Acidified KMnO₄.
3. When carboxylic acid is treated with an alcohol in the presence of dehydrating agent like conc.H₂SO₄, an ester is formed. This is called esterification reaction.

Question 4.
What happens when primary, secondary and tertiary alcohols are passed over red hot copper? Give equations.
Answer:
1° alcohol hot copper Aldehyde
2° alcohol over hot copper – Ketone
3° alcohol over hot copper- alkene

Question 5
Fill in the blanks.

Answer:

1. HVZ reaction
2. CH₃CH₃
3. Hoffmann bromamide degradation reaction
4. HCOONa + CH₃OH
5. Reimer-Tiemann reaction
6. C₆H₅Cl

Question 6.
Draw the structure of the following derivatives.
(i) The 2,4-dinitrophenylhydrazone of benzaldehyde
(ii) Cyclopropanone oxime
(iii) Acetaldehydedimethylacetal
(iv) The semicarbazone of cyclobutanone
(v) The ethylene ketal of hexan-3-one
(vi) The methyl hemiacetal of formaldehyde
Answer:

Question 7.

i) 2HCHO + NaOH → CH₃OH + HCOONa
ii)

(a) Identify Cannizzaro and Aldol condensation.
(b) What is the difference between the above two reactions?

Answer:
(a) Cannizzaro reaction:
2HCHO + NaOH CH₃OH + HCOONa
Aldol condensation:

(b) Cannizzaro reaction is given by aldehydes having no α-H atom whereas aldol condensation is given by aldehydes containing α-H atom.

Question 8.
a) In a practical class a group of students heated ethanal with NaOH. Another group heated methanal with conc.NaOH.
i) Identify the products in each reaction with equation.
ii) Name the reactions.
b) Aldehydes are more reactive than ketones. Comment on the statement.
Answer:
a)

ii) The first reaction is Cannizarro reaction and the second reaction is aldol condensation.

b) Due to the ‘+l’ effect and steric hindrance of surrounding alkyl group ketones are less reactive than aldehydes.

Plus Two Chemistry Aldehydes, Ketones and Carboxylic Acids Four Mark Questions and Answers

Question 1.
i) Arrange the following in the increasing order of acidic strength and justify your answer.
CH₃COOH, CHCI₂COOH, CH₂CICOOH, CCI₃COOH

ii) Suggest a method to convert acetic acid to chloroacetic acid. Name the reaction and write the chemical equation.
Answer:
i) CH₃COOH < CH₂CI-COOH < CHCI₂COOH < CCI₃COOH This is due to the electron withdrawing character of chlorine.

ii) HVZ reaction – When a carboxylic acid is treated with red P and halogen, the a-H atoms are successively replaced by halogen.

This reaction has great synthetic importance as the halogen atom can be replaced by a number of other groups giving useful products.

Question 2.
a) Which of the following carbonyl compounds answer aldol condensation reaction and give equation.
HCHO, CH₃CHO, CCI₃CHO, C₆H₆-CHO

b) Arrange the following compounds in the increasing order of acidity:
CH₃COOH, CH₂CICOOH, CH₃-CH₂-COOH, C₆H₅-COOH
Answer:
a) Among these compounds only CH₃CHO answer aldol condensation reaction. Others will not answer this reaction because they have no α – hydrogen atom.

Question 3.

1. Aldehydes and ketones are carbonyl compounds Give a test for identification of aldehydes.
2. Acidic strength is related to the stability of carboxylate anion. Which acid of each pair is stronger?

Answer:
1. Benedict test. Benedict reagent is a mixture of sodium carbonate and sodium citrate. This on reaction with aldehyde gives red precipitate of Cu₂S.

2. Acidic strength is related to the stability of carboxylate anion. Acid of each pair is stronger:
i) CH₂FCOOH. This is due to the high electron with drawing effect of F.

This is due to the high electron with drawing effect of the -CF₃ group.

Question 4.
Substituents on carboxylic acids have much effect on their acidity. Substantiate the statement with the following examples.
a) HCOOH, CH₃COOH, CH₂CICOOH

Answer:
CH₂CICOOH > HCOOH > CH₃COOH
The methyl group will intensify the negative charge on the carboxylate ion and destabilise it as compared to formate ion. So HCOOH is stronger than CH₃COOH. The electron withdrawing effect of a Cl makes chloroacetic acid stronger than HCOOH and CH₃COOH.

In the case of aromatic carboxylic acids, presence of electron withdrawing groups at ortho and para position increases their acidity while presence of electron donating groups decreases their acidity.

In 4-nitro benzonic acid acid strength is greater than that of benzoic acid due to the electron withdrawing nature of -NO₂ group while in 4-methoxy benzoic acid acid strength decreases due to the electron donating nature of the methoxy group.

Question 5.
Give a chemical test to distinguish between each of
the following pair of organic compounds.

1. Propanal and Propanol
2. Propanone and Propanal

Answer:

1. Propanal is an aldehyde and it gives a silver mirror with Tollens’ reagent while propanol is an alcohol and will not answer Tollens’test.

2. Propanone gives yellow precipitate of iodoform on reaction with I₂ and NaOH while propanal does not give iodoform test. OR Propanal gives silver mirror with Tollens’ reagent while propanone does not give silver mirror test.

Question 6.
What is meant by the following terms? Give an example of the reaction in each case.

1. Cyanohydrin
2. Acetal
3. Semicarbazone
4. Aldol
5. Hemiacetal
6. Oxime
7. Ketal
8. Imine
9. 2,4-DNP-derivative
10. Schiff’s base

Answer:
1. Cyanohydrin – It is a compound which contain both OH and CN groups. For example, Lactic acid can be obtained by hydrolysis of cyanohydrin.

2. Acetal – compounds formed by the reaction of aldehydes with monohydric alcohols in presence of dry HCI gas.

3. Semicarbazone – the product of carbonyl compounds with semicarbazide is known as semicarbazone.

4. Aldol – It is a condensation product of aldehydes or ketones having atleast one α – hydrogen atom with dilute alkali as catalyst.

5. Hemiacetal – It is a compound which contains an ether as well as alcohol functional group. For example, methoxyethanol is a hemiacetal.

6. Oxime – Addition compound formed by the reaction of aldehyde or ketone with hydroxylamine.

7. Ketal – It is a cyclic compound obtained by reaction of aceone with ethylene glycol.

8. Imine – Addition compound formed by the reaction of aldehyde or ketone with ammonia.

9. 2, 4-DNP derivative – 2, 4-phenylhydrazone (DNP) is the addition compound formed by the reaction of aldehydes and ketones with 2, 4-dinitrophenylhydrazine.

10. Schiff’s base – Addition compound formed by the reaction of aldehyde or ketone with amine.

Question 7.
Name the following compounds according to IUPAC system of nomenclature.

1. CH₃CH(CH₃)CH₂CH₂CHO
2. CH₃CH₂COCH(C₂H₅)CH₂CH₂CI
3. CH₃CH=CHCHO
4. CH₃COCH₂COCH₃
5. CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃
6. (CH₃)₂CCH₂COOH
7. OHCC₆H₅CHO-p

Answer:

1. 4-Methylpentanal
2. 6-Chloro-4-ethylhexan-3-one
3. But-2-enal
4. Pentane-2,4-dione
5. 3, 3, 5-Trimethylhexane-3-one
6. 3, 3-Dimethylbutanoicacid
7. Benzene 1, 4-dicarbaldehyde

Question 8.

1. What is the relation between an electron donating group and acidic character?
2. How carboxylic acids maintain their acid character?

Answer:

1. Electron donating group decreases the acid character.
2. Carboxylic acid decomposes to give proton and carboxylate ion and is stabilized by resonance. This explains the acidic character of carboxylic acid.

Question 9.
Predict the product formed when cyclohexane carbaldehyde reacts with following reagents:
i) PhMgBr and then H₃O⁺
ii) Tollens’reagent
iii) Semicarbazide and weak acid
iv) Excess ethanol and acid
v) Zinc amalgam and dilute hydrochloric acid
Answer:

Plus Two Chemistry Aldehydes, Ketones and Carboxylic Acids NCERT Questions and Answers

Question 1.
What is meant by the following terms? Give an example of the reaction in each case

1. Cyanohydrin
2. Acetal
3. Semicarbazone
4. Aldol
5. Hemiacetal
6. Oxime
7. Ketal
8. Imine
9. 2,4-DNP-derivative
10. Schiff’s base

Answer:
1. Cyanohydrin – It is a compound which contain both OH and CN groups. For example, Lactic acid can be obtained by hydrolysis of cyanohydrin.

2. Acetal – compounds formed by the reaction of aldehydes with monohydric alcohols in presence of dry HCI gas.

3. Semicarbazone – the product of carbonyl compounds with semicarbazide is known as semicarbazone.

4. Aldol – It is a condensation product of aldehydes or ketones having atleast one α – hydrogen atom with dilute alkali as catalyst.

5. Hemiacetal – It is a compound which contains an ether as well as alcohol functional group. For example, methoxyethanol is a hemiacetal.

6. Oxime – Addition compound formed by the reaction of aldehyde or ketone with hydroxylamine.

7. Ketal – It is a cyclic compound obtained by reaction of aceone with ethylene glycol.

8. Imine – Addition compound formed by the reaction of aldehyde or ketone with ammonia.

9. 2, 4-DNP derivative – 2, 4-phenylhydrazone (DNP) is the addition compound formed by the reaction of aldehydes and ketones with 2, 4-dinitrophenylhydrazine.

10. Schiff’s base – Addition compound formed by the reaction of aldehyde or ketone with amine.

Question 2.
Name the following compounds according to IUPAC system of nomenclature.

1. CH₃CH(CH₃)CH₂CH₂CHO
2. CH₃CH₂COCH(C₂H₅)CH₂CH₂CI
3. CH₃CH = CHCHO
4. CH₃COCH₂COCH₃
5. CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃
6. (CH₃)₃CCH₂COOH
7. OHCC₆H₅CHO-p

Answer:

1. 4-Methylpentanal
2. 6-Chloro-4-ethylhexan-3-one
3. But-2-enal
4. Pentane-2,4-dione
5. 3, 3, 5-Trimethylhexane-3-one
6. 3, 3-Dimethylbutanoicacid
7. Benzene 1, 4-dicarbaldehyde

Question 3.
Draw the structure of the following compounds
(i) 3-Methylbutanal
(ii) p-Nitropropiophenone
(iii) p-methylbenzaldehyde
(iv) 4-methylpent-3-en-2-one
(v) 3-bromo-4-phenylpentanoicacid
(vi) 4-Chloropentan-2-one
(vii) p, p-Dihydroxybenzophenone
(viii) Hex-2-en-4-ynoicacid
Answer:

Question 4.
An organic compound with molecular formula C₉H₁₀O forms 2, 4-DNP derivative, reduces Tollens’reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1, 2-benzene dicarboxylic acid. Identify the compound.
Answer:
From the given data it is clear that as the compound forms 2, 4-DNP derivative it has >CO group. Since it reduces Tollens’ reagent -CHO group is present. As it can also undergo Cannizzaro reaction α- hydrogen is absent.

The oxidation product suggests that the compound has a benzene ring. One of the – COOH groups have been obtained by the oxidation of – CHO group and the other from alkyl group. Hence on these basis, the structure of C₉H₁₀O is

Question 5.
Write structural formulas and names of the four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde served as nucleophile and which as electrophile.
Answer:
i) Propanal as electrophile as well as nucleophile

ii) Propanal as electrophile and butanal as nucleophile

iii) Butanal as electrophile and propanal as nucleophile

iv) Butanal as electrophile as well as nucleophile

Question 6.
Predict the product formed when cyclohexane carbaldehyde reacts with following reagents:
i) PhMgBr and then H₃O⁺
ii) Tollens’reagent
iii) Semicarbazide and weak acid
iv) Excess ethanol and acid
v) Zinc amalgam and dilute hydrochloric acid
Answer:

Question 7.
Give simple chemical tests to distinguish between

1. Propanal and propanone
2. Acetophenone and Benzophenone
3. Phenol and Benzoic acid
4. Benzaldehyde and acetophenone
5. Ethanal and propanal

Answer:
1. Propanal and propanone:
Propanal and propanone can be distinguished by iodoform test as it is given by propanone and not by propanal
Propanone reacts with hot NaOH/I₂ to give yellow precipitate of iodoform.

2. Acetophenone and benzophenone:
Acetophenone gives iodoform test but benzophenone does not respond.

3. Phenol and benzoic acid:
This can be distinguished by treating FeCI₃ solution. Phenol gives violet colour with FeCI₃ solution while benzoic acid gives buff colured precipitate.

4. Benzaldehyde and acetophenone:
Acetophenone responds to iodoform test while benzaldehyde does not.

5. Ethanal and propanal:
Ethanal gives yellow ppt. of iodoform with an alkaline solution of iodine (iodoform test)

Propanal does not give yellow ppt.

Students can Download Chapter 5 Magnetism and Matter Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 5 Magnetism and Matter

Plus Two Physics Magnetism and Matter NCERT Text Book Questions and Answers

Question 1.
Answer the following questions regarding earth’s magnetism:

1. A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.
2. The angle of dip at a location in southern India is about 18°. Would you expect a greater or lesser dip angle in Britain?
3. If you make a map of magnetic field lines at Melbourne in Australia, would the lines seen to go into the ground or come out of the ground?
4. In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole?
5. The earth’s field is claimed roughly approximately, the field due to a dipole magnetic moment 8 × 10²² JT^-1 located at the centre. Check your or¬der of magnitude of this number same way.
6. Geologists claim that besides the main magnetic NS poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all?

Answer:
1. Magnetic declination, angle of dip, horizontal component of earth’s magnetic field.

2. Greater angle of dip in Britain, (it is about 70°), because Britain is closer to the magnetic north pole.

3. Field lines of B due to the earth’s magnetism would seem to come out of the ground.

4. A compass is free to move in a horizontal plane, while the earth’s field is exactly vertical at the magnetic poles. So the compass can point in any direction there.

5. Use the formula for field B on the normal bisector of a dipole of magnetic moment m.
B = \(\frac{\mu_{0}}{4 \pi} \frac{M}{r^{3}}\)
Take M = 8 × 10²² JT^-1
r = 6.4 × 10⁶ m;
one gets B = 0.3G
which checks with the orde of magnitude of the observed field on the earth.

6. The earth’s field is only approximately a dipole field. Local N-S poles may arise due to, for instance, magnetized minerals deposits.

Question 2.
A short bar magnet of magnetic moment m = 0.32JT^-1 is placed in a uniform external magnetic field of 0.15T. If the bar is free to rotate in the plane of the field, which orientations would correspond to its

• stable and
• unstable equilibrium? What is the potential energy of the magnet in each case?

Answer:
Given M=0.32JT^-1 B = 0.15T, U = ?

• If \(\overrightarrow{\mathrm{M}} \| \overrightarrow{\mathrm{B}}\), then we have stable equilibrium and U = -MB = -0.32 × 0.15T = -4.8 × 10^-2 J
• If \(\overrightarrow{\mathrm{M}}\) is anti-parallel to \(\overrightarrow{\mathrm{B}}\) , we have unstable equilibrium and
  U = MB = 0.32 × 0.15J = 4.8 × 10^-2J

Question 3.
A closely wound solenoid of 800 turns and area of cross-section 2.5 × 10^-4 m² carries a current of 3.0A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
Answer:
Given N = 800, A = 2.5 × 10^-4 m², I = 3.0A, M = ?
Since M = NIA
∴ M = 800 × 3 × 2.5 × 10^-4
or M = 0.60JT^-1 along the axis of the solenoid.

Question 4.
A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10^-4 m², carrying a current of 4.0A, is suspended through its centre allowing it to turn in a horizontal plane.

1. What is the magnetic moment associated with the solenoid?
2. What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10^-2 is set up at an angle of 30° with the axis of the sole¬noid?

Answer:
1. Magnetic dipole moment,
M = nIA
= 2000 × 4.0 × 1.6 × 10^-4 = 1.28JT^-1

2. Net force = 0
Torque, τ = MB sinθ
= 1.28 × 7.5 × 10^-2 × sin 30°
= 1.28 × 7.5 × 10^-3 × 1/2 = 4.8 × 10^-2 Nm.

Question 5.
Answer the following questions:

1. Why does a paramagnetic sample display greater magnetization (for the same magnetizing field) when cooled?
2. Why is diamagnetism, in contrast, almost independent of temperature?
3. If a toroid used bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?
4.  If the permeability of a ferromagnetic material independent of the magnetic field? If not is it more for lower or higher fields?
5. Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point.
6. Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet?

Answer:
1. The tendency to disrupt the alignment of dipoles (with the magnetising field) arising from random thermal motion is reduced at lower temperatures.

2. The induced dipole moment in a diamagnetic sample is always opposite to the magnetising field, no matter what the internal motion of the atom is.

3. Slightly less, since bismuth is diamagnetic.

4. No, as is evident from the magnetisation curve. From the slope of the magnetisation curve, it is clear that μ is greater for lower fields.

5. Proof of the important fact (of much practical use) is based on boundary conditions of magnetic fields (B & H) at the interface of two media. (When one of the media has μ>>1, the field lines meet this medium nearly normally).

6. Yes. Apart from minor differences in the strength of the individual atomic dipoles of two different materials, a paramagnetic sample with saturated magnetisation will have the same order of magnetisation. But of course, saturation requires impractically high magnetising fields.

Plus Two Physics Magnetism and Matter One Mark Questions and Answers

Question 1.
Nickel shows ferromagnetic property at room temperature. If the temperature is increased beyond Curie temperature, then it will show
(a) anti ferromagnetism
(b) no magnetic property
(c) diamagnetism
(d) paramagnetism
Answer:
(d) paramagnetism

Question 2.
According to Curie’s law, the magnetic susceptibility of a substance at an absolute temperature T is proportional to
(a) 1/T
(b) T
(c) 1/T²
(d) T²
Answer:
(a) 1/T
Explanation : According to Curie’s law X ∝ \(\frac{1}{T}\)

Question 3.
Above Curie temperature
(a) a paramagnetic substance becomes ferromagnetic substance
(b) a ferromagnetic substance becomes paramagnetic
(c) a paramagnetic substance becomes diamagnetic
(d) a diamagnetic substance becomes paramagnetic
Answer:
(b) a ferromagnetic substance becomes paramagnetic

Plus Two Physics Magnetism and Matter Two Mark Questions and Answers

Question 1.
Classify the following properties into diamagnetic, paramagnetic and ferromagnetic.

1. Susceptibility -1≤x<0
2. In uniform magnetic field it acquires a large magnetisation in the direction of the field.
3. When it is suspended in a magnetic field, it will come to rest perpendiculartothe magnetic field.
4. Susceptibility of the substance varies inversely as temperature of the substance upto curie temperature ie. x_m∝ \(\frac{1}{T}\)

Answer:

• Dia – a,c
• Para – d
• Ferro – b

Question 2.
Classify the following properties into diamagnetism and ferro magnetism.

1. In non uniform magnetic field, it more from high to law field.
2. Magnetic field lines are repelled from this material, If we place in a external magnetic field.
3. susceptibility greater than one, <0, +ve
4. Iron, Nickel, Cobalt are the examples

Answer:

1. Diamagnetism
2. Diamagnetism
3. Ferro magnetism
4. Ferro magnetism

Question 3.
Fill in the blanks.

