QJEEM25Ph2ac5: JEE Main 2025 Physics - Class 11 JEEM25Ph2ac5 Solution: A force acts on an object in the x -direction. The work done by …

The work done by a force is given by the integral of the force over the displacement. The force is given by:

$F(x) = \alpha + \beta x^2$

To find the work done when the object is displaced from $x = 0$ to $x = 1$ , we compute:

$W = \int_{0}^{1} (\alpha + \beta x^2) \, dx$

Substituting $\alpha = 1 \, \text{N}$ :

$W = \int_{0}^{1} (1 + \beta x^2) \, dx = \int_{0}^{1} 1 \, dx + \int_{0}^{1} \beta x^2 \, dx$

Calculating each integral separately:

$\int_{0}^{1} 1 \, dx = [x]_{0}^{1} = 1$

$\int_{0}^{1} \beta x^2 \, dx = \beta \left[ \frac{x^3}{3} \right]_{0}^{1} = \beta \left[\frac{1^3}{3} - \frac{0^3}{3}\right] = \frac{\beta}{3}$

Thus, the total work done is:

$W = 1 + \frac{\beta}{3}$

We are given that the work done is 5 J, so:

$1 + \frac{\beta}{3} = 5$

Subtract 1 from both sides:

$\frac{\beta}{3} = 4$

Multiply both sides by 3 to solve for $\beta$ :

$\beta = 12 \, \text{N/m}^2$

Therefore, the value of $\beta$ is $\boxed{12 \, \text{N/m}^2}$ . Thus, Option C $12 \, \text{N/m}^2$ is the correct answer.