JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Alternating Current

The quality factor of a series $LCR$ circuit is $Q=\frac{1}{R}\sqrt{\frac{L}{C}}$, which is the same as $\frac{{\omega }_{0}L}{R}$ only when one remembers that the resonant frequency ${\omega }_0=\frac{1}{\sqrt{LC}}$ itself changes with $L$. Writing the first form, $Q_1=\frac{1}{R}\sqrt{\frac{L}{C}}=100$. When the inductance is doubled to $2L$ and the resistance halved to $\frac{R}{2}$, with the capacitance unchanged, the new value is Hence $Q_2=2\sqrt{2}\times 100=200\sqrt{2}\approx 282.84$. Holding ${\omega }_0$ fixed and using $Q\propto \frac{L}{R}$ would wrongly give $400$, but ${\omega }_0$ falls by a factor $\sqrt{2}$ when $L$ doubles, which leaves the overall factor $2\sqrt{2}$.