21. Which of the following Committees does NOT consist of Members from the

Which of the following Committees does NOT consist of Members from the Upper House of the Parliament ?

  • 1. Public Accounts Committee
  • 2. Estimates Committee
  • 3. Committee on Public Undertakings

Select the answer using the code given below :

1 and 2
1 and 3
2 only
3 only
This question was previously asked in
UPSC NDA-2 – 2024
The correct answer is 2 only.
The Estimates Committee consists exclusively of members from the Lok Sabha. It has 30 members, all of whom are elected by the Lok Sabha from amongst its members every year by proportional representation. The Public Accounts Committee (PAC) and the Committee on Public Undertakings (COPU) both consist of members from both Lok Sabha and Rajya Sabha (15 from Lok Sabha and 7 from Rajya Sabha for each).
These three committees are financial committees of the Parliament. The Estimates Committee scrutinizes the estimates included in the budget and suggests economies in public expenditure. The PAC examines the annual audit reports of the Comptroller and Auditor General (CAG), while COPU examines the reports of the CAG on public undertakings and also the working of public sector undertakings.

22. Consider the following statements: 1. The duties of the Attorney Gen

Consider the following statements:

  • 1. The duties of the Attorney General of India are to give advice to the Government of India upon such legal matters, and to perform such other duties of a legal character, as may from time to time be referred to or assigned to him by the President of India
  • 2. The Solicitor General of India is the secondary Law Officer of the country, who assists the Attorney General, and is himself assisted by several Additional Solicitor Generals of India

Which of the statements given above is/are correct ?

1 only
2 only
Both 1 and 2
Neither 1 nor 2
This question was previously asked in
UPSC NDA-2 – 2024
Both 1 and 2 are correct.
Statement 1 is correct. Article 76 of the Constitution of India defines the duties of the Attorney General, which include giving advice to the Government of India on legal matters referred by the President and performing other legal duties assigned by the President. Statement 2 is also correct as it accurately describes the functional hierarchy and roles within the Law Officers of the Union Government. The Solicitor General is the second-highest law officer and assists the Attorney General, and they are assisted by Additional Solicitor Generals. While the posts of SG and Addl. SG are statutory/executive, their description of function is accurate.
The Attorney General is the highest law officer of the country and is appointed by the President. The Solicitor General and Additional Solicitor Generals are appointed by the government to assist the Attorney General in their duties. The Attorney General is the only one of these posts created by the Constitution itself (Article 76).

23. In which Schedule of the Constitution of India, provisions as to disqu

In which Schedule of the Constitution of India, provisions as to disqualification on grounds of defection are given?

Ninth Schedule
Tenth Schedule
Eleventh Schedule
Sixth Schedule
This question was previously asked in
UPSC NDA-2 – 2024
The correct answer is the Tenth Schedule.
The Tenth Schedule of the Constitution of India contains provisions regarding the disqualification of Members of Parliament and State Legislatures on the grounds of defection. It was added to the Constitution by the 52nd Amendment Act, 1985, and is popularly known as the ‘Anti-Defection Law’.
The other schedules mentioned relate to different aspects: the Ninth Schedule was added to protect certain laws from judicial review; the Eleventh Schedule contains provisions regarding Panchayats; and the Sixth Schedule contains provisions for the administration of tribal areas in the states of Assam, Meghalaya, Tripura, and Mizoram.

24. Domingo Paes describes a king in these words: ‘Of medium height, and o

Domingo Paes describes a king in these words: ‘Of medium height, and of fair complexion and good figure, rather fat than thin; he has on his face signs of smallpox’. Identify the king from the options given below.

Krishnadeva Raya
Ram Raya
Devaraya II
Devaraya I
This question was previously asked in
UPSC NDA-2 – 2024
The correct answer is Krishnadeva Raya.
The description provided is from the account of Domingo Paes, a Portuguese traveller who visited the Vijayanagara Empire during the reign of Krishnadeva Raya (1509-1529). Paes provided vivid descriptions of the king’s appearance, court, and the empire’s prosperity.
Krishnadeva Raya was the most famous ruler of the Tuluva dynasty of Vijayanagara. His reign is considered the zenith of the empire. Domingo Paes and another Portuguese traveler, Duarte Barbosa, provide valuable historical accounts of this period, describing the grandeur of the city and the efficiency of the administration.

25. Vijayanagara rulers claimed that they were ruling on behalf of

Vijayanagara rulers claimed that they were ruling on behalf of

Lord Murugan
Lord Venkateshwara
Shri Virupaksha
Shri Vallabha
This question was previously asked in
UPSC NDA-2 – 2024
The correct answer is Shri Virupaksha.
Vijayanagara rulers claimed to rule on behalf of the deity Shri Virupaksha. The Virupaksha Temple in Hampi was central to the empire’s identity, and the rulers performed worship and ceremonies associating themselves with the divine authority vested in Virupaksha.
The Vijayanagara Empire (14th-17th centuries) was a major power in South India. Its rulers fostered a close relationship with religious institutions, particularly the Virupaksha temple, which was the patron deity of the capital city, Vijayanagara (modern Hampi). This practice of ruling on behalf of a deity was common in South Indian kingdoms, lending legitimacy and religious sanction to the rulers’ authority.

