Physics · Mechanical Properties of Matter

JEE Main 2026 — 2 April, Evening Shift — Question 6

The surface tension of a soap bubble is $0.03 \mathrm{N/m}$ . The work done in increasing the diameter of bubble from $2\mathrm{cm}$ to $6\mathrm{cm}$ is $\alpha \pi \times 10^{-4} \mathrm{J}$ . The value of $\alpha$ is ______. (Take $\pi=3.14$ )

Option A:
0.86
Option B:
0.64
Option C:
1.92
Correct
Option D:
7.68

Answer: C

Step-by-step solution

$\mathrm{N} = \mathrm{m}\omega^{2}\mathrm{R}$ , $\mathrm{f}_{\mathrm{s}} = \mathrm{mg}$ , $\mathrm{mg}\leq \mu \mathrm{m}\omega^{2}\mathrm{R}$ $\Rightarrow \mu \geq \frac{\mathrm{g}}{\omega^{2}\mathrm{R}} = \frac{10}{5^{2}\times 4} = 0.1$

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Exam: JEE Main 2026
Paper: 2 April, Evening Shift
Subject: Physics
Chapter: Mechanical Properties of Matter
Topic: Surface Tension and Surface Energy

If an air bubble of diameter 2 mm rises steadily through a liquid of density 2000 kg/m^3 at a rate of 0.5 cm/s , then the coefficient of…

Answer: C

Step-by-step solution

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