Question

If the radius of curvature of the path of two particles of same masses are in the ratio $1:3$, then in order to have same centripetal force, their velocity should be in the ratio of

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

Answer 1

The correct option is D $1:\sqrt{3}$

Given, $m_1=m_2=m$ (say)
Given that same centripetal force $F_1=F_2=F$ (say) and
$\frac{r_1}{r_2}=\frac{1}{3}$

For $1^{st}$ particle:
Centripetal force $\left(F_1\right)=\frac{m_1{v}_1^2}{r_1}$

For $2^{nd}$ particle:
Centripetal force $\left(F_2\right)=\frac{m_2{v}_2^2}{r_2}$

$\frac{F_1}{F_2}=\frac{\frac{m_1{v}_1^2}{r_1}}{\frac{m_2{v}_2^2}{r_2}}=\frac{m_1{v}_1^2{r}_2}{m_2{v}_2^2{r}_1}$

$(\because F_1=F_2=F;m_1=m_2=m)$

We get,
$\frac{v_1^2}{v_2^2}=\frac{r_1}{r_2}$
$\Rightarrow \frac{v_1}{v_2}=\sqrt{\frac{r_1}{r_2}}=\frac{1}{\sqrt{3}}$

Hence option D is the correct answer