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Students can Download Chapter 3 Money and Banking 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 Money and Banking

Plus Two Economics Money and Banking One Mark Questions and Answers

Question 1.
At liquidity trap, speculative demand for money becomes:
(a) zero
(b) unity
(c) infinity
(d) negative
Answer:
(c) infinity

Question 2.
RBI was established in:
(a) 1930
(b) 1935
(c) 1947
(d) 1949
Answer:
(b) 1935

Question 3.
Money Multiplier = ?
(a) M – H
(b) M + H
(c) \(\frac{M}{H}\)
(d) \(\frac{H}{M}\)
Answer:
(c) \(\frac{M}{H}\)

Question 4.
High powered money is equal to:
(a) DD + CU
(b) CU + R
(c) DD + R
(d) None of these
Answer:
(b) CU + R

Question 5.
Identify the agency responsible for issuing Rs. 1 currency note in India.
(a) RBI
(b) Ministry of Finance
(c) Ministry of Commerce
(d) Niti Aayog
Answer:
(b) Ministry of Finance

Question 6.
When r = min Speculative Demand for Money becomes
(a) Zero
(b)One
(c) Infinity
(d)GreaterthanOne
Answer:
(c) Infinity

Plus Two Economics Money and Banking Two Mark Questions and Answers

Question 1.
M₁ = CU + DD.
M₂ = M₁+ Savings deposits with Post Office savings banks.
M₃= M₁ + Net time deposits of commercial banks.
M₄ = M₃ + Total deposits with Post Office savings.
Answer:
M₄ and M₂ are narrow money.
M₃ and M₄ are broad money.

Question 2.
The RBI is known as the lender of last resort. Give reason for this.
Answer:
The Reserve Bank of India plays a crucial role here. In case of a crisis, it stands by the commercial banks as a guarantor and extends loans to ensure the solvency of the latter. This system of guarantee assures individual account holders that their banks will be able to pay their money back in case of a crisis and there is no need to panic thus avoiding bank runs. This role of the monetary authority is known as the lender of last resort.

Question 3.
What do you mean by double coincidence of wants? To which system is it associated with?
Answer:
By double coincidence of wants, we mean the need to have mutually exchangeable goods and the compulsion for exchange. This system is associated with the barter system.

Question 4.
Make pairs.
Credit money, gold coins, full bodied money, representative full-bodied money, cheques, paper money.
Answer:

• Credit money – cheques
• Full bodied money – gold coins
• Representative full bodied money – paper money

Question 5.
Give examples for fiat money and legal tender money.
Answer:

• Fiat money: Money with no intrinsic value Eg. Currency notes, coins.
• Full-bodied money: Money whose face value is equal to the intrinsic value Eg. Gold coins, silver coins.
• Legal tender money: Money issued by the monetary authority or the government which cannot be refused by anyone. Eg. Currency notes.

Question 6.
The speculative demand function is infinitely elastic. Do you agree?
Answer:
Yes, speculative demand for money is infinitely elastic.

Question 7.
Mr. Ramu sells his coconut to purchase tapioca from his neighbour Mr. Bhaskara.

1. What type of transaction is this?
2. Suggest other types of transaction in an economy

Answer:

1. Barter System
2. Money transaction

Question 8.
The price of 1 kg rice is ₹50. Identity the functions of money mentioend in the above statement.
(a) Medium of exchange
(b) Unit of account
(c) Store of value
(d) means of deffered payment
Answer:
(b) Unit of account

Question 9.
An individual exchanges coconut for rice.

1. Identify the system of exchange mentioned in the above statement.
2. List out anyone drawback of such system

Answer:

1. Barter system
2. Absence of double coincidence of wants

Question 10.
Explain the relationship between the aggregate transaction demand for money and the nominal GDP.
Answer:
There is a positive relationship between the value of transaction and nominal GDP. AN increase in the nominal GDP implies and increase in the total value of transaction and a greater transaction demand for money.

Plus Two Economics Money and Banking Three Mark Questions and Answers

Question 1.
Match columns B and C with A.

Answer:

Question 2.
Classify the following into the functions of central bank and commercial banks.
Note issue, accepting deposits, lender of last resort, lending money, credit creation, discounting bill of exchange, adviser to the government, publication of reports.
Answer:

Question 3.
‘Every loan creates a credit’. How can you connect this with the function of commercial bank?
Answer:
Banks can create credit. When a person goes to the bank and asks for a loan by providing required security, the bank grants the loan. The loan amount is not directly paid to the borrower in cash. Instead, it is deposited into an account operated by the borrower. Thus, every loan creates a deposit. This is known as credit creation.

Question 4.
How do changes in cash reserve ratio affect the availability of credit? Explain.
Answer:
The commercial banks have to keep a fraction of its deposits as reserves in the form of cash with the central bank. When this cash reserve ratio is increased, the liquidity of banks reduces and so does their credit creating capacity. This lowers excess demand in an economy. To increase demand, the cash reserve ratio is lowered so that the banks can create more credit.

Question 5.
Point out the instruments of monetary policy.
Answer:
The instruments of monetary policy of RBI are:

• Open market operations
• Bank rate
• Varying reserve requirements
• Margin requirements
• Moral suation
• Direct action

Question 6.
The transaction demand for money is written as, \(\mathrm{M}_{\mathrm{T}}^{\mathrm{d}}\)= K.T. What does \(\mathrm{M}_{\mathrm{T}}^{\mathrm{d}}\), K and T denote?
Answer:
In the equation\(\mathrm{M}_{\mathrm{T}}^{\mathrm{d}}\)= K.T.

• \(\mathrm{M}_{\mathrm{T}}^{\mathrm{d}}\) stands for stock of demand for money.
• K stands for a positive fraction.
• T stands for total value of transactions in the economy.

Question 7.
Define money. What are the main functions of money?
Answer:
Functions of Money:
Nowadays, no economies can sustain without money. Money performs the functions of medium of exchange, measure of value, store of value and the standard of deferred payments. The functions of money are classified into primary functions, secondary functions, and contingent functions.

1. Primary functions
2. Secondary functions

Question 8.
Write down the equation of speculative demand for money. Prove that when r = r_max the speculative demand for money is zero.
Answer:

Question 9.
Identify and explain the situation shown in the diagram.

Answer:
Liquidity Trap:
A situation in the economy may arise when everyone will hold their wealth in money balance. If additional money is injected into the economy, it will not be used to purchase bonds. It will be used to satisfy the people’s drawing for money balance without lowering the rate of interest. Such a situation is called ‘liquidity trap’.

Question 10.
K = 1/2, GDP Deflator =1.2, Real GDP = 1000. Cal¬culate Transaction Demand for Money.
Answer:
\(\begin{aligned}
&=\frac{1}{2} \times 1.2 \times 1000\\
&=\frac{1200}{2}=600
\end{aligned}\)

Plus Two Economics Money and Banking Four Mark Questions and Answers

Question 1.
Make a tabular presentation of the measure of money supply in India.
Answer:
The total stock of money in circulation among the public at a particular point of time is called money supply. RBI publishes figures for four alternative measures of money supply, viz. M₁ M₂ M₃and M₄. They are defined as follows.

Question 2.
Define money multiplier. Derive the formula of money multiplier.
Answer:
We define money multiplier as the ratio of the stock of money to the stock of high powered money in an economy, viz. MIH. Clearly, its value is greater than 1. We need not always go through the round effects in order to compute the value of the money multiplier. By definition, money supply is equal to currency plus deposits.

M = CU + DD = (1 + cdr) DD where, cdr = CU/DD. Assume, for simplicity, that treasury deposit of the Government with RBI is zero. High powered money then consists of currency held by the public and reserves of the commercial banks, which include vault cash and banks’ deposits with RBI. Thus H = CU + R = cdr.DD + rdr.DD = (cdr+ rdr)DD Thus the ratio of money supply to high powered money.

\(\mathrm{M} / \mathrm{H}=\frac{1+\mathrm{cdr}}{\mathrm{cdr}+\mathrm{rdr}}>1\) as rdr < 1
This is precisely the measure of the money multiplier.

Question 3.
Classify the following into appropriate titles.
Answer:

• national clearinghouse
• agency function
• credit creation
• banker’s bank
• lending money
• custodian of foreign exchange reserves oil
• discounting bills of exchange
• lender of last resort
• adviser of government

Answer:
1. Functions of central bank

• Banker’s bank
• Custodian of foreign exchange reserves
• Lender of last resort
• Adviser of government Functions of

2. commercial banks

• National clearing house
• Agency function
• Credit creation
• Lending money
• Discounting bills of exchange

Question 4.
Prepare a note on High Powered Money.
Answer:
The total liability of the monetary authority of the country, RBI, is called the monetary base or high powered money. It consists of currency (notes and coins in circulation with the public and vault cash of commercial banks) and deposits held by the Government of India and commercial banks with RBI.

If a member of the public produces a currency note to RBI the latter must pay her value equal to the figure printed on the note. Similarly, the deposits are also refundable by RBI on demand from deposit – holders. These items are claims which the general public, government or banks have on RBI and hence are considered to be the liability of RBI.

High powered money then consists of currency held by the public and reserves of the commercial banks, which include vault cash and banks’ deposits with RBI. Thus H = CU + R.

The following figure shows the High Powered Money in relation to Total Money Supply.

Question 5.
Rewrite statement if they are wrong.

1. The most liquid form of asset is shares of companies.
2. Demand deposit^usually carry high interest rates.
3. The responsibility of note issue in India is with the State Bank of India.
4. Canara Bank in India is a private sector bank.

Answer:

1. The most liquid form of asset is money
2. Demand deposits usually carries no interest
3. The responsibility of note issue in India is with RBI
4. Canara Bank is a private sector bank.

Question 6.
Suppose a bond promises ₹500 at the end of two years with no intermediate return. If the rate of interest is 5 percent per annum, what is the price of the bond?
Answer:
Let the price of the bond = x
Then,

Question 7.
Identify the condition for money market equilibrium.
Answer:
Money market is in equilibrium when money supply and money demand are equal, that is Ms = Md Here money supply is determined by RBI ie. Ms = Ms.

Money demand has 2 components namely Transaction demand (Mtd) and Speculative demand (Msd)
Therefore, the equilibrium condition is, Ms = Mtd + msd

Question 8.
Explain why speculative demand for money is inversely related to the rate of interest.“The speculative money demand function is infinitely elastic”. Prove this statement with the help of a diagram showing the relation between the speculative demand for money and liquidity trap.
Answer:
Total demand for money in an economy is composed of transaction demand and speculative demand. The speculative demand for money is inversely related to the market rate of interest. When the rate of interest is high then everyone expects it to fall in future as there is surety about future capital gain.

Thus everyone becomes ready to convert the speculative money balance into bonds. When rate of interest falls and reach its minimum level, everyone put whatever wealth they acquire in the form of money and the speculative demand for money is infinite.

Question 9.
Distinguish between the CDR and RDR, CDR, RDR
Answer:
Currency Deposit Ratio (CDR) is the ratio of money held by the public in currency to that they hold in bank deposit.
Cdr= CU\DD

The Reserve Deposit Ration (RDR) banks hold a part of the money people keep in this bank deposit as reserve money. Reserve money consists of two things. Vault cash in banks and deposits of commercial banks with RBI. It is the proportion of total deposits commercial banks keep as reserve

Plus Two Economics Money and Banking Five Mark Questions and Answers

Question 1.
The central bank adjusts the monetary instruments in relations with the economic situations.

1. Do you agree to this statement?
2. Also, define the terms

Answer:

1. Yes, I do agree to this statement
2. The terms

1. Bank Rate:
The rate of interest payable by commercial banks to RBI if they borrow money from the latter in case of a shortage of reserves.

2. Cash Reserve Ratio (CRR):
The fraction of their deposits which the commercial banks are required to keep with RBI.

3. Statutory Liquidity Ratio (SLR):
The fraction of their total demand and time deposits in which the commercial banks are required by RBI to invest in specified liquid assets.

4. Repo rate:
The rate at which commercial banks borrow from the central bank.

5. Reverse Repo Rate:
The rate at which the central bank borrow from the commercial banks.

Question 2.
Give appropriate terms for the following.

1. The total liability of the monetary authority of India.
2. System of exchange of goods for goods.
3. My = K.T.
4. The proportion of the total deposits commercial banks keep as resources in RBI
5. Revenue expenditure – Revenue Receipts on lays
6. Total expenditure – (Revenue receipt + Nondebt creating capital receipts)

Answer:

1. High powered money
2. Barter system
3. Transaction demand for money
4. Cash reserve ratio
5. Revenue deficit
6. Fiscal deficit

Question 3.
Draw a flow chart showing broad classification of commercial banks in India.
Answer:

Question 4.
Prepare a table pointing out major differences between commercial banks and central bank.
Answer:

Question 5.
Anand, a 12^th class student went to SBT and met the manager. He asked manager about major functions performed by the bank. What may be the answer given by the manager?
Answer:
Functions of commercial banks
A. Accepting deposits of 3 kinds

• Current account deposits
• Fixed-term deposits
•  Saving account deposits

B. Giving loans and advances

• Cash credit
• Demand loans
• Short term loans

C. Agency functions

• Transfer of funds
• Collection of funds
• Payments of various items
• Purchase and sale of shares and securities
• Collection of dividends
• Trustees and executors of property
• Letter of references

D. Financing of foreign trade
E. Supply of liquidity
F. Performing general utility functions

Question 6.
The data is related to the money supply in India

1. Distinguish M1 and M3.
2. Identify the aggregate monetary resources in 2002 and calculate the percentage change in 2003.

Answer:
1. M1 = Cu + DD
Cu = Currency and coin held by the people.
DD = Demand deposit in commercial banks. M3 = M1 + Net time deposit of commercial banks,

2. M3 is the aggregate monetary resources that is in 2002 it is 1498355.
The percentage change in 2003
\(=\frac{1725222-1498355}{1498355} \times 100=15.14\)

Plus Two Economics Money and Banking Eight Mark Questions and Answers

Question 1.
Prepare a seminar paper on functions of Central Bank.
Respected teacher and dear friends,
I am supposed to present a seminar paper on ‘the functions of the central bank’. In this paper, I would like to include the definition of a central bank in the first part and its functions in the second part of my seminar paper. In the words of R.P. Kent, ‘a central bank is an institution charged with the responsibility of maintaining the expansion and contraction of money in the interest of general public welfare”.

Functions of central bank The central bank in every country performs some common functions which are pointed out below (explain them in detail).

1. Note issue
2. Banker’s bank
3. Banker, agent, and adviserto the government
4. Custodian of foreign exchange reserves
5. Lender of last resort
6. Controller of credit and money supply
7. Publication of reports.

Thus it is clear that a central bank performs various functions in order to control the economy as a whole.

Question 2.
Consumer prices in India increased 4.88% in November 2017.

1. Name the policy and its three instruments that are used to control inflation.
2. Explain how these three policy instruments are used to control the inflation.

Answer:
Monetary policy is the policy used to control inflation. Its three instruments are
1. Open market operation:
The RBI sells and buys government securities in the open market. This is known as open market operation. At the time of high inflation, RBI sells government securities to the public. The money from the economy flow to the RBI. Since the money in the economy falls the demand for goods and services too will fall and the price will come down. So the inflation is controlled.

2. Changes in the bank rate:
When inflation is high RBI will raise the bank rate. At higher bank rate people will demand less loans, they will save more. This will lead to a reduction in the demand, the price level too will fall. An increased bank rate also will lead to a fall in the money supply.

3. Varying the reserve requirement:
At the time of high inflation, the RBI can make necessary changes in the various reserve requirements like CRR, CDR, SLR etc. At the time of inflation RBI raises the CRR, that means more money has to be kept in the RBI. This will reduce the money supply in the economy. So the demand will come down and the general price level too.

Question 3.
Mr. Ramu is a farmer. He has a surplus of rice which he wishes to exchange for clothing. But he may not be able to find any person who demand rice with a surplus of clothing. This is one of the problems in Barter System.

1. Can you identify other drawbacks of the Barter System?
2. How did invention of money help to overcome these drawbacks of Barter System?

Answer:
1. drawbacks of Barter System.

• Lack of double coincidence of wants
• Difficulty in store of value
• Difficulty in measuring value

2. Functions of Money:
Nowadays, no economy can sustain without money. Money performs the functions of medium of exchange, measure of value, store of value and the standard of deferred payments. The functions of money are clas¬sified into primary functions, secondary functions, and contingent functions.

A. Primary functions:
1. Medium of exchange:
The most important function of money is that it acts as a medium of exchange. It acts as a link between the buyer and the seller. Money mostly solves the issues of barter economy.

2. Standard of value:
Money acts as a conve-nient unit of account. When we purchase or sell goods, its value can be expressed in monetary terms. We express the value of goods in terms of money. Examples are : price per Kg of sugar is Rs. 30, price per meter of cloth is Rs. 25 and price per liter of milk is Rs. 20, etc.

B. Secondary functions:
1. Standard of deferred payments:
Deferred ‘ payments are those payments which are to be made in the future. Money is used as a unit of deferred payments. It helps in credit transactions. The money helps the consumer to purchase goods and services when they require it and the payment can be made in the future.

2. Store of value:
The money can be stored as an instrument for storing the value. The purchasing power of money can be transferred to the future from the current period. Thus, wealth can be stored in liquid form.

3. Transfer of value:
Transfer of value indicates the movement of value of goods from one place to another which are durable and immovable in nature. Money helps in the movement of transfer of value. For instance, the goods or property of a person can transfer to another person in terms of money.

C. Contingent function:
Money performs contingent functions as well. The initiative of the money in terms of development of a nation is termed as contingent function of money. The following are the contingent functions of money.

1. Distribution of national income:
The contribution of various sectors of the economy towards national income is measured in terms of money. Money is used to measure national income and the productivity of various sectors of the economy.

2. Basis of credit:
The industrial activities of a nation are totally depended on the availability of credit facilities. The credit facilities are to-tally depended on the availability of money.

3. Solvency:
The ability or the capacity of a firm or an institution to repay its debt is known as solvency. If a person is financially sound enough to clear off his debt is known as a solvent. On the other hand, if a person fails to clear off his debt is known as a bankrupt.

4. Utilization of resources:
Money is a common measuring of rod. Money helps the producer to measure the value of his output. Money helps the fuller utilization of resources.

Question 4.
a. Identify and explain the theory behind the diagram.

b. When the economy reaches to Liquidity trap, how does it affect the Speculative demand for money?
Answer:
a. Liquidity Preference theory Keynes defines the rate of interest as the reward for parting with liquidity for a specified period of time. According to him, the rate of interest is determined by the demand for and supply of money.

Demand for money:
Liquidity preference means the desire of the public to hold cash. According to Keynes, there are three motives behind the desire of the public to hold liquid cash:

1. the transaction motive,
2. the precautionary motive, and
3. the speculative motive.

1. Transactions Motive:
The transactions motive relates to the demand for money or the need of cash for the current transactions of individual and business exchanges. Individuals hold cash in order to bridge the gap between the receipt of income and its expenditure. This is called the income motive.

The businessmen also need to hold ready cash in order to meet their current needs like payments for raw materials, transport, wages, etc. This is called the business motive.

2. Precautionary motive:
Precautionary motive for holding money refers to the desire to hold cash balances for unforeseen contingencies. Individuals hold some cash to provide for illness, accidents, unemployment and other unforeseen contingencies.

Similarly, businessmen keep cash in reserve to tide over unfavorable conditions or to gain from unexpected deals.
Keynes holds that the/ransaction and precautionary motives are relatively interesting inelastic, but are highly income elastic. The amount of money held under these two motives (M₁ is a function (L)) of the level of income (Y) and is expressed as M₁ = L₁ (Y)

3. Speculative Motive:
The speculative motive relates to the desire to hold one’s resources in liquid form to take advantage of future changes in the rate of interest or bond prices. Bond prices and the rate of interest are inversely related to each other.

If bond prices are expected to rise, i.e., the rate of interest is expected to fall, people will buy bonds to sell when the price later actually rises. If, however, bond prices are expected to fall, i.e., the rate of interest is expected to rise, people will sell bonds to avoid losses.

According to Keynes, the higher the rate of interest, the lower the speculative demand for money, and lower the rate of interest, the higher the speculative demand for money. Algebraically, Keynes expressed the speculative demand for money as M₂ = L₂(r)
Where,
L₂ is the speculative demand for money, and r is the rate of interest. Geometrically, it is a smooth curve which slopes dovynward from left to right. Now, if the total liquid money is denoted by M, the transactions plus precautionary motives by Mi and the speculative motive by M₂, then M = M₁ + M₂.

Since M₁= (Y) and M₂= L₂ (r), the total liquidity preference function is expressed as M = L (Y, r). Supply of Money : The supply of money refers to the total quantity of money in the country. Though the supply of money is a function of the rate of interest to a certain degree, yet it is considered to be fixed by the monetary authorities. Hence the supply curve of money is taken as perfectly inelastic represented by a vertical straight line.

Determination of the Rate of Interest: Like the price of any product, the rate of interest is determined at the level where the demand for money equals the supply of money. In the following figure, the vertical line QM represents the supply of money and L the total demand for money curve. Both the curve intersect at E2 where the equilibrium rate of interest OR is established. ,

If there is any deviation from this equilibrium position an adjustment will take place through the rate of interest, and equilibrium E₂will be re-established. At the point, E₁the supply of money OM is greater than the demand for money OM₁.

