A 10 μF capacitor is connected to a 210 V, 50 Hz source as shown in figure. The peak current in the circuit is nearly (π = 3.14) :

The circuit consists of a single capacitor \(C = 10\;\mu\text{F}\) connected to an AC source that supplies a sinusoidal voltage with an RMS value of \(V_{\text{rms}} = 210\;\text{V}\) at a frequency of \(f = 50\;\text{Hz}\) (as shown in the provided figure).

1. **Find the angular frequency \(\omega\)**
\[ \omega = 2\pi f \]
Substituting \(f = 50\;\text{Hz}\):
\[ \omega = 2 \times 3.14 \times 50 = 314\;\text{rad/s} \]

2. **Calculate the capacitive reactance \(X_C\)**
\[ X_C = \frac{1}{\omega C} \]
Convert the capacitance to farads: \(C = 10\;\mu\text{F} = 10 \times 10^{-6}\;\text{F}\).
\[ X_C = \frac{1}{314 \times 10 \times 10^{-6}} = \frac{1}{3.14 \times 10^{-3}} \approx 318.5\;\Omega \]

3. **Determine the peak (maximum) source voltage \(V_{\text{peak}}\)**
The given 210 V is an RMS value, so the peak value is:
\[ V_{\text{peak}} = \sqrt{2}\; V_{\text{rms}} = 1.414 \times 210 \approx 296.9\;\text{V} \]

4. **Compute the peak current \(I_{\text{peak}}\)** using Ohm’s law for the reactive circuit:
\[ I_{\text{peak}} = \frac{V_{\text{peak}}}{X_C} \]
\[ I_{\text{peak}} = \frac{296.9\;\text{V}}{318.5\;\Omega} \approx 0.933\;\text{A} \]

5. **Round to the appropriate number of significant figures** – the answer is given to two decimal places, yielding \(0.93\;\text{A}\).

Hence, the peak current in the circuit is **0.93 A**, which matches the correct option provided.