JEE Physics Circular Motion MCQs Set 15

Answer: (b) A body can have constant speed but a varying velocity

Explanation:
1. Velocity is a vector: if only its direction changes the speed stays the same, so uniform circular motion has constant speed and varying velocity.
2. If velocity is constant then speed is constant. Constant speed on a curve still means centripetal acceleration. A force perpendicular to motion does no work.

Teacher's Note:
1. Separate speed (scalar) from velocity (vector).
2. A force perpendicular to the displacement does zero work.

Answer: (d) 0 and 10 m/s

Explanation:
1. Displacement over one loop is zero, so average velocity is 0.
2. Average speed = distance/time = \( \frac{2\pi\times100}{62.8} = 10 \) m/s.

Teacher's Note:
1. Average velocity uses displacement; average speed uses path length.
2. After a full loop the displacement vanishes.

Answer: (c) 25/π

Explanation:
1. \( \alpha = 20/5 = 4 \) rad/s², \( \theta = \tfrac12\alpha t^2 = \tfrac12\times4\times25 = 50 \) rad.
2. Revolutions = \( \frac{50}{2\pi} = \frac{25}{\pi} \).

Teacher's Note:
1. Angular displacement ÷ 2π gives revolutions.
2. Use θ = ½αt² from rest.

Answer: (c) \( 1.4\pi \) rad/s²

Explanation:
1. 210 r.p.m. = \( \frac{210\times2\pi}{60} = 7\pi \) rad/s.
2. \( \alpha = \frac{7\pi}{5} = 1.4\pi \) rad/s².

Teacher's Note:
1. Convert r.p.m. to rad/s first.
2. The diameter is not needed for angular acceleration.

Answer: (d) \( \frac{2\pi}{60\times60\times24} \) rad/s

Explanation:
1. The earth makes one rotation in 24 hours = 86400 s, so \( \omega = \frac{2\pi}{T} = \frac{2\pi}{60\times60\times24} \) rad/s.

Teacher's Note:
1. The spin period is one day, not one year.
2. ω = 2π/T.

Answer: (b) 12 : 1

Explanation:
1. Minute hand: period 1 h; hour hand: period 12 h. \( \omega \propto 1/T \), so the ratio is 12 : 1.

Teacher's Note:
1. Angular speed is inversely proportional to the period.
2. Hour hand takes 12 times as long.

Answer: (c) 10 cm/s and 10 cm/s²

Explanation:
1. The angular velocity ω of the turntable is unchanged, so \( v = \omega r \) and \( a = \omega^2 r \) both halve.
2. New values: 10 cm/s and 10 cm/s².

Teacher's Note:
1. Points on a rigid rotating body share the same ω.
2. v ∝ r and a ∝ r at fixed ω.

Answer: (a) 0.6 m/s

Explanation:
1. Position along the wall \( x = d\tan\theta \), so \( v = d\sec^2\theta\,\dot\theta \).
2. \( v = 3\times\frac{1}{\cos^2 45^\circ}\times0.1 = 3\times2\times0.1 = 0.6 \) m/s.

Teacher's Note:
1. Differentiate x = d tan θ.
2. sec² 45° = 2.

Answer: (b) \( \tan\alpha = \frac{2s}{R} \)

Explanation:
1. \( a_t = v\frac{dv}{ds} = \frac{a^2}{2} \) and \( a_c = \frac{v^2}{R} = \frac{a^2 s}{R} \).
2. The tangential component is along the velocity, so \( \tan\alpha = \frac{a_c}{a_t} = \frac{2s}{R} \).

Teacher's Note:
1. Use a_t = v dv/ds.
2. The velocity lies along the tangent.

Answer: (d) \( \frac{40\pi}{7} \) s

Explanation:
1. During the increase \( \omega^2 = 900 + \frac{700}{200\pi}\theta \), so \( 2\omega\alpha\,\) gives \( \alpha = \frac{1}{2}\frac{d(\omega^2)}{d\theta} = \frac{7}{4\pi} \) rad/s².
2. The angular velocity changes from 30 to 40 rad/s, so \( t = \frac{10}{7/4\pi} = \frac{40\pi}{7} \) s.

Teacher's Note:
1. Differentiate ω² with respect to θ to get 2α.
2. 100 rev = 200π rad.

Answer: (c) 12 m/s

Explanation:
1. \( T = \frac{mv^2}{r} \Rightarrow v^2 = \frac{16\times144}{16} = 144 \), so \( v = 12 \) m/s.

Teacher's Note:
1. Tension supplies the centripetal force.
2. v = √(Tr/m).

Answer: (c) 3 : 5 : 6

Explanation:
1. Outer section: \( T_3 = m\omega^2(3\ell) \). Middle: \( T_2 = T_3 + m\omega^2(2\ell) = 5m\omega^2\ell \). Inner: \( T_1 = T_2 + m\omega^2\ell = 6m\omega^2\ell \).
2. Ratio \( 3:5:6 \).

Teacher's Note:
1. Each section carries the centripetal force of all particles beyond it.
2. Start from the outermost particle.

