MHT CET · Physics · Mechanical Properties of Fluids
If the work done in blowing a soap bubble of volume ' \(\mathrm{V}\) ' is ' \(\mathrm{W}\) ', then the work done in blowing a soap bubble of volume ' \(2 \mathrm{~V}\) ' will be
- A \(2 \mathrm{~W}\)
- B \(4^{1 / 3} \mathrm{~W}\)
- C W
- D \(\sqrt{2} \mathrm{~W}\)
Answer & Solution
Correct Answer
(B) \(4^{1 / 3} \mathrm{~W}\)
Step-by-step Solution
Detailed explanation
If \(\mathrm{V}\) is the volume then we have
\(\begin{aligned}
& \mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3 \\
& \mathrm{~V} \propto \mathrm{r}^3 \\
& \therefore \frac{\mathrm{V}_2}{\mathrm{~V}_1}=\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^3 \\
& \frac{\mathrm{r}_2}{\mathrm{r}_1}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{1 / 3}=(2)^{1 / 3} \\
& \mathrm{~W}_1=8 \pi \mathrm{r}_1^2 \cdot \mathrm{T} \text { and } \mathrm{W}_2=8 \pi \mathrm{r}_2^2 \mathrm{~T} \\
& \therefore \frac{\mathrm{W}_2}{\mathrm{~W}_1}=\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^2=(2)^{2 / 3}=(4)^{1 / 3} \\
& \mathrm{~W}_2=(4)^{1 / 3} \mathrm{~W}
\end{aligned}\)
\(\begin{aligned}
& \mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3 \\
& \mathrm{~V} \propto \mathrm{r}^3 \\
& \therefore \frac{\mathrm{V}_2}{\mathrm{~V}_1}=\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^3 \\
& \frac{\mathrm{r}_2}{\mathrm{r}_1}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{1 / 3}=(2)^{1 / 3} \\
& \mathrm{~W}_1=8 \pi \mathrm{r}_1^2 \cdot \mathrm{T} \text { and } \mathrm{W}_2=8 \pi \mathrm{r}_2^2 \mathrm{~T} \\
& \therefore \frac{\mathrm{W}_2}{\mathrm{~W}_1}=\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^2=(2)^{2 / 3}=(4)^{1 / 3} \\
& \mathrm{~W}_2=(4)^{1 / 3} \mathrm{~W}
\end{aligned}\)
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