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JEE Main 2021 (25 Feb, Shift 1) · Resonance, Bandwidth & Q Factor

Resonance, Bandwidth & Q Factor question from JEE Main 2021 (25 Feb, Shift 1)

Q93JEE Main 2021 · 25 FEB · SHIFT 1Medium

Resonance, Bandwidth & Q Factor

A transmitting station releases waves of wavelength 960 m. A capacitor of 2.56 μ\muF is used in the resonant circuit. The self inductance of coil necessary for resonance is __________ ×\times 10−-8 H.

Numerical answer type. Work it out, then open the answer.

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Answer: 10

  1. f=cλ=3×108960=3.125×105f = \frac{c}{\lambda} = \frac{3\times10^{8}}{960} = 3.125\times10^{5} Hz
  2. L=14π2f2C=139.48×9.77×1010×2.56×10−6L = \frac{1}{4\pi^2 f^2 C} = \frac{1}{39.48\times9.77\times10^{10}\times2.56\times10^{-6}}
  3. L=19.87×106≈1.0×10−7L = \frac{1}{9.87\times10^{6}} \approx 1.0\times10^{-7} H =10×10−8= 10\times10^{-8} H
  4. So the answer is 10.
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