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Question

The density ρ of a liquid varies with depth h from the free surface as ρ=kh. A small body of density ρ1 is released from the surface of the liquid. The body executes SHM with amplitude

A
2ρ1k
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B
ρ1k
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C
2ρ1k
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D
3ρ1k
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Solution

The correct option is B ρ1k
Let V be the volume of body which is released on the surface of liquid. When body reaches at depth h from the surface of liquid, then upward thrust on body due to liquid =V(kh)g

Weight of body =Vρ1g

At equilibrium, `V(kx)g=Vrho_(1)g or x=rho_(1)//k`

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