JEE Physics Circular Motion MCQs Set 11

Answer: (a) 12 : 1 and \( \frac{\pi}{30} \) rad/s

Explanation:
1. Minute hand: \( \omega=\frac{2\pi}{3600} \) rad/s. Hour hand: \( \omega=\frac{2\pi}{12\times3600} \) rad/s.
2. Ratio = 12 : 1.
3. Second hand: \( \omega=\frac{2\pi}{60}=\frac{\pi}{30} \) rad/s.

Teacher's Note:
1. The angular speed is 2 pi divided by the time for one turn, so the ratio of speeds is the inverse ratio of periods.
2. The minute hand turns 12 times as fast as the hour hand.

Answer: (b) 3

Explanation:
1. For constant \( \alpha \) from rest, \( \theta=\tfrac12\alpha t^2 \).
2. \( \theta_1=\tfrac12\alpha(2)^2=2\alpha \) and total in 4 s = \( \tfrac12\alpha(4)^2=8\alpha \).
3. \( \theta_2=8\alpha-2\alpha=6\alpha \), so \( \theta_2/\theta_1=3 \).

Teacher's Note:
1. From rest, the angles covered in successive equal intervals are in the ratio 1 : 3 : 5 : ...
2. Here the first two equal intervals give 1 : 3.

Answer: (c) 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. Angular velocity is the time derivative of angular displacement; differentiate first, then substitute.
2. The constant 0.5 only fixes the starting position.

Answer: (a) \( 4\pi^2 \) (about 39.5)

Explanation:
1. \( \omega=2\times2\pi=4\pi \) rad/s and \( r=0.25 \) m.
2. \( a=\omega^2r=16\pi^2\times0.25=4\pi^2 \) m/s².

Teacher's Note:
1. Convert revolutions per second to radians per second before using the formula.
2. Uniform motion means there is no tangential acceleration, only the centripetal part.

Answer: (c) 4, 2 and \( 2\sqrt5 \)

Explanation:
1. \( v=2 \) cm/s at t = 1 s, so \( a_r=\frac{v^2}{r}=\frac{4}{1}=4 \).
2. \( a_t=\frac{dv}{dt}=2 \).
3. \( a=\sqrt{16+4}=2\sqrt5\approx4.47 \) cm/s².

Teacher's Note:
1. Differentiate the speed to get a_t, and use the speed at that instant for a_r.
2. The total is the vector sum, not the algebraic sum.

Answer: (a) 0.4

Explanation:
1. Normal reaction supplies the centripetal force: \( N=\frac{mv^2}{r} \).
2. Friction balances weight: \( \mu N\ge mg \).
3. \( \mu_{min}=\frac{rg}{v^2}=\frac{5\times10}{125}=0.4 \).

Teacher's Note:
1. The wall pushes inward (centripetal) while friction acts upward against weight.
2. A higher speed means a larger normal force and so a smaller required coefficient.

Answer: (b) \( 8T_0 \)

Explanation:
1. \( T=m\omega^2r \), so \( T\propto\omega^2r \).
2. New tension \( =m(2\omega_0)^2(2r)=8m\omega_0^2r=8T_0 \).

Teacher's Note:
1. Tension grows with the square of the angular velocity and linearly with the radius.
2. Factor = 2^2 x 2 = 8.

Answer: (d) \( \frac{u^2\sin^2\alpha}{g} \)

Explanation:
1. Angle with the vertical is \( \alpha \), so the horizontal component is \( u\sin\alpha \).
2. At the top the velocity is horizontal and the whole acceleration g is normal to it.
3. \( R=\frac{v^2}{a_n}=\frac{u^2\sin^2\alpha}{g} \).

Teacher's Note:
1. At the highest point only the horizontal component of velocity remains.
2. Be careful: the angle is measured from the vertical, which swaps sine and cosine.

Answer: (b) 5 rad/s

Explanation:
1. Tension is greatest at the lowest point: \( T-mg=m\omega^2r \).
2. \( 30-5=0.5\times\omega^2\times2\Rightarrow\omega^2=25 \), so \( \omega=5 \) rad/s.

Teacher's Note:
1. In a vertical circle the thread is most stretched at the bottom, where weight and centripetal need add up.
2. Use the breaking tension in the bottom-point equation.

