Question

A metal sphere of radius ${}^{\prime }{R}^{\prime }$ and density ${}^{\prime }{\rho }_1^{\prime }$ is dropped in a liquid of density ${}^{\prime }{\sigma }^{\prime }$ moves with terminal velocity ${}^{\prime }{v}^{\prime }$. Another metal sphere of same radius and density ${}^{\prime }{\rho }_2^{\prime }$ is dropped in the same liquid, its terminal velocity will be

• A. $v\left[{\rho }_2+\sigma )/\left({\rho }_1+\sigma \right)\right]$
• B. $v\left[{\rho }_1+\sigma )/\left({\rho }_2+\sigma \right)\right]$
• C. $v\left[{\rho }_2-\sigma )/\left({\rho }_1-\sigma \right)\right]$
• D. $v\left[{\rho }_1-\sigma )/\left({\rho }_2-\sigma \right)\right]$

Answer 1

Answer: (C)

The terminal velocity of the sphere
$v=\frac{2}{9}{R}^2\frac{\left({\rho }_1-\sigma \right)g}{\eta }$...(i)
Here, $R=$ radius of sphere,
${\rho }_1=$ density of sphere and
$\sigma =$ density of liquid.
Now, if density of metal sphere is changed to ${\rho }_2$, then terminal velocity
$v_2=\frac{2}{9}\frac{R^2\left({\rho }_2-\sigma \right)g}{\eta }$...(ii)
Divide Eqs. (i) and (ii), we get
$\frac{v_2}{v}=\frac{\left({\rho }_2-\sigma \right)}{\left({\rho }_1-\sigma \right)}$
or $v_2=\frac{v\left({\rho }_2-\sigma \right)}{\left({\rho }_1-\sigma \right)}$
Hence, the terminal velocity is $\frac{v\left({\rho }_2-\sigma \right)}{{\rho }_1-\sigma }$.