The tuned circuit of an oscillator in a simple AM transmitter employs a 250 micro henry  Coil and 1 nf condenser. If the oscillator output is modulated by audio frequency upto 10 KHz, the frequency range occupied by the side bands in KHz is

$$\begin{gathered} Given : L=250\times {10}^{-6}H , C=1\times {10}^{-9}F and f_m=10KHz=10\times {10}^{3}Hz \\ Carrier Wave frequency, f_c=\frac{1}{2\pi \sqrt{LC}}=\frac{1}{2\pi \sqrt{250\times {10}^{-6}\times {10}^{-9}}} = =\frac{{10}^6\times 7}{22}=0.318\times {10}^6 \\ \Rightarrow f_c=318\times {10}^{3}Hz = 318 KHz \\ Hence, \\ Lower side band =f_c-f_m =318-10=308 KHz \\ Upper side band =f_c+f_m =318+10=328 KHz \end{gathered}$$