Consequently, the rate of interest will start declining from OR₁ till the equilibrium rate of interest OR is reached. Similarly at OR₂level of interest rate, the demand for money OM₂is greater than the supply of money OM. As a result, the rate of interest OR₂will starts rising till it reaches the equilibrium rate OR.

It may be noted that, if the supply of money is increased by the monetary authorities, but the liquidity preference curve L remains the same, the rate of interest will fall. If the demand for money increases and the liquidity preference curve shifts upward, given the supply of money, the rate of interest will rise.

b. Speculative demand for money becomes infinitely elastic.

Students can Download Chapter 10 Wave Optic 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 10 Wave Optic

Plus Two Physics Wave Optic NCERT Text Book Questions and Answers

Question 1.
What is the shape of the wavefront in each of the following cases:

1. Light diverging from a point source.
2. Light emerging out of a convex lens when a point source is placed at its focus.
3. The portion of the wavefront of light from a distant star intercepted by the earth.

Answer:

1. It is spherical wavefront,
2. It is plane wavefront.
3. Plane wavefront (a small area on the surface of a large sphere is nearly planar).

Question 2.
What is the Brewster angle for to glass transition? (refractive index glass = 1.5).
Answer:
Given µ = 1.5, θ = ?
Since µ = tan θ
∴ tan θ = 1.5
or θ = tan^-1 1.5 = 56.3°.

Question 3.
Light of wavelength 5000 Å falls on a reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?
Answer:
Given λ = 5000 Å = 5 × 10^-7m
The wavelength and frequency of reflected light remains same.
∴ Wavelength of reflected light,
λ = 5000 Å.
Frequency of reflected light,

= 6 × 10¹⁴HZ
The reflected ray is normal to incident if angle of incidence i = 45°.

Question 4.
Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.
Answer:
Given a = 4mm = 4 × 10^-3m.
λ = 400 nm = 4 × 10^-7m
Z_F = ?
∴ The minimum distance a beam of light has to travel before its deviation from straight line path becomes significant is called Fresnel distance Z_F
∴ Z_F = \(\frac{a^{2}}{\lambda}=\frac{16 \times 10^{-6}}{4 \times 10^{-7}}\) = 40 m.

Plus Two Physics Wave Optic One Mark Questions and Answers

Question 1.
The reddish appearance of the sun at sunrise and sunset is due to
(a) The scattering of light
(b) The polarisation of light
(c) The colour of the sun
(d) The colour of the sky
Answer:
(a) The scattering of light
Explanation: The reddish appearance of the sun at sunrise and sunset is due to the scattering of light.

Question 2.
Angle between the plane of vibration and plane of polarization is
(a) 30°
(b) 90°
(c) 60°
(d) 70°
Answer:
(b) 90°
Explanation: Angle between the plane of vibration and plane of polarization is 90°.

Question 3.
If yellow light emitted by sodium lamp in Young’s double-slit experiment is replaced by monochromatic blue of light of the same intensity
(a) fringe width will decrease
(b) fringe width will increase
(c) fringe width will remain unchanged
(d) fringes will becomes less intense
Answer:
(a) fringe width will decrease
Explanation: As β = \(\frac{\lambda D}{d}\) and λ_b < λ_γ
∴ Fringe width p will decrease.

Question 4.
A Young’s double-slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen
(a) straight line
(b) parabola
(c) hyperbola
(d) circle
Answer:
(a) straight line
Explanation: Straight line fringes are formed on screen.

Question 5.
Find the odd one and justify interference, diffraction, polarisation.
Answer:
Polarisation, because polarisation is possible only for transverse wave. So all other phenomenon are due to super position of waves.

Question 6.
State Malus law related to the intensity of light transmitted through the analyzer.
Answer:
The Malus law states that the intensity of the polarized light transmitted varies as the square of the cosine of the angle between the plane of transmission of the analyser and plane of polarizer.

Question 7.
What is the minimum thickness of a soap film needed for constructive interference in reflected light, if the light incident on the film is 750 nm? Assume that the refractive index for the film is µ = 1.33.
(a) 282 nm
(b) 70.5 nm
(c) 141 nm
(d) 387 nm
Answer:
(c) 141 nm
Explanation: Here, 2µt = \(\frac{\lambda}{2}\)

Question 8.
Young’s experiment is performed with light of wavelength 6000 Å wherein 16 fringes occupy a certain region on the screen’. If 24 fringes occupy the same region with another light, of wavelength λ, then λ is
(a) 6000 Å
(b) 4500 Å
(c) 5000 Å
(d) 4000 Å
Answer:
(d) 4000 Å
Explanation: n₁λ₁ = n₂λ₂

Plus Two Physics Wave Optic Two Mark Questions and Answers

Question 1.
Name the following wavefronts according to its nature.

1. Wave front due to point source.
2. Wave front due to fluorescent lamp
3. Emergent wavefront from a concave lens.
4. Emergent wavefront from a prism, when plane is incident on other face.

Answer:

1. Spherical wave front
2. Cylindrical wavefront
3. Diverging wavefront
4. Plane wavefront

Question 2.
Two coherent sources have intensities in the ratio 25: 16. Find the ratio of intensities of maxima to minima after the superposition of waves from the two source.
Answer:
I₁ = a₁² = 25, a₁ = 5
I₂ = a₂² = 16
a₂ = 4
Maximum intensity I_max = (a₁ + a₂)² = (5 + 4)² = 81
Minimum intensity I_min = (5 – 4)² = 1

Plus Two Physics Wave Optic Three Mark Questions and Answers

Question 1.
Fill in the blanks in three columns.

Answer:
(i) θ = 42°
(ii) P = 57°
(iii) Straight line
(iv) Planks constant
(v) µ_r = 1.2
(vi) Paramagnetic.

Question 2.
Match the following suitably.

Answer:

Question 3.
Match the following

Answer:

Question 4.
A plane wave front is entering a lens is given in the figure

1. What is meant by a wave front
2. What are different types of wave fronts
3. Complete the diagram and draw the refracted wave front.

Answer:

1. Locus of all points having same phase of vibration is called wavefront.
2. Spherical wavefront, cylindrical wavefront plane wave front
3. Wave frond through a thin convex lens:

Consider a plane wave passing through a thin convex lens. The central part of the incident plane wave travels through the thickest portion of lens. Hence central part get delayed. As a result the emerging wavefrond has a depression at the centre. Therefore the wave front becomes spherical and converges to a point F.

Plus Two Physics Wave Optic Four Mark Questions and Answers

Question 1.
Thomas young successfully conducted double-slit experiment and explained the interference phenomenon using Huygens principle.

1. State Huygens wave theory.
2. State Huygens principle arrive at Snell’s law of refraction.
3. In the word of Hyugens “Light propagates as longitudinal waves” comment on the above statement.

Answer:
1. According to Huygen’s principle

• Every point in a wavefront act as a source of secondary wavelets.
• The secondary wavelets travel with the same velocity as the original value.
• The envelope of all these secondary wavelets gives a new wavefront.

2. Huygen’s principle: According to Huygen’s principle:

• Every point in a wavefront acts as a source of secondary wavelets.
• The secondary wavelets travel with the same velocity as the original value.
• The envelope of all these secondary wavelets gives a new wavefront.

Refraction of a plane wave. (To prove Snell’s law):
AB is the incident wavefront and c₁ is the velocity of the wavefront in the first medium. CD is the refracted wavefront and c₂ is the velocity of the wavefront in the second medium. AC is a plane separating the two media.

The time taken for the ray to travel from Pto R is

O is an arbitrary point. Hence AO is a variable. But the time to travel a wavefront from AB to CD is constant. In order to satisfy this condition, the term containing AO in eq.(2) should be zero.

where ¹n₂ is the refractive index of the second medium w.r.t. the first. This is the law of refraction.

3. Light wave cannot be longitudinal as it exhibit polarisation.

Question 2.
A long narrow slit is illuminated by blue light and the diffraction pattern is obtained on a white screen.

1. How the width of bands change as the distance from the centre increases?
2. What happens to the width of pattern, if yellow light is used instead of blue light?
3. In a double-slit experiment, the slits are separated by 0.03cm and the screen is placed 1.5m away. The distance between the central fringe and the fourth bright fringe is 1 cm. Determine the wavelength of light used in the experiment.

Answer:
1. Width of bands decreases from the centre of the bands

2. The pattern expands or bandwidth increases

3. x_n = \(\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\)
d = 0.03cm = 0.03 × 10^-2 m
D = 1.5m, n = 4, x_n = 1cm = 1 × 10^-2m

Question 3.
Light of wavelength 589 nm is used to view an object under a microscope. The aperture of the objective has a diameter of 0.900cm.

1. What do you mean by limit of resolution of an optical instrument?
2. What is the limiting of resolution of the above microscope?
3. What happens to the limit of resolution if the objective is immersed in oil? Explain.

Answer:
1. The minimum distance between two objects at which they can be observed as separated by an optical instrument is called the limit of resolution of the instrument.

2. Limit of resolution ∆θ = \(\frac{1.22 \lambda}{D}=\frac{1.22 \times 589 \times 10^{-9}}{0.9 \times 10^{-2}}\)
= 7.98 × 10^-5 rad.

3. Decreases. Refractive index of oil is more than that of air. Hence wavelength of the light decreases.

Question 4.
The law of refraction \(\frac{\sin i}{\sin r}=\frac{V_{1}}{V_{2}}\)

1. This law is called
2. Prove this law based on Huygiens wave theory.

Answer:
1. Snells law

2. Refraction of a plane wave. (To prove Snell’s law):
AB is the incident wavefront and c₁ is the velocity of the wavefront in the first medium. CD is the refracted wavefront and c₂ is the velocity of the wavefront in the second medium. AC is a plane separating the two media.

The time taken for the ray to travel from Pto R is

O is an arbitrary point. Hence AO is a variable. But the time to travel a wavefront from AB to CD is constant. In order to satisfy this condition, the term containing AO in eq.(2) should be zero.

where ¹n₂ is the refractive index of the second medium w.r.t. the first. This is the law of refraction.

Question 5.
A long narrow slit is illuminated by blue light and the diffraction pattern is obtained on a white screen.

1. How the width of bands change as the distance from the centre increases? (1)
2. What happens to the width of pattern, if yellow light is used instead of blue light? (1)
3. In a double-slit experiment, the slits are separated by 0.03cm and the screen is placed 1.5m away. The distance between the central fringe and the fourth bright fringe is 1 cm. Determine the wavelength of light used in the experiment. (2)

Answer:
1. Width of bands decreases from the centre of the bands

2. The pattern expands or bandwidth increases

3. x_n = \(\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\)
d = 0.03cm = 0.03 × 10^-2 m
D = 1.5m, n = 4, x_n = 1cm = 1 × 10^-2m

Question 6.
Light of wavelength 589 nm is used to view an object under a microscope. The aperture of the objective has a diameter of 0.900cm.

1. What do you mean by limit of resolution of an optical instrument? (1)
2. What is the limiting of resolution of the above microscope? (1)
3. What happens to the limit of resolution if the objective is immersed in oil? Explain. (2)

Answer:
1. The minimum distance between two objects at which they can be observed as separated by an optical instrument is called the limit of resolution of the instrument.

2. Limit of resolution

= 7.98 × 10^-5rad.

3. Decreases. Refractive index of oil is more than that of air. Hence wavelength of the light decreases.

Question 7.
A beam of light, with intensity I₀, is passing through a polarizer and an analyzer as shown in figure.

1. State Malus law related to the intensity of light transmitted through the analyzer. (1)

2. If 6 = 45°, what is the relation of the intensities of original light and transmitted light after passing through the analyzer? (2)

3. Which of the following waves can be polarized

• X-rays
• sound waves. Why? (1)

Answer:
1. The Malus law states that the intensity of the polarized light transmitted varies as the square of the cosine of the angle between the plane of transmission of the analyser and plane of polarizer.

2.

3. X-rays. Because it is a transverse wave.

Plus Two Physics Wave Optic Five Mark Questions and Answers

Question 1.
Consider a point source emitting waves uniformly in all directions.

1. Draw two wave fronts very near to the point source. (1)
2. Using Huygen’s principle, prove that angle of incidence is equal to angle of reflection. (3)
3. What is the shape of a plane wave front after passing through a thin convex lens? (1)

Answer:
1.

2.

AB is the incident wavefront and CD is the reflected wavefront, ‘i’ is the angle of incidence and r’ is the angle of reflection. Let c₁ be the velocity of light in the medium. Let PO be the incident ray and OQ be the reflected ray.
The time taken for the ray to travel from P to Q is

O is an arbitrary point. Hence AO is a variable. But the time to travel for a wave front from AB to CD is a constant. So eq.(2) should be independent of AO. i.e., the term containing AO in eq.(2) should be zero.
∴ \(\frac{A O}{C_{1}}\)(sin i – sin r) = 0
sin i – sin r = 0
sin i – sin r
i = r
This is the law of reflection.

3. Spherical wavefront.

Question 2.
A slit S is illuminated by a monochromatic suorce of light to give two coherent sources P₁ and P₂. These give a dark band at the point R on the screen as shown in figure.

1. Write the formula to find out bandwidth. (1)
2. What relationship must exist between the length P₁R and P₂R? (1)
3. Can interference fringes be produced by using two identical bulbs? (1)
4. If the distance between P₁ and P₂ is 1 mm and the screen is placed in 1m away, what is the fringe separation for a light of wavelength 5 × 10^-7m?(2)

Answer:
1. β = \(\frac{\lambda D}{d}\)

2. The condition for dark fringe is P₂R – P₁R = (2n -1)\(\frac{\lambda }{2}\) where n = 1,2, 3, etc.

3. No. They are not coherent sources.

4.

Question 3.
In interference, when light energy superimpose with a phase difference 180°, darkness occurs.

1. Two sources which can give sustained interference pattern is said to be …….
2. Does interference phenomenon violate law of conservation of energy? Justify.
3. Modify the expression for bandwidth in terms of refractive index of medium between slit and screen.

Answer:
1. Coherent sources.

2. No. In interference, only energy is redistributed.

3. We know refractive index n = \(\frac{c}{v}=\frac{\lambda}{\lambda_{1}}\)
λ₁ = \(\frac{\lambda}{n}\) substituting this in the expression for bandwidth \(\left(\beta=\frac{\lambda, \mathrm{D}}{\mathrm{d}}\right)\)
we get \(\beta=\frac{\lambda D}{n d}\).

Question 4.
Figure below shows a version of Young’s Experiment performed by directing a beam of electrons on a double slit. The screen reveals a pattern of bright and dark fringes similar to an interference pattern produced when a beam of light is used.

1. Which property of electron is revealed in this observation.
2. If the electrons are accelerated by a p.d. of 54v, what is the value of wavelength associated with electrons.
3. In similar experiment, if the electron beam is re-placed by bullets fired from a gun, no interference pattern is observed. Give reason.

Answer:
1. Dual nature or wave nature.

2.

3. λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\). Since the mass ofthe bullet is very much greater than the mass of electron, the de Broglie wavelength is not appreciable.

Question 5.
A. In young’s double-slit experiment two slits are illuminated by real monochromatic light source.

1. If one of the slits is closed, what will be the observation on the screen?
2. Arrive at an expression for bandwidth of interference fringes, when both the slits are open.
3. What happens to the bandwidths, if the experimental arrangement is immersed in water?

Answer:
1. Single slit diffraction pattern.

2. Expression for band width:

S₁ and S₂ are two coherent sources having wave length λ. Let ‘d’ be the distance between two coherent sources. A screen is placed at a distance D from sources. ‘O’ is a point on the screen equidistant from S₁ and S₂.

Hence the path difference, S₁O – S₂O = 0
So at ‘O’ maximum brightness is obtained.
Let ‘P’ be the position of n^th bright band at a distance x_n from O. Draw S₁A and S₂B as shown in figure.
From the right angle ∆S₁AP
we get, S₁P² = S₁A² + AP²
S₁P² = D² + (X_n – d/2)²
= D² + X_n² – X_nd + \(\frac{\mathrm{d}^{4}}{4}\)
Similarly from ∆S₂BP weget,

S₂P² = S₂B² + BP²
S₂P² = D² + (X_n + d/2)²

S₂P² – S₁P² =2x_nd (S₂P + S₁P)(S₂P – S₁P) = 2x_nd
But S₁P ≈ S₂P ≈ D
∴ 2D(S₂P – S₁P) = 2x_nd
i.e., path difference S₂P – S₁P = \(\frac{x_{n} d}{D}\) ____(1)
But we know constructive interference takes place at P, So we can take
(S₂P – S₁P) = nλ
Hence eq(1) can be written as

Let x_n+1 be the distance of (n+1)th bright band from centre o, then we can write

∴ band width, b

β = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\)
This is the width of the bright band. It is the same for the dark band also.

3. Since wavelength decreases, the bandwidth decreases.

Question 6.
Green light is incident at the Polarizing angle on a certain glass plate as shown in figure.

1. What do you mean by polarizing angle?
2. Indicate the polarization components on the reflected and refracted rays, by arrows and dots.
3. Find the refractive index of glass.

Answer:
1. The angle of incidence, at which incident light on a transparent medium, become completely plane polarized is known as polarizing angle.

2.

3. Polarizing angle = 90° – 32° = 58°, Refractive Index, n = tan 58° = 1.6.

Question 7.
Allow to lights rays to incident on a screen after passing through two slits.

1. Why light is passed through two slits.
2. Find the expression for fringe width
3. What happens to the pattern on the screen when the whole apparatus is dipped in water

Answer:
1. Because light has wave nature. Two slits gives effect of coherent sources.

2. Expression for band width:

S₁ and S₂ are two coherent sources having wave length λ. Let ‘d’ be the distance between two coherent sources. A screen is placed at a distance D from sources. ‘O’ is a point on the screen equidistant from S₁ and S₂.

Hence the path difference, S₁O – S₂O = 0
So at ‘O’ maximum brightness is obtained.
Let ‘P’ be the position of n^th bright band at a distance x_n from O. Draw S₁A and S₂B as shown in figure.
From the right angle ∆S₁AP
we get, S₁P² = S₁A² + AP²
S₁P² = D² + (X_n – d/2)²
= D² + X_n² – X_nd + \(\frac{\mathrm{d}^{4}}{4}\)
Similarly from ∆S₂BP weget,
S₂P² = S₂B² + BP²
S₂P² = D² + (X_n + d/2)²

S₂P² – S₁P² =2x_nd (S₂P + S₁P)(S₂P – S₁P) = 2x_nd
But S₁P ≈ S₂P ≈ D
∴ 2D(S₂P – S₁P) = 2x_nd
i.e., path difference S₂P – S₁P = \(\frac{x_{n} d}{D}\) ____(1)
But we know constructive interference takes place at P, So we can take
(S₂P – S₁P) = nλ
Hence eq(1) can be written as

Let x_n+1 be the distance of (n+1)th bright band from centre o, then we can write

∴ band width, b

β = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\)
This is the width of the bright band. It is the same for the dark band also.

3. The pattern shrinks (Band width decreases) if whole apparatus is dipped in water. (Because of high refractive index, velocity of light decreases. The wavelength also decreases and hence fringe width also reduce).

Question 8.
The double slit in Young’s experiment is replaced by a single narrow slit (illuminated by a monochromatic source). Then,

1. How the pattern of bands on the screen differ from the pattern due to double slit? (1)
2. Derive an expression for the bandwidth of the central fringe. (3)
3. Draw a graph which shows the variation of intensity of light with distance. (1)

Answer:
1. A broad pattern with a central bright region is seen. On both sides, there are alternate dark and bright regions of decreasing intensity.

2. Consider a point P, on the screen at which wavelets travelling in a direction making an angle θ with CO are brought to focus by the lens are shown in figure.

These wavelets travel unequal distances in reaching the point P₁. Hence these waves are not in phase. The wavelets from the points A and B reaching P₁ are having a path difference, BP₁ – AP₁ = BN = a sinθ.

This path difference equals to λ. Then for each point in the upper half AC of the slit, there is a corresponding point in the lower half CB such that the wavelets from these two points reach at P₁ with a path difference of λ/2.

These wavelets interfere destructively to make the intensity at P₁ minimum. The point P₁corresponds to first minima. Condition for first minima is a sinθ = λ.

Thus in general, the minima will occur when the path difference = asinθ = λ where n = 1,2,3. Thus minima are formed on both sides of O, i.e. the central maxima. In between minima, other maxima called secondary maxima are formed. Secondary maxima will be at those points for which the path difference for the rays is asinθ = (2n + 1)\(\frac{\lambda}{2}\).

The width of the central maximum is defined as the distance between the first minima on either side of the central maximum. For the first minimum, a sinθ = λ when θ is small sinθ = θ. i.e.a θ = λ.

3.

Question 9.
A slit S is illuminated by a monochromatic suorce of light to give two coherent sources P₁ and P₂. These give a dark band at the point R on the screen as shown in figure.

