A body having volume V and density rho is att | vedclass

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(B) When a container is accelerating upwards with acceleration $a$,the effective acceleration due to gravity becomes $g_{eff} = (g+a)$.The weight of t

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$V(g+a)(d - \rho)$

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(B) When a container is accelerating upwards with acceleration $a$,the effective acceleration due to gravity becomes $g_{eff} = (g+a)$.The weight of the body in this frame is $W = m g_{eff} = (\rho V)(g+a)$.The buoyant force (upthrust) acting on the body is $F_B = V d g_{eff} = V d (g+a)$.Since the body is attached to the bottom by a string and $d > \rho$,the buoyant force is greater than the weight,so the string will be in tension.Applying Newton's second law for the body in the accelerating frame:$T + W = F_B$$T = F_B - W$$T = V d (g+a) - V \rho (g+a)$$T = V(g+a)(d - \rho)$

$A$ body having volume $V$ and density $\rho$ is attached to the bottom of a container as shown. The density of the liquid is $d$ (where $d > \rho$). The container has a constant upward acceleration $a$. The tension in the string is:

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