Question

A large hollow metal sphere of radius R has a small opening at the top through which drops of mercury each of radius r and charged to a potential V fall into the sphere and the potential becomes $V^{\prime }$ after N drops fall into it. Then:

• A. $V^{\prime }$ $<\vee$ for any value of N
• B. $V$ ${}^{\prime }=$V for N=1
• C. $V^{\prime }=\vee$ for $N$=$\frac{R}{r}$
• D. $V^{\prime }=\vee$ for $N$=${\left(\frac{R}{r}\right)}^{\frac{1}{3}}$

Answer 1

The correct option is C $V^{\prime }=\vee$ for $N$=$\frac{R}{r}$

Potential of each drop, $V=\frac{KQ}{r}$

charge of each drop$=Q=\frac{Vr}{K}$

Total charge for N drops$=Q_t=NQ=\frac{NVr}{K}$

So, $Q_t$ is the gain charge of the sphere of radius R.

Now potential, $V^{\prime }=\frac{K{Q}_t}{R}=\frac{NVr}{R}$

so, $V^{\prime }=V$ for $N=\frac{R}{r}$