Question

A spherical solid ball of volume $V$ is made of a material of density ${\rho _1}$ . it is falling through a liquid of density ${\rho _2}$ $({\rho _2} < {\rho _1})$ . Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed v i.e., ${F_{viscous}} = - k{V^2}(k > 0)$ , The terminal speed of the ball is
(A) $\sqrt {\dfrac{{Vg({\rho _1} - {\rho _2})}}{k}}$
(B) $\dfrac{{Vg{\rho _1}}}{k}$
(C) $\sqrt {\dfrac{{Vg{\rho _1}}}{k}}$
(D) $\dfrac{{Vg({\rho _1} - {\rho _2})}}{k}$

Answer 1

Hint : Write the equation of motion of the body to find the terminal velocity. Equation of motion of the body can be written as total force on the body when terminal velocity is achieved is zero, that can be written as, ${F_{net}} = 0$ . where, is the total force acting on the body , which is, viscous force, gravitation and buoyancy.

Complete Step By Step Answer:
We know that, when a body is falling with a terminal velocity inside a fluid the net force acting on it is zero, that means, ${F_{net}} = 0$ . Here, the forces which are acting on it are, the buoyant force of the liquid upwards, gravitation of the earth downwards and the viscous force given by, ${F_{viscous}} = - k{v_t}^2(k > 0)$ , which is acting upwards as velocity of the ball is downwards.
(let, terminal velocity of the ball is ${v_t}$ ) Here, we have given that the density of the liquid is ${\rho _1}$ ,and the volume of the ball is $V$ .
So, the buoyant force of the liquid is $V{\rho _2}g$ .
Force of gravitation on the spherical ball is $V{\rho _1}g$ .
Hence, equation of motion of the body at terminal speed given by,
${F_{net}} = 0$
$\Rightarrow V{\rho _1}g - V{\rho _2}g - kv_t^2 = 0$
Therefore,
$V{\rho _1}g = V{\rho _2}g + kv_t^2$
$\therefore kv_t^2 = V{\rho _1}g - V{\rho _2}g$
Therefore, simplifying we get, ${v_t} = \sqrt {\dfrac{{Vg\left( {{\rho _1} - {\rho _2}} \right)}}{k}}$
Hence, this is our solution. So, option ( A) is correct

Note :
$\bullet$ When the body has not achieved the terminal velocity then ${F_{net}} = V{\rho _1}g - V{\rho _2}g$ where, $g$ is the gravitational acceleration. The viscous force of the liquid cancels out this force then the net force on the body is zero.
$\bullet$ In general, viscous force is given by the Stokes’ law. It is given by, $F = 6\pi \eta rv$ .So, in general viscous force is directly proportional to velocity but here, the force is said to be proportional to the square of the velocity.