UPSC NDA-2 Archives

11. Match List-I with List-II and select the correct answer using the code

Match List-I with List-II and select the correct answer using the code given below the Lists :

List-I (Mahamatta)	List-II (Function)
A. Anta-mahamatta	1. Women’s welfare
B. Ithijhakha-mahamatta	2. Spread of Dhamma
C. Dhamma-mahamatta	3. Associated with city administration
D. Nagalaviyohalaka-mahamatta	4. In-charge of frontier areas

Code :

A-3, B-2, C-1, D-4

A-3, B-1, C-2, D-4

A-4, B-1, C-2, D-3

A-4, B-2, C-1, D-3

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The correct option is C: A-4, B-1, C-2, D-3.

During the Mauryan Empire, particularly under Emperor Ashoka, various officials known as Mahamattas were appointed to oversee different aspects of administration and implement the emperor’s policies. The question asks to match specific types of Mahamattas with their functions:
– **Anta-mahamatta:** Anta means end or border. These officials were in charge of the frontier areas of the empire. (Match A with 4)
– **Ithijhakha-mahamatta:** Also known as Stri-adhyaksha Mahamatta, these officials were responsible for supervising and promoting the welfare of women. (Match B with 1)
– **Dhamma-mahamatta:** These officials were specially appointed by Ashoka to propagate Dhamma, ensure its observance, and look after the welfare of different religious and social groups. (Match C with 2)
– **Nagalaviyohalaka-mahamatta:** These officials were judicial officers appointed in the cities (Nagara) to administer justice. (Match D with 3)

Ashoka appointed Dhamma-mahamattas in the 14th year of his reign. Their role was crucial in implementing Ashoka’s policy of Dhamma, which emphasized social responsibility, tolerance, and welfare. Other types of officials existed in the Mauryan administration, but these specific Mahamattas played distinct roles as indicated in Ashokan inscriptions.

12. Which one of the following is not a power of Panchayats under Articl

Which one of the following is not a power of Panchayats under Article 243G?

Land improvement

Implementation of land reforms

Land consolidation and soil conservation

Regulation of land revenue

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The correct option is D, Regulation of land revenue.

Article 243G of the Constitution of India enumerates the powers, authority, and responsibilities of Panchayats, enabling them to function as institutions of self-government. It states that the Legislature of a State may endow the Panchayats with powers and authority necessary to enable them to function as institutions of self-government and to prepare plans for economic development and social justice and implement schemes entrusted to them, including those related to the matters listed in the Eleventh Schedule.
The Eleventh Schedule lists 29 matters. Items 1 and 2 of the Eleventh Schedule are:
1. Agriculture, including extension.
2. Land improvement, implementation of land reforms, land consolidation and soil conservation.
Options A, B, and C (Land improvement, Implementation of land reforms, Land consolidation and soil conservation) are all explicitly listed in the Eleventh Schedule, falling under the purview of Panchayats as per Article 243G.
Regulation of land revenue, including assessment and collection of land tax, is typically a function performed by the state government’s revenue administration, although Panchayats might be involved in record-keeping or assistance in collection in some areas depending on state laws, the primary regulatory and revenue collection power for land revenue generally does not rest with the Panchayats under the Eleventh Schedule.

The 73rd Constitutional Amendment Act, 1992, which inserted Part IX (The Panchayats) and the Eleventh Schedule into the Constitution, aimed to decentralize power and strengthen local self-government. While Panchayats are empowered in many areas related to local infrastructure, welfare, and resource management as listed in the Eleventh Schedule, major revenue-generating powers like regulating land revenue assessment and collection broadly remain with the state government.

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13. In which one of the following devices, the light energy is converted i

In which one of the following devices, the light energy is converted into the electrical energy?

Light-emitting diode

Laser diode

Solar cell

Transistor

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The correct option is C, the Solar cell.

The question asks for a device where light energy is converted into electrical energy. This process is known as the photovoltaic effect.
– Light-emitting diode (LED) converts electrical energy into light energy.
– Laser diode also converts electrical energy into coherent light energy.
– Solar cell (photovoltaic cell) is specifically designed to absorb photons from sunlight and convert their energy into an electric current.
– Transistor is a semiconductor device used as an electronic switch or amplifier; it controls the flow of electrical current but does not primarily convert light energy to electrical energy (though some transistors can be light-sensitive, their main function isn’t energy conversion).
Therefore, the solar cell is the device that converts light energy into electrical energy.

