A ball rises to surface at a constant velocity in liquid whose density is 3 times greater than that of the material of the ball. The ratio of force of friction acting on the rising ball to its weight is

The ball is moving with constant velocity. Therefore, the net force acting on it is zero.
The weight of the ball \(=\mathrm{W}=\frac{4}{3} \pi \mathrm{r}^3 \rho_{\mathrm{b}} \mathrm{g}\) ...(i)
....(acting downwards)
As the ball is rising up, the viscous force \(F_v\) will be in the downward direction.
The buoyant force \(F_B=\frac{4}{3} \pi r^3 \rho_l g\)
....(acting upwards)
\(\begin{array}{ll}
\therefore & \mathrm{F}_{\mathrm{v}}+W=F_B \\
\therefore & \mathrm{~F}_{\mathrm{v}}=\mathrm{F}_{\mathrm{B}}-\mathrm{W}=\frac{4}{3} \pi \mathrm{r}^3 g\left(\rho_l-\rho_{\mathrm{b}}\right) \\
& \mathrm{F}_{\mathrm{v}}=\frac{4}{3} \pi \mathrm{r}^3 \mathrm{~g} \times 2 \rho_{\mathrm{b}} \quad \ldots\left(\because \rho_l=3 \rho_{\mathrm{b}}\right)...(ii)
\end{array}\)
\(\therefore \quad \frac{F_v}{W}=\frac{\frac{4}{3} \pi r^3 \mathrm{~g} \times 2 \rho_b}{\frac{4}{3} \pi r^3 \rho_b g}=\frac{2}{1} \quad \ldots\) [From(i) and (ii)]