Physics · Mechanical Properties of Matter

JEE Main 2024 — 4 April, Shift 1 — Question 54

A soap bubble is blown to a diameter of 7 cm . 36960 erg of work is done in blowing it further. If surface tension of soap solution is 40 dyne/cm then the new radius is $\qquad$ cm. Take $:\left(\pi=\frac{22}{7}\right)$

Answer: 7

Numerical answer — enter this value.

Step-by-step solution

$\omega=\Delta \mathrm{U}=\mathrm{S} \Delta \mathrm{A}$

$36960 \mathrm{erg}=\frac{40 \text { dyne }}{\mathrm{cm}} 8 \pi\left[(\mathrm{r})^{2}-\left(\frac{7}{2}\right)^{2}\right] \mathrm{cm}^{2}$

$\mathrm{r}=7 \mathrm{~cm}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Physics
Chapter: Mechanical Properties of Matter
Topic: Surface Tension and Surface Energy