Q 11-09-082JEE MainJEE Main 2023 (29 Jan, Shift 2)Easy
A metal block of base area $0.20\ \text{m}^2$ is placed on a table, as shown in figure. A liquid film of thickness $0.25\ \text{mm}$ is inserted between the block and the table. The block is pushed by a horizontal force of $0.1\ \text{N}$ and moves with a constant speed. If the viscosity of the liquid is $5.0\times10^{-3}\ \text{Pl}$, the speed of block is ______ $\times10^{-3}\ \text{m s}^{-1}$.
Numerical value type. Enter your answer.
Answer: 25
Viscous force $F=\eta A\dfrac{v}{d}$, so
$$v=\frac{Fd}{\eta A}=\frac{0.1\times0.25\times10^{-3}}{5\times10^{-3}\times0.20}=0.025\ \text{m s}^{-1}=25\times10^{-3}\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics