Question

A proton goes round in a circular orbit of radius 0.01 m under a centripetal force of $4\times 10^{-3}$ N. What is the frequency of revolution of protons?
A. $2.5\times 10^{12}$ Hz
B. $4\times 10^{13}$ Hz
C. $8 \times 10^{12}$ Hz
D. $16\times 10^{12}$ Hz

Answer 1

Consider $m = 1.67\times 10^{-27}$ . Since angular velocity(ω) occurs due to circular path, so covert $v = r\omega$ , and convert ω to frequency to get the final answer.
Formula: For Centripetal force $\left( {{F_c}} \right) = m{\omega ^2}R$

Complete step by step solution:
It is important to understand that the centripetal force is not a fundamental force, but just a label given to the net force which causes an object to move in a circular path.
Given,
R(radius) = 0.01 m= $10^{-2}$ m
Fc (centripetal force) = $4\times 10^{-3}$ N
$m = 1.67\times 10^{-27}$
Using formula,
$\left( {{F_c}} \right) = m{\omega ^2}R$
$\Rightarrow 4\times {10^{ - 3}} = \left( {1.67\times {{10}^{ - 27}}} \right) \times {\omega ^2}\times{10^{ - 2}} \\\ \Rightarrow {\omega ^2} = \dfrac{{(4\times{{10}^{ - 3}})}}{{(1.67 \times {{10}^{ - 27}})\times{{10}^{ - 2}}}} \;$
$\therefore \omega = 1.55 \times {10^{13}}$ (removing the squares)
Using formula,
$\omega = 2\pi f \\\ \therefore f = \dfrac{\omega }{{2\pi }} = \dfrac{{1.55\times {{10}^{13}}}}{{2\pi }} = 2.46\times {10^{12}} \;$
Thus, the required frequency is $2.5\times 10^{12}$ Hz (approx.)
So, the correct answer is “Option A”.

Note: A centripetal force is a net force that acts on an object to keep it moving along a circular path.
Here we will see the one common example -: centripetal force is the case in which a body moves with uniform speed along a circular path. The centripetal force is directed at right angles to the motion and also along the radius towards the centre of the circular path.