NDA Work, energy, power and gravitation previous year questions with solutions
Work, energy, power and gravitation is about one General Studies question in thirty, and its share has grown since 2023. The newer papers set numericals on energy conservation, g on other planets and escape speed beside the old one-fact items. Every question UPSC has set on it since 2017, solved. Grouped by the method each needs, newest first; open a question for its answer and worked solution.
2.1 Work: the definition, its sign and the joule
Answer and solution
Answer: (c) is in perpendicular direction of the direction of force
Method. W = Fs cosθ
Work is zero when θ = 90°, i.e. the displacement is perpendicular to the force (NCERT Class 9 Science ch 7 (Work and energy)).
Answer and solution
Answer: (d) negative
Method. Sign of work
The pulling force is anti-parallel to the displacement, θ = 180°, so W = Fs cos180° = -Fs, negative (NCERT Class 9 Science ch 7 (Work and energy)). (As worded, the stem is about a force opposing the displacement.)
Answer and solution
Answer: (a) 4 N moves an object by 25 cm.
Method. W = Fs; 1 J = 1 N·m
(a) 4 N × 0.25 m = 1 J. (b) gives 2 J, (c) 0.01 J, (d) 0.5 J (NCERT Class 9 Science ch 7 (Work and energy)).
Answer and solution
Answer: (c) anti-parallel to each other.
Method. Sign of work W = Fs cosθ
Work is negative when θ = 180°, i.e. force and displacement are anti-parallel; perpendicular gives zero, parallel positive (NCERT Class 9 Science ch 7 (Work and energy)).
2.2 Energy: its forms, kinetic and potential energy, conservative forces and conservation of energy
Answer and solution
Answer: (c) Work done depends on the path followed by the body
Method. Gravity is a conservative force
Work done by gravity depends only on the vertical separation of the end points and not on path; so (c) is wrong. For horizontal displacement the force is perpendicular, W = 0 (NCERT Class 11 Physics ch 5 (Work, energy and power)).
Answer and solution
Answer: (a) isolated only
Method. Conservation of energy for an isolated system
Energy of a system is conserved when no energy crosses its boundary, i.e. for an isolated system; a non-isolated one exchanges energy with the surroundings and its own energy can change, even though the total including surroundings is conserved (NCERT Class 11 Physics ch 5 (Work, energy and power)).
Keyed on the textbook sense that a system's energy is conserved only if isolated; (c) is arguable if 'system' includes surroundings.
Answer and solution
Answer: (b) potential energy
Method. Definition of potential energy
Energy due to position or configuration (shape) is potential energy; kinetic energy is due to motion (NCERT Class 9 Science ch 7 (Work and energy)).
Answer and solution
Answer: (d) conservation of energy, momentum and charge.
Method. Fundamental conservation laws
Physics requires conservation of energy, linear momentum and charge together; the other options each deny one of them.
Answer and solution
Answer: (a) Frictional force
Method. Conservative vs dissipative force
Friction is non-conservative (work depends on path, energy is dissipated); electric, gravitational and spring forces are conservative (NCERT Class 11 Physics ch 5 (Work, energy and power)).
Answer and solution
Answer: (b) Chemical energy
Method. Chemical energy lies in bonds
Energy stored in the bonds linking atoms in molecules is chemical energy; nuclear energy sits in the nucleus.
Answer and solution
Answer: (c) Gravitational potential energy → kinetic energy → heat energy to air → heat energy to ground and apple → sound energy
Method. Energy conversion in a fall
Stored gravitational PE becomes KE as it falls; air drag turns some into heat of the air; on impact the energy appears as heat in ground and apple and as sound (NCERT Class 9 Science ch 7 (Work and energy)).
Answer and solution
Answer: (b) -x26(2Ax − 3B)
Method. U(x) = -∫F dx with U(0) = 0
U = -∫(Ax2 - Bx)dx = -(Ax33 − Bx22) = -x26(2Ax − 3B), taking U = 0 at x = 0 (NCERT Class 11 Physics ch 5 (Work, energy and power)).
Answer and solution
Answer: (c) Energy can neither be created nor destroyed.
