Question

If a ball of steel (density $\rho =7.8gc{m}^{-3}$) attains a terminal velocity of $10cm{s}^{-1}$ when falling in a tank of water (coefficient of viscosity ${\eta }_{water}=8.5\times {10}^{-4}Pa.s$) then its terminal velocity in glycerine $(\rho =12gc{m}^{-3},\eta =13.2pa.s)$ would be nearly:-

• A. $1.6\times {10}^{-5}cm{s}^{-1}$
• B. $6.25\times {10}^{-4}cm{s}^{-1}$
• C. $6.45\times {10}^{-4}cm{s}^{-1}$
• D. $1.5\times {10}^{-5}cm{s}^{-1}$

Answer 1

Answer: (B)

$6.25\times {10}^{-4}cm{s}^{-1}$

Given,

$v_1=10cm/s$

$\rho =7.8g/c{m}^3$

${\rho }_w=1g/c{m}^3$

${\rho }_g=12g/c{m}^3$

${\eta }_g=13.2Pa.s$

${\eta }_w=8.5\times {10}^{-4}Pa.s$

$v_2=?$

Terminal velocity of any liquid,

$v=\frac{2r^2(\rho -{\rho }_0)g}{9\eta }$

$v\propto \frac{(\rho -{\rho }_0)}{\eta }$

$\frac{v_2}{v_1}=\frac{\rho -{\rho }_g}{{\eta }_g}\times \frac{{\eta }_w}{\rho -{\rho }_w}$

$\frac{v_2}{10}=\frac{7.8-1.2}{13.2}\times \frac{8.5\times {10}^{-4}}{7.8-1}=0.625\times {10}^{-4}$

$v_2=10\times 0.625\times {10}^{-4}$

$v_2=6.25\times {10}^{-4}cm/s$

The correct option is B.