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Updated on : 2022-09-05

Solution

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Correct option is A)

$v=wA_{2}−y_{2} ⟹v_{2}=w_{2}(A_{2}−y_{2})$ ..........(1)

At the detaching point, the block will continue to move upwards due to inertia and reach the height $h$ above the mean position. At that height the velocity of the block will be zero.

Thus $0−v_{2}=2(−g)s⟹s=2gv_{2} $

Now the total height $h=s+y=2gv_{2} +y$

$∴h=2gw_{2}(A_{2}−y_{2}) +y$

For $h$ to be maximum, $dydh =0$

Thus $2gw_{2} (−2y)+1=0$

$⟹y=w_{2}g $

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