Current Density, Drift Velocity and Mobility Questions in English

(D) The drift velocity $v_d$ is given by the formula $v_d = \frac{eE\tau}{m}$,where $e$ is the charge of the electron,$E$ is the electric field,$\tau$ is the relaxation time,and $m$ is the mass of the electron.
When the temperature of the metal wire increases,the atoms in the metal lattice vibrate with greater amplitude.
This leads to more frequent collisions between the electrons and the lattice ions,which decreases the relaxation time $\tau$.
Since $v_d \propto \tau$,the drift velocity $v_d$ decreases.
Additionally,the thermal velocity of the electrons is proportional to the square root of the absolute temperature $(v_{th} \propto \sqrt{T})$,so as the temperature increases,the thermal velocity of the electrons increases.

(B) The formula for drift velocity is given by $v_{d} = \frac{I}{n e A}$.
Here, $I = 5 \,A$, $n = 5 \times 10^{26} \,m^{-3}$, $e = 1.6 \times 10^{-19} \,C$, and $A = 4 \times 10^{-6} \,m^{2}$.
Substituting the values:
$v_{d} = \frac{5}{(5 \times 10^{26}) \times (1.6 \times 10^{-19}) \times (4 \times 10^{-6})}$
$v_{d} = \frac{5}{5 \times 1.6 \times 4 \times 10^{26-19-6}}$
$v_{d} = \frac{1}{6.4 \times 10^{1}}$
$v_{d} = \frac{1}{64} = 0.015625 \,ms^{-1} = 1.56 \times 10^{-2} \,ms^{-1}$.

(A) The diameter of the conductor is $d = 0.1 \text{ mm} = 10^{-4} \text{ m}$.
The cross-sectional area $A$ of the cylindrical conductor is given by $A = \frac{\pi d^2}{4}$.
Using $\pi \simeq 3$,we get $A = \frac{3 \times (10^{-4})^2}{4} = \frac{3 \times 10^{-8}}{4} = 0.75 \times 10^{-8} \text{ m}^2$.
The current flowing through the conductor is $I = 90 \text{ mA} = 90 \times 10^{-3} \text{ A}$.
The current density $J$ is defined as $J = \frac{I}{A}$.
Substituting the values,$J = \frac{90 \times 10^{-3}}{0.75 \times 10^{-8}} = \frac{90}{0.75} \times 10^{5} = 120 \times 10^{5} = 1.2 \times 10^{7} \text{ Am}^{-2}$.

(D) The current $I$ is given by $I = nAev_d$,where $n$ is the number density,$A$ is the cross-sectional area,$e$ is the charge of an electron,and $v_d$ is the drift velocity.
The total number of electrons $N$ in a conductor of length $L$ and volume $V = AL$ is $N = nAL$.
From the current formula,$nA = I / (ev_d)$.
Substituting this into the expression for $N$,we get $N = (I / (ev_d)) \times L = IL / (ev_d)$.
The total momentum $P$ of the electrons is $P = N \times m_e \times v_d$,where $m_e$ is the mass of an electron.
Substituting $N$ into the momentum equation: $P = (IL / (ev_d)) \times m_e \times v_d = (I \times L \times m_e) / e$.
Given $I = 80 \ A$,$L = 10 \ m$,$m_e = 9.1 \times 10^{-31} \ kg$,and $e = 1.6 \times 10^{-19} \ C$.
$P = (80 \times 10 \times 9.1 \times 10^{-31}) / (1.6 \times 10^{-19})$.
$P = (728 \times 10^{-30}) / (1.6 \times 10^{-19}) = 455 \times 10^{-11} \ Ns$.

