MHT CET · Physics · Mechanical Properties of Fluids
The radius \(R\) of the soap bubble is doubled under isothermal condition. If \(T\) be the surface tension of soap bubble. The work done in doing so it given by
- A \(32 \pi R{ }^{2} T\)
- B \(24 \pi R^{2} T\)
- C \(8 \pi R^{2} T\)
- D \(4 \pi R^{2} T\)
Answer & Solution
Correct Answer
(A) \(32 \pi R{ }^{2} T\)
Step-by-step Solution
Detailed explanation
Surface energy of a bubble
\(
\begin{array}{l}
E=\text { Tension } \times \text { surface area } \\
E=T \times A
\end{array}
\)
For soap bubble
\(
\begin{array}{l}
E=2 T \times A \\
E=2 T \times 4 \pi R^{2}
\end{array}
\)
Here, \(\quad R=\) radius of bubble
If radius is doubled then we have
\(
\begin{array}{l}
E^{\prime}=2 T \times 4 \pi R^{2} \\
E^{\prime}=2 T \times 4 \pi(2 R)^{2} \\
E^{\prime}=32 \pi R^{2} T
\end{array}
\)
\(
\begin{array}{l}
E=\text { Tension } \times \text { surface area } \\
E=T \times A
\end{array}
\)
For soap bubble
\(
\begin{array}{l}
E=2 T \times A \\
E=2 T \times 4 \pi R^{2}
\end{array}
\)
Here, \(\quad R=\) radius of bubble
If radius is doubled then we have
\(
\begin{array}{l}
E^{\prime}=2 T \times 4 \pi R^{2} \\
E^{\prime}=2 T \times 4 \pi(2 R)^{2} \\
E^{\prime}=32 \pi R^{2} T
\end{array}
\)
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