Question

Eight drops of mercury of equal radii and possessing equal charge combine to form a big drop. If the capacity of each drop is \( C \), then capacity of the big drop is:

• A. \( 4C \)
• B. \( 2C \)
• C. \( 8C \)
• D. \( 16C \)

Hint:

When small drops combine to form a bigger drop, the total charge is conserved, and the radius of the big drop is the cube root of the sum of the volumes of the individual drops.

Answer 1

The Correct Option is B

To determine the capacity of the big drop formed by combining eight smaller drops, we must first understand the relationship between the radii, volume, and capacity of spherical conductors.

Consider each small drop has a radius \( r \). Since the volume of a sphere is given by the formula \( V = \frac{4}{3}\pi r^3 \), the total volume of the eight drops is \( 8 \times \frac{4}{3}\pi r^3 \).

If these combine to form a single large drop with radius \( R \), the volume of this large drop also follows the same formula, \( V = \frac{4}{3}\pi R^3 \). Equating the total volume of the smaller drops with the volume of the larger drop, we get:

\[ 8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \]

This simplifies to:

\[ 8r^3 = R^3 \]

Taking the cube root on both sides, we find:

\[ R = 2r \]

The capacity of a spherical conductor is directly proportional to its radius. If the capacity of a small drop is \( C \), then its capacity is proportional to its radius \( r \). Thus, the capacity \( C_1 \) of the large drop is proportional to its radius \( R = 2r \):

\[ C_1 = 2C \]

Therefore, the capacity of the big drop is \( 2C \).

Answer 2

We are given that eight drops of mercury of equal radii and possessing equal charges combine to form a big drop. The task is to find the capacity of the big drop, given that the capacity of each individual drop is \( C \). 
Step 1: The total charge on a drop is given by \( Q = k \cdot r \), where \( k \) is a constant and \( r \) is the radius of the drop. The capacity of a drop \( C \) is proportional to the radius of the drop, i.e. \( C \propto r \). 
Step 2: Since the total charge is conserved, when eight drops combine to form a big drop, the total charge of the new drop is the sum of the charges of the individual drops: \[ Q_{{big}} = 8Q_{{individual}}. \] The volume of the big drop is the sum of the volumes of the individual drops. Since volume is proportional to the cube of the radius, the radius of the big drop \( r_{{big}} \) is given by: \[ r_{{big}}^3 = 8r_{{individual}}^3 \quad \Rightarrow \quad r_{{big}} = 2r_{{individual}}. \] 
Step 3: The capacity of the big drop is proportional to its radius: \[ C_{{big}} \propto r_{{big}} = 2r_{{individual}}. \] Thus, the capacity of the big drop is \( 2C \), where \( C \) is the capacity of each individual drop. Thus, the capacity of the big drop is \( 2C \).