A spherical solid ball of volume $V$ is made up of material of density $\rho$. It is falling through a…

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)
  1. $\left[\frac{V g \rho}{K}\right]^{\frac{1}{2}}$
  2. $\left[\frac{V g(\rho-\sigma)}{K}\right]^{\frac{1}{2}}$
  3. $\frac{V g(\rho-\sigma)}{K}$
  4. $\frac{V g \rho}{K}$

Solution

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 $F=K v_{\mathrm{T}}^2$ $\therefore \rho V g=\sigma V g+K v_{\mathrm{T}}^2$ $v_{\mathrm{T}}=\sqrt{\frac{(\rho-\sigma) V g}{K}}$

Asked in: MHT CET 2022 (10 Aug Shift 1)

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