21. If four balloons A, B, C and D are filled with hydrogen, oxygen, heliu

If four balloons A, B, C and D are filled with hydrogen, oxygen, helium and nitrogen gases respectively and left in air, which balloon reaches to the highest distance from the Earth?

Balloon A
Balloon B
Balloon C
Balloon D
This question was previously asked in
UPSC CAPF – 2012
A balloon filled with a gas lighter than air will rise. The higher it will reach (or the more buoyant it will be) depends on the density of the gas relative to the density of the surrounding air. The lighter the gas, the greater the buoyant force. The average molecular weight of air is approximately 29 g/mol.
Comparing the molecular/atomic weights of the gases: Hydrogen (H₂) ≈ 2 g/mol, Oxygen (O₂) ≈ 32 g/mol, Helium (He) ≈ 4 g/mol, Nitrogen (N₂) ≈ 28 g/mol. Oxygen is heavier than air and will sink. Nitrogen is slightly lighter than air. Hydrogen and Helium are significantly lighter than air. Hydrogen is the lightest among the given gases.
Since Hydrogen is the lightest gas among the options and significantly lighter than air, a balloon filled with hydrogen will experience the greatest buoyant force and is expected to rise to the highest altitude compared to balloons filled with Oxygen, Helium, or Nitrogen (assuming balloon material weight and other factors are equal).

22. To start your loaded trolley bag, you exert more force than when it is

To start your loaded trolley bag, you exert more force than when it is moving. This is an example of

first law of thermodynamics
second law of thermodynamics
Newton's second law of motion
Newton's first law of motion
This question was previously asked in
UPSC CAPF – 2012
This phenomenon is an example that relates to Newton’s first law of motion. Newton’s First Law states that an object at rest stays at rest, and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force. To start the loaded trolley bag from rest, you need to exert a force strong enough to overcome its inertia (resistance to change in state of motion) and the static friction between the trolley and the ground. Once the trolley is moving, the friction acting on it is kinetic friction, which is typically less than the maximum static friction. Therefore, less force is required to keep it moving than to start it from rest. The principle that a force is required to change the state of rest is central to the first law.
– Newton’s First Law describes inertia, the resistance of an object to changes in its state of motion.
– Static friction opposes the start of motion and is generally greater than kinetic friction, which opposes motion while it is occurring.
– Overcoming inertia and static friction requires more force than overcoming kinetic friction to maintain motion.
Newton’s Second Law (F=ma) describes the relationship between force, mass, and acceleration, which applies once a net force is acting. While the forces (applied force, friction) and resulting motion are governed by the Second Law, the observation about needing more force to *start* is fundamentally about overcoming the initial resistance to motion described by the First Law and the nature of static vs. kinetic friction. The First Law provides the foundational principle that a force is necessary to initiate motion from rest.

23. ‘Hydraulic brakes’ and ‘Hydraulic lift’ are devices in which fluids ar

‘Hydraulic brakes’ and ‘Hydraulic lift’ are devices in which fluids are used for transmitting

force
momentum
pressure
power
This question was previously asked in
UPSC CAPF – 2011
‘Hydraulic brakes’ and ‘Hydraulic lift’ are devices in which fluids are used for transmitting pressure.
These devices operate based on Pascal’s Principle, which states that pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. In hydraulic systems, a small force applied to a small area (piston) creates pressure in the fluid (Pressure = Force/Area). This pressure is transmitted through the fluid to a larger area (piston), resulting in a larger force output (Force = Pressure * Area). The fluid acts as the medium to transmit this pressure.
While force is transmitted and amplified in hydraulic systems, the fundamental principle at play, enabling the transmission and multiplication of force, is the uniform transmission of pressure throughout the fluid. Work done (Force x Distance) is ideally the same at both input and output ends, neglecting energy losses, but the force and distance are traded off (small input force over large distance results in large output force over small distance).

24. X is twice as massive as Y. X also runs twice faster than Y. Which one

X is twice as massive as Y. X also runs twice faster than Y. Which one among the following is the ratio of kinetic energy of X and Y ?

1 : 8
8 : 1
4 : 1
2 : 1
This question was previously asked in
UPSC CAPF – 2010
The ratio of kinetic energy of X and Y is 8 : 1.
The kinetic energy (KE) of an object is given by the formula KE = ½ * m * v², where m is the mass and v is the velocity. Given that the mass of X (m_X) is twice the mass of Y (m_Y), so m_X = 2 * m_Y. Also, the velocity of X (v_X) is twice the velocity of Y (v_Y), so v_X = 2 * v_Y.
KE_X = ½ * m_X * v_X² = ½ * (2 * m_Y) * (2 * v_Y)² = ½ * 2 * m_Y * 4 * v_Y² = 8 * (½ * m_Y * v_Y²)
KE_Y = ½ * m_Y * v_Y²
The ratio KE_X / KE_Y = [8 * (½ * m_Y * v_Y²)] / [½ * m_Y * v_Y²] = 8 / 1. Thus, the ratio is 8:1.
Kinetic energy is a scalar quantity representing the energy of motion. It is directly proportional to the mass of the object and the square of its velocity. Doubling the velocity has a much greater impact on kinetic energy than doubling the mass.

25. The Earth travels on its orbit at a speed of approximately 4400 km per

The Earth travels on its orbit at a speed of approximately 4400 km per hour. Why do we not feel this high speed ?

