JEE Mains · Physics · STD 11 - 9.2 surface tension
A large number of water drops, each of radius \(r\), combine to have a drop of radius \(R\). If the surface tension is \(T\) and mechanical equivalent of heat is \(J ,\) the rise in heat energy per unit volume will be
- A \(\frac{2 T }{ J }\left(\frac{1}{ r }-\frac{1}{ R }\right)\)
- B \(\frac{2 T }{ rJ }\)
- C \(\frac{3 T }{ rJ }\)
- D \(\frac{3 T }{ J }\left(\frac{1}{ r }-\frac{1}{ R }\right)\)
Answer & Solution
Correct Answer
(D) \(\frac{3 T }{ J }\left(\frac{1}{ r }-\frac{1}{ R }\right)\)
Step-by-step Solution
Detailed explanation
\(n \times \frac{4}{3} \pi r ^{3}=\frac{4}{3} \pi R ^{3}\) \(\therefore n ^{1 / 3} r = R\) \(\therefore\) Total change in surface energy \(=\left(n\left(4 \pi r^{2}\right)-4 \pi R^{2}\right) T\) \(\Rightarrow 4 \pi T \left( nr ^{2}- R ^{2}\right)\) \(\therefore\) Heat energy…
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