Q 11-05-060JEE MainJEE Main 2025 (24 Jan, Shift 1)Easy
A force $F = \alpha + \beta x^2$ acts on an object in the $x$-direction. The work done by the force is $5\ \text{J}$ when the object is displaced by $1\ \text{m}$. If the constant $\alpha = 1\ \text{N}$, then $\beta$ will be
Answer: (B) $12\ \text{N/m}^2$
$$W = \int_0^1(\alpha + \beta x^2)\,dx = \alpha + \frac{\beta}{3}$$
$$1 + \frac{\beta}{3} = 5 \Rightarrow \beta = 12\ \text{N/m}^2$$
Solution by Sreeraj P, M.Sc Physics