We are too small compared to the size of the Earth
Our relative speed with respect to the Earth along the Earth's orbit is zero
The gravity of the Earth constantly pulls us towards the Earth's centre
The solar system as a whole is also moving
This question was previously asked in
UPSC CAPF – 2010
We do not feel the high speed of Earth’s orbit because our relative speed with respect to the Earth along the Earth’s orbit is zero.
– We are moving along with the Earth at the same speed and in the same direction as it orbits the sun.
– In physics, we typically feel acceleration (changes in speed or direction), not constant velocity. Since our velocity relative to the Earth’s surface (for the orbital motion) is zero and remains zero, we don’t feel the orbital speed.
– This is related to the concept of inertial frames of reference. We are in the same inertial frame as the Earth’s surface for this motion.
While the Earth is constantly accelerating towards the sun due to gravity (causing its curved path), this acceleration is relatively small (about 0.006 m/s²) and is the same for everything on Earth, so we don’t feel it as a distinct force pulling us away from the vertical. Other motions, like Earth’s rotation, do have slight effects (e.g., centrifugal force), but the primary reason for not feeling the vast orbital speed is being co-moving with the Earth.

26. The following figure shows the displacement time (x-t) graph of a body

The following figure shows the displacement time (x-t) graph of a body in motion. The ratio of the speed in first second and that in next two seconds is :

1 : 2
1 : 3
3 : 1
2 : 1
This question was previously asked in
UPSC CAPF – 2009
The correct option is D) 2 : 1.
The speed of the body is given by the magnitude of the slope of the displacement-time (x-t) graph.
In the first second (from t=0 to t=1), the displacement changes from 0 to 3. The velocity is `(3 – 0) / (1 – 0) = 3/1 = 3` units/second. The speed in the first second (v1) is `|3| = 3`.
In the next two seconds (from t=1 to t=3), the displacement changes from 3 to 0. The velocity is `(0 – 3) / (3 – 1) = -3/2` units/second. The speed in the next two seconds (v2) is `|-3/2| = 3/2`.
The ratio of the speed in the first second and that in the next two seconds is `v1 : v2 = 3 : (3/2)`. To express this ratio in integers, we can multiply both parts by 2: `(3 * 2) : (3/2 * 2) = 6 : 3`. Dividing by 3, we get the simplified ratio `2 : 1`.
The displacement-time graph provides information about the position of the body over time. Velocity is the rate of change of displacement, and speed is the magnitude of velocity. A positive slope indicates movement in one direction, while a negative slope indicates movement in the opposite direction. In this graph, the body moves from x=0 to x=3 in the first second and then back from x=3 to x=0 in the subsequent two seconds.

27. A person moves along a circular path by a distance equal to half the c

A person moves along a circular path by a distance equal to half the circumference in a given time. The ratio of his average speed to his average velocity is :

0.5
0.5π
0.75π
1.0
This question was previously asked in
UPSC CAPF – 2009
The correct option is B.
Let the circular path have radius R. The circumference is $C = 2\pi R$.
The distance covered by the person is half the circumference, $d = \frac{1}{2} C = \pi R$.
Let the time taken be $t$.
Average speed is defined as the total distance traveled divided by the total time taken.
Average speed = $\frac{d}{t} = \frac{\pi R}{t}$.

The person moves along a circular path by a distance equal to half the circumference. This means the person starts at one point on the circle and ends at the diametrically opposite point.
Let the starting point be A and the ending point be B, where AB is a diameter of the circle.
The displacement is the shortest straight-line distance from the initial position to the final position. In this case, the displacement is the length of the diameter.
Displacement = $2R$.

Average velocity is defined as the total displacement divided by the total time taken.
Average velocity = $\frac{\text{Displacement}}{t} = \frac{2R}{t}$.

The ratio of average speed to average velocity is:
Ratio = $\frac{\text{Average speed}}{\text{Average velocity}} = \frac{\pi R / t}{2R / t} = \frac{\pi R}{t} \times \frac{t}{2R} = \frac{\pi}{2}$.
The value $\frac{\pi}{2}$ is equivalent to $0.5\pi$.

This question highlights the difference between speed (scalar, based on distance) and velocity (vector, based on displacement). Distance is the path length, while displacement is the change in position vector. For motion along a curved path, the distance is generally greater than the magnitude of the displacement. For a half circle, the distance is $\pi R$ and the displacement magnitude is $2R$.

28. The explanation of, why we get thrown back with a jerk when the statio

The explanation of, why we get thrown back with a jerk when the stationary bus we are sitting in starts moving forward is given by :

Zeroth law of gravity
Newton's first law
Newton's second law
Newton's third law
This question was previously asked in
UPSC CAPF – 2009
The explanation of why you get thrown back with a jerk when a stationary bus starts moving forward is given by Newton’s first law.
Newton’s first law of motion, also known as the Law of Inertia, states that an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force. When the bus is stationary, you are also at rest relative to the ground. When the bus suddenly accelerates forward, your body, due to inertia, tends to maintain its state of rest. This relative motion causes you to feel a backward jerk with respect to the accelerating bus.
Newton’s second law relates force, mass, and acceleration (F=ma). Newton’s third law states that for every action, there is an equal and opposite reaction. The concept of inertia directly explains the resistance to changes in motion, which is fundamental to the first law. The “Zeroth law of gravity” is not a standard physics term; gravity is described by Newton’s law of universal gravitation or Einstein’s theory of general relativity.

29. 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 :

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.

30. 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.