A metal block of base area 0.20 m^2 is placed on a table. A liqui… — Solution | JEE Physics

NVAEasyJEE 2023Viscosity & Stoke's Law

A metal block of base area $0.20 \, m^2$ is placed on a table. A liquid film of thickness $0.25 \, mm$ is inserted between the block and the table. The block is pushed by a horizontal force of $0.1 \, N$ and moves with a constant speed. If the viscosity of the liquid is $5.0 \times 10^{-3} \, Pl$ , the speed of the block is _____ $\times 10^{-3} \, m/s$ .

Answer

Correct answer:25

Step-by-step solution

Standard Method

Given: base area $A = 0.20 \, m^2$ , liquid film thickness $\Delta h = 0.25 \times 10^{-3} \, m$ , horizontal force $F = 0.1 \, N$ , viscosity $\eta = 5 \times 10^{-3} \, Pl$ .

Find: the speed of the block in the form _____ $\times 10^{-3} \, m/s$ .

For steady motion, the applied force is balanced by the viscous force:

F = \eta A \frac{\Delta v}{\Delta h}

Substituting the given values:

0.1 = (5 \times 10^{-3})(0.2)\frac{v}{0.25 \times 10^{-3}}

Solving for $v$ :

v = \frac{0.1 \cdot 0.25 \times 10^{-3}}{5 \times 10^{-3} \cdot 0.2}

Therefore,

v = 25 \times 10^{-3} \, m/s

Thus, the required numerical value is 25.

Common mistakes

Using the thickness directly as $0.25$ instead of converting $0.25 \, mm$ to $0.25 \times 10^{-3} \, m$ is incorrect. The SI unit must be used in the viscous force formula.
Treating the motion as accelerated and applying Newton's second law is incorrect. The block moves with constant speed, so the applied force is balanced by viscous drag.
Substituting the formula incorrectly as $F = \eta \frac{A \Delta h}{\Delta v}$ is wrong. The correct relation is $F = \eta A \frac{\Delta v}{\Delta h}$ .

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