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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2025
  • Easy
  • MCQ

A force \(\mathrm{F}=\alpha+\beta \mathrm{x}^2\) 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 \(\alpha=1 \mathrm{~N}\) then \(\beta\) will be ________.

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Answer & Solution

Correct Answer

(B) \(12 \mathrm{~N} / \mathrm{m}^2\)

Step-by-step Solution

Detailed explanation

\(F=\alpha+\beta x^2\)
Work done \(\int d w=\int F \cdot d x\)
\(\begin{aligned}
& \Rightarrow \quad \Delta W=\int F \cdot d x=\int\left(\alpha+\beta x^2\right) d x \\
& \Rightarrow \quad \Delta W=\left|\alpha x+\frac{\beta x^3}{3}\right|_0^1=\alpha+\frac{\beta}{3}=5
\end{aligned}\)
Given \(\alpha=1\)
So, \(\frac{\beta}{3}=4\)
\(\Rightarrow \beta=12 \mathrm{~N} / \mathrm{m}^2\)
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