NDA Units, measurement and motion previous year questions with solutions

Answer: (c) 2 × 10⁵ A.U.

Method. 1 parsec = 206265 A.U.

One parsec is the distance at which 1 A.U. subtends 1 arc second: 1 pc = 206265 A.U. ≈ 2 × 10⁵ A.U.
Source: standard astronomical result.

Answer: (b) angular momentum

Method. dimensional analysis of h = E/⁠f

[h] = [ML²T^-2]/⁠[T^-1] = [ML²T^-1], the same as angular momentum (r × p = L·MLT^-1). Linear momentum is MLT^-1, displacement L, torque ML²T^-2.
Source: dimensions, KP1-A54, KP1-A56.

Answer: (b) Strain

Method. dimensionless quantities

Strain = change in length/⁠original length, a ratio, so dimensionless. Stress and pressure are force per area, force is a vector with dimensions [MLT^-2].
Source: standard result; dimensions, KP1-A54.

Answer: (d) N-m²/⁠kg²

Method. unit of G from F = Gm₁m₂/⁠r²

G = Fr²/⁠(m₁m₂), so its unit is N m²/⁠kg². N is force, m/⁠s speed, joule energy.
Source: Newton's law of gravitation (Class 11 Physics ch 7).

Answer: (c) 1 4 2 3

Method. SI units

Temperature – kelvin (1), weight is a force – newton (4), mass – kilogram (2), pressure – pascal (3): A1 B4 C2 D3.
Source: SI units, KP1-A05, KP1-A07, KP1-A19.

Answer: (d) Ω m

Method. resistivity ρ = RA/⁠l

ρ = R A / l, unit Ω × m²/⁠m = Ω m: (d).

Answer: (b) M^-1L³T^-2

Method. dimensions from F = Gm₁m₂/⁠r²

G = F r²/⁠(m₁m₂). [F] = M L T^-2, so [G] = M L T^-2 · L² / M² = M^-1 L³ T^-2: (b).

Answer: (a) frequency

Method. unit of thrust ÷ impulse

Thrust (force) is in N; impulse is in N s. The ratio is N/⁠(N s) = s^-1, the unit of frequency: (a).

Answer: (d) 1 only

Method. light year

The light year is the distance light travels in one year, so it measures very large distances. It is not a unit of time (2) and not of intensity (3). Only 1 is correct: (d).

Answer: (d) kg-metre/⁠sec²

Method. unit of energy

Joule, watt-hour (3600 J) and newton-metre are all units of energy (work). kg m/⁠s² is the newton, the unit of force: (d) is not a unit of energy.

No correct option. This question is defective as printed in UPSC's paper.

Method. precision of a measurement against the least count of the scale

A scale of least count 1 mm can give a reading only to the nearest mm (±0.5 to 1 mm), so 0.50 mm (0.01 mm digit) and 29.07 cm (0.1 mm digit) claim finer digits than it can read, while 0.925 m = 925 mm and 910 mm are both read to the mm. If precision is taken from the number of decimal places printed, (a) 0.50 mm is the finest; if from the scale's own limit, (c) and (d) are equally precise (±1 mm) and (c) has the smaller relative error; if from significant figures, (b) has four. The question fixes none of these.

The question does not say whether 'more precise' means the finest printed digit (a), the most significant figures (b), or the smallest fractional error on a 1 mm scale (c), and options (a) and (b) quote digits finer than the stated 1 mm least count, so no single option is fixed; of the readings the scale can give, (c) 0.925 m has the smaller fractional error.

Answer: (a) very large distances

Method. light year is a unit of distance

A light year is the distance light covers in one year (≈ 9.46 × 10¹⁵ m): a very large distance, not a time or a light quantity (Class 11 Physics ch. 1, KP1-A40).

Answer: (d) 10^-5 kg m/⁠s²

Method. 1 dyne = 1 g cm/⁠s² in SI

1 dyne = 1 g cm/⁠s² = 10^-3 kg × 10^-2 m/⁠s² = 10^-5 kg m/⁠s² (= 10^-5 N): (d). Option (a) and (b) are in the same CGS unit but have wrong magnitudes (1 dyne is 1 g cm/⁠s²).

