TS EAMCET · Physics · Mechanical Properties of Fluids
The change in surface energy when a big spherical drop fo radius \(R\) is split into \(n\) spherical droplets of radius \(r\) is \((T=\) surface tension)
- A \(4 R^2\left(n^{2 / 3}-1\right) T\)
- B \(4 R^2 \pi\left(n^{1 / 3}-1\right) T\)
- 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) \(4 R^2 \pi\left(n^{1 / 3}-1\right) T\)
Step-by-step Solution
Detailed explanation
According to given situation, Volume of big drop \(=\) Volume of \(n\) small drops \(\frac{4}{3} \pi R^3=n \times \frac{4}{3} \pi r^3 \Rightarrow r^3=\frac{R^3}{n} \text { or } r=\frac{R}{n^{1 / 3}}\) \(n r^2=\frac{n R^2}{n^{2 / 3}}=n^{1 / 3} R^2\) \(\ldots(i)\) Surface area of…
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