Explore topic-wise InterviewSolutions in Current Affairs.

1. Human Resource 

2. Technical Know – how 

3. Political Freedom 

4. Social Organization 

5. Corruption free administration 

6. Desire for Development 

7. Moral, ethical and social values (viii) Casino Capitalism 

8. Patrimonial Capitalism

Non – Economic Factors: ‘Economic Development has much to do with human endowments, social attitudes, political conditions and historical accidents. Capital is a necessary but not a sufficient condition of progress.’

(I) Human Resources:

1. Human resource is named as human capital because of its power to increase productivity and thereby national income. 

2. There is a circular relationship between human development and economic growth. 

3. A healthy, educated and skilled labour force is the most important productive asset. 

4. Human capital formation is the process of increasing knowledge, skills and the productive capacity of people.

(II) Technical Know – how: As the scientific and technological knowledge advances, more and more sophisticated techniques steadily raise the productivity levels in all sectors.

(III) Political Freedom: The process of development is linked with the political freedom.

(IV) Social Organization: People show interest in the development activity only when they feel that the fruits of development will be fairly distributed.

(V) Corruption free administration:

1. Corruption is a negative factor in the growth process. 

2. Unless the countries root-out corruption in their administrative system, the crony capitalists and traders will continue to exploit national resources.

(VI) Desire for development: The pace of economic growth in any country depends to a great extent on people’s desire for development. 

(VII) Moral, ethical and social values:

1. These determine the efficiency of the market, according to Douglas C. North. 

2. If people are not honest, market cannot function.

(VIII) Casino Capitalism: If People spend larger propotion of their income and time on entertainment liquor and other illegal activities, productive activities may suffer, according to Thomas Piketty.

(IX) Patrimonial Capitalism: If the assets are simply passed on to children from their parents, the children would not work hard, because the children do not know the value of the assets.

Correct option is (d) product, promotion, price and place.

Criticisms of Say’s Law: 

The following are the criticisms against Say’s law:

1. According to Keynes, supply does not create its demand. It is not applicable where demand does not increase as much as production increases

2. Automatic adjustment process will not remove unemployment. Unemployment can be removed by increase in the rate of investment. 

3. Money is not neutral. Individuals hold money for unforeseen contingencies while businessmen keep cash reserve for future activities.

4. Say’s law is based on the proposition that supply creates its own demand and there is no over production. Keynes said that over production is possible.

5. Keynes regards full employment as a special case because there is under – employment in capitalist economies. 

6. The need for state intervention arises in the case of general over production and mass unemployment.

सही विकल्प है (A) सूचना प्रेषण

Short-term finance, usually, refers to finance required by a firm for a period of one year or less. It is the finance required for the purchase of raw materials, payment of wages and salaries and for meeting the other day – to -day manufacturing,-administrative, marketing and other expenses of a firm. It is called short-term finance, as it is required for a short period of one year or less.

The business finance is classified into three types. 

1. Short – Term Finance 

2. Medium – Term Finance 

3. Long – Term Finance.

When money is borrowed from banks or other financial institutions one needs to pay interest on it. The interest paid can be termed as the ‘cost of capital’.

(i) Loss = Rs 12

Explanation:

Given CP = Rs. 1265, SP = Rs. 1253

Loss = CP – SP

= Rs. (1265 – 1253)

= Rs. 12

(ii) C.P. = Rs 300

Explanation:

Given CP = 7, SP = Rs. 450, profit = Rs. 150

Profit = SP – CP

150 = 450 – CP

CP = Rs. (450 – 150)

= Rs. 300

(iii) Profit = Rs 4000

Explanation:

Given CP = Rs. 3355, SP = Rs. 7355,

Here SP > CP, so profit.