Answer: (b) 1 nm = 10 Å

Method. 1 nm = 10^-9 m, 1 Å = 10^-10 m

1 nm = 10^-9 m and 1 Å = 10^-10 m, so 1 nm = 10 Å (standard SI/⁠length units; Class 11 Physics ch. 1, KP1-A77). The reading prints 'A°' where UPSC printed the angstrom sign Å; the sense is unchanged.

Answer: (b) 3.6 × 10⁶ J

Method. 1 kWh = 1000 W × 3600 s

1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J (standard conversion; verified).

Shortcut. 1 kWh = 3.6 MJ, a fact worth remembering.

Answer: (d) very large distance

Method. light year is a unit of distance

A light year = distance travelled by light in vacuum in one year ≈ 9.46 × 10¹⁵ m, a very large distance (Class 11 Physics ch. 1, KP1-A40).

Answer: (c) Henry

Method. SI unit of inductance

H stands for henry, the SI unit of inductance (standard SI list; Class 11 Physics ch. 1, KP1-A77). Hertz is Hz, the unit of frequency.

Answer: (b) distance

Method. light year is a unit of distance

A light year is the distance light travels in one year, about 9.46 × 10¹⁵ m. It is a unit of distance, not of time or of light received (Class 11 Physics ch. 1, KP1-A40 uses the speed of light × one year).

Answer: (b) Stress

Method. unit of pressure = N/⁠m² = unit of stress

Pressure = force/⁠area, unit N/⁠m² (pascal). Stress is also force per unit area, so it has the same unit (Class 11 Physics ch. 1, KP1-A77, SI units). Angular momentum is kg m²/⁠s, strain has no unit, work is the joule (N m).

Answer: (c) u² − v² = 2as

Method. kinematic equations (constant acceleration)

The correct relation is v² − u² = 2as, i.e. u² − v² = −⁠2as, so (c) as printed is wrong. (a) s = ut + ½at², (b) u = v − at and (d) s_n = u + ½a(2n−⁠1) are all correct.
Source: NS4-A80, NS4-A81.

Answer: (c) Zero

Method. definition of acceleration for uniform motion in a straight line

Taking uniform speed along a straight line (uniform velocity) the velocity does not change, so acceleration is zero. (Uniform speed on a curved path would still have a centripetal acceleration; the question is read as motion on a straight path.)
Source: NS4-A43, NS4-A95.

Answer: (c) −⁠1.6 m/⁠s²

Method. v² = u² + 2as

0 = 12² + 2a(45), so a = −⁠144/⁠90 = −⁠1.6 m/⁠s². Negative because it is a deceleration.
Source: NS4-A80, NS4-A81.

Answer: (a) [graph (a): see figure]

Method. terminal velocity of a skydiver with a parachute

Before the parachute: speed rises with ever-smaller slope (air drag grows) and levels at terminal speed (~50 m/⁠s). When the parachute opens drag jumps, speed falls quickly to a much lower terminal speed (a few m/⁠s), stays steady, then reaches zero on landing. (b) and (c) show constant slopes, i.e. constant acceleration, which ignores drag; (d) has no steady low-speed stretch after the fall. Only (a) has all stages.
Source: standard result (terminal velocity).

Answer: (a) 10 s

Method. two stages of uniform acceleration

Stage 1: v = 2t, s₁ = ½(2)t² = t². Stage 2 (10 s, a = 5): s₂ = 2t·10 + ½·5.100 = 20t + 250. Total t² + 20t + 250 = 550, so t² + 20t − 300 = 0, giving t = 10 s (the root −⁠30 is rejected).
Source: equations of motion (NS4-A81).

