JEE Physics System of Particles and Rotational Motion MCQs Set 11

Answer: (b) \( \frac{R}{14} \)

Explanation:
1. The cavity has radius \( \frac{R}{2} \), so its mass is \( \frac{M}{8} \) and its centre is at distance \( \frac{R}{2} \) from the centre O of the sphere.
2. Mass left \( = \frac{7M}{8} \). Taking the removed mass as negative: shift \( = \frac{(M/8)(R/2)}{7M/8} = \frac{R}{14} \), directed away from the cavity.

Teacher's Note:
1. Treat the cavity as a negative mass at its own centre.
2. Shift = (removed mass × its distance) ÷ remaining mass.

Answer: (a) zero

Explanation:
1. Equal masses, so the centre of mass velocity is the vector mean of the three velocities.
2. The three velocities have equal magnitude along the sides AB, BC and CA of an equilateral triangle, which are at 120° to one another, so their vector sum is zero.
3. The centre of mass velocity is zero, so its displacement is zero.

Teacher's Note:
1. The vector sum of three equal vectors at 120° is zero.
2. Do not add the speeds as numbers.

Answer: (c) 1120 m

Explanation:
1. Components: \( v_x = 80 \) m/s, \( v_y = 60 \) m/s. Time to the top \( = 6 \) s, so the top is 480 m away horizontally and \( H = 180 \) m high.
2. Momentum conservation at the top: \( m(80) = \frac{3m}{4}v \), so the heavier part moves at \( v = \frac{320}{3} \) m/s horizontally.
3. It falls 180 m in \( \sqrt{2(180)/10} = 6 \) s, covering \( \frac{320}{3}\times 6 = 640 \) m more, so the distance is \( 480 + 640 = 1120 \) m.

Teacher's Note:
1. The vertical motion is unaffected by the break-up, so the fall time equals the rise time.
2. Momentum in the horizontal direction is conserved.

Answer: (d) \( \frac{16}{3}MR^2 \)

Explanation:
1. For a hollow sphere about a diameter, \( I = \frac{2}{3}MR^2 \).
2. The second sphere's centre is \( 2R \) from the axis, so by the parallel axis theorem \( I_2 = \frac{2}{3}MR^2 + M(2R)^2 = \frac{14}{3}MR^2 \).
3. Total \( I = \frac{2}{3}MR^2 + \frac{14}{3}MR^2 = \frac{16}{3}MR^2 \).

Teacher's Note:
1. Use the parallel axis theorem for the sphere not on the axis.
2. The distance between the centres of two touching spheres is 2R.

Answer: (b) aluminium at the interior and iron surrounding it

Explanation:
1. Moment of inertia depends on how far the mass is from the axis.
2. Iron is denser than aluminium, so placing the heavier iron at the outer rim (and light aluminium inside) puts more mass far from the axis and gives the largest I.

Teacher's Note:
1. Maximum I needs the heaviest material farthest from the axis.
2. Density of iron is about 7.9 g/cm³ and of aluminium about 2.7 g/cm³.

Answer: (c) the total mechanical energy remains constant in the absence of air resistance

Explanation:
1. Gravity exerts a torque about the point of projection, so the angular momentum changes.
2. Gravity changes the vertical component of momentum, so the linear momentum changes.
3. Only gravity does work (a conservative force), so total mechanical energy is constant without air resistance.

Teacher's Note:
1. Constant force: momentum changes. Constant total energy: yes if only gravity acts.
2. Torque is non-zero because the weight does not pass through the point of projection.

Answer: (c) \( \frac{4}{5}\omega \)

Explanation:
1. No external torque, so angular momentum is conserved.
2. \( I = \frac{1}{2}MR^2 \) and the added disc has \( I' = \frac{1}{2}\cdot\frac{M}{4}R^2 = \frac{I}{4} \).
3. \( I\omega = \left(I + \frac{I}{4}\right)\omega' \), so \( \omega' = \frac{4}{5}\omega \).

Teacher's Note:
1. The moment of inertia of a disc is proportional to its mass for the same radius.
2. Rotating speed drops because the moment of inertia increases.

Answer: (c) \( \sqrt{\frac{4}{3}gh} \)

Explanation:
1. Energy conservation: \( Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 \) with \( I = \frac{1}{2}MR^2 \) and \( \omega = \frac{v}{R} \).
2. \( Mgh = \frac{3}{4}Mv^2 \), so \( v = \sqrt{\frac{4}{3}gh} \).

Teacher's Note:
1. For rolling, include both translational and rotational energy.
2. The speed is less than the sliding-without-friction value √(2gh).

Answer: (a) its magnitude changes but the direction remains the same

Explanation:
1. The cylinder speeds up as it rolls down, so its angular momentum (about a fixed point or about its axis) grows in magnitude.
2. The axis of rotation stays horizontal and fixed in direction, so the direction of the angular momentum vector stays the same.

Teacher's Note:
1. Angular momentum is conserved only when no external torque acts; gravity and friction give a torque here.
2. Direction is along the axis of rotation.