64 identical drops of mercury are charged simultaneously to the same potential of 10 volt. Assuming the drops to be spherical, if all the charged drops are made to combine to form one large drop, then its potential will be (A) \(100 \mathrm{~V}\) (B) \(320 \mathrm{~V}\) (C) \(640 \mathrm{~V}\) (D) \(160 \mathrm{~V}\)

First, we need to determine the total charge and capacitance of a single mercury drop. We know that the potential (V) of a sphere is given by: \[V = \frac{Q}{C}\] Where Q is the charge and C is the capacitance of the sphere. Rearranging the equation to solve for charge, we have: \[Q = V \times C\] Since there are 64 identical mercury drops, each with a potential of 10 volts, their total charge is as follows: \[Q_{total} = 64 \times Q\] And the total capacitance is: \[C_{total} = 64 \times C\]

Next, we need to determine the capacitance of the large drop formed by combining 64 smaller mercury drops. The volume of a sphere is given by: \[V = \frac{4}{3} \pi r^3\] When 64 smaller mercury drops combine to form a single larger drop, the volumes are additive: \[V_{large} = 64 \times V_{small}\] Substituting the volume formula for both large and small drops, we have: \[\frac{4}{3} \pi R^3 = 64 \times \frac{4}{3} \pi r^3\] Where R is the radius of the large drop and r is the radius of each small drop. After simplifying, we get: \[R = 4r\] The capacitance of a sphere is given by: \[C = 4 \pi \epsilon_0 r\] Where \(\epsilon_0\) is the vacuum permittivity. Substituting the radius of the large drop into the capacitance formula, we have: \[C_{large} = 4 \pi \epsilon_0 R = 4 \pi \epsilon_0 (4r) = 16 \pi \epsilon_0 r\]

Now we know the charge on the large drop and its capacitance, we can find its potential using the formula: \[V_{large} = \frac{Q_{total}}{C_{large}}\] Substituting the expressions for total charge and capacitance of the large drop, we have: \[V_{large} = \frac{64 \times V \times C}{16 \pi \epsilon_0 r}\] Simplifying the expression, we get: \[V_{large} = 4 \times V\] Finally, substituting the given value of the potential of each identical small drop, we have: \[V_{large} = 4 \times 10 \mathrm{~V}\] \[V_{large} = 40 \mathrm{~V}\] None of the given options match the calculated potential of the large drop. There might be a typo in the given options. However, the correct answer is \(40 \mathrm{~V}\).