JEE Main 2025 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 3 April 2025, Shift 2, Physics Q38

Question 38 of 75 in this shift, Physics question 13 of 25, Section A.

Using a battery, a 100 pF capacitor is charged to 60 V and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is 20 V, its capacitance is: (in pF)

(1)100
(2)200Official answer
(3)600
(4)400

Official answer

Option 2

NTA final key.

Chapter: Electrostatic Potential and Capacitance
Topic: Parallel Plate Capacitors
Idea tested: Charge Sharing and Heat Loss

Same idea in other shifts

Asked 4× in all

8 Apr 2023, Shift 2 · Q55A 600 pF capacitor is charged by 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF…MediumNumerical value
11 Apr 2023, Shift 1 · Q41A parallel plate capacitor of capacitance 2 F is charged to a potential V, The energy stored in the capacitor is $E_1$ . The capacitor is…EasySingle correct
2 Sep 2020, Shift 1 · Q23A $5\ \mu\mathrm{F}$ capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to…MediumNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.