Answer:

1. negative
2. Weak magnetic field to strong magnetic field.
3. Individual atoms have tiny magnetic moments
4. µ_r >1

Plus Two Physics Magnetism and Matter Three Mark Questions and Answers

Question 1.
The figure shows hysteresis curves for soft iron and stell.

1. Which among the two is the hysteresis loop of soft iron?
2. Which one among the two materials is preferred for use in transformers and galvanometers?
3. When steel is once magnetized, the magnetiza-tion is not easily destroyed even if it is exposed to strong reverse fields. Give reason.

Answer:

1. Fig. b
2. Soft iron
3. When steel is magnetised the magnetic domains in the material is permanently set and magnetised permanently in the direction of the applied magnetic field.

Question 2.
The figure shows a liquid placed on the pole pieces of two magnets.

1. Which magnetic behaviour is exhibited by the liquid? (1)
2. Write any two characteristics of this magnetic behaviour? (1)
3. Does this behaviour transform with temperature. Why? (1)

Answer:
1. Diamagnetism
2. characteristics of this magnetic behaviour:

• Permeability of a diamagnetic material is less than one.
• Susceptibility is small and negative.

3. No. The magnetic dipole moment induced in the diamagnetic material is opposite to the magnetising field and hence does not affect the thermal motion of atoms. Hence change in temperature has no effect on diamagnetism.

Question 3.
A magnetic material contained in a curved glass plate, when placed in a nonuniform magnetic field, exhibits a property as shown in figure.

1. Which type of magnetic material is this? Explain the property
2. Write two examples for such a magnetic material. Explain how the property relates with temperature?
3. “The susceptibility of a material is small” what do you mean by this statement.

Answer:

1. Diamagnetism: Diamagnetic materials are repelled from external magnetic field.
2. Bismuth, Copper, Lead, Nitrogen, Water. Diamagnetism is independent of temperature.
3. Diamagnetism developed by external magnetic field is very small.

Question 4.
A place where dip = 90°, B_H = 0

1. This place is at…………
2. What is the value of B_u at this place?
3. Can a magnetic needle align in the N-S direction at this place?

Answer:
1. Magnetic pole.

2. B = \(\sqrt{B_{v}^{2}+B_{h}^{2}}\)
But B_h = 0
∴ B = B_v

3. No. Magnetic needle align vertically

Question 5.
Classify into diamagnet, Paramagnet and Fero magnet.

1. Feebly magnetized in the same direction
2. m_r slightly more than none
3. does not allow lines of force
4. temperature independent
5. exhibit hysteresis
6. strongly attracted by a bar magnet

Answer:

1. Paramagnet
2. Para magnet
3. Dia magnet
4. Diamagnet
5. Ferromagnet
6. Ferromagnet

Plus Two Physics Magnetism and Matter Four Mark Questions and Answers

Question 1.
Magnetic field lines are the visualization of magnetic field.

1. Write any three properties of magnetic field lines.
2. The arrangement shows two bar magnets placed near each other. Draw the magnetic field lines of the system.

Answer:
1. properties of magnetic field lines:

• The magnetic field lines of a magnet form continuous closed loops.
• The tangent to the field line at a given point represents the direction of magnetic field at that point.
• Flux density of magnetic field represents the strength of magnetic field.
• The magnetic field lines do not intersect

2.

Question 2.
The figure shows the magnetic field of earth.

1. Identify the labels A, B, C. (1)
2. The lines drawn on a map through places that have the same declination are called………(1)
3. The horizontal component of earth’s magnetic field at a place is 0.25 × 10^-4T and the resultant magnetic field is 0.5 × 10^-4 T. Find the dip and the vertical component of the earth’s magnetic field at the place. (2)

Answer:
1. the labels A, B, C:

• A – Magnetic equator
• B – Magnetic axis
• C – Declination

2. Isogonic lines

3. Horizontal component of earth’s magnetic field is
B_H = B cosδ
0.25 × 10^-4 = 0.5 × 10^-4 δ
δ = 60°
Vertical component of earth’s magnetic field is
Bv = B sin δ = 0.5 × 10^-4 sin 60 = 0.43 × 10^-4T.

Plus Two Physics Magnetism and Matter Five Mark Questions and Answers

Question 1.
When a magnetic needle is placed in a non-uniform magnetic field it experiences
1.

• a force but no torque
• a torque but no force
• force and torque
• neither a force nor a torque (1)

2. A bar magnet held perpendicular to a uniform magnetic field as in the figure. If the torque acting on it is to be reduced to 1/4^th by rotating the magnet towards the direction of the field, find the angle through which the magnet is to be rotated. (2)

3. State whether the potential energy of the magnet increases or decreases after rotation. Justify your answer. (2)
Answer:
1. force and torque

2. When the bar magnet is perpendicular to the field Torque is maximum
τ = MBsinθ = MBsin(90) = MB
When rotated through an angle θ, torque is

Angle through which the magnet is to be rotated is 90 – θ = 75.53°

3. Potential energy decreases. Potential energy is minimum when the magnet is parallel to the field. U = MBcosθ. When rotated to 0° (to make magnet parallel to the field) Potential energy decreases.

Students can Download Chapter 4 Moving Charges and Magnetism Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 4 Moving Charges and Magnetism

Plus Two Physics Moving Charges and Magnetism NCERT Text Book Questions and Answers

Question 1.
A Circular coil of wire consisting of 100 turns, each of radius 8.0cm carries a current of 0.40A. What is the magnitude of the magnetic field B at the centre of the coil?
Answer:
Given n = 100, r = 8.0cm = 8 × 10^-2
I = 0.4A, B = ?
At the centre of circular coil

= π × 10⁴T = 3.1 × 10^-4T.

Question 2.
A uniform field equal to 1.5T exists in a cylindrical region of radius 10.0cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if

1. The wire intersects the axis?
2. The wire is turned from N-S to north-east-north-west direction.
3. The wire in the N-S direction is lowered from the axis by a distance of 4.0cm?

Answer:
1. F = \(\mathrm{Bl} \ell\) = 1.5 × 7 × \(\frac{20}{100}\) or F = 2.1 N acting vertically downwards.

2. Force will again be 2.1N.

3. F = \(\frac{1.5 \times 7 \times 16}{100}\) = 1.68N.

Question 3.
Two long and parallel straight wires A and B carrying currents of 8.0A and 5.0A in the same direction are separated by a distance of 4.0cm. Estimate the force on a 10cm section of wire A.
Answer:
Given I₁ = 8.0A, l₂ = 5.0A, r = 4.0cm = 0.04m
l = 10cm = 0.10m
Since

(direction is given by Fleming left-hand rule).

Plus Two Physics Moving Charges and Magnetism One Mark Questions and Answers

Question 1.
A straight wire of length 0.5 metre and carrying a current of 1.2 ampere is placed in uniform magnetic field of induction 2 tesla. The magnetic field is perpendicular to the length of the wire. Find the force on the wire is
Answer:
F = i/B = 1.2 × 0.5 × 2 = 1.2 N

Question 2.
To convert a galvanometer into a ammeter, one needs to connect a.
(a) low resistance in parallel
(b) high resistance in parallel
(c) low resistance in series
(d) high resistance in series
Answer:
(a) low resistance in parallel

Question 3.
A coil carrying electric current is placed in uniform magnetic field.
(a) torque is formed
(b) e.m.f is induced
(c) both a and b are correct
(d) none of these
Answer:
(a) torque is formed

Question 4.
Direction of motion of unit the positive test charge gives direction electric field. Direction of motion of…….gives direction of magnetic field.
Answer:
Unit north pole.

Question 5.
A magnetic system with zero dipole moment
(a) Solenoid
(b) current carrying coil
(c) current loop
(d) toroid
Answer:
(d) Toroid.

Question 6.
Find odd one regarding polarity solenoid, torroid, current carrying loop, bar magnet.
Answer:
Torroid (In this case North and South pole are absent).

Question 7.
The nature of path when a charged particle is projected 30° to the direction of magnetic field. (Helix, cycloid, straight line, parabola).
Answer:
Helix

Question 8.
What is solenoid?
Answer:
An insulated copper wire wound in the form of cylinder is called solenoid.

Question 9.
Write mathematical expression for ampere’s theorem.
Answer:
∫B.dl = µ₀I

Plus Two Physics Moving Charges and Magnetism Two Mark Questions and Answers

Question 1.
The figure shows a long straight conductor carrying a current I. A magnetic field is produced around the conductor.

What is the magnitude of the magnetic induction at a point ‘P’ which is at a distant ‘x’ from the conductor?
What is the shape of the magnetic line of force?

Answer:
1. B = \(\frac{\mu_{0} I}{2 \pi x}\). Magnetic field is directed in to the plane of paper.

2. Circular.

Question 2.
A particle of mass 6.65 × 10^-27 kg having positive charge equal to two times of electron, moves with a speed of 6 × 10⁵ m/s in a direction perpendicular to that of a given magnetic field of flux density 0.4 weber/m². Find the acceleration of the particle.
Answer:

q = 2e, v = 6 × 10⁵ m/s
B = 0.4, m = 6.65 × 10^-27 kg
∴ acceleration,

a = 1.15 × 10¹³ m/s².

Question 3.
Classify in to true or false.

1. The magnetic field in the middle of current carrying solenoid depends up on cross sectional area.
2. The magnetic field depends up on current.
3. The magnetic field depends up on the material of the core.
4. The magnetic field depends up on total numbers of turns per unit length.

Answer:

1. False
2. True
3. True
4. True

Question 4.
A short straight conductor carries current I

1. Write the expression for magnetic field due to this conductor.
2. Represent graphically the variation of magnetic field with distance from the wire.

Answer:
1. dB = \(\frac{\mu_{0}}{4 \pi} \frac{|d| \sin \theta}{r^{2}}\)

2.

Plus Two Physics Moving Charges and Magnetism Three Mark Questions and Answers

Question 1.
The magnetic field along the axis of a circular coil is found to be B = \(\frac{\mu_{\mathrm{o}} \mathrm{Ia}^{2}}{2\left(\mathrm{r}^{2}+\mathrm{a}^{2}\right)^{3 / 2}}\)
1. What is the magnetic field along the axis if r>>a
2.

• Compare the above magnetic field with the electric field along the axis of an electric dipole
• What is the equation of magnetic dipole moment?

Answer:
1. B = \(\frac{\mu_{0} / a^{2}}{2 r^{3}}\)

2.

• Electric field due to electric dipole, E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{P}{r^{3}}\) magnetic field due to magnetic dipole B = \(\frac{\mu_{0}}{4 \pi} \frac{m}{x^{3}}\).
• Magnetic moment m = IA

Question 2.
The circuit diagram for verifying Ohm’s Law is given below, A student unknowingly connects a galvanometer in the place of the ammeter.

1. What will happen to the galvanometer?
2. What modification has to be made in the galvanometer if he still wants to use the galvanometer in place of the ammeter?
3. Assuming voltmeter to be an ideal one, what will happen if the student interchanges the position of the voltmeter and ammeter?

Answer:

1. Galvanometer will be damaged.
2. use a shunt resistance.
3. No current flows, because an ideal voltmeter has infinite resistance.

Question 3.
Consider a galvanometer with a full scale deflection of 1 m A and resistance 100Ω.

1. How is the device connected in the circuit?
2. How can it be converted to an ammeter with full scale deflection 1 ampere?

Answer:
1. Connected in series.

2.

by connecting a shunt resistance. 0.1Ω in parallel with galvanometer, we can convert galvanometer in to ammeter.

Question 4.
Two infinitely long straight parallel wires carry currents I each as shown in fig.

1. Which law helps to find direction of magnetic field around a current carrying conductor?
2. What is the magnitude, direction of the magnetic fields at A, and C?

Answer:
1. Right hand grip rule.

2. The magnitude, direction of the magnetic fields at A, and C:

• A – outward to the plane of paper
• C-Inward to the plane of paper.

Question 5.
Analyze the figure and answer the following questions.

1. What is the nature of force between these conductors is……….
2. What is the field due to I₁ at second conductor?
3. What is the force experienced per unit length of II^nd conductor?

Answer:

1. Attractive
2. B= \(\frac{\mu_{0} d_{1}}{2 \pi r}\)
3. f = \(\frac{\mu_{0} l_{1} l_{2}}{2 \pi r}\)

Question 6.
“I am very light and present in every matter. When I move along the equator from east to west, I am pushed up. When I am stationary, no force”.

1. Name the force and write its expression.
2. Who am I?
3. What would happen if moved at the poles?

Answer:

1. Lorentz force in earth’s magnetic field F =qvB
2. Electron
3. The earth’s magnetic field at poles is perpendicular to the earth surface. When an electron move, it is pushed to one side, parallel to earth’s surface.

Plus Two Physics Moving Charges and Magnetism Four Mark Questions and Answers

Question 1.
The internal connections of a moving coil galvanometer is given in the fig (i) and fig (ii)

1. What is the use of moving coil galvanometer?
2. Which figure(i) or figure(ii) is used to measure Voltage?
3. Write an expression for resistance required to convert the moving coil galvanometer in to voltmeter.
4. If a very small resistance (eg, copper wire) is used to convert moving coil galvanometer in to voltmeter, will it work properly?

Answer:

1. Moving coil galvanometer is used to detect the presence of current
2. The instrument shown in figure (2) is used to measure voltage.
3. Resistance is connected in series with galvanometer. R = \(\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}\) – G
4. No, A high resistance is required to convert galvanometer in to voltmeter.

Question 2.
When a charged particle enter normal to a uniform magnetic field, it take a circular path.

1. Name the particle accelerator using this principle.
2. Explain the working of that particle accelerator with relevant theory.
3. The neutrons can’t be accelerated using this partide accelerator. Why?

Answer:
1. Cyclotron.

2. Cyclotron:
a. Uses:
It is a device used to accelerate particles to high energy.

b. Principles:
Cyclotron is based on two facts

• An electric field can accelerate a charged particle.
• A perpendicular magnetic field gives the ion a circular path.

c. Working:

At certain instant, let D₁ be positive and D₂ be negative. Ion (+ve) will be accelerated towards D₂ and describes a semicircular path (inside it). When the particle reaches the gap, D₁ becomes negative and D₂ becomes positive.

So ion is accelerated towards D₁ and undergoes a circular motion with larger radius. This process repeats again and again. Thus ion comes near the edge of the dee with high K.E. This ion can be directed towards the target by a deflecting plate.

3. Charge of neutron is zero. Only charged particle can be accelerated using particle accelerated.

Question 3.
A current flows through a circular loop of radius r is shown in the figure.

1. What is the direction of magnetic field at ‘o’? (1)

2. Derive an equation for magnetic field at ‘o’ due to the circular loop carrying current i? (2)

3. If the loop splits into two equal halves as shown in figure. (1)

What will be the magnetic field at the center ‘o’?
Answer:
1. B = \(\frac{\mu_{0} I}{2 r}\) in t0 the plane of paper

2. Magnetic field on the axis of a circular current loop:
Consider a circular loop of radius ‘a’ and carrying current T. Let P be a point on the axis of the coil, at distance x from A and r from ‘O’. Consider a small length dl at A.
The magnetic field at ‘p’ due to this small element dl,
dB =\(\frac{\mu_{0} \mathrm{Idl} \sin 90}{4 \pi \mathrm{x}^{2}}\)

[since sin 90° -1]
The dB can be resolved into dB cosΦ (along Py) and dB sinΦ (along Px). Similarly consider a small element at B, which produces a magnetic field ‘dB’ at P. If we resolve this magnetic field we get.
dB sinΦ (along px) and dB cosΦ (along py¹)
dB cosΦ components cancel each other, because they are in opposite direction. So only dB sinΦ components are found at P, so total filed at P is

from ∆AOP we get x = (r² + a² )^1/2

Let there be N turns in the loop then,

Point at the centre of the loop:
When the point is at the centre of the loop, (r = 0) Then,

3. Zero

Question 4.
A boy connects a galvanometer directly to a cell of emf 1.5v to measure a current through a load 1Ω.

1. Which instrument can be used to measure the current in such a circuit?
2. What changes should be made in the galvanometer to measure such a high current? Explain using a circuit diagram.
3. The boy connected the galvanometer into a high current measuring device and he connected the device parallel to the load. What will be the observation. Justify.

Answer:
1. Ammeter

2.

A galvanometer can be converted into an ammeter by a low resistance (shunt) connected parallel to it.
Theory:
Let G be the resistance of the galvanometer, giving full deflection fora current I_g. To convert it into an ammeter, a suitable shunt resistance ‘S’ is connected in parallel. In this arrangement, Ig current flows through Galvanometer and remaining (I-I_g) current flows through shunt resistance.
Since G and S are parallel
P.d Across G = p.d across S
I_g × G =(I-I_g)S
S = \(\frac{\lg \mathrm{G}}{\left(1-\mathrm{l}_{\mathrm{g}}\right)}\)
Connecting this shunt resistance across galvanometer we can convert a galvanometer into ammeter.

3. Ammeter is a low resistance device. Hence it draws high current. This high current will damage it.

Question 5.
An electric charge will experience a force in uniform electric field. Similarly a moving charge experience a magnetic force (Lorentz) in magnetic field. The SI unit of magnetic field intensity is defined in terms of Lorentz Force.

1. Write the expression for magnetic Lorentz force.
2. Mention any two difference between electric field arid magnetic field.
3. Give an account of work done by Magnetic Lorentz force on moving charge and corresponding change in K.E.

Answer:
1.

2. Electric field is due to a charge, either in motion or at rest. Magnetic field is due to the motion of charge. Direction of electric force is colinear to electric field. Direction of magnetic force is perpendicularto magnetic field.

3. If velocity (displacement) is perpendicular to Lorentz force the work done will be zero and hence there will no change in K.E.

Question 6.
“Moving coil Galvanorrieter is a device used for detecting very feeble current”.

1. What is the working principle of a moving coil galvanometer?
2. Describe the construction and working of a moving coil galvanometer.
3. When a high current is passed through a moving coil galvanometer, it will get destroyed. How?

Answer:
1. Principle:
A conductor carrying current when placed in a magnetic field experiences a force, (given by Fleming’s left hand rule), τ = NIAB.

2.

A moving coil galvanometer consists of rectangular coil of wire having area ‘A’ and number of turns ‘n’ which is wound on metallic frame and is placed between two magnets. The magnets are concave in shape, which produces radial field.
Working :
Let T be the current flowing the coil, Then the torque acting on the coil. τ = NIAB, Where A is the area of coil and B is the magnetic field.

This torque produces a rotation on coil, thus fiber is twisted and angle (Φ). Due to this twisting a restoring torque (τ = KΦ) is produced in spring. Under equilibrium, we can write
Torque on the coil = restoring torque on the spring
NIAB = KΦ
Φ \(=\left(\frac{\mathrm{BAN}}{\mathrm{K}}\right)\) I
The quantity inside the bracket is constant for a galvanometer.
Φ ∝ I
The above equation shows that the deflection depends on current passing through galvanometer.

