Question

A proton of mass $1.6\times {10}^{-27}$ kg goes round in a circular orbit of radius 0.10 m under a centripetal force of $4\times {10}^{-13}N$. then the frequency of revolution of the proton is about

• A. $0.08\times {10}^{8}cyclespersec$
• B. $4\times {10}^{8}cyclespersec$
• C. $8\times {10}^{8}cyclespersec$
• D. $12\times {10}^{8}cyclespersec$

Answer 1

The correct option is A

$0.08\times {10}^{8}cyclespersec$

$m4{\pi }^2{n}^{2}r=4\times {10}^{-13}\Rightarrow n=0.08\times {10}^{8}cycles/sec$.