Question

If a ball of steel (density ${\rho }_s=7.8{\text{g/cm}}^3$) attains a terminal velocity of $10\text{cm/s}$ when falling in a tank of water (coefficient of viscosity, ${\eta }_w=8.5\times {10}^{-4}\text{Pa-s}$). Then, its terminal velocity in glycerine (${\rho }_g=1.2{\text{g/cm}}^3,{\eta }_g=13.2\text{Pa-s}$) would be nearly:

• A. $1.6\times {10}^{-5}\text{cm/s}$
• B. $1.5\times {10}^{-5}\text{cm/s}$
• C. $6.45\times {10}^{-4}\text{cm/s}$
• D. $6.25\times {10}^{-4}\text{cm/s}$

Answer 1

The correct option is D $6.25\times {10}^{-4}\text{cm/s}$

Given: ${\rho }_s=7.8{\text{g cm}}^{-3}$
${\rho }_w=1{\text{g cm}}^{-3}$
${\rho }_g=1.2{\text{g cm}}^{-3}$
${\eta }_g=13.2\text{Pa-s}$
${\eta }_w=8.5\times {10}^{-4}\text{Pa-s}$

We can say, at terminal velocity condition, net force on the ball must be zero.
i.e $W-F_B=F_v$
where $W=$ Weight
$F_B=$ Buoyant force
$F_v=$ Viscous force

In water:
$V_b({\rho }_s-{\rho }_w)=6\pi {\eta }_{w}r{v}_{T_w}$ $\to \dots \left(1\right)$
where $V_b$ is the volume of the ball, $r$ its radius and $v_T$ is the terminal velocity.

In glycerine:
$V_b({\rho }_s-{\rho }_g)=6\pi {\eta }_{g}r{v}_{T_g}\dots \left(2\right)$

$\frac{\left(2\right)}{\left(1\right)}$ gives:

$\frac{{\rho }_s-{\rho }_g}{{\rho }_s-{\rho }_w}=\frac{{\eta }_g{v}_{T_g}}{{\eta }_w{v}_{T_w}}$

$\Rightarrow \frac{7.8-1.2}{7.8-1}=\frac{13.2\times V_{T_g}}{8.5\times {10}^{-4}\times 10}$
$\Rightarrow V_{T_g}=\frac{6.6\times 8.5\times {10}^{-3}}{6.8\times 13.2}$
$\Rightarrow V_{T_g}=6.25\times {10}^{-4}{\text{cm s}}^{-1}$