1. Write the formula to find out bandwidth, (1)
2. What relationship must exist between the length P₁R and P₂₁R? (1)
3. Can interference fringes be produced by using two identical bulbs? (1)
4. If the distance between P₁ and P₂ is 1 mm and the screen is placed in 1m away, what is the fringe separation for a light of wavelength 5 × 10^-7m?(2)

Answer:
1. β = \(\frac{\lambda D}{d}\)

2. The condition for dark fringe is P₂R – P₁R = (2n -1)\(\frac{\lambda }{2}\) where n = 1,2, 3, etc.

3. No. They are not coherent sources.

4.

Question 10.
The Photograph given below is obtained by passing a LASER beam on a pain of closely spaced slits.

1. Identify this pattern.
2. Obtain an expression for bandwidth of the pattern.
3. In the double slit experiment using wavelength 5461 A⁰, the fringe width measured is 0.15mm. By keeping the same arrangement, the fringe width is measured for an unknown wavelength is 0.12mm. Find the unknown wave length.
4. If you change the LASER light from red to blue, what will happen to the space between the pattern shown in photograph. Justify.

Answer:
1. Interference

2. Expression for band width:

S₁ and S₂ are two coherent sources having wave length λ. Let ‘d’ be the distance between two coherent sources. A screen is placed at a distance D from sources. ‘O’ is a point on the screen equidistant from S₁ and S₂.

Hence the path difference, S₁O – S₂O = 0
So at ‘O’ maximum brightness is obtained.
Let ‘P’ be the position of n^th bright band at a distance x_n from O. Draw S₁A and S₂B as shown in figure.
From the right angle ∆S₁AP
we get, S₁P² = S₁A² + AP²
S₁P² = D² + (X_n – d/2)²
= D² + X_n² – X_nd + \(\frac{\mathrm{d}^{4}}{4}\)
Similarly from ∆S₂BP weget,
S₂P² = S₂B² + BP²
S₂P² = D² + (X_n + d/2)²

S₂P² – S₁P² =2x_nd (S₂P + S₁P)(S₂P – S₁P) = 2x_nd
But S₁P ≈ S₂P ≈ D
∴ 2D(S₂P – S₁P) = 2x_nd
i.e., path difference S₂P – S₁P = \(\frac{x_{n} d}{D}\) ____(1)
But we know constructive interference takes place at P, So we can take
(S₂P – S₁P) = nλ
Hence eq(1) can be written as

Let x_n+1 be the distance of (n+1)th bright band from centre o, then we can write

∴ band width, b

β = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\)
This is the width of the bright band. It is the same for the dark band also.

3.

λ₁ = 5461 A° = 5461 × 10^-10 m
β₁ = 0.15mm = 0.15 × 10^-3m
β₂ = 0.12mm = 0.12 × 10^-3m.

4. Wave length of blue is less than that of red. Hence β decreases.

Question 11.
According to a principle, at a particular point in a medium, the resultant displacement produced by a number of waves is the vector sum of the displacements produced by each of the waves.

1. Name of the principle.
2. Derive an expression for the bandwidth in Young’s double-slit experiment.

Answer:
1. Superposition principle.

2. Expression for band width:

S₁ and S₂ are two coherent sources having wave length λ. Let ‘d’ be the distance between two coherent sources. A screen is placed at a distance D from sources. ‘O’ is a point on the screen equidistant from S₁ and S₂.

Hence the path difference, S₁O – S₂O = 0
So at ‘O’ maximum brightness is obtained.
Let ‘P’ be the position of n^th bright band at a distance x_n from O. Draw S₁A and S₂B as shown in figure.
From the right angle ∆S₁AP
we get, S₁P² = S₁A² + AP²
S₁P² = D² + (X_n – d/2)²
= D² + X_n² – X_nd + \(\frac{\mathrm{d}^{4}}{4}\)
Similarly from ∆S₂BP weget,
S₂P² = S₂B² + BP²
S₂P² = D² + (X_n + d/2)²

S₂P² – S₁P² =2x_nd (S₂P + S₁P)(S₂P – S₁P) = 2x_nd
But S₁P ≈ S₂P ≈ D
∴ 2D(S₂P – S₁P) = 2x_nd
i.e., path difference S₂P – S₁P = \(\frac{x_{n} d}{D}\) ………1)
But we know constructive interference takes place at P, So we can take
(S₂P – S₁P) = nλ
Hence eq(1) can be written as

Let x_n+1 be the distance of (n+1)th bright band from centre o, then we can write

∴ band width, b

β = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\)
This is the width of the bright band. It is the same for the dark band also.

Students can Download Chapter 1 Introduction Macroeconomics 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 Macroeconomics

Plus Two Economics Introduction Macroeconomics One Mark Questions and Answers

Question 1.
Give two examples for macroeconomic studies.
Answer:

1. National income
2. Aggregate employment

Question 2.
The expenses which raise productive capacity is known as
Answer:
(a) Consumption expenditure
(b) Investment expenditure
(c) Export expenditure
(d) None of the above
Answer:
(b) Investment expenditure

Question 3.

The pictures are related to the impacts of recession in 2008. Can you identify some impacts of the Great Depression of the 1930’s?
Answer:
Unemployment, Fall in Aggregate Demand, etc.

Plus Two Economics Introduction Macroeconomics Two Mark Questions and Answers

Question 1.
Categorize the following into micro and macroeconomics.
GDP, study of a firm, individual consumption, price of rubber, wholesale price level, total employment, production of coconut, total imports.
Answer:

Question 2.
“Macroeconomics has got some limitations” Do you agree? Substantiate your answer.
Answer:
Yes, I agree.
Macroeconomics has got some limitations as pointed out below.

1. The generalisation from individual experience may not be true for the aggregates.
2. Aggregates are not homogeneous.
3. An aggregative tendency may not influence all the sectors of the economy in the same manner.
4. A study of the aggregates may lead us to believe that no new policy is needed and there is no change in the policy.

Question 3.
Name four sectors of the economy.
Answer:
Four major sectors of the economy are:

• firms
• households
• government
• external sector

Question 4.
List out any two features of a capitalist economy.
Answer:
Private ownership of the means of production. Production takes place for selling in the market.

Question 5.
Identify the sectors of the economy from the given below sentences.

1. A subsidy is announced
2. People buy more goods

Answer:

1. Government
2. Household

Question 6.
Some variables are given below. Classify them under two branches of economics.

• Utility
• GDP
• Inflation
• Demand for pen
• Aggregate Consumption
• Taxes

Answer:
Micro Economics

• Utility
• Demand for pen

Macro Economics

• GDP
• Inflation
• Aggregate Consumption
• Taxes

Plus Two Economics Introduction Macroeconomics Three Mark Questions and Answers

Question 1.
Prepare a note on Macroeconomics.
Answer:
Macroeconomics deals with the aggregate economic variables of an economy. It also takes into account various interlinkages which may exist between the different sectors of an economy. This is what distinguishes it from microeconomics, which mostly examines the functioning of the particular sectors of the economy, assuming that the rest of the economy remains the same.

Macroeconomics emerged as a separate subject in the 1930s due to Keynes. The Great Depression, which dealt a blow to the economies of developed countries, had provided Keynes with the inspiration for his writings. In this book, we shall mostly deal with the working of a capitalist economy.

Hence it may not be entirely able to capture the functioning of a developing country. Macroeconomics sees an economy as a combination of four sectors, namely households, firms, government and external sector.

Question 2.
Choose the appropriate answer
a. The Greek word‘macros’ means

1. small
2. large
3. very small

b. Partial equilibrium analysis is the methodology of

1. Microeconomics
2. macroeconomics
3. both microeconomics and macroeconomics

c. The words microeconomics and macroeconomics were coined by

1. J.M. Keynes
2. Ragner Frisch
3. Adam Smith

Answer:
a. 2. large
b. 1. Microeconomics
c. 2. Ragner Frisch

Question 3.
Point out the significance of macroeconomics.
Answer:
The study of macroeconomics is important because of the following reasons

1. It helps to understand the functioning of the economy
2. It is helpful in comparison
3. It is useful in planning
4. It is helpful in studying growth and development
5. It is helpful in studying economic fluctuations 0 It is helpful in formulating economic policies.

Question 4.
Match A with B and C.

Answer:

Question 5.
Below is given a few features of a phenomenon.

• Output and employment in the countries of Europe and North AmeruSft fell by huge amounts.
• Demand for goods was low.
• Factories remain idle.

1. Identify the phenomena
2. Name the book that explained a solution for this crisis.
3. Identify the branch of economics emerged after this

Answer:

1. The great depression of 1929
2. General theory of employment, interest, and money.
3. Macro Economics

Plus Two Economics Introduction Macroeconomics Five Mark Questions and Answers

Question 1.
Fill in the blanks.

1. Macroeconomics, as a separate branch of economics, emerged after the British economist ………..
2. The General Theory of Employment, Interest, and Money’ was written in The General Theory of Employment, Interest and Money’ ………….
3. ‘General Theory of Employment, Interest, and Money’ was written byThe General Theory of Employment, Interest and Money’ ………………
4. The Great Depression occurred in ……………
5. Macroeconomics finds an economy as a combination of four sectors, namely …………………

Answer:

1. John Maynard Keynes
2. 1936
3. John Maynard Keynes
4. 1929
5. Households, firms, government and external sector.

Question 2.
Find the odd one out. Justify your answer.

1. Income of a family, production of rice, gross domestic saving, profit of a firm
2. Price of rice, wholesale price index, total exports, inflation
3. Partial equilibrium analysis, Entire economy, Individual units, Worm’s eye view
4. Aggregate units, Worm’s eye view, Bird’s eye view, General equilibrium analysis.

Answer:

1. Gross domestic saving. Others are microeconomic indicators
2. Price of rice. Others are macroeconomic indicators
3. Entire economy. Others are features of microeconomics.
4. Worm’s eye view. Others are features of macroeconomics.

Question 3.
Arrange the following points as the features of microeconomics and macroeconomics and complete the table.

Answer:

Question 4.
Match the following

Answer:

Question 5.
Discuss the importance of macroeconomics.
Answer:
Importance of macroeconomics:
1. Information about the economy:
Macroeconomics provides insight into various interrelated aspects and the functioning of the economy.

2. Formulation of economic policy:
Macroeconomics plays a critical role in the formulation of economic policies.

3. Economic planning:
The main objective of economic planning is to implement various police and programmes for the welfare of the people on priority basis. Macroeconomic helps economists to examine the interrelationship between various macroeconomic variables and choose the programmes best suited for the welfare of the people.

4. Examine Economic fluctuations:
Macroeconomics examine various fluctuations in different variables such as national income, aggregate output, trade deficit, investment, etc.

Students can Download Chapter 6 Non-Competitive Markets 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 6 Non-Competitive Markets

Plus Two Economics Non-Competitive Markets One Mark Questions and Answers

Question 1.
Point out the value of the Marginal Revenue when the demand curve is elastic.
Answer:
When the demand curve is elastic, the value of Marginal Revenue (MR) is zero.

Question 2.
Identify from the following demand curve faced by a firm under monopolistic competition.
Answer:

Answer:
(C)

Question 3.
The form of the market having only 2 sellers is called:
(a) monopoly
(b) duopoly
(c) oligopoly
(d) monopolistic competition
Answer:
(b) duopoly

Question 4.
Differentiated products is a characteristic of
(a) Monopolistic competition only
(b) Oligopoly only
(c) Both Monopolistic competition & Oligopoly
(d) Monopoly
Answer:
(c) Both Monopolistic competition & Oligopoly

Plus Two Economics Non-Competitive Markets Two Mark Questions and Answers

Question 1.
Suppose there are only two firms manufacturing cars in India, namely, Maruti and Hyundai. What market form is this?
Answer:
lip This market form is known as duopoly market. Duopoly is a market structure in which there are only two firms producing a product.

Question 2.
What do you mean by cartel?
Answer:
Cartel is a kind of mutual agreement or coordination of the output and pricing policies of firms having then individual identities so as to act if it is a monopoly. This is a feature of oligopoly market.

Question 3.
Observe the following figures and identify the market situations.

Answer:

1. Monopoly
2. Monopolistic competition

Question 4.
Suggest any 2 examples of a monopolistically competitive market.
Answer:

1. Soap industry
2. Toothpaste industry.

Question 5.
The features of firms under different market structure is given below. Classify them into perfect competition and oligopoly.
(a) Collusion
(b) Free entry and exit
(c) Intertdependance
(d) Firms are price takers
Answer:
a & c Oligopoly, b & d Perfect competition

Plus Two Economics Non-Competitive Markets Three Mark Questions and Answers

Question 1.
The monopolist cannot determine the price and quantity simultaneously. Do you agree? Substantiate your answer.
Answer:
Yes, I agree to the statement that the monopolist cannot determine the price and quantity simultaneously. This is because, if the monopolist wants to sell more of the commodity, he needs to reduce the price. Therefore, he can change either of the price or the quantity.

Question 2.
Examine the relationship between Marginal Revenue and Price Elasticity of Demand.
Answer:
Whenever the MR is positive the price elasticity of AR (Demand Curve) isgreaterthan one, that is elastic. When the MR is zero the price elasticity of AR (Demand Curve) is 1, that is unitary elastic. When the MR is negative the price elasticity of demand curve is less than one, that inelastic.

Question 3.
Choose the correct answer
a. In monopoly:

1. There are many producers
2. There is no seller
3. There is no buyer
4. There is single seller

b. When two commodities are complementary to one another

1. They may be jointly demanded
2. They may be complementary goods
3. They may be substitutes
4. None of the above

c. Generally government fix control price

1. Equal to equilibrium price
2. Less than equilibrium price
3. More than equilibrium price
4. None of these

Answer:
a. 4. There is single seller
b. 1. They may be jointly demanded
c. 2. Less than equilibrium price

Question 4.
State whether the followirig satements are true or false.

1. The seller in a monopoly is a price maker.
2. Price leadership is an important feature of oligopoly.
3. Selling cost is the cost of producing the commodity.

Answer:

1. Time
2. True
3. False. Selling cost is the cost of selling a Product.

Question 5.
State the condition and long run equilibrium in a monopoly competitive industry.
Answer:
The long run equilibrium conditions in a monopolisti-cally competitive industry are:
MR = LMC
P = LAC but P > LMC MR = LMC
P = LAC, P > LMC

Question 6.
The diagram below shows the equilibrium condition of a zero cost monopolist. Find out the quantity produced by such a firm, explain

Answer:
The firm will produce oq1 level of output to maximise the profit. Because at this level of output, the firm satisfies a condition that is MR = MC. Since the firm faces zero cost its MR also will be zero. For this equilibrium, MR and MC should be equal. Since MC is zero in all level of output the firm will produce at a level where its MR is zero.

Question 7.
Examine the behaviour of average revenue and marginal revenue of a firm which can sell more units of a good only by lowering the price of that good. Explain with the help of a diagram.
Answer:
monopolist can sell more units of the good only by lowering the price. Therefore AR & MR will be downward sloping. Draws the appropriate diagram.

Question 8.
Categorize the following features under two headings Perfect Competition and Monopolistic Competition. 4 Large Number of Producers, Differentiated products, Some Pricing power, Productive Efficiency in the Long run, Low Barriers, Homogeneous products, Long run Price = MC, Many producers, Zero Barriers, Productive Inefficiency in the Long run, Price Takers, Long run Price >MC.
Answer:
1. Perfect competition

• Large Number of Producers.
• Productive Efficiency in the Long run
• Homogeneous products
• Long run Price = MC
• Zero Barriers
• Price Takers

2. Monopolistic competition

• Differentiated products
• Some Pricing power
• Low Barriers,
• Many producers
• Productive Inefficiency in the Long run
• Long run Price >MC.

Question 9.
Consider the commodities given below. Identify the most likely market situation in which they are produced. Substantiate.

1. Airline industry.
2. Potatoes.
3. Toilet soap.

Answer:

1. Oligopoly – only a few producers
2. Perfecly Competitive Market – large number of producers
3. Monopolistic Competition – Many sellers producing differentiated products

Question 10.
A table related to a particular market is given below:

1. Find AR & MR
2. Identify the market related to the table.
3. Establish the relationship between TR, AR & MR

Answer:
1.

2. Monopoly.

3. relationship between TR, AR & MR

• TR rises and then falls
• AR is always falling but lies above MR
• MR falls and becomes negative

Question 11.
The total revenue equation of a firm is given by the equation,TR = 20Q – 2Q²

1. Calculate TR, AR & MR.
2. Identify the market form related to this equation

Answer:
1. TR = 20Q – 2Q2
\(\begin{aligned}
&A R=\frac{T R}{Q}=\frac{20 Q-2 Q^{2}}{Q}=20-2 Q\\
&\mathrm{MR}=\frac{\delta \mathrm{TR}}{8 \mathrm{Q}}=\frac{\left(20 \mathrm{Q}-2 \mathrm{Q}^{2}\right)}{\mathrm{Q}}=20-4 \mathrm{Q}
\end{aligned}\)

2. Monopoly

Plus Two Economics Non-Competitive Markets Five Mark Questions and Answers

Question 1.
State whether the following statements are true or false. Rewrite the statements if they are wrong.

1. The products in perfect competition are heterogeneous
2. The seller in monopoly is a price maker
3. Price leadership is an important feature of oligopoly.
4. Duopoly is a market situation in which two buyers buy the commodity
5. Selling cost is the cost for producing the commodity.

Answer:

1. False. Products in perfect competition are homogenous
2. True
3. True
4. False. Duopoly is a market situation in which two sellers supply the commodity
5. False. Selling cost is the cost for selling or giving publicity for the commodity

Question 2.
Find odd one out.

1. Single seller, price maker, selling cost, control over supply
2. Fairly large number of firms, product differentiation, selling cost, price maker
3. A few firms, interdependence between firms, no transport cost, indeterminate demand curve
4. Tata steel, Reliance industries, Post and Telegraph

Answer:

1. Selling cost. Others are features of monopoly
2. Price maker. Others are features of monopolistic competition
3. No transport cost. Others are features of oligopoly
4. Post and Telegraph. Others are private sector companies.

Question 3.
Match column B and Q with column A.

Answer:

Question 4.
The market price, quantity and total cost of a firm are given in the following table. Find out.

1. MRandMC
2. Equilibrium price and equilibrium quantity
3. TR, TC and Total profit at equilibrium

Answer:
1.

2. Equilibrium price is Rs. 19 and quantity is 6.

3. At equilibrium
TR= 114
TC = 109
Total profit = 5

Question 5.
Categorize the following into different market forms.

1. Indian Railways
2. Toothpaste
3. Hero Honda
4. KSEB

Answer:

1. Indian Railways – monopoly
2. Toothpaste – monopolistic competition
3. Hero Honda – oligopoly
4. KSEB – monopoly

Question 6.
Prepare a note on price rigidity.
Answer:
Price rigidity is an important feature of oligopoly. Price rigidity means that price will remain rigid without much fluctuation. This is because, price increase by one firm will not be followed by other firms.

However, price reduction by one firm will be followed by other firms, due to this; the firm affecting price change will not get the benefits from the reduction of price. Therefore, no firm will reduce or increase the price. This leads to a situation of price rigidity in the oligopoly markets.

Question 7.
Assume that there are two firms A and B in a duopoly market. Firm B supplies zero output. Firm A realizes that maximum demand in the market is 20 units, and he supplies half of it, i.e., 10 units. Construct a table the different steps showing the quantity supplied by the firms.
Answer:

Question 8.
Classify the statements under the head features of oligopoly market.
Answer:

1. Large no. of buyers and sellers
2. Firm is a price maker
3. Few number of sellers
4. Products may be homogneous or differentiated.
5. There is no interdependence between firms
6. There are no barries to entry
7. Firm is price taker
8. Interdependence between firms
9. Entry restricted
10. Price rigidity prevails oil

Answer:
The features of oligopoly market are given below:

1. Few number of sellers
2. Product may be homogneous or differentiated
3. There are no barriers to entry
4. Interdependence between firms
5. Price rigidity prevails.

Question 9.
Price rigidity is an important feature of oligopoly. Can you explain what is price rigidity.
Answer:
Price rigidity is an important feature of oligopoly market. In oligopoly market, price does not change m easily in response to change in demand. If one firm decides to increase the price to earn high profit and the other firms do not do so, due to increase the price, the demand of product will fall and it causes fall in revenue and profit. Hence it is not rational for any firm of increase the price. Thus in an oligopoly market, price remain rigid.

Question 10.
Suppose that firm A enters in a duopoly market for production of commodity X at zero cost. He finds that the total demand forX in the market is 300 units. When he starts production, firm B enters in market. Find out the profit maximising quantity by each firm.
Answer:
In order to maximisejprofit each firm will produce 1/3 of the total markerdemand. In one example, total demand in the market is 300 units. Therefore, the profit maximising output is 1/3 x 300 = 100 units.

Question 11.
Identify the market structure.

Answer:

Question 12.
Make pairs.
Price maker, Price leadership, Monopoly, Monopsonist, single buyer, Oligopoly, monopolistic competition, selling cost.
Answer:

• Price maker – Monopoly
• Price leadership – Oligopoly
• Single buyer – Monopsonist
• Selling cost – Monopolistic competition.

Question 13.
The demand curves of different market situations are given below.

1. Identify market situations represented by each demand curve.
2. Give reasons for the different shapes of demand curves in these two market forms.

Answer:
1. Figure (1) represents perfect competition market Figure (2) represents monopoly market.

