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
- (a) 15 N / m²
- (b) 10 N / m²
- (c) 12 N / m²
- (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.
- 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
- Put in numbers. With W = 5 J and α = 1 N, you get 1 + β/3 = 5. So β/3 = 4.
- 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.