Charge passing through a conductor of cross-section $0.3 \mathrm{~m}^2$ is given by $\mathrm{q}=\left(3…

Charge passing through a conductor of cross-section $0.3 \mathrm{~m}^2$ is given by $\mathrm{q}=\left(3 \mathrm{t}^2+5 \mathrm{t}+2\right) \mathrm{C}$ where ' $\mathrm{t}$ ' is in seconds. The drift velocity at $t=2 \mathrm{~s}$ is (Concentration of electrons in the conductor $=2 \times 10^{25} \mathrm{~m}^{-3}$ )
  1. $0.77 \times 10^{-5} \mathrm{~ms}^{-1}$
  2. $0.93 \times 10^{-5} \mathrm{~ms}^{-1}$
  3. $1.77 \times 10^{-5} \mathrm{~ms}^{-1}$
  4. $2.08 \times 10^{-5} \mathrm{~ms}^{-1}$

Solution

Given, area of cross-section $\mathrm{A}=0.3 \mathrm{~m}^2$, $ \begin{aligned} & \mathrm{n}=2 \times 10^{25} \mathrm{~m}^{-3} \text { charge, } \mathrm{q}=\left(3 \mathrm{t}^2+5 \mathrm{t}+2\right) \mathrm{C}, \\ & \mathrm{t}=2 \mathrm{~s}, \mathrm{~V}_{\mathrm{d}}=? \\ & \mathrm{~V}_{\mathrm{d}}=\frac{\mathrm{i}}{\mathrm{neA}}=\frac{\mathrm{dq} / \mathrm{dt}}{\mathrm{neA}}\left[\because \mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}\right] \\ & \mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(3 \mathrm{t}^2+5 \mathrm{t}+2\right)=6 \mathrm{t}+5=6 \times 2+5 \\ & =17 \\ & \therefore \mathrm{V}_{\mathrm{d}}=\frac{\mathrm{i}}{\mathrm{neA}}=\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{aligned} $

Asked in: AP EAMCET 2023 (15 May Shift 1)

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