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Students can Download Chapter 13 Nuclei 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 13 Nuclei

Plus Two Physics Nuclei NCERT Text Book Questions and Answers

Question 1.
Obtain the binding energy of a nitrogen nucleus (\(_{ 7 }^{ 14 }{ N }\)) Given m \(_{ 7 }^{ 14 }{ N }\) = 14.00307 u.
Answer:
Here Z = 7 and A = 14, A – Z = 14 – 7 = 7
∴ Mass defect
= [Z m_H + (A – Z)m_n – M_n] u
= (7 × 1.00783 + 7 × 1.00867 – 14.00307) u
= (7.05481 + 7.06069-14.00307) u = 0.11243 u
Since I u = 931.5 MeV
∴ B.E. of 14N = 0.11243 × 931 Mev
= 104.7 MeV.

Question 2.
Obtain approximately the ratio of the nuclear radii of the gold isotope \(_{ 79 }^{ 197 }\)Au and the silver isotope \(_{ 47 }^{ 107 }\)Ag
Answer:
Here A₁ = 197 and A₂ = 107

Question 3.
How long an electric lamp of 100 W can be kept glowing by fusion of 2.0 kg of deuterium? The fusion reaction as.

Answer:
Number of deuterium atoms is 2 kg
= \(\frac{6.023 \times 10^{23}}{2}\) × 2000 = 6.023 × 10²⁶
Energy released when 6.023 × 10²³ nuclei of deuterium fuse together

= 15.42 × 10¹³J = 15.42Ws
Power of lamp = 100 W
If the lamp glows fortime t, then electric energy consumed = 100 t
∴ 100 t = 15.42 × 10³³
∴ t = 0.1542 × 10¹³s
\(=\frac{0.1542 \times 10^{13}}{365 \times 86400} y\)
or t = 4.0 × 10⁴y.

Question 4.
From the relation R = R₀ A^{\(\frac{1}{3}\)}, Where R₀ is a constant and A is the mass number of a nucleus show that the nuclear matter density is nearly constant (i.e. Independent of A).
Answer:
Density of nucleus(p) It is defined as the nuclear mass per unit volume.

Thus the nuclear density is of the order of 10¹⁷kgm^-3 and is independent of its mass number. Therefore, all nuclei have the same approximate density.

Plus Two Physics Nuclei One Mark Questions and Answers

Question 1.
Fusion reaction takes place at high temperature because
(a) nuclei break up at high temperature
(b) atoms get ionized at high temperature
(c) kinetic energy is high enough to overcome the coulomb repulsion between nuclei
(d) molecules break up at high temperature
Answer:
(c) kinetic energy is high enough to overcome the coulomb repulsion between nuclei.

Question 2.
Give the relation between half-life and mean life
Answer:
t_1/2 = 0.693τ.

Question 3.
Write down the expression for nuclear radius
Answer:
R = R₀A_1/3.

Question 4.
Complete the given nuclear reaction

Answer:

Plus Two Physics Nuclei Two Mark Questions and Answers

Question 1.
A beam of radio active radiation is unaffected in a combined electric and magnetic field in mutually perpendicular direction. A boy argues that, it is essentially a gamma ray. Do you agree with him. Justify your answer.
Answer:
It need not be gama ray. It can be α or β ray as the electric and magnetic field are in mutually perpendicular direction.

Question 2.
Pick the odd one out of the following.
a.

1. Curie
2. Roentgen
3. Becquerel
4. Rutherford

b.

1. γ – decay
2. β – decay
3. β⁺ – decay
4. α – decay

Answer:
a. odd one:
2. Roentgen.

b. odd one:
1. γ – decay

Question 3.
Figure below represents radiation coming out from a radioactive element.

1. Identify the radiation B. Give reason.
2. If the electric field is replaced by magnetic field perpendicular and into the plane of the paper, identify the particle deflecting towards right (C).

Answer:

1. Gamma-ray. It has no change
2. C which is β ray

Question 4.
Two radioactive substances P and Q have half-life 6 months and 3 months respectively. Find the ratio of the activity of these two materials after one year.
Answer:
Activity R = λN
R₁ = λ₁N₁
R₂ = λ₂N₂

R₁: R₂ = 1: 1

Question 5.
Match the following.

Answer:

Question 6.
Suppose you are a health physicist and you are being consulted about a spill occurred in a radio chemistry lab. The isotope spilled out in the lab was 500 micro cure of ¹³¹Ba which has half-life of 12 – days.

1. What is the decay constant of ¹³¹Ba
2. Your recommendation is to clear the lab until the radiation level is down to 1 micro curie. How length will the lab have to be closed.

Answer:
1.

2. 9 × 12 = 108 days.

Question 7.
Classify the following statement in to nuclear fission and nuclear fusion

1. Products are radio active
2. Can be controlled
3. Reaction is spontaneous
4. Can’t proceed as a chain reaction

Answer:

1. fission
2. fission
3. fission
4. fusion

Question 8.

Answer:

Plus Two Physics Nuclei Three Mark Questions and Answers

Question 1.
1. Nuclear fusion can liberate more energy than nuclear fission. But nuclear fusion is not commonly used in energy production. Why?
2. Match the following.

Answer:
1. Nuclear fusion is a thermonuclear reaction. It occurs at high temperature (10⁶k) and it is difficult to attain such a high temperature. Hence fusion is not commonly used in energy production.

2.

Question 2.
β – Particles does not exist inside a nucleus. But it is emitted from the nucleus!

1. What is β Particle
2. What happens to nucleus of a atom, when a particle is emitted? Explain
3. Why β particle is emitted from the nucleus?

Answer:

1. It is electron
2. When α is emitted, mass number decreases to 4 and atomic decreases by 2.
3. To get stability, β is emitted from the nucleus.

Question 3.
Nuclear radius depends on the mass number of the element.

1. Write down the expression for nuclear radius.
2. Prove that the density of the nucleus is independent of mass number A.

Answer:
1. R = R₀ A^1/3.

2. Density of nucleus = (mass of the nucleus)/ (volume of the nucleus)

This shows that nuclear density is independent of mass number.

Question 4.
Match the following.

Answer:

Question 5.
The fission of one nucleus of ₉₂U²³⁹ release 200 Mev of energy.

1. What is meant by fission
2. Express 200 Mev energy in joule
3. How many fission of ₉₂U²³⁹ should occur per send for producing a power of 1 Mev

Answer:
1. The splitting of heavy nucleus into two nucleus is called fission.

2. E = 200Mev
= 200 × 10⁶ev
E = 200 × 10⁶ × 1.6 × 10^-19J
= 3.2 × 10^-11J.

3. Number of fission

Question 6.
Match the following.

Answer:

Question 7.
Size of the nucleus increases with number of nucleons. As size increases, volume increase (ie mass number increases).

1. Based on the above facts find an expression for radius of nucleus.
2. Calculate radius of ₁₃Al²⁷ nucleus. The constant R₀ = 1.2 fermi.

Answer:

1. R = R₀A^1/3
2. R = 1.2 × (27)^1/3 fermi = 3.6 fermi.

Question 8.
Binding energy curve shows the variation of Binding energy per nucleon of nuclei with mass number.
1. Binding energy per nucleon is maximum for mass number……. (1)
2. The figure shows dis integration of Deuteron. (2)

What should be the frequency of the incident photon to break Deuteron into proton and neutron?
Mass of proton m_p = 1,007276u.
Mass of neutron m_n = 1,008665u
Mass of deuteron = 2.013553u
Answer:
1. 56

2. Mass defect = (1.007276 + 1.008665) – 2.013553 = 0.002388u
Binding Energy = 0.002388 × 931 MeV = 2.223
MeV Energy supplied to the photon
= 2.223 × 10⁶ × 1.6 × 10^-19
= 3.56 × 10^-13J
Frequency of photon

Question 9.
The figure shows the potential energy of a pair of nuclear particles and their distance of separation in Fermi (fm).

1. Fill in the blanks.

2. What conclusion do you obtain about the nature of nuclear force from the graph.
Answer:
1.

2. The nuclear force is a short range force.

Plus Two Physics Nuclei Four Mark Questions and Answers

Question 1.
Atomic mass of \(_{ 8 }^{ 16 }{ O }\) is found to be 16.0000u

1. what is the mass of \(_{ 8 }^{ 16 }{ O }\) nucleus
   (Hint: mass of an electron = 0.00055u)
2. determine the total mass of the constituents particles of the \(_{ 8 }^{ 16 }{ O }\) nuclei.
   (Hint : Mass of neutron = 1.00864 u, Mass of proton = 1.007274 u)
3. Give a general expression for mass defect and explain what is binding energy?
4. Binding energy per nucleon is lower for both very light nuclei (Z ≤ 10) and very heavy nuclei (Z ≥ 70) Justify a nuclear fission and fusion

Answer:
1. Mass of nucleus = (8 × 1.00864 + 8 × 1.007274) amu.

2. (8 × 1.00864 + 8 × 1.007214 + 8 × 0.0055)amu

3. ∆m = [ZM_p + (A-Z) M_n – M)
The energy equivalent to mass defect is called binding energy.

4. When two light nuclei are combined to form a heavy nucleus, the binding per nucleon increases. Hence the stability of atom increases. When heavy nucleus split into two light nuclei; B.E. for nucleon also increases.

Question 2.
Classify the following into alpha, beta, and gama

1. Similar to fast moving electron
2. It is an electromagnetic wave
3. Similar to helium nucleus
4. Travel with 1/10^th the velocity of light
5. Travel with 99/100^th the velocity of light
6. Travel with the velocity of light
7. Positively charged
8. Negatively charged

Answer:

1. beta particle – 1, 3, 8
2. Gama ray – 2, 6
3. alpha particle – 3, 4, 7

Question 3.
a. Two protons and two neutrons may bound to form a single particle it is called

1. α particle
2. β particle
3. deuteron
4. triton

b. If such a particle is emitted what changes will occur in the nucleus
c. The penetrating power of the particle is very small in air. Why?
d. If the particle is projected upward in a uniform magnetic field with direction perpendicular to plane in ward, towards which plate (A or B) it is deflected? To find this which law is applied?

Answer:
a.
1. α particle.

b. Mass number decreases to 4 and atomic number decreases to 2

c. penetrating power of the particle is very small in air:
(i) Mass of α particle is 4. Hence penetrating power of the particle is small.

d. Towards A (left) Flemings left hand rule is used.

Question 4.
Nuclear radius depends on the mass number of the element.

1. Write down the expression for nuclear radius. (1)
2. Prove that the density of the nucleus is independent of mass number A. (2)
3. Read the following statement and choose the correct option.

“Electric dipole moment is zero for nuclei in stationary state.” (1)
Assertion:
All nuclei have spherical symmetry about the centre of mass.
Reason:
The zero dipole moment for stationary nuclei is due to the symmetry about the centre of mass.

• Assertion and reason are true.
• Assertion is false reason is true
• Assertion is true reason is false
• Assertion and reason are false.

Answer:
1. R = R₀A^1/3.

2. Density of nucleus = (mass of the nucleus)/(volume of the nucleus)

This shows that nuclear density is independent of mass number.

3.
(ii) Assertion is false reason is true.

Question 5.

1. Which process is represented by the equation?
2. What happens to the parent nucleus after the process?
3. After this process the nucleus will be in a higher energy state. How will it come to ground state?
4. If a proton splits in a nucleus what are the changes. Represent it with an equation.