Answer: (c) 300 m

Method. displacement as vector sum of two perpendicular legs

East: v = 2 × 10 = 20 m/⁠s, s₁ = ½·2.100 = 100 m. South: s₂ = ½(4⁠√2)(100)⁠ = 200⁠√2⁠ m. Net displacement = ⁠√(100²⁠ + (200⁠√2)²)⁠ = ⁠√(10000⁠ + 80000) = 300 m.
Source: NS4-A81.

Answer: (a) 50 km/⁠h

Method. average speed = total distance / total time

Distance = 2400 − 2000 = 400 km (odometer difference); average speed = 400/⁠8 = 50 km/⁠h.
Source: Class 9 Science ch 4 (NS4-A19).

No correct option. This question is defective as printed in UPSC's paper.

Method. reading of a linear v–⁠t graph with intercept u

v = u + at with u and a non-zero is a straight line on a velocity–⁠time graph with intercept u. As printed every graph has the vertical axis labelled 'distance (m)'. Taken as v–⁠t graphs, (a) is excluded (u ≠ 0) and (d) is curved, but both (b) (a > 0) and (c) (a < 0) are straight lines with a positive intercept, so the 'all quantities non-zero' wording does not choose between them. Taken literally as distance–⁠time, none of them represents v = u + at.

Every option graph is labelled 'distance', not velocity, so as printed none represents v = u + at; with the axis read as velocity both (b) (a > 0) and (c) (a < 0) fit the equation. The evident intended answer is option b (a straight line with positive intercept u and positive slope a).

Answer: (c) Straight line

Method. v = u + at is linear in t

For uniform acceleration v = u + at, so the v–⁠t graph is a straight line (the position–⁠time graph would be the parabola): (c).

Answer: (d) Speed is scalar quantity and velocity is vector quantity.

Method. scalar and vector

Speed has magnitude only (scalar); velocity has magnitude and direction (vector): (d). (Class 9 Science ch. 4, NS4-A25.)

Answer: (c) 4 s

Method. v = u − gt, v = 0 at the top

t = u/⁠g = 40/⁠10 = 4 s: (c). (With g = 9.8 it is 4.08 s ≈ 4 s.) (Class 11 Physics ch. 1, KP1-A66.)

Answer: (d) 20 m/⁠s

Method. v² = u² − 2gh

At the top v = 0, so u = ⁠√(2gh)⁠ = ⁠√(2⁠ × 10 × 20) = ⁠√400⁠ = 20 m/⁠s: (d). With g = 9.8 the value is 19.8 m/⁠s ≈ 20.

Answer: (a) depends on its initial velocity.

Method. s = ut + ½at²

With constant a, the displacement in time t is s = ut + ½at², which depends on the initial velocity u (a). It is not independent of u, and it is not linear in t because of the t² term. It does not depend on the starting position (the distance covered is measured from where it started).

Answer: (a) is 0.

Method. average velocity = displacement/⁠time

The car returns to its start, so its displacement is 0 and average velocity = 0/⁠2 h = 0: (a). (The average speed would be 100 km ÷ 2 h = 50 km/⁠h.) (Class 9 Science ch. 4, NS4-A25.)

Answer: (b) V_A > V_B > V_C

Method. slope of a t–⁠x line is 1/⁠speed

Here time t is on the vertical axis and position x on the horizontal axis. Each line is straight through O, so each has constant speed v = x/⁠t, and the line's slope on the graph is t/⁠x = 1/⁠v. The steepest line (C) has the greatest 1/⁠v, so the smallest speed; the shallowest (A) has the greatest speed. V_A > V_B > V_C: (b).

Shortcut. With t vertical, a steeper line means slower motion.

Answer: (c) CD and AB

Method. slope of the velocity–⁠time graph gives acceleration

In AB velocity falls (negative slope): deceleration. BC is horizontal: uniform velocity. In CD velocity rises (positive slope): acceleration. So accelerated and decelerated motions are CD and AB: (c) (Class 9 Science ch. 4, NS4-A67). The graph is described as lying above the V-axis zero, so a falling V is a retardation.

