Question

A proton is rotating along a circular path of radius $0.1\text{m}$ under a centrifugal force of $4\times {10}^{-13}\text{N}$. If the mass of proton is $1.6\times {10}^{-27}\text{kg}$, the angular velocity of rotation is (in rad/s)

• A. $2.5\times {10}^7$
• B. $5\times {10}^7$
• C. $5\times {10}^{14}$
• D. $25\times {10}^{14}$

Answer 1

The correct option is B $5\times {10}^7$

Given centrifugal force $\left(F\right)=4\times {10}^{-13}\text{N}$

Mass of proton $\left(m\right)=1.6\times {10}^{-27}\text{kg}$
Radius of circular path $\left(r\right)=0.1\text{m}$

We know that,
$F=mr{\omega }^2\Rightarrow {\omega }^2=\frac{F}{mr}$;
$\omega =\sqrt{\frac{F}{mr}}=\sqrt{\frac{4\times {10}^{-13}}{1.6\times {10}^{-27}\times 0.1}}$

$\omega =5\times {10}^7\text{rad/s}$

Hence option B is the correct answer.