JEE Physics Circular Motion MCQs Set 18

Answer: (b) 24 rad/s

Explanation:
1. \( \omega = \frac{d\theta}{dt} = 6t^2 \).
2. At t = 2 s: \( \omega = 6\times4 = 24 \) rad/s.

Teacher's Note:
1. Differentiate θ with respect to t.
2. Substitute t = 2.

Answer: (a) \( -18\hat i-13\hat j+2\hat k \)

Explanation:
1. \( \hat i \): (−4)(6) − (1)(−6) = −18; \( \hat j \): (1)(5) − (3)(6) = −13; \( \hat k \): (3)(−6) − (−4)(5) = 2.
2. \( \vec v = -18\hat i-13\hat j+2\hat k \).

Teacher's Note:
1. Evaluate the determinant of ω × r.
2. Watch the sign of the j component.

Answer: (c) \( \sqrt{\left(\frac{v^2}{R}\right)^2+a^2} \)

Explanation:
1. Radial acceleration \( v^2/R \) and tangential acceleration a are perpendicular, so the net acceleration is \( \sqrt{(v^2/R)^2+a^2} \).

Teacher's Note:
1. Components are at right angles.
2. Decreasing speed only reverses the tangential direction.

Answer: (d) \( \tan^{-1}(2) \)

Explanation:
1. v = 72 km/h = 20 m/s. \( \tan\theta = \frac{v^2}{rg} = \frac{400}{20\times10} = 2 \).

Teacher's Note:
1. Convert km/h to m/s.
2. tan θ = v²/(rg).

Answer: (b) \( \frac{ML\omega^2}{2} \)

Explanation:
1. The liquid's centre of mass is at \( L/2 \), so the force needed is \( M\omega^2\frac L2 = \frac{ML\omega^2}{2} \).

Teacher's Note:
1. Use the centre of mass of the liquid.
2. F = Mω²r_cm.

Answer: (a) \( g\sqrt{3\sin^2\theta+1} \)

Explanation:
1. \( v^2 = 2g\ell\sin\theta \), so \( a_c = \frac{v^2}{\ell} = 2g\sin\theta \); \( a_t = g\cos\theta \).
2. \( a = g\sqrt{4\sin^2\theta+\cos^2\theta} = g\sqrt{3\sin^2\theta+1} \).

Teacher's Note:
1. Resolve gravity along and perpendicular to the string.
2. Use energy conservation for v².

Answer: (d) \( 40\ \mathrm{m/s^2} \)

Explanation:
1. Distance in two revolutions \( s = 2\times2\pi\times\frac{20}{\pi} = 80 \) m.
2. \( v^2 = 2a_ts \Rightarrow a_t = \frac{6400}{160} = 40 \) m/s².

Teacher's Note:
1. Distance covered = 2 × circumference.
2. Use v² = 2as.

Answer: (d) instantaneous and average accelerations are the same

Explanation:
1. |Δv| = 2v sin 30° = 60 m/s. Instantaneous a = v²/R = 3600/300 = 12 m/s².
2. Time \( t = \frac{R\theta}{v} = \frac{300\times\pi/3}{60} = 5.24 \) s, so average a = 60/5.24 = 11.5 m/s², which differs from 12 m/s².

Teacher's Note:
1. Average acceleration is Δv over time.
2. Direction of instantaneous acceleration changes continuously.

Answer: (c) 14 m/s²

Explanation:
1. \( v = \frac{dS}{dt} = 3t^2 = 12 \) m/s; \( a_t = 6t = 12 \) m/s².
2. \( a_c = \frac{v^2}{r} = \frac{144}{20} = 7.2 \) m/s². \( a = \sqrt{12^2+7.2^2} \approx 14 \) m/s².

Teacher's Note:
1. Find speed and tangential acceleration first.
2. Combine the two components.

Answer: (a) \( \frac{V}{2d} \)

Explanation:
1. Distance OA \( = \sqrt2d \), and the line OA makes 45° with the path.
2. Perpendicular velocity component \( V\sin45^\circ = \frac{V}{\sqrt2} \), so \( \omega = \frac{V/\sqrt2}{\sqrt2d} = \frac{V}{2d} \).

Teacher's Note:
1. ω = v⊥ / r.
2. The geometry gives 45°.

Answer: (b) 6 : 9 : 10

Explanation:
1. Outermost section: \( T_3 = m\omega^2(6L) = 6m\omega^2L \).
2. Middle: \( T_2 = T_3+m\omega^2(3L) = 9m\omega^2L \). Innermost: \( T_1 = T_2+m\omega^2L = 10m\omega^2L \).

Teacher's Note:
1. Each particle needs a net inward force mω²r.
2. Start from the outermost particle.

Answer: (c) 10 m/s

Explanation:
1. With top speed u: \( T_{min} = \frac{mu^2}{L}-mg \), \( T_{max} = \frac{mu^2}{L}+5mg \).
2. \( \frac{u^2/L+5g}{u^2/L-g} = 4 \Rightarrow u^2 = 3gL = 100 \Rightarrow u = 10 \) m/s.

Teacher's Note:
1. Tension at the bottom exceeds the top by 6mg.
2. Set the ratio equal to 4.