Method. Law of conservation of energy
Energy changes form but is neither created nor destroyed (NCERT Class 9 Science ch 7 (Work and energy)).
2.3 The work-energy theorem and conservation of mechanical energy
Answer and solution
Answer: (c) 2 m/s
Method. Energy conservation, mgh = ½mv2
v = √(2gh) = √(2.10.0.20) = √4 = 2 m/s (NCERT Class 11 Physics ch 5 (Work, energy and power)).
Answer and solution
Answer: (d) M = 2.8 J, H = 0.8 m
Method. Max height and PE at top
H = u2/2g = 16/20 = 0.8 m; M = mgH = 0.35.10.0.8 = 2.8 J (equal to ½mu2 = 0.5.0.35.16 = 2.8 J) (NCERT Class 11 Physics ch 5 (Work, energy and power)).
Answer and solution
Answer: (b) v = √(F0Lm)
Method. Work–energy theorem with variable force
W = ∫0LF0(1 − x/L)dx = F0(L − L/2) = F0L/2 = ½mv2, so v = √(F0L/m) (NCERT Class 11 Physics ch 5 (Work, energy and power)).
Answer and solution
Answer: (c) ΔK is larger in frame S′.
Method. Work–energy theorem in two frames; ΔK is frame dependent
S: t = √(2h/g), ΔK = mgh. In S′ the ball starts at speed u (v = u) downward and during t the ball's displacement is h + ut (the frame moves up). W = mg(h + ut) so ΔK′ = mgh + mgut > mgh. Check: ½m(u + gt)2 - ½mu2 = mugt + ½mg2t2 = mgut + mgh. So ΔK is larger in S′ (NCERT Class 11 Physics ch 5 (Work, energy and power)).
Answer and solution
Answer: (b) The net work equals the change in kinetic energy.
Method. Work–energy theorem
Work done equals change in kinetic energy ½mvf2 - ½mvi2, which depends only on initial and final speeds for any force; that gives (b). It does not show the force is conservative (a), constant (c), or that speed rises (d) (NCERT Class 11 Physics ch 5 (Work, energy and power)).
Answer and solution
Answer: (a) Kinetic Energy at A = Potential Energy at B
Method. Mechanical energy conservation
Taking PE zero at A and neglecting air, all KE at A converts to PE at B: KEA = PEB. At B the KE is zero, at A the PE is zero (NCERT Class 11 Physics ch 5 (Work, energy and power)).
Answer and solution
Answer: (a) 62.5 m/s
Method. Energy conservation, PE → KE
m = 0.32 kg; ½mv2 = 625 ⇒ v = √(2.625/0.32) = √3906.25 = 62.5 m/s (NCERT Class 11 Physics ch 5 (Work, energy and power)).
Answer and solution
Answer: (b) 3 N
Method. Work–energy theorem
F·s = ½mu2: F = (½·2.9)/3 = 3 N (NCERT Class 11 Physics ch 5 (Work, energy and power)). Check: a = u2/2s = 1.5 m/s2, F = 2.1.5 = 3 N.
Answer and solution
Answer: (c) P1 = 2P2
Method. PE of a liquid = m g × height of its centre of mass
Initially: mass m with centre of mass at h/2, P1 = mgh/2. Final: containers identical, so liquid stands at h/2 in each; each holds m/2 with centre of mass at h/4, P2 = 2·(m/2)g(h/4) = mgh/4. P1/P2 = 2, so P1 = 2P2; the rest of the energy is dissipated as heat (NCERT Class 11 Physics ch 5 (Work, energy and power)).
Answer and solution
Answer: (a) 10.0 m/s
Method. K = ½mv2
m = 2000 g = 2 kg; v = √(2K/m) = √(2.100/2) = √100 = 10.0 m/s (NCERT Class 11 Physics ch 5 (Work, energy and power)).
(acceleration due to gravity = 9.8 m/s2)
Answer and solution
Answer: (d) √980 m.s-1 and 980 J
Method. Energy conservation in free fall
v = √(2gh) = √(2.9.8.50) = √980 m/s. Total energy at release (at rest, 50 m above ground, ground as reference) = mgh = 2.9.8.50 = 980 J. Both match (d).
