Question

A proton of mass $1.6\times {10}^{-27}$ kg, revolves in a circular path of radius 0.1m. If it is acted upon by a centripetal force of $4\times {10}^{-13}$ N, then the angular velocity of the proton is

• A. $3\times {10}^7$
• B. $4\times {10}^7$
• C. $5\times {10}^6$
• D. $8\times {10}^7$

Answer 1

Answer: (C)

$5\times {10}^6$

$\begin{array}{cc}\text{Given}-& \ast \text{mass of proton}m=1.6\times {10}^{-27}kg\\ & \ast \text{Radius of Circular path}r=0.1m\\ & \ast \text{centripetal force}=4\times {10}^{-13}N\end{array}$

$\begin{array}{c}\text{We know that cuntrepetal foru}F\\ \begin{array}{cc}F& =\frac{m{v}^2}{r}v\to \text{velocity}\\ \Rightarrow F& =m\frac{(r\omega {)}^2}{r}=m{w}^2\mu -1\end{array}\\ \omega =\text{angular velocity where}(v=r\omega \omega )\text{in}\end{array}$

$\begin{array}{c}\text{a pure circular motion.}\\ \Rightarrow \omega =\sqrt{\frac{F}{mr}}\text{from}eq-\left(1\right)\\ \Rightarrow w=\sqrt{\frac{4\times {10}^{-13}}{1.6\times {10}^{-27}\times 0.1}}=5\times {10}^6\text{rad}18\\ \text{Hence, option (C) is correct.}\end{array}$