JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2021
- Hard
- Numerical
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 ....... . \(\left[{m}_{{p}}=1.67 \times 10^{-27} {kg}, {e}=1.6 \times 10^{-19} {C},\right.\) Speed of light \(\left.=3 \times 10^{8} {m} / {s}\right]\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(543\)
Step-by-step Solution
Detailed explanation
\({V}=12\, {kV}\)
Number of revolution \(={n}\)
\({n}\left[2 \times {q}_{{P}} \times {V}\right]=\frac{1}{2} {m}_{{p}} \times {v}_{{P}}^{2}\)
\({n}\left[2 \times 1.6 \times 10^{-19} \times 12 \times 10^{3}\right.\)
\(=\frac{1}{2} \times 1.67 \times 10^{-27} \times\left[\frac{3 \times 10^{8}}{6}\right]^{2}\)
\({n}\left(38.4 \times 10^{-16}\right)=0.2087 \times 10^{-11}\)
\({n}=543.4\)
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