3. High current will produce large amount of heat. This heat will destroy coils in the galvanometer.

Question 7.
When a current carrying conductor is placed in a magnetic field it experiences a force.
1. Arrive at the expression forthe force experienced by the conductor. (2)
2. A conductor carrying current I direct out of the plane of the paper is lying in the magnetic field as in Fig. Draw the direction of force experienced by the conductor. (1)

3. If the conductor is lying parallel to the field what will be the force? (2)
Answer:
1. Consider a rod of uniform cross section ‘A’ and length ‘l’ Let ‘n’ be the number of electrons per unit volume (number density). ‘v_d’ be the drift velocityof electrons for steady current ‘I’.
Total number of electrons in the entire volume of rod =nAl
Charge of total electrons = nA l .e
‘e’ is the charge of a single electron.
The Lorentz force on electrons,

2.

3. Zero

Question 8.
A long straight conductor carrying current is placed near a current carrying circular loop as in the figure.

1. If B₁ is the field of the ring and B₂ the field due to straight conductor what will be the direction of B₁ and B₂ at O. (1)

2. The current through the loop and the conductor are 2A and the conductor is at as distance 20cm from the centre of loop. What should be the diameter of the loop so that the net field at O is zero. (3)
Answer:
1. B₁ into the plane and B₂ out of the plane.

2. B₁ is the field of the ring and B₂ the field of due to straight conductor.
B = B₁ – B₂ = 0
B₁ = B₂

2a = 40π × 10^-2m = 1,256m
Diameter d = 1.256 m.

Question 9.
A charged particle is travelling in the figure.

1. Name the force experienced by the charge in the region II. (1)
2. Give the expression for the net force experienced by the charge in the region II. (1)
3. If the charge reaches the region III without any change in its initial direction of motion find the velocity of the charged particle in terms of E and B. (2)

Answer:
1. Lorentz Force

2. F = q (E + v × B)

3. Electric and magnetic fields are perpendicular to each other and also perpendicular to the velocity of the particle. If the total force on the charge is zero and the charge will move in the fields undeflected. This happens when
qE = qvB or v = \(\frac{E}{B}\).

Question 10.
The medical diagnostic technique called magnetic resonance imaging (MRI) requires that patient lie in a strong magnetic field. It consists of two large solenoids, placed above and below.

1. What is solenoid?
2. Which law help you to find magnetic field due to solenoid? State the law.
3. Obtain an expression for magnetic field due to solenoid using the above law.
4. If the diameter of one of the MRI coil is increased without changing the current, does the magnetic field that it produce at its centre increases, decreases or stay the same? Justify.

Answer:
1. An insulated conducting wire wound in the form of cylinder is called solenoid.

2. Ampere’s circuital law:
Ampere’s circuital theorem states that the line integral of the magnetic field around any closed path in free space is equal to µ₀ times the net current passing through the surface.

3.

Consider a solenoid having radius Y. Let ‘n’ be the number of turns per unit length and I be the current flowing through it.

In order to find the magnetic field (inside the solenoid ) consider an Amperian loop PQRS. Let ‘l‘ be the length and ‘b’ the breadth

Applying Amperes law, we can write

Substituting the above values in eq (1), we get
Bl = µ₀ l_enc (2)
But l_enc = n l I
where ’nl ’ is the total number of turns that carries current I (inside the loop PQRS)
∴ eq (2) can be written as
Bl = µ₀nIl
B = µ₀ nI
If core of solenoid is filled with a medium of relative permittivity µ_r, then
B = µ₀µ_rnl

4. No change. Magnetic field is independent of radius.

Plus Two Physics Moving Charges and Magnetism Five Mark Questions and Answers

Question 1.
You are supplied with a galvanometer, resistor, and some connecting wires.

1. Using a circuit diagram, show how will you convert the given galvanometer into an ammeter.
2. Find the expression for the shunt resistance in the circuit.
3. A galvanometer is to be converted into an ammeter of range 0 – 1 A. Galvanometer has resistance 100Ω and the current for full scale deflection is 10mA. Find the length of the nichrome wire to be used as shunt.

Given, Resistivity ρ= 1.1 x 10^-6Ωm
Diameter of the wire = 1mm
Answer:
1.

2. Let I_g be the current through the galvanometer of resistance R_G and the shunt resistance be r_s.
Let I be the current to be measured by the converted ammeter.
We can write,
I_gR_G = (I – I_g)r_s
∴ r_s = \(\frac{I_{g} R_{G}}{\left(1-I_{g}\right)}\)

3. Given I = 1A
I_g = 10mA
R_G = 100Ω
Diameter = 1mm

Question 2.
Consider a conductor carrying current ‘I’, P is a point at a distance away from the conductor.

1. What is the direction of magnetic field at P? (1)
2. What are the factors affecting magnetic field at P due the element dl carrying current i? (1)
3. Derive an expression for magnetic field at P, if the current carrying conductor has infinite length? (2)
4. Draw a graph connecting Intensity of magnetic
   field and distance. (1)

Answer:
1. In to the plane of paper

2. dB = \(\frac{\mu_{0}}{4 \pi} \frac{|d| \sin \theta}{r^{2}}\)

• Current I
• Length of element dl
• sin of angle between element and line joining mid point of element and point.
• Inversely proportional to square of the distance between element and point.

3. Long straight conductor:
Consider a long straight conductor carrying ‘I’ ampere current. To find magnetic field at ‘P’, we construct a circle of radius r (passing through P).

According to Ampere’s circuital law we can write

4.

Question 3.
A student placed a rectangular loop carrying current in between the pole pieces of two magnetics and found that the loop is rotating.

1. What is the net force on the loop?
2. Write the expression for the torque experienced by the loop in vector form.
3. At what position will the coil be in stable equilibrium?
4. A wire of length 4m is bent in the form of a circular coil of single turn. A current 1A is flowing through the wire. If the coil is placed in a magnetic field of 0.4T find the maximum torque experienced by the coil.

Answer:
1. Zero

2.

3. θ = 0, τ = 0, Potential energy is minimum (Area vector of the coil is parallel to the direction of magnetic field.)

4. I = 4m
2πr = 4
r = \(\frac{4}{2 \pi}\)
Area = A = πr²
Maximum torque = NIAB = 1 × 1 × π × \(\left(\frac{4}{2 \pi}\right)^{2}\) × 0.4
= 0.509 Nm

Question 4.
Ampere’s theorem helps to find the magnetic field in a region around a current carrying conductor.

1. Draw the variation of intensity of magnetic field with the distance from the axis of a current carrying conductor.
2. A conductor carrying a current I is bent as shown in the figure. Apply Ampere’s theorem at the regions 1 and 2 shown in the figure.

Answer:
1.

2.

Question 5.
The force acting on a moving charge in a magnetic field is called magnetic Lorents force.

1. Write the equation of force experienced by a charged particle moving in a magnetic field.
2. Using the equation derive an expression for the force acting on a current carrying conductor of length T in a magnetic field B.
3. “A charge can move in a helical path in a magnetic field”. Do you agree with it? Explain.

Answer:
1.

2. Consider a rod of uniform cross section ‘A’ and length ‘ l ’. Let ’n’ be the number of electrons per unit volume (number density). ‘v_d’ be the drift velocity of electrons for steady current ‘I’.
Total number of electrons in the entire volume of rod = nA l
Charge of total electrons = nA l .e
‘e’ is the charge of a single electron.
The Lorentz force on electrons,

3. Yes, When an electron moves in a magnetic field with an angle θ, the electron undergoes for helical motion. The velocity of electron has two components, usinθ and ucosθ.

The component usinθ produces circular motion and ‘ucosθ’ produces translational motion. The combined effect of circular motion and translation motion will be helical motion.

Question 6.
A long wire is bent into a circular coil of one turn having radius ‘R’ and a current T is passed through it

1. Name the law to find the direction of magnetic field due to this current loop
2. Find an expression for magnetic field produced by this current loop at its centre
3. If the same wire is bent to a smaller radius Y having ‘n’ turns and send same current through it. Find the ratio of magnetic field at the centre in two cases.

Answer:
1. Right hand screw rule.

2. The magnetic field at a distance × from centre of loop is given by

3.

If wire is bend into smaller radius of n turns.

B₂ = n² B₁

Question 7.

A gas chamber is filled with hydrogen and a magnetic field is applied to it, then exposed to γ-ray. The γ-ray hits the hydrogen atom and produces high energy electron, low energy electron and positron (electron having + ve charge). The above photograph represents the trajectory of the particles. [Here magnetic field is applied Out of the plane of photo graph]

1. Which force drives the particle in a circular path and write the mathematical from.
2. Obtain a general expression for radius of the circular path.
3. Analyse the figure and match the columns given below.

Answer:
1. Magnetic Lorentz force

2. The centripetal force required for rotation is given magnetic Lorentz force Hence we can write

3. Analyse the figure and match the columns given below:

• Low energy electrons – B
• High energy electron – C
• Positron – A

Question 8.
A proton, an electron, a neutron, and an alpha particle are entering a region of uniform magnetic field with same velocities. The tracks of these particles are labelled.

1. Identify the tracks of each particle. (1)
2. Write the expression for the force experienced by a charged particle in the magnetic field in vector form. (1)
3. If the proton is moving at 90° to the uniform magnetic field what will be the change in kinetic energy of the proton? Give reason. (1)
4. An electron with energy 1 keV is entering a uniform magnetic field of 0.04T at an angle 60° with the field. Predict the path of the electron and find the characteristics of the path. (2)

Answer:
1. Identify the tracks of each particle:

• Path 1 – proton
• Path 2 – alpha particle
• Path 3 – neutron
• Path 4 – electron

2.

3. Zero. Since the force is perpendicular to the direction of velocity work done is zero.

4. KE = 1 keV = 1 × 10³ × 1.6 × 10^-19 = 1.6 × 10^-16


= 8.39 × 10^-3.

Question 9.
A current carrying conductor is bent in the form of a circular ring and is placed in the plane of the paper.

1. What is the direction of the magnetic field at the centre of the ring? (1)
2. Arrive at the expression for the magnetic field at a point on the axis of the ring. (3)
3. Another identical ring carrying the same current is brought with its axis perpendicular to the axis of the first as in figure. Find the magneticfield at the common centre. Calculate the angle between the net magnetic field and the axis of any one of the coils. (1)

Answer:
1. Out of the plane of the ring.

2.

Consider a circular loop of radius ‘a’ and carrying current ‘I’. Let P be a point on the axis of the coil, at distance x from A and r from ‘O’. Consider a small length dl at A. The magnetic field at ‘p’ due to this small element dl,

[since sin 90° -1]
The dB can be resolved into dB cosΦ (along Py) and dB sinΦ (along Px). Similarly consider a small element at B, which produces a magnetic field ‘dB’ at P.

If we resolve this magnetic field we get. dB sinΦ (along px) and dB cosΦ (along py¹) dB cosΦ components cancel each other, because they are in opposite direction. So only dB sinΦ components are found at P, so total filed at P is
B = ∫dBsinΦ

but from ∆AOP we get, sinΦ = a/x
∴ We get,

Let there be N turns in the loop then,

At the center of the loop,
r= 0

3. Magnetic fields are as shown in the fig.

The diagonal gives the resultant field.

Question 10.
An electron revolving round the nucleus acts as a magnetic dipole.

1. Which force provides the centripetal force for electron? (1)
2. Write the expression for the magnetic dipole moment of electron in vector form. What is the angle between direction of the magnetic moment and direction of angular momentum? (2)
3. A charge 2mC is moving through a circular path of radius 0.15m with frequency 1000Hz. Find the magnetic moment associated with the charge path. (2)

Answer:
1. Electrostatic force between electron and the nucleus.

2. Vectorially

The magnetic moment of the electron is opposite in direction to the angular momentum.
∴ Angle = 180°.

3. Revolving charge behaves as a current loop. Hence Magnetic field is given as

I = \(\frac{q}{T}\) = qv = 2 × 10^-3 × 100 = 0.2A
Magnetic moment = IA = 0.2 × π × 0.15²
= 0.0141 Am.

Question 11.
You are supplied with a galvanometer, resistor, and some connecting wires.

1. Using a circuit diagram, show how will you convert the given galvanometer into an ammeter. (1)
2. Find the expression for the shunt resistance in the circuit. (2)
3. A galvanometer is to be converted into an ammeter of range 0 -1 A. Galvanometer has resistance 100Ω and the current for full-scale deflection is 10mA. Find the length of the nichrome wire to be used as shunt. (3)

Given, Resistivity ρ = 1.1 × 10^-6Ωm
Diameter of the wire = 1 mm
Answer:
1.

2. Let I_g be the current through the galvanometer of resistance R_G and the shunt resistance be r_s. Let I be the current to be measured by the converted ammeter.
We can write,

3. Given I = IA
I_g = 10mA
R_G = 100Ω
Diameter = 1mm

Students can Download Chapter 2 Theory of Consumer Behaviour Questions and Answers, Plus Two Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Economics Chapter Wise Questions and Answers Chapter 2 Theory of Consumer Behaviour

Plus Two Economics Theory of Consumer Behaviour One Mark Questions and Answers

Question 1.
Suppose a consumer’s preferences are monotonic. Which bundle of goods the consumer will select over the bundle (15,15), (10,12) and (12,12).
Answer:
Consumer prefers the bundle (15,15) over the other bundles.

Question 2.
In drawing an individual demand curve, all but one of the following are kept constant.
(a) Price of the commodity
(b) Price of other commodities
(c) Income of the consumer
(d) Taste and preference of the consumer
Answer:
(b) Price of other commodities

Question 3.
Find out economic terms.

1. Bundles that are on or below the budget line.
2. Consumer prefers the bundle which has more of the goods compared to the other bundle.
3. A group of indifference curves.

Answer:

1. Budget set
2. Monotonic preferences
3. Indifference map

Question 4.
Slope of indifference curve shows.
(a) Price ratio (rule)
(b) DMRS
(c) DMU
(d) None of these
Answer:
(b) DMRS

Question 5.
In the case of inferior goods, the relationship between income and quantity demanded in.
(a) negative
(b) positive
(c) constant
(d) cannot be predicted
Answer:
(a) negative

Question 6.
Elasticity in a rectangulas hyperbola is:
(a) 0
(b) a
(c) 1
(d) 0.5
Answer:
(c) 1

Question 7.
Rise in demand due to fall in price is called:
(a) Increase in demand
(b) Expansion of demand
(c) Contraction of demand
(d) Decrease in demand
Answer:
(b) Expansion of demand

Question 8.
If demand falls from 100 to 75 units due to rise in price from 10 to 15, the value of elasticity is…
(a) 1
(b) 0.5
(c) 0
(d) 2
Answer:
(b) 0.5

Plus Two Economics Theory of Consumer Behaviour Two Mark Questions and Answers

Question 1.
Suppose Raju is indifferent to bundles (8,7) and (7,7). Are the preferences of Raju are monotonic?
Answer:
No, if his preferences are monotonic, he will prefer the bundle (8,7) over (7,7).

Question 2.
Consider a market where there are 3 consumers and suppose their demands for the good are given as follows:

Calculate the market demand for the good.
Answer:

Question 3.
Complete the following table.

Answer:

Question 4.
Pick out the odd one and justify your answer. Bread and butter, Pen and Ink, Butter and Jam, Tennis ball and Tennis racket.
Answer:
Butter and Jam. Others are complementary goods.

Question 5.
State the law of demand. Test the applicability of the law in the status symbol goods.
Answer:
Law of demand – “other things remaining the same as the price of a commodity falls, its quantity demanded increases and vice versa”.

Certain ostentatious goods like luxury cars; diamonds, etc. are exceptions to the law of demand. These goods are considered as status symbol goods consumed by the rich. The status goes up as price increases. Therefore, the demand for these goods increases as their price increases.

Question 6.
The demand function of a commodity is Q = 30 – 2 P. If it is a free good, quantity demanded would be.
Answer:
If the commodity is a free good, the price = 0
Q = 30 – 2 × 0
= 30 – 0 = 30

Question 7.
When the elasticity of demand for a product is unitary,

1. Name the shape of the demand curve?
2. Give the value of price elasticity of demand?

Answer:

1. Rectangular hyperbola
2. Value of price elasticity of demand is unity

Question 8.
Consider the demand curve D(p) = 10 – 3p. What is the elasticity at price 2?
Answer:
D(P) = 10 – 3p
Since P = 2, we get
D(P) = 10 – 3 × 2
= 10 – 6 = 4
Σd = \(\frac{\Delta Q}{\Delta P}=\frac{4}{2}\) = 2
Thus there is elastic demand.

Plus Two Economics Theory of Consumer Behaviour Three Mark Questions and Answers

Question 1.
Suppose a consumer buys bundles of good 1 and good 2. His income is given as ‘M’ and it is fully spent. If the prices of good 1 and good 2 are P₁ and P₂ respectively, state the consumer’s budget constraint.
Answer:
We assume that the consumer buys bundles of good 1 and good 2. The consumer’s consumption expenditure is limited by his income. Given the prices of good 1 and good 2 as P₁ and P₂ respectively and his income as ‘M’, consumer’s budget constraint can be represented as P₁X₁ + P₂X₂ ≤ M

Question 2.
Match the following

Answer:

Question 3.
Given below an indifference curve.

1. State the meaning of an indifference curve.
2. Identify the points A, B, C and D on the 1C.

Answer:
1. An indifference curve shows different bundles that give the consumer the same level of satisfaction. In other words, an indifference curve joins all points representing bundles which are considered indifferent by the consumer.

2. Points A and B are on the indifference curve indicating the same level of satisfaction. Point C lies above the indifference curve representing the preferred bundles. Point D lies below the indifference curve showing the inferior bundles.

Question 4.
Two demand function equations are given below.
QD₁ = 60 – 10P
QD₂ = 80 -10P
a. Derive two demand schedules forthe above de-mand functions (Take the values of P as 1,2,3,4,5)
QD₁= 60- 10P
QD₂ = 80 – 10P
P = 1,2,3,4,5
Answer:

Question 5.
Fora linear demand curve,
d (p) = a – bp; 0 < p < a/b = 0; p> a/b

1. State the meaning of ‘a’ and -b?
2. What does the slope of the demand curve mean?

Answer:

1. In the equation of linear demand curve, ‘a’ is the vertical intercept and-b is the slope of the demand curve.
2. The slope of the demand curve measures the rate at which demand changes with respect to its price.

Question 6.
Observe the three budget lines drawn below.

If AB is The Intial Budget Line, What Causes The Shift in budget line.

1. from ABtoAB
2. from ABtoA

Answer:

1. Fall in price of good 1
2. A rise in income or fall in prices of both good 1 & good 2.

Question 7.
Choose the correct answer from the given multiple choices.
1. Which of the following goods has more elastic demand?

• Rice
• Computer
• Electricity
• Life-saving drugs
• Salt

2. Identify the nature of demand curve when elasticity of demand is equal to one.

• Perfectly elastic demand
• Rectangular hyperbola
• Parallel to Vertical axis.
• Perfectly inelastic demand
• Parallel to horizontal axis

Answer:

1. computer
2. rectangular hyperbola

Question 8.
Compare the slope of the budget line and slope of the indifference curve.
Answer:
The slope of the indifference curve shows the rate at which the consumer is willing to substitute with others. Here the substitution is in terms of satisfaction. The slope of the budget line shows the rate at which a consumer is able to substitute our good for others. Here the substitution is in terms of price.

Question 9.
Indicate for each of the following situations whether it would shift the demand curve upward or downward

1. The price of substitute falls.
2. Consumer’s income increases
3. There is sudden rise in population.
4. Complementary goods become more expensive.
5. There is possibility of further fall in price, nilei
6. New cheaper substitutes of the commodity appear in the market.

