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JEE Main 2023 (30 Jan, Shift 2) · Power

Power question from JEE Main 2023 (30 Jan, Shift 2)

Q21JEE Main 2023 · 30 JAN · SHIFT 2Medium

Power

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 $\frac{1}{3}\alpha^2\sqrt{P}\text{ m}$ . The value of $\alpha$ will be ______.

Numerical answer type. Work it out, then open the answer.

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Answer: 4

For constant power from rest, $x = \left(\frac{8P}{9m}\right)^{1/2} t^{3/2}$ .
With $m = 2$ kg, $t = 4$ s: $x = \sqrt{\frac{4P}{9}} \times 8 = \frac{16}{3}\sqrt{P}$
Compare with $\frac{1}{3}\alpha^2\sqrt{P}$ : $\alpha^2 = 16$
$\alpha = 4$

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