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

Answer 1

Answer: (C)

8

Answer 2

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 E (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 $N=8$