A current through a wire depends on time as $\mathrm{i}=\alpha_{0} \mathrm{t}+\beta \mathrm{t}^{2}$ where $\alpha_{0}=20 \mathrm{~A} / \mathrm{s}$ and $\beta=8 \mathrm{As}^{-2}$. Find the charge crossed through a section of the wire in $15 \mathrm{~s}$.
Correct Option: , 4
(4)
given : $i=\alpha_{0} t+\beta t^{2}$
$\alpha=20 \mathrm{~A} / \mathrm{s}$ and $\beta=8 \mathrm{As}^{-2}$
$\mathrm{t}=15 \mathrm{sec}$
we know that, $i=\frac{d q}{d t} \Rightarrow \int_{0}^{t} i d t=\int_{0}^{Q} d q$\
$\Rightarrow \int_{0}^{15}\left(\alpha_{0} t+\beta t^{2}\right) d t=\int_{0}^{Q} d q$
$\Rightarrow Q=\left[\frac{\alpha_{0} t^{2}}{2}+\frac{\beta t^{3}}{3}\right]_{0}^{15}$
$\Rightarrow Q=\frac{20 \times 15 \times 15}{2}+\frac{8 \times 15 \times 15 \times 15}{3}-0$
$\Rightarrow Q=11250 \mathrm{c}$