Question

$N$ identical drops of mercury are charged simultaneously to $ 10 \,V $ . When combined to form one large drop, the potential is found to be $ 40\, V $ , the value of $ N $ is

• A. 4
• B. 6
• C. 8
• D. 10
• E. 12

Answer 1

Answer: (C)

After combining, the volume remains same ie, volume of bigger drop
$ =N\times $ volume of smaller drop or $ \frac{4}{3}\pi {{R}^{3}}=N\times \frac{4}{3}\pi {{r}^{3}} $ or $ N={{\left( \frac{R}{r} \right)}^{3}} $ .... (i)
As charge is conserved, hence $ Q=Nq $ ...(ii)
Capacity of bigger drop $ =4\pi {{\varepsilon }_{0}}R $ Capacity of smaller drop $ =4\pi {{\varepsilon }_{0}}r $
From Eq. (ii), we have $ (4\pi {{\varepsilon }_{0}}R){{V}_{big}}=N(4\pi {{\varepsilon }_{0}}r){{V}_{small}} $
Or $ (4\pi {{\varepsilon }_{0}}R)\times 40=N(4\pi {{\varepsilon }_{0}}r)\times 10 $ Or $ 4R=Nr $
Or $ \frac{R}{r}=\frac{N}{4} $ .... (iii)
From Eqs. (i) and (iii),
$ N={{\left( \frac{N}{4} \right)}^{3}} $
$ N=\frac{{{N}^{3}}}{64} $ Or
$ N=\frac{{{N}^{3}}}{64} $ Or
$ {{N}^{2}}=64 $ Or $ N=8 $