Learn

Saturday, 31 October

JEE Main 2025 · PhysicsMultiple choiceSingle correctMediumMulti-step

JEE Main 2 April 2025, Shift 1, Physics Q33

Question 33 of 75 in this shift, Physics question 8 of 25, Section A.

A particle is subjected to two simple harmonic motions as : x1=7sin⁡5tx_1=\sqrt{7}\sin5t cm and x2=27sin⁡(5t+π3)x_2=2\sqrt{7}\sin\left(5t+\frac{\pi}{3}\right) cm where xx is displacement and tt is time in seconds. The maximum acceleration of the particle is x×10−2 ms−2x\times10^{-2}\ \mathrm{ms^{-2}}. The value of xx is :
  1. (1)25725\sqrt{7}
  2. (2)575\sqrt{7}
  3. (3)125
  4. (4)175Official answer

Official answer

Option 4

NTA final key.