The velocity of a small ball of mass m and de | vedclass

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(C) When the ball reaches a constant velocity (terminal velocity),the net force acting on it is zero.The forces acting on the ball are:$1$. Gravitatio

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$mg\left( 1 - \frac{\sigma}{\rho} \right)$

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(C) When the ball reaches a constant velocity (terminal velocity),the net force acting on it is zero.The forces acting on the ball are:$1$. Gravitational force $(mg)$ acting downwards.$2$. Buoyant force $(F_B)$ acting upwards.$3$. Viscous force $(F_v)$ acting upwards.At terminal velocity,the equilibrium condition is:$mg = F_B + F_v$$F_v = mg - F_B$The buoyant force $F_B$ is equal to the weight of the displaced liquid:$F_B = V \sigma g$,where $V$ is the volume of the ball.Since $m = V \rho$,we have $V = \frac{m}{\rho}$.Substituting $V$ into the buoyant force equation:$F_B = \left( \frac{m}{\rho} \right) \sigma g = mg \left( \frac{\sigma}{\rho} \right)$.Now,substitute $F_B$ into the equilibrium equation:$F_v = mg - mg \left( \frac{\sigma}{\rho} \right)$$F_v = mg \left( 1 - \frac{\sigma}{\rho} \right)$.

The velocity of a small ball of mass $m$ and density $\rho$ when dropped in a container filled with glycerine of density $\sigma$ becomes constant after some time. The viscous force acting on the ball in the final stage is

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