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Physics · Horizontal Circular Motion

JEE Main 2024 — 29 January, Shift 1 — Question 39

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:

  1. Option A:

    3:2\sqrt{3}: 2

    Correct
  2. Option B:

    1:31: \sqrt{3}

  3. Option C:

    3:1\sqrt{3}: 1

  4. Option D:

    2:32: \sqrt{3}

Answer: A

Step-by-step solution

Given m1=m2\mathrm{m}_{1}=\mathrm{m}_{2} and r1r2=34\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{3}{4}

As centripetal force F=m2rF=\frac{m^{2}}{r}

In order to have constant (same in this question) centripetal force F1=F2\mathrm{F}_{1}=\mathrm{F}_{2} m1v12r1=m2v22r2\frac{\mathrm{m}_{1} \mathrm{v}_{1}^{2}}{\mathrm{r}_{1}}=\frac{\mathrm{m}_{2} \mathrm{v}_{2}^{2}}{\mathrm{r}_{2}}

⇒v1v2=r1r2=32\Rightarrow \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\sqrt{\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}}=\frac{\sqrt{3}}{2}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
29 January, Shift 1
Subject
Physics
Chapter
Horizontal Circular Motion
Topic
Problems involving application of circular motion