Resistance of wire, Resistivity and Conductivity Questions in English

(D) From the given $V-I$ graph,the slope represents the resistance $R$ of the wire.
$R = \tan(45^{\circ}) = 1 \, \Omega$.
We know that resistance $R = \rho \frac{\ell}{A}$,where $\ell$ is the length and $A$ is the cross-sectional area.
The area $A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}$.
Given $\ell = 31.4 \, cm$ and $d = 2.4 \, cm$.
Substituting the values: $1 = \rho \frac{31.4}{\frac{\pi \times (2.4)^2}{4}}$.
Using $\pi = 3.14$: $1 = \rho \frac{31.4}{\frac{3.14 \times 5.76}{4}}$.
$1 = \rho \frac{31.4}{0.785 \times 5.76} = \rho \frac{31.4}{4.5216}$.
Alternatively,$1 = \rho \frac{31.4 \times 4}{3.14 \times 5.76} = \rho \frac{10 \times 4}{5.76} = \rho \frac{40}{5.76}$.
$\rho = \frac{5.76}{40} = 0.144 \, \Omega \cdot cm$.
Expressing this as $x \times 10^{-3} \, \Omega \cdot cm$,we get $\rho = 144 \times 10^{-3} \, \Omega \cdot cm$.
Therefore,$x = 144$.

(A) Let the original length be $L_{1}$ and the original area be $A_{1}$. The original resistance is $R_{1} = \rho \frac{L_{1}}{A_{1}}$.
When the length is increased by twice its original length,the new length $L_{2} = L_{1} + 2L_{1} = 3L_{1}$.
Since the volume of the wire remains constant,$A_{1}L_{1} = A_{2}L_{2}$.
Thus,$A_{2} = A_{1} \frac{L_{1}}{L_{2}} = A_{1} \frac{L_{1}}{3L_{1}} = \frac{A_{1}}{3}$.
The new resistance is $R_{2} = \rho \frac{L_{2}}{A_{2}} = \rho \frac{3L_{1}}{A_{1}/3} = 9 \rho \frac{L_{1}}{A_{1}}$.
Therefore,the ratio $\frac{R_{2}}{R_{1}} = \frac{9 \rho (L_{1}/A_{1})}{\rho (L_{1}/A_{1})} = 9:1$.

$(A)$ Standard resistance coils require a resistance value that remains stable despite changes in ambient temperature.
The temperature coefficient of resistance $(\alpha)$ determines how much the resistance changes with temperature according to the formula $R_t = R_0(1 + \alpha \Delta T)$.
Alloys like constantan and manganin are specifically chosen because they possess a very low temperature coefficient of resistance.
This property ensures that the resistance of the coils remains nearly constant even if the temperature fluctuates.
Therefore, Assertion $A$ is true, Reason $R$ is true, and $R$ is the correct explanation of $A$.

(B) Let the total length of the wire be $L = 1\,m$. Let the length of part $X$ be $\ell_X$ and the length of part $Y$ be $\ell_Y$. Thus,$\ell_X + \ell_Y = 1$.
The resistance of a wire is given by $R = \rho \frac{\ell}{A}$.
For part $X$,$R_X = \rho \frac{\ell_X}{A_X}$. For part $Y$,$R_Y = \rho \frac{\ell_Y}{A_Y}$.
When wire $X$ is stretched to length $\ell_W = 2\ell_X$,its volume remains constant $(A_X \ell_X = A_W \ell_W)$.
Since $\ell_W = 2\ell_X$,we have $A_W = \frac{A_X}{2}$.
The resistance of wire $W$ is $R_W = \rho \frac{\ell_W}{A_W} = \rho \frac{2\ell_X}{A_X/2} = 4 \left( \rho \frac{\ell_X}{A_X} \right) = 4R_X$.
Given that $R_W = 2R_Y$,we substitute $R_W = 4R_X$ to get $4R_X = 2R_Y$,which implies $R_Y = 2R_X$.
Since both parts $X$ and $Y$ were cut from the same original wire,they have the same cross-sectional area $A$ and resistivity $\rho$. Thus,$R \propto \ell$.
Therefore,$\frac{R_X}{R_Y} = \frac{\ell_X}{\ell_Y}$.
Substituting $R_Y = 2R_X$,we get $\frac{R_X}{2R_X} = \frac{\ell_X}{\ell_Y} = \frac{1}{2}$.

