Q 11-05-093JEE MainJEE Main 2023 (30 Jan, Shift 2)Medium
A body of mass $2\ \text{kg}$ is initially at rest. It starts moving unidirectionally under the influence of a source of constant power $P$. Its displacement in $4\ \text{s}$ is $\dfrac13\alpha^2\sqrt P\ \text{m}$. The value of $\alpha$ will be ______.
Numerical value type. Enter your answer.
Answer: 4
$\tfrac12mv^2=Pt\Rightarrow v=\sqrt{\dfrac{2Pt}{m}}=\sqrt{P}\,t^{1/2}$ (with $m=2$).
$$x=\int_0^4\sqrt P\,t^{1/2}dt=\frac23\sqrt P\,(4)^{3/2}=\frac{16}{3}\sqrt P$$
So $\alpha^2=16$, $\alpha=4$.
Solution by Sreeraj P, M.Sc Physics