MHT CET · Physics · Mechanical Properties of Fluids
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 \(\mathrm{W}_2=4 \mathrm{~W}_1\)
- C \(\mathrm{W}_2 \lt 4 \mathrm{~W}_1\)
- D \(\quad \mathrm{W}_2=\mathrm{W}_1\)
Answer & Solution
Correct Answer
(C) \(\mathrm{W}_2 \lt 4 \mathrm{~W}_1\)
Step-by-step Solution
Detailed explanation
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.
\(\therefore \quad \mathrm{W}_2 \lt 4 \mathrm{~W}_1\)
\(\begin{aligned}
& \mathrm{W}_1=8 \pi \mathrm{R}^2 \mathrm{~T}_1 \\
& \mathrm{~W}_2=8 \pi(2 \mathrm{R})^2 \mathrm{~T}_2=8 \pi\left(\mathrm{R}^2 4\right) \mathrm{T}^2
\end{aligned}\)
\(\therefore \quad \frac{\mathrm{W}_1}{\mathrm{~W}_2}=\frac{\mathrm{T}_1}{4 \mathrm{~T}_2}\)
When \(\mathrm{T}_1=\mathrm{T}_2, \mathrm{~W}_2=4 \mathrm{~W}_1\)
When temperature increases, surface tension decreases.
\(\therefore \quad \mathrm{W}_2 \lt 4 \mathrm{~W}_1\)
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