Q149 / 150 · MHT CET 2024 (11 May Shift 1) · Physics
The work done in blowing a soap bubble of radius R is $\mathrm{W}_1$ at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius 2 R is blown and the work done is $\mathrm{W}_2$. Then
- A$\mathrm{W}_2=0$
- B$\mathrm{W}_2=4 \mathrm{~W}_1$
- Correct answerC$\mathrm{W}_2 \lt 4 \mathrm{~W}_1$
- D$\quad \mathrm{W}_2=\mathrm{W}_1$
Solution
Work done in blowing a soap bubble,
$\begin{aligned}
& \mathrm{W}_1=8 \pi \mathrm{R}^2 \mathrm{~T}_1 \\
& \mathrm{~W}_2=8 \pi(2 \mathrm{R})^2 \mathrm{~T}_2=8 \pi\left(\mathrm{R}^2 4\right) \mathrm{T}^2
\end{aligned}$
$\therefore \quad \frac{\mathrm{W}_1}{\mathrm{~W}_2}=\frac{\mathrm{T}_1}{4 \mathrm{~T}_2}$
When $\mathrm{T}_1=\mathrm{T}_2, \mathrm{~W}_2=4 \mathrm{~W}_1$ When temperature increases, surface tension decreases. $\therefore \quad \mathrm{W}_2 \lt 4 \mathrm{~W}_1$
When $\mathrm{T}_1=\mathrm{T}_2, \mathrm{~W}_2=4 \mathrm{~W}_1$ When temperature increases, surface tension decreases. $\therefore \quad \mathrm{W}_2 \lt 4 \mathrm{~W}_1$
Hence, the correct answer is option: 'C'.