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Two blocks with masses M(1) and M(2) of ...

Two blocks with masses `M_(1)` and `M_(2)` of `10 kg` and `20 kg` respectively are placed as in fig. `mu_(s)=0.2` between all surfaces, then tension in string and acceleration of `M_(2)` block will be:

A

`250N,3m//s^(2)`

B

`200N,6m//s^(2)`

C

`306N,4.7m//s^(2)`

D

`400N,6.5m//s^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`therefore` The spring shall always remain in natural length. Since block `M_(1)` is attached to the string so it will remain stationary, `M_(2)` has tendency to move downward. So, friction on `M_(2)` will act in upward direction.
`N` is the force by which the both bolcks are pressed i.e., `N=T cos 30^(@)`.
for blocks `M_(1)`
`M_(1)g+mu(T cos 30^(@))-T sin 30^(@)=M_(1)(0)`....(1)
( `a=0` for `M_(1)` since attach to string )
for block `M_(2)`
`M_(2)g-(muN+muN)=M_(2)a` .....(2)
on solving `T=250 N` and `a=3m//s^(2)`
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