Fluid Mechanics

JEE Main level, single correct, medium

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Question 165 of 192

A solid steel ball of diameter 3.6 mm acquired terminal velocity 2.45×10−2 m/s2.45 \times 10^{-2} \mathrm{~m} / \mathrm{s} while falling under gravity through an oil of density 925 kg m−3925 \mathrm{~kg} \mathrm{~m}^{-3}. Take density of steel as 7825 kg m−37825 \mathrm{~kg} \mathrm{~m}^{-3} and g as 9.8 m/s2\mathrm{m} / \mathrm{s}^2. The viscosity of the oil in SI unit is

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Answer: (D) 1.991.99

vT⇒29(ρ0−ρℓ)r2 gη\mathrm{v}_{\mathrm{T}} \Rightarrow \frac{2}{9} \frac{\left(\rho_0-\rho_{\ell}\right) \mathrm{r}^2 \mathrm{~g}}{\eta}
η=29(7825−9252.45×10−2)×(1.8)2×10−6×9.8η≈1.99\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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