(B) Let the length of each wire be $\ell$ and the cross-sectional area be $A$.
For the equivalent wire,the total length is $2\ell$ and the cross-sectional area is $A$.
The resistance of a wire is given by $R = \frac{\ell}{\sigma A}$,where $\sigma$ is the conductivity.
Since the wires are connected in series,the equivalent resistance $R_{eq}$ is the sum of individual resistances:
$R_{eq} = R_{1} + R_{2}$
Substituting the formula for resistance:
$\frac{2\ell}{\sigma_{eq} A} = \frac{\ell}{\sigma_{1} A} + \frac{\ell}{\sigma_{2} A}$
Dividing both sides by $\frac{\ell}{A}$:
$\frac{2}{\sigma_{eq}} = \frac{1}{\sigma_{1}} + \frac{1}{\sigma_{2}}$
$\frac{2}{\sigma_{eq}} = \frac{\sigma_{1} + \sigma_{2}}{\sigma_{1} \sigma_{2}}$
Therefore,the effective conductivity is $\sigma_{eq} = \frac{2 \sigma_{1} \sigma_{2}}{\sigma_{1} + \sigma_{2}}$.

(A) The resistance of the rod is given by $R = \frac{\rho l}{A}$. Since the volume $V_{vol} = Al$ remains constant,we can write $R = \frac{\rho l^2}{V_{vol}}$.
When the rod is stretched by a small amount $\Delta l$,the change in length is $l' = l + \Delta l$. The new resistance is $R' = \frac{\rho (l + \Delta l)^2}{V_{vol}} \approx \frac{\rho (l^2 + 2l \Delta l)}{V_{vol}} = R + \frac{2 \rho l \Delta l}{V_{vol}}$.
The change in resistance is $\Delta R = R' - R = \frac{2 \rho l \Delta l}{V_{vol}} = \frac{2 \rho l^2}{V_{vol}} \cdot \frac{\Delta l}{l} = 2R \varepsilon$,where $\varepsilon = \frac{\Delta l}{l}$ is the strain.
The voltage across the rod is $V = iR$. The change in voltage is $\Delta V = i \Delta R = i(2R \varepsilon) = (2iR) \varepsilon$.
Since $i$ and $R$ are constants,$\Delta V$ is directly proportional to the strain $\varepsilon$ $(\Delta V \propto \varepsilon)$.
Therefore,the total voltage $V_{total} = V_{initial} + \Delta V = V_{initial} + (2iR) \varepsilon$. This represents a linear relationship with a positive slope,starting from the initial voltage $V_{initial}$.

(C) The material of the rod remains the same,so the resistivity $\rho$ is constant for both rods.
Since the volume of the material remains constant during melting and reshaping,we have:
$V_1 = V_2$
$A_1 L_1 = A_2 L_2$
Given $L_1 = L$ and $L_2 = 2L$,we get:
$A_1 L = A_2 (2L) \Rightarrow A_2 = \frac{A_1}{2}$
Now,using the resistance formula $R = \rho \frac{l}{A}$,the new resistance $R_2$ is:
$R_2 = \rho \frac{L_2}{A_2} = \rho \frac{2L}{A_1 / 2} = 4 \left( \frac{\rho L}{A_1} \right)$
Since $R = \rho \frac{L}{A_1}$,we find:
$R_2 = 4R$

(A) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity of the material.
The resistivity $\rho$ in terms of microscopic parameters is given by $\rho = \frac{m}{n e^2 \tau}$,where $m$ is the mass of the electron,$n$ is the electron density,$e$ is the charge of an electron,and $\tau$ is the relaxation time.
Substituting the expression for $\rho$ into the resistance formula:
$R = \left( \frac{m}{n e^2 \tau} \right) \frac{l}{A} = \frac{m l}{n e^2 \tau A}$.
Therefore,the correct option is $A$.

(C) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.
Given dimensions are $2 \,mm \times 2 \,mm \times 5 \,m$. The current flows between the two square faces,so the length $l = 5 \,m$.
The cross-sectional area $A = 2 \,mm \times 2 \,mm = 4 \,mm^2 = 4 \times 10^{-6} \,m^2$.
The resistance $R = 0.02 \,\Omega$.
Substituting these values into the formula:
$0.02 = \rho \frac{5}{4 \times 10^{-6}}$
$\rho = \frac{0.02 \times 4 \times 10^{-6}}{5}$
$\rho = \frac{0.08 \times 10^{-6}}{5}$
$\rho = 0.016 \times 10^{-6} \,\Omega \cdot m = 1.6 \times 10^{-8} \,\Omega \cdot m$.

