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Assuming all the surface to be frictionl...

Assuming all the surface to be frictionless. The smaller block m is moving horizontally with acceleration a and vertically downwards with acceleration a. Then magnitude of net acceleration of smaller block m with respect to ground

A

`(2sqrt5mg)/((5m+M))`

B

`(2mg)/((5m+M))`

C

`7sqrt5g`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
Free body diagram for m For m,
`mg-T=m2a…(i)`
`N=ma …(ii)`
Free body diagram for M
For M,
`2T-N=Ma…(iii)`
on solving, `a=(2mg)/((M+5m))`
Net acceleration of m,
`a_(m)=sqrt(4a^(2)+a^(2))=sqrt5a=(2sqrt5mg)/((5m+M))`
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