A coil of inductance 8 μ H is connected to a capacitor of capacitance 0.02 μ F . To what wavelength is this circuit tuned?

Correct answer: D. 7.54 × 10^2 ~m aligned & Here, L=8 μ H=8 ×10^ -6 H ; & C=0.02 μ F=0.02 ×10^ -6 F & Resonant frequency, & f_r= 1 2 π L C = 1 2 π 8 × 10^ -6 × 0.02 × 10^ -6 & =3.98 ×10^5 Hz & aligned If c (=3 ×10^8 m s^ -1 ) is the velocity of the electromagnetic wave, then, Wavelength, λ=(c)/(f)=(3 × 10^8)/(3.98 × 10^5)=7.54 ×10^2 m

AIIMS 2015

Exam: AIIMS
Year: 2015
Subject: Physics
Chapter: Alternating Current

AIIMS 2015 · Physics · Alternating Current

A coil of inductance 8 μ H is connected to a capacitor of...

A coil of inductance

8 \mu \mathrm{H}

is connected to a capacitor of capacitance

0.02 \mu \mathrm{F}

. To what wavelength is this circuit tuned?

Options

A $7.54 \times 10^3 \mathrm{~m}$
B $4.12 \times 10^2 \mathrm{~m}$
C $15.90 \times 10^3 \mathrm{~m}$
D $7.54 \times 10^2 \mathrm{~m}$

Correct answer

D. $7.54 \times 10^2 \mathrm{~m}$

Step-by-step solution

\begin{aligned}& \text{ Here, } L=8 \mu \mathrm{H}=8 \times10^{-6} \mathrm{H} \text{; }\\ & C=0.02 \mu \mathrm{F}=0.02 \times10^{-6} \mathrm{ F}\\ & \therefore \text{ Resonant frequency, }\\ & f_r=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \sqrt{8 \times 10^{-6} \times 0.02 \times 10^{-6}}}\\ & =3.98 \times10^5 \mathrm{ Hz}\\ & \end{aligned}

c

\left(=3 \times10^8 \mathrm{ m} \mathrm{ s}^{-1}\right)

is the velocity of the electromagnetic wave, then, Wavelength,

\lambda=\frac{c}{f}=\frac{3 \times 10^8}{3.98 \times 10^5}=7.54 \times10^2 \mathrm{ m}$

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