A body of density rho is dropped from rest fr | vedclass

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(C) Let $V$ be the volume of the body. When the body is dropped from height $h$,its velocity $v$ just before entering the water is given by $v = \sqrt

Vedclass · Accepted Answer

$\frac{h \rho}{\sigma-\rho}$

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(C) Let $V$ be the volume of the body. When the body is dropped from height $h$,its velocity $v$ just before entering the water is given by $v = \sqrt{2gh}$.As the body sinks to a maximum depth $h'$,its final velocity becomes zero. We can apply the work-energy theorem: the work done by gravity plus the work done by the buoyant force equals the change in kinetic energy.Work done by gravity = $(mg)h' = (V \rho g)h'$Work done by buoyant force = $-(F_B)h' = -(V \sigma g)h'$Change in kinetic energy = $K_f - K_i = 0 - \frac{1}{2}mv^2 = -\frac{1}{2}(V \rho)(2gh) = -V \rho gh$Equating the work done to the change in kinetic energy:$(V \rho g)h' - (V \sigma g)h' = -V \rho gh$$V g h' (\rho - \sigma) = -V \rho gh$$h' (\sigma - \rho) = h \rho$$h' = \frac{h \rho}{\sigma - \rho}$

$A$ body of density $\rho$ is dropped from rest from a height $h$ into a lake of density $\sigma$,where $\sigma > \rho$. Neglecting all dissipative forces,the maximum depth to which the body sinks before returning to float on the surface is:

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