JEE MAIN - Physics (2023 - 30th January Evening Shift - No. 24)
A body of mass $2 \mathrm{~kg}$ is initially at rest. It starts moving unidirectionally under the influence of a source of constant power P. Its displacement in $4 \mathrm{~s}$ is $\frac{1}{3} \alpha^{2} \sqrt{P} m$. The value of $\alpha$ will be ______.
Answer
4
Explanation
$$P = Fv$$
$$m{{vdv} \over {dt}} = P$$
$$m\int_0^v {vdv = \int_0^t {Pdt} } $$
$${{m{v^2}} \over 2} = Pt$$
$$v = \sqrt {{{2P} \over m}} {t^{1/2}}$$
$$\int_0^s {dx = \sqrt {{{2P} \over m}} \int_0^t {{t^{1/2}}dt} } $$
$$s = {2 \over 3}\sqrt {{{2P} \over m}} {t^{3/2}}$$
or $$s = {2 \over 3}\sqrt {{{2P} \over 2}} \times {4^{3/2}}$$
$$ = {{16} \over 3}\sqrt P ~m$$
So, $$\alpha = 4$$
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