Q. The natural frequency of an L – C circuit is 125000 cycle/s. Then the capacitor C is replaced by another capacitor with a dielectric medium of dielectric constant K. In this case, the frequency decreases by 25 kHz. The value of K is
A) 3.0
B) 2.1
C) 1.56
D) 1.7
Explanation
(c) $\ \ \ \ \ \ \ \ \ \ \ \ \ f = \frac{1}{2\pi\sqrt{LC}}$
or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ f \propto \frac{1}{\sqrt{C}}$
When capacitor C is replaced by another capacitor C’ of dielectric constant K, then
$C^{'} = KC$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{f'}{f} = \sqrt{\frac{C}{C'}}$
or $\ \ \frac{125000 - 25000}{125000} = \sqrt{\frac{C}{KC}}$
or $\frac{100}{125} = \frac{1}{\sqrt{K}}$
or $\ \ \ \ \ \ K = \left( \frac{125}{100} \right)^{2} = 1.56$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.