A) \[7.2\times {{10}^{3}}V\]
B) \[7.2\times {{10}^{2}}V\]
C) \[1.44\times {{10}^{2}}V\]
D) \[1.44\times {{10}^{3}}V\]
Correct Answer: A
Solution :
Volume of bigger drop = volume of 64 small drops \[ie,\] \[\frac{4}{3}\pi {{R}^{3}}=64\times \frac{4}{3}\pi {{r}^{3}}\] or \[R=4r\] Potential on small drop \[{{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r}\] Potential on bigger drop \[{{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{64q}{R}\] \[{{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{64q}{4r}\] \[=\frac{9\times {{10}^{9}}\times 64\times {{10}^{-9}}}{4\times 2\times {{10}^{-2}}}\] \[=7.2\times {{10}^{3}}V\]
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