Answer:

1. downward
2. upward
3. upward
4. downward
5. downward
6. downward

Question 10.
Classify the following goods into two based on their elasticity.
Petrol, medicine, tomatoes, car, garments, salt
Answer:

Question 11.
Imagine that you are the finance minister of Kerala. You want to raise more tax revenue. How will you use elasticity in your tax proposals?
Answer:
As finance minister, I will raise taxes on goods like cigarettes, liquor, and luxury products. These goods have inelastic demand and therefore, their demand will not decline in proportion to the price rise. This will help to raise more tax revenue.

Question 12.
Categorize the following into substitutes and complementaries.
Coffee and tea, pen and ink, bread and jam, scooter and petrol, shoes and chappels, airplane and train.
Answer:
1. Substitutes:

• Coffee and tea
• Shoes and chapels
• Airplane and train

2. Complementaries:

• pen and ink
• bread and jam
• scooter and petrol

Question 13.
The price of X falls from ₹8 per unit to t 6. Consequently, the quantity demanded increased from 80 to 100. Calculate the price elasticity of demand.
Answer:
\(\frac{\Delta Q}{\Delta P} \times \frac{P}{2}\)
ΔQ = 20
ΔP = 2
P = 8
Q = 80
ep = \(\frac{20}{2} \times \frac{8}{80}=\frac{160}{160}=1\)
ep = 1
This is unitary elastic demand

Question 14.
State whether true or falls

1. The demand curve is generally sloping upward
2. Demand for cosmetic is elastic
3. Utility is want satisfying power of a commodity

Answer:

1. False. Demand curve slops downward.
2. True
3. True

Question 15.
What do you mean by inferior goods? Give some example?
Answer:
Inferior goods are those goods whose demand decreases with rise in the income and demand increases with the fall in the income.

Question 16.
Give an account of price elasticity of demand.
Answer:
Price elasticity of demand is a measure used in economics to show the responsiveness, or elasticity, of the quantity, demanded of a good or service to a change in its price. It was devised by Alfred Marshall. The formula for the

The above formula usually yields a negative value, due to the inverse nature of the relationship between price and quantity demanded, as described by the “law of demand”.

This measure of elasticity is sometimes referred to as the own-price elasticity of demand for a good, i.e., the elasticity of demand with respect to the good’s own price, in order to distinguish it from the elasticity of demand for that good with respect to the change in the price of some other good, i.e., a complementary or substitute good. The latter type of elasticity measure is called a cross-price elasticity of demand.

Question 17.
Given the level of income and market prices, the rational consumer wants to attain the maximum level of satisfaction. Using the budget line and indifference curve that you have studied, answer the following questions.

1. Construct a diagram showing the consumer’s equilibrium.
2. What condition is satisfied at this equilibrium point?

Answer:
1.

2. Price ratio = Marginal Rate of Substitution

Question 18.
Two diagrams related to demand are given below. What do they represent?

Answer:

• Figure 1 – Movement along the demand curve due to change in price.
• Figure 2 – Shift in demand due to change in non price factor.

Question 19.
If other things remaining same, graphically explain what happens to the demand curve for chicken if there is

1. An increase in the price of fish.
2. A decrease in family income.
3. An increase in the price of chicken.

Answer:
1.

2.

Plus Two Economics Theory of Consumer Behaviour Four Mark Questions and Answers

Question 1.
“Price elasticity of demand is different at different points on the linear demand curve” Prove this point diagrammatically.
Answer:
On the linear demand curve, Price elasticity of demand varies from point to point. It can be seen from the linear demand given below. As we move from the higher point to lower point the value of elasticity goes on decreasing.

Question 2.
Locate the optimum bundle of the consumer in a diagram. Also, suggest the conditions for the consumer’s optimum.
Answer:
The optimum bundle of the consumer is located at the point where the budget line is tangent to one of the indifference curves. It is drawn below.

1. Condition I: Budget line should be tangent to the indifference Curve
2. Condition II: Slope of 1C (MRSxy) should be equal to slope of budget line (Price Ratio)

Question 3.
Draw a demand curve at all points of which price elasticity remains the same. Also, name the demand curve and give the equation for it.
Answer:

The demand curve at which all points represent same elasticity is called a rectangular hyperbola. The equation is xy = c, where, x and y are two variables and c is a constant. With such a demand curve, no matter at what point, the consumer consumes at a constant rate.

Question 4.
Gopan buys 8 kg of rice at price ₹15 per kg. It is found that the price elasticity of demand is 2. At what price he will buy 13 kg of rice?
Answer:
Price elasticity of demand means the degree of responsiveness of demand due to change in price. The formula for price elasticity of a product is,
Ep = \(\frac{\Delta Q}{\Delta P} \times \frac{P}{Q}\)
Here the price elasticity of demand = 2
\(\frac{\Delta Q}{\Delta P} \times \frac{P}{Q}\) = 2
\(\Delta P=\frac{8.12}{2}\)
ΔP = 4.06
new price is 13 – 4.06 = 8.96

Question 5.
Show how the following changes affect the budget line
Answer:

1.  Increase in income of the consumer
2. Decrease in income
3. Fall in the price of good 1
4. Rise in the price of good 1

Answer:

Question 6.
Suppose there are 2 consumers in the market for a good and their demand function are as follows Find out market demand function.
Answer:
First individuals demand function Second individual’s demand function = d₂ (P) = 30 – 2p
∴ Market demand function = d₁ (P) = d₂ (P)
= 20 – p + 30 – 2p
= 50-3p

Question 7.
Derive the slope of budget line, using symbols?
Answer:
The slope of the budget line measures the amount of change in good 2 required per unit of change in good 1 along the budget line. Consider any two points (X₁, x₂) and (x₁ + ΔX₁,x₂ + Δx₂) on the budget line.

Question 8.
When price of orange is ₹4, consumer buys 50 units of it. The price elasticity is -2. How many units will the consumer buy at ₹3 per unit of orange?
Answer:
Σ d = -2
When P = 4, q = 50
When P = 3, q = ?

100 = 4 Δ q
Δ q = 100/4 =25
∴ New quantity = q + Δ q
= 50 + 25 = 75

Plus Two Economics Theory of Consumer Behaviour Five Mark Questions and Answers

Question 1.
The consumer has an income of Rs. 100. He wants to consume two goods X and Y. Prices of X and Y are Rs. 20 and 10 respectively.

1. State the equation of the budget line.
2. How much of good X he can buy if he spends entire income on good X.
3. How much of good Y he can buy if he spends entire income on good Y.

Answer:
1. The equation of the budget line is given as P₁ + P₂X₂=M
Where P₁ and P₂ are prices of good 1 and good 2 respectively. X₁ and X₂ are quantities of two goods. ‘M’ denotes the income of the consumer.

2. 5 units of good X.

3. 10 units of good Y

Question 2.
A consumer wants to consume two goods. The prices of the two goods are ₹4 and ₹5 respectively. The consumer’s income is ₹20.

1. Write down the equation of the budget line
2. How much of good 1 can the consumer consume if he spends his entire income on the good?
3. How much of good 2 can the consumer consume if he spends his entire income on the good?
4. What is the slope of the budget line?

Answer:
1. the equation of the budget line is p₁x₁+ p₂x₂= M
where P₁ and P₂ are prices of good1 and good 2. x-s and x2 are quantities of goods and M is his money income,

2. X₁ = M/ P₁
= 20/4 = 5 units
X₂ = M/ P₂
= 20/5 = 4 units

3. slope of the budget line is
= -P₁/P₂
= -4/5
= – 0.8 units

Question 3.
One important factor influencing demand is price of the product.

1. Can you make a list of four other factors that influence the demand for a good?
2. Also, establish the relationship between the factor identified and the demand for the product.

Answer:

• Column 1. Income, price of substitutes, advertisement, tastes & preferences
• Column 2. Positive, Negative, Positive, Positive

Question 4.
A few goods are given below. State whether the demand for the product is elastic or inelastic. Justify your answer.
Rice, salt, car, life-saving drugs, computer, electricity
Answer:

1. Rice – Inelastic – Essential good Car – Elastic – Luxury good.
2. Salt – Inelastic- Insignificant share in total expenditure. Life saving drugs – Inelastic – Not possible to postpone purchase.
3. Computer- Elastic demand,Non-essential good. Electricity – Inelastic – No substitutes

Question 5.
As you know elasticity of demand is influenced by several factors. Observing the nature of good given in the first column, complete the following table. Write whether the demand for the product is elastic or in¬elastic and also the reason.

Answer:
Column 2 – Column 3
Elastic – luxury product
Elastic – close substitutes are available
Elastic – no substitute
Elastic – Essential good
Elastic – Essential good

Question 6.
The diagram below shows the demand curves of commodities and b which are complementary to each other.

1. What are complementary goods?
2. Write two examples each for commodity
3. The diagram shows the changes in the demand curve of commodity ‘b’ due to a fall in the price of commodity ‘a’.

Answer:

1. Complementary goods are those goods demanded jointly for the consumption of one good, it requires other good too.
2. Car, petrol, mobile phone, sim card, etc.
3. When the price of ‘a1 comes down more of ‘a’ will be demanded. This would result in an increase in the demand for its complementary ‘b’.

Plus Two Economics Theory of Consumer Behaviour Eight Mark Questions and Answers

Question 1.
Prepare a seminar paper on the measurement of price elasticity of demand.
Answer:
Methods of measuring elasticity
Elasticity can be measured by using methods such as percentage method, linear method/point method, and expenditure method.
The detailed descriptions of the methods are as follows:
1. Percentage method:
Percentage method is also known as the proportionate method. As per percentage method estimate the elasticity of demand by dividing the percentage change in quantity demanded by the percentage change in price as given earlier. Thus, the formula for estimating elasticity of demand through percentage method is given as The procedure of computing elasticity using percentage method is provided in the example: 1.

2. Linear method/point method:
It is popularly known as the mathematical method of measuring price elasticity of demand. It is also known as point method of measuring elasticity of demand. The elasticity would be different on different points in a straight-line demand curve.

The following points can be observed from the above figure:
a. When a straight-line demand curve cut the X-axis, the elasticity of demand would be zero (e_D = 0) (perfect inelastic demand).

b. The elasticity of demand at the midpoint on a straight-line demand curve would be one (e_D = 1) (unitary elastic demand).

c. When a straight-line demand curve cut the Y-axis, the elasticity of demand would be zero (e_D = a) (perfect elastic demand).

d. Between the midpoint and the point where the demand curve cuts the X-axis on a straight line would be less than one (inelastic demand) (e_D = <1).

e. Between the midpoint and the point where the demand curve cuts the Y-axis on a straight line would be more than one (elastic demand) (e_D>1).

3. Expenditure method:
The total expenditure on a commodity can be find out by multiplying the quantity of commodity with its price. As per expenditure method, the price elasticity is measured by comparing the change in price and the change in total expenditure. Three possibilities may occur in this context.

a. If the total expenditure does not change even if there is a price change, then elasticity would be 1 or unity (e_D = 1)

b. If the total expenditure decreases as a result of increase in price or total expenditure increases as a result of fall in price, it would be more elastic (e_D> 1).

c. If the total expenditure increases as a result of decrease in price or total expenditure decreases as a result of rise in price, it would be less elastic (e_D = <1).

Question 2.
The demand curve for apple is given below. Show the effect on the demand curve for apple due to the following factors.

1. A newspaper report stating the health benefit of apple
2. The price of apple increases.
3. What is price elasticity of demand? What do you think about the price elasticity of apple. Justify your answer.

Answer:
1. The demand curve for apple shifts rightward. Taste and preference arise in favour of apple.

2. There is a movement along the demand curve. This is due to an increase in price. Quantity demand of apple falls.

3. It is the responsiveness of quantity demanded to a change in price.
\(\mathrm{Ped}=\frac{\% \mathrm{Δad}}{\% \mathrm{ΔP}}\)

Students can Download Chapter 13 Amines Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 13 Amines

Plus Two Chemistry Amines One Mark Questions and Answers

Question 1.
The reaction of aniline with benzoyl chloride gives
(a) benzoin
(b) benzanilide
(c) benzalaniline
(d) benzamide
Answer:
(b) benzanilide

Question 2.
Nitromethane on reaction with H₂/Pd gives ____________
Answer:
Ethanamine (CH₃CH₂NH₂)

Question 3.
Say TRUE or FALSE
Only aliphatic primary amines give a foul smelling compound with CHCl₃ and alcoholic potash.
Answer:
False

Question 4.
When benzene diazonium salt solution is treated with KI ___________ is formed.
Answer:
lodo benzene (C₆H₅I)

Question 5.
Identify ‘Z’ in the sequence:

(a) Nitrobenzene
(b) Benzene
(c) Fluorobenzene
(d) Phenol
Answer:
(a) Nitrobenzene

Question 6.
Amine that cannot be prepared by Gabriel Phthalimide Synthesis is _________
Answer:
Aniline

Question 7.
Phenyl isocyanide, is prepared from aniline by ________
Answer:
Carbylamine reaction

Question 8.
Secondary amines can be prepared by ___________
Answer:
Reduction of nitro compounds

Question 9.
By which process aniline can be purified ___________
Answer:
Steam distillation

Question 10.
A primary amine that can be obtained both by the reduction of cyanides and amides is ___________
(a) methyl amine
(b) benzylamine
(c) aniline
(d) isopropylamine
Answer:
(b) benzylamine

Question 11.
The amine which will not liberate nitrogen on reaction with nitrous acid is __________
Answer:
t-butyl amine

Plus Two Chemistry Amines Two Mark Questions and Answers

Question 1.
Ammonia is less basic than aniline.

1. What is the reason for low basic character of aniline compared to ammonia?
2. Draw the resonating structures of aniline.

Answer:
1. In aromatic amines the lone pair on nitrogen is in conjugation with benzene ring due to resonance effect and thus making it less available for protonation. Hence basic character is less than that of ammonia.
2.

Question 2.
Two compounds are functional isomers of each other. On reduction one gives primary amine and the other gives secondary amine.

1. Identify the class of compounds.
2. Explain the reduction reaction.

Answer:

1. Cyanides and Isocyanides.
2. When cyanides are reduced using LiAlH₄ primary amines are obtained. Whereas when isocyanides are reduced using LiAlH₄ secondary amines are obtained.

Question 3.
When alkyl halide is treated with alkali metal cyanide, cyanides are obtained as the major product. Assume that AgCN is used instead of NaCN.

1. What will be the product?
2. Justify.

Answer:

1. Isocyanide
2. Silver cyanide is predominantly covalent. Hence nitrogen atom is free to donate electron pair forming isocyanide as the main product.

Question 4.
How is sulphanilic acid prepared from aniline?
Answer:
Aniline reacts with concentrated sulphuric acid to form anilinium hydrogensulphate which on heating with sulphuric acid at 453-473 K produces sulphanilic acid.

Question 5.
Write chemical equations for the preparation of benzene, fluorobenzene, and nitrobenzene from benzenediazonium chloride.
Answer:

Question 6.

1. What are Schiff s bases?
2. How are they formed?

Answer:
1. Schiff’s bases are substituted imines.
2. These are formed when 1° amines are treated with carbonyl compounds.

Question 7.
What are the products formed by the reaction of ethanolic NH₃ with C₂H₅CI? Write the chemical equation.
Answer:
A mixture of ethanamine, N-ethylethanamine and N,N-diethylethanamine are formed

Question 8.
Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Answer:
Gabriel phthalimide synthesis involves the conversion of alkyl halide (R – X) to 1° amine (R – NH₂). Ammonolysis of R – X, on the other hand give 2°, 3° and quaternary salt as other byproducts. Hence, for the production of pure 1° amines, Gabriel phthalimide synthesis is preferred to ammonolysis reaction.

Plus Two Chemistry Amines Three Mark Questions and Answers

Question 1.
Nitrogen containing functional groups are classified into different types.

1. Which are they?
2. Explain one method to prepare cyanide.
3. Give chemical equations.

Answer:

1. Nitro compounds, Amines, Cyanides, Isocyanides and Diazo compounds.
2. When alkyl halides on heating with alcoholic KCN or NaCN cyanide is obtained.
3. R – X + KCN → R-CN + KX. e.g. CH₃ – CH₂ -Br + KCN → CH₃CH₂CN + KBr

Question 2.
Consider the following chemical equations:

1. a) Identify ‘A’ and ‘B’.
2. b) Name the reactions.
3. c) Give any one method to prepare beznene diazonium chloride.

Answer:
1. A – LiAIH₄ B – Br₂/KOH
2.

• i) is reduction of acid amide.
• ii) is Hoffmann’s bromamide degradation reaction.

3. Benzene diazonium chloride can be prepared by diazotization of aniline. It can be done by treating aniline with sodium nitrite and HCI at 273-278 K.

Question 3.
Chlorobenzene can be prepared from benzene diazonium chloride in two ways.

1. a) Do you agree with it?
2. b) Which are the two ways?
3. c) Write the chemical equation for the reactions.

Answer:
1. Yes.
2. By Sandmeyer’s reaction and Gattermann’s reaction.
3. Sandmeyer’s reaction

Gattermann’s reaction

Question 4.
Match the following:

Answer:

Question 5.
Diazonium salts are very important class of compounds used for synthesis of variety of aromatic compounds.

1. How is nitrobenzene prepared from diazonium salt?
2. Give an example for coupling reaction.
3. Give an important use of diazonium salt.

Answer:

1. Benzene diazonium chloride is treated with fluoroboric acid to get benzenediazonium fluoroborate which when heated with aqueous sodium nitrite solution in the presence of copper, nitrobenzene is formed.

2. When benzenediazonium chloride is coupled with aniline in acid medium, p-aminoazobenzene (an yellow azo dye) is formed.

3. It can be used as intermediate for the synthesis for many organic compounds.

Question 6.
An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B, and C.
Answer:
Hoffman bromide reaction

Question 7.
Write the IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

1. C₆H₅NHCH₃
2. (CH₃CH₂)₂NCH₃
3. m-BrC₆H₄NH₂

Answer:

1. N-Methylbenzenamine (2° amine)
2. N-Ethyl-N-methylethanamine (3° amine)
3. 3-Bromobenzenamine (1° amine)

Plus Two Chemistry Amines Four Mark Questions and Answers

Question 1.
C₆H₅NH₂ gives a foul smelling compound ‘A’ with chloroform in presence of KOH (alcoholic).

1. What is compound ‘A’?
2. Write the chemical equation for the reaction.

Answer:

a) ‘A’ is phenyl isocyanide (or phenyl carbylamine)
b) Carbyl amine reaction

Question 2.
1. In a group discussion, a student argued that alkyl amines are more basic than NH₃.

• Is it correct?
• Justify your answer with suitable explanation.

2. Arrange the following amines in the decreasing order of basic strength in aqueous solution.
CH₃NH₂, (CH₃)₂NH, NH₃, (CH₃)₃N Justify.
Answer:
1.

• Yes.
• Electron realeasing inductive effect (+l) of alkyl groups increases the availability of lone pair on nitrogen.

2. (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃
Inductive effect is maximum for 3° amines. At the same time steric hindrance is maximum for 3° amines. An interplay of the inductive effect, solvation effect and steric hindrance of the alkyl group decides the basic strength of alkylamines.

