Physics · 2025
JEE Main · 24 January 2025, Shift 1 · Q29
A force F =α+β x^2 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 α=1…
A force $\displaystyle \mathrm{F}=\alpha+\beta \mathrm{x}^2$ acts on an object in the x -direction. The work done by the force is $\displaystyle 5$ J when the object is displaced by $\displaystyle 1$ m . If the constant $\displaystyle \alpha=1 \mathrm{~N}$ then $\displaystyle \beta$ will be
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle 12 \mathrm{~N} / \mathrm{m}^2$
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JEE Main 2025 Physics question, with the answer from NTA’s final answer key. Where our answers come from.