(A) The resistance of a wire is given by $R = \rho \frac{l}{A}$.
When a wire is stretched,its volume $V = A \times l$ remains constant.
Let the original length be $l$ and area be $A$. The new length is $l' = 2l$.
Since volume is constant,$A \times l = A' \times l' \implies A \times l = A' \times (2l) \implies A' = \frac{A}{2}$.
The new resistance $R'$ is given by $R' = \rho \frac{l'}{A'} = \rho \frac{2l}{A/2} = 4 \left( \rho \frac{l}{A} \right) = 4R$.
Given $R' = 20 \ \Omega$ and $R = x$,we have $20 = 4x$.
Therefore,$x = \frac{20}{4} = 5 \ \Omega$.

(A) The resistance of a conductor is given by $R = \frac{\rho L}{A}$.
Given,linear expansion coefficient is $\alpha$. Thus,length $L = L_0(1 + \alpha \Delta T)$ and area $A = A_0(1 + 2\alpha \Delta T)$.
Resistivity varies as $\rho = \rho_0(1 + \alpha_\rho \Delta T)$.
Substituting these into the resistance formula:
$R = \frac{\rho_0(1 + \alpha_\rho \Delta T) L_0(1 + \alpha \Delta T)}{A_0(1 + 2\alpha \Delta T)}$
$R = R_0(1 + \alpha_\rho \Delta T)(1 + \alpha \Delta T)(1 + 2\alpha \Delta T)^{-1}$
Using binomial approximation $(1+x)^n \approx 1+nx$ for small $\Delta T$:
$R \approx R_0(1 + \alpha_\rho \Delta T)(1 + \alpha \Delta T)(1 - 2\alpha \Delta T)$
Neglecting higher-order terms of $\Delta T$:
$R \approx R_0(1 + \alpha_\rho \Delta T + \alpha \Delta T - 2\alpha \Delta T)$
$R \approx R_0(1 + (\alpha_\rho - \alpha) \Delta T)$
Comparing this with $R = R_0(1 + \alpha_R \Delta T)$,we get $\alpha_R = \alpha_\rho - \alpha$.

(B) Let the length of both wires be $l$. Let the radius of wire $BC$ be $r$,so the radius of wire $AB$ is $2r$.
The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$.
For wire $BC$,$R_{BC} = \rho \frac{l}{\pi r^2} = R$ (let).
For wire $AB$,$R_{AB} = \rho \frac{l}{\pi (2r)^2} = \rho \frac{l}{4\pi r^2} = \frac{R}{4}$.
Since the wires are in series,the same current $I$ flows through both.
The potential gradient $x$ is given by $x = \frac{V}{l} = \frac{IR}{l}$.
Since $I$ and $l$ are the same for both wires,the potential gradient is proportional to the resistance: $x \propto R$.
Therefore,the ratio of potential gradients is $\frac{x_{AB}}{x_{BC}} = \frac{R_{AB}}{R_{BC}} = \frac{R/4}{R} = \frac{1}{4}$.
Thus,the ratio is $1: 4$.

(A) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{\ell}{A}$, where $\rho$ is the resistivity, $\ell$ is the length, and $A$ is the cross-sectional area.
For a hollow cylinder, the cross-sectional area $A = \pi(R_{outer}^2 - R_{inner}^2)$.
Given: $\ell = 3.14 \, m$, $\rho = 2.4 \times 10^{-8} \, \Omega \, m$, outer diameter $D_{out} = 8 \, mm \implies R_{outer} = 4 \times 10^{-3} \, m$, and inner diameter $D_{in} = 4 \, mm \implies R_{inner} = 2 \times 10^{-3} \, m$.
Calculating the area $A = \pi((4 \times 10^{-3})^2 - (2 \times 10^{-3})^2) = \pi(16 - 4) \times 10^{-6} = 12\pi \times 10^{-6} \, m^2$.
Substituting the values into the resistance formula:
$R = \frac{2.4 \times 10^{-8} \times 3.14}{12 \times 3.14 \times 10^{-6}}$
$R = \frac{2.4}{12} \times 10^{-2} = 0.2 \times 10^{-2} = 2 \times 10^{-3} \, \Omega$.
Comparing this with $n \times 10^{-3} \, \Omega$, we get $n = 2$.

