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Counter-Strike: Source (2004)

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Comprehension-I
The capacitor of capacitance \( \mathrm{C} \) can be charged (with the help of a resistance \( \mathrm{R} \) ) by a voltage source \( \mathrm{V} \), by closing switch \( \mathrm{S}_{1} \) while keeping switch \( \mathrm{S}_{2} \) open. The capacitor can be connected in series with an inductor ' \( L \) ' by closing switch \( \mathrm{S}_{2} \) and opening \( \mathrm{S}_{1} \).
Di)
If the total charge stored in the LC circuit is \( \mathrm{Q}_{0} \), then for \( \mathrm{t} \geq 0 \)
(A) the charge on the capacitor is \( \mathrm{Q}=\mathrm{Q}_{0} \cos \left(\frac{\pi}{2}+\frac{\mathrm{t}}{\sqrt{\mathrm{LC}}}\right) \)
(B) the charge on the capacitor is \( \mathrm{Q}=\mathrm{Q}_{0} \cos \left(\frac{\pi}{2}-\frac{\mathrm{t}}{\sqrt{\mathrm{LC}}}\right) \)
(C) the charge on the capacitor is \( \mathrm{Q}=-\mathrm{LC} \frac{\mathrm{d}^{2} \mathrm{Q}}{\mathrm{dt}^{2}} \)
(D) the charge on the capacitor is \( \mathrm{Q}=-\frac{1}{\sqrt{\mathrm{LC}}} \frac{\mathrm{d}^{2} \mathrm{Q}}{\mathrm{dt}^{2}} \)
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