JEE Mains · Physics · STD 11 - Friction
- JEE Mains 2024
- Hard
- MCQ
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:
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\sqrt{3}: 2\)
Step-by-step Solution
Detailed explanation
Given \(\mathrm{m}_1=\mathrm{m}_2\)
\(\text { and } \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{3}{4}\)
As centripetal force \(F=\frac{\mathrm{mv}^2}{\mathrm{r}}\)
In order to have constant (same in this question) centripetal force
\( \mathrm{F}_1=\mathrm{F}_2 \)
\( \frac{\mathrm{m}_1 \mathrm{v}_1^2}{\mathrm{r}_1}=\frac{\mathrm{m}_2 \mathrm{v}_2^2}{\mathrm{r}_2} \)
\( \Rightarrow \frac{\mathrm{v}_1}{\mathrm{v}_2}=\sqrt{\frac{\mathrm{r}_1}{\mathrm{r}_2}}=\frac{\sqrt{3}}{2}\)
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