Answer:

1. β – emission.
2. Atomic increases to one unit. But mass number remains constant
3. The parent atom comes to ground state by emitting gamma ray.
4. P → n + e⁺ + ν
   Atomic number decreases by one unit. But mass number remains constant.

Plus Two Physics Nuclei Five Mark Questions and Answers

Question 1.
Rutherford and Soddy’s laws of radioactivity explain the rate of decay of radioactive material.

1. Arrive at the expression for the number of radio active atoms of a radioactive material remaining after an interval of time. (2)
2. Draw the curve showing the variation of log\(\left(\frac{N}{N_{0}}\right)\) with time. (1)
3. Two radioactive substances P and Q have half life 6 months and 3 months respectively. Find the ratio of the activity of these two materials after one year. (2)

Answer:
1. According to Law of Radioactive decay,

Integrating
InN = -λt + C ………(1)
C is the constant of integration. To get value of C, let us assume that initially (t=0) the number of nuclei be N₀.
∴ C = In N₀
Substituting for C in equation (1) we get,
In N – In N₀ = -λt

\(\frac{\mathrm{N}}{\mathrm{N}_{0}}\) e^-λt
N = N₀e^-λt

2.

3. Activity R = λN
R₁ = λ₁N₁
R₂ = λ₂N₂

Question 2.
a. In the given figure a radioactive source is placed inside a lead block. Identify the rays incident on the photographic plates.

b. Which of the following statement is correct.

1. Gamma rays consist of high energy neutrons.
2. Alpha rays are equivalent to singly ionized He atoms.
3. Protons and neutrons have exactly the same mass.
4. Beta rays are same as cathode rays.

c. How many alpha and beta particles are emitted in the following reaction.

Answer:
a.

• 1 – Alpha
• 2-Gamma
• 3 – Beta

b.
4. Beta rays are same as cathode rays

c.

Difference in mass number = 32
Mass number of one alpha particle = 4
Hence number of alpha particles = 8
Change in atomic number = 10
Change in atomic number due to alpha particles = 16
Charge of beta particles = -1
Hence number of beta particles emitted = 6

Question 3.
The figure shows a nuclear reactor based on thermal neutron fission.

1. The energy of thermal neutrons is……..ev. (1)
2. Name the parts labelled X and Y in the figure. (1)
3. Write the function of X and Y (1)
4. The multiplication factor has great significance in nuclear reactor. Give reason. (2)

Answer:
1. 0.025eV.

2. Control rods.

3. Control rods are used in nuclear reactors to control the fission rate of Uranium and Plutonium.

4. The ratio, K, of number of fission produced by a given generation of neutrons to the number of fission of the preceding generation is called the multiplication factor, it is the measure of the growth rate of the neutrons in the reactor.

For K = 1, the operation of the reactor is said to be critical. Unless the factor K is brought down very close to unity, the reactor will become supercritical and can even explode.

Students can Download Chapter 5 The Government: Budget and The Economy 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 The Government: Budget and The Economy

Plus Two Economics The Government: Budget and The Economy One Mark Questions and Answers

Question 1.
Balanced budget multiplier will be always equal to:
(i) 0
(ii) 1
(iii) ∞
(iv) -1
Answer:
(ii) 1

Question 2.
Which among the following is a tax revenue?
(i) Fees
(ii) Fine
(iii) GST
(iv) Profit
Answer:
(iii) GST

Question 3.
IF MPC = 0.5,What is the value of tax multiplier
(i) 0.5
(ii) 0
(iii) 1
(iv) 2
Answer:
(iii) 1

Question 4.
Identify indirect tax from the following
(i) Income tax
(ii) Profession tax
(iii) Wealth tax
(iv) Customs duty
Answer:
(iv) Customs duty.

Question 5.
Which of the following is not a major item of revenue expenditure?
(i) defence service
(ii) subsidies
(iii) interest payment
(iv) acquisition of land
Answer:
(iv) acquisition of land

Plus Two Economics The Government: Budget and The Economy Two Mark Questions and Answers

Question 1.
Classify the following aspects into a table as Revenue Expenditure and Capital Expenditure
Subsidies, Grants to state, Loans to foreign government, Capital projects of plants, investment in shares, Loans to state government, Salary to government staff, interest payment.
Answer:

Question 2.
State the objectives of government budget.
Answer:

• To achieve economic growth
• To reduce inequalities of income and wealth
• To achieve economic stability

Question 3.
Find the odd man out and reason.

1. Excise duty, fees, fines, penalties.
2. Income tax, sales tax, customs duty, excise duty.

Answer:

1. Excise duty. Others are non-tax revenue
2. Income tax. Others are indirect taxes.

Question 4.
Which of the following is revenue receipt
(a) Taxes
(b) Revenue from fees, penalties
(c) All the above
(d) None of the above
Answer:
(c) All the above

Question 5.
Classify the following into revenue expenditure and capital expenditure. (2)
(a) Interest payment on debit.
(b) Investment in share
(c) Land and advances by central government to state.
(d) Grant given to state government.
Answer:

• a & d Revenue expenditure
• b & c Capital expenditure

Question 6.
MGNREGA aims to increase the purchasing power of the lower income group.

1. How does this affect the value of mpc and Aggregate Demand of our economy?
2. Can the government reduce economic inequality through this scheme? How?

Answer:

1. High value of mpc and AD increases
2. Yes. Income of poor sections increases

Question 7.
Which of the following diagrams represent proportional taxation? Substantiate your answer.

Answer:
C. Because of tax rate increases in proportion to the change in income.

Question 8.
Name the economic terms.

1. The receipts of the government which are non- redeemable
2. The receipts of the government which creates liability

Answer:

1. Revenue Receipts
2. Capital Receipts

Question 9.
Prove that the value of Balanced Budget Multiplier is equal to one.
Answer:
\(\frac{1}{1-c}+\frac{-c}{1-c}\)

Plus Two Economics The Government: Budget and The Economy Three Mark Questions and Answers

Question 1.
Do you agree to the saying the revenue expenditure should not exceed revenue receipts. Substantiate your answer.
Answer:
Yes, I do agree to the saying the revenue expenditure should not exceed revenue receipts. The fact is that the revenue expenditure is non-developmental expenditure. Therefore, borrowing for meeting revenue expenditure has to be avoided.

Question 2.
“The proportional income tax acts as an automatic stabilizer”. Do you agree? Substantiate.
Answer:
Yes, I agree with this statement. The proportional income tax acts as an automatic stabilizer – a shock absorber because it makes disposable income, and thus consumer spending, less sensitive to fluctuations in GDP. When GDP rises, disposable income also rises but by less than the rise in GDP because a part of it is siphoned off as taxes.

This helps limit the upward fluctuation in consumption spending. During a recession when GDP falls, disposable income falls less sharply, and consumption does not drop as much as it otherwise would have fallen had the tax liability been fixed. This reduces the fall in aggregate demand and stabilizes the economy.

Question 3.
‘Deficits are not desirable for a government’ discuss the issue of deficit reduction.
Answer:
Government deficit can be reduced by an increase in taxes or reduction in expenditure. In India, the government has been trying to increase tax revenue with greater reliance on direct taxes (indirect taxes are regressive in nature -they impact all income groups equally).

There has also been an attempt to raise receipts through the sale of shares in PSUs. However, the major thrust has been towards reduction in government expenditure. This could be achieved through making government activities more efficient through better planning of programmes and better administration.

Question 4.
The Government of India is spending huge amount of money on various activities. This paved the way for the growth of public expenditure and external borrowings. Do you support external borrowings for meeting public expenditure? Substantiate your answer.
Answer:
1. It is true that the Government of India is spending huge amount of money on various activities such as defence, urbanization, development programmes, welfare activities, etc.

2. It is desirable for the modem state to spend more for the welfare of the people. However, huge expenses have caused burden for the government.

3. Therefore, I do not support external borrowings for meeting public expenditure. This is because; borrowing from abroad will again bring another burden on the government in the form of interest payment.

Question 5.
Are fiscal deficits necessarily inflationary?
Answer:
Fiscal deficits are generally treated as inflationary. Increase in govt, expenditure and cuts in taxes both leads to government deficit. Increased govt, expenditure and reduced taxes tend to increase the aggregate demand. Generally firms are not able to produce higher quantities that are demanded at the going prices.

This leads to inflationary pressure. However, there is a solution to this inflationary pressure. Economy can utilizatfle unutilized resources and raise production. Therefore, the deficit cannot be inflationary when an economy has unutilized the resources.

Question 6.
Classify the following into public and private goods. List out two features of public goods. Road, cars, clothes, national defence.
Answer:

1. Road, national defence are public goods
2. Cars, clothes are private goods

Two features of public goods are nonrivalry in consumption and non excludability.

Question 7.
Reducing inequalities is an objective of Indian Five year plans. Observe the following tax proportions in the total tax revenue.

1. Which of the above tax proportions help to reduce the economic inequality in society? How?
2. Direct taxes are progressive and Indirect taxes are regressive in nature. Substantiate.

Answer:

1. A, The higher income group contribute more.
2. In progressive taxes more income contributed by the richer sections In regressive taxes more income contributed by the poor sections.

Question 8.
The government of India spent ₹27000 crores under the scheme MGNREGA in 2015-16. Suppose the mpc of India is 0.6. Calculate the impact of this spending on the equilibrium income of the economy.
Answer:
Equilibrium income = Autonomous Govt, expenditure × multiplier Here, multiplier (k) is
\(\frac{1}{1-\mathrm{mpc}}=\frac{1}{1-0.6}=\frac{1}{0.4}=2.5\)
∴ Equilibrium income = 27000 × 2.5 = 67500.

Plus Two Economics The Government: Budget and The Economy Five Mark Questions and Answers

Question 1.
Classify the following aspects into a table as tax revenue and nontax Revenue. Fees and fines, Dividends from PSU, Income tax, Special assessment, Escheats, Excise duties, Customs duties, wealth tax.
Answer:

Question 2.
If the marginal propensity to consume
a) Govt. Expenditure multiplier
b) Tax multiplier
Answer:

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

Answer:

Question 4.
Find the odd one out

1. income tax, sales tax, wealth tax, excise duty
2. fine, corporate tax, fees
3. dividends, fees and fines, borrowings, grants
4. balanced budget, supplementary budget, surplus, budget.

Answer:

1. Income tax. Others are indirect taxes
2. Corporate tax. Others are non-tax revenues
3. Borrowings. Others are non-tax revenues
4. Supplementary budget. This is prepared during abnormal times.

Question 5.
Assume that marginal propensity to consume is 0.5, and change in government expenditure is ₹400. Calculate
a) government expenditure multiplier
b) equilibrium income
Answer:

Question 6.
Discuss the issue of deficit reduction.
Answer:
if the government increases taxes or decreases expenditure then the fiscal deficit gets reduced. Indian government is trying to reduce the fiscal deficit by increasing tax revenue by selling the share of PSUs and by reducing the government expenditure.

The deficit reduction influences the different sectors of an economy in different ways. The government is trying to fill the gap of reduced fiscal deficit by making government activities more efficient through better planning of programmes and better administration.

The cutting back government programmes in vital areas like agriculture, education, health, poverty alleviation has adverse effect on the economy. The same fiscal measures can lead to a large or small deficit government by the state of the economy. During recession period GDP falls which reduces tax revenue which increase the fiscal deficit.

