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 $$x \times 10^{-8}$$ H. Find $$x$$.

The transmitting station releases waves of wavelength $$\lambda = 960\,\text{m}$$. The frequency of the wave is $$f = \frac{c}{\lambda} = \frac{3 \times 10^8}{960}\,\text{Hz}$$.

For a resonant circuit, the resonant frequency is $$f = \frac{1}{2\pi\sqrt{LC}}$$. At resonance, this must equal the frequency of the transmitted wave.

Therefore: $$\frac{c}{\lambda} = \frac{1}{2\pi\sqrt{LC}}$$, which gives $$\sqrt{LC} = \frac{\lambda}{2\pi c}$$.

Squaring both sides: $$LC = \frac{\lambda^2}{4\pi^2 c^2}$$.

Solving for $$L$$: $$L = \frac{\lambda^2}{4\pi^2 c^2 C}$$.

Substituting $$\lambda = 960\,\text{m}$$, $$c = 3 \times 10^8\,\text{m/s}$$, and $$C = 2.56 \times 10^{-6}\,\text{F}$$:

$$L = \frac{(960)^2}{4\pi^2 \times (3 \times 10^8)^2 \times 2.56 \times 10^{-6}}$$

Computing the numerator: $$960^2 = 921600$$. Computing the denominator: $$4\pi^2 \approx 39.478$$, $$(3 \times 10^8)^2 = 9 \times 10^{16}$$. So the denominator is $$39.478 \times 9 \times 10^{16} \times 2.56 \times 10^{-6} = 39.478 \times 23.04 \times 10^{10} \approx 9.096 \times 10^{12}$$.

Therefore: $$L = \frac{921600}{9.096 \times 10^{12}} \approx 1.013 \times 10^{-7}\,\text{H} \approx 10 \times 10^{-8}\,\text{H}$$.

Comparing with $$x \times 10^{-8}$$ H, we get $$x = 10$$.