Charge passing through a conductor of cross-section 0.3 ~m^2 is given by q= (3 t^2+5 t+2 ) C where ' t ' is in seconds. The drift velocity at t=2 ~s is (Concentration of electrons in the conductor =2 × 10^25 ~m^-3 )

Correct answer: (C) 1.77 × 10^-5 ~ms^-1. Given, area of cross-section A=0.3 ~m^2 , n=2 × 10^25 ~m^-3 charge, q= (3 t^2+5 t+2 ) C, t=2 ~s, ~V_d=? ~V_d=i/neA=(dq / dt)/neA [∵ i=dq/dt ] i=dq/dt=d/dt (3 t^2+5 t+2 )=6 t+5=6 × 2+5 =17 ∴ V_d=i/neA=frac172 × 10^25 × 1.6 × 10^-19 × 0.3 =17/(0.96 × 10^6)=1.77 × 10^-5 ~m / s

Q148 / 160 · AP EAMCET 2023 (15 May Shift 1) · Physics

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 $=2 \times 10^{25} \mathrm{~m}^{-3}$ )

A
$0.77 \times 10^{-5} \mathrm{~ms}^{-1}$
B
$0.93 \times 10^{-5} \mathrm{~ms}^{-1}$
Correct answer
C
$1.77 \times 10^{-5} \mathrm{~ms}^{-1}$
D
$2.08 \times 10^{-5} \mathrm{~ms}^{-1}$

Solution

Given, area of cross-section $\mathrm{A}=0.3 \mathrm{~m}^2$, $$ \begin{aligned} & \mathrm{n}=2 \times 10^{25} \mathrm{~m}^{-3} \text { charge, } \mathrm{q}=\left(3 \mathrm{t}^2+5 \mathrm{t}+2\right) \mathrm{C}, \\ & \mathrm{t}=2 \mathrm{~s}, \mathrm{~V}_{\mathrm{d}}=? \\ & \mathrm{~V}_{\mathrm{d}}=\frac{\mathrm{i}}{\mathrm{neA}}=\frac{\mathrm{dq} / \mathrm{dt}}{\mathrm{neA}}\left[\because \mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}\right] \\ & \mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(3 \mathrm{t}^2+5 \mathrm{t}+2\right)=6 \mathrm{t}+5=6 \times 2+5 \\ & =17 \\ & \therefore \mathrm{V}_{\mathrm{d}}=\frac{\mathrm{i}}{\mathrm{neA}}=\frac{17}{2 \times 10^{25} \times 1.6 \times 10^{-19} \times 0.3} \\ & =\frac{17}{0.96 \times 10^6}=1.77 \times 10^{-5} \mathrm{~m} / \mathrm{s} \end{aligned} $$

Hence, the correct answer is option: 'C'.

Solution

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