A capacitor of capacitance `900 mu F` is charged by a `100V` battery.The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor.The loss of energy in this process is measured as `x times10^(-2)J`.The value of `x` is

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Initial Setup
We have a capacitor \( C_1 \) of capacitance \( 900 \, \mu F \) charged to a voltage \( V_1 = 100 \, V \). The energy stored in this capacitor can be calculated using the formula:
\[ U_1 = \frac{1}{2} C_1 V_1^2 \]

### Step 2: Calculate the Initial Energy
Substituting the values into the energy formula:
\[ C_1 = 900 \times 10^{-6} \, F \]
\[ U_1 = \frac{1}{2} \times 900 \times 10^{-6} \times (100)^2 \]
\[ U_1 = \frac{1}{2} \times 900 \times 10^{-6} \times 10000 \]
\[ U_1 = \frac{1}{2} \times 900 \times 10^{-2} \]
\[ U_1 = 450 \times 10^{-2} \, J \]

### Step 3: Connecting the Capacitors
When the charged capacitor \( C_1 \) is connected to an uncharged identical capacitor \( C_2 \), the total capacitance of the system becomes:
\[ C_{total} = C_1 + C_2 = 900 \, \mu F + 900 \, \mu F = 1800 \, \mu F \]

### Step 4: Calculate the Final Voltage
The charge \( Q \) on the charged capacitor before connection is:
\[ Q = C_1 V_1 = 900 \times 10^{-6} \times 100 = 90000 \times 10^{-6} \, C = 0.09 \, C \]
After connecting, the total charge remains the same, but it is now distributed across the two capacitors:
\[ V_f = \frac{Q}{C_{total}} = \frac{0.09}{1800 \times 10^{-6}} = \frac{0.09}{0.0018} = 50 \, V \]

### Step 5: Calculate the Final Energy
The energy stored in the system after connecting the capacitors is:
\[ U_f = \frac{1}{2} C_{total} V_f^2 \]
Substituting the values:
\[ U_f = \frac{1}{2} \times 1800 \times 10^{-6} \times (50)^2 \]
\[ U_f = \frac{1}{2} \times 1800 \times 10^{-6} \times 2500 \]
\[ U_f = \frac{1}{2} \times 1800 \times 10^{-3} = 900 \times 10^{-3} \, J = 9 \times 10^{-1} \, J \]

### Step 6: Calculate the Energy Loss
The energy loss \( \Delta U \) in the process is given by:
\[ \Delta U = U_1 - U_f \]
Substituting the values:
\[ \Delta U = 450 \times 10^{-2} - 900 \times 10^{-3} \]
\[ \Delta U = 450 \times 10^{-2} - 90 \times 10^{-2} = 360 \times 10^{-2} \, J \]

### Step 7: Find the Value of \( x \)
From the problem statement, the loss of energy is measured as \( x \times 10^{-2} \, J \). Therefore:
\[ x = 360 \]

### Final Answer
The value of \( x \) is \( \boxed{360} \).