JEE Physics System Of Particles and Rotational Motion MCQs Set 24

Answer: (b) \( I_x + I_z = I_y \)

Explanation:
1. The axis perpendicular to the plane is Y.
2. The theorem says the moment about the perpendicular axis equals the sum of the other two: I_y = I_x + I_z.

Teacher's Note:
1. Identify the axis perpendicular to the lamina.
2. That axis stands alone on one side of the equation.

Answer: (b) \( L = \sqrt{3}R \)

Explanation:
1. About its axis: ½MR².
2. About a perpendicular axis through the centre: M(R²/4 + L²/12).
3. Equate: R²/2 = R²/4 + L²/12, so L² = 3R² and L = √3 R.

Teacher's Note:
1. Write both moments of inertia and equate them.
2. The perpendicular-axis formula for a cylinder is M(R²/4 + L²/12).

Answer: (d) a ring about its tangential axis perpendicular to the plane of the ring

Explanation:
1. Hollow sphere about a tangent: (2/3 + 1)MR² = (5/3)MR².
2. Solid sphere about a tangent: (2/5 + 1)MR² = (7/5)MR².
3. Disc about a perpendicular tangent: (1/2 + 1)MR² = (3/2)MR².
4. Ring about a perpendicular tangent: (1 + 1)MR² = 2MR², the largest.

Teacher's Note:
1. Add MR² to the central value for each body.
2. Compare 5/3, 7/5, 3/2 and 2.

Answer: (b) \( \dfrac{5}{4}MR^2 \)

Explanation:
1. About the diameter DD′ (in the plane): ¼MR².
2. Parallel-axis theorem with d = R: I = ¼MR² + MR² = (5/4)MR².

Teacher's Note:
1. A diameter of a disc gives ¼MR², not ½MR².
2. Add MR² for the shift of R.

Answer: (a) 1.25 kg m²

Explanation:
1. I = MK² = 20 × (0.25)² = 20 × 0.0625 = 1.25 kg m².

Teacher's Note:
1. Radius of gyration K is defined by I = MK².
2. Square K before multiplying.

Answer: (a) \( \dfrac{\ell}{\sqrt{2}} \)

Explanation:
1. Each corner is at distance ℓ/√2 from the centre.
2. I = 4m(ℓ²/2) = 2mℓ², total mass 4m.
3. K = √(I/M) = √(2mℓ²/4m) = ℓ/√2.

Teacher's Note:
1. The half-diagonal is ℓ/√2.
2. K = √(I/M).

Answer: (c) \( \dfrac{\ell}{\sqrt{3}} \)

Explanation:
1. I = Mℓ²/3 about the end.
2. K² = I/M = ℓ²/3, so K = ℓ/√3.

Teacher's Note:
1. Use K = √(I/M).
2. The end-axis value is Mℓ²/3.

Answer: (b) \( \sqrt{5} : \sqrt{6} \)

Explanation:
1. Disc about a tangent in its plane: I = ¼MR² + MR² = (5/4)MR², so K² = (5/4)R².
2. Ring about a tangent in its plane: I = ½MR² + MR² = (3/2)MR², so K² = (3/2)R².
3. Ratio K₁ : K₂ = √(5/4) : √(3/2) = √5 : √6.

Teacher's Note:
1. Use the diameter value plus MR² for each body.
2. Compare K², then take the square root.

Answer: (c) \( 17\hat i - 6\hat j - 13\hat k \)

Explanation:
1. τ = r × F, with r = (3, 2, 3) and F = (2, −3, 4).
2. τ_x = 2·4 − 3·(−3) = 17.
3. τ_y = 3·2 − 3·4 = −6.
4. τ_z = 3·(−3) − 2·2 = −13.
5. So τ = 17 î − 6 ĵ − 13 k̂ N m.

Teacher's Note:
1. Use the determinant rule for the cross product with r first.
2. Keep the order r × F, not F × r.

Answer: (d) \( (-1.5\hat i - 8\hat j - \hat k) \)

Explanation:
1. Position relative to the point: r′ = (0, 0.5, 2) − (2, 0, 3) = (−2, 0.5, −1).
2. τ = r′ × F with F = (2, 0, −3).
3. τ_x = 0.5(−3) − (−1)(0) = −1.5.
4. τ_y = (−1)(2) − (−2)(−3) = −2 − 6 = −8.
5. τ_z = (−2)(0) − 0.5(2) = −1.
6. τ = (−1.5, −8, −1) N m.

