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Saturday, 31 October

JEE Main 2026 · PhysicsMultiple choiceSingle correctMediumCalculation

JEE Main 6 April 2026, Shift 2, Physics Q40

Question 40 of 75 in this shift, Physics question 15 of 25, Section A.

A sphere of capacitance 100 pF is charged to a potential of 100 V. Another identical uncharged metal sphere is brought in contact with the charged sphere, then the change in the total energy stored on these spheres, when they touch is α×10−7\alpha\times10^{-7} J. The value of α\alpha is __________. (combined capacitance of spheres is 200 pF)
  1. (1)5
  2. (2)52\frac{5}{2}Official answer
  3. (3)72\frac{7}{2}
  4. (4)92\frac{9}{2}

Official answer

Option 2

NTA final key.

Chapter
Electrostatic Potential and Capacitance
Topic
Electric Potential
Idea tested
Heat Loss During Charge Sharing

Same topic in other shifts

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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 108 of the 174 JEE Main shifts held from 2017 to 2026.