A ball rises to surface at a constant velocity in liquid whose density is 3 times greater than that of the…
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
$2: 1$
$3: 1$
$4: 1$
$6: 1$
Solution
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)]