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

Hint: 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.