JEE Main 2025 · 24 January · morning · Single correct · +4 / −1

A force F=α+β x² 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 then β will be

  1. (a) 15 N / m²
  2. (b) 10 N / m²
  3. (c) 12 N / m²
  4. (d) 8 N / m²
Answer and solution

Answer (c) 12 N / m²

Solution

Physics sir · 3 steps

The force changes with x, so integrate it from 0 to 1 m. Do not multiply force by distance.

  1. Variable force. Here F depends on x, so W = F × d does not work. Add up F dx over the displacement, taking the object from x = 0 to x = 1 m.

    W = ∫₀¹ (α + β x²) dx = α + β/3

  2. Put in numbers. With W = 5 J and α = 1 N, you get 1 + β/3 = 5. So β/3 = 4.
  3. So. β = 12 N/m², which is option (c).

The idea to keep. Whenever the force depends on position, work is ∫ F dx. Integrate each term separately: xⁿ becomes xⁿ⁺¹/(n+1).

Answer from the NTA final answer key. Solution by PrepWiser.

Study Work, Energy & Power

Asked about 100 times in JEE Main since 2020. This question is from 2025.