(D) The electrical conductivity $\sigma$ is given by the formula $\sigma = \frac{ne^2\tau}{m}$, where $n$ is the charge carrier density, $e$ is the charge of an electron, $\tau$ is the relaxation time (average time between collisions), and $m$ is the mass of an electron.
Rearranging the formula to solve for $\tau$, we get $\tau = \frac{\sigma m}{ne^2}$.
Given values: $n = 9.1 \times 10^{28} \,m^{-3}$, $\sigma = 6.4 \times 10^7 \,S \,m^{-1}$, $m = 9.1 \times 10^{-31} \,kg$, and $e = 1.6 \times 10^{-19} \,C$.
Substituting these values into the equation:
$\tau = \frac{(6.4 \times 10^7) \times (9.1 \times 10^{-31})}{(9.1 \times 10^{28}) \times (1.6 \times 10^{-19})^2}$
$\tau = \frac{6.4 \times 10^7 \times 9.1 \times 10^{-31}}{9.1 \times 10^{28} \times 2.56 \times 10^{-38}}$
$\tau = \frac{6.4 \times 10^{-24}}{2.56 \times 10^{-10}}$
$\tau = 2.5 \times 10^{-14} \,s$.
Thus, the average time between two successive collisions is $2.5 \times 10^{-14} \,s$.

(D) The mobility $\mu$ of an electron is defined as the ratio of drift velocity $v_d$ to the electric field $E$: $\mu = \frac{v_d}{E}$.
Given:
Length $l = 20 \ cm = 0.2 \ m$.
Potential difference $V = 30 \ V$.
Mobility $\mu = 2 \times 10^{-6} \ m^2 \ V^{-1} \ s^{-1}$.
The electric field $E$ is given by $E = \frac{V}{l} = \frac{30}{0.2} = 150 \ V/m$.
Now,the drift velocity $v_d = \mu E$.
$v_d = (2 \times 10^{-6} \ m^2 \ V^{-1} \ s^{-1}) \times (150 \ V/m) = 300 \times 10^{-6} \ ms^{-1} = 3 \times 10^{-4} \ ms^{-1}$.
Thus,the correct option is $D$.

(D) The relationship between current $I$ and drift velocity $v_d$ is given by the formula: $I = n A e v_d$,where $n$ is the number density of electrons,$A$ is the cross-sectional area,and $e$ is the elementary charge $(1.6 \times 10^{-19} \,C)$.
Given:
$I = 6.4 \,A$
$A = 4 \times 10^{-7} \,m^2$
$n = 8 \times 10^{28} \,m^{-3}$
$e = 1.6 \times 10^{-19} \,C$
Rearranging the formula to solve for $v_d$:
$v_d = \frac{I}{n A e}$
Substituting the values:
$v_d = \frac{6.4}{(8 \times 10^{28}) \times (4 \times 10^{-7}) \times (1.6 \times 10^{-19})}$
$v_d = \frac{6.4}{32 \times 10^{21} \times 1.6 \times 10^{-19}}$
$v_d = \frac{6.4}{51.2 \times 10^2} = \frac{6.4}{5120} = 0.00125 \,m/s$
Expressing this in terms of $10^{-3} \,m/s$:
$v_d = 1.25 \times 10^{-3} \,m/s$.
Thus,the drift velocity is $1.25$.

(A) The drift velocity $(v_d)$ of electrons in a conductor is given by the formula:
$v_d = \frac{eE\tau}{m}$
where:
$e$ is the charge of an electron,
$E$ is the electric field intensity,
$\tau$ is the relaxation time,
$m$ is the mass of an electron.
Since $e$,$\tau$,and $m$ are constants for a given conductor at a constant temperature,the drift velocity is directly proportional to the electric field intensity.
Therefore,$v_d \propto E$.

(C) Given: Area of cross-section $A = 0.3 \, m^2$, electron concentration $n = 2 \times 10^{25} \, m^{-3}$, charge $q = (3t^2 + 5t + 2) \, C$, and time $t = 2 \, s$.
We know that current $i = \frac{dq}{dt}$.
$i = \frac{d}{dt}(3t^2 + 5t + 2) = 6t + 5$.
At $t = 2 \, s$, $i = 6(2) + 5 = 17 \, A$.
The formula for drift velocity is $V_d = \frac{i}{neA}$.
Substituting the values: $V_d = \frac{17}{2 \times 10^{25} \times 1.6 \times 10^{-19} \times 0.3}$.
$V_d = \frac{17}{0.96 \times 10^6} = 17.708 \times 10^{-6} \, m/s = 1.77 \times 10^{-5} \, m/s$.

