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IIT JEE Super Course in Physics: Mechanics II
A solid ball of diameter $11 \mathrm{~cm}$ is rotating about one of its horizontal diameters with an angular velocity of $120 \mathrm{rad} \mathrm{s}^{-1}$ It is released from a height $=1.8 \mathrm{~m}$ and falls freely to collide with the horizontal floor $\left(\mathrm{e}=\frac{5}{6}\right) \cdot \mu$ between the ball and ground is $0.2$. The fractional change in angular momentum after collision is nearly
(a) $0.4$
(b) $0.5$
(c) $0.6$
(d) $0.7$