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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 velocity `v` of the string when it slips out of the pulley (height of pulley from floor`gt2L`)

A

`sqrt((gL)/(2))`

B

`sqrt(2gL)`

C

`sqrt(gL)`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C

`v(dv)/(dx)=(x)/(L)g" "rArrint_(0)^(v)"v dv"=(g)/(L)int_(0)^(L)"x dx"rArrvsqrt(gL)`
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