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JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2021
- Hard
- Numerical
\(27\) similar drops of mercury are maintained at \(10 \,V\) each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ....... times that of a smaller drop.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(243\)
Step-by-step Solution
Detailed explanation
\((27)\left(\frac{4}{3} \pi r ^{3}\right)=\frac{4}{3} \pi R ^{3}\)
\(R =3 r\)
Potential energy of smaller drop :
\(U _{1}=\frac{3}{5} \frac{ kq ^{2}}{ r }\)
Potential energy of bigger drop :
\(U =\frac{3}{5} \frac{ kQ ^{2}}{ R }\)
\(U =\frac{3}{5} \frac{ k (27 q )^{2}}{ R }\)
\(U =\frac{3}{5} k \frac{(27)(27) q ^{2}}{3 r }\)
\(U =\frac{(27)(27)}{3}\left(\frac{3}{5} \frac{ kq ^{2}}{ r }\right)\)
\(U =243 U _{1}\)
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