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Saturday, 31 October

JEE Main 2022 · PhysicsMultiple choiceDiagram basedHardMulti-step

JEE Main 26 July 2022, Shift 2, Physics Q45

Question 45 of 90 in this shift, Physics question 15 of 30, Section A.

A source of potential difference VV is connected to the combination of two identical capacitors as shown in the figure. When key ‘KK’ is closed, the total energy stored across the combination is E1E_1. Now key ‘KK’ is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now E2E_2. The ratio E1/E2E_1/E_2 will be :

The figure, in words

Source V on the left connected to two capacitors each labelled C placed in parallel branches; the second (right) capacitor branch contains a key K in its top wire.
  1. (1)110\frac{1}{10}
  2. (2)25\frac{2}{5}
  3. (3)513\frac{5}{13}Official answer
  4. (4)526\frac{5}{26}

Official answer

Option 3

NTA final key.

Chapter
Electrostatic Potential and Capacitance
Topic
Parallel Plate Capacitors
Idea tested
Dielectric Insertion (Battery Connected)

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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. This question needs a figure we do not reproduce; it is described in words above. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.