A particle is projected| RECTILINEAR MOTION

JEE MAIN 2021

24-02-2021 S2

Question

A particle is projected with velocity $v_0$ along $x$-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e., $m a=-a x^2 \ldots$. The distance at which the particle stops :

Select the correct option:

$\left(\frac{3 v_0^2}{2 \alpha}\right)^{\frac{1}{2}}$

$\left(\frac{2 \mathrm{v}_0}{3 \alpha}\right)^{\frac{1}{3}}$

$\left(\frac{2 \mathrm{v}_0^2}{3 \alpha}\right)^{\frac{1}{2}}$

$\left(\frac{3 \mathrm{v}_0^2}{2 \alpha}\right)^{\frac{1}{3}}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Physics

Medium

Start Preparing for JEE with Competishun

Report Issue