Question 7.
Point out the main objectives of public expenditure.
Answer:
The main objectives of public expenditure are pointed out below

1. for satisfaction of collective needs of the people
2. forsmooth functioning of government machinery
3. for economic and social welfare of the people
4. for creation of infrastructure
5. for controlling depressionary tendencies
6. for accelerating the speed of economic development
7. for reducing regional disparities of growth.

Question 8.
Explain why the tax multiplier is smaller in absolute value than the government expenditure multiplier.
Answer:
The tax multiplier is smaller in absolute value compared to the govt, expenditure multiplier. This is because, the govt, expenditure directly affect the total expenditure and taxes enter the multiplier, process and put impact on the disposable income.

Disposable income influences the consumption expenditure of households. Therefore, the tax multiplier is always less in absolute value than the govt, expenditure multiplier.

For example, assume that MPC = 0.75, then
Govt, expenditure multiplier \(=\frac{1}{1-c}\), C stands for MPC
\(=\frac{1}{1-0.75}=\frac{1}{0.25}=4\)
Tax multiplier = \(\frac{-c}{1-c}=\frac{-0.75}{1-0.75}=\frac{-0.75}{0.25}\)
= -3 (absolute value is 3)
Thus, it is clear from the example, that tax multiplier is smallerthan govt. expenditure multiplier.

Question 9.

Answer:

Question 10.
Match column B and C with Column A.

Answer:

Question 11.
If MPC = 0.5, calculate
a) Tax multiplier
b) Govt, expenditure multiplier
Answer:

Question 12.
Classify the following budget receipts into revenue receipts and capital receipts.
Borrowing, tax revenue, fines and penalties, fees, Recovery of loans, disinvestment, special assessment, small savings.
Answer:
1. Revenue receipts

• Tax revenue
• Fees
• Fines and penalties
• Special assessment

2. Capital receipts

• Borrowings
• Disinvestment
• Recovery of loans
• Small Savings

Question 13.
Distinguish between deficit and debt. Asses to what extent a government can borrow.
Answer:
Budget deficit is the difference between the yearly revenue and expenditure of the government. Debt is the accumulated deficit. The government can finance the budgetary deficit by borrowing, taxation or printing new currencies. To what extent the government can borrow is based on two issues.

1. Whether the debt is a burden
2. How the debt is financed.

When the government borrows it transfers the burden of reduced consumption to future generation. The future tax would increase, hence a fall in disposable income and consumption in future. If the government borrows too much the saving available to private sector may fall.

It is also argued that the consumer spending is based not only on their current income but also on their expected future income. There is another argument, debt does not matter because we owe to ourselves. This is because although there is a transfer of resources between generation, purchasing power remain within the nation. Any debt from foreign involves a burden.

Question 14.
Calssify the following into allocation, distribution and stabilization function of the government. (5)

1. The government increases the defence expenditure. .
2. The government reduces the subsidies on chemical fertilizers.
3. The government increases the rate of interest.
4. The government increases investment.
5. The government spends on building new roads.
6. The government increases the income tax.

Answer:

• 1, 5 allocation function
• 2, 6 distribution function
• 3, 4 stabilization function

Allocation function refers to the provision of public goods by the government. Stabilization function refers to the reduction of fluctuations in the economy by influency of the aggregate demand.

Question 15.
“Provision of Public good is inevitable to maintain social justice.”
Evaluate the statement in light of the experience of our State in General Education and Public Health.
Answer:
Health and Education Indicators of Kerala. Government can achieve this which Market economy can’t.
\(=\frac{1-c}{1-c}=1\)

Plus Two Economics The Government: Budget and The Economy Eight Mark Questions and Answers

Question 1.
Calculate

1. fiscal deficit,
2. revenue deficit and
3. primary deficit

on the basis of the facts given below.
Borrowing – 40,000
Budget deficit – 60,000
Revenue receipts – 60,000
Interest payments – 20,000
Revenue expenditure – 80,000
Answer:
1. Fiscal deficit = Budget deficit + borrowing
= 60000 + 40000 = 1,00,000

2. Revenue deficit = Revenue expenditure – Revenue receipts
= 80000 – 60000 = 20,000

3. Primary deficit = Fiscal deficit – Interest payments
= 100000 – 20000 = 80,000

Question 2.
Construct a flow chart showing budget receipts and expenditure of the government.
Answer:
Budget Receipts and Expenditure of the government

Question 3.
Conduct a discussion on different concepts of deficits.
Answer:
When a government spends more than it collects by way of revenue, it incurs a budget deficit. There are various measures that capture government deficit and they have their own implications for the economy. The important concepts of deficits are discussed below.

1. Revenue Deficit:
The revenue deficit refers to the excess of government’s revenue expenditure over revenue receipts.

2. Fiscal Deficit:
Fiscal deficit is the difference between the government’s total expenditure and its total receipts excluding borrowing.
Gross fiscal deficit = Total expenditure – (Revenue receipts + Non-debt creating capital receipts

3. Primary Deficit:
We must note that the borrowing requirement of the government includes interest obligations on accumulated debt. To obtain an estimate of borrowing on account of current expenditures exceeding revenues, we need to calculate what has been called the primary deficit. It is simply the fiscal deficit minus the interest payments.
Gross primary deficit = Gross fiscal deficit – net interest liabilities.

Question 4.
Represent in the diagram the effect of the following on the equilibrium level of income

1. Effect of higher government expenditure
2. Effect of a reduction in taxes

Answer:
1. Effect of higher government expenditure
When there is higher government expenditure, the aggregate expenditure curve will shift upward. This will lead to an increase in equilibrium level of income. This effort is shown in the diagram.

2. Effect of a reduction in taxes
When there is a tax cut, there will be an increase in consumption and output. Therefore, the equilibrium level Of income increases as shown in the diagram.

Question 5.
The equilibrium of an economy is given below.

1. What is fiscal policy?
2. Identify the level of equilibrium in an economy and show the seperate diagram the effect of an increased govt, expenditure and an increased tax on the level of output.

Answer:
1. The policy of the government related to taxation and govt, expenditure is known as fiscal policy.
2. OY is the equilibrium level of output. The effects of an increased govt, expenditure is shown in the diagram below.

Govt, expenditure is one of the components of AD. AD = C + I + G + (x – m)
So the AD curve shifts upward when there is an increase in the govt, expenditure. As a result the equilibrium changes from e to e₁. The level of output increases from y to y₁.

It can be seen from the diagram that the increase in govt, expenditure is from A to A₁ But the output increased from y to y₁ This is because of multiplier effect. The effect of an increased tax on the level of output is given in the diagram below.

Students can Download Chapter 12 Atoms 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 12 Atoms

Plus Two Physics Atoms NCERT Text Book Questions and Answers

Question 1.
What is the shortest wavelength present in the Paschen series of spectral lines?
Answer:
Since

For Paschen series, n₁ = 3 and n₂ = ∞

Question 2.
The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10^-11 m. What are the radii of the n = 2 and m = 3 orbits?
Answer:
Given r₁ = 5.3 × 10^-11m
n₁ = 1, n₂ = 2, r₂ = ?
n₃ = 3, r₃ = ?
Since r ∝ n²
∴ r₁ ∝ n₁²
And r₂ ∝ n₂²

or r₂ = 4 × r₁ = 4 × 5.3 × 10^-11
or e₂ = 2.12 × 10^-10m
Also r₃ ∝ n₃²

or r₃ = 9r₁ = 9 × 5.3 × 10^-11
or r₃ = 4.77 × 10^-10m.

Question 3.
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelength will be emitted?
Answer:
First excitation energy
E = E₂ – E₁ = -3.4 – (-13.6) = 10.2 eV
Second excitation energy
E = E₃ – E₁ = -1.51 -(-13.6) = 12.09 eV
Third excitation energy
E = E₄ – E₁ = -0.85 – (-13.6) = 12.75 eV
Second incident beam has energy = 12.5 eV
So only first two lines in the Lyman series of wave-length 103 nm and 122 nm will be emitted.
Also E₃ – E₂ = -1.51 – (-3.4) = 4.91 eV
i.e, first line in the Balmer series of wavelength 656nm will also be emitted.

Question 4.
In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 10¹¹ m with orbital speed 3 × 10⁴ ms^-1.
Answer:
Here r= 1.5 × 10¹¹ m,
v = 3 × 10⁴ ms^-1, m = 6 × 10²⁴kg, n = ?

Plus Two Physics Atoms One Mark Questions and Answers

Question 1.
A radioactive nucleus emits beta particle. The parent and daughter nuclei are
(a) isotopes
(b) isotones
(c) isomers
(d) isobars
Answer:
(d) isobars

Question 2.
Consider an electron in the n^th orbit of a hydrogen atom in the Bohr model. The circumference of the orbit can be expressed in terms of de Broglie wavelength λ of that electron as
(a) (0.529) nλ
(b) \(\sqrt{n} \lambda\)
(c) (13.6)λ
(d) nλ
Answer:
(d) nλ

Question 3.
The ionization energy of hydrogen atom is 13.6 eV. Find the energy corresponding to a transition between 3^rd and 4^th orbit.
Answer:
E = E₄ – E₃

= -0.85 + 1.51 = 0.66 eV.

Question 4.
If 13.6 eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron from n = 2 is
(a) 10.2 eV
(b) 0 eV
(c) 3.4 eV
(d) 6.8 eV
Answer:
(c) 3.4 eV
Explanation:

= 3.4 eV.

Question 5.
In the Bohr model of the hydrogen atom, the lowest orbit corresponds to
(a) infinite energy
(b) maximum energy
(c) minimum energy
(d) zero energy
Answer:
(c) minimum energy.

Question 6.
Write down the Balmerformula for wavelength of H_a line.
Answer:

Plus Two Physics Atoms Two Mark Questions and Answers

Question 1.
Given Rydberg constant as 1.097 × 10^-7m^-1. Find the longest and shortest wavelength limit of Baler Series.
Answer:
\(\bar{v}=\frac{1}{\lambda}=R_{H}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\)
Longest wavelength n₁ = 2 and n₂ = 3

Shortest Wavelength n₁ = 2 and n₂ = α

Plus Two Physics Atoms Three Mark Questions and Answers

Question 1.
Bohr combined classical and early quantum concept and gave his theory in the form of three postulates.

1. The total energy of an electron in ground state of hydrogen atom is -13.6eV. What is the significance of negative sign?
2. The radius of innermost electron orbit of hydrogen atom is 5.3 × 10¹¹m. What are the radii of n = 2 and n = 3 orbits?

Answer:
1. Negative sign implies that the electrons are strongly bounded to the nucleus.

2. r_n = n²a₀ = 5.3 × 10^-11m
r₁ = a₀ = 5.3 × 10^-11m
r₂ = 4a₀ = 21.2 × 10^-11m
r₃ = 9a₀ = 47.7 × 10^-11m.

Students can Download Chapter 8 The d and f Block Elements 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 8 The d and f Block Elements

Plus Two Chemistry The d and f Block Elements One Mark Questions and Answers

Question 1.
The element of the 3d series which is not regarded as a transition element is ____________ .
Answer:
Zinc

Question 2.
Say TRUE or FALSE:
Cr²⁺ compounds are strong oxidising agents
Answer:
False

Question 3.
Which of the following has the least magnetic moment?
(a) Cu²⁺
(b) Ni²⁺
(c) Co²⁺
(d) Fe²⁺
Answer:
(a) Cu²⁺

Question 4.
The oxidation state of Cr in K₂Cr₂O₇ is _________
Answer:
+ 6

Question 5.
Lanthanoid contraction implies
(a) Decrease in density.
(b) Decrease in mass.
(c) Decrease in ionic radii.
(d) Decrease in radioactivity.
Answer:
(c) Decrease in ionic radii.

