N drops of mercury of equal radii and possessing equal charges combine to form a big spherical drop. Then the capacitance of the bigger drop compared to each individual drop isa)N timesb)N2/3 timesc)N1/3 timesd)N5/3 timesCorrect answer is option 'C'. Can you explain this answer?

Question

Answer

Answer

Explanation:

When N drops of mercury combine to form a bigger drop, the charges on each individual drop combine to give a net charge on the bigger drop. However, the volume of the bigger drop is N times the volume of each individual drop.

Hence, the charge density on the bigger drop is N times smaller than the charge density on each individual drop.

Therefore, the potential on the bigger drop is N times smaller than the potential on each individual drop.

Now, we know that capacitance is given by the ratio of charge to potential. Hence, the capacitance of the bigger drop is N times larger than the capacitance of each individual drop.

However, we need to compare the capacitance of the bigger drop to each individual drop. Since there are N individual drops, we need to divide the capacitance of the bigger drop by N to get the capacitance of each individual drop.

Therefore, the capacitance of each individual drop is N times larger than the capacitance of the bigger drop divided by N.

i.e., C_individual = (C_bigger/N)

Now, we know that the volume of the bigger drop is N times the volume of each individual drop. Hence, the radius of the bigger drop is N^(1/3) times the radius of each individual drop.

Since capacitance is inversely proportional to radius, the capacitance of each individual drop is N^(1/3) times larger than the capacitance of the bigger drop.

Answer: Hence, the correct option is (c) N^(1/3) times.

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