Consider a spherical drop of mercury of radius R with capacitance C=4πεR. If two such droplets combine to form a larger one, what would be its…

Question

Consider a spherical drop of mercury of radius R with capacitance C=4πεR. If two such droplets combine to form a larger one, what would be its capacitance in terms of C ?

MARKS Content Team · Accepted Answer

Correct answer: (D) 2^(1/3)C. The volume of the bigger drop should be the same as the sum of the volume of two smaller drops if the radius of the bigger drop is R' and volume of smaller drops is R each, then, 4/3πR'^3=2×4/3πR^3⇒R'=2^13R, so the capacitance after combination becomes, C'=C×2^13.

Consider a spherical drop of mercury of radius $R$ with capacitance $C = 4 π ε R .$ If two such droplets combine to form a larger one, what would be its capacitance in terms of $C$ ?

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