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

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Question

Given $\mathrm{m}_1=\mathrm{m}_2$

$	ext { and } \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{3}{4}$

As centripetal force $F=\frac{\mathrm{mv}^2}{\ma

Vedclass · Accepted Answer

$\sqrt{3}: 2$

Vedclass · Answer

Given $\mathrm{m}_1=\mathrm{m}_2$

$	ext { and } \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{3}{4}$

As centripetal force $F=\frac{\mathrm{mv}^2}{\mathrm{r}}$

In order to have constant (same in this question) centripetal force

$ \mathrm{F}_1=\mathrm{F}_2 $

$ \frac{\mathrm{m}_1 \mathrm{v}_1^2}{\mathrm{r}_1}=\frac{\mathrm{m}_2 \mathrm{v}_2^2}{\mathrm{r}_2} $

$ \Rightarrow \frac{\mathrm{v}_1}{\mathrm{v}_2}=\sqrt{\frac{\mathrm{r}_1}{\mathrm{r}_2}}=\frac{\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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