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(A) Let the radius of each small drop be $r$ and the radius of the big drop be $R$.Since the volume remains constant,the volume of $64$ small drops eq

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$4: 1$

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(A) Let the radius of each small drop be $r$ and the radius of the big drop be $R$.Since the volume remains constant,the volume of $64$ small drops equals the volume of the big drop:$64 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$$R^3 = 64 r^3$$R = 4r$The capacitance of a spherical drop is given by $C = 4 \pi \epsilon_0 r$.Therefore,the capacitance of the small drop is $C_{	ext{small}} = 4 \pi \epsilon_0 r$ and the capacitance of the big drop is $C_{	ext{big}} = 4 \pi \epsilon_0 R$.The ratio of the capacitance of the big drop to the small drop is:$\frac{C_{	ext{big}}}{C_{	ext{small}}} = \frac{4 \pi \epsilon_0 R}{4 \pi \epsilon_0 r} = \frac{R}{r} = \frac{4r}{r} = 4$.Thus,the ratio is $4: 1$.

$64$ small drops of water having same charge and same radius are combined to form one big drop. The ratio of capacitance of the big drop to the small drop is . . . . . . .

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