A small steel ball is dropped into a long cylinder containing glycerine. Which one of the following is the correct repre

Question

A small steel ball is dropped into a long cylinder containing glycerine. Which one of the following is the correct representation of the velocity-time graph for the transit of the ball?

Accepted Answer

Given: A small steel ball falls through glycerine, a viscous liquid.

Find: The correct nature of the velocity-time graph during its motion.

As the ball is dropped, it starts accelerating downward due to gravity. While moving through glycerine, it experiences:

Gravitational force $mg$ downward
Buoyant force $F_B$ upward
Viscous drag force $F_d$ upward, which increases with velocity

Initially, the speed increases, so the drag force also increases. After some time, the net force becomes zero and the ball attains terminal velocity $v_t$ . Hence the graph rises at first and then gradually levels off to a constant value.

Therefore, the velocity approaches terminal velocity exponentially, so the correct option is B.

Given: The ball moves through glycerine under gravity, buoyancy, and viscous drag.

Find: The form of $v$ as a function of $t$ .

The forces acting on the ball are described by

mg - F_v - F_b = ma

where

F_b = \left( \frac{4}{3} \pi r^3 \right) g \rho_L, \quad F_v = 6 \pi \eta r v

Rearranging the force equation gives

\left( \frac{4}{3} \pi r^3 \right) g (\rho - \rho_L) = m \frac{dv}{dt}

Using constants

K_1 = \frac{4}{3} \pi r^3 (\rho - \rho_L) g, \quad K_2 = \frac{6 \pi \eta r}{m}

the differential equation becomes

\frac{dv}{dt} = K_1 - K_2 v

Integrating,

\int_0^v \frac{dv}{K_1 - K_2 v} = \int_0^t dt

which yields

v = \frac{K_1}{K_2} \left(1 - e^{-K_2 t}\right)

Thus $v$ increases with time exponentially and approaches a constant terminal value. Therefore, the correct option is B (Exponential).

Velocity-time graph starting from the origin, rising rapidly at first and then gradually leveling off toward a constant terminal velocity as time increases.

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