Question 3.

a) Watch the above diagram and fill the labelled boxes A, B, C, and D.
b) If you are treating B with nitrous acid, predict the product that can be formed.
Answer:
a)

b) CH₃ CH₂ – OH

Question 4.
a) Benzenediazonium chloride is a very important compound in organic chemistry. Give the structure. How it is prepared?
b) How phenol and iodobenzene are prepared from the above compound?
Answer:

Question 5.

Answer:

Question 6.
a) Arrange the following in the increasing order of basic strength.
C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH, CH₃NH₂
b) How will you convert aniline to phenol?
Answer:
a) C₆H₅NH₂ < C₆H₅N(CH₃)₂ < CH₃NH₂ < (C₂H₅)₂NH

Question 7.

1. How is nitrous acid prepared in the laboratory?
2. A student treated methylamine and aniline separately with nitrous acid. What are the products formed in each case? Give chemical equations.

Answer:
1. By treating hydrochloric acid with sodium nitrite,
2. Methyl amine reacts with nitrous acid to form methyldiazonium salt which being unstable, liberates nitrogen gas quantitatively and forms methanol.

Aniline reacts with nitrous acid at low temperature (273-278 K) to form benzenediazonium salt, a very important compound used in synthetic organic chemistry.

Question 8.
Give one chemical test to distinguish between the following pairs of compounds.

1. Methylamine and dimethylamine
2. Secondary and tertiary amines
3. Ethylamine and aniline
4. Aniline and benzylamine
5. Aniline and N-methylaniline

Answer:
1. Methyl mine is a 1° amine and so it gives the carbylamine test. When methylamine is warmed with chloroform and alcoholic solution of KOH, foul smelling methyl isocyanide is formed.

Dimethylamine is a 2° amine and hence it does not answer carbylamine test. Or they can be distinguished by Hinsberg’stest.

2. Secondary and tertiary amines can be distinguished by Hinsberg’s test.
Secondary amines react with Hinsberg’s reagent (benzene sulphonyl chloride) to form N, N- dialkyl benzene sulphonamide which is insoluble in alkali.
Tertiary amines do not react with Hinsberg’s reagent.

3. Ethylamine is an aliphatic 1° amine. When treated with HNO₂ (NaNO₂ and HCI) it forms ethanol with liberation of N₂ gas.

CH₃ – CH₂ – NH₂ + HNO₂ → CH₃ – CH₂ – OH + N₂ + H₂O

Aniline is an aromatic 1° amine. When treated with HNO₂ at low temperature (273 – 278 K) it gives benzene diazonium chloride which undergoes coupling reaction with phenol to form an orange azo dye.

C₆H₅NH₂ + HNO₃ + HCI → C₆H₅N₂⁺ + Cl^– + 2H₂O

4. Aniline forms azodye with benzene diazonium chloride but benzyl amine does not.

5. Aniline being a primary amine gives carbylamine test while N-methylaniline being a 2° amine does not answer carbylamine test.

Question 9.
Write short notes on the following:

1. Carbylamine reaction.
2. Hoffman’s bromamide reaction.

Answer:
1. Methyl amine is a 1° amine and so it gives the carbylamine test. When methylamine is warmed with chloroform and alcoholic solution of KOH, foul smelling methyl isocyanide is formed.

Dimethyl amine is a 2° amine and hence it does not answer carbylamine test. Or they can be distinguished by Hinsberg’stest.

2. When an acid amide is heated with Br₂ and an aqueous or ethanolic solution of alkali, a primary amine containing one carbon less than the initial amide is obtained. This is called Hoffmann’s bromamide degradation reaction.

Question 10.
Write the IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

1. (CH₃)₂CHNH₂
2. CH₃(CH₂)₂NH₂
3. CH₃NHCH(CH₃)₂
4. (CH₃)₃CNH₂

Answer:

1. Propan-2-amine (1° amine)
2. Propan-1-amine (1° amine)
3. N-Methylpropan-2-amine (2° amine)
4. 2-Methylpropan-2-amine (1° amine)

Question 11.
Account for the following:

1. pK_b of aniline is more than that of methylamine.
2. Ethylamine is soluble in water whereas aniline is not.

Answer:

1. In CH₃– NH₂, the +l effect of methyl group increases the electron density around N atom and it increases the electron releasing tendency of the molecule. In C₆H₅ – NH₂ the resonance effect causes delocalisation of lone pair over the ring and thereby decreases its basic strength. Since aniline is less basic than methylamine, its pKb value is greater than that of CH₃NH₂.

2. Solubility of ethylamine in water is attributed to its ability to form hydrogen bonds with water molecules. In aniline the non-polar hydrocarbon part (the ring skeleton) is relatively larger and therefore, has no interactions with polar water molecules. It also decreases the tendency of hydrogen bonding with water molecules. Hence, aniline is not soluble in water.

Question 12.
Account for the following:

1. Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
2. Although amino group is o- and p- directing in anomatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.

Answer:
1. Methyl amine, being more basic than water exists in water as N-methylammonium hydroxide which reacts with ferric chloride to form hydrated ferric oxide.

2. Aniline being a base mostly gets protonated in the presence of acids to form anilinium ions (NH₃⁺). For electrophlic ring substitution, – NH₂ group is activating and ortho and para directing whereas – NH₃⁺ is deactivating and meta directing. In aniline, the probability of NO₂⁺ attack at para position is relatively more because of steric hinderance at ortho position. Anilinium ion, the attack of NO₃⁺ mostly occurs at meta position.

Question 13.
Account for the following:

1. Aniline does not undergo Friedel-Crafts reaction.
2. Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

Answer:
1. Aniline is a base due to electron donating nature of lone pair on N atom of -NH₂ goup. Aniline therefore, forms salt with AICI₃ which is Lweis acid and a catalyst used in Friedel Craft’s reaction.

So the catalyst will not be available to produce electrophile.

2. Arenediazonium ion is resonance stabilised whereas no resonance stabilisation occurs in alkyl diazonium ion.

Plus Two Chemistry Amines NCERT Questions and Answers

Question 1.
Write the IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

1. (CH₃)₂CHNH₂
2. CH₃(CH₂)₂NH₂
3. CH₃NHCH(CH₃)₂
4. (CH₃)₃CNH₂
5. C₆H₅NHCH₃
6. (CH₃CH₂)₂NCH₃
7. m-BrC₆H₄NH₂

Answer:

1. Propan-2-amine (1° amine)
2. Propan-1-amine (1° amine)
3. N-Methylpropan-2-amine (2° amine)
4. 2-Methylpropan-2-amine (1° amine)
5. N-Methylbenzenamine (2° amine)
6. N-Ethyl-N-methylethanamine (3° amine)
7. 3-Bromobenzenamine (1° amine)

Question 2.
Give one chemical test to distinguish between the following pairs of compounds.

1. Methylamine and dimethylamine
2. Secondary and tertiary amines
3. Ethylamine and aniline
4. Aniline and benzylamine
5. Aniline and N-methylaniline

Answer:
1. Methyl amine is a 1° amine and so it gives the carbylamine test. When methylamine is warmed with chloroform and alcoholic solution of KOH, foul smelling methyl isocyanide is formed.

Dimethyl amine is a 2° amine and hence it does not answer carbylamine test. Or they can be distinguished by Hinsberg’stest.

2. Secondary and tertiary amines can be distinguished by Hinsberg’stest.

Secondary amines react with Hinsberg’s reagent (benzene sulphonyl chloride) to form N, N- dialkyl benzene sulphonamide which is insoluble in alkali.
Tertiary amines do not react with Hinsberg’s reagent.

3. Ethyl amine is an aliphatic 1° amine. When treated with HNO₂ (NaNO₂ and HCI) it forms ethanol with liberation of N₃ gas.

Aniline is an aromatic 1° amine. When treated with HNO₂ at low temperature (273 – 278 K) it gives benzene diazonium chloride which undergoes coupling reaction with phenol to form an orange azo dye.
C₆H₅NH₂ + HNO₃ + HCI → C₆H₅N₂⁺+Cl^– + 2H₂O

4. Aniline forms azodye with benzene diazonium chloride but benzyl amine does not.

5. Aniline being a primary amine gives carbylamine test while N-methylaniline being a 2° amine does not answer carbylamine test.

Question 3.
How will you convert:
(i) Ethanoic acid into methanamine
(ii) Hexanenitrile into 1-aminopentane
Answer:

Question 4.
How will you convert:
(i) Methanol to ethanoic acid
(ii) Ethanamine into
Answer:

Question 5.
How will you convert:
(i) Ethanoic acid into propanoic acid
(ii) Nitromethane into dimethylamine
Answer:

Question 6.
An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B, and C.
Answer:

Question 7.
Write short notes on the following:

1. Carbylamine reaction
2. Diazotisation
3. Hoffmann’s bromamide reaction
4. Coupling reaction
5. Gabriel phthalimide synthesis.

Answer:
1. Carbylamine reaction
Primary amines (both aliphatic and aromatic) when heated with chloroform and ethanolic potassium hydroxide form isocyanides (also known as carbylamines) which have a foul smell. This reaction is called carbylamine reaction and is used as a test for primary amines. Secondary and tertiary amines do not show this reaction.

2. Diazotisation:
Aromatic amines react with nitrous acid (HNO₂) at low temperature (273-278 K) to form diazonium salts. The process is known as diazotisation eg.

3. Hofmann’s bromamide reaction:
In this reaction, an acid amide is heated with Br₂ and aq. NaOH when 1° amine having one carbon atom less is produced. This involves migration of alkyl or aryl group from carboxyl carbon of the amide to nitrogen atom of the amine. The amine so formed contains one carbon less than that present in amide.

4. Coupling reaction:
Diazonium salts react with phenol and amines to give azo compounds which have an extended conjugate system having both the aromatic rings joined by -N = N – bond. eg.

5. Gabriel phthalimide synthesis:
This method is used for preparing only 1° amines. In this method, phthalimide is treated with alcoholic KOH to give potassium phthalimide, which is treated with alkyl halide or benzyl halide to form N- alkyl or aryl phthalimide. The hydrolysis of N- alkyl phthalimide with 20% HCI under pres¬sure or refluxing with NaOH gives 1° amines.


The more convenient method is by the treatment of alkyl phthalimide with hydrazene.

Phthalic acid can again be converted into phthalimide and is used again and again. This method is very useful because it gives pure amines.

Question 8.
Give plausible explanation for each of the following:

1. Why are amines less acidic than alcohols of comparable molecular masses?
2. Why are primary amines higher boiling than tertiary amines?
3. Why are aliphatic amines stronger bases than aromatic amines?

Answer:
1. Loss of proton from amines give amide ion whereas loss of a proton from alcohol gives an alkoxide ion.

Since O is more electronegative than N, therefore, RO^– can accomodate the -ve charge more eaisly than RNH^–. Consequently, RO^– is more stable than RNH^–. Thus, alcohols are more acidic than amines.

2. Primary amines (RNH₂) have two hydrogen atoms on the N atoms and therefore, form intermolecular hydrogen bonding

Tertiary amines (R₃N) do not have hydrogen atoms on the N atom and therefore, these do not form hydrogen bonding in primary amines, they have higher boiling points than tertiary amines of comparable molecular mass. For example, boiling point of n-butylamine is 351 K while that of tert-butylamine is 319K.

3. Both arylamines and alkalamines are basic in nature due to the presence of lone pair on N-atom. But arylamines are less basic than alkyamines. For example, aniline is less basic than ethylamine as shown by K_b values:
Ethylamine: K_b = 4.7 × 10^-4
Aniline: K_b =4.2 × 10^-10

The less basic character of aniline can be explained on the basis of aromatic ring present in aniline. Aniline can have the following resonating structures:

It is clear from the above resonating structures that three of these (II, III and IV) acquire some positive charge on N atom. As a result, the pair of electrons become less available for protonation. Hence, aniline is less basic than ethyl amine in which there is no such resonance.

Students can Download Chapter 3 Current Electricity Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 3 Current Electricity

Plus Two Physics Current Electricity NCERT Text Book Questions and Answers

Question 1.
A storage battery of a car has an e.m.f. of 12V. If the internal resistance of the battery is 0.4W, what is the maximum current that can be drawn from the battery?
Answer:
E = 12 V, r = 0.4Ω, I = ?
Since I = \(\frac{E}{r}=\frac{12}{0.4}\) = 30A.

Question 2.
A battery of e.m.f. 10V and internal resistance 3W is connected to a resistor. If the current in the circuit is 0.5A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Answer:
Given E = 10V, r=3?, I = 0.5, R = ?
Since V = E – Ir
or V= 10 – 0.5 × 3
or V = 8.5 volt
And R = \(\frac{V}{I}=\frac{8.5}{0.5}\) = 17A.

Question 3.

1. Three resistors 1W, 2W and 3W are combined in series. What is the total resistance of the combination?
2. If the combination is connected to a battery of e.m.f. 12V and negligible internal resistance, obtain the potential drop across each resistor:

Answer:
Given

R₁ = 1Ω, R₂ = 2Ω , R₃= 3Ω
1. Total resistance of series combination
R = R₁ + R₂ + R₃
or R – 1 + 2 + 3 = 6Ω.

2. Since V=IR
I = \(\frac{V}{R}=\frac{12}{6}\)
∴ V₁ = IR₁ = 2 × 1 = 2V
V₂ = IR₂ = 2 × 2 = 4V
V₃ = IR₃ = 2 × 3 = 6V.

Question 4.
A silver wire has a resistance of 2.1W at 27.5°c, and a resistance of 2.7W at 100°C. Determine the temperature coefficient of resistivity of silver.
Answer:
R_t1 = R_27.5 = 2.1Ω and
R_t2 = R₁₀₀ = 2.7Ω
Applying the relation

a = 0.0039°C^-1

Question 5.
The number density of free electrons in a copper conductor estimated is 8.5 × 10²⁸m^-3. How long does an electron take in drifting from one end of a wire 3.0m long to its other end? The area of cross-section of the wire is 2.0 × 10^-6m² and it is carrying a current of 3.0A.
Answer:
n = 8.5 × 10²⁸m^-3; A = 2 × 10^-6m²
I – 3.0A; I = 3m
I = v_dneA
\(\frac{3}{t}=\frac{3}{8.5 \times 10 \times 1.6 \times 10 \times 2 \times 10}\)
t = 8.5 × 11²⁸ × 1 .6 × 10^-19 × 2 × 10^-6
= 27.2 × 10^-3sec.

Question 6.
Choose the correct alternative:

1. Alloys of metals usually have (greater/ less) resistivity than that of their constituent metals.
2. Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
3. The resistivity of alloy manganin (is nearly indepent of/ increases rapidly) with increase of temperature.
4. The resistivity of a typical insulator (e.g. amber) is greater than that of a metal by a factor of the order of (10²²/10²³).

Answer:

1. Greater
2. lower
3. nearly independent of
4. 10²².

Plus Two Physics Current Electricity One Mark Questions and Answers

Question 1.
n equal resistors are first connected in series and then connected in parallel. What is the ratio of the maximum to the minimum resistance?
(a) n
(b) 1/n²
(c) n²
(d) 1/ n
Answer:
(c) n²
Explanation: In series R_s = nR
In parallel \(\frac{1}{Rp}\) = \(\frac{1}{R}\) + \(\frac{1}{R}\) ….n terms, Rp = \(\frac{R}{n}\)
∴ R_s/R_p = n²/1.

Question 2.
A car battery of emf 12 V and internal resistance 5 × 10^-2Ω, receives a current of 60 amp, from external source, then potential difference of battery is.
(a) 12V
(b) 9V
(c) 15V
(d) 20 V
Answer:
(c) 15V
When cell is changed by an external source, terminal voltage, V = E + Ir
V= 12 + 60 × 5 × 10^-2
=15.

Question 3.
A flowing of 10⁷ electron persecond in a conducting wire constitutes a current of……….A
(a) 1.6 × 10^-12
(b) 1.6 × 10²⁶
(c) 1.6 × 10^-26
(d) 1.6 × 10¹²
Answer:
(a) 1.6 × 10^-12
Flow of electrons \(\frac{n}{t}\) = 10⁷/sec.
There fore, Current (I) = \(\frac{q}{t}\) = \(\frac{ne}{t}\) = \(\frac{n}{t}\) × e
= 10⁷ × (1.6 × 10^-19)
= 1.6 × 10^-12 A.

Question 4.
Copper and silicon is cooled from 300K to 60K, the specific resistance.
(a) decrease in copper but increase in silicon
(b) increase in copper but decrease in silicon
(c) increase in both
(d) decrease in both
Answer:
(a) decrease in copper but increase in silicon.

Question 5.
State the potentiometer principle.
Answer:
Potential difference between two points of a current carrying conductor is directly proportional to the length of the wire between two points.

Question 6.
Find the current following through the network shown in the figure

Answer:
Since given circuit is in the form of Wheatstone bridge,

Question 7.
Kirchhoff’s first and second laws of electrical circuits are consequences of……(1)….and…..(2).. respectively
Answer:

• Conservation of electric charge
• Energy respectively

Question 8.
Pick the odd one out the following.
(a) Ohm’s law
(b) Lenz’slaw
(c) Coulomb’s law
(d) Gauss’s law
(e) Energy conservation law
Answer:
(a) Ohm’s law (It is not a universal law).

Question 9.
“Ohms law is not a fundamental law” Comment on this.
Answer:
Ohms law is not a universal law because metals do not obey this law at high temperature. Moreover, certain materials (diode and transistors, etc.) does not obey ohms law.

Plus Two Physics Current Electricity Two Mark Questions and Answers

Question 1.
Figure below shows a diagram of a water circuit. In many ways it behaves like an electric circuit. Draw an equivalent electric circuit. (Hint – water-wheel can be replaced by motor)

Answer:

Question 2.
A, B, C, and D are four rings on a carbon resistor. A = yellow, B = violet, C = Yellow, D, Silver

1. What is the value of resistance of above resistor?
2. The combined resistance of the above two resistor is

• 120Ω
• 45Ω
• 165Ω
• 35Ω

Answer:
1. 47 × 10⁴ ± 10%.

2. 12 × 10¹ + 45 × 10°
120 + 45 = 165Ω.

Question 3.
A = Brown B = Black C= Red D= Gold

1. What is the value of resistance without considering variation?
2. If the fourth ring is silver coloured what will be the change in accuracy?

Answer:

1. 1000Ω ± 5%
2. 10%

Question 4.
A junction of a electrical circuit is given below. Analyze the figure and answer the following

1. What is the value of I₁ and I₂?
2. State the law that can be applied to find I₁ and I₂?

Answer:
1. According Kirchoff first rule
3 + 2 = I + I₁
I₁ = 4A
I₁ = 3 +I₂
4 = 3 + I₂
I₂ = 1A

2. Total current meeting at any junction is zero.

Question 5.
The following question consists of two statements each, printed as assertion and reason. While answering these questions you should choose any one of the following responses.
Assertion: In a simple battery circuit, the point at the lowest potential is positive terminal of the battery. Reasons: The electrons flows from higher potential to lower potential.
(a) Both assertion and reason are true and the reason is a correct explanation of the assertion.
(b) Both assertion and reason are true but the reason is not a correct explanation of the assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(d) Both assertion and reason are false.

Plus Two Physics Current Electricity Three Mark Questions and Answers

Question 1.
Match the following.

Answer:

Question 2.
Match the following.

Answer:

Question 3.
“When electric current is passed through a resistance wire, it get heated up”.