2.4 Power, efficiency and simple machines
Answer and solution
Answer: (b) v ∝ √t
Method. P = Fv = d(½mv2)/dt
Constant power: ½mv2 = Pt, so v = √(2Pt/m) ∝ √t (NCERT Class 11 Physics ch 5 (Work, energy and power)).
Answer and solution
Answer: (c) [sketch (c); see figure]
Method. Second-class lever: load between fulcrum and effort
In (c) the fulcrum is at the left end, the load acts in the middle and the effort at the right end, which is a second class lever. (a) fulcrum lies between load and effort (first class), (b) first class, (d) effort lies between fulcrum and load (third class).
Answer and solution
Answer: (b) 160 W
Method. P = mgh/t
Work = 8.10.4 = 320 J; P = 320/2 = 160 W (NCERT Class 9 Science ch 7 (Work and energy)).
Answer and solution
Answer: (b) A bottle opener
Method. Classes of lever
In a second-class lever the load lies between fulcrum and effort. A bottle opener rests on the cap edge (fulcrum), the cap is the load, and the hand applies effort further out. Scissors are first class; a cricket bat and a bow are third class (effort between).
2.5 Newton's law of gravitation: G, the inverse square law, field and potential
I. Gravitational field due to a spherical body
II. Electrostatic field due to a point charge
III. Strong nuclear force field due to a proton
IV. Magnetic field due to a bar magnet
How many of the above follow the inverse square law?
Answer and solution
Answer: (b) Two
Method. Which fields obey 1/r2
Gravitational field of a sphere (outside) and electrostatic field of a point charge follow 1/r2. The strong nuclear force is short range, not inverse-square. A bar magnet's field is dipolar, ∝ 1/r3 on axis. So two (NCERT Class 11 Physics ch 7 (Gravitation)).
Answer and solution
Answer: (d) No work is done by gravity in moving the object from point A to point B.
Method. W = m(VA - VB)
Work by gravity between A and B = -m·ΔV; equal potentials give zero work, so (d). Different fields at the two points do not matter; gravity is still conservative (NCERT Class 11 Physics ch 7 (Gravitation)).
Answer and solution
Answer: (c) 1 : 1
Method. Newton's third law
The two planets exert gravitational forces on each other that are equal in magnitude and opposite in direction whatever their masses and sizes, so ratio 1 : 1 (NCERT Class 11 Physics ch 7 (Gravitation)). (Direct check: F = Gm1m2/r2, same for both.)
Answer and solution
Answer: (d) General theory of relativity
Method. LIGO and gravitational waves
LIGO detected gravitational waves in 2015, predicted by Einstein's general theory of relativity.
Answer and solution
Answer: (a) 16 F
Method. Inverse-square law
F = GM2/R2; second system F′ = G(2M)2/(R/2)2 = 4.4·GM2/R2 = 16F (NCERT Class 11 Physics ch 7 (Gravitation)).
Answer and solution
Answer: (c) star which has collapsed into itself and has large acceleration due to gravity on its surface
Method. Definition of black hole
A black hole is a star collapsed into itself to very small radius, so g = GM/R2 on its surface is enormous, large enough that even light cannot escape; (a) and (d) wrongly say zero.
Answer and solution
Answer: (d) It is same for all pairs of bodies in our universe
Method. Dependence of gravitational force on masses and distance
F = Gm1m2/r2 differs from pair to pair with masses and separation, so (d) is not correct. (a) holds for all bodies, (b) it dominates between celestial bodies, (c) it is negligible between atoms because the masses are tiny (NCERT Class 11 Physics ch 7 (Gravitation)).
(All symbols have their usual meanings)
Answer and solution
Answer: (d) The quantity G is a universal constant.
Method. Universal gravitational constant
G is the same in the law of gravitation for any two masses anywhere, hence a universal constant; it does not depend on g (g is derived from it: g = GM/R2), does not vary with place, and is not restricted to the Earth (NCERT Class 11 Physics ch 7 (Gravitation)).