Answer: (d) -1 m/⁠s²

Method. average acceleration = Δv/⁠Δt from a velocity–⁠time graph

From the graph the speed falls from 8 m/⁠s at t = 8 s to 4 m/⁠s at t = 12 s. a = (4 − 8)/⁠(12 − 8) = −⁠4/⁠4 = −⁠1 m/⁠s²: (d) (Class 9 Science ch. 4, NS4-A36, NS4-A37). The opening sentence of the question was dropped in the retype but the graph is described in full, so the answer is fixed.

Answer: (c) θ = sin^-1(1/⁠80)

Method. s = ½at² on a frictionless incline, a = g sin θ

From rest: 100 cm = ½ a (4 s)², so a = 200/⁠16 = 12.5 cm/⁠s².
On a smooth incline a = g sin θ, so sin θ = 12.5/⁠1000 = 1/⁠80, θ = sin^-1(1/⁠80): (c). The options print the fractions 1/⁠9.8, 1/⁠20, 1/⁠80, 1/⁠100 and (c) is the only one that matches; (a) is 1/⁠9.8.

Answer: (b) Its speed changes with time.

Method. constant velocity means constant speed and direction

Constant velocity: straight-line motion, zero acceleration, displacement = vt, increasing linearly. Speed is the magnitude of velocity, so it does not change. The statement 'Its speed changes with time' is the incorrect one: (b) (Class 9 Science ch. 4, NS4-A25).

Answer: (c) with y-intercept u

Method. v = u + at is a straight line v against t

Compare v = at + u with y = mx + c: slope = a, intercept on the velocity (y) axis = u (the value at t = 0). So the line has y-intercept u: (c). It passes through the origin only if u = 0, and the slope is a, not u (Class 9 Science ch. 4, NS4-A67).

Answer: (a) 1, 2 and 3

Method. uniform acceleration from a speed table; s = ut + ½at²; average speed = distance/⁠time

Speed rises 2 m/⁠s every second (0, 2, 4, 6, 8), so a = 2 m/⁠s² and it is uniform: (1) is correct (Class 9 Science ch. 4, NS4-A36, NS4-A37).
Distance in 4 s: s = 0 + ½ × 2 × 4² = 16 m: (2) is correct.
Average speed = 16 m ÷ 4 s = 4 m/⁠s (also (0 + 8)/⁠2 = 4): (3) is correct. All three hold.

Answer: (c) R only

Method. circular motion on a level road: friction supplies the centripetal force

No slipping needs mv²/⁠r ≤ μmg, so v² ≤ μgr = 0.1 × 10 × 10 = 10, v ≤ 3.16 m/⁠s. P(2 m/⁠s) and Q (3 m/⁠s, v² = 9) are safe; R (5 m/⁠s, v² = 25) slips.
Source: NS4-A95 (circular motion needs a force), friction NS6-A15.

Answer: (d) 60°

Method. relative velocity of rain with respect to each observer

Take rain velocity (0, 0, −⁠v). Observer E moving (v, 0, 0): relative velocity (−⁠v, 0, −⁠v). Observer N moving (0, v, 0): (0, −⁠v, −⁠v). cos φ = (v²)/⁠(⁠√2⁠ v · ⁠√2⁠ v) = 1/⁠2, so φ = 60°.
Source: relative velocity, standard result.

Answer: (a) 5 m/⁠s

Method. condition for a complete vertical circle with a string

At the top: mv²/⁠L ≥ mg, so v_top ≥ ⁠√(gL)⁠. Energy: v_bottom² = v_top² + 4gL ≥ 5gL. v_min = ⁠√(5⁠ × 10 × 0.5) = ⁠√25⁠ = 5 m/⁠s.
Source: standard result (circular motion, Class 11 Physics).

Answer: (c) Both I and II

Method. vector addition of two equal forces

R² = F²+F²+2F²cos θ = F², so cos θ = −⁠1/⁠2 and θ = 2π/⁠3: II is correct. The resultant bisects the angle between equal forces, so each makes π/⁠3 with it: I is correct.
Source: standard result (parallelogram law).