2. In perfect competition, there are large number of buyers and sellers. Each firm is a price taker and there is uniform price prevailing in the market. Since each unit is sold at uniform price, P = MR = AR in the market. Therefore, demand curve is horizontal straight line. However, in a monopoly market, firm is a price maker. He can vary the price. If he wants to sell more of the product, he need to reduce the price. Therefore, the demand curve is falling downward.

Question 14.
Prepare a note on monopolistic competition.
Answer:
A market structure where the number of firms is large, there is free entry and exit of firms, but the goods produced by them are not homogeneous such a market structure is called monopolistic competition. Its features are as following:

1. There are large number of buyers and sellers.
2. There is free entry and exist in long run.
3. There is product differentiation.

The monopolist produces less and charges a higher price compared to perfect competiton. It is found in the industry where there is large number of sellers, selling differentiated but close substitute products. Monopolistic competition in a commodity market arises due to the commodity being non-homogeneous.

Question 15.
Compare the price and output of a firm under perfect competition and monopolistic competition.
Answer:
A firm under perfect competition is a price taker and have a horizontal demand, but a firm under monopolistic competition is a price maker and faces a demand curve that in downward sloping and elastic. Under perfect competition MR = AR. So the firm produces more output and charge less compared to monopolistic competition. Under monopolistic MR < AR.

Question 16.
Identify the market condition with following feature.

1. Interdependence
2. Price rigidity
3. Entry restriction

Explain why prices are rigid in such market situations.
Answer:
Oligopoly. Under such markets, the price is supposed to be rigid. The reason is that here the firms are interdependent. The actions of every firm will be determined by the actions and reactions of every other firm. If one firm increases the price none other follows. The customers of that firm may switch to other firms. The firm which increased the price may feel a fall in revenue and profit.

On the other hand, if one firm reduces the price everyone else will follow. All firm’s revenue and profit fall. So firms under oligopoly will always try to keep their price rigid.

Question 17.
Two tables are given below.

1. Find TR, AR&MR.
2. There are five variables in each of the above table. Which two variables in both the tables have the same values? Give reasons.
3. The two tables are related to two market forms. Identify the form of market for each table. Give reasons.

Answer:
1.

2. Table 1: Price and AR are same Table 2: Price, AR, and MR are same

3. Table 1: Monopoly market because by reducing price firm sells more
Table 2: Perfect competition. Firm is price takes and P=AR=MR

Question 18.
Does the statement below better describe a firm operating in a Perfectly Competitive market or a firm that is Monopoly?

1. The demand curve faced by the firm is downward sloping.
2. The demand curve and the MR curve are the same.
3. Entry and exit are relatively difficult.
4. Price Taker
5. Price Maker

Answer:

1. Monopoly
2. Perfectly Competitive Market
3. Monopoly
4. Perfectly Competitive Market
5. Monopoly

Plus Two Economics Non-Competitive Markets Eight Mark Questions and Answers

Question 1.
Prepare a seminar paper on ‘Non-Competitive Markets’.
Answer:
Respected teachers and dear friends,
The topic of my seminar paper is noncompetitive markets. As we know there are different kinds of markets depending upon the number of firms, nature of the product, freedom of entry and exit, etc. On the basis, of the above, we name the non competitive markets as monopoly, monopolistic competition and oligopoly. .

Introduction:
A market structure in which there is a singe seller is called monopoly. A market structure where the . number of firms is large, there is free entry and exit of firms, but the goods produced by them are not homogeneous. Such a market structure is called monopolistic competition. If the market of a particular commodity consists of more than one seller but the number of sellers is few, the market structure is termed oligopoly.

Contents:

1. Monopoly market
2. Monopolistic competition
3. Oligopoly

1. MONOPOLY MARKET:
Monopoly may be defined as a market situation in which there is only a single seller. He controls the entire market. The term monopoly has derived from two Greek words such as ‘mono’ means single and poly means ‘seller’. The meaning of the combined term is single seller. In a boardersense, a monopolist is single seller of a commodity which does not have close substitutes, e.g. KSEB

Features of Monopoly Market:
Some of the salient features of monopoly are as follows:

1. There is only a single firm producing the product
2. There is no close substitute for the product
3. Entry is denied for other producers
4. Since there is only one seller, the firm and the industry are same
5. The firm under monopoly is the price maker

2. MONOPOLISTIC COMPETITION:
Monopolistic competition is a market characterized by the elements of perfect competition and monopoly. It is a market situation characterized by large number of firms producing various kinds of goods and services. The products of a firm will be different from the products of other firms in terms of size, shape, smell, colour, etc.

Features
The salient features of perfect competition are as follows:
1. Large number of buyers and sellers:
Under monopolistic competition, there exists large number of buyers and sellers. But the number of sellers will be less compared to perfect competition.

2. Product differentiation:
One of the most important characteristic of monopolistic competition is the existence of product differentiation. Each firm has its own product with unique brand names. The products of one firm will be different from the products of other firms in terms of size, shape, smell, color, etc.

3. Freedom of entry and exit:
Under monopolistic competition, there is freedom of entry and exit.

4. Selling cost:
The cost incurred for sales promotion such as advertisement, coupons, gifts, etc. are known as selling cost. Under monopolistic competition, the selling costs would be relatively high.

3. OLIGOPOLY:
The term oligopoly has derived from two terms oligo (small) and poly (seller). Thus oligopoly is a market situation characterized by competition among few sellers. In simple terms, it is a competition among few sellers in the market selling either homogenous or differentiated product. The industries manufacturing car, motorcycle, scooter, etc. are some of the examples for oligopolistic competition.

The main features of oligopolistic competition are as follows:
1. Few sellers:
The number of sellers or producers would be few under oligopolistic competition.

2. Homogneous or differentiated products:
The products sold under oligopolistic competition would be either homogneous (e.g. gas, petrol) or differentiated (e.g. car, scooter)

3. Free entry and exit:
Free entry and exit persist under oligopolistic competition.

4. Selling cost:
Firms spend on advertisement and sales promotion.

5. Interdependence of the firms:
Since the number of firms under oligopoly are few, they are highly interdependent. The action of one firm will certainly have impact on other firms in terms of price, quality of the product, etc.

6. Price leadership:
Some of the firms may emerge as price leaders under oligopoly. The price leader could be the first firm in the industry or the firm with largest number of consumers. The price leader takes important decisions regarding vital decisions such as the price of the product or number of units to be produced in the market, etc.

Conclusion:
Thus it can be concluded that there are three kinds of non-competitive markets. This classification is made on the basis of the number of firms, nature of the product, freedom of entry and exit, etc. In contrast to perfect competition, we find that these market forms are more realistic.

Question 2.
Prepare a table to show the distinction between monopoly, monopolistic competition, and oligopoly. Major points of distinction are given below in the table.

Answer:

Question 3.
If the market of a particular commodity consists of more than one seller but the number of sellers is few, the market structure is termed oligopoly. The special case of oligopoly where there are exactly two sellers is termed duopoly. We shall explain the different ways in which the oligopoly firms may behave.

1. Firstly duopoly firms may collude together and decide not to compete with each other and maximize total profits of the two firms together. In such a case the two firms would behave like a single monopoly firm that has two different factories producing the commodity.

2. Secondly, take the case. of a duopoly where each of the two firms decide how much quantity to produce by maximizing its own profit assuming that the other firm would not change the quantity that it is supplying. We can examine the impact using a simple example where both the firms have zero cost.

3. Thirdly, some economists argue that oligopoly market structure makes the market price of the commodity rigid, i.e., the market price does not move freely in response to changes in demand.

Question 4.
Name important non-competitive markets and give the meaning of them.
Answer:
The important forms of non-competitive markets also:

1. Monopoly
2. Monopolistic competition
3. Oligopoly

1. Monopoly:
A monopoly is a market situation in which there is a single seller of the commodity and no close substitutes of the commodity are available. The single seller can influence the price by varying his sales.

2. Monopolistic Competition:
Monopolistic competition is a market situation in which both the monopolistic element and the competitive elements are present. Its basic features are large number of buyers and sellers in the market and existence of differentiated products.

3. Oligopoly:
Oligopoly is market situation in between monopolistic competition and monopoly. In this market form, there are only a few sellers of the commodity and each seller has a substantial share in the market.

Question 5.
The diagram below shows the level of output produced and price charged in monopoly and perfect competition.

1. Identify the levels of output and price charged in monopoly and perfect competition, explain.
2. Critically evaluate the merits and demerits of perfect competition.

Answer:
1. A firm, if it is under monopoly, will produce when its MR = MC and will charge a price which is equal to AR. It produces oq level of output and charges a price op. But if it is in perfect competition if the market would

2. Confirmed to monopoly perfect competition would charge higher prices and produce less quantity. It is argued that the monopoly firms benefit themselves at the cost of consumers. The monopolist may get a profit even in tire long run and the consumers pay more and get less quantity.

But it is another argument. The profit made by the monopolist would be used for research and development and it may be useful for society in the long run in terms of new technology and new products. Moreover, due to the economies of scale the cost of the monopolist may be much lower than the cost of a firm under perfect competition.

Students can Download Chapter 9 Coordination Compounds 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 9 Coordination Compounds

Plus Two Chemistry Coordination Compounds One Mark Questions and Answers

Question 1.
The ions or molecules bound to the central atom/ion in the coordination entity are called ___________.
Answer:
Ligands

Question 2.
In [Co(C₂O₄)₃]^3-, the coordination number of cobalt is _________.
Answer:
six

Question 3.
Which complex has a square planar structure?
(a) [Ni(CO)₄]
(b) [NiCI₄]^2-
(c) [Ni(H₂O)₆]²⁺
(d) [CU(NH₃)₄]²⁺
Answer:
(d) [CU(NH₃)₄]²⁺

Question 4.
Say TRUE or FALSE.
[Co(NO₃)(NH₃)₅]SO₄ and [Co(NO₃)(NH₃)₄(SO₄)](NH₃) are ionisation isomers.
Answer:
FALSE

Question 5.
The formula of Wilkinson’s catalyst is _________.
Answer:
[RhCl(PPh₃)₃]

Question 6.
The charge of Ni in [Ni(CO)₄] is
(a) +1
(b) +2
(c) 0
(d) +4
Answer:
(c) 0

Question 7.
The central metal ion present in chlorophyll?
(a) Fe²⁺
(b) Cu²⁺
(c) Mg²⁺
(d) CO²⁺
Answer:
(c) Mg²⁺

Question 8.
EDTA is a dentate ligand
(a) uni dentate
(b) bidentate
(c) Tridentate
(d) hexadentate
Answer:
(d) hexadentate

Question 9.
Which is an example for homoleptic complexes
(a) [Co(NH₃Cl₂]⁺
(b) [CoNH₃)₆]³⁺
(c) [Cr(NH₃)(H₂O)₃]Cl₃
(d) [CoCl₂(en)₂]
Answer:
(b) [CoNH₃)₆]³⁺

Question 10.
Ammonia will not form complex with
(a) Ag²⁺
(b) Pb²⁺
(c) Cu²⁺
(d) Cd²⁺
(e) Fe²⁺
Answer:
(b) Pb²⁺

Plus Two Chemistry Coordination Compounds Two Mark Questions and Answers

Question 1.
In a seminar, Jishnu argued that the “hexaflourocobaltate(III) ion is highly paramagnetic than hexacyanoferrate(III) ion.

1. Do you agree with this words?
2. Explain it.
3. Write the formulae of the given coordination compounds.

Answer:

1. Yes
2. CN^– is a strong field ligand so paring occurs. Number of unpaired electron decreases, paramagnetism decreases. F^– is a weak field ligand so no paring occurs, paramagnetism increases.
3. [CoF₆]^3-and [Co(CN)₆]^3-

Question 2.
Raju: Coordination compounds are coloured.
Ramu: No, co-ordination compounds are colourless.

1. Whose statement is correct?
2. Explain the reason for your answer.

Answer:

1. Raju’s statement is correct. Coordination compounds are usually coloured.
2. The colour of coordination compounds is due to d-d transition.

Question 3.
Why is geometrical isomerism not possible in tetrahedral complexes having two different types of unidentate ligands coordinated with the central metal ion?
Answer:
Tetrahedral complexes do not show geometrical isomerism because the relative positions of the unidentate ligands attached to the central metal atom are the same with respect to each other.

Question 4.
FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1:1 molar ratio gives the test of Fe²⁺ ion but CuSO₄ solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu²⁺ ion. Explain why?
Answer:
FeSO₄ does not form any complex with (NH₄)₂SO₄. Instead, it forms a double salt FeSO₄. (NH₄)₂SO₄.6H₂O which dissociates completely into ions. CuSO₄ when mixed with NH₃ forms a complex [CU(NH₃)₄]SO₄ in which the complex ion [CU(NH₃)₄]²⁺ does not dissociate to give Cu²⁺ ion.

Question 5.
Write the geometrical isomers of [Pt(NH₃)(Br)(Cl) (py)] and how many of these will exhibit optical isomerism?
Answer:
The complex [Pt(NH3)(Br)(Cl)(py)] will form three geometrical isomers:

Square planar complexes of this type will not show geometrical isomerism.

Question 6.
Aqueous copper sulphate solution (blue in colour) gives:

1. a green precipitate with aqueous potassium fluoride and
2. a bright green solution with aqueous potassium chloride. Explain these experimental results.

Answer:
Aqueous copper sulphate contains a coordination entity, [Cu(H₂O)₄]²⁺ which is blue in colour. Water molecule is a weaker ligand than Cl^– and F^–.
1. On addition of aqueous KF solution, a new complex entity is formed wich is green in colour.

2. On addition of aqueous solution of KCl, an another bright green complex entity is formed which is soluble in water.

Question 7.
What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H₂S(g) is passed through this solution?
Answer:
Aqueous solution of copper sulphate contains Cu²⁺ ions in the form of complex entity, [Cu(H₂O)₄]²⁺ and H₂O ligand is a weak ligand. When excess of KCN is added, a new coordination entity, [Cu(CN)₄]^2- is formed due to the following reaction:

Cyanide ligand, CN^– is a strong field lignad and stability constant of [Cu(CN)₄]^2- is quite large and thus practically no Cu²⁺ ions are left in solution. On passing H₂S gas, no CuS is formed due to non- availability of Cu²⁺ ions in solution.

Question 8.
Optical isomerism is usually exhibited by complexes containing polydentate ligand. What do you mean by ligand?
Answer:
Ligand is a neutral molecule or charged ion which can donate a lone pair of electron to the metal.

Question 9.
Coordination complexes are of different types. Name the compounds.

1. [Cr(H₂O)₅Cl₂]
2. K₃[Cr(C₂O₄)₃]

Answer:

1. Pentaaquadichloridochromium(II)
2. Potassiumtrioxalatochromate(III)

Question 10.
Write the IUPAC names of the following compounds.

1. K_3M[Fe(CN)₆]
2. [C0(NH₃)₅(CO₃)]Cl

Answer:

1. Potassiumhexacyanoferrate(III)
2. Pentaamminecarbanatocobalt(III) chloride

Question 11.
[Fe(CN)₆]^3- is paramagnetic, while [Fe(CN)₆]^4- is diamagnetic. Explain with the help of VB theory.
Answer:
In [Fe(CN)₆]^3- iron is in +3 state and in [Fe(CN)₆]⁴, iron is in +2 state. [Fe(CN)₆]^3- contains five electrons in d-level (3d⁵). In this complex iron undergoes d²sp³ hybridisation.

Due to the presence of one unpaired electron, [Fe(CN)₆]^3-, is paramagnetic. In [Fe(CN)₆]^4- iron contains six electrons in d-level (3d⁶). It undergoes d²sp³ hybridisation and has no unpaired electrons. Hence, [Fe(CN)₆]^4- is diamagnetic.

Plus Two Chemistry Coordination Compounds Three Mark Questions and Answers

Question 1.
Look at the following two diagrams.

1. Is the diagrams I and II correct? Justify. If the figure is not correct, redraw it.
2. Which theory is related to this?
3. Explain briefly, how this theory is applicable to octahedral complexes.

Answer:
1. No, Figure (II) is wrong.

2. Crystal field theory.

3. In the case of octahedral complexes, the ligands are approaching the ‘d’ orbitals through the axis. As a result of this the energy of d_x²-y² and d_z² orbitals increases and the energy of the remaining three orbitals decreases. The orbitals which possess high energy are represented as ‘e_g’ levels and the orbitals which possess less energy are represented as “t_2g” levels.

Question 2.
A list of coordination compounds are given below:
[Cr(H₂O)₆] Cl₃, [Co(NH₃)₅Br] SO₄, [Co(NH₃)₅NO₂]²⁺ and [Pt(NH₃)₂Cl₂]. Which type of isomerism do these compounds exhibit?
Answer:
Hydrate Isomerism, Ionisation Isomerism, Linkage Isomerism, Geometrical Isomerism.

Question 3.
The following are examples of coordination compounds. Identify the type of isomerism exhibited by each of them and write their possible isomers,

1. [Cr(NH₃)₅Br]SO₄
2. CrCl₃.6H₂O
3. [PtCl₂(NH₃)₂]

Answer:

1. Cr(NH₃)₅Br]SO₄ – Ionisation isomerism – [Cr(NH₃)₅SO₄]Br
2. CrCl₃.6H₂O – Hydrate isomerism – [Cr(H₂O)₅Cl]Cl₂.H₂₎
3. PtCl₂(NH₃)₂] – Geometrical isomerism

Question 4.
What will be the correct order for the wavelengths of absorption in the visible region for the following:
[Ni(NO₂)₆]^4-, [Ni(NH₃)₆]²⁺, [Ni(H₂O)₆]²⁺
Answer:
The order of the ligand in the spectrochemical series is H₂O < NH₃ < NO₂^–. Hence the wavelength of the light observed will be in the order:
[Ni(H₂O)₆]²⁺ < [Ni(NH₃)₆]²⁺ < [Ni(NO₂)₆]^4-

Thus, the wavelength of light absorbed (E = \(\frac{\mathrm{hc}}{\lambda}\)) will be in the opposite order:
[Ni(H₂O)₆]²⁺ > [Ni(NH₃)₆]²⁺ > [Ni(NO₂)₆]^4-

Question 5.
In a classroom discussion, Sajan argued that CN^–, OH^–, Cl^– etc. are examples for neutral ligands.

1. Do you agree with his argument?
2. If not, give a reason with the help of examples.
3. What do you mean by chelating ligand and chelation?

Answer:

1. No.
2. They are charged ligands.
3. If a polydentate ligand is coordinated to the metals, a ring structure is obtained. It is called chelate and the phenomenon is called chelation.

Question 6.
Ligands can be arranged according to the magnitude of Δ₀ and the arrangement is given below:
l- < Br^– < Cl^– < F^– < OH^– < H₂O < NH₃ < (en)

1. What is this series known as?
2. Is the sequence incorrect order?
3. Identify the weak field and strong field ligands.

Answer:

1. Spectrochemical series
2. Yes
3. The ligands above water are strong ligand and the ligands below water are weak ligands.

Weak filed ligands – l^–, Br^–, Cl^–, F^–, OH^–, H₂0 Strong filed ligands – NH₃, en

Question 7.
Consider the following coordination compounds.

• [Co(NH₃)₅ Cl] SO₄
• [Co(NH₃)₅ SO₄]Cl

1. Write down the IUPAC name of these compounds.
2. Name the isomerism exhibited by these compounds.

Answer:
1. The IUPAC name of compounds

• Pentaamminechloridocobalt(III) sulphate
• Pentaamminesulphatocobalt(III) chloride

2. Ionisation isomerism

Question 8.
Consider the following compounds:
[Co (NH₃)₅ NO₂] Cl₂ and [Co (NH₃)₅ ONO] Cl₂

1. Identify the isomerism exhibited by these compounds.
2. Explain ionisation isomerism with example.

Answer:

1. Linkage isomerism
2. This form of isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion.
   e.g. [Co(NH₃)₅Cl] SO₄ & [Co(NH₃)₅SO₄] Cl

These complexes ionises as:

• [Co(NH₃)₅Cl] SO₄ → [Co(NH₃)₅Cl]²⁺ + SO₄^2-
• [Co(NH₃)₅SO₄] Cl → [Co(NH₃)₅SO₄]+ + Cl^–

Question 9.
Consider the statement: Crystal Field Theory (CFT) is applicable to octahedral and tetrahedral complexes.

1. Is this statement true?
2. Explain the crystal field splitting in octahedral complexes with the help of a neat diagram.

Answer:

1. Yes
2. In an octahedral crystal field the ligands are approaching the metal along the axes. Hence, the energy of d_x²-y² and d_z² orbitals (e_g set) increases by 3/5Δ₀and that of d_xy, d_yz and d_xz orbitals (t_2g set) decreases by 2/5 Δ₀

Question 10.
When a ligand approaches to an octahedral complex, the degenerate ‘d’ orbitals undergoes splitting.

1. What will be the observation?
2. What are the factors influencing crystal field splitting energy?

Answer:
1. The energy of d_x²-y² and d_z² orbitals (e_g set) increases by 3/5Δ₀ and that of d_xy, d_yz and d_xz orbitals (t_2g set) decreases by 2/5 Δ₀

2. The factors influencing crystal field splitting energy.

• Nature of the ligand
• Geometry of the complex
• Valency of the metal

Question 11.
What will happen when a ligand approaches to a tetrahedral complex?
Answer:
the energy of d_xy, d_yz and d_xz orbitals (t_2g set) increases by 2/5Δ_t and that of d_x²-y² and d_z² orbitals (e_g set) decreases by 3/5 Δ_t.

