Q 12-03-194JEE MainJEE Main 2021 (24 Feb, Shift 1)Easy
A current through a wire depends on time as $i = \alpha_0 t + \beta t^2$, where $\alpha_0 = 20\ \text{A s}^{-1}$ and $\beta = 8\ \text{A s}^{-2}$. Find the charge crossed through a section of the wire in 15 s.
Answer: (B) 11250 C
$$q = \int_0^{15}(20t + 8t^2)\,dt = 10t^2 + \frac{8}{3}t^3\Big|_0^{15} = 2250 + 9000 = 11250\ \text{C}$$
Solution by Sreeraj P, M.Sc Physics