Answer: (c) R = 24 m

Method. horizontal projectile

Fall time from 20 m: t = ⁠√(2h/⁠g)⁠ = ⁠√(2⁠ × 20/⁠10) = 2 s. Range R = vt = 12 × 2 = 24 m: (c).

Answer: (c) 2v²πR

Method. average acceleration = Δv/⁠Δt in circular motion

Over half a circle the velocity reverses: |Δv| = v − (−⁠v) = 2v. The time taken is πR/⁠v. Average acceleration = 2v ÷ (πR/⁠v) = 2v²/⁠(πR): (c).

Answer: (a) 8.6 N

Method. resultant of two vectors, R = ⁠√(A²+B²+2AB⁠ cos θ)

R² = 5²+5²+2.5.5·cos 60° = 25+25+25 = 75, so R = 5⁠√3⁠ = 8.66 N ≈ 8.6 N.
Source: standard result (parallelogram law); NCERT Class 11 Physics treats the vector sum of forces.

Shortcut. For equal forces F at angle θ, R = 2F cos(θ/⁠2) = 10 cos 30° = 8.66 N.

Answer: (b) a change in velocity due to a change in its direction of motion.

Method. velocity is a vector

In uniform circular motion the speed is constant but the direction of motion keeps changing, so the velocity (and the momentum mv) changes continuously: (b). (a) is wrong because the speed does not change; (c) and (d) are contradicted.

Answer: (a) The centripetal acceleration of the object is smaller for a gentle curve (i.e., curve of larger radius) than that for a sharp curve (i.e., curve of smaller radius).

Method. centripetal acceleration a = v²/⁠r

At constant speed v, a_c = v²/⁠r. A gentle curve has a larger r, so a_c is smaller than on a sharp curve: (a). (b) and (c) contradict this; (d) is wrong because a_c is perpendicular to the velocity and changes only its direction, not the speed.

Answer: (d) The force acting on the particle is not parallel to the instantaneous momentum of the particle.

Method. differentiating the position vector

v = 2⁠√3⁠ t î + ⁠√2⁠ ĵ, a = 2⁠√3⁠ î (constant, non-zero), so F = ma = 2⁠√3⁠ m î. Momentum p = m(2⁠√3⁠ t î + ⁠√2⁠ ĵ) always has the ĵ component ⁠√2⁠ m, so p is never parallel to F; (a) false. F·p = 12 m² t, zero only at t = 0, not at every instant, so (b) fails. F ≠ 0, (c) false. So F is not parallel to the momentum.
Source: NS6-A45, NS4-A81.

Answer: (c) 588 N

Method. equilibrium at uniform velocity: net force zero

Uniform speed means a = 0, so upward force = weight = mg = 60 × 9.8 = 588 N (g = 9.8 m/⁠s², the value which gives an option; g = 10 would give 600 N, which is not offered).
Source: NS6-A48, NS6-A32.

Answer: (d) The net force on a body is proportional to that body’s momentum

Method. Newton's second law statements

F_net = ma = dp/⁠dt: the net force is proportional to acceleration, in the same direction, and equals the rate of change of momentum. It is not proportional to the momentum itself, so (d) is NOT correct.
Source: NS6-A45, NS6-A53.

Answer: (d) 1, 2 and 3

Method. characteristics of contact forces

All three statements hold: contact forces arise only when bodies touch (friction, normal reaction, tension), they obey the third law, and they act between a solid and a fluid too (buoyancy, drag, viscous force).
Source: Class 9 Science ch 6 (NS6-A64, NS6-A79).

Answer: (c) After the parachute opens, force P remains the same while force Q increases

Method. forces on a skydiver: weight P down, air drag Q up

P is the weight (constant, mg) and Q the upward air resistance, which grows with speed and jumps when the parachute opens (larger area). (b) is wrong: if P > Q the net force is downward and the speed increases. (d) is wrong: weight does not decrease. So (c).
Source: standard result; NS6-A18 (air resistance as a force opposing motion).

