JEE Mains · Physics · STD 11 - 4.2 friction
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:
- A \(\sqrt{3}: 2\)
- B \(1: \sqrt{3}\)
- C \(\sqrt{3}: 1\)
- D \(2: \sqrt{3}\)
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 \( \mathrm{F}_1=\mathrm{F}_2 \)…
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