26. The figure given below shows the direction of the two forces P and Q a

The figure given below shows the direction of the two forces P and Q acting on a skydiver :

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Force P is caused by the gravity and force Q is caused by the friction
When the force P is bigger than the force Q, the speed of the skydiver remains the same
After the parachute opens, force P remains the same while force Q increases
After the parachute opens, force P decreases while force Q increases
This question was previously asked in
UPSC NDA-2 – 2024
The figure shows two forces acting on a skydiver: Force P downwards and Force Q upwards. In the context of a skydiver falling through the air, the dominant downward force (P) is gravity (weight = mg), and the dominant upward force (Q) is air resistance (drag).
Let’s analyze the options:
A) Force P is caused by the gravity and force Q is caused by the friction. This statement correctly identifies the source of the forces. P is gravity, and Q is air resistance, which is a form of fluid friction. This statement is correct.
B) When the force P is bigger than the force Q, the speed of the skydiver remains the same. If P > Q, there is a net downward force (P – Q). According to Newton’s Second Law (F_net = ma), this net force causes acceleration in the downward direction, meaning the speed will increase, not remain the same. This statement is incorrect.
C) After the parachute opens, force P remains the same while force Q increases. Force P is gravity (mg), which depends on the skydiver’s mass and the acceleration due to gravity. Opening a parachute does not significantly change the skydiver’s mass or gravity. Thus, P remains essentially the same. Air resistance (Q) depends on the skydiver’s speed, the density of the air, and the skydiver’s shape and size (drag coefficient and area). Opening a parachute dramatically increases the surface area and drag coefficient, causing the air resistance force Q to increase significantly at the same speed. This statement is correct.
D) After the parachute opens, force P decreases while force Q increases. As explained above, P (gravity) remains essentially the same. Q increases dramatically. This statement is incorrect because it says P decreases.

Both A and C are factually correct statements about the forces. However, MCQs typically have a single best answer. Option C describes a crucial dynamic event in skydiving (parachute deployment) and its direct impact on the forces and resulting motion, which is a common physics concept tested. Option A is a static identification of the forces. Given the options and the nature of physics questions regarding skydiving, Option C is likely considered the intended answer as it addresses a key change in the system’s dynamics.

The forces acting on a skydiver are primarily gravity (downwards) and air resistance (upwards). Gravity is constant (for a constant mass). Air resistance depends on speed, shape, and size. Opening a parachute significantly increases air resistance.
When falling, a skydiver reaches a terminal velocity when the air resistance force (Q) equals the gravitational force (P), resulting in zero net force and zero acceleration (constant speed). Opening a parachute increases Q dramatically, making Q temporarily much larger than P, causing rapid deceleration until a new, much lower terminal velocity is reached.

27. Given below are the four cases in which certain heat transfer is takin

Given below are the four cases in which certain heat transfer is taking place :

  • 1. Ice is melting in a glass full of water
  • 2. Water is boiling in an open container
  • 3. A metal rod is heated in a furnace
  • 4. A cup of coffee is allowed to cool on a table

In which of the above cases, the Newton’s Law of Cooling is applicable?

1 only
4 only
1 and 4 only
1, 2 and 3
This question was previously asked in
UPSC NDA-2 – 2024
Newton’s Law of Cooling states that the rate of heat loss from an object is proportional to the temperature difference between the object and its surroundings. This law is generally applicable when heat transfer is primarily by convection and radiation, and the temperature difference is relatively small.
1. Ice melting: Involves a phase change at a constant temperature (0°C). The rate of heat absorption is constant if the surroundings are at a constant temperature, but the process is melting, not cooling according to the temperature-difference proportionality of Newton’s law.
2. Water boiling: Involves a phase change at a constant temperature (100°C at atmospheric pressure). This is a process of heating (adding heat to cause vaporization), not cooling.
3. Metal rod heated in a furnace: This is a process of heating the rod by transferring heat *from* the furnace *to* the rod. Newton’s Law of Cooling describes heat *loss* (cooling).
4. Cup of coffee cooling on a table: The hot coffee loses heat to the cooler surroundings (air and table) via convection and radiation, causing its temperature to decrease. This scenario directly fits the conditions and description of Newton’s Law of Cooling, especially as the temperature difference between the coffee and the surroundings decreases over time.
Newton’s Law of Cooling applies to the cooling of an object where the rate of heat loss is proportional to the temperature difference between the object and its surroundings. It is typically applicable for heat transfer by convection and radiation and for relatively small temperature differences.
The formula for Newton’s Law of Cooling is often given as dT/dt = -k(T – T_s), where T is the temperature of the object, T_s is the temperature of the surroundings, t is time, and k is a positive constant. This shows that the rate of temperature change (cooling if T > T_s) is proportional to the temperature difference (T – T_s).