(C) The resistance $R$ of a wire is given by $R = \rho \frac{L}{A}$.
Since the volume $V = A \times L$ remains constant when the wire is stretched,we have $A = \frac{V}{L}$.
Substituting this into the resistance formula,we get $R = \rho \frac{L^2}{V}$.
Since $\rho$ and $V$ are constants,$R \propto L^2$.
If the length increases by $20 \%$,the new length $L' = L + 0.20L = 1.2L$.
The new resistance $R' \propto (1.2L)^2 = 1.44L^2$.
Therefore,$R' = 1.44R$.
The percentage increase in resistance is $\frac{R' - R}{R} \times 100 = \frac{1.44R - R}{R} \times 100 = 0.44 \times 100 = 44 \%$.

(C) The initial resistance is given by $R_{\text{initial}} = \frac{\rho \ell}{A} = 5 \Omega$.
Since the volume of the wire remains constant during stretching:
$V_i = V_f$
$A_i \ell_i = A_f \ell_f$
Given $\ell_f = 5 \ell_i$,we have $A_i \ell_i = A_f (5 \ell_i)$,which implies $A_f = \frac{A_i}{5}$.
The new resistance $R_f$ is:
$R_f = \frac{\rho \ell_f}{A_f} = \frac{\rho (5 \ell_i)}{\left(\frac{A_i}{5}\right)}$
$R_f = 25 \left(\frac{\rho \ell_i}{A_i}\right)$
$R_f = 25 \times 5 = 125 \Omega$.

(D) The reciprocal of resistance is defined as conductance.
Mathematically, $\text{Conductance} = \frac{1}{\text{Resistance}}$.
The $SI$ unit of conductance is Siemens $(S)$ or $\Omega^{-1}$.

(D) The initial resistance of the wire is given by $R = \rho \frac{\ell}{A}$,where $\rho$ is the resistivity,$\ell$ is the length,and $A$ is the area of cross-section.
When the length is increased by $20 \%$,the new length $\ell' = \ell + 0.20\ell = 1.2\ell$.
When the area of cross-section is reduced by $4 \%$,the new area $A' = A - 0.04A = 0.96A$.
The new resistance $R'$ is given by $R' = \rho \frac{\ell'}{A'} = \rho \frac{1.2\ell}{0.96A}$.
Simplifying the expression: $R' = \left( \frac{1.2}{0.96} \right) \rho \frac{\ell}{A} = 1.25 R$.
The percentage change in resistance is calculated as $\frac{R' - R}{R} \times 100$.
Percentage change $= \frac{1.25R - R}{R} \times 100 = 0.25 \times 100 = 25 \%$.

(C) The dimensions of the rectangular parallelopiped are $1\,cm \times 1\,cm \times 100\,cm$. The current flows between the two opposite rectangular faces of size $1\,cm \times 100\,cm$.
Thus,the length of the path of the current is $\ell = 1\,cm = 10^{-2}\,m$.
The area of the cross-section through which the current flows is $A = 1\,cm \times 100\,cm = 10^{-2}\,m \times 1\,m = 10^{-2}\,m^2$.
The specific resistance (resistivity) is $\rho = 3 \times 10^{-7}\,\Omega\,m$.
The resistance $R$ is given by the formula $R = \rho \frac{\ell}{A}$.
Substituting the values: $R = (3 \times 10^{-7}) \times \frac{10^{-2}}{10^{-2}} = 3 \times 10^{-7}\,\Omega$.
Therefore,the resistance is $3 \times 10^{-7}\,\Omega$.

(A) The volume of the wire remains constant during the process of melting and redrawing.
$V_1 = V_2 \implies A_1 L_1 = A_2 L_2$
Given $L_2 = \frac{L_1}{4}$,we have $A_1 L_1 = A_2 (\frac{L_1}{4})$,which simplifies to $A_2 = 4 A_1$.
The resistance of a wire is given by $R = \rho \frac{L}{A}$.
Thus,the ratio of the new resistance $R_2$ to the original resistance $R_1$ is:
$\frac{R_2}{R_1} = \frac{\rho L_2 / A_2}{\rho L_1 / A_1} = \frac{L_2}{L_1} \times \frac{A_1}{A_2}$
Substituting the values: $\frac{R_2}{R_1} = (\frac{1}{4}) \times (\frac{A_1}{4 A_1}) = \frac{1}{16}$.
Therefore,$R_2 = \frac{R_1}{16} = \frac{160}{16} = 10\,\Omega$.

