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A uniform chain of length 2L is hanging ...

A uniform chain of length 2L is hanging in equilibrium position, if end `B` is given a slightly downward displacement the imbalance causes an acceleration. Here pullley is small and smooth & string is inextensible

The acceleration of end `B` when it has been displaced by distance `x`, is

A

`(x)/(L)g`

B

`(2x)/(L)g`

C

`(x)/(2)g`

D

g

Text Solution

Verified by Experts

The correct Answer is:
A
`lambda(l+x)g-T=lambda(l+x)a`…(i)
`T-lambda(l-x)g=lambda(l-x)a`…(ii)
after solving we get.
`a=(x)/(L)g`
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