Answer: (d) 1, 2 and 3

Method. fundamental forces of nature

Gravitational, electromagnetic, strong nuclear and weak nuclear are the four fundamental forces, so 1, 2 and 3 are all fundamental.
Source: standard physics (Class 11 Physics ch 5 treats the fundamental forces).

Answer: (c) 8 N

Method. equilibrium: net force zero

Leftward: 10 + 15 = 25 N. Rightward: 2 + F + 15 = 17 + F. For rest, 17 + F = 25, so F = 8 N.
Source: NS6-A32 (zero net force).

Answer: (d) F = – mg

Method. F = ma, with a from differentiating y twice

v = dy/⁠dt = u − gt, a = d²y/⁠dt² = −⁠g. So F = ma = −⁠mg (downward, the weight). (a) and (c) are not forces; (b) has the wrong sign for this y-axis (up positive).
Source: Newton's second law (NS6-A45).

Answer: (d) W₁ = 10 kg and W₂ = 10 kg

Method. a series of spring balances

The tension is the same all the way down the chain (10 kg weight, taking the balances' own weight as negligible): each balance carries the full 10 kg load, so W₁ = W₂ = 10 kg: (d).

Answer: (d) The force of gravity of the Earth on the Moon and of the Moon on the Earth are equal in magnitude but are in opposite directions.

Method. Newton's third law applied to gravity

The Earth–⁠Moon pull is an action–⁠reaction pair: equal in magnitude, opposite in direction, acting on different bodies. So (a), (b) fail on magnitude and (c) fails on direction.
Source: Newton's third law, Class 9 Science ch 6 (NS6-A71, NS6-A80).

Answer: (b) 0.5 m/⁠s²

Method. F = ma

a = F/⁠m = 5 N / 10 kg = 0.5 m/⁠s²: (b). (0.5 m/⁠s² = 50 cm/⁠s², so (a) and (c) are wrong.)

Answer: (b) Mass is a constant of proportionality.

Method. F = ma

Newton's second law F = ma has mass m as the constant of proportionality between force and acceleration (and W = mg ties weight to it). Weight is a force that varies with g, not a constant: (b). (Class 9 Science ch. 6, NS6-A52, NS6-A53.)

Answer: (a) always true.

Method. contact and non-contact forces

Friction acts at the touching surfaces, so it is a contact force; the magnetic force acts through a field with no contact. This is a general fact and does not depend on temperature: always true (a).

Answer: (a) Frictional force

Method. central and conservative forces

Electric and gravitational forces are central (directed along the line joining the centres) and conservative. Friction acts along the surface against the relative motion, not along a line to a centre, and the work done depends on the path, so it is non-central and non-conservative: (a). 'Mechanical force' is not a type of force.

Answer: (d) A cricket ball

Method. inertia is measured by mass

Inertia of a body is measured by its mass (Class 9 Science ch. 6, NS6-A30). The cricket ball (about 0.16 kg) is by far the most massive of the four; a one-rupee coin is a few grams, a molecule or atom ~10^-26 kg.

Answer: (b) unstable equilibrium

Method. type of equilibrium by the effect of a small displacement

A ball balanced on top of a vertical rod: after a small displacement its weight acts to move it further from the position, and its centre of mass falls. That is unstable equilibrium (stable would need it to return; neutral would need it to stay put).

Answer: (b) 0.60 N

Method. average force = impulse / time

Impulse on the coin = change of its momentum = 0.005 × 0.36 = 1.8 × 10^-3 N s. F = 1.8 × 10^-3/⁠(3 × 10^-3) = 0.60 N. The striker's mass is not needed.
Source: NS6-A53.

Answer: (d) The centre of mass remains fixed.

Method. centre of mass of an isolated system

No external force acts, so the centre of mass of the system (X, Y and the tool) keeps its initial state, at rest. Internal forces (the throw) cannot move it.
Source: Newton's first law (NS6-A32) and standard result for the centre of mass.

Answer: (b) As the rope starts moving, every point on the rope has the same speed, but not the same acceleration.

