JEE MAIN - Physics (2021 - 25th February Morning Shift - No. 27)
A transmitting station releases waves of wavelength 960 m. A capacitor of 2.56 $$\mu$$F is used in the resonant circuit. The self inductance of coil necessary for resonance is __________ $$\times$$ 10$$-$$8 H.
Answer
10
Explanation
$$\lambda$$ = 960 m
C = 2.56 $$\mu$$F = 2.56 $$\times$$ 10$$-$$6 F
c = 3 $$\times$$ 108 m/s
L = ?
Now at resonance, $${\omega _0} = {1 \over {\sqrt {LC} }}$$
[Resonant frequency]
$$2\pi {f_0} = {1 \over {\sqrt {LC} }}$$
On substituting $${f_0} = {c \over \lambda }$$, we have $$2\pi {c \over \lambda } = {1 \over {\sqrt {LC} }}$$
Squaring both sides : $$4{\pi ^2}{{{c^2}} \over {{\lambda ^2}}} = {1 \over {LC}}$$
$$ = {{4 \times 10 \times {{(3 \times {{10}^8})}^2}} \over {{{(960)}^2}}} = {1 \over {L \times 2.56 \times {{10}^{ - 6}}}}$$
$$ \Rightarrow {1 \over L} = {{4 \times 10 \times 9 \times {{10}^{16}} \times 2.56 \times {{10}^{ - 6}}} \over {960 \times 960}}$$
$$ \Rightarrow L = 10 \times {10^{ - 8}}$$ H
C = 2.56 $$\mu$$F = 2.56 $$\times$$ 10$$-$$6 F
c = 3 $$\times$$ 108 m/s
L = ?
Now at resonance, $${\omega _0} = {1 \over {\sqrt {LC} }}$$
[Resonant frequency]
$$2\pi {f_0} = {1 \over {\sqrt {LC} }}$$
On substituting $${f_0} = {c \over \lambda }$$, we have $$2\pi {c \over \lambda } = {1 \over {\sqrt {LC} }}$$
Squaring both sides : $$4{\pi ^2}{{{c^2}} \over {{\lambda ^2}}} = {1 \over {LC}}$$
$$ = {{4 \times 10 \times {{(3 \times {{10}^8})}^2}} \over {{{(960)}^2}}} = {1 \over {L \times 2.56 \times {{10}^{ - 6}}}}$$
$$ \Rightarrow {1 \over L} = {{4 \times 10 \times 9 \times {{10}^{16}} \times 2.56 \times {{10}^{ - 6}}} \over {960 \times 960}}$$
$$ \Rightarrow L = 10 \times {10^{ - 8}}$$ H
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