(D) Since the conductor is connected across the same source,the voltage $V$ remains constant. Thus,$V = i_0 R_0 = i_{100} R_{100} = i_{50} R_{50}$.
Given $i_0 = 2\,A$ and $i_{100} = 1.2\,A$.
Using the resistance-temperature relation $R_t = R_0(1 + \alpha t)$:
$2 R_0 = 1.2 R_0(1 + 100 \alpha)$
$1 + 100 \alpha = \frac{2}{1.2} = \frac{20}{12} = \frac{5}{3}$
$100 \alpha = \frac{5}{3} - 1 = \frac{2}{3} \Rightarrow 50 \alpha = \frac{1}{3}$.
Now,for $t = 50^{\circ}C$:
$i_{50} = \frac{V}{R_{50}} = \frac{i_0 R_0}{R_0(1 + 50 \alpha)} = \frac{2}{1 + \frac{1}{3}} = \frac{2}{4/3} = \frac{6}{4} = 1.5\,A$.
Converting to $mA$: $1.5\,A = 1500\,mA = 15 \times 10^2\,mA$.

(D) The resistance of a conductor at temperature $T$ is given by the formula: $R_T = R_0[1 + \alpha(T - T_0)]$,where $R_T$ is the resistance at temperature $T$,$R_0$ is the resistance at reference temperature $T_0$,and $\alpha$ is the temperature coefficient of resistance.
Given: $R_0 = 2\,\Omega$ at $T_0 = 0^{\circ}C$,and $R_T = 6.8\,\Omega$ at $T = 80^{\circ}C$.
Substituting the values into the formula:
$6.8 = 2[1 + \alpha(80 - 0)]$
$6.8 = 2 + 160\alpha$
$4.8 = 160\alpha$
$\alpha = \frac{4.8}{160} = \frac{48}{1600} = \frac{3}{100} = 0.03\,^{\circ}C^{-1}$.
Thus,$\alpha = 3 \times 10^{-2}\,^{\circ}C^{-1}$.

(C) Given: Resistance at room temperature $R_0 = 60 \ \Omega$ at $T_0 = 27^{\circ} C$.
Voltage $V = 220 \ V$ and current $I = 2.75 \ A$.
The resistance at the final temperature $T$ is given by $R_T = \frac{V}{I} = \frac{220}{2.75} = 80 \ \Omega$.
The temperature dependence of resistance is given by $R_T = R_0(1 + \alpha \Delta T)$,where $\Delta T = T - T_0$.
Substituting the values: $80 = 60[1 + 2 \times 10^{-4}(T - 27)]$.
$80/60 = 1 + 2 \times 10^{-4}(T - 27)$.
$1.333 - 1 = 2 \times 10^{-4}(T - 27)$.
$0.333 = 2 \times 10^{-4}(T - 27)$.
$T - 27 = \frac{0.333}{2 \times 10^{-4}} = 1665$.
$T = 1665 + 27 = 1692^{\circ} C$.
Rounding to the nearest option,the temperature is approximately $1694^{\circ} C$.

(B) Let the resistance of each conductor at $0^{\circ} C$ be $R$.
For series combination:
$R_{eq} = R_1 + R_2$
$2R(1 + \alpha_{eq} \Delta \theta) = R(1 + \alpha_1 \Delta \theta) + R(1 + \alpha_2 \Delta \theta)$
$2 + 2\alpha_{eq} \Delta \theta = 2 + (\alpha_1 + \alpha_2) \Delta \theta$
$\alpha_{eq} = \frac{\alpha_1 + \alpha_2}{2}$
For parallel combination:
$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$
$\frac{1}{\frac{R}{2}(1 + \alpha_{eq} \Delta \theta)} = \frac{1}{R(1 + \alpha_1 \Delta \theta)} + \frac{1}{R(1 + \alpha_2 \Delta \theta)}$
$\frac{2}{1 + \alpha_{eq} \Delta \theta} = \frac{1}{1 + \alpha_1 \Delta \theta} + \frac{1}{1 + \alpha_2 \Delta \theta}$
Using the binomial approximation $(1+x)^{-1} \approx 1-x$ for small $\Delta \theta$:
$2(1 - \alpha_{eq} \Delta \theta) = (1 - \alpha_1 \Delta \theta) + (1 - \alpha_2 \Delta \theta)$
$2 - 2\alpha_{eq} \Delta \theta = 2 - (\alpha_1 + \alpha_2) \Delta \theta$
$\alpha_{eq} = \frac{\alpha_1 + \alpha_2}{2}$

