A capacitor of capacitance `1 mu F` is charged to potential 2 V and is connected to an inductor of 1 mH. At an instant when potential difference across. The capacitor is 1 V, the current in the circuit is `10^(-2) sqrt((10x)/(3))` ampere. Find out the value of x.

Answer

To solve the problem, we will use the conservation of energy principle in an LC circuit, where the energy stored in the capacitor is equal to the sum of the energy stored in the capacitor and the inductor at any instant.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Capacitance, \( C = 1 \mu F = 1 \times 10^{-6} F \)
- Initial Voltage across the capacitor, \( V_0 = 2 V \)
- Voltage across the capacitor at the instant of interest, \( V = 1 V \)
- Inductance, \( L = 1 mH = 1 \times 10^{-3} H \)
- Current at that instant, \( I = 10^{-2} \sqrt{\frac{10x}{3}} \, A \)

2. **Calculate Initial Energy Stored in the Capacitor:**
\[ U_C = \frac{1}{2} C V_0^2 = \frac{1}{2} \times (1 \times 10^{-6}) \times (2^2) = \frac{1}{2} \times (1 \times 10^{-6}) \times 4 = 2 \times 10^{-6} \, J \]

3. **Calculate Energy Stored in the Capacitor at the Instant:**
\[ U_C' = \frac{1}{2} C V^2 = \frac{1}{2} \times (1 \times 10^{-6}) \times (1^2) = \frac{1}{2} \times (1 \times 10^{-6}) \times 1 = 0.5 \times 10^{-6} \, J \]

4. **Using Conservation of Energy:**
The total energy initially stored in the capacitor is equal to the energy stored in the capacitor at the instant plus the energy stored in the inductor at that instant:
\[ U_C = U_C' + U_L \]
Where \( U_L = \frac{1}{2} L I^2 \).

5. **Substituting Values:**
\[ 2 \times 10^{-6} = 0.5 \times 10^{-6} + \frac{1}{2} (1 \times 10^{-3}) I^2 \]

6. **Rearranging to Find \( I^2 \):**
\[ 2 \times 10^{-6} - 0.5 \times 10^{-6} = \frac{1}{2} (1 \times 10^{-3}) I^2 \]
\[ 1.5 \times 10^{-6} = \frac{1}{2} \times 10^{-3} I^2 \]
\[ 1.5 \times 10^{-6} = 0.5 \times 10^{-3} I^2 \]
\[ I^2 = \frac{1.5 \times 10^{-6}}{0.5 \times 10^{-3}} = 3 \times 10^{-3} \]

7. **Substituting the Expression for Current:**
\[ I^2 = \left(10^{-2} \sqrt{\frac{10x}{3}}\right)^2 \]
\[ 3 \times 10^{-3} = 10^{-4} \frac{10x}{3} \]
\[ 3 \times 10^{-3} \times 3 = 10^{-4} \times 10x \]
\[ 9 \times 10^{-3} = 10^{-4} \times 10x \]
\[ 9 \times 10^{-3} = 10^{-3} x \]
\[ x = 9 \]

### Final Answer:
The value of \( x \) is \( 9 \).