Answer: (d) \( \lambda R^2\omega^2 \)

Explanation:
1. Take a small element subtending \( 2\theta \) at the centre: \( 2T\sin\theta = (\lambda R\,2\theta)\omega^2R \).
2. For small θ this gives \( T = \lambda R^2\omega^2 \).

Teacher's Note:
1. Use a small element and resolve tensions towards the centre.
2. Tension equals λv² as v = ωR.

Answer: (b) \( \frac{m\omega^2}{2\ell}(\ell^2-x^2) \)

Explanation:
1. The part of the rod beyond x needs the centripetal force: \( T = \int_x^{\ell}\frac{m}{\ell}\omega^2 r\,dr \).
2. \( T = \frac{m\omega^2}{2\ell}(\ell^2-x^2) \).

Teacher's Note:
1. Tension at x supplies the centripetal force for the outer portion.
2. T falls to zero at the free end.

Answer: (c) \( \mu_k\sqrt{g^2+\frac{v_0^4}{r^2}} \)

Explanation:
1. The wire exerts a vertical component mg and a horizontal component \( \frac{mv_0^2}{r} \), so \( N = m\sqrt{g^2+\frac{v_0^4}{r^2}} \).
2. Friction \( \mu_kN \) is the only tangential force, so \( a_t = \mu_k\sqrt{g^2+\frac{v_0^4}{r^2}} \).

Teacher's Note:
1. Add the two perpendicular normal-force components vectorially.
2. Friction = μN opposes the motion.

Answer: (a) 21 m/s

Explanation:
1. \( v_{max} = \sqrt{\mu rg} = \sqrt{0.75\times60\times9.8} = \sqrt{441} = 21 \) m/s.

Teacher's Note:
1. Friction supplies the centripetal force.
2. v_max = √(μrg).

Answer: (b) 1 cm

Explanation:
1. Slipping: \( \mu g = \omega^2 r \), so \( r \propto 1/\omega^2 \).
2. Doubling ω gives \( r = 4/4 = 1 \) cm.

Teacher's Note:
1. The limiting friction is fixed.
2. r ∝ 1/ω².

Answer: (d) 200 m

Explanation:
1. \( \tan\theta = \frac{v^2}{rg} \) is fixed, so \( r \propto v^2 \).
2. Doubling v multiplies r by 4: \( 4\times50 = 200 \) m.

Teacher's Note:
1. Banking angle fixed means v²/r fixed.
2. r ∝ v².

Answer: (c) 5/6 rad/s

Explanation:
1. \( \mu m\omega^2R = mg \Rightarrow \omega = \sqrt{\frac{g}{\mu R}} = \sqrt{\frac{10}{0.8\times18}} = \frac56 \) rad/s.

Teacher's Note:
1. Friction balances weight, normal force gives centripetal force.
2. ω = √(g/μR).

Answer: (c) \( g\sqrt{10} \)

Explanation:
1. To just complete the circle, \( v_{bottom}^2 = 5gR \). At the level of the centre \( v^2 = 5gR-2gR = 3gR \), so \( a_c = 3g \).
2. The tangential acceleration is g (weight), so \( a = \sqrt{(3g)^2+g^2} = g\sqrt{10} \).

Teacher's Note:
1. Add radial and tangential components in quadrature.
2. At the horizontal level gravity is entirely tangential.

Answer: (d) T is directly proportional to h

Explanation:
1. At depth h the speed is \( v^2 = 2gh \) and the string makes angle φ with the vertical, \( \cos\phi = h/\ell \).
2. \( T - mg\cos\phi = \frac{mv^2}{\ell} \Rightarrow T = \frac{3mgh}{\ell} \), a straight line through the origin.

Teacher's Note:
1. Use energy conservation for v² and resolve along the string.
2. T = 3mg cos φ.

Answer: (b) Only (1) and (3)

Explanation:
1. Speed is the magnitude of velocity: a speed change forces a velocity change, hence acceleration. So (1) is true.
2. A velocity change can be only in direction, leaving speed unchanged (3 true, 2 false). A speed change with fixed direction still changes velocity (4 false).

Teacher's Note:
1. Velocity change can be in magnitude, direction, or both.
2. Any change of velocity means acceleration.

Answer: (c) Only (2) and (4)

Explanation:
1. \( a_t = 2 \) constant and \( a_c = \frac{4t^2}{R} \) grows, so \( |a| = \sqrt{4+\frac{16t^4}{R^2}} \) increases.
2. The angle between a and v satisfies \( \tan\alpha = \frac{a_c}{a_t} = \frac{2t^2}{R} \), which increases with t.

Teacher's Note:
1. Total acceleration has a constant tangential and a growing radial part.
2. tan α = a_c / a_t.

Answer: (a) Only (1) and (2)

Explanation:
1. \( \frac{dv}{dt} = -\frac{v^2}{R} \Rightarrow \frac1v-\frac1{v_0} = \frac tR \), giving (1).
2. With \( v\frac{dv}{ds} = -\frac{v^2}{R} \), \( v = v_0e^{-s/R} \), giving (2).

Teacher's Note:
1. Write a_t = −v²/R and integrate.
2. Use v dv/ds for distance dependence.