Solar cells are the fundamental components of solar panels, which are widely used for generating renewable electricity. They are made of semiconductor materials, typically silicon, which release electrons when struck by photons of light, creating an electric current. This phenomenon is the basis of solar power generation.

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14. A simple pendulum having bob of mass m and length of string l has time

A simple pendulum having bob of mass m and length of string l has time period of T. If the mass of the bob is doubled and the length of the string is halved, then the time period of this pendulum will be

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T /√2

√2 T

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The correct option is B, stating that the new time period will be T/√2.

The time period (T) of a simple pendulum is given by the formula T = 2π√(l/g), where l is the length of the string and g is the acceleration due to gravity. This formula shows that the time period depends only on the length of the pendulum and the acceleration due to gravity, and it is independent of the mass of the bob.
In the given problem, the initial time period is T for a pendulum with mass m and length l. When the mass of the bob is doubled (to 2m) and the length of the string is halved (to l/2), the new time period T’ will be T’ = 2π√((l/2)/g) = 2π√(l/(2g)). We can rewrite this as T’ = 2π√(l/g) * (1/√2). Since T = 2π√(l/g), the new time period T’ is T/√2.

The independence of the time period from the mass of the bob is a key characteristic of a simple pendulum, provided the amplitude of oscillation is small (usually less than 15 degrees). This property allows pendulums to be used as reliable timekeeping devices. Real-world factors like air resistance and the rigidity of the string can affect the period slightly, but for an ideal simple pendulum, mass is irrelevant.

15. The volume of a sealed packet is 1 litre and its mass is 800 g. The pa

The volume of a sealed packet is 1 litre and its mass is 800 g. The packet is first put inside water with density 1 g cm^-3 and then in another liquid B with density 1.5 g cm^-3. Then which one of the following statements holds true?

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The packet will float in both water and liquid B.

The packet will sink in both water and liquid B.

The packet will sink in water but will float in liquid B.

The packet will float in water and sink in liquid B.

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The correct option is A, stating that the packet will float in both water and liquid B.

The principle of buoyancy states that an object will float in a fluid if its density is less than the density of the fluid. If its density is greater, it will sink. The packet has a volume of 1 litre (1000 cm³) and a mass of 800 g. Its density is mass/volume = 800 g / 1000 cm³ = 0.8 g/cm³. The density of water is 1 g/cm³. Since 0.8 g/cm³ < 1 g/cm³, the packet will float in water. The density of liquid B is 1.5 g/cm³. Since 0.8 g/cm³ < 1.5 g/cm³, the packet will also float in liquid B.

When an object floats, the buoyant force exerted by the fluid is equal to the weight of the object. The buoyant force is also equal to the weight of the fluid displaced by the object. If the object sinks, the buoyant force is less than the weight of the object. The proportion of the object submerged when floating is equal to the ratio of the object’s density to the fluid’s density. In water, 0.8/1 = 80% of the packet’s volume will be submerged. In liquid B, 0.8/1.5 ≈ 53.3% of the packet’s volume will be submerged.

16. The refractive index of crown glass is close to 3/2. If the speed of l

The refractive index of crown glass is close to 3/2. If the speed of light in air is c, then the speed of light in the crown glass will be close to

(3/2)c

(4/9)c

(2/3)c

(9/4)c

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The refractive index (n) of a medium is defined as the ratio of the speed of light in a vacuum (or approximately in air) to the speed of light in that medium (v). The formula is n = c / v, where c is the speed of light in vacuum/air.
Given that the refractive index of crown glass is close to 3/2, we have n = 3/2.
Using the formula n = c / v, we can rearrange it to solve for the speed of light in the glass (v): v = c / n.
Substituting the given refractive index: v = c / (3/2).
To divide by a fraction, we multiply by its reciprocal: v = c * (2/3) = (2/3)c.

The refractive index indicates how much the speed of light is reduced when it passes through a medium compared to its speed in a vacuum. A higher refractive index means a lower speed of light in the medium.

The speed of light is highest in a vacuum (approximately 3 x 10⁸ m/s) and slows down when it enters any medium, causing light to bend (refract). The refractive index is a dimensionless quantity.

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17. Which one of the following terms cannot represent electrical power in

Which one of the following terms cannot represent electrical power in a circuit?