Question 12.
Consider the complex ion [Ti (H₂O)₆]³⁺ In the case of an octahedral complex, what is the condition for the pairing of forth electron in the d- level?
Answer:
If the crystal field splitting energy is greater than the pairing energy, the fourth electron will pair at the t_2g level and if the pairing energy is greater than the crystal field splitting energy the electron will go to the on e_g level.

Question 13.
Is bidentate ligands same as the amidentate ligands? Justify.
Answer:
A bidentate ligand like (en), can form two coordinate bonds with the metal at the same time.
An amidentate ligand like -NO₂ can form only one coordinate bond with the metal at a time. But it can ligate through two different atoms.

Plus Two Chemistry Coordination Compounds Four Mark Questions and Answers

Question 1.
Match the following table:

Answer:

Question 2.
1. Write the IUPAC names of the following coordination compounds.

• [Pt(NH₃)Cl(NO₂]₂)
• K₃[Cr(C₂O₄)₃]

2. Identify the type of isomerism exhibited by the following complexes and distinguish them. [Co(NH₃)₅SO₄] Br and [Co(NH₃)₅Br] SO₄
Answer:
1. The IUPAC names of the coordination compounds:

• Amminechloridonitrito-N platinum(II)
• Potassiumtrisoxalatochromate(III)

2. Ionisation isomerism.
The first compound gives pale yellow ppt. with AgNO₃ solution whereas the second compound gives white ppt. with BaCl₂ solution.

Question 3.
1. Write the IUPAC name of the following compounds:

• [Pt(NH₃)₂Cl₂]
• K₄[Fe(CN)₆]

2. A list of coordination compounds are given below:

• [Cr(H₂O)₆]Cl₃,
• [CO(NH₃)₅ Br]SO₄,
• [CO(NH₃)₅ NO₂]²⁺,
• [CO(NH₃)₆] [Cr(CN)₆]

Which type of isomerism do these compounds exhibit?
Answer:
1. The IUPAC name of the coordination compounds:

• Diamminedichloridoplatinum(II)
• Potassium hexacyanoferrate(II)

2. Type of isomerism

• [Cr(H₂O)₆]Cl₃ – Hydrate isomerism
• [CO(NH₃)₅ Br]SO₄ – Ionisation isomerism
• [CO(NH₃)₅ NO₂]²⁺ – Linkage isomerism
• [CO(NH₃)₆] [Cr(CN)₆] – Coordination isomerism

Question 4.
1.. Write down the IUPAC name of

• K₄[Fe(CN)₆]
• [Pt(NH₃)₂Cl₂]

2. On the basis of VBT, explain why [Fe(H₂O)₆]²⁺ is strongly paramagnetic while [Fe(CN)₆]^3- is weakly paramagnetic.
Answer:
1. The IUPAC names are:

• Potassium hexacyanoferrate(II)
• Diamminedichloridoplatinum(II)

2.

In [Fe(H₂0)₆]^2-, iron undergoes sp³d² hybridisation (inner orbital complex). It has four unpaired electrons and hence it is highly paramagnetic.

In [Fe(CN)₆]³⁺. iron undergoes d²sp³ hybridisation (outer orbital complex). It has only one unpaired electron. Hence, it is less paramagnetic.

Question 5.
The names of some co-ordination compounds are given below:

• EDTA
• Haemoglobin
• cis-platin
• Vitamin B¹²
• D-penicillamine
• Chlorophyll
• Ni(CO)₄

a. Classify the above compounds on the basis of application of coordination compounds?
b. There are given some of the coordination compounds Name them.

1. K₃[Fe(C₂O₄)₃]
2. [Cr(CN)₃]³⁺
3. [CoSO₄(NH₃)₄]NO₃
4. [CO(NO₂)₃(NH₃)₃]

Answer:
a. On the basis of application of coordination compounds:

• In biological system – Haemoglobin, Vitamin B₁₂, Chlorophyll.
• Estimation of hardness of water – EDTA.
• Extraction of metals – Ni(CO)₄
• In medicine – D-penicillamine, cis-platin

b. The coordination compounds are:

1. K₃[Fe(C₂O₄)₃] – Potassiumtrioxalatoferrate(III)
2. [Cr(CN)₃]³⁺ – Trisethylenediaminechromium(III) ion
3. [CoSO₄(NH₃)₄]NO₃ – Tetraamminesulphato- cobalt(III) nitrate
4. [CO(NO₂)₃(NH₃)₃] – Triamminetrinitrito-N-cobalt(III)

Question 6.
1. Name the following compounds.

• [Pt(NH₃)₄ ] [CuCl₄]
• [PtCl₂ (NH₃)₄] Br₂

2. What type of isomerism is shown by the following coordination compounds?

• [Pt(NH₃)₄ ] [CuCl₄]
• [Cr(en)₃]³⁺

Answer:
1.

• Tetraammineplatinum(II) tetrachlorocuprate(II)
• Tetraamminedichloridoplatinum(IV) bromide

2. Type of isomerism:

• Coordination isomerism
• Optical isomerism

Question 7.

1. Write the d-orbital configuration of [Ti(H₂O₆]²⁺
2. Ti⁴⁺ is colourless. Why?
3. Write the possible isomers of [Co(NH₃)₅ Br] SO₄ and name them.

Answer:
1.

2. In Ti⁴⁺ there are no electrons in 3d orbitals. Hence ‘d-d’ transition cannot take place. Hence it is colourless.

3. Possible isomers of [Co(NH₃)₅ Br] SO₄ and names

• [CO(NH₃)₅ Br] SO₄ – Pentamminebromido- cobalt(III) sulphate
• [Co(NH3)5SO4]Br-Pentamminesulphatecobalt(III) bromide

Question 8.
‘A’ and ‘B’ are isomers. They have the same composition. But ‘B’ cannot give the test for sulphate.

1. Write two suitable coordination compounds which give the test for sulphate.
2. What are the two major classes of isomerism exhibited by coordination compounds?
3. Draw the structure of an octahedral complex that show optical isomerism.

Answer:
1. Two suitable coordination compounds which give the test for sulphate

• [CO(NH₃)₅Cl]SO₄
• [Co(NH₃)₅Br] SO₄

2. Structural isomerism, Stereoisomerism
3. [PtCl₂(en)₂]²⁺

Question 9.

1. Write the formula of the complex ion Chloridonitro tetramminecobalt(III)?
2. Identify the ligands, coordination number and coordination sphere.
3. Explain the structure of Tetracarbonylnickel(O) with the help of Valence Bond Theory.

Answer:

1. [CO(NH₃)₄Cl(NO₂^–)
2. Ligands-NH₃, Cl^– , NO₂^– Coordination number -4
3. Tetracarbonylnickel(0) – [Ni(CO)]₄] – Structure

Ni(CO₄) is diamagnetic as it does not contain any unpaired electron.

Question 10.
Consider the complex ion [Ti (H₂O)₆]³⁺

1. What is its outer electronic configuration and its shape?
2. What do you mean by crystal field splitting theory?

Answer:

1. Ti³⁺ – 3d¹ 4s⁰ Octahedral
2. In the case of an isolated gasesous metal atom/ ion all the five d-orbitals have the same energy (degenerate). Due to the presence of ligands are splitted the degeneracy of the d-orbitals is lifted.

Question 11.
Coordination compounds are those compounds which retain their identity even in solution and it is essential for all living matter.

1. Name a coordination compound containing Magnesium, which is essential for plants.
2. When we coordinate EDTAwith any metal, we get a ring structure. What is this process called?
3. Explain.

Answer:

1. Chlorophyll.
2. Chelation.
3. When a poly dentate ligand is coordinated to the metal, a ring structure is obtained, called chelate complex. Such complexes are more stable than similar complexes containing unidentate ligands.

Question 12.
Some ligands are given below. Arrange them in suitable headings.
[H₂O, NH₃, CN^–, CO, Cl^–, OH^– (en)]
1. What do you mean by the term ligand?
2. Write down the nomenclature of the coordination compounds given below.

• K₄[Fe(CN)₆]
• [Ag (NH₃)₂]Cl

Answer:
Neutral ligands – H₂O, NH₃, CO, en
Charged ligands – CN^–, OH^–, Cl^–
1. Ligand is a neutral molecule or charged ion which can donate atleast one lone pair of electron to the metal.
2. nomenclature of the coordination compounds

• Potassium hexacyanoferrate(II)
• Diamminesilver(I) chloride

Question 13.

1. What do you mean by optically active compounds? Give two examples.
2. Draw the ‘d’ and T forms of [Co(en)₃]³⁺.

Answer:
1. Optically active compounds are formed by chiral moneluces i.e., molecules which do not have plane of symmetry. These isomers are non- superimposable mirror images of each other. They are optically active and rotate the plane of polarised light equally but in opposite directions.

The isomer which rotates the plane of polarised light towards left is called leavorotatory (-) while that which rotate plane towards right is called dextrorotatory (+).
e.g. [Co(en)₃]³⁺, [PtCl₂(en)₂]²⁺
Dextro and laevo forms of these compounds are possible.

2.

Question 14.
What is the importance of the following coordination compounds is different fields?

1. EDTA
2. Gold cyanide [AU(CN)₂]
3. cis-platin
4. [Ag(S₂O₃)]^3-

Answer:

1. EDTA → Estimation of hardness of water
2. AU(CN)₂ → Metallurgy
3. cis-platin → Cancer therapy
4. [Ag(S₂O₃)]^3- → Photography

Question 15.
The d-block elemetns forms coordination compounds.

1. Name the coordination compound K₃ [CoF₆].
2. Write the electronic configuration of the central metal atom of the above complex by using CFT
3. Draw the figure to show the splitting of degenerate, ‘d’ orbitals in an octahedral field.

Answer:
1. K₃[CoF₆] – Potassiumhexafluridocobaltate(III)
2. The electronic configuration of Co (Z = 27) is [Ar]3d⁷4s² In K₃ [CoF₆], Co is in +3 state. The configuration of Co³⁺ is 3d⁶ 4s°.
3.

Question 16.

1. Name the compound K₃[Cr(C₂O₄)₃]
2. Explain on the basis of VB theory that [Ni(CN)₄]^2- ion with square planar structure is diamagnetic and [Ni(Cl)₄]^2- ion with tetrahedral structure is paramagnetic.

Answer:

1. K₃[Cr(C₂O₄)₃] – Potassiumtrioxalatochromate(III)
2. Ni = 1s²2s²2p63s²3p64s²3d²

It is diamagnetic due to absence of unpaired electrons.

∴ Due to presence of unpaired electrons it is paramagnetic.

Plus Two Chemistry Coordination Compounds NCERT Questions and Answers

Question 1.
FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1:1 molar ratio gives the test of Fe²⁺ ion but CuSO₄ solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu²⁺ ion. Explain why?
Answer:
FeSO₄ does not form any complex with (NH₄)₂SO₄. Instead, it forms a double salt FeSO₄.(NH₄)₂SO₄.6H₂O which dissociates completely into ions. CuSO₄ when mixed with NH₃ forms a complex [CU(NH₃)₄]SO₄ in which the complex ion [CU(NH₃)₄]²⁺ does not dissociate to give Cu²⁺ ion.

Question 2.
Write the geometrical isomers of [Pt(NH₃)(Br)(Cl) (py)] and how many of these will exhibit optical isomerism?
Answer:
The complex [Pt(NH₃)(Br)(Cl)(py)] will form three geometrical isomers:

Square planar complexes of this type will not show geometrical isomerism.

Question 3.
Aqueous copper sulphate solution (blue in colour) gives:

1. a green precipitate with aqueous potassium fluoride and
2. a bright green solution with aqueous potassium chloride. Explain these experimental results.

Answer:
Aqueous copper sulphate contains coordination entity, [Cu(H₂O)₄]²⁺ which is blue in colour. Water molecule is a weaker ligand than Cl^– and F^–.
1. On addition of aqueous KF solution, a new complex entity is formed wich is green in colour.

2. On addition of aqueous solution of KCl, another bright green complex entity is formed which is soluble in water.

Question 4.
What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H₂S(g) is passed through this solution?
Answer:
Aqueous solution of copper sulphate contains Cu²⁺ ions in the form of complex entity, [Cu(H₂O)₄]²⁺ and H₂O ligand is a weak ligand. When excess of KCN is added, a new coordination entity, [Cu(CN)₄]^2- is formed due to the following reaction:

Cyanide ligand, CN^– is a strong field lignad and stability constant of [Cu(CN)₄]^2- is quite large and thus practically no Cu²⁺ ions are left in solution. On passing H₂S gas, no CuS is formed due to non-availabiliy of Cu²⁺ ions in solution.

Question 5.
What will be the correct order for the wavelengths of absorption in the visible region for the following:
[Ni(NO₂)₆]^4-, [Ni(NH₃)₆]²⁺, [Ni(H₂O)₆]²⁺
Answer:
The order of the ligand in the spectrochemical series is H₂O < NH₃ < NO₂^–. Hence the wavelength of the light observed will be in the order:
[Ni(H₂O)₆]²⁺ < [Ni(NH₃)₆]²⁺ < [Ni(NO₂)₆]^4-

Thus, the wavelength of light absorbed (E = \(\frac{\mathrm{hc}}{\lambda}\)) will be in the opposite order:
[Ni(H₂O)₆]²⁺ > [Ni(NH₃)₆]²⁺ > [Ni(NO₂)₆]^4-

Students can Download Chapter 9 Ray Optics and Optical Instruments 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 9 Ray Optics and Optical Instruments

Plus Two Physics Ray Optics and Optical Instruments NCERT Text Book Questions and Answers

Question 1.
A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is

1. a convex lens of focal length 20 cm, and
2. a concave lens of focal length 16 cm?

Answer:
Here the object is virtual and the image is real.
u = 12cm object on right and virtual.
1. f = +20 cm

i.e., v = 7.5 cm. (image on right and real). It is located 7.5 cm from the lens.

2. f = -16 cm

i.e., v = 48 cm. (image on right and real). Image will be located 48 cm from the lens.

Question 2.
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length of the lens is to be 20 cm?
Answer:
\(\frac{\mu_{2}}{\mu_{1}}\) = µ = 1.55
R₁ = R₂ = R
f = 20 cm

R = 0.55 × 2 × 20 = 22 cm.

Question 3.
A small telescope has an objective lens of focal length 144 cm and eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
Answer:
1. For normal adjustment.
M.P. of telescope = \(\frac{f_{0}}{f_{e}}=\frac{144}{6}\) = 24

2. The length of the telescope in normal adjustment
L = f_o + f_e
= 144 + 6 = 150 cm.

Plus Two Physics Ray Optics and Optical Instruments One Mark Questions and Answers

Question 1.
Fora total internal reflection, which of the following is correct?
(a) Light travel from rarer to denser medium.
(b) Light travel from denser to rarer medium.
(c) Light travels in air only.
(d) Light travels in water only.
Answer:
(b) Light travel from denser to rarer medium.
Explanation: In total internal reflection, light travel from denser to rarer medium.

Question 2.
Focal length of a convex lens of refraction index 1.5 is 2 cm. The focal length of lens, when immersed in a liquid of refractive index of 1.25, will be.
(a) 10 cm
(b) 2.5 cm
(c) 5 c
(d) 7.5 cm
Answer:
(c) 5 c

Question 3.
If the refractive index of a material of equilateral prism is \(\sqrt{3}\), then angle of minimum deviation of the prism is
(a) 60°
(b) 45°
(c) 30°
(d) 75°
Answer:
(a) 60°
Explanation: A = 60°, n = \(\sqrt{3}\), D = ?

D = 60°.

Question 4.
Which of the following is correct for the beam which enters the medium?
(a) Travel as a cylindrical beam
(b) Diverge
(c) Converge
(d) Diverge near the axis and converge near the periphery
Answer:
(c) Converge.
Explanation: Since the refractive index is less at the beam boundary, the ray at the edges of the beam move faster compared to the axis of beam. Hence, the beam converges.

Question 5.
A beam of monochromatic light is refracted from vacuum into a medium of refraction index 1.5. the wavelength of refracted light will be.
(a) Depend on intensity of refracted light
(b) Same
(c) smaller
(d) larger
Answer:
(c) smaller
Explanation: velocity of light decreases in a medium. Hence λ decrease in a medium (v ∝ λ).

Question 6.
A convex lens and a concave lens, each having same focal length of 25 cm, are put in contact to form a combination of lenses. What is the power in diopters of the combination is
Answer:
Focal length of convex lens f₁ = 25 cm
Focal length of concave lens f₂ = -25 cm
Power of combination in diopters,

Question 7.
Fill in the blanks

Answer:
(i) n = \(\frac{c}{v}=\frac{3 \times 10^{8}}{2 \times 10^{8}}\) = 1.5

(ii) Optical fibre

Plus Two Physics Ray Optics and Optical Instruments Two Mark Questions and Answers

Question 1.
Match the following.

Answer:

Plus Two Physics Ray Optics and Optical Instruments Three Mark Questions and Answers

Question 1.
A hemispherical transparent paperweight of radius 5m and refractive index 1.5 is placed on a table. A beam of lazar, at a distance of 2m from the centre is directed as shown in the figure.
1. Name the law which is related to refraction.
2. Locate the position of the image by completing the ray diagram.

3. With the source of laser at the centre of the hemisphere, redraw the ray diagram.
Answer:
1. Snell’s law
2.

3. The refracted ray is undeviated.

Question 2.
Figure (a) below snows the image observed at the near point of eye by a boy through a simple microscope. Eye focused on near point.

1. Draw ray diagram which shows the image formation at infinity, so that the boy can observe it with a relaxed eye.
2. Distinguish between linear magnification and angular magnification.

Answer:
1.

2. Linear magnification is ratio of image height to object height. Angular magnifications is the ratio of angle subtended by the image and the object on the eye when both are at the least distance of distinct vision.

Question 3.
Figure shows the path of the light rays through a glass slab.

1. Name the phenomena involved here.
2. Relate the values of n_1, n₂, i and r on the basis of one figure.
3. Copy the figure of glass and draw the path of ray when n₂ < n₁.

Answer:
1. Refraction

2. \(\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}}\)

3.

Question 4.
A light ray travelling from one medium to another medium is given in the figure.

1. Write a mathematical relation for this refraction.

• n₂ < n₁
• n₂ > n₁
• n₂ = n₁

2. What is a relation between angle of incidence, angle of refraction and refractive index of medium.

3. A flint glass rod when immersed in carbon disulfide is nearly invisible why?

Answer:
1. n₂ < n₁

2. \(\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}}\)

3. Refractive index of flint glass rod and carbon disulfide are nearly equal. Hence no refraction (or reflection) take place.

Question 5.
A convex lens and concave lens are placed as shown in figure. For convex lens f = 10cm for concave it is 5 cm

1. Is it converging or diverging why?
2. If f₁ = 5cm and f₂ =10cm What change will occur in the optical nature of system?

Answer:
1.

f = -10 cm
Effective focal length is negative. Hence this lens is diverging.

2. Effective focal length becomes positive- Hence the lens will act as converging.

Plus Two Physics Ray Optics and Optical Instruments Four Mark Questions and Answers

Question 1.
The maximum possible magnification for a simple microscope is 10

1. How do you increase the magnification further(1)
2. Draw the ray diagram for compound microscope and find an expression for magnification (3)
3. What is the advantage of forming image at infinity? (1)

Answer:
1. Use two convex lens instead of single lens.

2.

The magnification produced by the compound microscope

Multiplying and dividing by I₁M₁ we get,

Where m₀ & m_e are the magnifying power of objective lens and eyepiece lens.
∴ m = m_e × m₀ ______(1)
Eyepiece acts as a simple microscope.
Therefore \(\mathrm{m}_{\mathrm{e}}=1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\) ______(2)
We know magnification of objective lens
m₀ = \(\frac{V_{0}}{u_{0}}\) ______(3)
Where v₀ and u₀ are the distance of the image and object from the objective lens.
Substituting (2) and (3) in (1), we get

for compound microscope, u₀ » f₀ (because the object of is placed very close to the principal focus of the objective) and v₀ ≈ L, length of microscope (because the first image is formed very close to the eye piece).

where L is the length of microscope, f₀ is the focal length of objective lens.

3. Strain for eye, will be minimum when image is at infinity.

Question 2.
The refraction of light travelling from glass to water is shown in the figure.
1. The snells law in the above case can be written as………..

2. Show that c = \(\sin ^{-1}\left(_{g} n_{w}\right)\). Where C is the critical angle of glass water interface. (2)

3. Three light rays, (Red, blue and yellow) incident at one side and its refractions are shown in the figure. Copy the figure and mark Red, blue and yellow in the figure. (1)

Answer:
1.

2. In this i = c using snell law, we can write

3.

Question 3.
When a point object is placed in front of a spherical refracting surface an image is formed in the refracting medium.
1.

Complete the ray diagram to locate the position of the image.

2. Obtain the expression \(\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{n_{2}-n_{1}}{R}\) for the position of image inside refracting medium.

3. If the refracting surface is concave in nature, with the same set up, locate the position of the image by drawing a ray diagram.
Answer:
1.

