Q 11-09-085JEE MainJEE Main 2023 (24 Jan, Shift 2)Medium
A spherical ball of radius $1\ \text{mm}$ and density $10.5\ \text{g cc}^{-1}$ is dropped in glycerine of coefficient of viscosity $9.8$ poise and density $1.5\ \text{g cc}^{-1}$. Viscous force on the ball when it attains constant velocity is $3696\times10^{-x}$ N. The value of $x$ is ______. (Given, $g=9.8\ \text{m s}^{-2}$ and $\pi=\frac{22}{7}$)
Numerical value type. Enter your answer.
Answer: 7
At terminal velocity, viscous force $=$ weight $-$ buoyancy:
$$F=\frac43\pi r^3(\rho-\sigma)g=\frac43\times\frac{22}{7}\times10^{-9}\times9000\times9.8$$
$$F=3.696\times10^{-4}\ \text{N}=3696\times10^{-7}\ \text{N}$$
So $x=7$.
Solution by Sreeraj P, M.Sc Physics