JEE Main 2021 — Electromagnetic Induction Question with Solution
From: JEE Main 2021 (Online) 26th August Evening Shift
Question
If the maximum value of accelerating potential provided by a ratio 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 ...............
[mp = 1.67 1027 kg, e = 1.6 1019C, Speed of light = 3 108 m/s]
[mp = 1.67 1027 kg, e = 1.6 1019C, Speed of light = 3 108 m/s]
Enter your answer
Show full solutionCorrect answer: 543.4
Correct answer
543.4
Step-by-step explanation
V = 12 kV
Number of revolution = n
n(38.4 1016) = 0.2087 1011
n = 543.4
Number of revolution = n
n(38.4 1016) = 0.2087 1011
n = 543.4
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