Question

A metal sphere of radius $'R'$ and density $'\rho'_{1}$ is dropped in a liquid of density $'\sigma'$ moves with terminal velocity $'v'$. Another metal sphere of same radius and density $'\rho'_{2}$ 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 $\frac{v\left(\rho_{2}-\sigma\right)}{\rho_{1}-\sigma}$.