Question

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 :

• A. $3:1$
• B. $4:1$
• C. $1:3$
• D. $1:4$

Answer 1

The correct option is A $3:1$

Frictional Force$=\frac{4\pi r^3}{3}(d_1-d_2)g$

Weight$=\frac{4\pi r^3}{3}\left(d_{1}g\right)$

where $d_1$ is density of ball

$d_2$ is density of liquid

Given:$d_2=4d_1$

Frictional Force$=\frac{4\pi r^3}{3}(d_1-4d_1)g=\frac{4\pi r^3}{3}(-3d_1)g$

Considering only magnitude of frictional force because its already known that both forces act in opposite directions.

Frictional Force$=3\frac{4\pi r^3}{3}\left(d_{1}g\right)=3\times$ Weight

Therefore the correct opton is (A).