Question

Two protons of masses $1.6\times {10}^{-27}\text{kg}$, goes round in a concentric circular orbits of radii $0.50\text{nm}$ and $1\text{nm}$ experience centripetal forces in the ratio of $1:4$, then the ratio of frequency of revolution of the protons is (Assume no force of interaction between protons )

• A. $1:2$
• B. $2:1$
• C. $1:\sqrt{2}$
• D. $\sqrt{3}:1$

Answer 1

The correct option is C $1:\sqrt{2}$

Given, mass of proton $=1.6\times {10}^{-27}\text{kg}$

Centripetal force experienced by $1^{st}$ proton $F_1=m_1{r}_1{\omega }_1^2$

Centripetal force experienced by $2^{nd}$ proton $F_2=m_2{r}_2{\omega }_2^2$

$\therefore \frac{F_1}{F_2}=\frac{m_1{r}_1{\omega }_1^2}{m_2{r}_2{\omega }_2^2}$,

Since $m_1=m_2=m$ (mass of proton)
And, $\frac{r_1}{r_2}=\frac{0.5\text{nm}}{1\text{nm}}=\frac{1}{2};\frac{F_1}{F_2}=\frac{1}{4}$
We get,
$\frac{1}{4}=\frac{1}{2}\times \frac{{\omega }_1^2}{{\omega }_2^2}$

$\Rightarrow \frac{{\omega }_1}{{\omega }_2}=\sqrt{\frac{1}{2}}$

We also know that the frequency is given by $(f=\frac{\omega }{2\pi })$, hence $f\propto \omega$

Hence $\frac{f_1}{f_2}=\frac{{\omega }_1}{{\omega }_2}=\frac{1}{\sqrt{2}}$

Hence option C is the correct answer