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Physics · Work, Power & Energy

JEE Main 2025 — 24 January, Morning Shift — Question 65

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 \mathrm{~N} then β\beta will be

  1. Option A:

    15 N/m215 \mathrm{~N} / \mathrm{m}^{2}

  2. Option B:

    10 N/m210 \mathrm{~N} / \mathrm{m}^{2}

  3. Option C:

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

    Correct
  4. Option D:

    8 N/m28 \mathrm{~N} / \mathrm{m}^{2}

Answer: C

Step-by-step solution

F=α+βx2F=\alpha+\beta x^{2}

Work done =∫Fdx=\int F d x

5=∫(α+βx2)dx5=\int\left(\alpha+\beta x^{2}\right) d x

5=αx+βx33∣015=\alpha x+\left.\frac{\beta x^{3}}{3}\right|_{0} ^{1}

5=α+β3[α=1]5=\alpha+\frac{\beta}{3}[\alpha=1]

4=β3⇒β=12 N/m24=\frac{\beta}{3} \Rightarrow \beta=12 \mathrm{~N} / \mathrm{m}^{2}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
24 January, Morning Shift
Subject
Physics
Chapter
Work, Power & Energy
Topic
Work Done by a Constant and Variable Force