1. Name the law associated with this phenomenon.
2. What happens to the heat energy developed, if the current through the wire is doubled.

Answer:

1. Joules law of heating
2. Heat developed in the conductor H = 1²R If current is doubled, Heat developed becomes 4 times.

Question 4.

1. Name the pd between terminals of the cell when

• key K is open
• K is closed

2. What is the reason for the difference in potential in the above two cases?
Answer:
1. key K is open

• e.m.f
• voltage.

2. Emf in a circuit
E = Ir + V
When key is open, I = 0
∴ E = V
When key is closed I ? 0
∴ V = E – Ir.

Question 5.
The variation of resistivity (ρ) with temperature (T) of a conductor, semiconductor and super conductor are given in the figure.

1. Identify them from the graph. (1)
2. Identify the figure in which the temperature coefficient of resistance of the material is positive. (1)
3. Write the equation connecting resistivity of the material with relaxation time. (1)

Answer:
1. From the graph:

• Fig (i) – Conductor
• Fig (ii) – Superconductor
• Fig (iii) – Semiconductor

2. Fig(i)

3. We have resistivity (ρ) from equation.
ρ = \(\frac{m}{n e^{2} \tau}\), where t is the relaxation time, m mass of electron, n number density of electron and e charge of electron.

Question 6.
When an animal touches an ‘electric-fence’ the animal gets a shock by completing an electrical circuit.

1. Draw an equivalent circuit using symbols, which shows the completed electrical circuit. (Assume that the animal has a resistance)
2. How much charge passes through the animal, if it receives a current pulse of 20mA for 0.1 seconds.
3. When a bird perches on the electric fence, the bird will not get a shock. Why?

Answer:
1.

2. 1 = Q/t
Q = 1 × t = 20 × 10^-3 × 0.1 = 20 × 10^-4C

3. Since the bird does not come in contact with the earth, its body is at same potential. So there is no potential difference and current. Hence the bird will not get a shock.

Question 7.
1. Pd between terminals when k is open is called

• emf
• lost voltage
• terminal voltage
• induced voltage

2. Give an equation connecting R, r, I and E What is the value of E for open circuit and ideal cell.

3. What happens to terminal voltage if current increases

• for an ideal cell
• for an ordinary cell.

Answer:
1. e.m.f

2. E = I(r+R).

3. For ideal cell, internal resistance is zero. Hence E = v. ie. terminal voltage does not change with current.
For ordinary cell, when current increases, V decreases.

Question 8.

1. What is the current through this circuit?
2. What is p.d. across 2Ω and 3Ω?
3. What is potential at A?
4. What is potential at B?

Answer:
1. Current I = \(\frac{\text { total voltage }}{\text { total resistance }}\)
I = \(\frac{5}{2+3}\) = 1A

2. Voltage across 2Ω
V = 1 × 2
V₂ = 2 V
Voltage across 3Ω
V = 1 × 3 = 3V

3. 5V

4. 3V

Question 9.
The relation between Voltage V (across the conductor) and current I through the conductor is given in the graph.

1. Which law establishes the relation between voltage and current?
2. A metal wire of resistivity 6.4 × 10^-5 ohm-cm and length 1.98 m has a resistance of 7Ω. Find radius of wire

Answer:
1. Ohms law

2. Resistance of wire R = \(\frac{ρl}{A}\)

= 2 × 10^-3m.

Question 10.
A circuit diagram of an instrument is given below.

1. Identify the instrument and state the principle of this instrument
2. Modify the circuit diagram to compare the emf of two cells.
3. How can increase the sensitivity of this instrument?

Answer:
1. Potentio meter

2. Comparison of e.m.f of two cells using potentiometer

a. Principle:
Potential difference between two points of a current carrying conductor (having uniform thickness) is directly proportional to the length of the wire between two points.

b. Circuit details :
A battery (B₁), Rheostat and key are connected in between A and B. This circuit is called primary circuit. Positive end of E₁ and E₂ are connected to A and other ends are connected to a two way key. Jockey is connected to a two key through galvanometer. This circuit is called secondary circuit.

c. Working and theory :
Key in primary circuit is closed and then E₁ is put into the circuit and balancing length l₁ is found out.
Then E_1∝ l₁ …………(1)
Similarly E₂ is put into the circuit and balancing length (l₂) is found out.
Then, E_{2 ∝} l₂ (2)
Dividing Eq(1)byEq(2),
\(\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}\)………..(3)

3. Increase the length of potentio meter wire.

Question 11.
A long resistance wire is stretched between two iron nails. A battery of 2V is applied across the wire. One end of a torch bulb is connected to nail and other end is made in contact as shown in figure.
1. If this wire slides over the resistance wire from nail 1 to nail 2, what happens to the brightness of the bulb.

• Increases
• Decreases
• Remains constant
• First increases then decreases.

2. How this principle is used to determine internal resistance of cell.

3. What happens to the reading, if we change 2V with 3V during the time of reading.

Answer:
1. Increases.

2. Measurement of internal resistance using potentiometer Principle:
[Same as before]
a. Circuit details :
Battery B₁ Rheostat and key K₁ are connected in between A and B. This circuit is called primary.

In the secondary circuit a battery E having internal resistance ‘r’ is connected. A resistance box (R) is connected across the battery through a key (K₂). Jockey is connected to battery through galvanometer.

b. Working and theory :
The key (K₁) in the primary circuit is closed and the key is the secondary (K₂) is open. Jockey is moved to get zero deflection in galvanometer. The balancing length l₁, (from A) is found out.
Then we can write.
E_1∝ l₁ ____(1)
Key K₂ is put in the circuit, corresponding balancing length (l₂) is found out. Let V be the applied voltage, then
V_{1 ∝} l₁ ____(2)
‘V’ is the voltage across resistance box. Current through resistance box ie, voltage across resistance,
V = \(\frac{E R}{(R+r)}\) _____(3)
Substituting eq (3) in eq (2),
\(\frac{E R}{(R+r)}\) _∝ l₂ ______(4)
Dividing eq (1) by eq (4),

r = \(\frac{\mathrm{R}\left(l_{1}-l_{2}\right)}{l_{2}}\)

3. Primary Voltage should not change while doing experiment. When we use 3v instead of 2v, potential gradient will change. Hence balancing length will change.

Question 12.
“Electric current has both direction and magnitude”

1. What is meant electric current
2. What is the conventional direction of electric current
3. Even though current has both magnitude and direction it is not a vector quantity. Why?
4. Thermal Motion of electrons in a conductor cannot constitute an electric current. Why?

Answer:

1. Rate of flow of charge is called current I = dq/dt
2. Direction of motion of positive charges
3. It does not obey vector law of addition
4. Average velocity of thermal motion is zero. Hence thermal motion does not produce current.

Question 13.
Match the following

Answer:

1. Metals – electrons
2. Semiconductors – electrons and holes
3. Superconductors – cooper pairs

Plus Two Physics Current Electricity Four Mark Questions and Answers

Question 1.
A cell arid two resistors R₁ and R₂ are provided to you.

1. Draw different combinations of resistors using R₁, R_2, and the cell.
2. Derive an expression for the effective resistance of the circuit in which current is the same in both resistors
3. lf R₁= 4Ω and R₂ = 6Ω, in which combination effective resistance is minimum? Find its value?

Answer:
1.

2. Derive an expression for effective resistance in series:
Consider three resistors R₁, R₂ and R₃ connected in series and a pd of V is applied across it.

In the circuit shown above the rate of flow of charge through each resistor will be same i.e. in series combination current through each resistor will be the same. However, the pd across each resistor are different and can be obtained using ohms law.
pd across the first resistor V₁ = I R₁
pd across the second resistor V₂ = I R₂
pd across the third resistor V₃ = I R₃
If V is the effective potential drop and R is the effective resistance then effective pd across the combination is
V = IR
Total pd across the combination = the sum pd across each resistor, V = V₁ + V₂ + V₃
Substituting the values of pds we get IR = IR₁ + IR₂ + IR₃
Eliminating I from all the terms on both sides we get
R = R₁ + R₂ + R₃ ………(1)
Thus the effective resistance of series combination of a number of resistors is equal to the sum of resistances of individual resistors.

3. The effective resistance becomes minimum in parallel connection.

R = 2.4Ω.

Question 2.
1. State whether the following statement is true or false “The value of resistance of a metal increase with the rise of temperature”.

2. Explain the reason.

3. With the help of the graph, match the following

4. Alloys like manganin, eureka, constantan, etc. are used in making standard resistance coils. Why?
Answer:
1. True.

2. When temperature increases, the amplitude of vibration of atom increases. Hence relaxation time decreases. Hence resistivity of metal increases according to the equation

3.
A_____Carbon
B_____Manganin
C_____Iron.

4. The temperature coefficient of resistance of manganin, eureka and constantan, etc are zero. Hence they are used in making standard resistance coils.

Question 3.

1. Identify the above device and give the principle behind it.

2. Obtain the mathematical condition for the galvanometer current to be zero.

3. If the balancing length T obtained fora resistance wire in the arrangement is 40cm. Find the new balancing length if the same resistance wire is folded to half its length and connected to the same gap.
Answer:
1. Meter bridge.

2. We get galvanometer current as zero, when P/Q = R/S. For derivation of P/Q=R/S
Wheatstone’s Bridge:
Four resistances P, Q, R, and S are connected as shown in figure. Voltage ‘V’ is applied in between A and C. Let I₁, I₂, I₃ and I₄ be the four currents passing through P, R, Q, and S respectively.

Working:
The voltage across R
When key is closed, current flows in different branches as shown in figure. Under this situation
The voltage across P, V_AB = I₁P
The voltage across Q, V_BC = I₃Q ……(1)
The voltage across R, V_AD = I₂R
The voltage across S, V_DC = I₁S
The value of R is adjusted to get zero deflection in galvanometer. Under this condition,
I₁ = I₃ and I₂ = I₄……(2)
Using Kirchoff’s second law in loopABDA and BCDB, weget
V_AB = V_AD ……….(3)
and V_BC = V_DC …….. (4)
Substituting the values from eq(1) into (3) and (4), we get
I₁P = I₂R……….(5)
and I₃Q = I₄S……..(6)
Dividing Eq(5) by Eq(6)
\(\frac{I_{1} P}{I_{3} Q}=\frac{I_{2} R}{I_{4} S}\)
\(\frac{P}{Q}=\frac{R}{S}\) [since I₁ = I₃ and I₂ = I₄]
This is called Wheatstone condition.

3. When we apply this condition in meter bridge, we get

If wire is folded, New resistance x₁ = \(\frac{x}{2}\)
Substituting this in P/Q = R/S we get

Question 4.
1. Potentiometer is better than voltmeter for measuring emf because

• It is cheap
• Easy to handle
• Its measurement uses null method

2. Give the basic principle of potentiometer,

3.

If K₂ is open balancing length is 600cm, if K₂ is closed 350cm is balancing length. Calculate the internal resistance.

Answer:
1. Its measurement uses null method.

2. Principle:
Potential difference between two points of a current carrying conductor (having uniform thickness) is directly proportional to the length of the wire between two points.

3. I = 600cm, I = 350 cm

Question 5.
Under an external electric field electrons drift slowly inside the conductor.
1. The velocity of drift is

• 1 mm/s
• 10⁵ m/s
• 3 × 10⁸ m/s
• 3 × 10⁹ m/s

2. What is meant by relaxation time?

3. Write an expression for drift velocity in terms of relaxation time.

4. When temperature increases what happens to drift velocity?
Answer:
1. 1mm/s.

2. The average time between two successive collision is called relaxation time.

3. V_d = \(\frac{\mathrm{eE}}{\mathrm{m}}\) t

4. We know drift velocity
V_d = \(\frac{\mathrm{eE}}{\mathrm{m}}\) t
When temperature increases, relaxation time decreases. As a result drift velocity decreases.

Question 6.
1. State whether the following statement is correct. “If the current through the cell is from its positve to negative, against the direction of emf, then the potential drop across the internal resistance aids the emf of the cell. (1)

2. You are given two cells. Group them so that they give more voltage. Arrive at the expression for effective emf and internal resistance. (3)
Answer:
1. True. Internal resistance and lost volt always opposes current flow through the cell”.

2.

Considertwo cells in series. Let ε₁, r₁ be the emf and internal resistance of first cell. Similarly ε₂, r₂ be the emf and internal resistance of second cell. Let I be the current in this circuit.

From the figure, the P.d between A and B
V_A – V_B = ε₁ – 1 ………(1)
Similarly P.d between B and C
V_B – V_C = ε₂ – 1 ……..(2)
Hence, P.d between the terminals Aand C
V_AC = V_A – V_C = V_A – V_B + V_B – V_C
V_AC = [V_A – V_B] + [V_B – V_C]
when we substitute eqn. (1) and (2) in the above equation.
V_AC = ε₁ – Ir₁ + ε₂ – Ir₂
V_AC = (ε₁ – ε₂) – I(r₁ + r₂)
V_AC = ε_eq – Ir_eq
where ε_eq = ε₁ + ε₂, and r_eq = r₁ + r₂

Question 7.
The resistance value of a conductor depends on its physical dimensions.
1. Give the expression for resistance of a conductor in terms of its physical dimension.

2.

(l length and d diameter) A potential V is applied between the ends of two conductor of same material shown in the figure.

• Express the resistance of the second conductor in terms of the resistance of the first conductor. (1)
• Find the ratio of electric field across the two conductors.

Answer:
1. R = ρ \(\frac{1}{A}\)
l – length of the conductor and
A – Area of cross section

2.
a.

b.

Question 8.
Three resistors R₁, R₂, R₃ are to be combined as shown in the figure.

1. Identify the series and parallel combinations. (1)
2. Which combination has lowest resistance. Arrive at the expression for the effective resistance of this combination. (3)

Answer:
1. Fig(i) – parallel Fig. (ii) – series

2. Fig. (i)
Let V be the potential between A and B.

I = I₁ + I₂ + I₃
and applying Ohm’s law to R₁, R₂ and R₃ we get,
V= I₁R₁,V=I₂R₂,V=I₃R₃
So that
I = I₁ + I₂ + I₃ = V\(\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\right)\)
An equivalent resistance R_eq that replace the combination, and hence

Question 9.
An electric circuit is given in the Figure.

The potential difference between A and D is 40V.

1. Find the effective resistance between A and D in terms of R. (1)
2. Calculate the potential difference between A and O.

Answer:
1.

2. V_AD = V_AO + V_OD
But V_OD = 3V_AO
∴ V_AD = V_AO + 3V_AO
40 = 4V_AO
V_AO = \(\frac{40}{4}\) = 10V.

Plus Two Physics Current Electricity Five Mark Questions and Answers

Question 1.
Resistance are used to reduce the current flow in a circuit.
1. A carbon resistor has coloured strip and shown in the figure. What is its resistance?

2. Resistance can be connected in series and parallel to obtain the required value of resistance. Derive an expression for the effective resistance when three resistors are connected in parallel.

3. Kirchhoff’s rules are used to analyses the electric circuit. Use it to analyze the Wheatstone Bridge and arrive at Wheatstone’s condition for balancing the bridge.
Answer:
1. blue, Gray, yellow, gold
1st – Blue – 6
IInd – Gray – 8
IIIrd – Yellow – 10⁴
Iv – gold – 5%
68 × 10⁴ ± 5%.

2. Derive an expression for effective resistance in series:
Consider three resistors R₁, R₂ and R₃ connected in series and a pd of V is applied across it.

In the circuit shown above the rate of flow of charge through each resistor will be same i.e. in series combination current through each resistor will be the same. However the pd across each resistor are different and can be obtained using ohms law.
pd across the first resistor V₁ = I R₁
pd across the second resistor V₂ = I R₂
pd across the third resistor V₃ = I R₃
If V is the effective potential drop and R is the effective resistance then effective pd across the combination is
V = IR
Total pd across the combination = the sum pd across each resistor, V = V₁ + V₂ + V₃
Substituting the values of pds we get IR = IR₁ + IR₂ + IR₃
Eliminating I from all the terms on both sides we get
R = R₁ + R₂ + R₃ ………(1)
Thus the effective resistance of series combination of a number of resistors is equal to the sum of resistances of individual resistors.

3. Wheatstone’s Bridge:
Four resistances P,Q,R and S are connected as shown in figure. Voltage ‘V’ is applied in between A and C. Let I₁, I₂, I₃ and I₄ be the four currents passing through P,R,Q and S respectively.

Working:
The voltage across R
When key is closed, current flows in different branches as shown in figure. Under this situation
The voltage across P, V_AB = I₁P
The voltage across Q, V_BC = I₃Q ……(1)
The voltage across R, V_AD = I₂R
The voltage across S, V_DC = I₁S
The value of R is adjusted to get zero deflection in galvanometer. Under this condition,
I₁ = I₃ and I₂ = I₄……(2)
Using Kirchoff’s second law in loopABDA and BCDB, weget
V_AB = V_AD ……….(3)
and V_BC = V_DC …….. (4)
Substituting the values from eq(1) into (3) and (4), we get
I₁P = I₂R……….(5)
and I₃Q = I₄S……..(6)
Dividing Eq(5) by Eq(6)
\(\frac{I_{1} P}{I_{3} Q}=\frac{I_{2} R}{I_{4} S}\)
\(\frac{P}{Q}=\frac{R}{S}\) [since I₁ = I₃ and I₂ = I₄]
This is called Wheatstone condition.

Question 2.
A rectangular conductor of length I and area of cross section A and electron density n; is shown below.

1. When the face Y is given positive potential and X negative potential what will happen to the electrons inside the block
2. What is meant by drift velocity? How is it related to the field inside the metal?
3. Deduce an expression connecting intensity of electric field and drift velocity.
4. Under the application of an electric field do all the electrons move in a same direction? Explain

Answer:
1. Electrons will be accelerated towards the side Y.

2. Drift Velocity (v_d):
The average velocity acquired by an electron under the applied electric field is called drift velocity.
Explanation :
When a voltage is applied across a conductor, an electric filed is developed. Due to this electric field electrons are accelerated. But while moving they collide with atoms, lose their energy and are slowed down. This acceleration and collision are repeated through the motion. Hence electrons move with a constant average velocity. This constant average velocity is called drift velocity.

3. Expression for drift velocity :
Let ‘V’ be the potential difference across the ends of a conductor. This potential difference makes an electric field E. Under the influence of electric field E, each free electron experiences a Coulomb force.
F = -eE
or ma = -eE
a = \(\frac{-e E}{m}\) ……….(1)
Due to this acceleration, the free electron acquires an additional velocity. A metal contains a large number of electrons.
For first electron, additional velocity acquired in a time τ,
v₁ = u₁ + aτ₁
where u₁ is the thermal velocity and τ is the relaxation time.
Similarly the net velocity of second, third……electron
v₂ = u₂ + aτ₂
v₃ = u₃ + aτ₃
v_n = u_n + aτ_n
∴ Average velocity of all the ‘n’ electrons will be

V_av = 0+ aτ (∴ average thermal velocity of electron is zero)
where

where V_av is the average velocity of electron under an external field. This average velocity is called drift velocity.
ie. drift velocity V_d = aτ……(2)
Substitute eq (1) in eq (2)
V_d = \(\frac{-e E}{m}\) τ

4. Electrons will continues its random thermal motion even in the presence of electric field.

Question 3.
To study the relation between potential difference and current in an electrical circuit, a student is provided with a resistance wire, a cell, and a key.
1. Draw a circuit which allows current flow through the resistance wire.

