Acidified water from certain reservoir kept at a potential V falls in the form of small droplets each of radius r through a hole into a hollow conducting sphere of radius a. The sphere is insulated and is initially at zero potential. If the drops continue to fall untill the sphere is half full, the potential acquired by the sphere is

Options

A. `(a^(2)V)/(2r^(2))`
B. `sqrt(a/r) V/2`
C. `(a^(3)V)/(2r^(3))`
D. `(aV)/r`

Answer

To solve the problem step-by-step, we need to analyze the situation involving the droplets falling into the hollow conducting sphere and how this affects the potential of the sphere.

### Step 1: Understand the Problem
We have a hollow conducting sphere of radius \( a \) that is initially at zero potential. Acidified water droplets of radius \( r \) fall into this sphere from a reservoir at potential \( V \). We need to find the potential of the sphere when it is half full of these droplets.

### Step 2: Calculate the Volume of the Sphere
The volume \( V_s \) of the hollow conducting sphere when it is half full can be calculated using the formula for the volume of a sphere:
\[ V_s = \frac{2}{3} \pi a^3 \]

### Step 3: Calculate the Volume of One Droplet
The volume \( V_d \) of one droplet (which is spherical) is given by:
\[ V_d = \frac{4}{3} \pi r^3 \]

### Step 4: Determine the Number of Droplets
To find the number of droplets \( n \) that fit into the half-full sphere, we equate the total volume of the droplets to the volume of the sphere:
\[ n \cdot V_d = V_s \]
Substituting the volumes we calculated:
\[ n \cdot \frac{4}{3} \pi r^3 = \frac{2}{3} \pi a^3 \]
Now, simplifying this equation:
\[ n \cdot \frac{4}{3} r^3 = \frac{2}{3} a^3 \]
\[ n = \frac{2 a^3}{4 r^3} = \frac{a^3}{2 r^3} \]

### Step 5: Calculate the Charge of the Droplets
Each droplet has a charge \( q \) due to the potential \( V \) of the reservoir. The charge \( q \) can be expressed as:
\[ q = V \cdot V_d = V \cdot \frac{4}{3} \pi r^3 \]

### Step 6: Calculate the Total Charge in the Sphere
The total charge \( Q \) in the sphere when it is half full is:
\[ Q = n \cdot q = \left(\frac{a^3}{2 r^3}\right) \cdot \left(V \cdot \frac{4}{3} \pi r^3\right) \]
Simplifying this gives:
\[ Q = \frac{a^3}{2 r^3} \cdot V \cdot \frac{4}{3} \pi r^3 = \frac{2 \pi a^3 V}{3} \]

### Step 7: Calculate the Potential of the Sphere
The potential \( V' \) of the sphere when it is half full can be calculated using the formula for the potential due to a charge \( Q \) at a distance \( a \):
\[ V' = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{a} \]
Substituting \( Q \):
\[ V' = \frac{1}{4 \pi \epsilon_0} \cdot \frac{\frac{2 \pi a^3 V}{3}}{a} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2 \pi a^2 V}{3} \]
This simplifies to:
\[ V' = \frac{2 a^2 V}{12 \epsilon_0} = \frac{a^2 V}{6 \epsilon_0} \]

### Final Answer
Thus, the potential acquired by the sphere when it is half full is:
\[ V' = \frac{a^2 V}{6 \epsilon_0} \]