JEE MAIN - Physics (2023 - 6th April Morning Shift - No. 6)
A small ball of mass $$\mathrm{M}$$ and density $$\rho$$ is dropped in a viscous liquid of density $$\rho_{0}$$. After some time, the ball falls with a constant velocity. What is the viscous force on the ball ?
$$\mathrm{F}=\mathrm{Mg}\left(1-\frac{\rho_{\mathrm{O}}}{\rho}\right)$$
$$\mathrm{F}=\mathrm{Mg}\left(1+\frac{\rho}{P_{o}}\right)$$
$$\mathrm{F}=\mathrm{Mg}\left(1+\frac{\rho_{\mathrm{o}}}{\rho}\right)$$
$$F=M g\left(1 \pm \rho \rho_{0}\right)$$
Explanation
When the ball is falling with a constant velocity, it means the net force acting on the ball is zero. This is because it's in a state of dynamic equilibrium - the downward force equals the upward force.
The downward force is the gravitational force (weight of the ball), which is given by $F_g = Mg$.
The upward force is the sum of buoyant force and the viscous drag. The buoyant force is the weight of the fluid displaced by the ball, which is given by $F_b = Vg\rho_0 = Mg\rho_0/\rho$ where $V = M/\rho$ is the volume of the ball.
The viscous force, $F_v$, is the force that we need to find.
Since the net force is zero, we have:
$F_g = F_b + F_v$
or
$Mg = Mg\rho_0/\rho + F_v$
which simplifies to
$F_v = Mg - Mg\rho_0/\rho = Mg(1 - \rho_0/\rho)$
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