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JEE Main 2024 (29 Jan, Shift 1) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from JEE Main 2024 (29 Jan, Shift 1)

Q14JEE Main 2024 · 29 JAN · SHIFT 1Easy

Centripetal Acceleration & Force

If the radius of curvature of the path of two particles of same mass are in the ratio 3:43 : 4, then in order to have constant centripetal force, their velocities will be in the ratio of:

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Answer: (c)

  1. F=mv2rF = \frac{mv^2}{r} with same mm and same FF gives v2∝rv^2 \propto r.
  2. v1v2=r1r2=34=32\frac{v_1}{v_2} = \sqrt{\frac{r_1}{r_2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}
  3. So the answer is (c) 3:2\sqrt{3} : 2.
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