(D) The resistance of a wire is given by $R = \frac{\rho \ell}{A}$.
Since the mass $m = \text{density} \times \text{volume} = \rho_d \times A \times \ell$ is constant,and the material is the same (same density $\rho_d$),the volume $V = A \ell$ is constant.
Thus,$\ell = \frac{V}{A}$.
Substituting this into the resistance formula: $R = \frac{\rho V}{A^2}$.
Since $\rho$ and $V$ are constant,$R \propto \frac{1}{A^2}$.
Since $A = \pi r^2$,we have $R \propto \frac{1}{r^4}$.
Therefore,$\frac{R_A}{R_B} = \left( \frac{r_B}{r_A} \right)^4$.
Given $r_A = 2.0 \ mm$,$r_B = 4.0 \ mm$,and $R_B = 2 \ \Omega$:
$\frac{R_A}{2} = \left( \frac{4.0}{2.0} \right)^4 = (2)^4 = 16$.
$R_A = 16 \times 2 = 32 \ \Omega$.

(B) The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$,which implies $R \propto \frac{l}{r^2}$.
Since the volume of the wire remains constant during stretching,$V = A \cdot l = \pi r^2 l = \text{constant}$.
Let the initial radius be $r$ and length be $l$. After stretching,the new radius is $r' = r/2$ and the new length is $l'$.
Equating the volumes: $\pi r^2 l = \pi (r/2)^2 l'$.
$\pi r^2 l = \pi (r^2/4) l' \implies l' = 4l$.
The new resistance $R'$ is given by $R' = \rho \frac{l'}{\pi (r')^2} = \rho \frac{4l}{\pi (r/2)^2} = \rho \frac{4l}{\pi r^2 / 4} = 16 \left( \rho \frac{l}{\pi r^2} \right) = 16R$.
Given $R' = xR$,we find $x = 16$.

(D) The resistance of a conductor at temperature $T$ is given by $R = R_0(1 + \alpha \Delta T)$,where $R_0$ is the resistance at $0^{\circ} C$ and $\alpha$ is the temperature coefficient of resistance.
For the first case,at $100^{\circ} C$:
$10.2 = 10(1 + \alpha(100 - 0))$
$10.2 = 10 + 1000\alpha$
$0.2 = 1000\alpha \Rightarrow \alpha = \frac{0.2}{1000} = 2 \times 10^{-4} /^{\circ} C$.
For the second case,at $t^{\circ} C$:
$10.95 = 10(1 + \alpha(t - 0))$
$10.95 = 10 + 10\alpha t$
$0.95 = 10 \times (2 \times 10^{-4}) \times t$
$0.95 = 2 \times 10^{-3} \times t$
$t = \frac{0.95}{0.002} = 475^{\circ} C$.
To convert the temperature to the Kelvin scale:
$T(K) = t(^{\circ} C) + 273$
$T(K) = 475 + 273 = 748 \ K$.

(B) The resistance of a conductor at a temperature $T$ is given by the formula: $R = R_0(1 + \alpha \Delta T)$.
Here,$R_0 = 50 \Omega$ at $T_0 = 27^{\circ} C$,$R = 62 \Omega$,and $\alpha = 2.4 \times 10^{-4} { }^{\circ} C^{-1}$.
Substituting the values into the equation:
$62 = 50(1 + 2.4 \times 10^{-4} \Delta T)$
$1.24 = 1 + 2.4 \times 10^{-4} \Delta T$
$0.24 = 2.4 \times 10^{-4} \Delta T$
$\Delta T = \frac{0.24}{2.4 \times 10^{-4}} = 1000^{\circ} C$.
Since $\Delta T = T - T_0$,we have $T = T_0 + \Delta T = 27^{\circ} C + 1000^{\circ} C = 1027^{\circ} C$.