Teacher's Note:
1. Shift the origin to the given point before taking the cross product.
2. Be careful with signs in the middle component.

Answer: (a) \( 14\hat i - 38\hat j + 16\hat k \)

Explanation:
1. τ = r × F with r = (7, 3, 1), F = (−3, 1, 5).
2. τ_x = 3·5 − 1·1 = 14.
3. τ_y = 1·(−3) − 7·5 = −38.
4. τ_z = 7·1 − 3·(−3) = 16.

Teacher's Note:
1. Order matters: r × F.
2. Compute each component separately.

Answer: (b) \( mK^2\alpha \)

Explanation:
1. I = mK².
2. τ = Iα = mK²α.

Teacher's Note:
1. Torque = moment of inertia × angular acceleration.
2. I = mK² defines the radius of gyration.

Answer: (c) 4.5 N m

Explanation:
1. τ = F R = 15 × 0.3 = 4.5 N m.

Teacher's Note:
1. The string pulls along the tangent, so the lever arm is R.
2. Torque = force × perpendicular distance.

Answer: (a) 2.25 rad/s²

Explanation:
1. α = τ/I = 4.5/2 = 2.25 rad/s².

Teacher's Note:
1. Use τ = Iα.
2. Torque is already 4.5 N m.

Answer: (c) 4.5 rad/s

Explanation:
1. ω = ω₀ + αt = 0 + 2.25 × 2 = 4.5 rad/s.

Teacher's Note:
1. Rotational equations mirror linear ones.
2. ω₀ = 0.

Answer: (b) 4.5 rad

Explanation:
1. θ = ½αt² = ½ × 2.25 × 4 = 4.5 rad.

Teacher's Note:
1. Use θ = ω₀t + ½αt² with ω₀ = 0.
2. Square the time before multiplying.

Answer: (a) 750 rad

Explanation:
1. α = τ/I = 30/2 = 15 rad/s².
2. θ = ½αt² = ½ × 15 × 100 = 750 rad.

Teacher's Note:
1. The mass is not needed; use the moment of inertia.
2. Find α first.

Answer: (b) \( 3.6\pi \)

Explanation:
1. ω = 2π × 30 = 60π rad/s.
2. α = 60π/5 = 12π rad/s².
3. τ = Iα = 3 × 10⁻³ × 12π = 3.6π × 10⁻² N m.

Teacher's Note:
1. Convert revolutions per second to rad/s first.
2. τ = Iα.

Answer: (a) \( \dfrac{\pi}{12} \)

Explanation:
1. ω = 30 rpm = 30 × 2π/60 = π rad/s.
2. α = π/60 rad/s².
3. τ = Iα = 5π/60 = π/12 N m.

Teacher's Note:
1. Convert rpm to rad/s by multiplying by 2π/60.
2. Use time in seconds.

Answer: (b) 120 N m

Explanation:
1. α = 45/15 = 3 rad/s².
2. τ = Iα = 40 × 3 = 120 N m.

Teacher's Note:
1. τ = Iα with α = Δω/Δt.
2. Magnitude only.

Answer: (d) \( \dfrac{3g}{2L} \)

Explanation:
1. Torque of the weight about the hinge: Mg(L/2).
2. I about the end = ML²/3.
3. α = τ/I = (MgL/2)/(ML²/3) = 3g/2L.

Teacher's Note:
1. The weight acts at the centre, L/2 from the hinge.
2. Use I = ML²/3 for the end hinge.

Answer: (a) \( \dfrac{15g}{16} \)

Explanation:
1. α = 3g/(2L) = 3g/(2 × 1.6) = 3g/3.2.
2. 3/3.2 = 15/16, so α = 15g/16 rad/s².

Teacher's Note:
1. Use α = 3g/2L for a rod hinged at an end.
2. The mass cancels.

Answer: (d) 4 N m

Explanation:
1. P = τω.
2. τ = 1600/400 = 4 N m.

Teacher's Note:
1. Rotational power is torque times angular velocity.
2. Convert kW to W.

Answer: (d) \( \dfrac{n}{3} \)

Explanation:
1. First stage: (ω₀/2)² = ω₀² − 2α(2πn), so 3ω₀²/4 = 4παn.
2. Second stage: 0 = (ω₀/2)² − 2α(2πn′), so n′ = ω₀²/(16πα).
3. From the first stage ω₀²/(πα) = 16n/3, so n′ = n/3.

Teacher's Note:
1. Use ω² = ω₀² − 2αθ with θ = 2π × rotations.
2. Eliminate α between the two stages.