2. Modify the circuit by introducing an ammeter, Voltmeter and a rheostat for varying the potential difference across the resistance and to measure that potential difference and the corresponding current.

3. Let in the above experiment the student obtained the following data.

Draw a graph connecting V and I using above data. Then establish the relation between V and I as a law.

4. Instead of the resistance wire if the student uses a p-n junction diode in the forward biased condition how the relation between V and I changes? Justify.
Answer:
1.

2.

3.

The above graph shows that, current flowing through a conductor is directly proportional to potential difference across it ends.

4. The relation between V and I becomes nonlinear. Because, Pn diode does not obey ohms law.

Question 4.
In all metallic conductors, electric conduction is due to drifting of free electrons. But the resistivity of different metals are different.

1. Write the expression for resistivity of a conductor in terms of its dimensions.
2. Name the factors on which resistivity of a metal depends.
3. Arrive at an expression for electrical resistivity of a metal in terms of relaxation time.
4. Using the above expression explain the variation of resistivity with temperature.

Answer:
1. ρ = \(\frac{RA}{l}\).

2. Temperature and Nature of metal

3. We know current density
J = nv_de
But V_d = \(\frac{e E}{m}\) τ

4. When temperature increases, the amplitude of oscillation of atom increases. This will decrease . the relaxation time and hence resistivity increases.

Question 5.
You are supplied with a 1m long uniform resistance wire of resistance 3Ω and a cell of emf 1.5v.

1. Can you construct a potentiometer using 1m wire? If no, give reason; if yes, what is the least count of the arrangement?
2. Draw the connection diagram to compare the emf of Leclanche cell and Daniel cell using this arrangement.
3. How can you modify above arrangement to measure a p.d. in the range 0-1 mV with a least count of 0.15m V.

Answer:

1. Yes, LC is 1.5mV

2.

3. To get least count of 0.15mV, 10m wire has to be used.

Question 6.
A circuit diagram is given below. Analyze the figure and answer the following questions.

1. The above circuit is a modification of……..
2. What is the value of balancing length?
3. If 1.5 V cell is replaced by a 3V cell what will be the balancing length?
4. Calculate the value of X.

Answer:
1. Wheatstone bridge.

2. 100-20=80cm

3. No change in balancing length

4. P/Q = R/S
ie X/5 = 80/20
x = 80/20 × 5 = 20Ω.

Question 7.

1. Define electric power. What is its SI unit? (2)
2. Two bulbs of 50W, 220V and 100W, 220V are given. How will you connect the bulbs so that 50W, 220V bulb will glow brighter than 100W, 220V bulb. (1)
3. When the bulb 50W, 220V is connected to an 110V supply calculate the power generated. (2)

Answer:
1. The energy dissipated per unit time is the power.
P = \(\frac{\Delta \mathrm{W}}{\Delta \mathrm{t}}\)
P = IV =I²R = \(\frac{V^{2}}{R}\)
Unit is watt.

2. In series

3.

Question 8.
The figure shows the diagram of a potentiometer.

1. Give the principle of a potentiometer. (1)

2. The length of AB is 3m and resistance per unit length of the potentiometer wire is 4Ω/m. If E₁ = 4V, R = 20Ω and E₂ = 1V find the length of the potentiometer wire that balance E₂. (3)

3. If E₂>E₁ can we get the null deflection in galvanometer. Give reason. (1)
Answer:
1. When a constant current is flowing through a wire having uniform area of cross section and uniform composition the potential difference across any length of the wire is directly proportional to its length.
V ∝ l.

2. The resistance of potentiometer
R_p = 4 × 3 = 12Ω
Current through the potentiometer

Potential across potentiometer wire
V_AB = IR_p = 0.125 × 12 = 1.5V
Potential gradient k = \(\frac{V_{A B}}{I}=\frac{1.5}{3}\) = 0.5 V/m
Balancing length for cell E₂ is given from equation
E₂ = kI₂ = 0.5I₂

3. If E₂ > E₁, we will not get null deflection. The potential difference across the potentiometer wire AB should be higher than the emf of E₂.

Question 9.

1. When the switch is closed will all the bulbs glow? Give reason. (2)
2. Identify the underlying principle. Deduce the principle for a resistance network. (3)

Answer:
1. No. Since all the bulbs are identical the bridge is balanced. So potential B and C is same and no current flows through the bulb connected between B and C. So all other bulbs except R₅ will glow.

2. Four resistances P, Q, R, and S are connected as shown in figure. Voltage ‘V’ is applied in between A and C. Let I₁, I₂, I₃, and I₄ be the four currents passing through P, R, Q, and S respectively.

Students can Download Chapter 1 Introduction Microeconomics Questions and Answers, Plus Two Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Economics Chapter Wise Questions and Answers Chapter 1 Introduction Microeconomics

Plus Two Economics Introduction Microeconomics One Mark Questions and Answers

Question 1.
The diagram shows:
(a) A movement from ‘a’ to ‘b’ has no opportunity cost.
(b) A movement from ‘f’ to ‘b’ has an opportunity cost.
(c) Higher is the production of good 2 greater is the opportunity cost of reducing its production.
(d) Higher is the production of good 2 lesser is the opportunity cost of reaching its output.

Answer:
The concave shape of PPC shows that higher the production of goods 1 and 2. Goods 2 higher will be the opportunity cost of reducing production.

Question 2.
State economic terms. The allocations of scarce resources and the distribution of the final goods and services.
Answer:
The central problem of an economy.

Question 3.
As a result of liberalisation policy, the inflow of foreign capital has increased. What is its impact on production possibility frontier?
Answer:
PPC shifts upward as a result of the increase in the availability of capital resources.

Question 4.
What is the shape of a production possibility curve?
(i) Convex to origin
(ii) Concave to origin
(iii) Horizontal
(iv) Vertical
Answer:
(ii) Concave to origin

Question 5.
How does a market economy solve central economic problems?
(i) Central planning
(ii) Price mechanism
(iii) Both (i) and (ii)
(iv) None of the above
Answer:
(ii) Price mechanism

Question 6.
Scarcity definition was given by:
Answer:
(i) Adam Smith
(ii) Alfred Marshall
(iii) Lionel Robbins
(iv) Samuelson
Answer:
(iii) Lionel Robbins

Plus Two Economics Introduction Microeconomics Two Mark Questions and Answers

Question 1.
Fill in the blanks.

Answer:

Question 2.
Classify the following statement into two branches of economics.

1. Indian economy grew by 9.2% GDP in the financial year 2006.
2. An unexpected lorry strike caused the price of vegetables to rise.
3. Recently the RBI reduced the Cash Reserve Ratio to 5.5%.
4. Madras Cement LTD is planning to add 40 lakh tonne to its existing production capacity of 60 lakh tonnes.

Answer:

1. Macroeconomics
2. Microeconomics
3. Macroeconomics
4. Microeconomics

Question 3.
Distinguish between centrally planned economy and a market economy.
Answer:
In a centrally planned economy, the government or the central authority plan all the important activities in the economy. All important decisions regarding production, exchange, and consumption of goods and services are made by the government.

On the other hand in a market economy, all the important decisions are made on the basis of demand and supply conditions. The central problems regarding what and how much to produce are solved through the coordination of economic activities brought about by the price signals.

Question 4.
Give two examples of underutilization of resources.
Answer:

1. Improper distribution of scarce resources leads to underutilisation of capacities,
2. Due to technological backwardness, industrial workers’ capacity is underutilized.

Question 5.
Give a few examples of resources in economics.
Answer:
By the term ‘resources’, we mean land, labour, tools, machinery, etc. in economics.

Question 6.
The government should increase tax on tobacco products. Explain whether the statement is positive or normative.
Answer:
This is a normative statement because it says how the government should tax. It is only an opinion. This cannot be proved. Since it contains a value judgment it is a normative statement.

Question 7.
“Study of aggregates is equally important to study individual units.”
Substantiate the above statement by distinguishing the two branches of Economics. Give two examples for each.
Answer:

1. Microeconomics which is the study of individual units is helpful in analysing a micro-economy, whereas macroeconomics is helpful in under-standing the working of macroeconomy.
2. Microeconomics – Individual income, price of apple Macroeconomics – Inflation, national income.

Plus Two Economics Introduction Microeconomics Three Mark Questions and Answers

Question 1.
Classify the following into Microeconomics and Macroeconomics.
Answer:
Demand analysis, Consumption function, Theory of international trade, Income determination, Pricing of factors of production, Investment function.

Question 2.
Match Column B and C With Column A.

Answer:

Question 3.
“Since resources are limited, they should be properly used”. Comment on this statement in the light of utilization of energy resources in Kerala.
Answer:
This statement relates to the problem of scarcity of resources and thus connected to the scarcity definition of Lionel Robbins. According to this statement, the scarcity of resources forces the economy to choose the most urgent need that is to be satisfied.

Since energy resource in Kerala is limited in supply, it has to be judiciously utilized. Control over the utilization of energy is necessary in states like Kerala so that this scarce resource can be protected.

Question 4.
Classify the following features under the title centrally planned economy and market economy. Price mechanism, comprehensive planning, welfare motive, profit motive, public sector, private sector.
Answer:

Question 5.
“Labour intensive technique is the best technique of production”. Give arguments in favour of and against this statement.
Answer:
Arguments in favour of:

1. Labour intensive technique provides more employment opportunities.
2. Labour intensive technique needs less capital.
3. Labour intensive technique requires less skill only.

Arguments against:

1. Labour intensive technique is less productive.
2. Labour intensive technique prevents development
3. Labour intensive technique makes the economy less productive.

Question 6.
Given below a Production Possibility Curve of an economy. Compare the points in the context of production possibilities of the economy.

Answer:
The Production Possibility Curve is an analytical tool presenting the alternative production possibilities of an economy. It is used to explain the central problems of an economy and how they are solved.
In the diagram:
1. Point A indicates the efficient utilization of available resources.

2. Point B shows that the available resources of the economy are not fully utilized. In other words, it is an indication of the underutilization of resources.

3. Point C is outside the Production Possibility frontier. This means that the economy cannot produce at this point using the available resources.

Question 7.
Suppose there is growth of resources in an economy. How does it affect the PPC?
Answer:
When there is growth of resources the PPC shifts outwards as shown, below.

The PPC shifts from PP to P due to growth of resources in the economy. Therefore, economy produces more of food and cloth.

Question 8.
With the ₹500 cash award received by a student prepares a list of goods. She writes to have each of the goods priced @₹500.

1. An Economics textbook.
2. A movie with her friends.
3. An outing.
4. A dinner with her parents.

Explain opportunity cost. Identify the opportunity cost of buying one economic textbook.
Answer:
Opportunity cost is the next best alternative forgone. The opportunity cost of buying an economic textbook is the foregone movie with friends.

Question 9.
Different economic systems solve basic economic problems using different mechanisms. Complete the following table by writing the basic economic problems as well as the mechanisms used. Also, give one example for each economic system.

Answer:

Question 10.
Read the following statements and write the terms used in economics.
Answer:

1. The curve representing various combinations of any two goods the economy can produce with the available resources and technology.
2. An investigation in economics concerned with it is rather than what ought to be.
3. An economic system in which basic problems are solved through planning.

Answer:

1. PPC
2. Positive economics
3. Socialism

Plus Two Economics Introduction Microeconomics Five Mark Questions and Answers

Question 1.
Classify the following statements into positive and normative statements.

1. Statement I: India introduced a new economic policy in 1991.
2. Statement II: Globalization badly affected India’s agricultural sector.
3. Statement III: The number of people living below poverty line has to be reduced from the present level of 21%.
4. Statement IV: Mean, median and mode are the measures of central tendency.

Answer:

1. Statement I: positive statement
2. Statement II:normative statement
3. Statement III:normative statement
4. Statement IV:positive statement.

Question 2.
“The implementation of Vizhinjam project will shift our PPC rightward”. Suggest two points in favour and against this statement.
Answer:
A production Possibility Curve is an analytical tool presenting the alternative production possibilities of an economy.
Points in favour of:

1. Since the Vizhinjam project needs abundant skilled manpower, it would shift our PPC rightward.
2. Kerala has abundant skilled manpower. They will get more employment and hence ppc will shift rightward.

Points against:

1. The abundant skilled labour available in Kerala is lying unutilized only partially.
2. The project need not be a continuous success as it will badly affect environment.

Question 3.
Prepare a production possibility schedule showing constant marginal opportunity cost.
Answer:

It is clear from the tables that in each production possibility to increase the production of wheat by one tonne, 20 kgs of rubber have to be sacrificed. This shows that there is constant marginal opportunity cost operating in this case. In this case, the PPC of rubber and wheat becomes a straight line.

Question 4.
Differentiate between micro economics and macro economics.
Answer:

Question 5.
A few statements are given below. Classify them under two branches of economics.

1. RBI formulated its new monetary policy.
2. National Income recorded the highest growth last year.
3. Shyam purchased a new mobile phone.
4. Inflation adversely affects the fixed income of people.
5. Total Fixed Cost of a firm remains constant even if output increases.

Answer:

1. Macro economics
2. Macro economics
3. Microeconomics
4. Macroeconomics
5. Microeconomics

Plus Two Economics Introduction Microeconomics Eight Mark Questions and Answers

Question 1.
Suppose a country uses its entire resources to provide educational and health facilities required for the people. Given the resources, the country can provide various combinations of number of schools and hospitals as shown in the table below.

a. Define PPC and represent the above schedule on a diagram.

b. Suppose the country has already attained near-total literacy. If so, will the country prefer points on upper portion of PPC or points on the lower portion of PPC? Substantiate your answer.

c. Which mechanism will you advice to utilise the resources to provide more health facilities planning or market? Give reasons.
Answer:
a. Production Possibility Curve (PPC) is a graphical representation of all possible combinations of two goods or services that can be produced in an economy with given level of resources and technology. It is also known as production possibility frontier (PPF). The shape of PPC is concave to the origin.

b. Combination D. Because more resources will be spent for health and less on education.
c. Planning. Because only the government can ensure public services.

Question 2.
Prepare a seminar paper on “central problems of an economy.
Answer:
Respected teachers and dear friends,
The topic of my seminar paper is “central problems of an economy”. As we know the central problems arise due to the fact that the human wants are unlimited and the means to satisfy those wants are limited. In this seminar paper, I would like to present the central problems such as what to produce, how to produce and for whom to produce.

Introduction:
The resources available to the consumer are limited but their wants are unlimited. Due to scarcity of resources, the economy faces the problem of choice. It is this mismatch between unlimited wants and the limited resources that gives rise to three central problems faced by every economy.

Contents:
a. What to produce and in what quantities?
An economy faces the problems of what to produce because the resources available to an economy are scarce. As resoruces are scarce, an economy cannot produce all those goods and services the society needs. Therefore, society has to take the crucial decision of what goods and services to be produced in an economy.

For example, the resources of an economy can be used for the production of food, defense equipment or luxury goods. It can also be used for education, health or entertainment. A national society has to make a priority list of items to be produced and allocate the available resources accordingly.

Once the decision regarding what to produce is taken, the next problem is to decide in what quantities the goods and services are to be produced. It is important because the production of one good may lead to the withdrawal of the production of some other goods.

b. How to produce?
After taking the decision regarding the type and quantity of goods to be produced, the next question is ‘how to produce goods arid services’. This problem is related to the method or technology of production. Goods can be produced using different technologies.

There are mainly two technologies for producing goods, viz., labor-intensive technology and capital intensive technology. Labour intensive technology uses more labour compared to capital. On the other hand, capital intensive technology uses more caiptal compared to labour. The choice of technique depends upon various factors like the availability of labour force and capital resources and its prices.

c. For whom to produce?
The goods and services produced once should be distributed among the people of the economy. Whether it should be distributed equally among the people? Should the distribution of the goods be in such a way that at least minimum consumption level has to be attained by everyone in the economy? Should everyone get primary health and education?

Conclusion:
Thus it can be concluded that every economic system faces three basic problems. The solution to these economic problems depends upon the nature of the economic system.

Students can Download Chapter 14 Biomolecules Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 14 Biomolecules

Plus Two Chemistry Biomolecules One Mark Questions and Answers

Question 1.
Which of the following is an example of globular protein?
(a) Myosin
(b) Collagen
(c) Insulin
(d) Keratin
Answer:
(c) Insulin

Question 2.
The vitamin essential for blood clotting is
Answer:
Vitamin K

Question 3.
Which base is present in RNA but not in DNA?
(a) Uracil
(b) Thymine
(c) Guanine
(d) ytosine
Answer:
(a) Uracil

Question 4.
Say TRUE or FALSE:
The coagulation of egg white on boiling is an example of protein denaturation.
Answer:
True

Question 5.
The linkage that holds monosaccharide units together in a polysaccharide is called …………………….
Answer:
Glycosidic linkage

Question 6.
Proteins are essential for growth in animals. They are build up of amino acid molecules. How are different amino acid molecules linked in a proteins?
Answer:
By peptide linkage.

Question 7.
What are the different types of RNA found in the cell?
Answer:
m-RNA, t-RNA, r-RNA

Question 8.
Lactose is made of ………………….
Answer:
β -D galactose and β -D glucose

Question 9.
Glucose does not react with ………………..
a) Br₂/H₂O
b) NH₂OH
c) HI
d) NaHSO₃
e) CH₃-CO-O-CO-CH₃
Answer:
(d) NaHSO₃

Question 10.
Anaemia is caused by the deficiency of vitamin ………………………
Answer:
B₁₂

Question 11.
The number of chiral C atoms on glucose and fructose are
Answer:
4 in glucose and 3 in fructose

Question 12.
Glucose on oxidation with bromine water give …………………………
Answer:
Gluconic acid

Question 13.
Name the vitamin responsible for the coagulation of blood.
Answer:
Vitamin K.

Plus Two Chemistry Biomolecules Two Mark Questions and Answers

Question 1.
Explain the denaturation of protein.
Answer:
When a protein is treated with acid, alkali or heated or subjected to change in pH, the secondary and primary structure of protein gets ruptured. Denaturation does not change the primary structure of proteins.

Question 2.
Why cannot vitamin C be stored in our body?
Answer:
Vitamin ‘C’ is a water soluble vitamin and they are readily excreted in urine and cannot be stored in our body.

Question 3.
Classify the following into monosaccharides and disaccharides.
Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
Answer:

• Monosaccharides – Ribose, 2-deoxyribose, galactose and fructose.
• Disaccharides – Maltose and lactose.

Question 4.
What do you understand by the term glycosidic linkage?
Answer:
The two monosaccharide units are linked together by an oxide or either linkage formed by the loss of water molecules. Such a linkage called glycosidic linkage.

Question 5.
Distinguish between essential and non-essential amino acids. Give examples.
Answer:

1. The amino acids which can be synthesized in our body are known as non-essential amino acids, eg. Glycine, Alanine.
2. Those amino acids which can not be synthesized in our body and must be obtained through diet are known as essential amino acids. eg. valine, Lysine.

Question 6.
What is ‘peptide linkage’ as related to proteins?
Answer:
The linkage, – CO – NH – which unites various amino acid units in a peptide molecule is called peptide linkage.

