1.

A large hollow metal sphere of radius R has a small opening at the top. Small drops of mercury each of radius r and charged to a potential V fall into the sphere. The potential of the sphere becomes V' after N drops fall into it. Then, (a) V' < V for all N (b) V' = V for N = 1 (c) V' = V for N = R/r (d) V' = V for N = (R/r)1/3

Answer»

Correct Answer is: (a) V' < V for all N 

Capacitance of each mercury drop is 4πɛ0r.

Charge on each mercury drop is 4πɛ0r VN.

Capacitance of the hollow sphere is 4πɛ0R.

When it acquires Q charge, its potential is V' = Q/4πɛ0R.

For V' = V,  4πɛ0RV = 4πɛ0rVN

or N = R/r.



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