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2. Electric Potential and Capacitance
hard
$27$ 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.
A
$256$
B
$144$
C
$324$
D
$243$
(JEE MAIN-2021)
Solution
$(27)\left(\frac{4}{3} \pi r ^{3}\right)=\frac{4}{3} \pi R ^{3}$
$R =3 r$
Potential energy of smaller drop :
$U _{1}=\frac{3}{5} \frac{ kq ^{2}}{ r }$
Potential energy of bigger drop :
$U =\frac{3}{5} \frac{ kQ ^{2}}{ R }$
$U =\frac{3}{5} \frac{ k (27 q )^{2}}{ R }$
$U =\frac{3}{5} k \frac{(27)(27) q ^{2}}{3 r }$
$U =\frac{(27)(27)}{3}\left(\frac{3}{5} \frac{ kq ^{2}}{ r }\right)$
$U =243 U _{1}$
Standard 12
Physics
