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JEE Main 202524 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Morning Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2025 · 24 January · Morning Shift · Physics · Work, Power and Energy

A force F=α+β x^2 acts on an object in the x -direction. The work...

A force F=α+βx2\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 α=1 N\alpha=1 \mathrm{~N} then β\beta will be

Options

  1. A15 N/m215 \mathrm{~N} / \mathrm{m}^2
  2. B12 N/m212 \mathrm{~N} / \mathrm{m}^2
  3. C8 N/m28 \mathrm{~N} / \mathrm{m}^2
  4. D10 N/m210 \mathrm{~N} / \mathrm{m}^2

Correct answer

B. 12 N/m212 \mathrm{~N} / \mathrm{m}^2

Step-by-step solution

F=α+βx2F=\alpha+\beta x^2 Work done ∫dw=∫F⋅dx\int d w=\int F \cdot d x⇒ΔW=∫F⋅dx=∫(α+βx2)dx⇒ΔW=∣αx+βx33∣01=α+β3=5\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}$GivenGiven\alpha=1So,So,\frac{\beta}{3}=4 \Rightarrow \beta=12 \mathrm{~N} / \mathrm{m}^2$
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