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A spherical solid ball of volume $V$ is made up of material of density $\rho$. It is falling through a liquid of density $\sigma(\sigma<\rho)$. Assume that the liquid applies a viscous force on the ball that is proportional to square of the terminal speed $v_{\mathrm{T}}, F=-K v_{\mathrm{T}}^2, \forall(K>0)$, then the terminal speed of the ball is ( $g=$ acceleration due to gravity)
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The correct answer is:
$\frac{V g(\rho-\sigma)}{K}$
The condition for terminal speed $v_{\mathrm{T}}$ is given by,
weight $(W)=$ buoyant force $(f)+$ viscous force $(F)$
We know, $W=\rho V g, f=\sigma V g$ and $F=K v_{\mathrm{T}}^2$
$\therefore \rho V g=\sigma V g+K v_{\mathrm{T}}^2$
$v_{\mathrm{T}}=\sqrt{\frac{(\rho-\sigma) V g}{K}}$
weight $(W)=$ buoyant force $(f)+$ viscous force $(F)$
We know, $W=\rho V g, f=\sigma V g$ and $F=K v_{\mathrm{T}}^2$
$\therefore \rho V g=\sigma V g+K v_{\mathrm{T}}^2$
$v_{\mathrm{T}}=\sqrt{\frac{(\rho-\sigma) V g}{K}}$
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