I^2/R

I^2R

V^2/R

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Electrical power (P) in a circuit can be calculated using different formulas derived from the relationship between voltage (V), current (I), and resistance (R), as defined by Ohm’s Law (V = IR). The standard formulas for power are:
1. P = VI (Power = Voltage × Current)
2. P = I²R (Substituting V = IR into P = VI: P = (IR)I = I²R)
3. P = V²/R (Substituting I = V/R into P = VI: P = V(V/R) = V²/R)
Looking at the options:
A) VI – This is a correct formula for power.
B) I²/R – This is incorrect. The correct formula involving I and R is I²R.
C) I²R – This is a correct formula for power.
D) V²/R – This is a correct formula for power.

Understanding the relationships between power, voltage, current, and resistance through Ohm’s Law is fundamental to electrical circuit analysis.

The power dissipated by a resistor represents the rate at which electrical energy is converted into heat energy. These power formulas are applicable to resistive components in both DC and AC circuits (though for AC, power can also involve a power factor).

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18. Two convex lenses have focal lengths of 50 cm and 25 cm, respectively.

Two convex lenses have focal lengths of 50 cm and 25 cm, respectively. If these two lenses are placed in contact, then the net power of this combination will be equal to

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+2 dioptre

+6 dioptre

-6 dioptre

+3 dioptre

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The power of a lens (P) is the reciprocal of its focal length (f) in meters (P = 1/f). For a combination of thin lenses in contact, the total power (P_net) is the sum of the individual powers (P_net = P1 + P2 + …).
Given focal lengths: f1 = 50 cm = 0.5 m and f2 = 25 cm = 0.25 m. Both are convex lenses, so focal lengths are positive.
Power of the first lens: P1 = 1 / f1 = 1 / 0.5 m = +2 Dioptre (D).
Power of the second lens: P2 = 1 / f2 = 1 / 0.25 m = +4 Dioptre (D).
Net power of the combination: P_net = P1 + P2 = +2 D + +4 D = +6 D.

The power of a lens is a measure of its ability to converge or diverge light, and for lenses in contact, powers are additive.

The unit of power is the Dioptre (D), defined as the reciprocal of the focal length in meters. Convex lenses have positive power (converging), and concave lenses have negative power (diverging).

19. In January 2020, the administration of which of the following Union Te

In January 2020, the administration of which of the following Union Territories has been merged together?

Daman and Diu and Puducherry

Puducherry and Dadra and Nagar Haveli

Puducherry and Andaman and Nicobar Islands

Dadra and Nagar Haveli and Daman and Diu

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In January 2020, the two Union Territories of Dadra and Nagar Haveli and Daman and Diu were merged into a single Union Territory named Dadra and Nagar Haveli and Daman and Diu. The merger was enacted through the Dadra and Nagar Haveli and Daman and Diu (Merger of Union Territories) Act, 2019, passed by the Parliament of India.

This merger aimed to improve administrative efficiency and streamline services for the citizens of these two previously separate Union Territories.

The Act came into effect on January 26, 2020. With this merger, the number of Union Territories in India reduced from nine to eight (at that time, before the reorganization of Jammu and Kashmir). The capital of the new merged Union Territory is Daman.

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20. Which of the following statements is/are correct? 1. The Earth’s cru

Which of the following statements is/are correct?

1. The Earth’s crust is brittle in nature.
2. The mean thickness of the oceanic crust is 15 km whereas that of the continental crust is around 30 km.

Select the correct answer using the code given below.

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1 only

2 only

Both 1 and 2

Neither 1 nor 2

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Let’s analyze each statement:
1. The Earth’s crust is brittle in nature: This statement is correct. The upper part of the Earth’s crust behaves in a brittle manner, meaning it tends to break and fracture under stress rather than deform plastically. This brittle behavior is responsible for phenomena like faulting and earthquakes.
2. The mean thickness of the oceanic crust is 15 km whereas that of the continental crust is around 30 km: This statement is incorrect. The oceanic crust is significantly thinner than the continental crust. The mean thickness of the oceanic crust is typically around 5-10 km, while the mean thickness of the continental crust is around 30-40 km (and can be up to 70 km under major mountain ranges).

The Earth’s crust is the outermost solid shell of the planet and exhibits brittle characteristics, particularly near the surface. Oceanic and continental crusts differ significantly in their thickness and composition.

The oceanic crust is primarily composed of basalt and gabbro and is denser than the continental crust, which is mainly composed of granitic and andesitic rocks. This difference in density and thickness is fundamental to plate tectonics and the formation of Earth’s surface features.