A small rigid spherical ball of mass M is dropped in a long vertical tube containing glycerine. The velocity of the ball becomes constant after some time. If the density of glycerine is half of the density of the ball, then the viscous force acting on the ball will be (consider g as acceleration due to gravity)

Properties of Matter PYQ — JEE Main 2025 · 22 Jan · Shift 2 · Q70 with Solution

View written solutionFree

Correct answer: C

When the ball falls through glycerine and its velocity becomes constant, it has reached terminal velocity.

At terminal velocity, the net force on the ball is zero.

So,

\text{Weight} = \text{Buoyant force} + \text{Viscous force}

W = Mg

F_b = \rho_{g} V g

where $\rho_g$ is density of glycerine and $V$ is volume of the ball.

Since mass of the ball is $M$ and density of the ball is $\rho_b$ ,

M = \rho_b V \quad \Rightarrow \quad V = \frac{M}{\rho_b}

Given:

\rho_g = \frac{1}{2}\rho_b

Therefore,

F_b = \rho_g V g = \left(\frac{1}{2}\rho_b\right)\left(\frac{M}{\rho_b}\right)g = \frac{Mg}{2}

Let viscous force be $F_v$ .

Then,

Mg = F_b + F_v

Mg = \frac{Mg}{2} + F_v

So,

F_v = Mg - \frac{Mg}{2} = \frac{Mg}{2}

F_v = \frac{Mg}{2}

Hence, the correct option is:

C. $\dfrac{Mg}{2}$

Stored correct answer: C

This matches our derived answer.