Question 6.
Which of the following species are paramagnetic?
(a) Fe²⁺
(b) Zn
(c) Hg²⁺
(d) Ti⁴⁺
Answer:
(a) Fe²⁺

Question 7.
The approximate percentage of iron in misch metal is____
Answer:
5

Question 8.
The maximum oxidation state shown by Mn in its compounds is _____
Answer:
+ 7

Question 9.
The acidic, basic or amphoteric nature of Mn₂O₇, V₂O₅ and CrO are respectively
Answer:
acidic, amphoteric and basic

Question 10.
The correct order of oxidising power of the following is
(a) Cr₂O₇^2- > MnO₄^– > VO₂⁺
(b) MnO₄^– > Cr₂O₇^2- > VO₂⁺
(c) VO₂⁺ >MnO₄^– > Cr₂O₇^2-
(d) MnO₄^– > VO₂⁺ > Cr₂O₇^2-
(e) Cr₂O₇^2- > VO₂⁺ > MnO₄^–
Answer:
(b) MnO₄^– > Cr₂O₇^2- > VO₂⁺

Plus Two Chemistry The d and f Block Elements Two Mark Questions and Answers

Question 1.
Transition metals can form complexes.

1. Is the statement true?
2. Justify.
3. Give two examples of complexes.

Answer:

1. Yes
2. Transition elements form complex compounds due to the comparatively smaller size of the metal ion, their high ionic charges and the availability of d-orbital for the bond formation.
3. [Fe(CN)₆]^3-, [Cr(NH₃)₆]³⁺

Question 2.
Consider the statement: Scandium is a transition element but zinc is not.

1. Do you agree with his statement?
2. Justify your answer.

Answer:
1. Yes.

2. Scandium has incompletely filled 3d orbitals in its ground state (3d⁰), it is regarded as a transition element. On the other hand, zinc atom has completely filled d orbitals (3d¹⁰) in its ground state as well as in its oxidised state. Hence it is not regarded as a transition element.

Question 3.
Transition elements show variable oxidation state.

1. Give reason.
2. Illustrate with an example.

Answer:

1. The variable oxidation states of transition elements is due to the participation of ns and (n-1) d electrons in bonding,
2. E.g.: The outer configuration of Mn is 3d⁵,4s² it exhibits all the oxidation states from +2 to +7.

Question 4.
Cu+ is colourless while Cu²⁺ is coloured in aqueous solution.

1. Is it true?
2. What are the reasons for this?

Answer:

1. Yes.
2. In Cu+ it is 3d¹⁰,4s⁰ and there is no d-d transition. Hence no colour. But in Cu²⁺ it is 3d⁹, 4s⁰ it has d-d transition and hence it has colour.

Question 5.
The highest oxidation state of a metal is exhibited in its oxide or fluoride only.

1. Do you agree?
2. If yes, give reason.

Answer:

1. Yes.
2. Due to very high electronegativity and small size oxygen and fluorine can oxidise the metal to its highest oxidation state.

Question 6.
What is misch metal? Give its use.
Answer:
‘Misch metal’ is an alloy which consists of a lanthanoid metal (~ 95%) and iron (~ 5%) and traces of S, C, Ca and Al. It is used in Mg-based alloy to produce bullets, shell and lighter flint.

Question 7.
Some ions of transition metals are given below. Categorize them into those which are coloured or not. Justify your answer.
Ti³⁺, V²⁺, Sc³⁺, Mn²⁺, Ti⁴⁺, Cu²⁺, Zn²⁺, Cu⁺
Answer:

• In Ti³⁺, V²⁺, Mn²⁺, Ca²⁺ d-d transition is possible. Hence they have colour.
• In Sc³⁺ and Ti⁴⁺(3d⁰), Zn²⁺ and Cu⁺(3d¹⁰) d-d transition is not possible.

Question 8.
Why are Mn²⁺ compounds more stable than Fe²⁺ toward oxidation to their +3 state?
Answer:
The electronic configuration of Mn²⁺ is 3d⁵ , which is stable due to half-filled sub-shell. Hence, Mn²⁺ is not easily oxidised to Mn³⁺. On the other hand, Fe²⁺ has electronic configuration 3d⁶ . After losing one electron it changes to Fe³⁺ which has stable 3d⁵ electronic configuration. Hence, Fe²⁺ is relatively easily oxidised to Fe³⁺

Question 9.
Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
Answer:
Except Sc, the most common oxidation state of the first row transition elements is +2 which arises from the loss of two 4s electrons. The +2 state becomes more and more stable in the first half of first transition elements with increasing atomic number because 3d orbitals acquires only one electron in each of the five 3d orbitals.

Question 10.
In what way is the electronic configuration of the transition elements different from that of the nontransition elements?
Answer:
Transition elements have partially filled d-dubshell belonging to penultimate energy level whereas non-transition elements do not have any partially filled d-subshell. In non-transition elements, the last electron enters the s or p-subshell whereas in transition elements the last electron enters the d-subshell of penultimate energy level.

Question 11.
Predict which of the following will be coloured aqueous solution?
Ti³⁺ , V³⁺ cu⁺, Sc³⁺ , Mn²⁺, Fe³⁺ and CO²⁺ and MnO₄^–. Give reasons for each.
Answer:
Any ion that has partially filled d-orbitals is coloured due to d-d transition in visible light. MnO₄^– has purpule colour due to charge-transfer. Cu⁺ (d¹⁰ ) and Sc³⁺ (d⁰) are white since there is no d-d transition.

Question 12.
Use Hund’s rule to derive the electronic configuration of Ce³⁺ ion and calculate its magnetic moment on the basis of ‘spin-only’ formula.
Answer:
Ce (Z = 58) = ₅₄[Xe]4f¹5d¹6s²
Ce³⁺ = ₅₄[Xe]4f¹
It has one unpaired electron.
µ = \(\sqrt{n(n+2)}\) B.M.
µ = \(\sqrt{1(1+2)}=\sqrt{3}\) B.M = 1.732 B.M.

Question 13.
Among the following transition metal compounds which are coloured. Why?
TiO, TiO₂, ZnSO₄, MnCl₂, CrCl₃
Answer:

• TiO → Coloured
• MnCl₂ → Coloured
• CrCl₃ → Coloured

Question 14.
In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest. Why?
Answer:
Zinc has weaker interparticle forces due to fully filled d-subshell (3d¹⁰).

Question 15.
Fill in the blanks:

Answer:
i. 5

ii. 4.90 BM

Plus Two Chemistry The d and f Block Elements Three Mark Questions and Answers

Question 1.
Explain why

1. Transition metals have higher density than alkaline earth metals?
2. Transition metals are less electropositive than alkaline earth metals?
3. Transition elements exhibit higher enthalpies of atomisation?

Answer:
1. Transition metals have smaller atomic size and stronger interparticle attractive forces (metallic bond) than alkaline earth metals and hence have higher density than alkaline earth metals.

2. Transition metals due to their smaller atomic size and greater effective nuclear charge, have higher ionisation energy than alkaline earth metals and hence are less electropositive.

3. Transition elements have very strong interatomic bonds due to their small size and also due to the presence of large number of unpaired electrons in their atoms. As a result, they have very high enthalpies of atomisation.

Question 2.
Two students wrote the electronic configuration of Chromium (Z = 24) as shown below:
Student A: [Ar] 4s², 3d⁴
Student B : [Ar] 4s¹ 3d⁵

1. Pick out the correct electronic configuration and justify.
2. Calculate the spin only magnetic moments of Sc³⁺, Mn²⁺, Fe³⁺ and find out which is more paramagnetic.

Answer:
1. [Ar] 3d⁵4s¹ . Due to the extra stability of the half-filled 3d level.

2. Sc³⁺ – n = 0 µ = 0 BM
Mn²⁺ – n = 5 µ = \(\sqrt{5(5+2)}=\sqrt{35}\) = 5.92 BM
Fe³⁺ – n = 5 µ = \(\sqrt{5(5+2)}=\sqrt{35}\) = 5.92 BM

Plus Two Chemistry The d and f Block Elements Four Mark Questions and Answers

Question 1.
Some information regarding a particular compound is given below.
A deep purple black crystal which is moderately soluble in water at room temperature and it is prepared from pyrolusite.

1. Identify the compound.
2. Give the chemical reactions involved in the preparation of this compound.
3. Upon heating to 746 K, what happens to this compound? Give the equation.

Answer:

1. KMnO₄
2. 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O 3MnO₄^2- + 4H⁺→ 2MnO₄^– + MnO₂ + 2H₂O
3. Potassium permanganate on strong heating gives potassium manganate, manganese dioxide and oxygen.
   2KMnO₄→ K₂MnO₄ + MnO₂ + O₂

Question 2.
Consider the statement: All transition elements form coloured compounds.

1. Do you agree?
2. Name three transition elements which cannot form coloured compounds.
3. What is the reason for colour of transition metal compounds?

Answer:

1. No.
2. Zn, Cd and Hg.
3. The colour of transition compound is due to d-d transition. The colour observed corresponds to the complementary colour of the light absorbed.

Question 3.
Complete the table:

Answer:

Question 4.
The size of the atoms from Lanthanum to Lutetium, shows a steady and slow decrease in atomic size.

1. Do you agree with it?
2. What are the causes for this?
3. What are the consequences of this phenomenon?

Answer:
1. Yes.

2. It is due to Lanthanoid contraction. In Lanthanoid series the 4f electrons are being added in the antipenultimate shell. The 4f electrons have very poor shielding effect because of diffused shape of 4f orbitals and the effective nuclear charge experienced by each 4f electrons increases. Hence there is a regular decrease in radii with increase in atomic number.

3. Consequences of lanthanoid contraction are:

• Difficulty in separation of lanthanoids due to similarity in chemical properties.
• Similarity in size of elements belonging to same group of second & third transition series.

Question 5.
Zn, Cd, Hg are pseudo transition elements.

1. What do you mean by pseudo transition elements?
2. Write the outer electronic configuration of transition elements.
3. Write any four properties of transition elements,

Answer:
1. Pseudo transition elements have completely filled d orbitals and will not exhibit the characteristic properties of true transition elements.

2. The outer electronic configuration of transition elements is (n-1) d^{1 – 10} ns^{1 – 2}.

3. Four properties of transition elements.

• They form coloured compounds.
• They exhibit variable oxidation state.
• They form complex.
• They are good catalysts.

Question 6.
In a demonstration experiment, the teacher prepared the following solutions.
Solution 1: ZnSO₄ in water
Solution 2: CuSO₄ in water

1. Which of the above solutions is coloured?
2. Explain the Chemistry behind it.

Answer:
1. Solution 2: CuSO₄ in water is coloured.

2. In CuSO₄, Cu is in +2 oxidation state. Cu²⁺ has 3d⁹ electronic configuration and it can undergo d-d transition and hence has colour(blue). In ZnSO₄, zinc is in +2 oxidation state. Zn²⁺ has the 3d¹⁰ configuration. It has no d-d transition and hence it is colourless.

Question 7.

1. What is the effect of heat on KMnO₄
2. Give any two examples to show the oxidising property of KMnO₄.

Answer:
1. On heating, KMnO₄ decomposes at 513 K into potassium manganate and oxygen gas is evolved.

