A) \[\sqrt{\frac{T}{\rho }\left( \frac{1}{r}-\frac{1}{R} \right)}\]
B) \[\sqrt{\frac{2T}{\rho }\left( \frac{1}{r}-\frac{1}{R} \right)}\]
C) \[\sqrt{\frac{4T}{\rho }\left( \frac{1}{r}-\frac{1}{R} \right)}\]
D) \[\sqrt{\frac{6T}{\rho }\left( \frac{1}{r}-\frac{1}{R} \right)}\]
Correct Answer: D
Solution :
When drops combine to form a single drop of radius R. Then energy released, \[E=4\pi {{R}^{3}}T\left[ \frac{1}{r}-\frac{1}{R} \right]\] \[\frac{1}{2}\times \left[ \frac{4}{3}\pi {{R}^{3}}\rho \right]{{v}^{2}}=4\pi {{R}^{3}}T\left[ \frac{1}{r}-\frac{1}{R} \right]\] \[{{v}^{2}}=\frac{6T}{\rho }\left[ \frac{1}{r}-\frac{1}{R} \right]\] \[v=\sqrt{\frac{6T}{\rho }\left[ \frac{1}{r}-\frac{1}{R} \right]}\] If this energy is converted into kinetic energy then\[\frac{1}{2}m{{v}^{2}}=4\pi {{R}^{3}}T\left[ \frac{1}{r}-\frac{1}{R} \right]\] \[\frac{1}{2}\times \left[ \frac{4}{3}\pi {{R}^{3}}\rho \right]{{v}^{2}}=4\pi {{R}^{3}}T\left[ \frac{1}{r}-\frac{1}{R} \right]\] \[{{v}^{2}}=\frac{6T}{\rho }\left[ \frac{1}{r}-\frac{1}{R} \right]\]\[v=\sqrt{\frac{6T}{\rho }\left[ \frac{1}{r}-\frac{1}{R} \right]}\]
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