(B) The drift velocity $v_{d}$ is given by the formula $v_{d} = \frac{I}{neA}$.
Since $I = \frac{V}{R}$ and $R = \frac{\rho \ell}{A}$,we substitute these into the equation:
$v_{d} = \frac{V}{neAR} = \frac{V}{neA (\frac{\rho \ell}{A})} = \frac{V}{ne \rho \ell}$.
Here,$V$ is the potential difference,$n$ is the electron density,$e$ is the charge of an electron,$\rho$ is the resistivity,and $\ell$ is the length of the wire.
From the formula,we see that $v_{d} \propto \frac{1}{\ell}$.
If the length $\ell$ is doubled $(\ell' = 2\ell)$,the new drift velocity $v_{d}'$ becomes:
$v_{d}' = \frac{V}{ne \rho (2\ell)} = \frac{1}{2} \left( \frac{V}{ne \rho \ell} \right) = \frac{V_{d}}{2}$.
Note that the area of cross-section $A$ cancels out in the expression for drift velocity when the potential difference $V$ is kept constant.

(A) Statement $(A)$ is true: In the absence of an external electric field,free electrons move randomly in all directions due to thermal energy. Between two successive collisions,there is no external force acting on the electron,so it moves in a straight line at a constant velocity.
Statement $(B)$ is true: When an electric field $E$ is applied,the electrons experience a force $F = -eE$. This force causes a constant average acceleration $a = -eE/m$. The drift velocity is defined as $v_d = a\tau$,where $\tau$ is the average relaxation time. Since $a$ and $\tau$ are constant for a given conductor and temperature,the drift velocity $v_d$ is independent of time.

(A) The current density $J$ is given as $J(r) = \beta(r + r_0)^2$.
Consider a small elemental area $dA$ at a radial distance $r$ with thickness $dr$ and angular width $d\theta$. The area element is $dA = r dr d\theta$.
The current $di$ through this element is $di = J(r) dA = \beta(r + r_0)^2 r dr d\theta$.
The shaded region consists of two sectors,each with an angular width of $\pi/6$. The total angular width is $\Delta \theta = \pi/6 + \pi/6 = \pi/3$.
The total current $I$ through the shaded region is obtained by integrating $di$ over $r$ from $0$ to $R$ and $\theta$ over the total angular width $\Delta \theta$:
$I = \int_0^{\pi/3} d\theta \int_0^R \beta(r^2 + 2rr_0 + r_0^2) r dr$
$I = \frac{\pi}{3} \beta \int_0^R (r^3 + 2r_0r^2 + r_0^2r) dr$
$I = \frac{\pi \beta}{3} \left[ \frac{r^4}{4} + \frac{2r_0r^3}{3} + \frac{r_0^2r^2}{2} \right]_0^R$
$I = \frac{\pi \beta}{3} \left[ \frac{R^4}{4} + \frac{2r_0R^3}{3} + \frac{r_0^2R^2}{2} \right]$
$I = \pi \beta \left[ \frac{R^4}{12} + \frac{2r_0R^3}{9} + \frac{r_0^2R^2}{6} \right]$

(D) The drift velocity $v_d$ is given by the ratio of length $L$ to the time $t$ taken to drift across the wire:
$v_d = \frac{L}{t} = \frac{2 \ m}{40 \times 10^3 \ s} = 0.5 \times 10^{-4} \ m/s = 5 \times 10^{-5} \ m/s$.
The current $I$ in a conductor is related to the number density $n$ of free electrons by the formula:
$I = n e A v_d$
where $e = 1.6 \times 10^{-19} \ C$ is the charge of an electron and $A$ is the cross-sectional area.
Given $A = 4 \ mm^2 = 4 \times 10^{-6} \ m^2$ and $I = 1.6 \ A$,we rearrange for $n$:
$n = \frac{I}{e A v_d}$
$n = \frac{1.6}{(1.6 \times 10^{-19}) \times (4 \times 10^{-6}) \times (5 \times 10^{-5})}$
$n = \frac{1.6}{1.6 \times 10^{-19} \times 20 \times 10^{-11}}$
$n = \frac{1}{20 \times 10^{-30}} = \frac{1}{2} \times 10^{29} = 5 \times 10^{28} \ m^{-3}$.

