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Friday, 9 October

JEE Main 2024 · PhysicsMultiple choiceSingle correctMediumCalculation

JEE Main 30 January 2024, Shift 2, Physics Q43

Question 43 of 90 in this shift, Physics question 13 of 30, Section A.

An alternating voltage V(t)=220sin⁡100πtV(t) = 220\sin 100\pi t volt is applied to a purely resistive load of 50 Ω50\ \Omega. The time taken for the current to rise from half of the peak value to the peak value is:
  1. (1)3.3 ms3.3\ msOfficial answer
  2. (2)2.2 ms2.2\ ms
  3. (3)7.2 ms7.2\ ms
  4. (4)5 ms5\ ms

Official answer

Option 1

NTA final key.