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A block of mass 1 kg is at rest relative...

A block of mass `1 kg` is at rest relative to a smooth wedge moving leftwards with constant acceleration `a = 5 m//s^(-2)` .Let `N` be the normal reation between the block and the wedge . Then `(g = 10 m//s^(-2))`

A

`N = 5sqrt5` newton

B

`N = 15` newton

C

`tan theta = (1)/(2)`

D

`tan theta = 2`

Text Solution

Verified by Experts

The correct Answer is:
A, C

`N cos theta = mg = 10` …(i)
`N sin theta = ma = 5` …(ii)
On solving these two equations, we get

`N = 5 sqrt5 N` and `tan theta = (1)/(2)`
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