Profit = SP – CP

Profit = Rs. (7355 – 3355)

= Rs. 4000

(iv) C. P. = Rs 2395.50

Explanation:

Given CP = 7, SP = Rs. 2390, loss = Rs. 5.50

Loss = CP – SP

5.50 = CP – 2390 CP

= Rs. (5.50 + 2390)

= Rs. 2395.50

(i) Given CP = Rs. 4560, SP = Rs. 5000

Here, clearly SP > CP. So, profit.

Profit = SP – CP

= Rs. (5000 – 4560)

= Rs. 440

Profit % = {(Profit/CP) x 100} %

= {(440/4560) x 100} %

= {0.0965 x 100} %

Profit % = 9.65%

(ii) Given CP = Rs. 2600, SP = Rs. 2470.

Here, clearly CP > SP. So, loss.

Loss = CP – SP

= Rs. (2600 – 2470)

= Rs. 130

Loss % = {(Loss/CP) x 100} %

= {(130/2600) x 100} %

= {0.05 x 100} %

Loss % = 5%

(iii) Given CP = Rs. 332, SP= Rs. 350.

Here, clearly SP > CP. So, profit.

Profit = SP – CP

= Rs. (350 – 332)

= Rs. 18

Profit% = {(Profit/CP) x 100} %

= {(18/332) x 100} %

= {0.054 x 100} %

Profit % = 5.4%

(iv) Given CP = Rs. 1500, SP = Rs. 1500

Here clearly SP = CP.

So, neither profit nor loss.

We know that

(i) f (x) = cos 2x is a trigonometric function which is continuous

Hence, f (x) = cos 2x is continuous on [0, π]

(ii) f’(x) = – 2sin 2x exist in [0, π]

Hence, f (x) = cos 2x is differentiable on (0, π)

(iii) We know that

f (0) = cos 2(0) = 0

Similarly f (π) = cos 2(π) = 0

Here f (0) = f (π)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (0, π) where f’(c) = 0

– 2sin 2c = 0

Which gives 2c = π

We know that value of c = π/2 ϵ (0, π)

Therefore, Rolle’s Theorem is satisfied.

(i) Given CP = Rs. 4000, gain = Rs. 40

Gain % = {(Gain/CP) x 100) %

= {(40/4000) x 100} %

= (0.01 x 100) %

Gain % = 1%

(ii) Given SP = Rs. 1272, loss = Rs. 328

Loss = CP – SP

Hence, CP = Loss+ SP

= Rs. 328 + Rs. 1272

= Rs. 1600

Loss % = {(Loss/CP) x 100} %

= {(328/1600) x 100%

Loss % = 20.5%

(iii) Given SP = Rs. 1820, gain = Rs. 420

Gain = SP – CP

CP = 1820 – 420

= Rs. 1400

Gain % = {(Gain/CP) x 100} %

= {(420/1400) x 100 %

Gain % = 30%

Given Cost of 1 quintal of rice = Rs. 250

Cost of 600 quintals of rice = 600 x 250 = Rs. 150000

Overhead expenses = Rs. 1000

CP = Rs. (150000 + 1000) = Rs. 151000

Profit % = (Profit/CP) x 100

7 = (Profit /151000) x 100

Profit = 1510 x 7

Profit = Rs. 10570

Now SP = CP + profit

= Rs. (151000 + 10570)

SP = Rs. 161570

We know that

(i) f (x) = cos x is a trigonometric function which is continuous

Hence, f (x) = cos x is continuous on [-π/2, π/2]

(ii) f’(x) = – sin x exist in [-π/2, π/2]

Hence, f (x) = cos x is differentiable on (-π/2, π/2)

(iii) We know that

f (-π/2) = cos (-π/2) = 0

Similarly f (π/2) = cos (π/2) = 0

Here f (-π/2) = f (π/2)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (-π/2, π/2) where f’(c) = 0

– sin c = 0

Which gives c = 0

We know that value of c = 0 ϵ (-π/2, π/2)

Therefore, Rolle’s Theorem is satisfied.

