QAIPMT-2007-Ph-2331: NEET 2007 Chemistry - Class 12 AIPMT-2007-Ph-2331 Solution: Which one of the following on treatment with 50% aqueous sodium …

Step-by-Step Derivation 1. Condition for Maximum Current: In a series LCR alternating current circuit, the current ($I$) is maximized when the total impedance ($Z$) of the circuit is at its absolute minimum. This state is known as electrical resonance. At resonance, the inductive reactance ($X_{L}$) perfectly cancels out the capacitive reactance ($X_{C}$): $X_{L}=X_{C}$ 2. Relate reactances to Inductance ($L$) and Capacitance ($C$): $\omega L=\frac{1}{\omega C}$ Rearranging the formula to isolate the inductance ($L$): $L=\frac{1}{\omega ^{2}C}$ 3. Substitute the given parameters: � Capacitance ($C$) = $10\ \mu\text{F} = 10 \times 10^{-6}\ \text{F} = 10^{-5}\ \text{F}$ � Angular Frequency ($\omega$) = $1000\ \text{s}^{-1} = 10^3\ \text{s}^{-1}$ $L=\frac{1}{(10^{3})^{2}\times 10^{-5}}$ $L=\frac{1}{10^{6}\times 10^{-5}}=\frac{1}{10^{1}}=0.1\ \text{H}$ 4. Convert Henries to milliHenries: $L=0.1\times 1000\ \text{mH}=100\ \text{mH}$ (Note: While the value of the resistance $R$ dictates the actual peak height of the maximum current, it has absolutely no effect on the specific resonant frequency or the choice of $L$ needed to hit that resonance. Therefore, $R$ is not required to solve the problem).