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

Correct answer: D. 1.99 v_ T (2)/(9) ( _0- _ ) r^2 ~g aligned & =(2)/(9)( 7825-925 2.45 × 10^ -2 ) ×(1.8)^2 × 10^ -6 × 9.8 & 1.99 aligned

JEE Main 20253 AprilEvening Shift

JEE Main 2025 · 3 April · Evening Shift · Physics · Mechanical Properties of Fluids

A solid steel ball of diameter 3.6 mm acquired terminal velocity...

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

D. $1.99$

\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}