A) 4
B) 6
C) 8
D) 10
E) 12
Correct Answer: C
Solution :
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 \[N=8\]
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