A transmitting station releases… - JEE Main 2021 Physics PYQ

JEE Main 2021 — Physics Electromagnetism

2021integerhard

A transmitting station releases waves of wavelength $960m$ . A capacitor of $2.56\mu F$ is used in the resonant circuit. The self-inductance of coil necessary for resonance is $x\times {10}^{-8}H$ . find $x$

Official previous-year question

Held on 25 Feb 2021 · Verified 6 Jul 2026.

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Solution

$\lambda =960m$

$C=2.56\mu F=2.56\times {10}^{-6}F$

$c=3\times {10}^{8}m{s}^{-1}$

$L=?$

Now at resonance, ${\omega }_{0}=\frac{1}{\sqrt{LC}}$ [Resonant frequency]

$2\pi {f}_{0}=\frac{1}{\sqrt{LC}}$

On substituting ${f}_{0}=\frac{c}{\lambda },$ we have $2\pi \frac{c}{\lambda }=\frac{1}{\sqrt{LC}}$

Squaring both sides: $4{\pi }^{2}\frac{{c}^{2}}{{\lambda }^{2}}=\frac{1}{LC}$

$=\frac{4\times 10\times {(3\times {10}^{8})}^{2}}{{(960)}^{2}}=\frac{1}{L\times 2.56\times {10}^{-6}}$

$\Rightarrow \frac{1}{L}=\frac{4\times 10\times 9\times {10}^{16}\times 2.56\times {10}^{-6}}{960\times 960}$

$\Rightarrow L=10\times {10}^{-8}H$

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JEE Main 2021 — Physics Electromagnetism

2021integerhard

Official previous-year question

Held on 25 Feb 2021 · Verified 6 Jul 2026.

Did you get this right?

Solution

$\lambda =960m$

$C=2.56\mu F=2.56\times {10}^{-6}F$

$c=3\times {10}^{8}m{s}^{-1}$

$L=?$

Now at resonance, ${\omega }_{0}=\frac{1}{\sqrt{LC}}$ [Resonant frequency]

$2\pi {f}_{0}=\frac{1}{\sqrt{LC}}$

On substituting ${f}_{0}=\frac{c}{\lambda },$ we have $2\pi \frac{c}{\lambda }=\frac{1}{\sqrt{LC}}$

Squaring both sides: $4{\pi }^{2}\frac{{c}^{2}}{{\lambda }^{2}}=\frac{1}{LC}$

$=\frac{4\times 10\times {(3\times {10}^{8})}^{2}}{{(960)}^{2}}=\frac{1}{L\times 2.56\times {10}^{-6}}$

$\Rightarrow \frac{1}{L}=\frac{4\times 10\times 9\times {10}^{16}\times 2.56\times {10}^{-6}}{960\times 960}$

$\Rightarrow L=10\times {10}^{-8}H$

Your note

JEE Main Electromagnetism in other years

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Mechanics

Modern Physics

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