In the pulley-block arrangement shown in figure , find the relation between acceleration of blocks `A` and `B`

Options

A. `a_(B)=-3a_(A)`
B. `a_(B)=-a_(A)`
C. `a_(B)=-2a_(A)`
D. `a_(B)=-4a_(A)`

Answer

A

Explanation

`x_(1)+x_(3) =l_(1)`
differentiating with respect to time
we get `v_(1) +v_(3) =0`
Again differentiating w.r.to to time
`a_(1) +a_(3) = 0 rArr a_(10) = - a_(3), a_(3) = - a_(1)`
`(x_(1) -x_(3)) + (x_(2) -x_(3)) =l_(2)`
`x_(1) +x_(2) -2x_(3) =l_(2)`
differentiating w.r.to time `v_(1) + v_(2) -2v_(3) =0`
Again differentiating w.r.to time `a_(1) +a_(2) -2a_(3) =0`
`a_(1) +a_(2) +2a-(1) =0, 3a_(1) +a_(2) =0`
`a_(2) =-3a_(1), a_(B) = - 3a_(A)`
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