16:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 17:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 15:T4eb,{"@context":"https://schema.org","@type":"Quiz","name":"Mht Cet 2025 Capacitor PYQ with Solution","url":"https://pyq-test-series-platform-alsjdcxu2-sameers-projects-41c5a71e.vercel.app/mht-cet/questions/2025/physics/capacitor/mf67qper","educationalLevel":"Entrance exam preparation","about":"Mht Cet / Physics / Capacitor","hasPart":{"@type":"Question","name":"Mht Cet 2025 Capacitor PYQ with Solution","text":"A 4 \\mu \\mathrm{~F} capacitor is charged to 10 V . The battery is then disconnected and a pure 10 mH coil is connected across the capacitor so that LC oscillations are set up. The maximum current in the coil is","eduQuestionType":"mcq","acceptedAnswer":{"@type":"Answer","text":"A"},"comment":{"@type":"Comment","text":"We have an LC oscillation setup: Capacitance C = 4 \\mu F = 4 \\times 10^{-6} F Initial capacitor voltage V = 10\\ \\text{V} Inductance L = 10\\ \\text{mH} = 10 \\times 10^{-3} H = 0.01\\ H Step 1: Initial energy stored in the capacitor E_C = \\frac{1}{2} C V^2 E_C = \\tfrac{1}{2} \\cdot 4 \\times 10^{-6} \\cdot (10^2) E_C = 2 \\times 10^{-6} \\cdot 100 = 2 \\times 10^{-4} \\ \\text{J} Step 2: Maximum energy in the inductor At peak current, all energy resides in the inductor : E_L = \\frac{1}{2} L I_{\\text{max}}^..."}}}18:T47d,

We have an LC oscillation setup:

Capacitance $C = 4 \mu F = 4 \times 10^{-6} F$
Initial capacitor voltage $V = 10\ \text{V}$
Inductance $L = 10\ \text{mH} = 10 \times 10^{-3} H = 0.01\ H$

Capacitor

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