Answer: (d) \( \frac{9}{2}mg \)

Explanation:
1. At the slack point \( mg\cos60^\circ=\frac{mv^2}{a}\Rightarrow v^2=\frac{ga}{2} \).
2. Energy from the bottom: \( u^2=v^2+2ga(1+\cos60^\circ)=\frac{ga}{2}+3ga=\frac{7ga}{2} \).
3. \( T-mg=\frac{mu^2}{a}=\frac72mg\Rightarrow T=\frac92mg \).

Teacher's Note:
1. The string goes slack when tension is zero, so the radial component of weight alone gives the centripetal force.
2. Height gained to this point is a(1 + cos 60).

Answer: (a) \( \sqrt{5gl} \)

Explanation:
1. At the top the minimum speed satisfies \( mg=\frac{mv_T^2}{l} \) so \( v_T^2=gl \).
2. Energy: \( \tfrac12mv_B^2=\tfrac12mv_T^2+2mgl \).
3. \( v_B^2=5gl \).

Teacher's Note:
1. The limiting case is zero tension at the top.
2. Add the kinetic energy needed at the top to the potential energy 2mgl gained.

Answer: (c) \( 2\sqrt{30} \) m/s (about 10.95 m/s)

Explanation:
1. \( v_{max}=\sqrt{\mu rg}=\sqrt{0.4\times30\times10}=\sqrt{120}=2\sqrt{30} \) m/s.

Teacher's Note:
1. Take the square root: the answer is about 11 m/s, which is roughly 39 km/h.
2. A value of 120 m/s would be absurdly large for a car on a 30 m curve.

Answer: (d) \( \frac{2}{45} \) m (about 4.4 cm)

Explanation:
1. \( v=\frac{48\times1000}{3600}=\frac{40}{3} \) m/s.
2. \( \tan\theta=\frac{v^2}{rg}=\frac{1600/9}{4000}=\frac{2}{45} \).
3. For a small angle \( h=l\tan\theta=\frac{2}{45} \) m.

Teacher's Note:
1. Convert 48 km/h to m/s first.
2. Because the angle is tiny, height over gauge equals tan theta.

Answer: (b) \( 100\sqrt3 \) m/s

Explanation:
1. \( v_{max}^2=rg\,\frac{\tan\theta+\mu}{1-\mu\tan\theta}=10^4\times\frac{1.5}{0.5}=3\times10^4 \).
2. \( v_{max}=100\sqrt3\approx173 \) m/s.

Teacher's Note:
1. Use the banked-with-friction formula for the upper limit.
2. Substitute tan 45 = 1.

Answer: (c) 10 m/s²

Explanation:
1. Period \( T=2\pi\sqrt{\frac{h}{g_{eff}}} \).
2. \( 2=2\pi\sqrt{\frac{2}{g_{eff}}}\Rightarrow g_{eff}=2\pi^2=20 \) m/s².
3. A lift moving down and retarding accelerates upward: \( g_{eff}=g+a\Rightarrow a=10 \) m/s².

Teacher's Note:
1. Inside an accelerating lift the effective gravity is g plus the upward acceleration.
2. Downward motion with retardation means an upward acceleration.

Answer: (a) 1 : 1

Explanation:
1. Period \( T=\frac{2\pi}{\omega} \) depends only on the angular speed.
2. Equal angular speeds give equal periods, ratio 1 : 1.

Teacher's Note:
1. Time for one turn depends on omega alone, not on radius or mass.
2. The faster linear speed on the larger circle compensates for the longer path.

Answer: (b) 200 rad

Explanation:
1. \( \theta=\frac{\omega_0+\omega}{2}t=\frac{0+80}{2}\times5=200 \) rad.

Teacher's Note:
1. For uniform acceleration the average angular velocity is the mean of initial and final values.
2. Displacement = average x time.

Answer: (d) Both velocity and acceleration change (in direction)

Explanation:
1. Velocity is tangent to the path and its direction keeps changing.
2. Acceleration points to the centre, and its direction also keeps changing, only its magnitude is constant.

Teacher's Note:
1. Vectors are constant only if both magnitude and direction are constant.
2. In uniform circular motion only the magnitudes stay fixed.

Answer: (a) P has greater acceleration than Q

Explanation:
1. \( a=\omega^2r \) with the same \( \omega \) for both, so \( a_P>a_Q \).

Teacher's Note:
1. All points of a rigid disc turn with the same angular velocity.
2. Centripetal acceleration then grows in proportion to the distance from the axis.

Answer: (c) 12 m/s

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

Teacher's Note:
1. The tension is the only force providing the centripetal force.
2. Solve T = mv^2/r for v at the breaking tension.