Q147 / 160 · TS EAMCET 2024 (11 May Shift 1) · Physics
The relation between the current $i$ (in ampere) in a conductor and the time $t$ (in second) is $i=12 t+9 t^2$. The charge passing through the conductor between the times $t=2 \mathrm{~s}$ and $t=10 \mathrm{~s}$ is
- A3720 C
- B3648 C
- C3600 C
- Correct answerD3552 C
Solution
$\begin{aligned} & \text {Current, } i=12 \mathrm{t}+9 \mathrm{t}^2, \mathrm{t}_1=2 \mathrm{~s}, \mathrm{t}_2=10 \mathrm{~s} \\ \therefore \quad & \text { Charge, } \mathrm{q}=\int_{\mathrm{t}_1}^{\mathrm{t}_2} \mathrm{idt}=\int_2^{10}\left(9 \mathrm{t}^2+12 \mathrm{t}\right) \mathrm{dt} \\ = & {\left[9 \frac{\mathrm{t}^3}{3}+12 \cdot \frac{\mathrm{t}^2}{2}\right]_2^{10}=3552 \mathrm{C} }\end{aligned}$
Hence, the correct answer is option: 'D'.