2. Two examples to show the oxidising property of KMnO₄

• In acid medium it oxidises iodide to iodine.
  10I^– + 2MnO₄^– + 16H⁺ → 2Mn²⁺ + 8H₂O + 5I₂
• In neutral or faintly alkaline solutions it oxidises iodide to iodate.
  2MnO₄^– + H₂O + I^– → 2MnO₂ + 2OH^– + IO₃^–

Question 8.
(a) Hf and Zr have similar ionic radii though they belongs to the 3^rd and 4^th transition series. Explain the reason for the anomalous behaviour.
(b) MnO₄^– & MnO₄^2- are the ions of manganese. They differ in colour and magnetic nature.

1. How do this difference occured?
2. Support your answer with their structure.

Answer:
(a) This is a consequence of lanthanide contraction.

(b) Differ in colour and magnetic nature.
1. This is due to difference in number of electrons in d- subshell.
2.

Question 9.
Pyrolusite on heating with KOH in the presence of air gives a dark green compound (A). The solution of (A) on treatment with H₂SO₄ gives a purple coloured compound (B).
a) Identify ‘A’ and ‘B’ and write the balanced chemical equation.
b) Write two reactions to show the oxidising nature of ‘B’.
Answer:

Question 10.
Give the method of preparation and two examples for the oxidizing property of K₂Cr₂O₇.
OR
Some d-block elements are given below:
Cr, Mn, Fe, Co, Ni
Identify the element which shows maximum paramagnetic behaviour.
Answer:
K₂ Cr₂ O₇ is prepared from chromate ore Fe Cr₂ O₄.
Step I: The powdered ore is heated with molten alkali to form soluble sodium chromate.

4 Fe Cr₂ O₄ + 16 NaOH + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8H₂O

Step II: Sodium chromate (Na₂CrO₄) is filtered and acidified with dil.H₂SO₄ to form sodium dichromate.

2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O

Step III: Na₂Cr₂O₇ solution is treated with KCI to form

K₂ Cr₂ O₇.
Na₂Cr₂O₇ + 2 KCI —» K₂ Cr₂ O₇ + 2NaCI

OR

Magnetic moment, µ = \(\sqrt{n(n+2)}\) BM
Cr; n=6 µ = \(\sqrt{48}\) = 6.93 BM
Mn; n=5 µ = \(\sqrt{35}\) = 5.92 BM
Fe; n=4 µ = \(\sqrt{24}\) = 4.90 BM
Co; n=3 µ = \(\sqrt{15}\) = 3.87BM
Ni; n=2 µ = \(\sqrt{8}\) = 2.83 BM
∴ Cr has maximum paramagnetic behaviour.

Question 11.

1. Even though usually transition elements exhibit +2 oxidation states, it changes from 0 to +8 in different compounds. What is the reason for it?
2. In the first transition series the atomic radius is decreasing from Sc to Cr and remains constant from Cr to Cu. From Cu to Zn it increases. Why?
3. The atomic radii of second and third transition series are same. Why?

Answer:
1. The variable valency is due to the following reason. The energy difference between inner ‘d’ level and outers level is so small. Therefore electrons from ‘d’ level as well as ‘s’ level can take part in reaction.

2. Sc to Cr – Atomic radius decreases due to increase in nuclear charge. Cr to Cu – Due to a balance between the increased nuclear charge and increased screening effect atomic radii become almost constant.

Cu to Zn – Repulsive interactions between the paired electrons in d-orbitals become very dominant towards the end of the period and cause the expansion of electron cloud and thus result in increased atomic size.

3. Lanthanoid, Contraction.

Question 12.
a) Calculate the magnetic moments of the following ions and predict its magnetic property.
Ti³⁺ (Z = 22)
Ni²⁺ (Z = 25)
Fe³⁺ (Z = 26)
Zn²⁺ (Z = 30)
Answer:

Question 13.
TiO, TiO₂, ZnSO₄, MnCl₂, CrCl₃
Arrange the above metal ions in the increasing order of their magnetic moment. (3)
Answer:

The increasing order of magnetic moment is TiO₂ = ZnSO₄ < TiO < CrCl₃, MnCl₂

Question 14.
How magnetic moment is calculated? How will you identify magnetism of an element from its electronic configuration?
Answer:
By using the ‘spin-only’ formula, n = \(\sqrt{n(n+2)}\) BM where ‘n’ is the number of unpaired electrons and ‘µ’ is the magnetic moment in Bohr Magneton (BM). If the element contains an unpaired electron it is paramagnetic. If all the electrons are paired and

Plus Two Chemistry The d and f Block Elements NCERT Questions and Answers

Question 1.
Why are Mn²⁺ compounds more stable than Fe²⁺ toward oxidation to their +3 state?
Answer:
The electronic configuration of Mn²⁺ is 3d⁵, which is stable due to half-filled sub-shell. Hence, Mn²⁺ is not easily oxidised to Mn³⁺. On the other hand, Fe²⁺ has electronic configuration 3d⁶. After losing one electron it changes to Fe³⁺ which has stable 3d⁵ electronic configuration. Hence, Fe²⁺ is relatively easily oxidised to Fe³⁺

Question 2.
Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
Answer:
Except Sc, the most common oxidation state of the first row transition elements is +2 which arises from the loss of two 4s electrons. The +2 state becomes more and more stable in the first half of first row transition elements with increasing atomic number because 3d orbitals acquire only one electron in each of the five 3d orbitals (i.e., each d-orbital remains half-filled) and inter electronic repulsion is the least and nuclear charge increases.

In second half of first row transition elements, electron starts pairing up in 3d orbitals and hence there is inter electronic repulsion. (Ti²⁺ to Mn²⁺ electronic configuration changes from 3d² to 3d⁵ but in second half i.e., Fe²⁺ to Zn²⁺ it changes from 3d⁶ to 3d¹⁰.)

Question 3.
In what way is the electronic configuration of the transition elements different from that of the non-transition elements?
Answer:
Transition elements have partially filled d-subshell belonging to penultimate energy level whereas non-transition elements do not have any partially filled d- subshell. In non-transition elements, the last electron enters the s or p-subshell whereas in transition elements the last electron enters the d-subshell of penultimate energy level.

Question 4.
Predict which of the following will be coloured in aqueous solution?
Ti³⁺, V³⁺, Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺ and CO²⁺ and MnO^4-. Give reasons for each.
Answer:
Any ion that has partially filled d-orbitals is coloured due to d-d transition in visible light. Ti³⁺ (d¹), V³⁺ (d²) Mn²⁺ (d⁵), Co²⁺ (d⁸) are coloured. MnO^4- has purpule colour due to charge-transfer. Cu⁺ (d¹⁰) and Sc³⁺ (d⁰) are white since there is no d-d transition.

Question 5.
Use Hund’s rule to derive the electronic configuration of Ce³⁺ ion and calculate its magnetic moment on the basis of ‘spin-only’ formula.
Answer:
Ce (Z = 58) = ₅₄[Xe]4f¹5d¹6s²
Ce³⁺ = ₅₄[Xe]4f¹
It has one unpaired electron.
µ = \(\sqrt{n(n+2)}\) B.M.
where n = number of unpaired electrons.
∴ Magnetic moment of Ce³⁺,
μ = \(\sqrt{1(1+2)}\) = \(\sqrt{3}\) BM
= 1.732 BM

Students can Download Chapter 11 Dual Nature of Radiation and Matter Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 11 Dual Nature of Radiation and Matter

Plus Two Physics Dual Nature of Radiation and Matter NCERT Text Book Questions and Answers

Question 1.
Find the

1. Maximum frequency, and
2. Minimum wavelength of X-rays produced by 30 kV electrons.

Answer:
Given V_o = 30 kV = 30 × 10³ V
v_max = ?
λ_max = ?
1. Since k_max = eV_o
So hv_max = eV_o
or

= 7.24 × 10¹⁸ Hz

2.

= 0.041 × 10^-9
or λ_min = 0.041 nm.

Question 2.
The work function of caesium metal is 2.14 eV. When light of frequency 6 × 10¹⁴Hz is incident on the metal surface, photoemission of electrons occurs. What is the

1. maximum kinetic energy of the emitted electrons,
2. maximum speed of the emitted photoelectrons?

Answer:
Given W₀ = 2.14 eV
= 2.14 × 1.6 × 10^-19 J
= 3.424 × 10^-19J
v = 6 × 10¹⁴ Hz

1. K_max = hv – W₀
= 6.62 × 10^-34 × 6 × 10¹⁴ – 3.424 × 10¹⁹
= 3.972 × 10^-19 – 3.424 × 10^-19
= 0.54 × 10^-19 J.

2. Since eV₀ =k_max

or v₀ = 0.34 V
Since \(\frac{1}{2}\) mV²_max = K_max

or v_max = 0.344 × 10⁶ ms-1 = 344 kms^-1

Question 3.
The photoelectronic cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Answer:
Given V₀ = 1.5 V
Since k_max = eV₀
= 1.5 eV
= 1.5 × 1.6 × 10^-19 J = 2.4 × 10^-19 J.

Question 4.
In an experiment on photoelectric effect, the slope of the cutoff voltage versus frequency of incident light is found to be 4.12 × 10^-15Vs. Calculate the value of Planck’s constant.
Answer:
Given, slope of graph = 4.12 × 10^-15 Vs
since, slope of graph = \(\frac{h}{e}\)
∴ h = e × slope of graph
= 1.6 × 10^-19 × 4.12 × 10^-15 = 6.59 × 10^-34 Js.

Question 5.
The threshold frequency fora certain metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.
Answer:
Given v₀ = 3.3 × 10¹⁴Hz
v = 8.2 × 10¹⁴Hz
Since eV₀ = hv – hv₀
So V₀ = \(\frac{h}{e}\) (v – v₀)
\(=\frac{6.62 \times 10^{-34}}{1.6 \times 10^{-19}}\) × (8.2 × 10¹⁴ – 3.3 × 10¹⁴)
= 4.14 × 10^-15 × 4.9 × 10¹⁴
= 2.02 V = 2.0 V

Question 6.
Light of frequency 7.21 × 10¹⁴Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10⁵m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Answer:
Given v = 7.21 × 10¹⁴Hz
u_max = 6.0 × 10¹⁴ms^-1
v₀ = ?
Since K_max = hv – hv₀
∴ v₀ = v – K_max

Plus Two Physics Dual Nature of Radiation and Matter One Mark Questions and Answers

Question 1.
Find out the wrong statement
(i) As frequency increases photo current increases
(ii) As frequency increase KE increases
(iii) As frequency increase velocity of electrons increases
(iv) As frequency increase stopping potential increases
(v) As frequency is below a certain value photo electrons are not emitted.
Answer:
(i) As frequency increases photo current increases.

Question 2.
Read the following statements and write whether true or false.

1. During photo electric effect photon share its energy with a group of electrons.
2. Intensity is directly proportional to square of amplitude

Answer:

1. False
2. True

Question 3.
In photoelectric emission the number of photoelectrons emitted per second depends on
(a) wavelength of incident light
(b) frequency of incident light
(c) intensity of incident light
(d) work function of the material
Answer:
(c) intensity of incident light

Question 4.
What is de-Broglie wave?
Answer:
The wave associated with material particle is called de-Broglie wave.

Question 5.
In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is
(a) 1.8 V
(b) 1.3 V
(c) 0.5 V
(d) 2.3 V
Answer:
(c) 0.5 V
Explanation:
The stopping potential V_s is related to the maximum kinetic energy of the emitted electrons K_max through the relation
K_max = eV_s
0.5 eV = eV_s or V_s = 0.5 V

Question 6.
Name the experiment, which establish the wave nature of moving electrons.
Answer:
Davisson Germer experiment.