28. A block of mass 2·0 kg slides on a rough horizontal plane surface. Let

A block of mass 2·0 kg slides on a rough horizontal plane surface. Let the speed of the block at a particular instant is 10 m/s. It comes to rest after travelling a distance of 20 m. Which one of the following could be the magnitude of the frictional force ?

10 N
20 N
40 N
50 N
This question was previously asked in
UPSC NDA-2 – 2024
Given mass m = 2.0 kg, initial velocity u = 10 m/s, final velocity v = 0 m/s, distance s = 20 m. The block is brought to rest by the frictional force. We can calculate the work done by friction using the work-energy theorem, which states that the net work done on an object is equal to its change in kinetic energy. Work done by friction (W_friction) = Change in KE = (1/2)mv² – (1/2)mu². W_friction = (1/2)(2.0 kg)(0 m/s)² – (1/2)(2.0 kg)(10 m/s)² = 0 – 100 J = -100 J. Since the frictional force (F_friction) opposes the motion, the work done by friction is -F_friction * s. So, -F_friction * 20 m = -100 J. This gives F_friction = 100 J / 20 m = 5 N.

However, 5 N is not among the options. Let’s check the options against possible work done. If the force was 10N, the work done by friction over 20m would be -10N * 20m = -200 J. This would correspond to a change in KE of -200 J, meaning the initial KE would have to be 200 J. (1/2) * 2kg * u² = 200 J => u² = 200 => u = sqrt(200) ≈ 14.14 m/s.
If the distance was 10m instead of 20m, and the force was 10N, the work done would be -10N * 10m = -100J, matching the initial KE calculated from u=10m/s.
Given the discrepancy between the calculated value (5 N) and the provided options, and the common structure of such problems, it is highly probable there is a typo in the question’s distance value, and it was intended to be 10 m, leading to a force of 10 N (Option A). Alternatively, the options might be based on a different initial speed or mass. Assuming the question and options are from a single source and intended to have a correct answer among the choices, and that Option A is the intended answer, the underlying scenario would involve a force of 10N. While the provided data leads to 5N, Option A (10N) is the most plausible intended answer given the multiple choices.

The work-energy theorem states that the net work done on an object equals its change in kinetic energy. Frictional force does negative work as it opposes motion.
The magnitude of kinetic friction is often modeled as F_friction = μ_k * N, where μ_k is the coefficient of kinetic friction and N is the normal force. For a horizontal surface, N = mg.

29. The AC mains domestic supply current in India changes direction in

The AC mains domestic supply current in India changes direction in every

50 s
$ rac{1}{50}$ s
100 s
$ rac{1}{100}$ s
This question was previously asked in
UPSC NDA-2 – 2024
The standard frequency of AC mains supply in India is 50 Hz. The frequency (f) is the number of complete cycles per second. The time period (T) of one complete cycle is given by T = 1/f. For f = 50 Hz, T = 1/50 s. In one complete cycle of alternating current, the current starts from zero, reaches a maximum in one direction, passes through zero, reaches a maximum in the opposite direction, and returns to zero. The current changes direction twice within one complete cycle (at the points where it crosses the zero line). Thus, the time taken for the current to change direction is half the time period. Time for direction change = T/2 = (1/50 s) / 2 = 1/100 s.
The frequency of AC supply is the number of cycles per second. The time period is the duration of one cycle (T = 1/f). AC current changes direction twice in every complete cycle.
In many other countries, such as the USA, the standard AC mains frequency is 60 Hz. In a 60 Hz system, the current changes direction every 1/120 s.

30. Which one among the following is the correct focal length of a combina

Which one among the following is the correct focal length of a combination of lenses of power 2·5 D and –2·0 D ?

+0·5 m
–0·5 m
+2·0 m
–2·0 m
This question was previously asked in
UPSC NDA-2 – 2024
For thin lenses in contact, the total power of the combination (P_total) is the algebraic sum of the individual powers (P₁, P₂). Given P₁ = 2.5 D and P₂ = -2.0 D, the total power is P_total = P₁ + P₂ = 2.5 D + (-2.0 D) = 0.5 D. The focal length (f) of a lens or lens combination is the reciprocal of its power (P), with power in dioptres (D) and focal length in meters (m). So, f_total = 1 / P_total = 1 / 0.5 D = 1 / (0.5 m⁻¹) = 2 m. The focal length is positive, indicating a converging lens combination.
Power of a lens (P) is the reciprocal of its focal length (f) in meters (P = 1/f). The power of a combination of thin lenses in contact is the sum of their individual powers. Positive power indicates a converging lens, and negative power indicates a diverging lens.
The unit of power is the dioptre (D), which is equal to m⁻¹. A combination of a converging lens (+P) and a diverging lens (-P’) can result in a net converging or diverging combination depending on the relative magnitudes of their powers.


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