Method. constraint motion of a rope over a pulley

The rope is inextensible, so every point moves with the same speed. The velocity directions differ (down on one side, up on the other, turning along the pulley), so the acceleration vectors are not the same: (a) false, (b) true. Tension varies along the rope (it must supply different accelerating forces), so (c) is false; the heavier side goes down so the centre of mass moves, (d) false.
Source: Newton's second law (NS6-A45); standard constraint method.

Answer: (c) The sand increases the impact time.

Method. impulse = Δp = F·Δt

Landing on soft sand makes the stopping time longer. Same change in momentum over a longer time gives a smaller force, so less injury. The change in momentum is the same, so (b) is false.
Source: NS6-A55, NS6-A56.

Answer: (d) 40 m/⁠s

Method. impulse = area under the F–⁠t graph; Δp = FΔt

Area = ½(5)(20) + (15)(20) + ½(5)(20) = 50 + 300 + 50 = 400 N s, so Δv = 400/⁠10 = 40 m/⁠s; the force starts from zero at t = 0, so the object is taken to start from rest and the final speed is 40 m/⁠s: (d).

the question gives no initial speed; it is read as starting from rest, the only reading that yields an option.

Answer: (a) To reduce the acceleration of the ball

Method. F = Δp/⁠Δt: a longer stopping time lowers the force and the deceleration

Pulling the hands back lengthens the time over which the ball's momentum changes to zero. The same Δp over a longer Δt gives smaller force and smaller deceleration (acceleration of the ball), so hands are hurt less. (b) and (c) are the opposite.
Source: NS6-A55.

Answer: (d) 100 kN

Method. F = m·Δv/⁠Δt

v = 72 km/⁠h = 20 m/⁠s, m = 1000 kg (the 'weighs 1000 kg' means mass). a = 20/⁠0.2 = 100 m/⁠s², F = 1000 × 100 = 1,00,000 N = 100 kN.
Source: NS6-A45, NS6-A53.

Answer: (b) 6.0 N

Method. impulse–⁠momentum theorem on the bouncing ball

Speed just before impact ⁠√(2gh)⁠ = ⁠√(2.10.0.45)⁠ = 3 m/⁠s down; just after, ⁠√(2.10.0.20)⁠ = 2 m/⁠s up. Change of momentum = 0.1(2 − (−⁠3)) = 0.5 kg m/⁠s. Average net force on the ball = 0.5/⁠0.1 = 5 N upward; this is N − mg, so the floor pushes the ball with N = 5 + 1 = 6 N, and by Newton's third law the ball presses the floor with 6 N.
The question asks for the force the ball applies on the floor, which is 6 N; 5 N is only the net force on the ball.
Source: NS6-A53, NS6-A71.

Answer: (d) It stops at position B

Method. elastic collision of equal masses exchanges velocities

X reaches B with some speed and strikes the identical, stationary Y elastically (head-on, along the table). Momentum and kinetic energy conservation give v_X' = 0 and v_Y' = v. So X stops at B and Y moves off with X's speed.
Source: standard result for equal masses.

Answer: (d) 40 m/⁠s

Method. conservation of momentum

0.1 × 20 = 0.05 × v, so v = 40 m/⁠s: (d).

the data are not consistent with an elastic collision, since the kinetic energy would rise from 20 J to 40 J; the answer follows from momentum conservation alone.

Answer: (c) M₁v₁ = M₂v₂

Method. conservation of horizontal momentum (rain adds no horizontal momentum)

No horizontal external force acts, and the falling rain has zero horizontal momentum, so M₁v₁ = M₂v₂. Then v₂ < v₁, and kinetic energy falls (M₂v₂² = (M₁v₁)v₂ < M₁v₁²), so (b) is false; (d) is false because momentum is equal, not increased.
Source: Newton's laws, Class 9 Science ch 6 (NS6-A45, NS6-A53).