(B) Given: Length of wire $l = 3 \,m$,Cross-sectional area $A = 6.14 \times 10^{-6} \,m^2$,Current $I = 6 \,A$,Atomic weight $M = 108 \,g/mol = 0.108 \,kg/mol$,Density $\rho = 10500 \,kg/m^3$,Avogadro number $N_A = 6.023 \times 10^{23} /mol$,Charge of electron $e = 1.6 \times 10^{-19} \,C$.
The number of atoms per unit volume $n$ is given by $n = \frac{\rho N_A}{M}$.
Since each atom contributes one electron,the number of free electrons per unit volume is $n = \frac{10500 \times 6.023 \times 10^{23}}{0.108} \approx 5.86 \times 10^{28} \,m^{-3}$.
The drift velocity $v_d$ is given by the formula $I = n e A v_d$.
Rearranging for $v_d$: $v_d = \frac{I}{n e A}$.
Substituting the values: $v_d = \frac{6}{(5.86 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (6.14 \times 10^{-6})}$.
$v_d = \frac{6}{5.86 \times 1.6 \times 6.14 \times 10^{3}} \approx \frac{6}{57.56 \times 10^3} \approx 1.04 \times 10^{-4} \,m/s$.
Thus,the drift velocity is close to $10^{-4} \,m/s$.

(A) The current $I$ in a conductor is given by $I = nAev_d$,where $n$ is the number density of electrons,$A$ is the cross-sectional area,$e$ is the charge of an electron,and $v_d$ is the drift velocity.
The total number of electrons in a length $l$ of the conductor is $N = nAl$.
The total momentum $P$ of these electrons is $P = Nmv_d = (nAl)mv_d$,where $m$ is the mass of an electron.
We can rewrite the expression for current as $v_d = I / (nAe)$.
Substituting $v_d$ into the momentum equation: $P = nAlm \times (I / (nAe)) = (lmI) / e$.
The specific charge $s$ is defined as $s = e/m$,which implies $m/e = 1/s$.
Therefore,the momentum per unit length is $P/l = (mI) / e = I / s$.

(A) The drift velocity $v_d$ is given by the formula $v_d = \frac{I}{neA}$.
Since the current $I = \frac{V}{R}$ and the resistance $R = \rho \frac{l}{A}$, we can substitute these into the expression:
$v_d = \frac{V/R}{neA} = \frac{V}{(\rho l/A)neA} = \frac{V}{\rho l ne}$.
Here, $V$ is the potential difference of the cell, $\rho$ is the resistivity of the material, $l$ is the length of the rod, $n$ is the free electron density, and $e$ is the charge of an electron.
As seen from the final expression, $v_d$ is independent of the cross-sectional area $A$.
Therefore, when a hole is made along the axis, the cross-sectional area changes, but the drift velocity remains unchanged.

(B) The drift velocity $v_d$ is given by the formula $v_d = \frac{eE\tau}{m}$,where $E = \frac{V}{l}$.
Substituting $E$,we get $v_d = \frac{eV\tau}{ml}$.
Since the potential difference $V$ is constant,$v_d \propto \frac{1}{l}$.
If the length $l$ is increased to $4l$,the new drift velocity $v_{d'}$ becomes $v_{d'} = \frac{eV\tau}{m(4l)} = \frac{v_d}{4}$.
Therefore,the drift velocity decreases by $4$ times.

