JEE Mains · Physics · STD 11 - Surface tension
- JEE Mains 2024
- Hard
- Numerical
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 \(40 \mathrm{dyne} / \mathrm{cm}\) then the new radius is ______ \(\mathrm{cm}\). Take \(:\left(\pi=\frac{22}{7}\right)\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(7\)
Step-by-step Solution
Detailed explanation
\(\omega=\Delta \mathrm{U}=\mathrm{S} \Delta \mathrm{A}\)
\(36960\) \(\mathrm{erg}=\frac{40 \text { dyne }}{\mathrm{cm}} 8 \pi\left[(\mathrm{r})^2-\left(\frac{7}{2}\right)^2\right] \mathrm{cm}^2\)
\(\mathrm{r}=7 \mathrm{~cm}\)
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