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A uniform chain of length 2L is hanging in equilibrium position, if end B is given aslightly downward displacement the imbalance causes an acceleration. Here pulley issmall and smooth& string is inextensibleQ.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

A uniform chain of length 2L is hanging in equilibrium position, if end B is given aslightly downward displacement the imbalance causes an acceleration. Here pulley issmall and smooth& string is inextensibleQ.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

Grade:11

1 Answers

Vishwajit
21 Points
6 years ago
\lambda(L+x)g-T=\lambda(L+x)a
T-\lambda(L-x)g=\lambda(L-x)a
Where lambda is mass per unit length of both ends 
Solve these equations by adding them give you
2xg=2La
(x/L)g=a
And  these question is specially check you concept in NLM

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