A steel ball of radius $0.05 \mathrm{~cm}$ and density $7.8 \mathrm{~g} \mathrm{~cm}^{-3}$ is dropped into a…
A steel ball of radius $0.05 \mathrm{~cm}$ and density $7.8 \mathrm{~g} \mathrm{~cm}^{-3}$ is dropped into a tank of water. The terminal velocity of the steel ball is
(Density of water $=1 \mathrm{~g} \mathrm{~cm}^{-3}$ and viscosity of water $=$ $0.001 \mathrm{~Pa} \mathrm{~s}$ )
$3.42 \mathrm{~ms}^{-1}$
$1.81 \mathrm{~ms}^{-1}$
$5.11 \mathrm{~ms}^{-1}$
$3.77 \mathrm{~ms}^{-1}$
Solution
Radius of steel ball, $\mathrm{r}=0.05 \times 10^{-2} \mathrm{~m}$
Density of steel, $\rho=7.8 \mathrm{~g} \mathrm{~cm}^{-3}=\frac{7.8 \times 10^{-3}}{10^{-6}}$ $=7800 \mathrm{~kg} / \mathrm{m}^3$
Density of water, $\rho_\omega=1 \mathrm{~g} \mathrm{~cm}^{-3}$
$=\frac{10^{-3}}{10^{-6}}=1000 \mathrm{~kg} / \mathrm{m}^3$
Viscosity of water, $\eta=0.001$ pas Terminal velocity is given by
$\begin{aligned} & \mathrm{v}_{\mathrm{T}}=\frac{2}{9} \frac{\mathrm{r}^2\left(\rho-\rho_\omega\right)}{\eta} \\ & =\frac{2 \times\left(0.05 \times 10^{-2}\right)^2 \times(7800-1000)}{9 \times 0.001} \\ & =3.77 \mathrm{~m} / \mathrm{s}\end{aligned}$