(C) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length of the conductor in the direction of current flow,and $A$ is the cross-sectional area perpendicular to the current flow.
In this case,the current flows between the two opposite shaded faces.
The distance between these two faces is the side length $L$,so $l = L$.
The area of each shaded face is $A = L \times t$.
Substituting these values into the resistance formula,we get:
$R = \rho \frac{L}{L \times t} = \frac{\rho}{t}$.
Since the expression for resistance $R = \frac{\rho}{t}$ does not contain $L$,the resistance is independent of $L$.

(A) Standard resistors are typically made from alloys like Manganin, Constantan, or Nichrome.
These materials are chosen because they have a very low temperature coefficient of resistivity, meaning their resistivity $(\rho)$ remains nearly constant over a wide range of temperatures.
Among the given options, the graph that represents a nearly constant resistivity with respect to temperature is the one where the line is almost horizontal (or has a very small slope).
However, in the context of standard textbook questions, the curve representing a very weak dependence of resistivity on temperature is the one that is nearly flat.
Looking at the provided options, none of the graphs show a perfectly constant line, but in standard physics curricula, the material used for standard resistors (like Manganin) is characterized by a very small, almost negligible change in resistivity with temperature.
Therefore, the graph that shows the least variation (the most horizontal line) is the most suitable representation.

(D) The speed of light in vacuum is given by $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.
Also,the characteristic impedance of free space is given by $Z_0 = \sqrt{\frac{\mu_0}{\epsilon_0}}$.
The unit of impedance is $\text{ohm} (\Omega)$,which is the same as the unit of resistance.
Therefore,the dimension of $\sqrt{\frac{\mu_0}{\epsilon_0}}$ is the same as the dimension of resistance.

(A) The resistivity $(\rho)$ of materials at room temperature is as follows:
$1$. Copper $(Cu)$ (Conductor): $\approx 1.7 \times 10^{-8} \ \Omega \cdot m$
$2$. Nichrome (Alloy): $\approx 1.0 \times 10^{-6} \ \Omega \cdot m$ (or $100 \times 10^{-8} \ \Omega \cdot m$)
$3$. Germanium (Semiconductor): $\approx 0.46 \ \Omega \cdot m$
$4$. Silicon (Semiconductor): $\approx 2300 \ \Omega \cdot m$
Comparing these values,the increasing order of resistivity is: Copper < Nichrome < Germanium < Silicon.
Therefore,the correct option is $A$.

(C) For a conductor,as the temperature increases,the thermal vibrations of the lattice ions increase,which leads to more frequent collisions of free electrons. This results in an increase in the resistivity (specific resistance) of the conductor.
For a semiconductor,as the temperature increases,more charge carriers (electrons and holes) are generated due to the breaking of covalent bonds. This increase in the number density of charge carriers significantly reduces the resistivity (specific resistance) of the semiconductor.
Therefore,the specific resistance of a conductor increases and that of a semiconductor decreases.

(B) According to Ohm's Law,the ratio of voltage $(V)$ to current $(i)$ is equal to the resistance $(R)$: $\frac{V}{i} = R$.
For a metallic wire,the resistance depends on temperature according to the relation: $R = R_0(1 + \alpha \Delta T)$,where $\alpha$ is the temperature coefficient of resistance.
For metals,$\alpha$ is positive.
Therefore,as the temperature increases,the resistance $(R)$ of the metallic wire increases.
Since $\frac{V}{i} = R$,the ratio of voltage to current also increases with a rise in temperature.

(A) The resistance of a wire is given by $R = \rho \frac{L}{A}$,where $A = \frac{\pi d^2}{4}$.
Since the volume of the wire remains constant during stretching,$V = A_1 L_1 = A_2 L_2$,which implies $L_1 d_1^2 = L_2 d_2^2$,or $\frac{L_1}{L_2} = \frac{d_2^2}{d_1^2}$.
The ratio of resistances is $\frac{R_1}{R_2} = \frac{L_1}{L_2} \times \frac{A_2}{A_1} = \frac{L_1}{L_2} \times \frac{d_2^2}{d_1^2}$.
Substituting $\frac{L_1}{L_2} = \frac{d_2^2}{d_1^2}$ into the equation,we get $\frac{R_1}{R_2} = \left(\frac{d_2^2}{d_1^2}\right) \times \left(\frac{d_2^2}{d_1^2}\right) = \frac{d_2^4}{d_1^4}$.

