The capacity of a condenser A is `10muF` and it is charged to a battery of 100 volt. The battery is disconnected and the condenser A is connected to a condenser B the common potential is 40V. The capacity of B is

Options

A. `8muF`
B. `15muF`
C. `2muF`
D. `1muF`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the initial charge on condenser A
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \times V \]
where \( C \) is the capacitance and \( V \) is the voltage.

For condenser A:
- Capacitance \( C_A = 10 \, \mu F = 10 \times 10^{-6} \, F \)
- Voltage \( V_A = 100 \, V \)

Calculating the charge:
\[ Q_A = C_A \times V_A = (10 \times 10^{-6}) \times 100 = 10 \times 10^{-4} \, C = 10^{-3} \, C \]

### Step 2: Set up the conservation of charge equation
When the battery is disconnected and condenser A is connected to condenser B, the total charge is conserved. Let \( Q_B \) be the charge on condenser B after they are connected.

The total charge before connection is:
\[ Q_{total} = Q_A = 10^{-3} \, C \]

After connection, the common potential \( V \) is given as \( 40 \, V \).

### Step 3: Express the final charges on both capacitors
The charge on condenser A after connection is:
\[ Q_A' = C_A \times V = 10 \times 10^{-6} \times 40 = 4 \times 10^{-5} \, C \]

Let the capacitance of condenser B be \( C_B \). The charge on condenser B after connection is:
\[ Q_B' = C_B \times V = C_B \times 40 \]

### Step 4: Apply the conservation of charge
According to the conservation of charge:
\[ Q_A' + Q_B' = Q_{total} \]
Substituting the known values:
\[ 4 \times 10^{-5} + C_B \times 40 = 10^{-3} \]

### Step 5: Solve for \( C_B \)
Rearranging the equation gives:
\[ C_B \times 40 = 10^{-3} - 4 \times 10^{-5} \]
Calculating the right side:
\[ 10^{-3} = 10^{-3} \, C = 1000 \times 10^{-6} \, C \]
\[ 4 \times 10^{-5} = 40 \times 10^{-6} \, C \]
Thus,
\[ 10^{-3} - 4 \times 10^{-5} = 1000 \times 10^{-6} - 40 \times 10^{-6} = 960 \times 10^{-6} \, C \]
Now substituting back:
\[ C_B \times 40 = 960 \times 10^{-6} \]
Dividing both sides by 40:
\[ C_B = \frac{960 \times 10^{-6}}{40} = 24 \times 10^{-6} \, F = 24 \, \mu F \]

### Final Answer
The capacitance of condenser B is \( 24 \, \mu F \).

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