A) \[16~\pi \times {{10}^{-7}}J\]
B) \[4.15~\pi \times {{10}^{-7}}J\]
C) \[2.0.75~\pi \times {{10}^{-7}}J\]
D) \[8~\pi \times {{10}^{-7}}J\]
Correct Answer: B
Solution :
Volume of 8 drops of water = Volume of big drop of water Given that, Radius of small drop r = 0.6 mm \[\therefore \] \[8\times \frac{4}{3}\pi {{(0.6)}^{3}}=\frac{4}{3}\pi {{R}^{3}}\] or \[2\times 0.6=R\] or \[R=1.2\,mm\]where R is the radius of big drop. Change in surface area \[\Delta A=4\pi {{r}^{2}}\times 8-4\pi {{R}^{2}}\] \[=4\pi [{{(0.6)}^{2}}\times 8-{{(1.2)}^{2}}]\times {{10}^{-6}}\] \[=4\pi (0.36\times 8-1.44)\times {{10}^{-6}}\] \[=4\pi (2.88-1.44)\times {{10}^{-6}}\] \[\Delta A=4\pi (1.44)\times {{10}^{-6}}{{m}^{2}}\] Energy dissipated in this process \[E=T\times \Delta A\] \[=0.072\times 4\pi \times 1.44\times {{10}^{-6}}\] \[=28.8\times {{10}^{-2}}\times \pi \times 1.44\times {{10}^{-6}}\] \[E=4.15\pi \times {{10}^{-7}}J\]
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