A ball rises to surface at a constant velocity in liquid whose density is 4 times greater than that of the…
A ball rises to surface at a constant velocity in liquid whose density is 4 times greater than that of the material of the ball. The ratio of the force of friction acting on the rising ball and its weight is
$3: 1$
$4: 1$
$1: 3$
$1: 4$
Solution
Frictional force $=$ Viscous Force $=V\left(\rho_1-\rho_2\right) g$
Weight $=\mathrm{mg}=\mathrm{V} \rho_1 \mathrm{~g}$
Where $\rho_1$ is the density of ball and $\rho_2$ is density of liquid.
$\begin{array}{ll}
& \rho_2=4 \rho_1 \\
\therefore \quad & V\left(\rho_1-4 \rho_1\right) g=V\left(-3 \rho_1\right) g . \\
& \text { Frictional force }=3 V\left(\rho_1 g\right) \\
\therefore \quad & \frac{\text { Frictional force }}{\text { Weight }}=\frac{3}{1}=3: 1 \text { ratio }
\end{array}$