A soap bubble is blown to a diameter of $7 \mathrm{~cm}$. $36960 \mathrm{~erg}$ of work is done in blowing it further. If surface tension of soap solution is 40 dyne/$\mathrm{cm}$ then the new radius is ________ cm Take $(\pi=\frac{22}{7})$.
Answer (integer)
7
Solution
<p>$$\begin{aligned}
& \Delta W=8 \pi\left(R_2^2-R_1^2\right) T \\
& 36960=8 \times \frac{22}{7} \times 40\left(R_2^2-\frac{49}{4}\right) \\
& R_2=7 \mathrm{~cm}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
This question is part of PrepWiser's free JEE Main question bank. 183 more solved questions on Properties of Solids and Liquids are available — start with the harder ones if your accuracy is >70%.