Question 7.
Amino acids show amphoteric behavior. Why?
Answer:
Amino acids contain both NH₂ & -COOH group and hence they exhibit amphoteric character.
NH₃⁺ – CH₂ – COO^–

Question 8.
What are reducing sugars? Give one example.
Answer:
Sugar units which are having free – CHO groups are reducing sugars, eg. Maltose.

Question 9.
What is primary structure of proteins?
Answer:
The primary structure gives an idea regarding the sequence in which amino acids are arranged.

Question 10.
Differentiate between fibrous & globular proteins.
Answer:
Fibrous proteins have threads lying side by side to form a fiber-like structure, e.g. Keratin. Globular proteins have molecules which are folded into compact units that often approach spherical shape.

Plus Two Chemistry Biomolecules Three Mark Questions and Answers

Question 1.
D-glucose is obtained in two different forms, α- D- glucose, and β-D-glucose.

1. In which name the α and β forms of glucose are known?
2. Explain the difference in their configuration with diagram.

Answer:

1. They are known as anomers.
2. Anomers are a pair of stereo isomeric ring forms of a sugar which differ in configuration only around first carbon atom.

Question 2.
Complete the following table:

Answer:

1. Reducing
2. Fructose
3. Disaccharide
4. Reducing
5. Disaccharide
6. Non-reducing

Question 3.
There are two types of nucleic acids, DNA and RNA.

1. Identify the sugar that is present in DNA and RNA.
2. Give the differences between DNA and RNA.

Answer:

1. Sugar that is present in DNA and RNA:

• DNA – β-D-2-deoxyribose
• RNA – β-D-ribose

2.

Question 4.
Classify the given vitamins as fat soluble and water soluble.

1. Vitamin A, D, B, E, K, C
2. Name a deficiency disease caused by the deficiency of Vitamin A.
3. The deficiency of which vitamin is responsible for the disease Rickets?

Answer:

1. Fat soluble – Vitamins A, D, E and K. Water soluble – Vitamins B and C
2. Night blindness
3. Vitamin D

Question 5.
Write the products obtained when glucose is treated with the following reagents?

• HF
• Bromine water
• HNO₃

Answer:

• When glucose is heated with HI, n-Hexane is formed.
• When glucose is treated with bromine water, it is oxidised to gluconic acid.
• Glucose is oxidised by HNO₃ to saccharic acid.

Question 6.
What is the basic structural difference between starch and cellulose?
Answer:
Starch has 2 components namely amylose and amylopectin. Amylose is a long unbranched chain made of α-D-(+)-glucose units held by C1 – C4 glycosidic linkage.

Amylopectin is a branched chain polymer of α-D-glucose units in which chain is formed by C1 – C4 glycosidic linkage whereas branching occurs by C1 – C6 glycosidic linkage. Cellulose is a straight chain polysaccharide composed of only β-D-glucose units joined by C1 – C4 glycosidic linkage.

Plus Two Chemistry Biomolecules Four Mark Questions and Answers

Question 1.
Amino substituted carboxylic acids are called amino acids.

1. Analyse the statement and explain what are essential amino acids?
2. How many essential amino acids are there?
3. Give two examples of essential amino acids.

Answer:

1. The amino acids which can be synthesized in our body are known as non-essential amino acids, eg. Glycine, Alanine
   Those amino acids which can not be synthesized in our body and must be obtained through diet are known as essential amino acids. eg. valine, Lysine
2. Ten
3. Valine and Lysine

Question 2.
Starch on enzymatic hydrolysis by diastase gives a reducing disaccharide ‘A’ which undergoes hydrolysis by enzyme maltase to form ‘B’ which is also a reducing sugar.

1. Identify the compound ‘A’ and ‘B’ with suitable chemical equations.
2. Explain the term reducing sugar.

Answer:
1. Compound A-Maltose
Compound B – Glucose

b) All those carbohydrates which contain free aldehydic or ketonic group to reduce Tollens reagent and Fehling’s solution and are called reducing sugars, e.g. Glucose, Fructose.

Question 3.
In the following table, the names of Vitamins, their sources, and deficiency diseases are tabulated in the wrong order. Match them in the correct order.

Answer:

Question 4.

1. Draw the pyranose structures of the α and β forms of glucose.
2. Draw the furanose structures of the α and β

Answer:
1.

2.

Question 5.
When egg is boiled its physical structure changes.

1. Is there any change in its chemical nature?
2. Mention the peculiar type of bond present in proteins.
3. When egg is boiled it become hard. Why? Explain.

Answer:

1. No
2. Peptide bond
3. This is due to denaturation of protein. On boiling egg, the soluble form of globular proteins undergo coagulation to give fibrous proteins which are insoluble in water.

Question 6.

1. What type of bonding helps in stabilising the α – helix structure of proteins?
2. What is nucleotide?

Answer:
1. Intermolecular hydrogen bonding between -NH group of each amino acid and carbonyl group of an adjacent turn of the helix.

2. The repeating structural units of nucleic acids are called nucleotides.

• Pentose sugar + Base → nucleoside
• Nucleoside + Phosphoric acid → nucleotide

Question 7.

1. What do you mean by isoelectric point of amino acids?
2. What are Zwitter ions? Give one example.

Answer:

1. The pH at which the Zwitter ions do not migrate neither towards cathode nor towards anode is known as isoelectric point of the amino acids.

2. Amino acids possess both acidic and basic group, they generally exist as dipolar ions called Zwitter ions.
e.g. glycine H₃N⁽⁺⁾ – CH₂ – COO^(-)

Plus Two Chemistry Biomolecules NCERT Questions and Answers

Question 1.
Classify the following into monosaccharides and disaccharides.
Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
Answer:

1. Monosaccharides – Ribose, 2-deoxyribose, galactose and fructose.
2. Disaccharides – Maltose and lactose.

Question 2.
What do you understand by the term glycosidic linkage?
Answer:
In oligosaccharides and polysaccharides, the two monosaccharide units are linked together by an oxide or ether linkage formed by the loss of water molecules. Such a linkage between two monosaccharide units through oxygen atom is called glycosidic linkage.

Question 3.
What is the basic structural difference between starch and cellulose?
Answer:
Both starch and cellulose contain a large number of α -D(+)-glucose units. Starch consists of two components:

1. Amylose which is a linear polymer and
2. Amylopectin which is a branched polymer but in both the D-glucose units are linked through α -glycosidic linkage between C1 of one glucose with C4 of next glucose unit. In amylopectin branching occurs by C1-C6 glycosidic linkage.

Cellulose is only a linear polymer of D-glucose units joined through β -glycosidic linkage between C1 of one glucose with C4 of next glucose unit.

Question 4.
Enumerate the reactions of D-glucose which cannot obe explained by its open chain structure.
Answer:

1. Glucose does not give 2,4-DNP test, Schiff’s test and it does not form the hydrogen sulphite addition product with NaHSO₃.
2. The pentaacetate of glucose does not react with hydroxylamine.
3. Glucose exists in two different crystalline forms such as α – form and β -form.

Question 5.
What is the difference between a nucleoside and a nucleotide?
Answer:
Nucleoside is formd by condensation of a purine or pyrimidine base with pentose sugar at position 1. When nucloeside is linked to phosphoric acid at 5 position of sugar moeity, we get a nucleotide. So a nucleoside has two units: pentose sugar and a base while a nucleotide has three units: phosphate group, pentose sugar and a base.

Students can Download Chapter 15 Polymers Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 15 Polymers

Plus Two Chemistry Polymers One Mark Questions and Answers

Question 1.
Orion is a polymer of
(a) Styrene
(b) Tetrafluoroethylene
(c) Caprolactam
(d) Acrylonitrile
Answer:
(d) Acrylonitrile

Question 2.
Buna-S is a copolymer of styrene and ……………
Answer:
1,3-Butadiene

Question 3.
F₂C = CF₂ is a monomer of
(a) Glyptal
(b) Teflon
(c) Nylon 6
(d) PVC
Answer:
(b) Teflon

Question 4.
Say TRUE or FALSE :
Chloroprene is an addition polymer of Neoprene.
Answer:
False

Question 5.
Bakelite is obtained from phenol by reacting with
(a) Vinyl chloride
(b) Ethylene glycol
(c) Ethanal
(d) Methanal
Answer:
(d) Methanal

Question 6.
Which one of the following is an example of a biode-gradable polyester
(a) PHBV
(b) PET
(c) Nylon
(d) Bakelite
Answer:
(a) PHBV

Question 7.
Novalac is ……………
Answer:
Phenol – formaldehyde resin

Question 8.
Ziegler-Natta catalyst is used in the preparation of …………….
Answer:
HDP

Question 9.
Nylon is a ………………….
Answer:
Polyamide

Question 10.
The polymer used in the manufacture of lacquers is ……………….
Answer:
Glyptal

Plus Two Chemistry Polymers Two Mark Questions and Answers

Question 1.
PVC is commonly used to make pipes.

1. What is the monomer of PVC?
2. What is the purpose of PVC covering of electrical connecting wire?

Answer:

1. Vinyl chloride
2. To provide electrical insulation.

Question 2.
Name the monomer units of natural rubber.
Answer:
2-Methyl-1,3-butadiene or Isoprene

Question 3.
Name two commercially important synthetic polymers. Also name their monomers.
Answer:

1. Nylon 6,6 – Hexamethylene diamene and Adipic acid
2. Bakelite – Phenol and Formaldehyde

Question 4.
What are polymers? Based on structure, how they are classified?
Answer:
Polymers are compounds of higher molecular mass formed by the combination of large number of small molecules.
Based on structure polymers are classified into 3 types:

1. Linear polymers
2. Branched polymer
3. Cross linked or network polymers

Question 5.
How are the polymers classified on the basis of molecular forces?
Answer:
Based on molecular forces polymers are classified into four subgroups.

1. Elastomers
2. Fibres
3. Thermoplastic polymers
4. Thermosetting polymers

Question 6.
What are bio-degradable polymers?
Answer:
These are polymers which undergo degradation by the action of microorganisms, eg. Nylon 2-nylon 6

Question 7.
Name the monomers of Terylene and Buna-N?
Answer:
Ethylene glycol and terephthalic acid are the monomers of terylene. Buna-N is a copolymer of 1, 3-Butadiene and Vinyl cyanide.

Question 8.
Write the names of monomers of the following polymers.

1. Glyptal
2. Bakelite

Answer:

1. Glyptal → Ethylene glycol and Phthalic acid
2. Bakelite → Phenol and Formaldehyde

Question 9.
Explain the term copolymerisation and give two examples.
Answer:
Copolymerisation is a process in which a mixture of more than one monomeric species is allowed to polymerise. The copolymer contains multiple units of each monomer in the chain. The examples are copolymers of 1,3-butadiene and styrene (Buna-S) and 1, 3-butadiene and acrylonitrile (Buna-N).

Question 10.
How does the presence of double bonds in rubber molecules influence their structure and reactivity?
Answer:
In this polymer the double bonds are located between C₂ and C₃ of isoprene units. This cis-configuration about double bonds do not allow the chains to come closer for effective attraction due to weak intermolecular attractions. Hence, the natural rubber has a coiled structure and shows elasticity.

Question 11.
Write the monomers of Teflon and Neoprene.
Answer:

• Teflon – Tetrafluoroethene
• Neoprene – 2-Chloro-1,3-butadiene

Question 12.
What is PHBV? What is its importance in polymer chemistry?
Answer:
PHBV is polyhydroxy butyrate – co – β – hydroxy valerate. PHBV is a biodegradable polymer.

Question 13.
What is meant by vulcanisation?
Answer:
The process of heating natural rubber with sulphur to improve its properties is called vulcanisation.

Plus Two Chemistry Polymers Three Mark Questions and Answers

Question 1.

• What is bakelite?
• What is its use?
• Why bakelite is said to be an example for thermosetting polymer?

Answer:

• Bakelite is the condensation polymer of phenol and formaldehyde.
• It is used for the manufacture of electrical plugs and switches.
• Bakelite contains cross linked molecules which on heating undergo extensive cross linking in moulds and again become infusible. It cannot be reused. Hence, it is a thermosetting ploymer.

Question 2.
Classify the following compounds into natural and synthetic polymers.
(Starch, Nylon, Butadiene, Styrene rubber, Natural rubber, PVC, Cellulose)
Answer:
Natural polymers: Starch, Natural rubber, Cellulose Synthetic polymers: Nylon, Butadiene, Styrene rubber, PVC.

Question 3.
Write notes on

1. Linear polymer
2. Branched polymer
3. Cross linked polymer

Answer:

1. Here small monomer units are arranged one behind the other so as to form a chain.
2. Small monomer units combine together in such a way that branched arrangement is possible in the polymer.
3. A rigid arrangement is possible when different monomer units are connected by cross linking.

Question 4.
Some polymers are given below:

1. PVC
2. Teflon
3. Nylon 6,6.

Make a table using the above representing name of polymer, name of monomer and type of polymerisation.
Answer:

Question 5.
Distinguish between Buna-N and Buna-S.
Answer:
S Buna N: It is a synthetic rubber and it is a copolymer of 1,3-Butadiene and Vinyl cyanide.

is a synthetic polymer formed by the copolymerisation of 1,3-butadiene and styrene.

Question 6.
In a science exhibition, one student mixed two transparent liquids in a beaker and got a sticky material from the interface of two liquids and he claims that the material formed is nylon 6,6.

1. Name the two liquids he mixed.
2. Name the monomer of nylon 6.
3. Suggest two uses of nylon 6, 6.

Answer:
1. Hexamethylene diammine & adipic acid.
2. Caprolactum
3. Uses of nylon 6, 6:

• For making sheets, bristles for brushes.
• In textile industry.

Question 7.
Rubber is a polymer.

1. Name the monomer of natural rubber.
2. Discuss the importance of the vulcanisation of rubber.
3. How is Neoprene prepared?

Answer:

1. 2-Methyl-1, 3-butadiene (or Isoprene) is the monomer.

2. Vulcanised rubber has high elasticity, low water absorption, good resistance to oxidation and organic solvents.
3. Neoprene is formed by the free radical polymerisation of chloroprene.

Plus Two Chemistry Polymers Four Mark Questions and Answers

Question 1.
In rubber industry, natural rubber is processed using sulphur.

• What is the need of processing rubber in this way?
• Name this process.
• What is the monomer unit of natural rubber?

Answer:

• Natural rubber is a gummy substance which has poor elasticity when it is heated with sulphur, it becomes non sticky and more elastic.
• Vulcanisation
• 2-Methyl-1, 3-butadiene or Isoprene

Question 2.
Fill in the blanks:
Table

Answer:

1. Vinyl chloride
2. Glyptal
3. 1,3-butadiene and acrylonitrile
4. Neoprene

Question 3.

1. Give the monomeric repeating units of nylon 6 and nylon 6, 6.
2. Differentiate between addition polymerisation and condensation polymerization.

Answer:

1. monomeric repeating units:

1. Nylon – 6 – Caprolactam
2. Nylon – 66 – Adipic acid, Hexamethylene diamine

2. Difference between addition polymerisation and condensation polymerization:

• Addition polymer: Simple monomer units combine together to form polymer.
  Example: polythene.
• Condensation polymer: Simple monomer units combine together to form polymer followed by the removal of small molecules like H₂O, NH₃ etc.
  Example: Nylon-6,6,

Question 4.

1. What is polymerisation?
2. What is a homo-polymer?

Answer:

1. Polymerisation is the process of formation of polymers from respective monomers,
2. Homopolymers are polymers formed by the addition polymerisation of a single monomeric species.

Plus Two Chemistry Polymers NCERT Questions and Answers

Question 1.
What are natural and synthetic polymers? Give two examples of each type.
Answer:
Natural polymers are high molecular mass macromolecules and are found in plants and animals, e.g. proteins, nucleic acids.
Synthetic polymers are man-made high molecular mass macromolecules. These include synthetic- plastics, fibres and rubbers, e.g. polythene, dacron.

Question 2.
In which classes, the polymers are classified on the basis of molecular forces?
Answer:
On the basis of molecular forces present between the chains of various polymers, the classification of polymers is give as follows:

1. Elastomers
2. Fibres
3. Thermoplastics and
4. Thermosetting plastics

Question 3.
How do you explain the functionality of a monomer?
Answer:
Functionality of a monomer is the number of bonding sites in a monomer.

Question 4.
Is (NH-CHR-CO)_n, a homopolymer or copolymer?
Answer:
Since the unit (NH-CHR-CO)_n is obtained from a single monomer unit, it is a homopolymer.

Question 5.
How does the presence of double bonds in rubber molecules influence their structure and reactivity?
Answer:
From the structural point of view, the natural rubber is a linear cis-1, 4-polyisoprene. In this polymer, the double bonds are located between C₂ and C₃ of isoprene units. This cis-configuration about double bonds do not allow the chains to come closer for effective attraction due to weak intermolecular attractions. Hence, the natural rubber has a coiled structure and shows elasticity.

Students can Download Chapter 2 Electric Potential and Capacitance Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 2 Electric Potential and Capacitance

Plus Two Physics Electric Potential and Capacitance NCERT Text Book Questions and Answers

Question 1.
Two charges 5 × 10^-8C and -3 × 10^-8C are located 16 cm apart. At what point on the line joining the two charges is the electric potential zero?
Take the potential at infinity to be zero.
Answer:
Given q₁ = 5 × 10^-8C, r=16cm = 0.16m q₂= -3 × 10^-8C Let potential be zero at a distance × metre from positive charge q₁.
∴ r₁ = x meter
r₂ = (0.16 – x) metre
S0 V = \(\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q_{1}}{r_{1}}+\frac{q_{2}}{r_{2}}\right]\)

or 0.8 – 5x = 3x
or x = 0.1m = 10cm.

Question 2.
A regular hexagonal of side 10cm has a charge 5mC at each of its vertices. Calculate the potential at the centre of the hexagon.

Answer:
From the figure, we have
OP = OQ = OR = OS = OT = OU
= r = 10cm = 0.1m
And given q = 5µC = 5 × 10^-6C
∴ Potential at O due to all the charges

= 2.7 × 10⁶volt.

Question 3.
Two charges 2mC and -2mC are placed at points A and B 6cm apart.

1. Identify an equipotential surface of the system.
2. What is the direction of the electric field at every point on this surface?

Answer:

1. The plane normal to AB and passing through its mid-point has zero potential everywhere hence the plane is equipotential.
2. Normal to the plane is the direction AB.

Question 4.
A spherical conductor of radius 12cm has a charge of 1.6 × 10^-7C distributed uniformly on its surface. What is the electric field

1. Inside the sphere.
2. Just outside the sphere.
3. At point 18cm from the centre of the sphere?

Answer:
1. Zero

2.

3.

Question 5.
A parallel plate capacitor with air between the plates has a capacitance of 8pF (1 pF=10^-12F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant s?
Answer:
The capacitance of capacitor with air as dielectric is given by
C = \(\frac{\varepsilon_{0} A}{d}\)
Given C = 8pF = 8 × 10^-12F …………(1)
If C₁ is new capacitance when d₁ = \(\frac{d}{2}\) and space is filled with a substance of dielectric constant k=6.
Then

Using Eq.(1)
C₁ = 12 × 8 × 10^-12F
or C₁ = 96pF.

Question 6.
Three capacitors each of capacitance 9pF are connected in series.

1. What is the total capacitance of the combination?
2. What is the potential difference across each capacitor, if the combination is connected to a 120 volt supply?