JEE Mains · Physics · STD 11 - Fluid mechanics
- JEE Mains 2025
- Medium
- MCQ
A solid steel ball of diameter 3.6 mm acquired terminal velocity \(2.45 \times 10^{-2} \mathrm{~m} / \mathrm{s}\) 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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(1.99\)
Step-by-step Solution
Detailed explanation
\(\mathrm{v}_{\mathrm{T}} \Rightarrow \frac{2}{9} \frac{\left(\rho_0-\rho_{\ell}\right) \mathrm{r}^2 \mathrm{~g}}{\eta}\)
\(\begin{aligned} & \eta=\frac{2}{9}\left(\frac{7825-925}{2.45 \times 10^{-2}}\right) \times(1.8)^2 \times 10^{-6} \times 9.8 \\ & \eta \approx 1.99\end{aligned}\)
\(\begin{aligned} & \eta=\frac{2}{9}\left(\frac{7825-925}{2.45 \times 10^{-2}}\right) \times(1.8)^2 \times 10^{-6} \times 9.8 \\ & \eta \approx 1.99\end{aligned}\)
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