Q. A spherical drop of mercury having a potential of 2.5V is obtained as a result of merging 125 droplets. The potential of a constituent droplets would be
A) 1.0V
B) 0.5V
C) 0.2V
D) 0.1V
Explanation
(d) Potential of big drop $V = n^{2/3}V'$
V’= potential of small drop
$n$=no. of drops=125
V=2.5
$\therefore\ \ \ \ \ \ 2.5 = (125)^{2/3}\ V'$
2.5=25V’
V’=0.1 volt
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.