A solid steel ball of diameter 3.6 mm acquired terminal velocity $2.45 imes 10^{-2} \mathrm{~m} / \mathrm{s}$ while falling under gravity through an oil of density $925 \mathrm{~kg} \mathrm{~m}^{-3}$. Take density of steel as $7825 \mathrm{~kg} \mathrm{~m}^{-3}$ and g as 9.8 $\mathrm{m} /…

Question

A solid steel ball of diameter 3.6 mm acquired terminal velocity $2.45 	imes 10^{-2} \mathrm{~m} / \mathrm{s}$ while falling under gravity through an oil of density $925 \mathrm{~kg} \mathrm{~m}^{-3}$. Take density of steel as $7825 \mathrm{~kg} \mathrm{~m}^{-3}$ and g as 9.8 $\mathrm{m} / \mathrm{s}^2$. The viscosity of the oil in SI unit is

Aicharya · Accepted Answer

$\mathrm{v}_{\mathrm{T}} \Rightarrow \frac{2}{9} \frac{\left(\rho_0-\rho_{\ell}\right) \mathrm{r}^2 \mathrm{~g}}{\eta}$
$\begin{aligned} & \eta=\frac{2}{9}\left(\frac{7825-925}{2.45 \times 10^{-2}}\right) \times(1.8)^2 \times 10^{-6} \times 9.8 \\ & \eta \approx 1.99\end{aligned}$

A solid steel ball of diameter 3.6 mm acquired terminal velocity $2.45 \times 10^{-2} \mathrm{~m} /…

Solution