• question_answer 64 small drops of water having same charge and same radius are combined to form one big drop. The ratio of capacitance of big drop to small drop is A)  2 : 1                            B)  1 : 2                C)  4 :1                             D)  1 : 4

Let the radius of each small drop is $r$and the radius of big drop is Ft. When 64 small drops of water are combined to form one big drop, then the volume remains constant. So, the volume of 64 small drops = the volume of big drop. i.e.,      $64\times \frac{4}{3}\pi {{r}^{3}}=\frac{4}{3}\pi {{R}^{3}}$ $\Rightarrow$            $64{{r}^{3}}={{R}^{3}}$ $\Rightarrow$            $4r=R$ $\Rightarrow$            $R=4r$                                   ?(i) Now, the capacitance of a spherical conductor is $C=4\pi {{\varepsilon }_{0}}a$ [a is the radius of the conductor] Now, the capacitance of small drop ${{C}_{1}}=4\pi {{\varepsilon }_{0}}r$                           ...(ii) and the capacitance of big drop is ${{C}_{2}}=4\pi {{\varepsilon }_{0}}R$ On putting the value of fffrom Eq. (i), then ${{C}_{2}}=4\pi {{\varepsilon }_{0}}\,(4r)$ $\Rightarrow$            ${{C}_{2}}=16\pi {{\varepsilon }_{0}}r$               ...(iii) On dividing the Eq. (iii) by Eq. (ii) $\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{16\pi {{\varepsilon }_{0}}r}{4\pi {{\varepsilon }_{0}}r}$ $\Rightarrow$            $\frac{{{C}_{2}}}{{{C}_{1}}}=\frac{4}{1}$ $\Rightarrow$            ${{C}_{2}}:{{C}_{1}}=4:1$