2. Refraction at a spherical surface:

Consider a convex surface XY, which separates two media having refractive indices n₁ and n₂. Let C be the centre of curvature and P be the pole. Let an object is placed at ‘O’, at a distance ‘u’ from the pole.

I is the real image of the object at a distance V from the surface. OA is the incident ray at angle ‘i’ and Al is the refracted ray at an angle ‘r’. OP is the ray incident normally. So it passes without any deviation. From snell’s law,

\(\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}}\)
If ‘i’ and ‘r’ are small, then sin i » i and sin r » r.

n₁i = n₂r ……….(1)
From the Δ OAC, exterior angle = sum of the interior opposite angles
i-e., i = α + θ ………(2)
Similarly, from Δ IAC,
a = r + β
r = α – β ……..(3)
Substituting the values of eq(2) and eq(3)in eqn.(1) we get,
n₁(α + θ) = n₂(α – β)
n₁α + n₁β = n₂α – n₂β
n₁θ + n₂β = n₂α – n₁α
n₁θ + n₂β = (n₂ – n₁)α ………….(4)
From OAP, we can write,

From IAP, β = \(\frac{\mathrm{AP}}{\mathrm{PI}}\), From CAP, α = \(\frac{\mathrm{AP}}{\mathrm{PC}}\)
Substituting θ, β and α in equation (4) we get,

According to New Cartesian sign convection, we can write,
OP=-u, PI = +v and PC = R
Substituting these values, we get


Case -1:
If the first medium is air, n₁ = 1, and n₂ = n,

3.

Question 4.
A compound microscope consists of an objective lens of focal length 2cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image at

1. Least distance of distinct vision.
2. infinity

Answer:
1. v_e = -25cm

∴ u_o = 5cm
Length of the tube, L= |v_o| + |u_e|
∴ v_o = 15-5 = 10

u_e = -2.5cm

2. ∴ v_o = 15 – f_e = 15 – 6.25 = 8.75

u_o = -2.59cm

Question 5.
You may be observed that, the fish inside the aquarium appears to be raised.
1. What is the reason for this phenomenon?
2. Obtain an expression for apparent shift offish.
3. What happens to the height of the object, (That vertically stands in the aquarium) when it is observed by the fish.

• becomes taller
• becomes smaller
• Does not change the height. Justify your answer.

Answer:

1. Refraction
2. Expression for apparent shift is not included in the syllabus
3. Becomes taller. When light enters from rare to denser medium, it deviates towards the normal.

Question 6.
High precision optical instruments uses prisms instead of mirror to reflect light.

1. Name the phenomena used for reflecting light using prism.
2. What is the advantage of using prism instead of mirror for reflecting light?
3. The critical angle of water is 52°. Calculate the refractive index of water.

Answer:
1. Total internal reflection

2. Prism can be used for total internal reflection. Mirrors can’t be used for total internal reflection.

3.

n = 1.26.

Question 7.
Two lenses of focal lengths f₁ and f₂ are placed in contact

1. If the object is at principal axis, draw ray diagram of the image formation by this lens.
2. Obtain a general expression for effective focal length in terms of f₁ and f₂.
3. How will you combine a convex lens of focal length f₁ and concave lens f₂ such that combination acts as

• Converging
• diverging
• plane glass plate

Answer:
1.

2. Obtain an expression for the effective focal length of the combination of two thin convex lenses in contact.

3.

• Keep in a medium of refractive index lower than that of lens.
• Keep in a medium of refractive index higher than that of lens.
• Keep in a medium of refractive index equal to refractive index of lens.

Question 8.
The light rays travelling from rarer to denser medium is given in the figure
1. Redraw the diagram and correct it

2. State the law relating i and r for retracted ray.
3. Velocity of light in water is 2.25 × 10⁸ m/s, If angle of incidence is 30° calculate angle of refraction.
Answer:
1.

2. Snells law:
The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media and for the given colour of light used. This constant is known as the refractive index of second medium w.r. t. the first medium.

Explanation: If ‘i’ is the angle of incidence in the first medium and ‘r’ is the angle of refraction in the second medium, then by Snell’s law,

Where ₁n₂ is the refractive index of the second medium with respect to the first medium. If the first medium is air, then sin i/sin r is known as absolute refractive index of the second medium.

where ‘n’ is the refractive index of the second medium.

3.

Question 9.

1. An air bubble inside an ice block shine brilliant by……… (Refraction, Reflection, total internal reflection)
2. Explain the above phenomenon.
3. The light ray incident at one face of the prism is shown in figure. Copy this figure complete the path of the ray. (Take critical angle of prism C = 42°)

Answer:
1. Total internal reflection.

2. Whenarayoflightpassesfromadenserto rarer medium, after refraction the ray bends away from the normal. If we increases the angle of incidence beyond the critical angle, the ray is totally reflected back to the denser medium itself. This phenomenon is called total internal reflection.

3.

Question 10.
A convex lens produces an inverted image of size 1.4cm The size of object is 0.7cm

1. What is magnification in the case
2. What is the nature of image
3. If the object is at distance 30 cm from the lens calculate focal length of the lens

Answer:
1.

2. Real, inverted, magnified

3.

Question 11.
A ray of light parallel to the principal axis of a spherical mirror falls at a point M as shown in the figure.

1. Identify the type of the mirror used in the diagram. (1)
2. If focal length of the mirror is 10cm, what is the distance CF in the figure? (1)
3. Complete the ray diagram and mark the angle of incidence and angle of reflection. (2)
4. If the mirror is immersed in water its focal length will be

• less than 10cm
• 10cm
• greater than 10cm
• 20cm

Answer:
1. Concave mirror.

2. 10 cm (CF = PC – PF = R = f – 2f – f = f = 10cm).

3.

4. 10cm. Focal length of mirror is independent of medium.

Question 12.
The given figure shows a compound microscope with two lenses PQ and RS.

1. Identify Objective and eyepiece in the microscope.
2. A compound microscope has a magnification of 30. The focal length of its eyepiece is 5cm. Assuming the final image to be formed at the least distance of distinctive vision, calculate the magnification produced by the objective.
3. What is the length of a compound microscope in normal adjustment?

Answer:
1. Objective – PQ, eyepiece – RS.
2. Magnification, M = m₀ × m_e

3. The length of a compound microscope in normal adjustment is f₀ + f_e.

Question 13.
A compound microscope consists of an objective lens of focal length 2cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image at

1. Least distance of distinct vision.
2. infinity

Answer:
1. v_e = -25cm

∴ u₀ = 5cm
Length of the tube, L= |v₀| + |u_e|
∴ v₀ = 15 – 5 = 10

u_e = -2.5 cm.

2. ∴ v₀ = 15 – f_e = 15 – 6.25 = 8.75

u₀ = -2.59cm

Plus Two Physics Ray Optics and Optical Instruments Five Mark Questions and Answers

Question 1.
A point object at a distance of 36 cm from the convex lens of focal length 10cm, is moved by 10cm in 2 sec along principle axis towards the lens. Then image will also change its position.

1. Write the law which relates object and image distance from the lens.
2. Find the initial and final position of the image and calculate average speed of image.
3. A man argues that the image will move uniformly at the same speed as that of object. What is your opinion? Justify.

Answer:
1. The lens equation \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

2. u = -36, f = 10
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

If object is moved 10cm towards the lens we can find position
u = -26, f=10
v = 7.2 cm
Speed = \(\frac{7.8-7.2}{2}\) = 3cm/sec.

3. Comparing speed of object and image we can arrive at conclusion that the argument of man is false, speed of image is different from speed of object.

Question 2.
A ray of light parallel to the principal axis of a spherical mirror falls at a point M as shown in the figure.

1. Identify the type of the mirror used in the diagram.
2. If focal length of the mirror is 10cm, what is the distance CF in the figure?
3. Complete the ray diagram and mark the angle of incidence and angle of reflection.
4. If the mirror is immersed in water its focal length will be

• less than 10cm
• 10cm
• greater than 10cm
• 20cm

Answer:
1. Concave mirror.

2. 10 cm (CF = PC – PF = R = f – 2f – f = f = 10cm).

3.

4. 10cm. Focal length of mirror is independent of medium.

Question 3.
The image formed by a thin lens is shown in the figure.

1. What is the nature of the image
2. Find out the power of the image
3. Draw the ray diagram showing above lens forming a magnified erect, virtual image
4. If a convex lens of focal length 20cm is kept in contact with above lens. What is the focal length and power of the combination

Answer:
1. Inverted.

2. P= \(\frac{1}{1}\) =ID.

3.

4. P = P₁ + P₂
in this case f₁ = 1 m, f₂ = 0.2m

Question 4.
A beam of light passing from one transparent medium to another obliquely, undergoes an abrupt change in direction. This phenomenon is known as refraction of light.

1. Name the law which satisfies during this refraction.
2. Draw a figure, which shows refraction through a parallel sided glass slab (Ray passing from air)
3. Using the figure obtained in (b), show that the incident ray and the emergent ray are parallel to each other. Redraw the same figure, if the light is entering from a medium denser than glass. Justify your answer.

Answer:
1. Snell’s law.

2.

3. This derivation is out of syllabus.

4. The light bends away from the normal if light enter from glass to water.

Question 5.
A group of students are given a project for constructing a telescope and they were supplied with two biconvex lens of power 1 diopter and 0.1 dioptre.

1. Of the two lenses, which can be used as objective?
2. Draw the ray diagram for the formation of the image by a telescope.
3. Arrive at an expression for magnification of a telescope.
4. Prepare a notice/ label about the precaution to be taken while using the telescope and limitations of the telescope constructed.

Answer:
1. Biconvex lens of power 0.1 dioptre.

2.

3. Magnification:
The magnifying power of a telescope is the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the objective.
\(\mathrm{m}=\frac{\text { angle subtented by the image at eye (eye piece) }}{\text { angle subtended by the object at the objective }}\)
i.e. m = \(\frac{β}{α}\) …….(1) [from figure]
But from ∆CIM, tanα = \(\frac{IM}{IC}\), α = \(\frac{IM}{IC}\)
(For small values tan α ≈ α)
from ∆C¹IM, tanβ = \(\frac{IM }{\mathrm{IC}^{1}}\) , β = \(\frac{IM }{\mathrm{IC}^{1}}\)
substituting α and β in eq (1) we get

But IC = f_o (the focal length objective lens) and IC^l = f_e (the focal length eyepiece lens.)

In this case the length of the telescope tube is (f₀ + f_e).
Case 1 :
When the image formed by the objective is within the focal length of the eyepiece, Then the final image is formed at the least distant of distinct vision. In this case, magnifying power.

4.

• As magnifying power is negative, the final image in an astronomical telescope is inverted.
• To have large magnifying power, f_o must be as large as possible and f_e must be as small as possible.
• As intermediate image is between the two lens, cross wire (ora measuring device) can be used.
• In normal setting of telescope, the final image is at infinity.

Question 6.
The following graph represent id curve of a optical instrument placed in air.

1. Name the device which give the above i-d curve.
2. Obtain an expression for deviation produced by such a device.
3. What is the relevance of the value ‘D’? Arrive at an expression for refractive index in terms of this value from (b).
4. How is the deviation affected if the above arrangement is immersed in a liquid of refractive index less than that of the above device.

Answer:
1. Prism.

2. Refraction through a prism:

ABC is a section of a prism. AB and AC are the refracting faces, BC is the base of the prism. ∠A is the angle of prism. Aray PQ incidents on the face AB at an angle i₁. QR is the refracted ray inside the prism, which makes two angles r₁ and r₂ (inside the prism). RS is the emergent ray at angle i₂.

The angle between the emergent ray and incident ray is the deviation ‘d’.
In the quadrilateral AQMR,
∠A + ∠R = 180°
[since N₁M are normal]
ie, ∠A + ∠M = 180°____(1)
In the ∆ QMR,
∴ r₁ + r₂ + ∠M = 180° ____(2)
Comparing eq (1) and eq (2)
r₁ + r₂ = ∠A ____(3)
From the ∆ QRT,
(i₁ – r₁) + (i₂ – r₂) = d
[since exterior angle equal sum of the opposite interior angles]
(i₁ + i₂) – (r₁ + r₂) = d
but, r₁ + r₂ = A
∴ (i₁ + i₂) – A = d

3. At minimum deviation D = 2i – A, r₁ = r₂ = r

4. Deviation decreases.

Question 7.
\(\frac{1}{f}=\left(\frac{n_{2}}{n_{i}}-t\right)\left(\frac{1}{R_{i}}-\frac{1}{R_{2}}\right)\) is lens maker’s formula.
1. Write down lens maker’s formula for a convex lens.
2. “If a convex lens is immersed in water its converging power decrease. Do you agree with it? Justify your answer.
3. A convex lens of refractive index n₂ is placed in different media. Explain optic behavior in each. If n₁ is refractive index of surrounding media.

• in medium with n₂>n₁
• in a medium with n₂ < n₁
• in a medium n₁ = n₂

Answer:
1. For convex lens R₁ = +ve and R₂ = -ve

2. Yes.

From the above equation it is clear that, \(\mathrm{P} \alpha \frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}\)

In water, \(\frac{n_{2}}{n_{1}}\) is less. Hence power decreases.

3. P = +ve, converging
P = -ve, diverging
P = 0, Plane glass

Question 8.
Two convex lense are given in the figure A and figure B

1. Which has more curvature
2. Which has more power
3. Which lens produce more magnification
4. which lens has less focal length
5. Can these lenses act as diverging lenses in any condition?

Answer:

1. A
2. A
3. A
4. A
5. Yes, If we place this lens in a medium of higher refractive index than lens.

Question 9.
In the figure given below, PQ represents an incident ray falling in the side AB of a prism, when monochromatic light is used

1. Draw the refracted ray, emergant ray and mark the angle of deviation
2. Derive an equation for refractive index of the material of the prism in terms of angle of minimum deviation
3. Draw the incident ray and refracted ray, at the angle of minimum deviation

Answer:
1.

2. Refraction through a prism:

ABC is a section of a prism. AB and AC are the refracting faces, BC is the base of the prism. ∠A is the angle of prism. Aray PQ incidents on the faceAB at an angle i₁. QR is the refracted ray inside the prism, which makes two angles r₁ and r₂ (inside the prism). RS is the emergent ray at angle i₂.

The angle between the emergent ray and incident ray is the deviation ‘d’.
In the quadrilateral AQMR,
∠A + ∠R = 180°
[since N₁M are normal]
ie, ∠A + ∠M = 180°____(1)
In the ∆ QMR,
∴ r₁ + r₂ + ∠M = 180° ____(2)
Comparing eq (1) and eq (2)
r₁ + r₂ = ∠A ____(3)
From the ∆ QRT,
(i₁ – r₁) + (i₂ – r₂) = d
[since exterior angle equal sum of the opposite interior angles]
(i₁ + i₂) – (r₁ + r₂) = d
but, r₁ + r₂ = A
∴ (i₁ + i₂) – A = d
(i₁ + i₂) = d + A ____(4)
It is found that for a particular angle of incidence, the deviation is found to be minimum value ‘D’.
At the minimum deviation position,
i₁ = i₂ =i, r₁ = r₂ = r and d = D
Hence eq (3) can be written as,
r + r = A
or r = \(\frac{A}{2}\) ____(5)
Similarly eq (4) can be written as,
i + i = A + D
n = \(\frac{A + D}{2}\) ____(6)
Let n be the refractive index of the prism, then we can write,
n = \(\frac{sin i}{sin r}\) ____(7)
Substituting eq (5) and eq (6) in eq (7),
\(n=\frac{\sin \frac{A+D}{2}}{\sin \frac{A}{2}}\)
i – d curve:

It is found that when the angle of incidence increases deviation (d) decreases and reaches a minimum value and then increases. This minimum value of the angle of deviation is called the angle of minimum deviation.

3. Refracted ray is parallel to base

Question 10.
Refraction of a ray of light at a spherical surface separating two media having refractive indices n₁ and n₂ is shown in the figure.

1. Which of the two media is more denser?
2. In the figure, show that\(\frac{\mathrm{n}_{1}}{\mathrm{OA}}+\frac{\mathrm{n}_{2}}{\mathrm{AI}}=\frac{\mathrm{n}_{2}-\mathrm{n}_{1}}{\mathrm{AC}}\).
3. Using the above relation arrive at the thin lens formula.
4. An object is placed on the principal axis of a convex lens at a distance 8 cm from it. Find the magnification of the image if the focal length of the lens is 4 cm.

Answer:
1. Refractive index of medium 2 is greater than medium 1.

2. Refraction at a spherical surface

Consider a convex surface XY, which separates two media having refractive indices n₁ and n₂. Let C be the centre of curvature and P be the pole. Let an object is placed at ‘O’, at a distance ‘u’ from the pole.

I is the real image of the object at a distance V from the surface. OA is the incident ray at angle ‘i’ and Al is the refracted ray at an angle ‘r’. OP is the ray incident normally. So it passes without any deviation. From snell’s law,

\(\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}}\)
If ‘i’ and ‘r’ are small, then sin i » i and sin r » r.

n₁i = n₂r ……….(1)
From the Δ OAC, exterior angle = sum of the interior opposite angles
i-e., i = α + θ ………(2)
Similarly, from Δ IAC,
α = r + β
r = α – β……..(3)
Substituting the values of eq(2) and eq(3)in eqn. (1) we get,
n₁(α + θ) = n₂(α – β)
n₁α + n₁β = n₂α – n₂β
n₁θ + n₂β = n₂α – n₁α
n₁θ + n₂β = (n₂ – n₁)α ………….(4)
From OAP, we can write,

From IAP, β = \(\frac{\mathrm{AP}}{\mathrm{PI}}\), From CAP, a= \(\frac{\mathrm{AP}}{\mathrm{PC}}\)
Substituting θ, β, and α in equation (4) we get,

According to New Cartesian sign convection, we can write,
OP=-u, PI = +v and PC = R
Substituting these values, we get


Case -1:
If the first medium is air, n₁ = 1, and n₂ = n,

3. Consider a thin lens of refractive index n₂ formed by the spherical surfaces ABC and ADC. Let the lens is kept in a medium of refractive index n₁. Let an object ‘O’ is placed in the medium of refractive index n₁ Hence the incident ray OM is in the medium of refractive index n₁ and the refracted ray MN is in the medium of refractive index n₁.

The spherical surface ABC (radius of curvature R₁) forms the image at I₁ Let ‘u’ be the object distance and ‘v₁’ be the image distance.
Then we can write,

This image I₁ will act as the virtual object for the surface ADC and forms the image at v.
Then we can write,

Dividing throughout by n₁, we get

if the lens is kept in air, \(\frac{n_{2}}{n_{1}}\) = n
So the above equation can be written as,

From the definition of the lens, we can take, when
U = ∞, f = v
Substituting these values in the eq (3), we get

This is lens maker’s formula

For convex lens,
f = +ve, R₁ = +ve, R₂ = – ve

For concave lens,
f = -ve, R₁ = -ve, R₂ = +ve

Lens formula From eq(4),

4. u = -8cm, f = +4cm
m = \(\frac{f}{f+u}=\frac{4}{4+-8}\)
m = -1.

Question 11.
Two lenses L₁ and L₂ are placed in contact as shown in figures. The focal length of each lens is 10cm

1. What is power of L₁
2. What is power of L₂
3. What is effective focal length of combination
4. “The power of convex is greater than that of concave and combination can act as a diverging lens”. Is this statement true in any situation? Explain?

Answer:
1.

2.

3.

The combination will act as plane glass.

4. This is true statement. If we place the above combination in a medium of refractive index greater than this condition.

Question 12.
A lens of particular focal length is made from a given glass by adjusting radius of curvature. The formula applied in this case is lens maker’s formula
1. Write down lens maker’s formula
2. Derive lens maker’s formula considering refraction at a spherical surface
3. Explain the following facts based on lens maker’s formula

• power of sun glasses is zero even though they are curved
• if a lens is immersed in water focal length increases

Answer:
1.

2. Refraction by a lens:
Lens Maker’s Formula (for a thin lens): Consider a thin lens of refractive index n₂ formed by the spherical surfaces ABC and ADC. Let the lens is kept in a medium of refractive index n₁. Let an object ‘O’ is placed in the medium of refractive index n₁. Hence the incident ray OM is in the medium of refractive index n₁ and the refracted ray MN is in the medium of refractive index n₂.

The spherical surface ABC (radius of curvature R,) forms the image at l₁. Let ‘u’ be the object distance and ‘v₁’ be the image distance.
Then we can write,

Adding eq (1) and eq (2) we get

Dividing throughout by n₁, we get

if the lens is kept in air, \(\frac{n_{2}}{n_{1}}\) = n
So the above equation can be written as,

From the definition of the lens, we can take, when u = ∞, f = v
Substituting these values in the eq (3), we get

This is lens maker’s formula
\(\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) ____(5)
For convex lens.
f = +ve, R₁ = +ve, R₂ = – ve

For concave lens,
f = -ve, R₁ = -ve, R₂ = +ve

Lens formula
From eq(4),

From eq(5)

From these two equations, we get

Linear magnification :
If h_o is the height of the object and h_i is the height of the image, then linear magnification.

