A) \[{{\left( 2 \right)}^{3/2}}V\]
B) \[{{\left( 2 \right)}^{2/3}}V\]
C) \[{{\left( 3 \right)}^{2/3}}V\]
D) \[{{\left( 3 \right)}^{1/2}}V\]
Correct Answer: C
Solution :
Potential on small drop \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r}\](charge on small drop = q, radius of small drop = r) \[\because \] Volume of 2 small drops = volume of one big drop \[\Rightarrow \] \[2\times \frac{4}{3}\pi {{r}^{3}}=\frac{4}{3}\pi {{R}^{3}}\] or \[R={{(2)}^{1/3}}r\] Potential on big drop \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(2q)}{R}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2q}{{{(2)}^{1/3}}r}\] \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r}{{(2)}^{1-\frac{1}{3}}}=V{{(2)}^{2/3}}\]
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