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JEE MAIN 2021
27-07-2021 S2
Question
A $100 \Omega$ resistance, a $0.1 \mu \mathrm{~F}$ capacitor and an inductor are connected in series across a 250 V supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is 60 Hz
Select the correct option:
A
0.70 H
B
70.3 mH
C
$7.03 \times 10^{-5} \mathrm{H}$
D
70.3 H
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$$ \mathrm{C}=0.1 \mu \mathrm{~F}=10^{-7} \mathrm{~F} $$ Resonant frequency $=60 \mathrm{~Hz}$ $$ \begin{aligned} & \omega_0=\frac{1}{\sqrt{L C}} \\ & 2 \pi f_0=\frac{1}{\sqrt{L C}} \Rightarrow L=\frac{1}{4 \pi^2 f_0^2 C} \end{aligned} $$ by putting values $\mathrm{L}=70.3 \mathrm{~Hz}$
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