3.
a. R₁ = R, R₂ = +R

power of lens, P = 0

b. We know

The above equation shows when n₁ increases f decreases the refractive index of water is greater than air. Hence when we place a lens in water, focul length decreases.

Students can Download Chapter 5 Market Equilibrium 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 5 Market Equilibrium

Plus Two Economics Market Equilibrium One Mark Questions and Answers

Question 1.
Complete the statement given below. Free entry and of firms imply that the market price will always be equal to ……………
Answer:
Minimum average cost (P = Min. AC)

Question 2.
Choose the correct answer. The imposition of price ceiling below the equilibrium price leads to ……….
Answer:
Excess demand

Question 3.
Market equilibrium of a commodity shows,
(a) excess demand
(b) quantity demanded greater than quantity supplied
(c) quantity demanded equals quantity supplied
(d) excess supply
Answer:
(c) quantity demanded equals quantity supplied

Question 4.
When there is increase in demand, the demand curve.
(a) shifts reight ward
(b) shifts leftward
(c) shifts downward
(d) remains constant
Answer:
(a) shifts reight ward

Question 5.
The government imposing upper limit on the price of a good or service is called:
(a) price floor
(b) price ceiling
(c) equilibrium price
(d) fair price
Answer:
(b) price ceiling

Plus Two Economics Market Equilibrium Two Mark Questions and Answers

Question 1.
At what price – higher or lower than the equilibrium price, there will be excess demand?
Answer:
When the market price is lower than the equilibrium price, there will be excess demand.

Question 2.
Make pairs.
Price floor, below equilibrium price, above equilibrium price, price ceiling
Answer:

• Price floor – above equilibrium price.
• Price ceiling – below equilibrium price.

Question 3.
Point out the consequences of price ceiling.
Answer:

1. Black marketing
2. Malpractices by fair price shops
3. Sale of inferior quality goods.

Question 4.
What do you mean by control price?
Answer:
Fixation of price of a commodity at a lower level than equilibrium price is called control price. Control price is determined to protect the interest of the consumers.

Question 5.
Market equilibrium of apple is given below in the diagram below.

1. Define market equilibrium
2. Find out the market price and equilibrium quantity

Answer:

1. market equilibrium is a situation where quantity demanded is exactly equal to the quantity supplied.
2. The price of apple is ₹40 and the equilibrium quantity is 30kg.

Plus Two Economics Market Equilibrium Three Mark Questions and Answers

Question 1.
Match the following.

Answer:

Question 2.
The demand function and supply function of a product are given as q^D = 60 – P for 0 = P = 60 q^S = 30 + P for P > 10 Calculate equilibrium price.
Answer:
The demand and supply functions are given as
q^D= 60 – P for 0 = P = 60
q^S = 30 + P for P > 10
Equilibrium price is considered as the price at which quantity demanded is exactly equal to quantity supplied. Therefore we get.
60 – P = 30 + P
60-30 = 2 P
30 = 2P
P = 30/2 = 15
Therefore equilibrium price is ₹15.

Question 3.
Complete the following statements.

1. Long-run price under perfect competition will be equal to ……….
2. Minimum price fixed by government for a product is known as ……….
3. Maximum price fixed by government for a product is known as ………

Answer:

1. average cost
2. floor price
3. price ceiling

Question 4.
Demand curve for labour is downward sloping. Explain
Answer:
Demand curve for labour is downward sloping indicating that more and more labour is demanded at lower wages. This is due to the operation of declining marginal productivity of labour. Marginal productivity of labodr declines due to the operation of diminishing returns. That is why the demand curve for labour slopes downward.

Question 5.
The market demand function and market supply functions are given as, Find the equilibrium price and equilibrium quantity.
q^D = 200 – P for 0 = P = 200
q^S = 120 + P for P > 10
Answer:
We find equilibrium price by equating market demand function and market supply functions as shown below.
q^D = q^S
200-P = 120 + P
2P = 80
P = 80/2 = 40
Therefore equilibrium price is ₹40. Equilibrium quantity is obtained by substituting the equilibrium price into either the demand or supply function equations. Applying the value of price ₹40 in demand equation we have.
q^D= 200 – P
q^D =200 – 40 = 160
Therefore equilibrium price is ₹40 and equilibrium quantity is 160.

Question 6.
Prepare a note on market equilibrium.
Answer:
Equilibrium is defined as a situation where the plans of all consumers and firms in the market match and the market dears. In equilibrium, the aggregate quantity that all firms wish to sell equals the quantity that all the consumers in the market wish to buy; in other words, market supply equals market demand. The price at which equilibrium is reached is called equilibrium price and the quantity bought and sold at this price is called equilibrium quantity.

Therefore, q^D (P*) = q^S (P*) where P* denotes the equilibrium price and q^D (P*) and q^S (P*) denote the market demand and market supply of the commodity respectively at price P*

Question 7.
Suppose the demand and supply functions of commodity X are given by, Qd = 500 + 3P and Qs = 700 – P Qd = 500 + 3P Qs = 700 – P Find out the equilibrium price and quantity demanded and supplied.
Answer:
Equilibrium price and quantity can be determined by equating the demand and supply functions
Qd = Qs
500 + 3P = 700 – P
4P = 200
\(P=\frac{200}{4}=50\)
Equilibrium price is ₹50. Applying the price in the demand function, we get
500 + 3 × 50
500 + 150 =650
Therefore, equilibrium price is ₹50 and quantity is 650.
Qd = Qs
500 + 3P = 700 – P
4P = 200
\(P=\frac{200}{4}=50\)
Qd = 500 + 3 × 50
= 500 + 150 = 650

Question 8.
The diagram below illustrates the supply and demand for television sets. The original demand curve is D₂

Using the diagram, state the new demand curve (D₁ or D₃) which will apply after each of the following changes taken place. (The same answer may be used more than once)

1. A successful advertising campaign for television sets occurs
2. Income decreases
3. An increase in the population

Answer:

1. Demand increases (D₂ curve shifts right to D₃)
2. Demand decreases (D₂ curve shifts left to D₁)
3. Demand increases (D₁ curve shifts right to D₃)

Question 9.
Calculate equilibrium price and quantity based on the following information.
q^d= 400 – P (1)
q^s= 240 + 3 (p – 4) (2)
Answer:
At equilibrium,
q^d = q^s
Putting the values,
400 – p = 240 + 3(p – 4)
400 – p = 240 + 3p – 12
400-240 + 12 = 3p + p
172 = 4p
\(p=\frac{172}{4}\)
p =43
Putting p = 43 in the first equation, we get,
q^d =400 – 43 = 357 Therefore, equilibrium price = 43 and
equilibrium quantity is = 357 units

Question 10.
The diagram shows relationship between two commodities A and B.

1. Identify the commodities A and B
2. Explain what happens to the price and quantity demanded of A when the price of A falls.

Answer:

1. A and B are substitutes.
2. When the price of A falls people will demand more A. So the demand for B will fall. That will result in a decrease in the price of B.

Question 11.
The diagram below shows one of the government intervention programmes in the market.

1. Identify the programme and calculate the excess supply.
2. Explain how the government is monitoring the higher price fixed.

Answer:
1. Minimum price/floor price, 40 unit excess supply.

2. The government announces the minimum price above the market price. As a result of this intervention there occurs excess supply in the market. The government has to remove the excess from the market to maintain the price. So the government store the excess supply in the warehouses and redistribute it at the time of shortage.

Question 12.
Under perfect competition, a market for a good is in equilibrium. There is simultaneous “decrease” both in demand and supply, but there is no change in market price. Explain with the help of a diagram how it is possible.
Answer:
The decrease in demand and supply is the same, and hence the price remain the same. Shows this by drawing appropriate diagram

Question 13.
The diagrams below indicate four possible shifts in demand or in supply that could happen in particular markets. Relate each of the events described below to one of them. Also, give reason for the shift.

1. How does the lorry strike in Karnataka and Tamil Nadu affect the market for vegetables in Kerala?
2. People become aware of the fact that Birds Eye Chilly is very much helpful to prevent Cholesterol. What happens to the market for Birds Eye Chilly?
3. How do you think the rising income affect the market for fish?
4. A new technique is discovered for manufacturing computer that greatly lowers their production cost. What happens to the market for computers?

Answer:

1. figure C, supply falls and price rises.
2. figure A, demand increases and price rises.
3. figure B, demand increases and price rises.
4. figure D, supply increases and price falls.

Plus Two Economics Market Equilibrium Five Mark Questions and Answers

Question 1.
Mention the impact of the following.

1. Imposition of price ceiling below equilibrium price
2. Imposition of price floor above the equilibrium price

Answer:

1. Imposition of price ceiling below equilibrium price leads to excess demand.
2. Imposition of price floor above the equilibrium price leads to an excess supply

Question 2.
Complete the following table to show the impact of simultaneous shifts of demand and supply on equilibrium price and quantity.

Answer:

Question 3.
What will happen if the price prevailing in the market is?

1. above the equilibrium price
2. below the equilibrium price.

Answer:
1. If the price prevailing in the market is above equilibrium price, supply will exceed demand. Under such a situation some firms will not be able to sell their desired quantity; so they will lower their price. All other things remaining constant as price falls quantity demanded rises quantity supplied falls, and finally equilibrium price P* will be restored. At P* quantity demanded will be equal to quantity supplied.

2. If the price prevailing in the market is above equilibrium price, demand will exceed supply. Under such a situation some consumers will be ready to pay more prices to get the commodity. This will tend to increase the price. All other things remaining constant as price rises quantity demanded falls, quantity supplied rises, and finally, equilibrium price P* will be restored. At P* quantity demanded will be equal to quantity supplied.

Question 4.
Draw distinction between floor pricing and price ceiling.
Answer:
Floor price means minimum price. Floor price is fixed to protect producers like farmers from price crashes. It ensures a remunerative price to producers. In India, floor prices are fixed for a variety of agricultural commodities like paddy, wheat, coconut, rubber etc.

On the other hand, price ceiling mean maximum price. It is the maximum price fixed by the government. The aim of price ceiling is to protect consumers. Government fixes price ceiling for essential products and medicines to protect the interests of the consumers.

Question 5.

1. Identify the situations depicted in the following figures in panel A and B.
2. Why do such policies are followed and explain the impact of such policies?

Answer:
1. PANEL A – Price ceiling PANEL B –Price floor.

2. Price ceiling is fixed below equilibrium price. Imposition of price ceiling at ‘Pg’ gives rise to excess demand in the market price floor is fixed above equilibrium price. Imposition of floor price at ‘pg’ gives rise to excess supply.

Question 6.
Mention the factors that cause shift in the supply curve.
Answer:
The factors that cause shift in the supply curves are:

1. The change in the number of firms
2. The change in the price of factor inputs
3. Change in production technology
4. Change in the prices of related goods
5. Change in production tax.

Question 7.
How will a change in price of coffee affect the equilibrium price of tea? Explain the effect on equilibrium quantity through a diagram.
Answer:
Coffee and tea are substitutes. If prices of coffee are increased then its demand will decrease and demand tea would increase.lt will shift the demand curve of tea upwards. The equilibrium price and quantity will increase. This is shown in the following diagram.

In the diagram, when the price of coffee increased then the demand for tea increases. This has resulted in equilibrium price and quantity of tea.

Question 8.
How is price determined in labour market?
Answer:
The price of labour is determined by the forces of demand and supply of labour. The households are the suppliers of labour and demand for labour comes from firms. Labour means the hours of work provided by labourers. The wage rate is determined at the intersection of the demand and supply curve of labour.

The firm being a profit maximiser will always employ labour up to the point where the extra cost it incurs for the last labour is equal to the additional benefit he earns from employment that labour. The extra cost of hiring one more labour is the wage rate. For each extra unit of labour, he gets a benefit equal to marginal revenue product of labour.

Thus firm employs labour up to a point where: W = MRPL Where MRPL = MR x MPL As long as MRPL is greater than the wage rate the firm will earn more profit by hiring one more labour and if at any level of labour employment MRPL is less than the wage rate the firm can increase here profit by reducing labour employed.

Question 9.
Prepare a table showing differences between price ceiling and price floor.
Answer:

Question 10.

1. With the help of a diagram show how the wage rate is determind in a free market.
2. Analyse the impact of an increased entrance of foreign migrant labourers into the labour market.

Answer:
1. The diagram below shows how the wage rate of labour in a free market is determined. DL is the demand for labour and SL is the supply of labour, ‘e’ is the point of equilibrium, ‘ow’ is the wage rate and ‘oq’ is the quantity of labours.

2. When the foreign migrant labour enters into the labour market, the supply of labour will shift rightward and the wage rate will come down as shown in the figure. ‘ow’ is the original wage rate and ‘OQ’ is the original quantity of labour. ow1 is the new wage rate and OQ, is the new quantity of labourers.

Question 11.
Suppose we have two equations, one for demand and other for supply.
Demand equation: Qx^d = 100 – 10P_x
Supply equation : Qx^s = 60 + 10P_x

1. Calculate equilibrium price and quantity using the equations.
2. Construct demand and supply schedules by assigning various prices. Obtain equilibrium price and quantity graphically.

Answer:
1. Equilibrium price =2, equilibrium quantity=80
2. Demand & supply schedules

Question 12.
Let us take market of commodity ‘X’, which is in equilibrium. Suppose demand for the commodity increases. Explain the chain of effects of this change till the market again reaches equilibrium. Use diagram.
Answer:
Increase in demand leads to disequilibrium-price in-creases – super profit – new firms enter the industry – or existing firms expand production – increase in output – supply increases – supply curve shifts – the process continue until price returns the to the equilibrium level. Draws the diagram, and explains the process.

Plus Two Economics Market Equilibrium Eight Mark Questions and Answers

Question 1.
Observe the following table.

1. Find equilibrium price.
2. Fill the fourth column
3. Why ₹35 and ₹40 are not equilibrium prices?
4. Product surplus drives prices up and shortage drives them down. Do you agree?
5. Draw a diagram of the above table showing the equilibrium price determination

Answer:
1. The equilibrium price is ₹37. At this price both demand and supply are equal.
2.

3. At ₹35, demand exceeds the supply causing a shortage in the market. At ₹40, supply exceeds demand causing surplus. Therefore, these prices are equilibrium prices.
4. No. I do not agree with this argument.
5.

Question 2.
Discuss the impact of the factors mentioned below on equilibrium price and quantity.

1. shift in demand to right
2. shift in demand to left
3. shift in supply to right
4. shift in supply to left.

Answer:
1. When demand curve shifts to right (increase in demand), there will be increase in equilibrium price and increase in equilibrium quantity. This change is shown in the diagram.

2. When demand curve shift to left (decrease in demand), both equilibrium quantity and equilibrium price falls. This is shown in the diagram.

3. When supply curve shift to right (increase in supply), the equilibrium price deceases and the equilibrium quantity increases. This is given in the following diagram.

4. When supply curve shifts to left (decrease in supply), the equilibrium price increases and the equilibrium quantity decreases. This is given in the following diagram.

Question 3.
Suppose the demand and supply curves of salt are given by. q^D = 1000 – P q^S = 700 + 2P

1. Find the equilibrium price and quantity
2. Suppose that the price of input used to produce salt has increased so that the supply curve is q^S = 400 + 2P How does the equilibrium price and quantity change? Does the change conform to your expectation?
3. Suppose the government has imposed a tax of 3 per unit of salt. How does it affect the equilibrium price and quantity?

Answer:
1. equilibrium price and quantity
q^D = 1000 – P
q^S = 700 + 2P
For equilibrium
q^D = q^S
1000 – P = 700 + 2P
1000 – 700 = 3 P
3P = 300
P = 300/3 = 100
Put the value of P in supply equation
q^S = 700 + 2P
q^S = 700 + 2×100
q^S =700 + 200 = 900
Therefore the equilibrium price = ? 100 and the equilibrium quantity is = 900 units

2. For equilibrium
q^D = q^S
1000 – P = 400 + 2P
1000 – 400 = 3P
600 = 3 P
P = 600 / 3 = 200
Put the value of P in demand equation
Q^D= 1000 – P
Q^D = 1000  – 200 = 800
Therefore the equilibrium price = ₹200 and the equilibrium quantity is = 800 units This change confirms to our expectations, i.e., rise in input prices raises prices and lowers supply.

3. q^D= 1000 – P
q^S= 700 + 2P
When ₹3 as tax is imposed on sale of salt the new demand and supply function will change
q^D= 1000 – (P + 3)
q^S= 700 + (2P +3)
In part A equilibrium price was ₹100 which goes up to ₹103 with imposition of tax
q^D = 1000 – (100 + 3) = 1000 – 103 =897
q^S = 700 + (2P + 3)
= 700 + 2(100 + 3)
= 700 + 2×103
= 700 + 206 = 906
q^D < q^S
Therefore, new price and quantity has to be adjusted.

Question 4.
The diagram below shows how the price of wheat is determined in a free market.
a. Show in a seperate diagram the changes on price and quantity demanded of wheat due to the following factors

1. The price of fertilizers increases.
2. The price of rice a substitute of wheat increases.

b. Assess the impact of an increased demand for wheat and an increase in its production.

Answer:
a. price and quantity demanded of wheat.
1. When the price of fertilizers increases the supply of wheat decreases. Its price increases and the quantity falls.

2. When the price of rice, a substitute of wheat increases people may switch to consume wheat, this will increase the demand for wheat. Its price will increase and quantity also will increase.

b. When the demand for heat increases its demand curve will shift rightward. When production increases its supply curve will shift rightward as shown in the diagram below.

Due to these shifts, the quantity will increase anyhow. But the effect on the price will be in different forms. The price may fall, will be constant or even may increase. Whether the price will increase, decrease or remain constant is determined by the respective shifts in demand and supply.

If both demand and supply shift in the same magnitude the price will be same. If the shift in the demand is more than the supply the price will increase. And if the shift in the supply is more than the shift in the demand the price will fall.

Students can Download Chapter 10 Haloalkanes and Haloarenes 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 10 Haloalkanes and Haloarenes

Plus Two Chemistry Haloalkanes and Haloarenes One Mark Questions and Answers

Question 1.
Alkolic KOH is a specific reagent for
(a) dehydration
(b) dehalogenation
(c) dehydrohalogenation
(d) dehydrogenation
Answer:
(c) dehydrohalogenation

Question 2.
Chloroform is slowly oxidised by air in the presence of light to an extremely poisonous gas called ………………
Answer:
carbonyl chloride/phosgene

Question 3.
Which of the following represents a gem di halide?
(a) ethylene dichloride
(b) 2,2-dichloropropane
(c) 1,3-dichloropropane
(d) 1,2-dichloropropane
Answer:
(b) 2,2-dichloropropane

Quesiton 4.
Here ‘A’ is
(a) phenol
(b) sodium phenoxide
(c) benzene
(d) cyclohexyl chloride
Answer:
(b) sodium phenoxide

Question 5.
There are _________ structural isomers of C₄H₉Br.
Answer:
Four

Question 6.
Name the insecticide prepared from chloral and chloro benzene.
Answer:
DDT

Question 7.
The reaction between arylhalide and alkylhalide in the presence of sodium and dry ether.
Answer:
Wurtz-Fittig reaction

Question 8.
Name the substance which is used as anaesthetic.
Answer:
Chloroform

Question 9.
Name an alkyl magnesium halide.
Answer:
Methyl magnesium chloride (CH₃MgCI) – Grignard reagent

Question 10.
Compounds in which the halogen atom is directly attached to an aromatic ring carbon.
Answer:
Aryl halides

Question 11.
Even though alkylhalides are polar in nature, they are insoluble in water. Comment on it.
Answer:
This is because alkylhalides can neither make or break hydrogen bonds with water molecules.

Question 12.
From the following select those compounds which are used for the preparation of alkyl halide?

Answer:
SOCI₂, HCI, and ZnCI₂

Question 13.
When (-) 2methyl butan-1-ol is heated with con. HCI +1-chloro-2 methyl butane is obtained. This reaction is an example of
(a) retension
(b) invission
(c) racemisation
(d) resolution
Answer:
(a) retension

Question 14.
If alkaline hydrolysis of a tertiary alkyl halide by aqeous alkali, if concentration of alkali is doubled then the reaction rate at constant temperature will be ___________
Answer:
will be tripled.

Question 15.
The organic compound used as feedstock in the synthesise of chlorofluorocarbon is ___________
Answer:
CCI₄

Question 16.
DDT is prepared from _____________
Answer:
Chlorobenzene and B.H.C

Question 17.
Chlorination of benzene in the presence of halogen is an example of ____________
Answer:
aromatic electrophilic substitution

Plus Two Chemistry Haloalkanes and Haloarenes Two Mark Questions and Answers

Question 1.
Write the preparation of extra pure alkyl halide from ethyl alcohol.
Answer:
When ethyl alcohol is heated with thionyl chloride, chloroethane is obtained. Here the byproducts are in gaseous state and hence this method is used for the preparation of extra pure alkyl halide.
CH₃-CH₂-OH + SOCI₂ → CH₃CH₂CI + HCI + SO₂

Question 2.
A student was treating iodoform with silver nitrate and he got yellow precipitation. Then he used chloroform instead of Iodoform.

1. Will he get the earlier result?
2. Why?

Answer:

1. No
2. C-I bond in iodoform is weaker than C-CI bond of chloroform. So C-I bond of iodoform is easily broken to form yellow precipitate of AgI when heated with AgNO₃ solution.