2.6 Acceleration due to gravity, free fall, mass and weight
I. The weight of an object is the same everywhere on the surface of the Earth, but its inertial mass could be different.
II. The inertial mass and gravitational mass of an object are proportional and the proportionality constant is the same everywhere.
III. The inertial mass and weight of an object are the same at different places on Earth.
Which of the statements given above is/are correct?
Answer and solution
Answer: (b) II only
Method. Inertial mass, gravitational mass, weight
I is false: weight mg varies slightly with place, mass does not. II is true: gravitational and inertial mass are proportional with one universal constant (equivalence principle). III is false: weight changes with g (NCERT Class 11 Physics ch 7 (Gravitation)).
Answer and solution
Answer: (a) ω = √(4πGρ3)
Method. SHM inside a uniform planet
Inside, g(r) = 43πGρ r, so acceleration = -(4πGρ3)x, and ω = √(4πGρ/3) (NCERT Class 11 Physics ch 7 (Gravitation)).
Answer and solution
Answer: (b) Acceleration
Method. Free fall
Both fall with a = g regardless of mass. Momentum, PE and KE all scale with mass so differ (NCERT Class 9 Science ch 6 (Gravitation)).
Answer and solution
Answer: (a) g1 > g2
Method. g = 43πGρR
With equal density g ∝ R, so R1 > R2 gives g1 > g2 (NCERT Class 11 Physics ch 7 (Gravitation)).
Answer and solution
Answer: (c) g1/2
Method. g = GM/R2
g2 = G(2M1)/(2R1)2 = 24g1 = g1/2 (NCERT Class 11 Physics ch 7 (Gravitation)).
Answer and solution
Answer: (b) It is same everywhere
Method. Mass is an invariant property of matter
Mass is a measure of the quantity of matter/inertia and is the same everywhere, unlike weight, which changes with g (b) (NCERT Class 9 Science ch 6 (Gravitation)). Option (d) is true only in classical physics: strictly a heated body gains mass E/c2, so (b) is the one sound statement; (a) and (c) are false.
Keep (b); note that (d) is also true at school level, so this rests on the strict mass-energy reading.
Answer and solution
Answer: (d) t1 = t2 = t3
Method. Free fall in vacuum is independent of mass
With no air, every body falls with a = g, so t = √(2h/g) is the same: t1 = t2 = t3 (NCERT Class 9 Science ch 6 (Gravitation)).
2.7 Satellites, Kepler's laws, escape speed and weightlessness
Answer and solution
Answer: (d) The semimajor axis of the planet's orbit is four times as compared to that of the Earth.
Method. Kepler's third law T2 ∝ a3
a3 = T2 = 64 (in Earth units), a = 4 times the Earth's (NCERT Class 11 Physics ch 7 (Gravitation)).
Answer and solution
Answer: (b) normal reaction of the space-station floor on the astronaut is zero
Method. Weightlessness = zero normal reaction
The station and astronaut are in free fall with the same acceleration g', so the floor exerts no reaction (apparent weight is zero). Gravity still acts (nearly its surface value), the acceleration is not zero, and no centrifugal force is needed in the inertial view (NCERT Class 11 Physics ch 7 (Gravitation)).
Answer and solution
Answer: (a) 11.2
Method. ve = √(2GM/R) = R√(8πGρ/3)
ve ∝ R√ρ, so v'/v = (1/2)·√4 = 1; escape speed stays 11.2 km/s (NCERT Class 11 Physics ch 7 (Gravitation)).
Answer and solution
Answer: (b) 1/8
Method. Kepler's law of periods T2 ∝ R3
T1/T2 = (R1/R2)3/2 = (1/4)3/2 = 1/8 (NCERT Class 11 Physics ch 7 (Gravitation)).
Answer and solution
Answer: (d) Satellite does not require any energy for orbiting.
Method. Orbit needs only gravity as centripetal force
Gravity supplies the centripetal force m v2/r, and no drag acts in space, so no energy is spent to keep it in orbit; fuel and solar panels serve manoeuvres and instruments, not orbiting (NCERT Class 11 Physics ch 7 (Gravitation)).
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