Question

A Proton of mass $1.6\times {10}^{-27}$ kg goes round in a circular orbit of radius 0.1 metre under a centripetal force of $6\times {10}^{-14}N$ then the frequency of revolution of proton is about ?

• A. $1.25\times {10}^{6}rps$
• B. $2.50\times {10}^{6}rps$
• C. $3.75\times {10}^{7}rps$
• D. $5\times {10}^{6}rps$

Answer 1

Answer: (C)

$3.75\times {10}^{7}rps$

We know,

$\omega =\frac{2\pi r}{v}$

$f=\frac{2\pi }{\omega }=\frac{v}{r}$

Also,

$F_c=\frac{m{v}^2}{r}$

numerator and denominator is multily by $r$

$F_c=\frac{m{v}^{2}r}{r^2}=mr{f}^2$

$6\times {10}^{-14}=1.6\times {10}^{-27}\times 0.1\times f^2$

$f^2=\frac{6}{1.6}\times {10}^{14}$

$f=3.75\times {10}^7$

Option $\text{C}$ is the correct answer