We know that

(i) f (x) = (x – 2)⁴ (x – 3)³ is a polynomial which is continuous for all x ϵ R

Hence, f (x) = (x – 2)⁴ (x – 3)³ is continuous on [2, 3]

(ii) f’(x) = 4 (x – 2)³ (x – 3)³ + 3 (x – 2)⁴ (x – 3)² exist in [2, 3]

Hence, f (x) = (x – 2)⁴ (x – 3)³ is differentiable on (2, 3)

(iii) We know that

f (2) = (2 – 2)⁴ (2 – 3)³ = 0

Similarly f (3) = (3 – 2)⁴ (3 – 3)³ = 0

Here f (2) = f (3)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (2, 3) where f’(c) = 0

So we get

4 (c – 2)³ (c – 3)³ + 3 (c – 2)⁴ (c – 3)² = 0

We know that

(c – 2)³ (c – 3)² (7c – 18) = 0

Which gives c = 2 or 3 or 18/7

We know that value of c = 18/7 ϵ (2, 3)

Therefore, Rolle’s Theorem is satisfied.

We know that

(i) f (x) = (x – 1) (x – 2)² is a polynomial which is continuous for all x ϵ R

Hence, f (x) = (x – 1) (x – 2)² is continuous on [1, 2]

(ii) f’(x) = (x – 2)² + 2 (x – 1) (x – 2) exist in [1, 2]

Hence, f (x) = (x – 1) (x – 2)² is differentiable on (1, 2)

(iii) We know that

f (1) = (1 – 1) (1 – 2)² = 0

Similarly f (2) = (2 – 1) (2 – 2)² = 0

Here f (1) = f (2)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (1, 2) where f’(c) = 0

So we get

(c – 2)² + 2 (c – 1) (c – 2) = 0

We know that

(3c – 4) (c – 2) = 0

Which gives c = 2 or 4/3

We know that value of c = 4/3 ϵ (1, 2)

Therefore, Rolle’s Theorem is satisfied.

Condition (1):

Since, f(x) = x³ - 7x²+16x -12 is a polynomial and we know every polynomial function is continuous for all x ϵ R.

⇒ f(x) = x³ - 7x²+16x -12 is continuous on [2,3].

Condition (2):

Here, f’(x) = 3x² - 14x+16 which exist in [2,3].

So, f(x) = x³ - 7x²+16x -12 is differentiable on (2,3).

Condition (3):

Here, f(2) = 2³ - 7(2)²+16(2) -12 = 0

And f(3) = 3³ - 7(3)²+16(3) -12 = 0

i.e. f(2) = f(3)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one c ϵ (2,3) such that f’(c) = 0

i.e. 3c² - 14c+16 = 0

i.e. (c - 2)(3c - 7) = 0

i.e. c = 2 or c = 7 ÷ 3

Value of c = 7/3 ϵ (2, 3)

Thus, Rolle’s theorem is satisfied.

We know that

(i) f (x) = x³ + 3x² – 24x – 80 is a polynomial which is continuous for all x ϵ R

Hence, f (x) = x³ + 3x² – 24x – 80 is continuous on [- 4, 5]

(ii) f’(x) = 3x² + 6x – 24 exist in [- 4, 5]

Hence, f (x) = x³ + 3x² – 24x – 80 is differentiable on (- 4, 5)

(iii) We know that

f (- 4) = (- 4)³ + 3 (4)² – 24(4) – 80 = 0

Similarly f (5) = 5³ + 3(5)² – 24 (5) – 80 = 0

Here f (- 4) = f (5)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (- 4, 5) where f’(c) = 0

3c² + 6c – 24 = 0

Which gives c = – 4 or 2

We know that value of c = 2 ϵ (- 4, 5)

Therefore, Rolle’s Theorem is satisfied.

