A sphere of mass $m$ and radius $r$ is released from a wedge of mass $2 m$ as shown. $A B C$ is hemispherical position of radius $R$ . The impulse imparted to the system consisting wedge and sphere by the vertical wall $w_{1} w_{2}$ till the time sphere reaches at the bottom most position of spherical portion for the first time is: (Friction between wedge and horizontal surface is absent and between sphere and wedge friction is sufficient to avoid slipping between them.)

m \sqrt{\frac{2 g (R - r)}{7}}

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m \sqrt{\frac{10 g (R - r)}{7}}

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2 m \sqrt{\frac{5 g (R - r)}{7}}

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3 m \sqrt{\frac{2 g (R - r)}{7}}

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None of these

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The correct option is B $m \sqrt{\frac{10 g (R - r)}{7}}$

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Similar questions

Q. A uniform solid sphere of mass m and radius r is released from a wedge of mass 2m as shown. ABC is hemispherical position of radius R. Impulse imparted to the system consisting wedge and sphere by the vertical wall

w_{1} w_{2}

till the time sphere reaches at the bottom most position of spherical portion for the first time is

m \sqrt{\frac{10 g (R - r)}{δ}}

. Friction between wedge and horizontal surface is absent and between sphere and wedge friction is sufficient to avoid slipping between them. Here

δ

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The correct option is B m√10g(R−r)7

The correct option is B $m \sqrt{\frac{10 g (R - r)}{7}}$