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A series $L C R$ 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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Verified Answer
The correct answer is:
$\left(\frac{1}{3}\right)^{1 / 4} f_0$
We knows in LCR circuits
and
$$
\begin{aligned}
& f=\frac{1}{2 \pi \sqrt{L C}} \\
& f \propto \frac{1}{\sqrt{L}}
\end{aligned}
$$
Here, in given condition
$$\begin{aligned}
& \frac{f_1}{f_2}=\sqrt{\frac{L_1}{L_2}}=\sqrt{\frac{L}{\sqrt{3} L}} \\
& \Rightarrow \quad f_1=\left(\frac{1}{3}\right)^{1 / 4} f_2 \\
&
\end{aligned}
$$
and
$$
\begin{aligned}
& f=\frac{1}{2 \pi \sqrt{L C}} \\
& f \propto \frac{1}{\sqrt{L}}
\end{aligned}
$$
Here, in given condition
$$\begin{aligned}
& \frac{f_1}{f_2}=\sqrt{\frac{L_1}{L_2}}=\sqrt{\frac{L}{\sqrt{3} L}} \\
& \Rightarrow \quad f_1=\left(\frac{1}{3}\right)^{1 / 4} f_2 \\
&
\end{aligned}
$$
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