We know that

(i) f(x) = x² – 3x – 18 is a polynomial which is continuous for all x ϵ R

Hence, f(x) = x² – 3x – 18 is continuous on [-3, 6]

(ii) f’(x) = 2x – 3 exist in [-3, 6]

Hence, f(x) = x² – 3x – 18 is differentiable on (- 3, 6)

(iii) We know that

f(-3) = (-3)² – 3(-3) – 18 = 0

Similarly f(6) = 6² – 5(6) – 18 = 0

Here f(-3) = f(6)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (- 3, 6) where f’(c) = 0

2c – 3 = 0 which gives c = 3/2

We know that value of c = 3/2 ϵ (-3, 6)

Therefore, Rolle’s Theorem is satisfied.

We know that

(i) f (x) = x³ – 7x² + 16x – 12 is a polynomial which is continuous for all x ϵ R

Hence, f (x) = x³ – 7x² + 16x – 12 is continuous on [2, 3]

(ii) f’(x) = 3x² – 14x + 16 exist in [2, 3]

Hence, f (x) = x³ – 7x² + 16x – 12 is differentiable on (2, 3)

(iii) We know that

f (2) = (2)³ – 7 (2)² + 16(2) – 12 = 0

Similarly f (3) = 3³ – 7(3)² + 16 (3) – 12 = 0

Here f (2) = f(3)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (2, 3) where f’(c) = 0

3c² – 14c + 16 = 0

It can be written as

(c – 2) (3c – 7) = 0 which gives c = 7/3

We know that value of c = 7/3 ϵ (2, 3)

Therefore, Rolle’s Theorem is satisfied.

We know that

(i) f (x) = x (x – 4)² is a polynomial which is continuous for all x ϵ R

Hence, f (x) = x (x – 4)² is continuous on [0, 4]

(ii) f’(x) = (x – 4)² + 2x (x – 4) exist in [0, 4]

Hence, f (x) = x (x – 4)² is differentiable on (0, 4)

(iii) We know that

f (0) = 0(0 – 4)² = 0

Similarly f (4) = 4 (4 – 4)² = 0

Here f (0) = f(4)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (0, 4) where f’(c) = 0

(c – 4)² + 2c (c – 4) = 0

It can be written as

(c – 4) (3c – 4) = 0

Here c = 4 or c = 3/4

We know that value of c = 3/4 ϵ (0, 4)

Therefore, Rolle’s Theorem is satisfied.

We know that

(i) f (x) = x² – 4x + 3 is a polynomial which is continuous for all x ϵ R

Hence, f (x) = x² – 4x + 3 is continuous on [1, 3]

(ii) f’(x) = 2x – 4 exist in [1, 3]

Hence, f (x) = x² – 4x + 3 is differentiable on (1, 3)

(iii) We know that

f (1) = (1)² – 4(1) + 3 = 0

Similarly f (3) = 3² – 4(3) + 3 = 0

Here f (1) = f(3)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (1, 3) where f’(c) = 0

2c – 4 = 0 which gives c = 4/2 = 2

We know that value of c = 2 ϵ (1, 3)

Therefore, Rolle’s Theorem is satisfied.

Condition (1):

Since, f(x) = x² - x -12 is a polynomial and we know every polynomial function is continuous for all x ϵ R.

⇒ f(x) = x² - x -12 is continuous on [-3,4].

Condition (2):

Here, f’(x) = 2x -1 which exist in [-3,4].

So, f(x) = x² - x -12 is differentiable on (-3,4).

Condition (3):

Here, f(-3) = (-3)² - 3 -12 = 0

And f(4) = 4² - 4 -12 = 0

i.e. f(-3) = f(4)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one c ϵ (-3,4) such that f’(c) = 0

i.e. 2c-1 = 0

i.e. c = 1/2

Value of c = 1/2 ϵ (-3,4)

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x) = x² is a polynomial and we know every polynomial function is continuous for all x ϵ R.

⇒ f(x) = x² is continuous on [-1,1].

Condition (2):

Here, f’(x) = 2x which exist in [-1,1].

So, f(x) = x² is differentiable on (-1,1).