Question 3.

From the box, write the raw materials of chloroform and Iodoform.
Answer:
Chloroform – ethyl alcohol and bleaching powder Iodoform – Sodium carbonate, ethyl alcohol and iodine

Question 4.
During a class room discussion a student 1 argued: “When chlorobenzene is allowed to react with metallic sodium in the presence of dry ether medium, diphenyl is obtained”. Then student 2 countered: “This reaction will take place in the presence of ale. KOH”.

1. Whom you will support?
2. Name the reaction? Explain it.

Answer:

1. Student 1
2. Fittig reaction

When aryl halide is allowed to heat with sodium in dry ether medium, diaryl is obtained, or

Question 5.
Analyse the following statements:
Statement 1: Alkyl halides are polar compounds.
Statement 2: Alkyl halides are insoluble in water because alkyl halides are non-polar compounds. What is your opinion? Explain it.
Answer:
Even though alkylhalides are polar compounds, they are insoluble in water. Because they can neither form hydrogen bonds with water nor break the hydrogen bonds existing between water molecules.

Question 6.
Which of the following has the highest dipole moment?

1. CH₂CI₂
2. CHCI₃
3. CCI₄

Answer:
Dipole moment of CH₂CI₂ (1.60 D) is the highest. The dipole moment of CCI₄ is zero while that of CHCI₃ is 1.03 D. The dipolement of CHCI₃ is less than that of CH₂CI₂ because the bond dipole of third C-CI bond opposes the resultant of bond dipoles of the other two C-CI bonds.

Question 7.
A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₉CI in bright sunlight. Identify the hydrocarbon.
Answer:
Since the hydrocarbon gives only one monochloro compound, it indicates that all the hydrogen atoms in the hydrocarbon are equivalent. Thus, the compound is cyclopentane.

Question 8.
Which of the following has the highest dipole moment? CH₂CI₂, CHCI₃, CCI₄. Justify.
Answer:
The dipole moment of CH₂CI₂ (1.6 D) is the highest. The dipole moment of CCI₄ is zero while that of CHCI₃ is 1.03 D. The dipole moment of CHCI3 is less than that of CH₂CI₂ because the bond dipole of third C-CI bond opposes the resultant of bond dipoles of the other two C-CI bonds.

Question 9.
Which alkyl halide from the following pair would you expect to react more rapidly by SN² reaction mechanism? Justify your answer. CH₃-CH₂-CH₂-CH₂-Br or CH₃-CH₂-CH(Br)-CH₃
Answer:

Presence of bulky groups around carbon atoms induce an inhibitory effect. Bulky group hinders the approaching nucleophiles in SN² mechanisam.

Question 10.
Explain the stereochemical aspects of SN¹ and SN² reactions selecting suitable example.
Answer:
S_N¹ reaction:
In the case of optically active alkyl halides, S_N¹ reactions are accompanied by racemisation. This is because the attack of the nucleophile may be accomplished from either side resulting in a mixture of products, one having the same configuration and the other having opposite configuration.

S_N² reaction:
In the case of optically active alkyl halides, there is inversion of configuration. This is because the nucleophile attaches itself on the side opposite to the one where the halogen atom is present

Question 11.
Arrange the following compounds in the order of reactivity towards SN² displacement. 2-Bromo-2-methyl butane, 1-Bromopentane 2-Bromopentane
Answer:
1-Bromopentane > 2-Bromo-pentane > 2-Bromo-2- methyl butane

Question 12.
What happens when methyl bromide is treated with sodium in presence of dry ether? Write the chemical equation and name the reaction.
Answer:
Ethane is formed.

called Wurtz reaction.

Question 13.
Chloroform is stored in black coloured bottles. Why?
Answer:
In presence of sunlight chloroform undergoes oxidation to form carbonyl chloride (phosgene).

Plus Two Chemistry Haloalkanes and Haloarenes Three Mark Questions and Answers

Question 1.
Raju heated the test tube containing ale. KOH and primary amine with one compound. A foul smell is obtained.

1. What is the compound?
2. Name the foul smelling product obtained.
3. Name the reaction.

Answer:

1. Chloroform
2. Isocyanide/Carbyl amine
3. Carbyl amine reaction/lsocyanide test

Question 2.
Nitration is one of the ring substitution reactions.

1. How it will be carried out?
2. What are the products obtained when chloro-benzene is heated with the nitrating mixture?
3. What is the difference if the same compound is heated with fuming H₂SO₄?

Answer:
1. By heating with nitrating mixture (mixture of conc.HN0₃ and conc.H₂O₄)

2. A mixture of 1-Chloro-2-nitrobenzene (minor product) and 1-Chloro-4-nitrobenzene (major product) is obtained.

3. If chloro benzene is heated with fuming H₂So₄ sulphonation will take place resulting in the formation of a mixture of 2-Chlorobenzene sulphonic acid (minor product) and 4- Chlorobenzene sulphonic acid (major product).

Question 3.
Consider the reaction: RX + Mg → RMgX

1. Identify the compound ‘RMgX’.
2. Explain the reaction.

Answer:

1. Grignard Reagent
2. When alkyl halides are treated with magnesium in the presence of dry ether, alkyl magnesium halide is obtained.

Question 4.
Chloroform kept in brown coloured bottles filled up to the neck.

1. What is the reason for this?
2. Few drops of 1% ethyl alcohol is added to chloroform to be kept for long. Give reason.

Answer:
1. In the presence of sunlight chloroform undergoes oxidation to form carbony chloride or phosgene, a highly poisonous gas.

This reaction can be avoided by storing it in dark bottles, completely filled upto brim. The use of brown bottles cuts off active light radiations and filling upto brim keeps out air. So chloroform is kept in brown bottles.

2. Addition of a little ethanol fixes the toxic COCI₂ as non-poisonous diethyl carbonate.

Question 5.
Fill in the blanks:

Answer:

1. CH₃CH₂I
2. Ethyl Iodide
3. CH₃-CHBr-CH₃
4. 2-Bromopropane
5. Iso-butyl chloride
6. 1-Chloro-2-methylpropane

Question 6.
In a Chemistry class, teacher explained that when benzene diazonium chloride is allowed to react with cuprous chloride and HCI, Chloro benzene is obtained. Then teacher asked

1. If we use copper powder instead of cuprous chloride. What will be the product?
2. Write the name of the reaction.
3. Write the equation which shows the reaction between benzene diazonium chloride with copper powder and HCI.

Answer:
1. Chlorobenzene is obtained.

2. Gattermann reaction.

3.

Question 7.
In a Lab, one student took a compound in test tube and he added iodine and alkali. He notices a yellow precipitate.

1. Write the name of the test.
2. Which type of compounds give this test?
3. According to the above answer, ethyl alcohol or methyl alcohol, which one gives this test?

Answer:

1. Iodoform test
2. Iodoform test is given by those compounds which are having CH₃CO – group or CH₃CHOH group.
3. Ethyl alcohol gives Iodoform test becuase it has CH₃CHOH group.

Question 8.

For the preparation of chlorobenzene, Nikhil wrote the equation ‘A’ and Nishanth wrote the equation ‘B’.

1. Which of the above equation is correct? Why?
2. Write the name of the reactions ‘A’ and ‘B’
3. Explain any one reaction.

Answer:

1. 1. Both the reactions (A) and (B) are correct.
   2. (A) -Sandmeyer reaction (B) – Gattermann reaction
   3. Sandmeyer reaction – When benzene diazonium chloride is allowed to react with cuprous chloride and HCI, chlorobenzene is obtained, benzene diazonium chloride + cuprous chloride
      

Question 9.
a. In the following pairs of compounds which would undergo SN₂ reaction faster?
(CH₃)₃C-Brand CH₃-CH₂-Br
b. i) CH₃ – CH₂ – CH₂ – OH + SOCI₂ → ……..

Answer:
a. CH₃CH₂Br. SN₂ reaction is faster in the case of primary alkyl halides since the transition state is more stable.
b. i) CH₃CH₂CH₂CI
ii) CH₃-CH₂ -CH₂ – CH₂Br (Anti-Markownikoff’s addition)

Question 10.

1. Nucleophilic substitution of haloalkane takes place through two different mechanisms, SN¹ and SN². Why do inversion of configuration take place in SN²?
2. What is racemic mixture?
3. Comment on the optical activity of recemic mixture.

Answer:

1. is because the nucleophile attaches itself on the side opposite to the one where the halogen atom is present.
2. Equimolar mixture of ‘d’ and T forms of an optically active compound is called racemic mixture.
3. Racemic mixture is optically inactive due to external compensation. A mixture containing two enantiomers in equal proportions will have zero optical rotation, as the rotation due to one isomer will be cancelled by the rotation due to the other isomer.

Question 11.
Write equations for the preparation of 1-iodobutane from
i) 1 – butanol
ii) 1 – chlorobutane
iii) but -1- ene
Answer:
i) CH₃CH₂CH₂CH₂OH+ HI → CH₃CH₂CH₂CH₂I + H₂O

Question 12.
Which compound in each of the following pairs will react faster in S_N2 reaction with – OH?

1. CH₃Br or CH₃I
2. (CH₃)₃CCI or CH₃CI

Answer:

1. CH₃I will react faster because C-I bond undergoes cleavage more easily as compared to C-Br bond.
2. CH₃CI will react faster because in it the carbon carrying halogen is sterically less hindered.

Question 13.
Identify the product X, Y, and Z in the following reaction.

Answer:

• X = CH₃CH₂CI
• Y = CH₃CH₂CN
• Z = CH₃CH₂CONH₂

Question 14.
Identify the compounds X, Y, and Z in the following.

Answer:

• X = CH₃ – CH₂Br
• Y = CH₃CH₂OH
• Z = CHI₃

Plus Two Chemistry Haloalkanes and Haloarenes Four Mark Questions and Answers

Question 1.
CH₃-CH = CH₂ + HCI

1. What are the possible products?
2. Of these which one is the major product?
3. Name the rule which helps you to answer the above question.
4. Explain the rule.

Answer:
1.

2. 2-Chloropropane is major.
3. Markownikoffs rule.
4. When a hydrogen halide is added to an unsymmetrical alkenethe halogen atom of alkyl halide will go to double-bonded carbon atom containing lesser number of hydrogen atom.

Question 2.
Two compounds are given

1. What is the difference between these two compounds?
2. Write the name of reaction they undergo and explain it.

Answer:

1. 1^st compound is cyanide. 2^nd compound is isocyanide
2. Nucleophilic substitution reactions.

The reactions in which a stronger nucleophile substitutes or displaces a weaker nucleophile are called nucleophilic substitution reactions.

Question 3.

1. Name ‘A’.
2. What is the role of ‘A’?
3. Name the product obtained.
4. What is this reaction called?

Answer:

1. Dry ether
2. Dry ether is used to prevent explosion
3. Diphenyl
4. Fittig reaction

Question 4.


Answer:

Question 5.
a) Haloalkanes give β-elimination.
i) Prepare CH₂ = CH₂ from CH₃ – CH₂X and alchoholic KOH.
ii) Give the major product of the β-elimination of

Name the related rule.

b) In the following pairs of halogen compounds which would undergo SN² reaction faster? Justify.

Answer:


It is primary halide and therefore undergoes S_N² reaction faster.

An iodide is a better leaving group because of its large size. It will be released at a faster rate in the presence of attacking nucleophile.

Question 6.

1. Which among the following compounds undergo SN¹ substitution easily – 3° or 1° alkyl halide? Give reason. Justify.
2. Grignard reagents should be prepared under anhydrous conditions. Give reason.

Answer:

1. 3°- alkyl halide. Because the 3° carbocation is more stable than a 1° carbocation.
2. Grignard reagents react with water and get decomposed (hydrolysed). Hence they should be prepared under anhydrous conditions.

Question 7.
a. In the following pair of halogen compounds which undergo SN² reactions faster.

1. C₆H₂ – CH₂ – CI and C₆ H₃ -CI
2. CH₃ – CH₂ – CH₂ – CI and CH₃ – CH₂ – CH₂ – I

b.

Answer:
a. Undergo SN² reactions faster

1. C₆H₅CH₂CI
2. CH₃ -CH₂ -CH₂ -I (R-I bond is weaker than R-CI bond)

b.

i) CH₃ -CH₂ -CH₂ -CH₂ Br
ii) CH₃-CH₂-CN

Question 8.

Complete the reaction.
b) R- CH₂– CI, R₂CHCI, R₃CCI. Arrange these alkylhalide in the order of reactivity towards SN¹ and SN² mechanism.

Answer:

Question 9.
Write the isomers of compound having molecular formula C₄H₉Br.
Answer:
For the molecular formula C₄H₉Br four isomers are possible.
(i) CH₃-CH₂ – CH₂ – CH₃ (1-Bromobutane)

Plus Two Chemistry Haloalkanes and Haloarenes NCERT Questions and Answers

Question 1.
A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₁₀CI in bright sunlight. Identify the hydrocarbon.
Answer:
Since the hydrocarbon gives only one monochloro compound, it indicates that all the hydrogen atoms in the hydrocarbon are equivalent. Thus, the compound is cyclopentane.

Question 2.
Write the isomers of compound having molecular formula C₄H₉Br.
Answer:
For the molecular formula C₄H₉Br four isomers are possible.
(i) CH₃-CH₂ – CH₂ – CH₃ (1-Bromobutane)

Question 3.
Write equations for the preparation of 1-iodobutane from
i) 1 – butanol
ii) 1 – chlorobutane
iii) but – 1 – ene
Answer:
i) CH₃CH₂CH₂CH₂OH+ HI → CH₃CH₂CH₂CH₂I + H₂O

Question 4.
Which compound in each of the following pairs will react faster in S_N2 reaction with OH^–?
Answer:

• CH₃I will react faster because C-I bond undergoes cleavage more easily as compared to C-Br bond.
• CH₃CI will react faster because in it the carbon carrying halogen is sterically less hindered.

Question 5.
Out of C₆H₅CH₂CI and C₆H₅CHCIC₆H₅which is more easily hydrolysed by aqueous KOH?
Answer:
C₆H₅CH₂CIC₆H₅ is more easily hdrolysed because in this case the reaction proceeds through more stable intermediate carbocation.

The intermediate carbocation in this case is stabilised by resonance effect of two phenyl groups whereas in the other case it is stabilised by resonance effect of only one phenyl group.

Students can Download Chapter 8 Electro Magnetic Waves 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 8 Electro Magnetic Waves

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

Question 1.
What physical quantity is the same for X-rays of wavelength 10^-10, red light of wavelength 6800 Å and radio waves of wavelength 500m?
Answer:
The speed in vacuum is the same for all.
c = 3 × 10⁸ms^-1

Question 2.
A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?
Answer:
\(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{B}}\) are in x-y plane and are mutually perpendicular. Given v = 30 MHz = 30 × 10⁶ Hz, c = 3 × 10⁸ms^-1.
∴ \(\frac{c}{v}=\frac{3 \times 10^{8}}{30 \times 10^{6}}\) = 10m.

Question 3.
A radio can tune into any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Answer:
For v₁ = 7.5 MHz = 7.5 × 10⁶ Hz

For v₂ = 12 MHz = 12 × 10⁶ Hz

So the wavelength band: 40m – 25m.

Question 4.
Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1N/C) cos [(1.8 rad/m) y + (5.4 × 10⁶ rad s^-1t] \(\hat{i}\)

1. What is the direction of propagation?
2. What is the wavelength X?
3. What is the frequency v?
4. What is the amplitude of the magnetic field part of the wave?
5. Write an expression for the magnetic field part of the wave.

Answer:

1. \(-\hat{j}\)
2. λ = \(\frac{2 \pi}{k}=\frac{2 \times 3.14}{1.8}\) = 3.48 = 3.5m
3. v = C/λ = 3 × 10⁸/ 3.5 = 85.7 × 10⁶ = 86 MHz.
4. B₀ = \(\frac{E_{0}}{c}=\frac{31}{3 \times 10^{8}}\) = 10 × 10^-8T = 100nT.
5. 100nT) cos (1.8 rad/m) y + (5.4 × 10⁶ rad/s)t \(\hat{k}\)

Plus Two Physics Electro Magnetic Waves One Mark Questions and Answers

Question 1.
The structure of solids is investigated by using
(a) cosmic rays
(b) X-rays
(c) y-rays
(d) infra-red radiations
Answer:
(b) X-rays

Question 2.
Wavelength of light of frequency 100Hz.
(a) 4 × 10⁶m
(b) 3 × 10⁶m
(c) 2 × 1o⁶m
(d) 5 × 10^-5m
Answer:
(b) 3 × 10⁶m
λ = \(\frac{3 \times 10^{8}}{100}\) = 3 × 10⁶m.

Question 3.
What is the biological importance of ozone layer
(a) It stops ultraviolet rays
(b) ozone layer reduces green house effect
(c) ozone layer reflects radio waves
(d) ozone layer controls O₂/H₂ ratio in atmosphere
Answer:
(a) It stops ultraviolet rays
The ozone layer absorbs the harmful ultraviolet rays coming from sun.

Question 4.
Pick the odd one out from the following x-rays, visible light, matter waves, radio waves.
Answer:
Matter waves, (matter-wave is not a em wave).

Question 5.
What physical quantity is the same for X-rays of wavelength 10^-10, red light of wavelength 6800 Å and radio waves of wavelength 500m?
Answer:
The speed in vacuum is the same for all.
c = 3 × 10⁸ms^-1.

Plus Two Physics Electro Magnetic Waves Two Mark Questions and Answers

Question 1.
State whether True or False.

1. Electromagnetic waves propagate in the direction of electric field.
2. For an electromagnetic wave the ratio of E to B is equal to speed of light.
3. In an electromagnetic wave, electric field leads by π/2.
4. Electromagnetic waves can be produced by accelerating electric charge.

Answer:

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

Plus Two Physics Electro Magnetic Waves Three Mark Questions and Answers

Question 1.
Match the following:

Answer:

• Radiowave – accelerated motion of charges in wires – cellular phones.
• Infrared waves – Hot bodies and molecule – Green house effect.
• X-rays – High energy electrons – diagnostic purpose
• Gama-rays – Radio active nuclei – destroy cancer cells.

Plus Two Physics Electro Magnetic Waves Four Mark Questions and Answers

Question 1.
In an electro magnetic wave electric and magnetic field vectors are given by E = 120 sin (ωt + kz) i, B = 40 × 10^-8sin (ωt +kz)j

1. What is the direction of propagation of electro magnetic wave?
2. Determine the ratio of amplitude of electric field to magnetic field in the case of the above em wave?
3. How can you relate the above ratio with m₀ and e₀.

Answer:

1. A direction or perpendicular to the direction of variation of electric field and magnetic field, ie.z direction.
2. \(\frac{E_{0}}{B_{0}}=\frac{120}{40 \times 10^{-8}}\) = 3 × 10⁸m/s.
3. \(\frac{E_{0}}{B_{0}}=\sqrt{\frac{1}{\mu_{0} \varepsilon_{0}}}\) = C.

Question 2.
Match the following.

Answer:

• γ-ray – Radioactivity – Nucleus
• X-ray – Diagnosis- Photon emission by fast moving e’s
• uv-ray- Sunburn – Electronicde excitation
• Microwave – Remote sensing – Oscillating current.

Question 3.
The magnetic field of a plane electromagnetic wave is given by B = 2 × 10^-7 sin (0.5 × 10³x + 1.5 × 10¹¹t)

1. What is the maximum amplitude of this magnetic field?
2. What is wave length and frequency?
3. If this electro magnetic move in z-direction. Write the equation of it’s electric field.

Answer:
1. 2 × 10^-7.

2. Comparing with standard equation
B = B_x Sin (kx + ωt)
we get, kx = 0.5 × 10³x.

m = 12.56 × 10^-3m
ω = 1.5 × 10¹¹, 2πf = 1.5 × 10¹¹
f = 2.3 × 10¹¹hz

3. Amplitude E₀ = B₀ C
= 2 × 10^-7 × 3 × 10⁸ = 60 N/C
If magnetic field vibrate in x direction,
then E_y = 60 sin (0.5 × 10³z + 1.5 × 10¹¹t).

Question 4.
The diagram shows the main regions of the electro magnetic spectrum.

1. Write the names of the blank regions.
2. Which radiation shown in the table has the shortest wavelength?
3. Which ray among these are responsible for green house effect? Explain green house effect.

Answer:
1. Ultra violet, Infrared.

2. Gamma-ray

3. Infrared
Earth is heated by infrared radiations from the sun. At the same time earth radiates longer wavelength to the space. This radiation (outgoing) of longer wavelength is reflected back by CO₂ and other gases.

Hence radiation (from earth) can’t escape from earth’s surface. Thus the temperature at earth’s surface increases. This phenomenon is called Green house effect.

Question 5.

1. Sun is a very powerful source of ultraviolet rays. But we do not receive much ultraviolet rays on the surface of the earth. Why?
2. The electric field of an ultraviolet ray is given by E = 60 sin (0.5 × 10³ – 1.5 × 10¹¹t) What is the amplitude of the magnetic field vector in the ultraviolet ray?

Answer:
1. Ozone prevents the UV ray.

2. \(B=\frac{E}{C}=\frac{60}{3 \times 10^{8}}\)
= 2 × 10^-7wb/m².

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.