(B) When a resistor is heated due to the flow of current,its temperature increases.
$1$. Drift speed $(v_d)$ is given by $v_d = \frac{eE\tau}{m}$. As temperature increases,the relaxation time $(\tau)$ decreases,so the drift speed changes.
$2$. Resistivity $(\rho)$ of a conductor is given by $\rho = \frac{m}{ne^2\tau}$. As temperature increases,$\tau$ decreases,so resistivity increases.
$3$. Resistance $(R)$ is given by $R = \rho \frac{l}{A}$. Since resistivity changes with temperature,resistance also changes.
$4$. The number of free electrons $(n)$ per unit volume is a property of the material and remains invariant with temperature changes.
Therefore,only the number of free electrons $(D)$ remains unchanged.

(A) The current $I$ flowing through a conductor is given by the formula: $I = n e v_d A = n e v_d \pi R^2$.
Here,$n$ is the number density of free charge carriers,$e$ is the elementary charge,$v_d$ is the drift velocity,and $R$ is the radius of the wire.
Since both wires are made of the same material,the number density $n$ is the same for both.
Given that the current $I$ is the same for both wires $(I_A = I_B)$,we can write:
$n e v_A \pi R_A^2 = n e v_B \pi R_B^2$
Dividing both sides by $n e \pi$,we get:
$v_A R_A^2 = v_B R_B^2$
Rearranging to find the ratio $\frac{v_A}{v_B}$:
$\frac{v_A}{v_B} = \left(\frac{R_B}{R_A}\right)^2$
Given $R_A = 2R_B$,substitute this into the equation:
$\frac{v_A}{v_B} = \left(\frac{R_B}{2R_B}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} = 0.25$.

(A) The mobility $\mu$ of charge carriers is defined as the ratio of drift velocity $v_d$ to the applied electric field $E$.
Formula: $\mu = \frac{v_d}{E}$
Given:
Drift velocity $v_d = 0.3 \ m s^{-1}$
Electric field $E = 2 \ V m^{-1}$
Calculation:
$\mu = \frac{0.3}{2} = 0.15 \ m^2 V^{-1} s^{-1}$
Therefore,the correct option is $A$.

(C) The drift velocity $v_d$ is given by the formula $v_d = \frac{I}{neA}$,where $I$ is the current,$n$ is the electron density,$e$ is the charge of an electron,and $A$ is the cross-sectional area.
Since the wires are connected in parallel to the same battery,the potential difference $V$ across both wires is the same.
Using Ohm's law,$I = \frac{V}{R}$,where $R = \rho \frac{L}{A}$. Thus,$I = \frac{VA}{\rho L}$.
Substituting this into the drift velocity formula: $v_d = \frac{VA}{neA\rho L} = \frac{V}{ne\rho L}$.
Since both wires are made of the same material,$n$,$e$,and $\rho$ are constant. Also,$V$ is constant.
Therefore,$v_d \propto \frac{1}{L}$.
The ratio of lengths is $L_1 : L_2 = 2 : 3$.
Thus,the ratio of drift velocities is $\frac{v_{d1}}{v_{d2}} = \frac{L_2}{L_1} = \frac{3}{2}$.

(D) Given:
Length of conductor,$\ell = 20 \ cm = 0.2 \ m$
Potential difference,$V = 16 \ V$
Drift speed,$V_d = 2.4 \times 10^{-4} \ ms^{-1}$
Electric field,$E = \frac{V}{\ell} = \frac{16}{0.2} = 80 \ V/m$
Mobility of electron is defined as $\mu = \frac{V_d}{E}$
Substituting the values:
$\mu = \frac{2.4 \times 10^{-4}}{80}$
$\mu = \frac{2.4}{80} \times 10^{-4} = 0.03 \times 10^{-4} = 3 \times 10^{-6} \ m^2 \ V^{-1} \ s^{-1}$

(D) Given: Length $l = 1.5 \ m$,Area $A = 3 \times 10^{-5} \ m^2$,Resistance $R = 15 \ \Omega$,Electric field $E = 21 \ Vm^{-1}$.
We know that current density $J = \sigma E$.
Since conductivity $\sigma = \frac{l}{RA}$,we substitute this into the equation:
$J = \left( \frac{l}{RA} \right) E$
$J = \frac{1.5 \times 21}{15 \times 3 \times 10^{-5}}$
$J = \frac{31.5}{45 \times 10^{-5}}$
$J = 0.7 \times 10^5 \ Am^{-2}$.

