Question

A transmitting station releases waves of wavelength $960m$. 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 1

Answer: 10

$\lambda =960m$
$C=2.56\mu F=2.56\times 10^{-6}F$
$c=3\times 10^{8}m/s$
$L=?$
Now at resonance, ${\omega }_0=\frac{1}{\sqrt{LC}}$
[Resoant frequency]
$2\pi f_0=\frac{1}{\sqrt{LC}}$
On substituting $f_0=\frac{c}{\lambda },$ we have $2\pi \frac{c}{\lambda }=\frac{1}{\sqrt{LC}}$
Squaring both sides : $4{\pi }^2\frac{c^2}{{\lambda }^2}=\frac{1}{LC}$
$=\frac{4\times 10\times {\left(3\times 10^8\right)}^2}{(960{)}^2}=\frac{1}{L\times 2.56\times 10^{-6}}$
$\Rightarrow \frac{1}{L}=\frac{4\times 10\times 9\times 10^{16}\times 2.56\times 10^{-6}}{960\times 960}$
$\Rightarrow L=10\times 10^{-8}H$