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JEE MAIN 2019
12-04-2019 S2
Question
A smooth wire of length $2 \pi r$ is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed $\omega$ about the vertical diameter AB, as shown in figure, the bead is at rest with respect to the circular ring at position P as shown. Then the value of $\omega^2$ is equal to:
Question Image
Select the correct option:
A
$\frac{(\mathrm{g} \sqrt{3})}{\mathrm{r}}$
B
$\frac{2 \mathrm{~g}}{\mathrm{r}}$
C
$\frac{2 \mathrm{~g}}{(\mathrm{r} \sqrt{3})}$
D
$\frac{\sqrt{3} g}{2 r}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Question Tags
JEE Main
Physics
Medium
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