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JEE Main 2025 (24 Jan, Shift 1) · Work by a Variable Force

Work by a Variable Force question from JEE Main 2025 (24 Jan, Shift 1)

Q2JEE Main 2025 · 24 JAN · SHIFT 1Medium

Work by a Variable Force

A force F=α+βx2F = \alpha + \beta 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\text{ N} then β\beta will be

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Answer: (c)

  1. Taking the displacement from x=0x = 0 to x=1x = 1 m: W=∫01(α+βx2) dx=α+β3W = \int_0^1 (\alpha + \beta x^2)\,dx = \alpha + \frac{\beta}{3}
  2. 5=1+β3⇒β=125 = 1 + \frac{\beta}{3} \Rightarrow \beta = 12
  3. So the answer is (c) 12 N/m212\text{ N/m}^2.
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