(A) The expression for electrical resistivity $\rho$ is given by $\rho = \frac{m}{ne^2 \tau}$.
Rearranging the formula to solve for the relaxation time $\tau$:
$\tau = \frac{m}{ne^2 \rho}$
Given values:
$m = 9 \times 10^{-31} \ kg$
$n = 9 \times 10^{28} \ m^{-3}$
$e = 1.6 \times 10^{-19} \ C$
$\rho = 1 \times 10^{-8} \ \Omega \cdot m$
Substituting these values into the equation:
$\tau = \frac{9 \times 10^{-31}}{(9 \times 10^{28}) \times (1.6 \times 10^{-19})^2 \times (1 \times 10^{-8})}$
$\tau = \frac{9 \times 10^{-31}}{9 \times 10^{28} \times 2.56 \times 10^{-38} \times 10^{-8}}$
$\tau = \frac{9 \times 10^{-31}}{23.04 \times 10^{-18}}$
$\tau \approx 0.39 \times 10^{-13} \ s = 3.9 \times 10^{-14} \ s$
Rounding to the nearest value, we get $\tau \approx 4 \times 10^{-14} \ s$.

(B) The power dissipated in a resistor is given by $P = I^2 R$, where $I$ is the current and $R$ is the resistance.
Resistance $R$ is given by $R = \rho \frac{l}{A}$, where $\rho = 10^{-6} \, \Omega \cdot m$, $l = 0.04 \, m$, and $A = \pi r^2 = \pi (7 \times 10^{-3})^2 \, m^2$.
Substituting $R$ in the power equation: $P = I^2 \left( \frac{\rho l}{A} \right) \Rightarrow I = \sqrt{\frac{P A}{\rho l}}$.
Current density $J$ is defined as $J = \frac{I}{A}$.
Substituting $I$: $J = \frac{1}{A} \sqrt{\frac{P A}{\rho l}} = \sqrt{\frac{P}{\rho l A}}$.
Substituting the values: $J = \sqrt{\frac{1.54}{10^{-6} \times 0.04 \times \pi \times (7 \times 10^{-3})^2}}$.
$J = \sqrt{\frac{1.54}{10^{-6} \times 0.04 \times \pi \times 49 \times 10^{-6}}} = \sqrt{\frac{1.54}{1.96 \times 10^{-9} \times \pi}} \approx 5 \times 10^5 \, A/m^2$.

(B) The mobility $\mu$ of an electron is defined as the ratio of drift velocity $V_d$ to the electric field $E$,given by the formula: $\mu = \frac{V_d}{E} = \frac{e \tau}{m}$.
Given values are:
Charge of electron $e = 1.6 \times 10^{-19} \,C$
Mass of electron $m = 9.1 \times 10^{-31} \,kg$
Average collision time (relaxation time) $\tau = 9.1 \times 10^{-15} \,s$.
Substituting these values into the formula:
$\mu = \frac{(1.6 \times 10^{-19} \,C) \times (9.1 \times 10^{-15} \,s)}{9.1 \times 10^{-31} \,kg}$
$\mu = \frac{1.6 \times 10^{-19} \times 9.1 \times 10^{-15}}{9.1 \times 10^{-31}}$
$\mu = 1.6 \times 10^{-19 - 15 + 31} \,m^2/V \cdot s$
$\mu = 1.6 \times 10^{-3} \,m^2/V \cdot s$.

