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Friday, 9 October

JEE Main 2021 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 25 July 2021, Shift 1, Physics Q24

Question 24 of 90 in this shift, Physics question 24 of 30, Section B.

An inductor of 10 mH is connected to a 20 V battery through a resistor of 10 kΩ\Omega and a switch. After a long time, when maximum current is set up in the circuit, the current is switched off. The current in the circuit after 1 μ\mus is x100\frac{x}{100} mA. Then xx is equal to _______. (Take e−1=0.37e^{-1}=0.37)

Official answer

74

NTA final key.

Topic
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Idea tested
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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 42 of the 174 JEE Main shifts held from 2017 to 2026.