A ball rises to the surface of a liquid with constant velocity. The density of the liquid is four times the density of the material of the ball. The viscous force of the liquid on the rising ball is greater than the weight of the ball by a factor of

The ball is moving with constant velocity.
Hence the net force acting on it is zero.
The weight of the ball, \(\mathrm{W}=\frac{4}{3} \pi \mathrm{r}^3 \rho_{\mathrm{b}} \mathrm{g}\) (in downward direction)
The viscous force \(F_v\) is in downward direction
The buoyant force \(\mathrm{F}_{\mathrm{B}}=\frac{4}{3} \pi \mathrm{r}^3 \rho_{\ell}\) g (in upward direction)
\(
\begin{aligned}
& \therefore \mathrm{F}_{\mathrm{v}}+\mathrm{W}=\mathrm{F}_{\mathrm{B}} \\
& \therefore \mathrm{F}_{\mathrm{v}}=\mathrm{F}_{\mathrm{B}}-\mathrm{W} \\
& =\frac{4}{3} \pi \mathrm{r}^3 \mathrm{~g}\left(\rho_{\ell}-\rho_{\mathrm{b}}\right)=\frac{4}{3} \pi \mathrm{r}^3 \mathrm{~g} \times 3 \rho_{\mathrm{b}} \quad\left(\because \rho_{\ell}=4 \rho_{\mathrm{b}}\right) \\
& \therefore \frac{\mathrm{F}_{\mathrm{v}}}{\mathrm{W}}=3
\end{aligned}
\)