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Wedge of 10 kg is free to move on horizo...

Wedge of `10 kg` is free to move on horizontal surface. At the given instant, acceleration of wedge is (string and pulleys are ideal)

A

`2 m//s^(2)` towards right

B

`2 m//s^(2)` towards left

C

`1 m//2^(2)` towards left

D

`1 m//s^(2)` toward right

Text Solution

Verified by Experts

The correct Answer is:
D
FBD of wedge FBD :

`T = F = 50 N`
`F cos 37^(2)-T cos 53^(@) =10a`
`rArra =1 m//s_(2)`
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