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JEE Advanced 2017 Paper 1 Offline
MCQ (More than One Correct Answer)
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A block of mass $$M$$ has a circular cut with a frictionless surface as shown. The block resets on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at $$x=0,$$ in a co-ordinate system fixed to the table. A point mass $$m$$ is released from rest at the topmost point of the path as shown and it slides down.

When the mass loses contact with the block, its position is $$x$$ and the velocity is $$v.$$ At that instant, which of the following options is/are correct?

JEE Advanced 2017 Paper 1 Offline Physics - Rotational Motion Question 45 English
A
The position of the point mass $$m$$ is :

$$x = - \sqrt 2 {{mR} \over {M + m}}$$
B
The velocity of the point mass $$m$$ is :

$$v = \sqrt {{{2gR} \over {1 + {m \over M}}}} $$
C
The $$x$$ component of displacement of the center

of mass of the block $$M$$ is: $$ - {{mR} \over {M + m}}$$
D
The velocity of the block $$M$$ is:

$$V = - {m \over M}\sqrt {2gR} $$
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