(A) The current density $J$ is related to the electric field $E$ by the relation $J = \frac{E}{\rho}$, where $\rho$ is the resistivity of the material.
We know that current density $J = \frac{I}{A}$.
Therefore, $E = J \cdot \rho = \frac{I \cdot \rho}{A}$.
Given values are $I = 5 \, A$, $A = 1 \, mm^2 = 1 \times 10^{-6} \, m^2$, and $\rho = 3.0 \times 10^{-8} \, \Omega \cdot m$.
Substituting these values into the formula:
$E = \frac{5 \times 3.0 \times 10^{-8}}{1 \times 10^{-6}} = 15 \times 10^{-2} = 0.15 \, V/m$.

(C) Given, cross-sectional area, $A = 1 \,cm^2 = 10^{-4} \,m^2$.
Charge density, $n = 3 \times 10^{23} \,m^{-3}$.
Current through wire, $I = 24 \,mA = 24 \times 10^{-3} \,A$.
We know that for conductors, the relation between current and drift velocity is $I = n A e v_d$, where $e = 1.6 \times 10^{-19} \,C$.
Rearranging for drift velocity, $v_d = \frac{I}{n A e}$.
Substituting the values: $v_d = \frac{24 \times 10^{-3}}{3 \times 10^{23} \times 10^{-4} \times 1.6 \times 10^{-19}}$.
$v_d = \frac{24 \times 10^{-3}}{4.8 \times 10^{0}} = \frac{24}{4.8} \times 10^{-3} = 5 \times 10^{-3} \,m/s$.

(B) Given: Area $A = 0.3 \,m^2$,charge density $n = 2 \times 10^{25} / m^3$,and charge $q = 3t^2 + 5t + 2$.
First,calculate the current $i$ at $t = 2 \,s$ using $i = \frac{dq}{dt}$.
$i = \frac{d}{dt}(3t^2 + 5t + 2) = 6t + 5$.
At $t = 2 \,s$,$i = 6(2) + 5 = 17 \,A$.
The relationship between current and drift velocity $v_d$ is $i = n e A v_d$,where $e = 1.6 \times 10^{-19} \,C$.
Rearranging for $v_d$: $v_d = \frac{i}{n e A}$.
Substituting the values: $v_d = \frac{17}{2 \times 10^{25} \times 1.6 \times 10^{-19} \times 0.3}$.
$v_d = \frac{17}{0.96 \times 10^6} = 17.708 \times 10^{-6} \,m/s = 1.77 \times 10^{-5} \,m/s$.

(C) Specific resistance (resistivity) depends on both the temperature and the nature of the material. Therefore,Statement $(I)$ is incorrect.
When a wire is stretched to $n$ times its original length,its new resistance $R'$ becomes $n^2 R$. Here,$n = 4$ and $R = 6 \ \Omega$,so $R' = 4^2 \times 6 = 16 \times 6 = 96 \ \Omega$. Therefore,Statement $(II)$ is incorrect.
Drift velocity is defined as the average velocity with which free electrons get drifted in the direction opposite to the applied electric field. Therefore,Statement $(III)$ is correct.
Thus,only Statement $(III)$ is true.

(D) The resistivity $\rho$ is given by $\rho = \frac{m}{ne^2\tau}$,where $\tau$ is the relaxation time.
Also,the mean free path $\lambda$ is given by $\lambda = v_d \tau$,where $v_d$ is the drift speed.
From the resistivity formula,$\tau = \frac{m}{ne^2\rho}$.
Substituting this into the mean free path formula:
$\lambda = v_d \left( \frac{m}{ne^2\rho} \right) = \frac{m v_d}{ne^2\rho}$.
Given: $m = 9 \times 10^{-31} \ kg$,$v_d = 1.6 \times 10^6 \ m/s$,$n = 9 \times 10^{28} \ m^{-3}$,$e = 1.6 \times 10^{-19} \ C$,and $\rho = 1 \times 10^{-8} \ \Omega \cdot m$.
$\lambda = \frac{(9 \times 10^{-31}) \times (1.6 \times 10^6)}{(9 \times 10^{28}) \times (1.6 \times 10^{-19})^2 \times (1 \times 10^{-8})}$.
$\lambda = \frac{14.4 \times 10^{-25}}{9 \times 10^{28} \times 2.56 \times 10^{-38} \times 10^{-8}} = \frac{14.4 \times 10^{-25}}{23.04 \times 10^{-18}} = 0.625 \times 10^{-7} \ m = 62.5 \ nm$.

