JEE Main 2021 · PhysicsNumerical answerNumerical valueMediumMulti-step

JEE Main 26 August 2021, Shift 2, Physics Q25

Question 25 of 90 in this shift, Physics question 25 of 30, Section B.

If the maximum value of accelerating potential provided by a radio frequency oscillator is 12 kV. The number of revolution made by a proton in a cyclotron to achieve one sixth of the speed of light is __________.

[\mathrm{m_p}=1.67\times10^{-27}\ \mathrm{kg},\ \mathrm{e}=1.6\times10^{-19}\ \mathrm{C},\ \text{Speed of light}=3\times10^{8}\ \mathrm{m/s}]

Official answer

543

NTA final key.

Chapter: Moving Charges and Magnetism
Topic: Lorentz Force
Idea tested: Cyclotron Frequency and Energy

Same idea in other shifts

Asked 4× in all

4 Apr 2025, Shift 2 · Q49A particle of charge 1.6 $\mu$ C and mass 16 $\mu$ g is present in a strong magnetic field of 6.28 T. The particle is then fired…EasyNumerical value
27 Jul 2022, Shift 2 · Q43A cyclotron is used to accelerate protons. If the operating magnetic field is 1.0 T and the radius of the cyclotron 'dees' is 60 cm, the…EasySingle correct
30 Jun 2022, Shift 1 · Q43A cyclotron is working at a frequency of 10 MHz. If the radius of its dees is 60 cm. The maximum kinetic energy of accelerated proton will…MediumSingle correct

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.