Answer: (b) 0.2 N

Method. force = rate of change of momentum, F = v·(dm/⁠dt)

Each second 0.1 kg of sand, initially without horizontal speed, is brought to 2 m/⁠s: F = (dm/⁠dt)·v = 0.1 × 2 = 0.2 N.
Source: Newton's second law in momentum form (NS6-A53).

Answer: (c) −⁠3 m s^-1

Method. conservation of momentum, recoil

Initial momentum is 0, so 0.010 × 300 + 1 × V = 0, giving V = −⁠3 m/⁠s. The pistol recoils opposite to the bullet, so with the bullet's direction as positive the recoil velocity is −⁠3 m/⁠s: (c). (Class 9 Science ch. 6, NS6-A81.)

Answer: (b) 1.89 m/⁠s

Method. conservation of momentum, perfectly inelastic

Horizontal momentum is conserved: 52 × 2 = (52 + 3) v, so v = 104/⁠55 = 1.89 m/⁠s: (b).

Answer: (a) 1 only

Method. internal forces and the centre of mass

(1) Chemical reactions give internal forces only, and internal forces cannot change the velocity of the centre of mass: true. (2) The reaction converts chemical energy into the kinetic energy of the particles relative to the centre of mass (an explosion is the extreme case), so it can change that kinetic energy: false. Answer: 1 only (a).

Answer: (b) Its momentum

Method. momentum is a vector

With no energy lost, the ball leaves with the same speed, so speed, KE and PE (position at the instant of contact is the same) do not change. The velocity reverses, so the momentum mv, a vector, changes suddenly: (b).

No correct option. This question is defective as printed in UPSC's paper.

Method. work–⁠energy for a sliding block

Work–⁠energy: μmg·20 = ½·2.10² = 100 J, so a constant friction is 5 N (equivalently a = v²/⁠2s = 2.5 m/⁠s², F = 2 × 2.5). 5 N is not among the options. Friction that varies along the path could take other values at some instant, but then the data fix no option.

A constant friction force stopping a 2 kg block moving at 10 m/⁠s in 20 m is 5 N (f × 20 m = ½ × 2 × 10² = 100 J), which is not among 10, 20, 40, 50 N; with variable friction no single option follows from the data.

Answer: (c) Static friction > Kinetic friction > Rolling friction

Method. ordering of frictional forces

Limiting static friction is greater than kinetic (sliding) friction, which is greater than rolling friction (rolling is the easiest). So static > kinetic > rolling.
Source: Class 9 Science ch 6 (NS6-A15, NS6-A16, NS6-A20).

Answer: (c) 4 N

Method. limiting static friction on the upper block

The upper block is accelerated only by friction from the lower one: f = m₁a ≤ μ m₁g. Maximum f = 0.2 × 2.0 × 10 = 4 N (the common acceleration is then 2 m/⁠s²). The 3.0 kg block's weight does not enter.
Source: friction, Class 9 Science ch 6 (NS6-A15, NS6-A16).

Answer: (c) N₁ = 0.6 N and N₂ = 1.8 N

Method. equilibrium: sum of forces and sum of torques about the centre of mass

Weight 0.24 × 10 = 2.4 N acts at the 0.5 m mark. W1 is at 0.2 m, W2 at 1 − 0.4 = 0.6 m. Distances from the centre: 0.3 m and 0.1 m. Torques: N₁(0.3) = N₂(0.1), so N₂ = 3N₁. With N₁+N₂ = 2.4: N₁ = 0.6 N, N₂ = 1.8 N.
Source: standard result (principle of moments).

Answer: (a) The ring has higher kinetic energy

Method. rotational KE = ½Iω²

Same M, R and ω. I_ring = MR², I_disc = ½MR² (axis through the centre, perpendicular to the plane). KE = ½Iω² is therefore larger for the ring: (a). KE of rotation is not zero (d).

Answer: (a) The disc

Method. moment of inertia of a disc and a sphere

About the symmetry axis through the centre: I_disc = ½MR² = 0.5 MR², I_sphere = (2/⁠5)MR² = 0.4 MR². The disc has the higher moment of inertia: (a).