Condition (3):

Here, f(-1) = (-1)² = 1

And f(1) = 1¹ = 1

i.e. f(-1) = f(1)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one c ϵ (-1,1) such that f’(c) = 0

i.e. 2c = 0

i.e. c = 0

Value of c = 0 ϵ (-1,1)

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x) = x(x - 4)² is a polynomial and we know every polynomial function is continuous for all x ϵ R.

⇒ f(x) = x(x - 4)² is continuous on [0,4].

Condition (2):

Here, f’(x) = (x - 4)²+2x(x - 4) which exist in [0,4].

So, f(x) = x(x - 4)² is differentiable on (0,4).

Condition (3):

Here, f(0) = 0(0 - 4)² = 0

And f(4) = 4(4 - 4)² = 0

i.e. f(0) = f(4)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one c ϵ (0,4) such that f’(c) = 0

i.e. (c - 4)²+2c(c - 4) = 0

i.e. (c - 4)(3c - 4) = 0

i.e. c = 4 or c = 3 ÷ 4

Value of c = 3/4 ϵ (0, 4)

Thus, Rolle’s theorem is satisfied.

Answer is: 3/√1-x²

Given that y = sin^-1(3x - 4x³)

Let x = sin θ

⇒ θ = sin^-1x

Then, y = sin^-1(3sinθ - 4sin³θ)

Using sin3θ = 3sinθ - 4sin 3θ, we get

y = sin^-1(sin3θ) = 3θ = 3sin^-1x

Differentiating with respect to x, we get

dy/dx = 3/√1-x²

Condition (1):

Since, f(x) = x² - 4x+3 is a polynomial and we know every polynomial function is continuous for all x ϵ R.

⇒ f(x) = x² - 4x+3 is continuous on [1,3].

Condition (2):

Here, f’(x) = 2x - 4 which exist in [1,3].

So, f(x) = x² - 4x+3 is differentiable on (1,3).

Condition (3):

Here, f(1) = (1)² - 4(1)+3 = 0

And f(3) = (3)² - 4(3)+3 = 0

i.e. f(1) = f(3)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one c ϵ (1,3) such that f’(c) = 0

i.e. 2c - 4 = 0

i.e. c = 2

Value of c = 2 ϵ (1,3)

Thus, Rolle’s theorem is satisfied.

(a) The forecast temperature was same as the actual temperature on Tuesday, Friday, and Sunday.

(b) The maximum forecast temperature during the week was 35°C.

(c) The minimum actual temperature during the week was 15°C.

(d) The actual temperature differs the most from the forecast temperature on Thursday.

Correct Answer is: (a, b, & c)

The velocity of efflux = v = √2gy.

The emerging liquid moves as a projectile and reaches the ground in time t, where, 

H - y = 1/2 gt² or  t =√(2(H - y))/g

∴ x = vt= 2 √(H - y)y.

For x to be maximum,dx/dy = 0 or y = H/2

∴ x_max = H. 

B) Algae and fungi

Increasing order of atomic mass and similarity in chemical properties i.e. the nature of oxide and hydride formed.

(i) Biomass is a renewable source of energy only if we plant trees in a planned manner which is not the case with hydroelectricity.
(ii) The energy from bio-mass can be obtained by using a chulha or a gobar gas plant whereas hydroelectricity requires construction of dams on rivers.
(iii) Biomass provides pollution-free energy only, when converted into bio-gas whereas hydroelectricity.
is totally pollution-fi'ee.

The stranger seems to be very simple and foolish. He cannot understand Taffy’s body language and the non-verbal communication. He interprets Tegumai’s behavior too in the wrong manner. He seems to be in awe of Tegumai and Taffy, thinking that Tegumai is a haughty chief.

1. B = \(\frac{\mu_0/a^2}{2r^3}\)

2.

