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

Options

A. 100 units
B. 320 units
C. 640 units
D. 160 units

Answer

To solve the problem of finding the potential of a large drop formed by combining 64 identical charged mercury drops, we can follow these steps:

### Step 1: Understand the Given Information
We have 64 identical mercury drops, each charged to a potential of 10 volts. We need to find the potential of a single large drop formed by combining these 64 drops.

### Step 2: Calculate the Charge on One Small Drop
The potential \( V \) of a spherical conductor is given by the formula:
\[ V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{R} \]
where \( Q \) is the charge on the drop and \( R \) is the radius of the drop.

Given that the potential \( V = 10 \) volts, we can express the charge \( Q \) on one small drop as:
\[ Q = V \cdot 4 \pi \epsilon_0 R \]
Substituting \( V = 10 \) volts, we have:
\[ Q = 10 \cdot 4 \pi \epsilon_0 R \]

### Step 3: Calculate the Total Charge from All Drops
Since there are 64 identical drops, the total charge \( Q_{total} \) when all drops are combined is:
\[ Q_{total} = 64 \cdot Q = 64 \cdot (10 \cdot 4 \pi \epsilon_0 R) = 640 \cdot 4 \pi \epsilon_0 R \]

### Step 4: Volume Conservation
When the 64 drops combine to form one large drop, the volume of the large drop must equal the total volume of the 64 small drops. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the volume of 64 small drops is:
\[ V_{small} = 64 \cdot \frac{4}{3} \pi R^3 = \frac{256}{3} \pi R^3 \]
Let \( R' \) be the radius of the large drop. The volume of the large drop is:
\[ V_{large} = \frac{4}{3} \pi (R')^3 \]
Setting these equal gives:
\[ \frac{256}{3} \pi R^3 = \frac{4}{3} \pi (R')^3 \]
Cancelling \( \frac{4}{3} \pi \) from both sides:
\[ 256 R^3 = 4 (R')^3 \]
Dividing both sides by 4:
\[ 64 R^3 = (R')^3 \]
Taking the cube root:
\[ R' = 4R \]

### Step 5: Calculate the Potential of the Large Drop
Now we can find the potential \( V' \) of the large drop using the charge \( Q_{total} \) and the new radius \( R' \):
\[ V' = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_{total}}{R'} \]
Substituting \( Q_{total} \) and \( R' \):
\[ V' = \frac{1}{4 \pi \epsilon_0} \cdot \frac{640 \cdot 4 \pi \epsilon_0 R}{4R} \]
The \( 4R \) in the denominator cancels with the \( 4R \) in the numerator:
\[ V' = \frac{640}{4} = 160 \text{ volts} \]

### Final Answer
The potential of the large drop formed by combining the 64 identical drops is **160 volts**.

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