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(C) Let $V$ be the volume of the stone, $\sigma$ be the density of the stone, and $\rho$ be the density of water.The relative density $K$ of the stone

Vedclass · Accepted Answer

$g\left(1-\frac{1}{K}\right)$

Vedclass · Answer

(C) Let $V$ be the volume of the stone, $\sigma$ be the density of the stone, and $\rho$ be the density of water.The relative density $K$ of the stone is given by $K = \frac{\sigma}{\rho}$, so $\sigma = K\rho$.The forces acting on the stone as it sinks are:$1$. Weight of the stone acting downwards: $W = V\sigma g = V(K\rho)g$.$2$. Buoyant force acting upwards: $F_B = V\rho g$.According to Newton's second law, the net force $F_{net} = W - F_B = ma$, where $m = V\sigma$ is the mass of the stone.$V\sigma g - V\rho g = (V\sigma)a$Dividing by $V\sigma$:$a = g - \frac{V\rho g}{V\sigma} = g - \frac{\rho}{\sigma}g$Since $K = \frac{\sigma}{\rho}$, we have $\frac{\rho}{\sigma} = \frac{1}{K}$.Therefore, the acceleration $a = g - \frac{g}{K} = g\left(1 - \frac{1}{K}\right)$.

$A$ stone of relative density $K$ is released from rest on the surface of a lake. If viscous effects are ignored, the stone sinks in water with an acceleration of

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