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 answerA$\sqrt{\frac{\operatorname{Vg}\left(\rho_1-\rho_2\right)}{\mathrm{k}}}$
- 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'.