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Charge passing through a conductor of crosssection area $\mathrm{A}=0.3 \mathrm{~m}^{2}$ is given by $\mathrm{q}=3 \mathrm{t}^{2}+5 \mathrm{t}+2$ in coulomb, where $\mathrm{t}$ is in second. What is the value of drift velocity at $t=2 s ?$ $\left(\right.$ Given, $\left.\mathrm{m}=2 \times 10^{25} / \mathrm{m}^{3}\right)$
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Verified Answer
The correct answer is:
$1.77 \times 10^{-5} \mathrm{~m} / \mathrm{s}$
$$
\begin{array}{l}
\text { Given: } \mathrm{A}=0.3 \mathrm{~m}^{2} \mathrm{n}=2 \times 10^{25} / \mathrm{m}^{3} \\
\mathrm{q}=3 \mathrm{t}^{2}+5 \mathrm{t}+2 \\
\mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}=6 \mathrm{t}+5=17
\end{array}
$$
Drift velocity, $v_{\mathrm{d}}=\frac{\mathrm{i}}{\mathrm{neA}}$
$$
\begin{array}{l}
=\frac{17}{2 \times 10^{25} \times 1.6 \times 10^{-19} \times 0.3} \\
=\frac{17}{0.96 \times 10^{6}}=1.77 \times 10^{-5} \mathrm{~m} / \mathrm{s}
\end{array}
$$
\begin{array}{l}
\text { Given: } \mathrm{A}=0.3 \mathrm{~m}^{2} \mathrm{n}=2 \times 10^{25} / \mathrm{m}^{3} \\
\mathrm{q}=3 \mathrm{t}^{2}+5 \mathrm{t}+2 \\
\mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}=6 \mathrm{t}+5=17
\end{array}
$$
Drift velocity, $v_{\mathrm{d}}=\frac{\mathrm{i}}{\mathrm{neA}}$
$$
\begin{array}{l}
=\frac{17}{2 \times 10^{25} \times 1.6 \times 10^{-19} \times 0.3} \\
=\frac{17}{0.96 \times 10^{6}}=1.77 \times 10^{-5} \mathrm{~m} / \mathrm{s}
\end{array}
$$
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