JEE Main 202126 AugustEvening Shift

Exam: JEE Main
Year: 2021
Date: 26 August
Shift: Evening Shift
Subject: Physics
Chapter: Magnetic Effects of Current

JEE Main 2021 · 26 August · Evening Shift · Physics · Magnetic Effects of Current

If the maximum value of accelerating potential provided by a radio...

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

$\left[{m}_{p} = 1 . 67 \times {10}^{- 27}kg, e = 1 . 6 \times {10}^{- 19}C\right, Speed of$ light $\left= 3 \times {10}^{8}m{s}^{- 1}\right]</p>$

Correct answer

543

Step-by-step solution

From the conservation of energy,

$2neV = \dfrac{1}{2}m{v}^{2}$

$⇒ 2n \times 12 \times {10}^{3} \times 1 . 6 \times {10}^{- 19} = \dfrac{1}{2} \times 1 . 67 \times {10}^{- 27} \times {\left(5 \times {10}^{7}\right)}^{2}$

$⇒ n = 543$

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