GATE 2025 IN Q.62: solution and answer

Official answer: 4.5 to 5.5Source: Official answer key, GATE 2025 IN (IIT Roorkee)

1. The quality factor of a series RLC circuit at resonance is $Q=\frac{{\omega }_{0}L}{R}$. The resonant frequency ${\omega }_0=7500\text{rad/s}$ is given as a fixed value; only $L$ and $R$ are measured with uncertainty. The nominal value is $Q=\frac{7500\times 0.02}{10}=15$.

2. Relative uncertainty of each measured quantity: $\frac{\delta L}{L}=\frac{0.8\text{mH}}{20\text{mH}}=0.04=4\%$ and $\frac{\delta R}{R}=\frac{0.3\Omega }{10\Omega }=0.03=3\%$.

3. Since $Q=k{L}^1{R}^{-1}$ with $k={\omega }_0$ constant, $\ln Q=\ln k+\ln L-\ln R$. The sensitivity of $\ln Q$ to $\ln L$ is $+1$ and to $\ln R$ is $-1$, so each relative uncertainty passes into $Q$ with weight 1 (the sign does not matter once it is squared).

4. Independent uncertainties combine by root-sum-square (the Kline–McClintock rule): $$\frac{\delta Q}{Q}=\sqrt{{\left(\frac{\delta L}{L}\right)}^2+{\left(\frac{\delta R}{R}\right)}^2}=\sqrt{(4\%{)}^2+(3\%{)}^2}=\sqrt{25}\%=5\%$$

5. So the uncertainty in $Q$ is $\pm 0.05\times 15=\pm 0.75$, and the percentage uncertainty in the quality factor is $5.0\%$.