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JEE Advance 2017
Paper-1
Multiple correct answers - Select all that apply
Question
A block of mass $M$ has a circular cut with a frictionless surface as shown. The block rests 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?
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Select ALL correct options:
A
The velocity of the point mass $m$ is : $v=\sqrt{\frac{2 g R}{1+\frac{m}{M}}}$
B
The $x$ component of displacement of the center of mass of the block $M$ is ; $-\frac{m R}{M+m}$
C
The position of the point mass is : $x=-\sqrt{2} \frac{m R}{M+m}$
D
The velocity of the block $M$ is : $V=-\frac{m}{M} \sqrt{2 g R}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
⚠ Partially correct. Some answers are missing.
Solution
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Question Tags
JEE Advance
Physics
Easy
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