Rotational Motion and Gravitation — Practice

Question 1

The moment of inertia of a ring of mass M and radius R about its central axis (perpendicular to its plane) is:
(a) MR² (b) ½MR² (c) ⅖MR² (d) ¼MR²

Accepted Answer

(a) MR².

Every element of a ring lies at the same distance R from the central axis, so I = Σmr² = R²Σm = MR². A disc, whose mass is spread inward, has only ½MR².

Question 2

The moment of inertia of a solid sphere of mass M and radius R about a diameter is:
(a) ⅖MR² (b) ⅔MR² (c) MR² (d) ½MR²

Accepted Answer

(a) ⅖MR².

A solid sphere has I = (2/5)MR² about a diameter. A thin hollow sphere, with all its mass on the surface, has the larger (2/3)MR² — a distinction NEET likes to test.

Question 3

A force of 20 N acts at 15 cm from a bolt, at 30° to the spanner. The torque about the bolt is:
(a) 1.5 N·m (b) 3.0 N·m (c) 300 N·m (d) 0.75 N·m

Accepted Answer

(a) 1.5 N·m.

τ = rF sinθ = 0.15 × 20 × sin30° = 0.15 × 20 × 0.5 = 1.5 N·m. Applying the force perpendicular (θ = 90°) would double this to 3 N·m.

Question 4

A thin rod of mass M and length L has moment of inertia ML²/12 about its centre. About an axis through one end (perpendicular to the rod), it is:
(a) ML²/3 (b) ML²/6 (c) ML²/12 (d) ML²/4

Accepted Answer

(a) ML²/3.

Parallel axis theorem with d = L/2: I = ML²/12 + M(L/2)² = ML²/12 + ML²/4 = ML²/12 + 3ML²/12 = 4ML²/12 = ML²/3.

Question 5

A disc has moment of inertia ½MR² about its central axis. Its moment of inertia about a diameter is:
(a) ¼MR² (b) ½MR² (c) MR² (d) ⅖MR²

Accepted Answer

(a) ¼MR².

By symmetry the two in-plane diameters are equal (I_x = I_y = I_d). Perpendicular axis theorem: I_z = I_x + I_y, so ½MR² = 2I_d, giving I_d = ¼MR².

Question 6

A skater spinning with moment of inertia I at angular velocity ω pulls in her arms so her moment of inertia halves. Her new angular velocity is:
(a) 2ω (b) ω/2 (c) ω (d) 4ω

Accepted Answer

(a) 2ω.

No external torque acts, so L = Iω is conserved: I·ω = (I/2)·ω′, giving ω′ = 2ω. Reducing the moment of inertia speeds up the spin.

Question 7

A disc of moment of inertia I rotating at angular velocity ω has an identical stationary disc dropped coaxially onto it, so they rotate together. The common angular velocity is:
(a) ω/2 (b) ω (c) 2ω (d) ω/4

Accepted Answer

(a) ω/2.

Angular momentum is conserved (no external torque): Iω = (I + I)ω′, so ω′ = ω/2. The rotational KE halves — the lost energy becomes friction heat, so KE is not conserved even though L is.

Question 8

A constant torque of 5 N·m acts for 4 s on a body of moment of inertia 2 kg·m², initially at rest. Its final angular velocity is:
(a) 10 rad/s (b) 40 rad/s (c) 2.5 rad/s (d) 20 rad/s

Accepted Answer

(a) 10 rad/s.

Angular impulse = change in angular momentum: τΔt = IΔω, so 5 × 4 = 2 × ω, giving ω = 20/2 = 10 rad/s.

Question 9

A solid sphere, a disc and a ring (same mass and radius each) roll from rest down the same incline. Which reaches the bottom first?
(a) Solid sphere (b) Disc (c) Ring (d) All together

Accepted Answer

(a) Solid sphere.

Rolling acceleration is a = g sinθ/(1 + I/mR²). The shape factor I/mR² is ⅖ (sphere) < ½ (disc) < 1 (ring), so the sphere has the largest acceleration. Mass and radius do not appear, so they are irrelevant.

Question 10

For a solid sphere rolling without slipping, the fraction of its total kinetic energy that is rotational is:
(a) 2/7 (b) 5/7 (c) 1/2 (d) 2/5

Accepted Answer

(a) 2/7.

KE_rot = ½Iω² = ½(⅖mR²)ω² = ⅕mR²ω². With v = ωR, KE_trans = ½mR²ω². Total = (7/10)mR²ω², so rotational fraction = (1/5)/(7/10) = 2/7.

Question 11

If the distance between two point masses is doubled, the gravitational force between them becomes:
(a) one-quarter (b) one-half (c) double (d) four times

Accepted Answer

(a) one-quarter.

F ∝ 1/r². Doubling r multiplies the force by 1/2² = 1/4.

Question 12

At a height equal to the Earth's radius above the surface, the acceleration due to gravity is (surface value g):
(a) g/4 (b) g/2 (c) g (d) g/9

Accepted Answer

(a) g/4.

g_h = g(R/(R+h))². With h = R: g_h = g(R/2R)² = g(1/2)² = g/4.

Question 13

At a depth equal to half the Earth's radius below the surface, the acceleration due to gravity is (surface value g, uniform Earth):
(a) g/2 (b) g/4 (c) g (d) zero

Accepted Answer

(a) g/2.

g_d = g(1 − d/R). With d = R/2: g_d = g(1 − ½) = g/2. Unlike the inverse-square fall with height, depth gives a linear decrease to zero at the centre.

Question 14

The escape velocity from the surface of the Earth is approximately:
(a) 11.2 km/s (b) 7.9 km/s (c) 3.0 km/s (d) 22.4 km/s

Accepted Answer

(a) 11.2 km/s.

v_e = √(2GM/R) = √(2gR) ≈ 11.2 km/s for Earth — √2 times the low-orbit orbital speed of about 7.9 km/s, and independent of the escaping body's mass.

Question 15

Two satellites orbit the Earth at radii r and 4r. The ratio of their periods (T at 4r : T at r) is:
(a) 8 : 1 (b) 4 : 1 (c) 16 : 1 (d) 2 : 1

Accepted Answer

(a) 8 : 1.

T² ∝ r³, so (T₂/T₁)² = (4r/r)³ = 64, giving T₂/T₁ = 8. The 3/2 power turns a 4× radius into an 8× period.

Practice — Rotational Motion and Gravitation

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