• Electric field due to electric dipole,

E = \(\frac{1}{4\pi ε_0}\frac{P}{r^3}\) magnetic field due to magnetic dipole

B = \(\frac{\mu_0}{4\pi}\frac{m}{x^3}\)

• Magnetic moment m = IA

Helen looked forward to her second year at Gilman’s school. However, she was confronted with unexpected difficulties that year which caused her a great deal of frustration. She had to study mathematics without the needed tools. The classes were larger and it was not possible for the Cambridge teachers to give her special instructions. Anne Sullivan had to read all the books to her. Helen had to wait in order to buy a Braille writer so that she could do her algebra, geometry and physics. When the embossed books and the other apparatus arrived, Helen’s difficulties began to disappear and she began to study with confidence. However, Mr. Gilman thought that Helen was overworked and was breaking down. He insisted that she was overworked, and that she should remain at his school three years longer. He made changes in her studies. A difference of opinion between Mr. Gilman and Miss Sullivan resulted in Helen’s mother withdrawing Helen and Mildred from the Cambridge school. Helen went on to continue her studies under a tutor. Helen found it easier to study with a tutor than receive instructions in class. When Helen took her exam in June 1899, she faced many difficulties, as the administrative board of Radcliffe did not realize how difficult they were making her examinations. They did not understand the peculiar difficulties Helen had to go through. However, Helen, with her great and determination, overcame them all.

Correct Option Is (C) mass of the body

\(\theta = \sigma AT^4\)

Where,

\(\theta = \) Energy Radiated

A = Area

T = Temperature

\(\sigma =\) Conductivity

(C) mass of the body

Correct Option is (C) 6166 K

Given, \(\lambda m = 470 \ nm\)

T = ?

\(\lambda m T = b\)

\(T = \frac{b}{\lambda m}\)

\(T = \frac{2.898 \times 10^{-3}}{470 \ nm}\)

\(T = \frac{2.898 \times 10^{-3}}{470 \times 10^{-9}}\)

\(T = 0.0061659 \times 10^6\)

T = 6165 K

Correct option is (C) 6166 K

Correct option is (D) dry air.

Correct option is (C) Quartz

Correct Option is (C) Benzene

Benzene is a very stable hudrocarbon . Hence very opaque to radiant heat or thermal radiation.

Correct option is (C) Benzene

Correct answer is: (B) 16 : 1

The degrees of freedom depend upon 

(i) the number of atoms forming a molecule 

(ii) the structure of the molecule 

(iii) the temperature of the gas.

 Correct answer is: (C) \(\cfrac{1}{\sqrt{2\pi nd^2}}\)

Correct answer is: (B) 0.04

Correct answer is: (A) only kinetic energy

Data: a = 0.4, R_b = 5 × 10⁴ W/m²

As the emissivity, e = a, 

we have, e = 0.4

∴ R = eR_b = (0.4) (5 × 10⁴ W/m²)

= 2 × 10⁴ W/m²

This is the required quantity.

In the usual notation, a + r +1 = 1 

Hence, the coefficient of transmission of the body = 1 – (a + r) = 1 – (0.2 + 0.5) = 1 – 0.7 = 0.3.

The three modes of heat transfer are conduction, convection and radiation.

1. Conduction is the mode of heat transfer within a body or between two bodies in contact, from a region of high temperature to a region of lower temperature without the migration of the particles of the medium.

2. Convection is the mode of heat transfer from one part of a fluid to another by the migration of the particles of the fluid. 

3. Radiation is the mode of heat transfer by electromagnetic waves / quanta.

An ideal or perfect gas is a gas which obeys the gas laws (Boyle’s law, Charles’ law and Gay-Lussac’s law) at all pressures and temperatures. An ideal gas cannot be liquefied by application of pressure or lowering the temperature.

A molecule of an ideal gas is an ideal particle having only mass and velocity. Its structure and size are ignored. Also, intermolecular forces are zero except during collisions

1. A substance which is largely opaque to thermal radiations, i.e., a substance which does not transmit heat radiations incident on it, is known as an athermanous substance.

2. A substance through which heat radiations can pass is known as a diathermanous substance.

Answer is (a) 1