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

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(D) The centripetal force is given by $F = \frac{mv^2}{r}$.Since the masses $m$ and the centripetal force $F$ are constant,we have $F = \frac{mv^2}{r}

Vedclass · Accepted Answer

$1 : \sqrt 2$

Vedclass · Answer

(D) The centripetal force is given by $F = \frac{mv^2}{r}$.Since the masses $m$ and the centripetal force $F$ are constant,we have $F = \frac{mv^2}{r} \implies v^2 \propto r$.Taking the square root on both sides,we get $v \propto \sqrt{r}$.Therefore,the ratio of their velocities is $\frac{v_1}{v_2} = \sqrt{\frac{r_1}{r_2}}$.Given the ratio of the radii of curvature is $\frac{r_1}{r_2} = \frac{1}{2}$,we substitute this into the equation:$\frac{v_1}{v_2} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.Thus,the ratio of their velocities is $1 : \sqrt{2}$.

If the radius of curvature of the path of two particles of same masses are in the ratio $1 : 2$,then in order to have constant centripetal force,their velocity,should be in the ratio of

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