Question 7.
Pick the odd one out of the following,
(a) Interference
(b) Diffraction
(c) Polarization
(d) Photoelectric effect
Answer:
(d) Photoelectric effect

Question 8.
Find out the wrong statement

1. If two particles have same momentum then they have same de-Broglie wave length.
2. If two particles have same KE the lighter particle has smallerwave length
3. As velocity of a given mass decreases wave-length increases

Answer:

1. True
2. False
3. True

Question 9.
Pick the odd one out from the following x-rays, visible light, matter waves, radio waves
Answer:
Matter waves.

Question 10.
If the electrons are accelerated by a potential of 50V, calculate the de-Broglie wavelength of electrons.
Answer:

Plus Two Physics Dual Nature of Radiation and Matter Two Mark Questions and Answers

Question 1.
Classify the following properties of the waves into de Broglie wave, em wave, and sound wave.

1. Associated with the moving particle.
2. Longitudinal wave
3. Electric field and magnetic field are perpendicular to each other.
4. Can produce photo electric effect.
5. Wave length is inversely proportional to mass of the moving particle.
6. Velocity in vacuum is 3×10⁸ m/s.

Answer:

1. de-Broglie wave – 1,5
2. Em wave – 3,4,6
3. Sound wave – 2

Question 2.
Table given below gives the work function of certain elements.

Identify the element in which photoelectric effect occurs easily. Justify your answer.
Answer:
Na – 2.70eV, work function is least.

Question 3.
“Louis De Broglie suggested existence of matter waves based on a hypothesis”

1. What do you mean by matter wave?
2. The objects in our daily life do no exhibit wave like properties. Why?

Answer:

1. The wave associated with material particle is called matterwave.
2. objects in ourdaily life have large mass. Hence λ is very small (In the order of 10^-34 cm) to be neglected.

Question 4.
A body of mass 1 Kg is moving with a velocity 1 m/s, a wave is associated with this body

1. Name the wave
2. Can you measure wave length of this wave. Explain?

Answer:

1. Matterwave
2. No. wave length of matterwave is very small.

Plus Two Physics Dual Nature of Radiation and Matter Three Mark Questions and Answers

Question 1.
In figure below represents the variation of current with potential fora metal

1. Identify the law governing it.
2. Even when the potential is zero, there is current. Explain.
3. Current is zero for a particular potential. How does this potential help in determining the velocity of electrons.

Answer:
1. Laws of photoelectric emission

• For a given frequency of radiation, number of photoelectrons emitted is proportional to the intensity of incident radiation.
• The K.E. of photoelectrons depends on the frequency of incident light but is independent of the light intensity.

2. When radiation falls on metal, photo electrons are emitted with certain velocity even if accelerating potential is zero.

3. At slopping potential (v₀), photocurrent is zero.
ie. 1/2 mv²_max = ev₀

Question 2.
A metal whose work function is 2 eV is illuminated by light of wavelength 3 × 10^-7 m. Calculate

1. threshold frequency
2. maximum energy of photoelectrons
3. the stopping potential.

Answer:
F₀ = h ν₀ – 2eV

1. F₀ = h ν₀

2. 1/2mv² = h(ν – ν₀)
= 6.63 × 10^-34 (10¹⁵ – 4.8 × 10¹⁴)
= 3.44 × 10^-19J

3. eV₀ = h(ν – ν₀)

Question 3.
The wave nature of electron was experimentally verified by diffraction of electron by Nickel crystal.

1. Name the experiment which establish wave nature of moving electron.
2. An electron and a proton have same kinetic energy which of these particles has shortest de-Broglie wave length?

Answer:
1. Davisson and Gemner experiment

2.

Mass of alpha particle is more than that of proton, hence it has shortest wavelength.

Question 4.
“Moving particles of matter shows wave like properties under suitable conditions’’

1. Who put forward this hypothesis?
2. A proton and an electron have been accelerated through same potential. Which one have higher matter wave length. Write the reason

Answer:
1. De broglie

2. \(\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}\)
The mass of proton is higher than electron. Hence wave length of proton is less than electron.

Question 5.
Three light beams of same frequency and different intensity I₁, I₂ and I₃ are incident on the same metal. I₁ >I₂ >I₃.

1. Which beam produce maximum photocurrent?
2. Which beam produce electrons of maximum speed and KE?
3. Draw a graph showing variation of photocurrent with intensity in same speed.

Answer:
1. I₃.

2. Frequency is same. Hence electron emitted from the metals will be same.

3.

Plus Two Physics Dual Nature of Radiation and Matter Four Mark Questions and Answers

Question 1.
The graph shows photoelectric current with anode potential.

1. The potential at ‘O’ is called

• accelerating potential
• retarding potential
• stopping potential
• saturation potential

2. Why current becomes constant in the region BC?

3.

Is the above graph possible? Justify your answer
Answer:

1. Stopping potential
2. The whole electrons emitted from the cathode will reach at anode. Hence current becomes saturation at BC.
3. This graph is not possible. Stopping potential is directly proportional to frequency of incident light, ie. when frequency of incident light increases, stopping potential also increases. Hence we expect a high stopping potential for v₂.

Question 2.
The magnification of an electron microscope is much larger than that of an optical microscope, because electron beams are used instead of light beams in an electron microscope.

1. Which property of electrons is used in the construction of the electron microscope?
2. Obtain expression for the wavelength of de-Droglie waves associated with an electron accelerated through a potential of V volts.

Answer:
1. Wave nature.

2. ev = 1/2mv²
mv² = 2 eV
m²v² = 2eVm

Question 3.
The wavelengths of violet and red ends of visible spectrum are 390nm and 760 nm respectively.
1. Evaluate the energy range of the photons of the visible light in electron volts.
2. The work function of four different materials is given in the table below.

Pick out the suitable metal/ metals for the construction of the photo cell which is to operate with visible light.
3. Calculate the threshold frequency of the selected metal/metals.
Answer:
1. 1.65ev to 3.1ev

2. Cs

3.

Question 4.
“To emit a free electron from a metal surface a minimum amount of energy must be supplied”.

1. It is called……..
2. Give three method to supply energy to a free electron
3. For metal A (tungsten) – work function is 4.52 eV for metal B (thoriated tungsten) it is 2.6 ev, for metallic (oxide coated tungsten) it is 1 eV. Which will you prefer as a good electron emitter and why?

Answer:

1. Threshold energy
2. Give light energy or heat energy
3. Work function for metallic oxide coated tungsten is small (1 ev.) Hence this material is good electron emitter.

Question 5.
Louis de Broglie argued that electron in circular orbit as proposed by Bohr, must be seen as a Particl wave.

1. From Bohr’s postulate of angular quantization momentum, arrive at an expression for wave length of an orbital electron. (2)
2. Comment on the above result (2)

Answer:
1.

2. Since λ = \(\frac{2 \pi r}{n}\), length of the first orbit is the de-Broglie wavelength of the orbit.

Plus Two Physics Dual Nature of Radiation and Matter Five Mark Questions and Answers

Question 1.
Schematic diagram of an experimental set up to study the wave nature of electron is shown below.

1. Identify the experiment.
2. In the experiment the intensity of electron beam is measured for different values of ‘q’. At 54V accelerating potential and q = 50°, a sharp diffraction maximum is obtained. What is the wave length associated with the electron.
3. A particle is moving three times as fast as electron. The ratio of debroglie wavelength of the particle to that electron is 1.813 × 10^-14. Calculate the mass of the particle.

Answer:
1. Davisson and German experiment

2. \(\lambda=\sqrt{\frac{150}{v}}=\sqrt{\frac{150}{54}}=1.66 \mathrm{A}^{0}\)

3.

Question 2.
An electron moves under a potential difference of 300V

1. The wave associated with electron is called……….
2. Derive an expression for its wave in terms of charge of particle.
3. Calculate the wavelength of above electron.

Answer:
1. matter wave

2.

3.

Question 3.
Figure below shows a version of Young’s Experiment performed by directing a beam of electrons on double slit. The screen reveals a pattern of bright and dark fringes similar to an interference as 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 replaced by bullets fired form 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 of the bullet is very large compared to the mass of electron, the de Broglie wavelength is not considerable.

Question 4.
A particles moving with KE 5Mev. Its mass is 1.6 × 10^-27 Kg

1. What is the energy of particle in joule?
2. Derive an equation of find De Broglie wave length in terms of KE
3. Calculate De-Broglie wave length of above particle.

Answer:
1. KE = 5 Mev = 5 × 10⁶ × 1.6 × 10^-19J

2.

3.

Question 5.
Einstein got Nobel Prize in 1921 for his explanation of photoelectric effect.

1. In order to start photoelectric emission, the minimum energy acquired by free electron in the metal is called as…… (1)
2. The minimum energy forthe emission of an electron from metallic surface is given below Na: 2.75eV K: 2.3eV Mo:4.17eV Ni:5.15eV Select the metal which is more photo sensitive. Why? (1)
3. Draw variation of photoelectric current with applied voltage for radiation of intensities I₁ and I₂ (I₁ > I₂). Comment on the relatiion between intensity of light and photoelectric current. (2)
4. Does Light from a bulb falling on an iron table emit photoelectron? Justify your answer. (1)

Answer:
1. Work function/ Threshold energy.

2. K is more photosensitive because it has less work function.

3.

As intensity increases photoelectric current also increases.

4. No. The work function of iron is very large.

Question 6.
Lenard and Hallwachs investigated the phenomenon of photoelectric effect in details during 1886-1902 through experiments
1. Write any two characteristic features observed in the above experiment? (2)
2. Explain with reason

• Green light emit electron from certain metal surface while yellow light does not
• When the wavelength of incident light is decreased, the velocity of emitted photo electrons increases (2)

3. Complete the following statement about photoelectric effect.
The radiations having minimum frequency called…….falls on a metallic surcace, electrons are emitted from it. The metal which emits photoelectrons are called………The kinetic energy of photoelectrons emitted by a metal depends on………of the radiations, while intensity of the incident radiations depends on……… (1)
Answer:
1. Any two statement of laws of photoelectric effect.

2. Explain with reason:

• Energy of incident photon is inversely proportional to its wavelength. Since λ of green light is less than that of yellow, it has larger energy. So it can emit photoelectrons
• As the wavelength decreases, frequency and hence energy of incident radiation increases and hence kinetic energy of photo electrons increases.

3. Threshold frequency, photosensitive frequency, number of photons.

Question 7.
Figure below shows variation of stopping potential (V₀) with frequency(?) of incident radiations for two different metals A and B.

1. Write down the values of work function A and B.

2. What is the significance of slope of the above graph? (1)

3. The value of stopping potential for A and B for a frequency γ₀₁ (which is greaterthan γ₀₂) of incident radiations are V₁ and V₀ respectively. Show that the slopes of the lines is equal to \(\frac{v_{1}-v_{2}}{\gamma_{01}-\gamma_{02}}\). (3)
Answer:
1. Work function of A, Φ₀₁ – hν₀₁
Work function of B, Φ₀₁ – hν₀₂.

