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

Solution

Physics sir · 4 steps

The wavelength gives you the frequency, and that frequency must be the circuit's resonant frequency.

1. Frequency of the wave. Radio waves travel at c = 3 × 10⁸ m s⁻¹, so f = c/λ = (3 × 10⁸)/960 = 3.125 × 10⁵ Hz.
2. Resonance condition. An LC circuit resonates at f = 1/2π√LC. Square both sides and solve for L.

   L = 1/(4π² f² C)

3. Put in numbers. f² C = (3.125 × 10⁵)² × 2.56 × 10⁻⁶ = 9.766 × 10¹⁰ × 2.56 × 10⁻⁶ = 2.5 × 10⁵. Then 4π² f² C = 39.48 × 2.5 × 10⁵ = 9.87 × 10⁶, so L = 1.01 × 10⁻⁷ H.
4. So. L ≈ 10.1 × 10⁻⁸ H, which rounds to 10. Taking π² ≈ 10 gives exactly 10 × 10⁻⁸ H.

The common slip: 400. Dropping the 4π² and using L = 1/(f² C) gives 4 × 10⁻⁶ H, that is 400, not 10.

The idea to keep. A station tuned to a wavelength means an LC circuit resonating at f = c/λ. Get f first, then use f = 1/2π√LC for whichever of L or C is missing.

Answer from the NTA final answer key. Solution by PrepWiser.