Q.29. A capacitor C is first charged fully with potential difference of and disconnected from the b

Q: What is the correct answer for JEE Main Physics Q29?

The correct answer is B. The initial energy stored in the capacitor is $U_C = \frac{1}{2}CV_0^2$.When 25% of the initial energy is transferred to the inductor, the energy in the inductor is $U_L = 0.25 U_C$.By energy conservation, the energy remaining in the capacitor is $U_C' = U_C - U_L = U_C - 0.25 U_C = 0.75 U_C$.So, $\frac{1}{2}CV^2 = 0.75 (\frac{1}{2}CV_0^2)$, which means $V^2 = 0.75 V_0^2$, or $V = \sqrt{0.75} V_0

Q: Q.29. A capacitor C is first charged fully with potential difference of and disconnected from the battery. The charged capacitor is connected across an inductor having inductance L. In t s 25% of…

The correct answer is B. The initial energy stored in the capacitor is $U_C = \frac{1}{2}CV_0^2$.When 25% of the initial energy is transferred to the inductor, the energy in the inductor is $U_L = 0.25 U_C$.By energy conservation, the energy remaining in the capacitor is $U_C' = U_C - U_L = U_C - 0.25 U_C = 0.75 U_C$.So, $\frac{1}{2}CV^2 = 0.75 (\frac{1}{2}CV_0^2)$, which means $V^2 = 0.75 V_0^2$, or $V = \sqrt{0.75} V_0

Q: What are the options for this JEE Main Physics question?

A. $$\frac{\pi\sqrt{LC}}{3}$$ · B. $$\frac{\pi\sqrt{LC}}{6}$$ · C. $$\frac{\pi\sqrt{LC}}{2}$$ · D. $$\pi\sqrt{LC}$$

Question

Q.29. A capacitor C is first charged fully with potential difference of  and disconnected from the battery. The charged capacitor is connected across an inductor having inductance L. In t s 25% of the initial energy in the capacitor is transferred to the inductor. The value of t is .......... s.

FirstInTest · Accepted Answer

The initial energy stored in the capacitor is $U_C = \frac{1}{2}CV_0^2$.When 25% of the initial energy is transferred to the inductor, the energy in the inductor is $U_L = 0.25 U_C$.By energy conservation, the energy remaining in the capacitor is $U_C' = U_C - U_L = U_C - 0.25 U_C = 0.75 U_C$.So, $\frac{1}{2}CV^2 = 0.75 (\frac{1}{2}CV_0^2)$, which means $V^2 = 0.75 V_0^2$, or $V = \sqrt{0.75} V_0 = \frac{\sqrt{3}}{2} V_0$.The charge on the capacitor at time t is $Q = Q_0 \cos(\omega t)$, where $Q_0 = CV_0$.So, $V = V_0 \cos(\omega t)$.Comparing the two expressions for V: $V_0 \cos(\omega t) = \frac{\sqrt{3}}{2} V_0$.$\cos(\omega t) = \frac{\sqrt{3}}{2}$.This implies $\omega t = \frac{\pi}{6}$.The angular frequency for an LC circuit is $\omega = \frac{1}{\sqrt{LC}}$.So, $t = \frac{\pi}{6\omega} = \frac{\pi\sqrt{LC}}{6}$.

Answer

$$\frac{\pi\sqrt{LC}}{3}$$

Answer

$$\frac{\pi\sqrt{LC}}{6}$$

Answer

$$\frac{\pi\sqrt{LC}}{2}$$

Answer

$$\pi\sqrt{LC}$$

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