A) \[89.5\times {{10}^{-5}}J\]
B) 0.895 J
C) \[895\times {{10}^{-5}}J\]
D) None of these
Correct Answer: B
Solution :
The volume of big drop\[={{10}^{6}}\times \]volume of small drop \[\Rightarrow \] \[\frac{4}{3}\pi {{R}^{3}}={{10}^{6}}\times \frac{4}{3}\pi {{r}^{3}}\] \[\Rightarrow \] \[R=100r\] Therefore, work done \[W=T.\Delta A\] \[=T[n.4\pi {{r}^{2}}-4\pi {{R}^{2}}]\] \[=4\pi T[{{10}^{6}}{{r}^{2}}-{{10}^{4}}{{r}^{2}}]\] \[=4\pi \times 72\times {{10}^{-3}}\times 990000\times {{10}^{-6}}\] \[=0.895J\]
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