Question

N identical drops of mercury are charged simultaneously to 10 volt. when combined to form one large drop, the potential is found to be 40 volt, the value of \(\mathrm{N}\) is \(\ldots \ldots\) (A) 4 (B) 6 (C) 8 (D) 10

Short answer

The value of N is approximately 6. Answer: (B) 6.

Step-by-step solution

Write the capacitance and charge equations for the small drops

Since there are N identical small drops, each with a potential of 10 volts, we can express their capacitance as: \(C_{small} = k \cdot R_{small}\) and their charge as: \(Q_{small} = N \cdot q = C_{small} \cdot V_{small} = k \cdot R_{small} \cdot 10\)

Write the capacitance and charge equations for the large drop

When the N small drops combine to form a large drop, we can express the capacitance of the large drop as: \(C_{large} = k \cdot R_{large}\) and its charge as: \(Q_{large} = q' = C_{large} \cdot V_{large} = k \cdot R_{large} \cdot 40\)

Relate the charges of the small and large drops

Since the charges of the small drops are summed up to form the charge of the large drop, we can write: \(Q_{large} = N \cdot Q_{small}\)

Substitute the charge equations and express the radii in terms of volume

Using the expressions for charges, we can write: \(k \cdot R_{large} \cdot 40 = N \cdot k \cdot R_{small} \cdot 10\) Note that the total volume of small drops is equal to the volume of the large drop: \(N \cdot \frac{4}{3} \pi R_{small}^3 = \frac{4}{3} \pi R_{large}^3\) From this, we can express the radii in terms of their volumes: \(R_{large} = \sqrt[3]{N} \cdot R_{small}\)

Substitute the radii relation and find the value of N

Substituting the radii relation into our previous equation, we get: \(k \cdot \sqrt[3]{N} \cdot R_{small} \cdot 40 = N \cdot k \cdot R_{small} \cdot 10\) Now, we can solve for N: \(4 \cdot \sqrt[3]{N} = N\) Since the problem provides options to choose from, we can test each option: - (A) N = 4: \(4 \cdot \sqrt[3]{4} = 4 \cdot 1.587 = 6.348 \neq 4\) - (B) N = 6: \(4 \cdot \sqrt[3]{6} = 4 \cdot 1.817 = 7.268 \approx 6 \) Therefore, the value of N is approximately: Answer: (B) 6