2. The slope of the graph gives value of h/e.

3. For the metalA, hν₁ = hν₀₁ + eV₁………..(1)
For the metal B, hν₁ = hν₀₂ + eV₂…………..(2)
From equation (1) and (2)
hν₀₁ + eV₁ = hν₀₂ + eV₂
e(V₁ – V₂) = h(ν₀₂ – ν₀₁)
\(\frac{h}{e}=\frac{V_{1}-V_{2}}{v_{02}-v_{01}}\)

Question 8.
Albert Einstein proposed a radically new picture of electromagnetic radiation to explain photoelectric effect.
1. Identify Einstein’s photoelectric equation? (1)
2. With the help of Einstein’s photoelectric equation explain the following facts.

• Kinetic energy of photoelectrons is directly proportional to frequency not on intensity.
• Existence of threshold frequency for a given photosensitive material. (2)

3. A metal whose work function is 2 eV is illuminated by light of wavelength 3 × 10^-7m. Calculate

• threshold frequency
• maximum energy of photoelectrons
• the stopping potential. (3)

Answer:
1. hν = hν₀ + 1/2 mv²

2. Einstein’s photoelectric equation:
a. hν ∝ 1/2 mv²
1/2 mv² ∝ ν
Hence kinetic energy is proportional to frequency.

b. hν – hν₀ = 1/2 mv²
h(ν – ν₀) = 1/2 mv²
ν should be greater than ν₀ otherwise h(ν – ν₀) is negative and is not possible.

3.

a. Φ₀ = hν₀
ν₀ = \(\frac{2 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}}\)
= 4.8 × 10¹⁴ Hz

b. 1/2 mv² = h(ν – ν₀)
= 6.63 × 10^-34 (10¹⁵ – 4.8 × 10¹⁴)
= 3.44 × 10^-19J

c. ev₀ = h(ν – ν₀)

Question 9.
The wave nature of electron was experimentally verified by diffraction of electron by Nickel crystal.

1. Name the experiment which establish wave nature of moving electron. (1)
2. With a neat diagram explain the existence of matter wave associated with an electron. (3)
3. An electron and a proton have same kinetic . energy which of these particles has shortest de-Broglie wave length? (2)

Answer:
1. Davisson and Germer experiment.

2.

Aim:
To confirm the wave nature of electron.
Experimental setup:
The Davisson and Germer Experiment consists of filament ‘F’, which is connected to a low tension battery. The Anode Plate (A) is used to accelerate the beam of electrons. A high voltage is applied in between A and C. ’N’ is a nickel crystal. D is an electron detector. It can be rotated on a circular scale. Detector produces current according to the intensity of incident beam.

Working:
The electron beam is produced by passing current through filament F. The electron beam is accelerated by applying a voltage in between A (anode) and C. The accelerated electron beam is made to fall on the nickel crystal.

The nickel crystal scatters the electron beam to different angles The crystal is fixed at an angle of Φ = 50° to the incident beam. The detector current for different values of the accelerating potential ‘V’ is measured. A graph between detector current and voltage (accelerating) is plotted. The shape of the graph is shown in figure.

Analysis of graph:

The graph shows that the detector current increases with accelerating voltage and attains maximum value at 54V and then decreases. The maximum value of current at 54 V is due to the constructive interference of scattered waves from nickel crystal (from different planes of crystal). Thus wave nature of electron is established.

Experimental wavelength of electron:
The wave length of the electron can be found from the formula
2d sinθ = n λ ………..(1)
From the figure, we get
θ + Φ + θ = 180°
2θ = 180 – Φ, 2θ = 180 – 50°
θ = 65°
for n = 1
equation (1) becomes
λ = 2d sinθ ………(2)
for Ni crystal, d = 0.91 A°
Substituting this in eq. (2), we get
wavelength λ = 1.65 A°

Theoretical wave length of electron:
The accelerating voltage is 54 V
Energy of electron E = 54 × 1.6 × 10^-19J
∴ Momentum of electron P = \(\sqrt{2 \mathrm{mE}}\)
P = \(\sqrt{2 \times 9.1 \times 10^{-31} \times 54 \times 1.6 \times 10^{-19}}\)
= 39.65 × 10^-25 Kg ms^-1
∴ De- Broglie wavelength λ = \(\frac{h}{P}\)

The experimentally measured wavelength is found in agreement with de-Broglie wave length. Thus wave nature of electron is confirmed.

3.

Mass of alpha particle is more than that of proton, hence it has shortest wavelength.

Question 10.
In Geiger-Marsden Scattering experiment alpha particles of 5.5 MeV is allowed to fall on a thin gold foil of thickness 2.1 × 10^-7m.
1. Draw Schematic diagram of above experimental arrangement.
2.

In the above graph nearly 10⁷ particles were detected when scattering angle is Zero. What do you understand by it?
3. Why gold foil is used in this experiment?
4. Does there exist any relation between impact parameter and scattering angle? If yes, explain your answer.
Answer:
1.

2. Most of the alpha particles get unscattered means that most of the space in an atom is empty.

3. Atomic number of gold 79, so number of protons is very high. Hence scattering between alpha and nucleons is larger. Gold foil can be made very thin so that the alpha particles suffer not more than one scattering.

4. Yes.
As impact Parameter increases, scattering angle decreases.

Question 11.
The study of emission line spectra of a material serve as a fingerprint for identification of the gas.

1. Name different series of lines observed in hydrogen spectrum. (1)
2. Draw energy level diagram of hydrogen atom? (2)
3. Write down the Balmer formula for wavelength of H_α line. (1)
4. Given Rydberg constant as 1.097 × 10⁷m^-1. Find the longest and shortest wavelength limit of Baler Series. (2)

Answer:
1. Lyman series, Balmer series, Paschen series, Bracket series, Pfund series

2.

3.

4.

Longest wavelength n₁ = 2 and n₂ = 3

Shortest Wavelength n₁ = 2 and n₂ = α

Question 12.
Bohr combined classical and early quantum concept and gave his theory in the form of three postulates.

1. State three postulates of Bohr Model of atom? (2)
2. The total energy of an electron in ground state of hydrogen atom is -13.6eV. What is the significance of negative sign? (1)
3. The radius of innermost electron orbit of hydrogen atom is 5.3 × 10¹¹m. What are the radii of n = 2 and n = 3 orbits? (2)

Answer:
1. Bohr postulates:
Bohr combined classical and early quantum concepts and gave his theory in the form of three postulates.

• Electrons revolve round the positively charged nucleus in circular orbits.
• The electron which remains in a privileged path cannot radiate its energy.
• The orbital angular momentum of the electron is an integral multiple of h/2π.
• Emission or Absorption of energy takes place when an electron jumps from one orbit to an other.

2. Negative sign implies that the electrons are strongly bounded to the nucleus.

3.

• r_n = n²a₀ = 5.3 × 10^-11m
• r₁ = a₀ = 5.3 × 10^-11m
• r₂ = 4a₀ = 21.2 × 10^-11m
• r₃ = 9a₀ = 47.7 × 10^-11m.

Students can Download Chapter 4 Income Determination 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 Income Determination

Plus Two Economics Income Determination One Mark Questions and Answers

Question 1.
The ratio of total additional planned savings in an economy to the total additional income of the economy is …….
(а) APS
(b) MPS
(c) MPC
(d) APC
Answer:
(b) MPS

Question 2.
If MPS is 0.7, then MPC is
(a) 1
(b) 0
(c) 0.7
(d) 0.3
Answer:
(d) 0.3

Question 3.
In an economy, C = 40, 1 = 10, MPC = 0.8, Y = 200. Then equilibrium level of income in
(a) 200
(b) 210
(c) 160
(d) 250
Answer:
(b) 210

Question 4.
Multiplier is equal to
(a) \(\frac{1}{1-\mathrm{MPC}}\)
(b) \(\frac{1}{1+M P C}\)
(c) \(\frac{1}{1-\mathrm{MPS}}\)
(d) \(\frac{1}{M P C+M P S}\)
Answer:
(a) \(\frac{1}{1-\mathrm{MPC}}\)

Question 5.
In the aggregate demand function y = A + c.y, MPC is represented by
(a) y
(b) A
(c) c
(d) None of these
Answer:
(c) c

Question 6.
The world wars adversely affected the European economy. Hence consumption and savings were low. But the ratio of consumption to income was high.
a. Can you connect the above situation with any Keynesian proposition? Name it.
Answer:
a. Consumption function with high mpc.

Question 7.
One among the following is not a characteristic of Keynesian consumption function.
(a) The aggregate real consumption expenditure is a stable function of real income.
(b) The mpc must lie in between zero and One
(c) The consumption is a function of rate of interest
(d) The mpc = 1 – mps
Answer:
(c) The consumption is a function of rate of interest

Question 8.
Suppose that there is an increase in autonomous investment. If so which of the following situations represent greater multiplier effect on income.

1. a relatively high MPC or
2. a relatively low MPC? Substantiate.

Answer:

1. A relatively high MPC
2. Larger size of Multiplier.

Plus Two Economics Income Determination Two Mark Questions and Answers

Question 1.
State whether the following statements are true or false and justify your answer.

1. If MPC = 0.8, MPS will be 0.8
2. The amount of consumption when income is zero is called negative consumption.

Answer:

1. False – MPS = 0.2.
2. False – Autonomous consumption.

Question 2.
Calculate Multiplierwhen MPS = 0.8
Answer:
K = 1 / MPS
= 1/ 0.8 =1.2

Question 3.
Give the meaning of Ex ante, Ex post, Ex ante consumption, and Ex ante investment.
Answer:
Ex-ante and Ex-post:
Consumption, savings, and investment can be classified into Ex-ante and Ex-post variables. The terms Ex-ante and Ex-post have been derived from the Latin word. Ex-ante means planned or desired. Ex-post means actual or realized. In national income accounting, the variables such as consumption, investment and savings are considered as ex-post variables. The rate at which consumption, savings, and investment are presented in the ex-post sense.

Question 4.
Define output multiplier.
Answer:
The ratio of the total increment in equilibrium value of final goods output to the initial increment in autonomous expenditure is called the output multiplier of the economy.
\(\mathrm{K}=\frac{1}{1-\mathrm{MPC}}\)

Question 5.
Represent the parametric shift of a graph.
Answer:

Question 6.
Classify the following sentences ex-ante and ex-post.

1. Government made an investment of ₹20,000 excess in 2017-2018.
2. A firm has decided to set up a new factory.
3. The government expects an increase in revenue receipts by 25%.
4. Cement consumption increased by 18%.

Answer:

• Ex-ante 2 and 3.
• Ex post 1 and 4.

Question 7.
Some values of mpc are given below. Choose the possible mpc value of a lower income group. Justify mpc value : 0.4, 0.2, 0.9, 0.6
Answer:
The lower-income group have the highest mpc (0.9). They spend a major portion of their income for consumption.

Plus Two Economics Income Determination Three Mark Questions and Answers

Question 1.
Find the odd one out. Justify your answer.

1. Consumption function, investment function, government demand, derived demand, net export.
2. K = 1/MPS, K = 1/1-MPC, K = 1/ MPC + MPS
3. Taxation, reserve ratio, bank rate, open market operations

Answer:

1. Derived demand. Others are components of aggregate demand.
2. K = 1/ MPC + MPS. Others are right formula.
3. Taxation. Others are instruments of monetary policy

Question 2.
State whether the following statements are true or false. If false, correct the statement.

1. Once the level of full employment is reached, the Keynesian AS curve becomes a downward sloping curve.
2. The rate of increase in exante consumption due to a unit increment in income is called marginal.
3. An increase in autonomous spending causes aggregate output of final goods to increase by a larger amount through the multiplier process.

