Q59 / 75 · JEE Main 2025 (24 Jan Shift 1) · Physics
A force $\mathrm{F}=\alpha+\beta \mathrm{x}^2$ 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 $\alpha=1 \mathrm{~N}$ then $\beta$ will be
- A$15 \mathrm{~N} / \mathrm{m}^2$
- Correct answerB$12 \mathrm{~N} / \mathrm{m}^2$
- C$8 \mathrm{~N} / \mathrm{m}^2$
- D$10 \mathrm{~N} / \mathrm{m}^2$
Solution
$F=\alpha+\beta x^2$
Work done $\int d w=\int F \cdot d x$
$\begin{aligned}
& \Rightarrow \quad \Delta W=\int F \cdot d x=\int\left(\alpha+\beta x^2\right) d x \\
& \Rightarrow \quad \Delta W=\left|\alpha x+\frac{\beta x^3}{3}\right|_0^1=\alpha+\frac{\beta}{3}=5
\end{aligned}$
Given $\alpha=1$
So, $\frac{\beta}{3}=4$
$\Rightarrow \beta=12 \mathrm{~N} / \mathrm{m}^2$
Hence, the correct answer is option: 'B'.