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(D) The resonant frequency of a series $LCR$ circuit is given by the formula: $f = \frac{1}{2\pi\sqrt{LC}}$.From this formula,we can see that the reso

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$\left(\frac{1}{3}\right)^{1/4} f_0$

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(D) The resonant frequency of a series $LCR$ circuit is given by the formula: $f = \frac{1}{2\pi\sqrt{LC}}$.From this formula,we can see that the resonant frequency $f$ is inversely proportional to the square root of the inductance $L$,provided the capacitance $C$ remains constant: $f \propto \frac{1}{\sqrt{L}}$.Let the initial resonant frequency be $f_0$ with inductance $L$. Thus,$f_0 = \frac{1}{2\pi\sqrt{LC}}$.When the inductance is increased by $\sqrt{3}$ times,the new inductance $L' = \sqrt{3}L$. The resistance $R$ does not affect the resonant frequency.The new resonant frequency $f'$ is given by: $f' = \frac{1}{2\pi\sqrt{L'C}} = \frac{1}{2\pi\sqrt{(\sqrt{3}L)C}}$.Dividing $f'$ by $f_0$: $\frac{f'}{f_0} = \frac{\frac{1}{2\pi\sqrt{\sqrt{3}LC}}}{\frac{1}{2\pi\sqrt{LC}}} = \frac{1}{\sqrt{\sqrt{3}}} = \left(\frac{1}{3^{1/2}}\right)^{1/2} = \left(\frac{1}{3}\right)^{1/4}$.Therefore,the new resonant frequency is $f' = \left(\frac{1}{3}\right)^{1/4} f_0$.

$A$ series $LCR$ circuit is connected across a source of alternating emf of changing frequency and resonates at frequency $f_0$. Keeping capacitance constant,if the inductance $(L)$ is increased by $\sqrt{3}$ times and resistance is increased $(R)$ by $1.4$ times,the resonant frequency now is

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