Answer:

1. False. AS curve will be still rising
2. True
3. True

Question 3.
Calculate the levels of consumption at different levels of income, if consumption is worth ₹200 when income is zero, MPC is 0.8, and income 100, 200, 300, 400, 500.
Answer:
We shall define the consumption function equation as C = a + MPC.Y
where ‘a’ is autonomous consumption.

Question 4.
“Increase in saving leads to decrease in saving”- comment.
Answer:
It is usually accepted that an increase in saving leads to a decrease in saving. This is because, if all the people of the economy increase the proportion of income they save (i.e. if the mps of the economy increases) the total value of savings in the economy will not increase – it will either decline or remain unchanged. This result is known as the ‘Paradox of Thrift’ – which states that as people become more thrifty they end up saving less or same as before.

Question 5.
What is deflationary gap? State two measures to remove it.
Answer:
When aggregate demand falls short of aggregate supply at full employment, it gives rise to deflationary gap. Thus,
deficient demand = aggregate supply – aggregate demand. Measures to remove deflationary gap,

1. Reduce the bank rate.
2. Reduce the cash reservice ratio.

Question 6.
Depict ‘inflationary gap’ in a diagram.
Answer:
The excess of aggregate demand over aggregate supply at full employment level is known as inflationary gap. The following diagram depicts inflationary gap.

Question 7.
What are the components of aggregate demand?
Answer:
The components of aggregate demand are:

1. Household consumption expenditure
2. Government consumption expenditure
3. Private Investment expenditure

Question 8.
Using data given below construct and aggregate demand schedule for the level of income 500,1000, 1500 and 2000.
\(\overline{\mathrm{C}}=50, \mathrm{c}=0.8, \overline{\mathrm{I}}=60\)
Answer:
\(A D=\bar{C}+\overline{1}+c \cdot y\)

Question 9.
The diagram below shows parametric shifts. Explain parametric shift and identify the reason for shifts in the following diagram.

Answer:
parametric shifts are those shifts in the lines of a diagram, due to changes in slope or intercept. In diagram (a) the line shifts upward due to increase in the slope. In diagram (b) the shift in the line is due to the change in the intercept. The intercept increases here.

Question 10.
According to Keynesian model, as saving propensity increases, the equilibrium income decreases. Since the level of income is reduced, the volume of savings also comes down automatically.

1. Can you associate the above statement with any Keynesian proposition? Name it.
2. Substantiate with an example.

Answer:
1. Paradox of Thrift
2. example.

Question 11.
The following diagram shows the equilibrium income situation of a hypothetical economy with mpc 0.6 and a =70

Analyse the following situations and represent the resulting change in the diagram.

1. Income of the economy increases due to a good harvest. Consequently, the mpc increased to 0.7
2. Income of the economy increases due to a good harvest. Consequently, the mpc increased to 0.7
3. An amount of 50 million is planned to invest for the construction of a new multipurpose dam in the economy.
4. Calculate the new equilibrium income of the economy.

Answer:
1. Change in the slope parameter. AD curve swings upward

2. Change in the intercept parameter. AD curve shift upward

3.

Plus Two Economics Income Determination Five Mark Questions and Answers

Question 1.
Match the following.

Answer:

Question 2.
From the following data, calculate

1. MPC
2. APC
3. MPS
4. APS

Consumption = 400, Income = 500, Change in income 600, change in consumption = 450
Answer:
From the data, we get, C = 400,Y = 500, ΔC = 50, ΔY = 100

1. MPC = ΔC/ΔY = 50/100 = 0.5
2. APC = C/Y = 400/500 = 0.8
3. MPS = ΔS/ΔY = 50/100 = 0.5
4. APS = S/Y = 100/500 = 0.2

Question 3.
Suppose income increases by 10 and the MPC is 0.8. Explain the multiplier mechanism with the help of a table.
Answer:
When income increases by 10, consumption expenditure goes up by (0.8)10 since people spend 0.8 (= MPC) fraction of their additional income on consumption. Hence, in the next round, aggregate demand in the economy goes up by (0.8)10 and there again emerges an excess demand equal to (0.8)10.

Therefore, in the next production cycle, producers increase their planned output further by (0.8)10 to restore equilibrium. When this extra output is distributed among factors, the income of the economy goes up by (0.8)10 and consumption demand increases further by (0.8)210, once again creating excess demand of the same amount.

This process goes on, round after round, with producers increasing their output to clear the excess demand in each round and consumers spending a part of their additional income from this extra production on consumption items-thereby creating further excess demand in the next round.

In order to find out the total increase in output of the final goods, we must add up the infinite geometric series in the last column, i.e.
10 + (0.8)10 + (0.8)210 + ……… ∝
= 10(1 + (0.8) + (0.8)2 + ……….. ∝}
= 10/1 – 0.8 = 50
The increment in equilibrium value of total output thus exceeds the initial increment in autonomous expenditure. The ratio of the total increment in equilibrium value of final goods output to the initial increment in autonomous expenditure is called the output multiplier of the economy.

Question 4.
Explain the meaning of the break-even point. Illustrate the concept.
Answer:
Break-even point refers to that point in the level of income at which consumption expenditure becomes exactly equal to income and there is no saving. In other words, whole of income is spent on consumption and as a result, there is no saving.

It happens when income increases from low level to high level, it becomes equal to consumption expenditure at some level and that level is called break even point. Below this level of income, consumption is greater than income but above this level, income is greater than consumption.

This concept can be illustrated as follows. A hypothetical consumption and saving schedule is given.

Question 5.
State the relationship between MPC and MPS.
Answer:
1. Marginal Propensity to Consume refers to the ratio of change in consumption to change in income. MPC = AC / AY.

2. Marginal Propensity to Save refers to the ratio of change in saving to change in income. MPS = ΔS / ΔY.

3. The sum of MPC and MPS is always one and equal to unity. That is MPC + MPS = 1 MPS = ΔS / ΔY.

Question 6.
MPC of Indian economy is 0.5 and if the level of investment in the economy is increased by 100.

1. Give the equation for multiplier.
2. Calculate the value of multiplier.
3. If the level of investment declines what will be the effects on total income.

Answer:

1. The equation for multiplier is, \(k = \frac{1}{1-M P S}\)
2. Since MPC = 0.5, we get the value of multiplier as,
   \(K = \frac{1}{1-0.5} = \frac{1}{0.5} = 2\)
3. Since investment is increased by ₹100,
   AY = K. Δl ΔY =2 x 100 = 200

When the level of investment declines, the total in-come in the economy will decline.

Question 7.
From the following data, calculate
a) APC
b) APS
c) MPC
d) MPS

• Income = ₹2500
• Consumption = ₹1000
• Change in income = ₹750
• Change in consumption =₹250

Answer:
In the given example,

• Income = 2500 (Y)
• Consumption = 1000 (C)
• Change in income = 750 (ΔY)
• Change in consumption =750 (ΔC)

Question 8.
In an economy investment increases by ₹120 crores. The value of multiplier is 4. Calculate the MPC.
Answer:

Question 9.
In an economy investment increases by ₹120 crore. The value of multiplier is 4. Calculate the MPC.
What do you understand by “parameters of line”? How does a line shifts when its

1. slope increases and
2. its intercept increase?

Answer:
Consider the equation of a straight line of form b = ma + e where, m 0 is called the slope of the straight line, e is called the intercept on the vertical axis. When a increased by 1 unit the value b increases by m units. These are called movements of variables along the line. The entities e and m are called the parameters of the line. As the value of m increases the straight line swings upwards. This is called a parametric shift of line.

1. A positively sloping straight line swings downward as its slope decreases.
2. A positively sloping straight line shifts upwards parallel when its intercept increases.

Question 10.
What is effective demand? How will you derive the autonomous expenditure multiplier when price of goods and the rate of interest are given?
Answer:
If the elasticity of supply is infinite, then the output will be solely determined by the aggregate demand at this price in the economy. This called effective demand. The equilibrium output and aggregate demand at the given price of goods and rate of interest is derived by solving the equation.
Y = AD.
Y = A + CY
or Y – CY = A
or Y(1 – C) = A
A is the total value of autonomous expenditure. The value of Y is dependent on the value of parameters A and C.

Question 11.
State true or falls. Correct if they are wrong.
(a) v = c + s
(b) MPC + MPS ≥ 1
(c) \(K = \frac{1}{1-M P S}\)
(d) ΔY = KΔl
Answer:
(c) \(K = \frac{1}{1-M P S}\)

Question 12.
Graphically determine the following situations.

1. Full employment situation or equilibrium
2. Deficient demand
3. Excess demand

Answer:
1. Full employment situation or equilibrium

2. Deficient demand

3. Excess demand

Question 13.
Suggest the fiscal policy measures to correct excess demand and deficient demand
Answer:
a. Excess demand

• Increase taxes
• Decrease Government expenditure
• Reduce deficit financing
• Increase public borrowing

b. Deficient demand

• Decrease taxes
• Increase government expenditure
• Increase deficit financing
• Reduce public borrowing

Question 14.
identify the relation between multiplier and MPC.
Answer:
The value of the multiplier is determined by marginal propensity to consume. Higher the MPC, greater the size of multiplier lower the MPC, smaller the size of multiplier.

When income of consumers rises they spend more the value of increase in income, i.e., multiplier depends on MPC, greater the value of multiplier depends on greater size of MPC. Thus there is direct relation between multiplier and MPC. The relation can be expressed in terms of an equation as under

Thus it is clear from the above equation that the value of MPC and multiplier are positively related.

Question 15.
In the function AD = \(\overline{\mathrm{A}}\)+ c. y what happens to the aggregate demand curves when,

1. \(\overline{\mathrm{A}}\)
2. C (mpc) increases.

Answer:
1. \(\overline{\mathrm{A}}\)

When \(\overline{\mathrm{A}}\) increases, the AD curve shifts upwards parallel to the original AD curve. As a result, income increases from Y, to Y2 and economy reaches to new equilibrium position E2.
2. C (mpc) increases.

When c increases, the AD curve shifts upwards. But it is not a parallel shift, it only swings upward.

Question 16.
Nowadays, all governments of third world economies try to attract domestic and foreign investors by providing financial concessions and establishing Special Economic Zones.

1. Do you think that investment is an essential component of economic growth? Why?
2. Suppose an investor proposed to invest Rs.300crore in an economy with the mpc value of 0.7. Calculate the impact of this investment on the equilibrium income (Multiplier Effect) of the economy?

Answer:
1. Yes. Investment has a multiplier effect on income. Hence investment is necessary for economic growth.
2. impact of this investment on the equilibrium income

Plus Two Economics Income Determination Eight Mark Questions and Answers

Question 1.
Diagram shows the equilibrium of an economy.

1. Identify the equilibrium level of output and explain it.
2. The diagram show what happens to the equilibrium level of output when investment increases.
3. Analyze the change in investment and the change in output.

Answer:
1. The economy will be in equilibrium when AD = Y₁ \(A D=\bar{C} + \bar{I} + c, y\) The two components of aggregate demand in a two-sector model is consumption and investment. In the diagram, OY is the equilibrium level of output.

2. When the investment increases the AD curve will shift upward. This is because invest is one of the components of AD. As a results the initial equilibrium changes from e to e₁ The output increases from y to y₁.

3. When the investment increased the level of output also increased. In other words, the initial change in investment has brought out a more than proportionate change in the level of income. This is due to the multiplier effect. lt shows the rate of an initial change in investment to the final change in national income.
\(\text { Multiplier } K = \frac{1}{1-M P C}\)

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.