If the radius of curvature of the path of two | vedclass

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(A) Given that the masses of the two particles are equal,so $m_1 = m_2 = m$.The ratio of their radii of curvature is given as $\frac{r_1}{r_2} = \frac

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$\sqrt{3}: 2$

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(A) Given that the masses of the two particles are equal,so $m_1 = m_2 = m$.The ratio of their radii of curvature is given as $\frac{r_1}{r_2} = \frac{3}{4}$.The centripetal force $F$ acting on a particle is given by the formula $F = \frac{mv^2}{r}$.Since the centripetal force is constant for both particles,we have $F_1 = F_2$.Substituting the formula,we get $\frac{m_1 v_1^2}{r_1} = \frac{m_2 v_2^2}{r_2}$.Since $m_1 = m_2$,the equation simplifies to $\frac{v_1^2}{r_1} = \frac{v_2^2}{r_2}$.Rearranging the terms to find the ratio of velocities,we get $\frac{v_1^2}{v_2^2} = \frac{r_1}{r_2}$.Taking the square root on both sides,we get $\frac{v_1}{v_2} = \sqrt{\frac{r_1}{r_2}}$.Substituting the given ratio $\frac{r_1}{r_2} = \frac{3}{4}$,we get $\frac{v_1}{v_2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.Therefore,the ratio of their velocities is $\sqrt{3}:2$.

If the radius of curvature of the path of two particles of same mass are in the ratio $3:4,$ then in order to have constant centripetal force,their velocities will be in the ratio of:

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