Q75 / 105 · JEE Main 2008 · Physics

A spherical solid ball of volume $V$ is made of a material of density $\rho_1$. It is falling through a liquid of density $\rho_2\left(\rho_2 < \rho_1\right)$. Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed v, i.e., $F_{\text {viscous }}=-k v^2(k>0)$. The terminal speed of the ball is

  • Correct answer
    A
    $\sqrt{\frac{\operatorname{Vg}\left(\rho_1-\rho_2\right)}{\mathrm{k}}}$
  • B
    $\frac{\mathrm{Vg} \rho_1}{\mathrm{k}}$
  • C
    $\sqrt{\frac{V g \rho_1}{\mathrm{k}}}$
  • D
    $\frac{\operatorname{Vg}\left(\rho_1-\rho_2\right)}{\mathrm{k}}$

Solution

$$ \begin{aligned} & \rho_1 V g-\rho_2 V g=k v_T^2 \\ & \Rightarrow \quad V_T=\sqrt{\frac{V g\left(\rho_1-\rho_2\right)}{k}} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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