JEE Main 2025 — Work Power Energy Question with Solution
JEE Main 2025 (24 Jan Shift 1)
Question
A force acts on an object in the x -direction. The work done by the force is 5 J when the object is displaced by 1 m . If the constant then will be
Choose an option
Show full solutionCorrect option: B
Correct answer
B
Step-by-step explanation
Work done
$\begin{aligned}
& \Rightarrow \quad \Delta W=\int F \cdot d x=\int\left(\alpha+\beta x^2\right) d x \\
& \Rightarrow \quad \Delta W=\left|\alpha x+\frac{\beta x^3}{3}\right|_0^1=\alpha+\frac{\beta}{3}=5
\end{aligned}$
Given
So,
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This is a previous-year question from JEE Main 2025, covering the Work Power Energy chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.