(B) The motion of an electron in a wire is governed by the electric field (due to potential difference) and the magnetic field (if applicable). These forces are categorized under the electromagnetic force. Gravitational force is negligible at the atomic scale,while strong and weak nuclear forces act only within the nucleus. Therefore,the dominant force is the electromagnetic force.

(B) Mobility $\mu$ is defined as the ratio of drift velocity $v_d$ to the applied electric field $E$, given by the formula $\mu = \frac{e \tau}{m}$.
Here, $e$ is the charge (unit: $A \cdot s$), $\tau$ is the relaxation time (unit: $s$), and $m$ is the mass (unit: $kg$).
Substituting these units into the formula: $\text{Unit of } \mu = \frac{(A \cdot s) \cdot s}{kg} = \frac{A \cdot s^2}{kg}$.
This can be written as $kg^{-1} \,s^2 \,A$.

(C) The drift velocity $v_{d}$ is given by the formula $v_{d} = \frac{I}{neA}$,where $I$ is the current,$n$ is the electron density,$e$ is the charge of an electron,and $A$ is the cross-sectional area.
Since $I = \frac{E}{R}$ and $R = \rho \frac{l}{A}$,we have $I = \frac{E A}{\rho l}$.
Substituting this into the drift velocity formula:
$v_{d} = \frac{E A}{\rho l n e A} = \frac{E}{\rho l n e}$.
From this expression,we see that $v_{d} \propto \frac{1}{l}$.
If the length is changed to $l' = 2l$,the new drift velocity $v_{d}'$ is:
$v_{d}' = \frac{E}{\rho (2l) n e} = \frac{1}{2} \left( \frac{E}{\rho l n e} \right) = \frac{v_{d}}{2}$.

(A) Given: Length $l = 2 \ m$,Area $A = 0.2 \ mm^{2} = 0.2 \times 10^{-6} \ m^{2}$,Current $I = 1.6 \ A$,Voltage $V = 2 \ V$,Concentration $n = 5 \times 10^{28} \ m^{-3}$,Charge $e = 1.6 \times 10^{-19} \ C$.
The drift velocity $V_{d}$ is given by $V_{d} = \frac{I}{neA}$.
The mobility $\mu$ is defined as $\mu = \frac{V_{d}}{E}$,where $E = \frac{V}{l}$ is the electric field.
Substituting $V_{d}$ and $E$,we get $\mu = \frac{I}{neA} \times \frac{l}{V}$.
Substituting the values: $\mu = \frac{1.6 \times 2}{5 \times 10^{28} \times 1.6 \times 10^{-19} \times 0.2 \times 10^{-6} \times 2}$.
$\mu = \frac{3.2}{5 \times 10^{28} \times 1.6 \times 10^{-19} \times 0.2 \times 10^{-6} \times 2} = \frac{3.2}{3.2 \times 10^{3}} = 1 \times 10^{-3} \ m^{2}/V \cdot s$.
Comparing with $\alpha \times 10^{-3} \ m^{2}/V \cdot s$,we get $\alpha = 1$.

(D) The relation between current $I$ and drift velocity $v_d$ is given by the formula $I = n e A v_d$,where $n$ is the electron density and $e$ is the charge of an electron.
From this,the drift velocity is expressed as $v_d = \frac{I}{neA}$.
Given that the new current $I' = 2I$ and the new area of cross-section $A' = 2A$,the new drift velocity $v'_d$ is calculated as:
$v'_d = \frac{I'}{neA'} = \frac{2I}{ne(2A)} = \frac{I}{neA} = v_d$.
Therefore,the drift velocity remains unchanged. Option $(D)$ is correct.