Question

The mass of a lead ball is M. It falls down in a viscous liquid with terminal velocity V. The terminal velocity of another lead ball of mass 8M in the same liquid will be:

$$A.\;64V \\\ B.4V \\\ C.\;8V \\\ D.V \\\$$

Answer 1

Terminal velocity is directly proportional to the square of the radius of the ball$V \propto {r^2}$. The mass of the second ball is given from which its radius can also be obtained by evaluating the respective cases. Now substitute the values in to find the terminal velocity of another lead ball.

Complete step-by-step answer

If a spherical body of radius r is dropped in a viscous fluid, it is first accelerated and then becomes zero and it attains a constant velocity called terminal velocity. It is given by,
$V = \dfrac{{2{r^2}(\rho - \sigma )g}}{{9\eta }}$
From the above equation we know that
$V \propto {r^2}$
Let ${r_1}$and ${r_2}$be the radius of the first and second ball respectively.
Given that mass of the lead ball is M and for the second ball is 8M. The terminal velocity for the first case is V.
Case (1), since density is mass by volume,
$\rho \times \dfrac{4}{3}\pi {r_1}^3 = M$
${V_1} = \dfrac{{2{r_1}^2(\rho - \sigma )g}}{{9\eta }}$
Case (2), for the second ball,
$\rho \times \dfrac{4}{3}\pi {r_2}^3 = 8M$
Divide Case (1) by (2)
$\dfrac{{\rho \times \dfrac{4}{3}\pi {r_1}^3}}{{\rho \times \dfrac{4}{3}\pi {r_2}^3}} = \dfrac{M}{{8M}}$
${r_2}^3 = 8{r_1}^3$
${r_2} = 2{r_1}$
Since,
$V \propto {r^2}$
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{r_1}^2}}{{{r_2}^2}}$
Substitute the values of radius one first and second ball
${V_2} = \dfrac{{V \times 4r_1^2}}{{r_1^2}}$
${V_2} = 4V$

The terminal velocity for the second case is 4V and the correct option is B.

Note: Terminal velocity is directly proportional to the density of the ball. In the terminal velocity expression, if$\rho > \sigma$ , $V$ is positive and hence the body will attain constant velocity is the downward direction. If$\rho < \sigma$ ,$V$ is negative and the spherical body will attain constant velocity in upward direction.