A hollow sphere A and a solid sphere B, both having same mass and radius are released simultaneously from the top of a right triangular fixed wedge as shown. They roll down the wedge without slipping. Choose the correct statements.

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They reach the bottom of the wedge simultaneously

A reaches the bottom of wedge before B

They reach the bottom of wedge with the same kinetic energy

A reaches the bottom of wedge with speed more than that of B

answer is B, C.

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Detailed Solution

$a = \frac{g \sin θ}{1 + \frac{k^{2}}{R^{2}}}, t = \sqrt{\frac{2 s}{a}} = \sqrt{\frac{2 h}{a \sin θ}} (∵ s \sin θ = h)$
For $A, \frac{2}{3} m R^{2} = m k^{2} \Rightarrow \frac{k^{2}}{R^{2}} = \frac{2}{3}$
$∴ a_{A} = \frac{g \sin 60^{o}}{1 + \frac{2}{3}} = \frac{3 \sqrt{3} g}{10}$
$t_{A} = \sqrt{\frac{\frac{2 h}{3 \sqrt{3} g}}{10}} \times \frac{\sqrt{3}}{2} = \sqrt{\frac{40 h}{9 g}}$
For B, $\frac{2}{5} m R^{2} = m k^{2} \Rightarrow \frac{k^{2}}{R^{2}} = \frac{2}{5}$
$∴ a_{B} = \frac{g \sin 30^{o}}{1 + \frac{2}{5}} = \frac{5 g}{14}$
$t_{B} = \sqrt{\frac{2 h}{\frac{5 g}{14} \times \frac{1}{2}}} = \sqrt{\frac{56 h}{5 g}} ∴ t_{B} > t_{A}$
$k_{A} = k_{B} = m g h$
$k_{A} = \frac{1}{2} m v_{A}^{2} + \frac{1}{2} \frac{2}{3} m R^{2} {(\frac{v_{A}}{R})}^{2} = \frac{5}{6} m v_{A}^{2}$
$k_{B} = \frac{1}{2} m v_{B}^{2} + \frac{1}{2} \frac{2}{3} m R^{2} {(\frac{v_{B}}{R})}^{2} = \frac{7}{10} m v_{B}^{2}$
$∴ \frac{5}{6} m v_{A}^{2} = \frac{7}{10} m v_{B}^{2} \Rightarrow v_{B} = \sqrt{\frac{25}{21} v_{A}}$