MHT CET · Physics · Mechanical Properties of Fluids
The change in energy when a big dropis split in sma \(\| n\) droplets is
- A \(4 R^{2}\left(n^{2 / 3}-1\right) I\)
- B \(4 \pi R^{2}\left(n^{1 / 3}-1\right) T\)
- C \(4 \pi R^{2}\left(n^{-1 / 3}-1\right) T\)
- D \(4 \pi R^{2}\left[n^{-2 / 3}-1\right] T\)
Answer & Solution
Correct Answer
(B) \(4 \pi R^{2}\left(n^{1 / 3}-1\right) T\)
Step-by-step Solution
Detailed explanation
When a drop of radius \(R\) split into \(n\) smaller drops (each of radius \(r\) ), then surface area of liquid increases. Hence, the work is to be done against surface tension. Since, the volume of liquid remains constant
\(
\begin{array}{l}
\therefore \quad \frac{4}{3} \pi R^{3}=n \frac{4}{3} \pi r^{3} \quad \therefore R^{3}=n r^{3} \\
\text { Work done }=T \times \Delta A \\
=T\left[n 4 \pi r^{2}-4 \pi R^{2}\right] \\
=4 \pi R^{2} T\left(n^{1 / 3}-1\right)
\end{array}
\)
\(
\begin{array}{l}
\therefore \quad \frac{4}{3} \pi R^{3}=n \frac{4}{3} \pi r^{3} \quad \therefore R^{3}=n r^{3} \\
\text { Work done }=T \times \Delta A \\
=T\left[n 4 \pi r^{2}-4 \pi R^{2}\right] \\
=4 \pi R^{2} T\left(n^{1 / 3}-1\right)
\end{array}
\)
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