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(C) The resonance frequency of an $LC$ circuit is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.When the container is evacuated,the capacitance is $C_0$,so

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$1.01$

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(C) The resonance frequency of an $LC$ circuit is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.When the container is evacuated,the capacitance is $C_0$,so $f_0 = \frac{1}{2 \pi \sqrt{LC_0}} = 10,000\, Hz$.When the container is filled with a gas of dielectric constant $K$,the new capacitance becomes $C_g = K C_0$.The new resonance frequency is $f_g = \frac{1}{2 \pi \sqrt{L(K C_0)}} = \frac{f_0}{\sqrt{K}}$.Given that the frequency changes by $50\, Hz$,the new frequency is $f_g = 10,000 - 50 = 9,950\, Hz$.Thus,$\frac{f_g}{f_0} = \frac{1}{\sqrt{K}} = \frac{9,950}{10,000} = 0.995$.Squaring both sides,$\frac{1}{K} = (0.995)^2 \approx 0.990025$.$K = \frac{1}{0.990025} \approx 1.010075 \approx 1.01$.

The capacitor of an oscillatory circuit is enclosed in a container. When the container is evacuated,the resonance frequency of the circuit is $10\, kHz$. When the container is filled with a gas,the resonance frequency changes by $50\, Hz$. The dielectric constant of the gas is

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