(B) The resistance $R$ of a wire is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area.
Since the wire is reshaped,its volume $V = A \times L$ remains constant.
We can express the area $A$ in terms of diameter $D$ as $A = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4}$.
Substituting $A$ into the resistance formula: $R = \rho \frac{L}{\pi D^2 / 4} = \frac{4 \rho L}{\pi D^2}$.
Since the volume $V = A \times L = \frac{\pi D^2 L}{4}$ is constant,we have $L = \frac{4V}{\pi D^2}$.
Substituting this into the resistance formula: $R = \frac{4 \rho}{\pi D^2} \times \left(\frac{4V}{\pi D^2}\right) = \frac{16 \rho V}{\pi^2 D^4}$.
To minimize $R$,we need to maximize $D$. Since volume is constant,increasing $D$ will result in a decrease in $L$. Therefore,we should decrease $L$ and increase $D$.

(C) The resistivity of a conductor (like copper) is directly proportional to temperature. As the temperature decreases,the resistivity of copper decreases.
Conversely,the resistivity of a semiconductor (like silicon) is inversely related to temperature because the number of charge carriers increases with temperature. As the temperature decreases,the number of free charge carriers decreases,causing the resistivity of silicon to increase.
Therefore,when cooled from $300 \ K$ to $100 \ K$,the resistivity of copper decreases and the resistivity of silicon increases.

(A) Given:
Resistance at $T_0 = 27.5^{\circ} C$ is $R_0 = 2.1 \Omega$.
Resistance at $T = 100^{\circ} C$ is $R = 2.7 \Omega$.
The formula for temperature dependence of resistance is $R = R_0[1 + \alpha(T - T_0)]$.
Substituting the values:
$2.7 = 2.1[1 + \alpha(100 - 27.5)]$
$2.7 = 2.1[1 + \alpha(72.5)]$
$\frac{2.7}{2.1} = 1 + \alpha(72.5)$
$\frac{9}{7} = 1 + \alpha(72.5)$
$\frac{9}{7} - 1 = \alpha(72.5)$
$\frac{2}{7} = \alpha(72.5)$
$\alpha = \frac{2}{7 \times 72.5} = \frac{2}{507.5} \approx 0.00394 \approx 3.9 \times 10^{-3} {}^{\circ} C^{-1}$.
Thus,the correct option is $A$.

(B) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length of the conductor in the direction of current flow,and $A$ is the cross-sectional area perpendicular to the current flow.
To maximize the resistance $R$,we need to maximize the ratio $\frac{l}{A}$. This means we need the largest possible length $l$ and the smallest possible cross-sectional area $A$.
The dimensions of the rod are $10 \ cm$,$1 \ cm$,and $0.5 \ cm$.
If the battery is connected across the $1 \ cm \times 0.5 \ cm$ faces,the current flows through the length $l = 10 \ cm$. The area of cross-section is $A = 1 \ cm \times 0.5 \ cm = 0.5 \ cm^2$.
In this case,the ratio $\frac{l}{A} = \frac{10}{0.5} = 20 \ cm^{-1}$.
Comparing this with other configurations,this configuration yields the maximum value for the ratio $\frac{l}{A}$,hence the resistance is maximum.

(A) The relationship between resistance and temperature is given by the formula: $R_\theta = R_0 [1 + \alpha (\theta - \theta_0)]$
Given values are: $R_0 = 100 \ \Omega$,$\theta_0 = 27^\circ \text{C}$,$R_\theta = 137 \ \Omega$,and $\alpha = 1.35 \times 10^{-4} \ ^\circ \text{C}^{-1}$.
Substituting these values into the equation:
$137 = 100 [1 + 1.35 \times 10^{-4} (\theta - 27)]$
$1.37 = 1 + 1.35 \times 10^{-4} (\theta - 27)$
$0.37 = 1.35 \times 10^{-4} (\theta - 27)$
$\theta - 27 = \frac{0.37}{1.35 \times 10^{-4}}$
$\theta - 27 \approx 2740.74$
$\theta \approx 2740.74 + 27 = 2767.74^\circ \text{C}$
Rounding to the nearest integer,we get $\theta \approx 2767^\circ \text{C}$.

(A) The ratio of conductivity $(\sigma)$ to resistivity $(\rho)$ is given by $\frac{\sigma}{\rho}$.
Since $\sigma = \frac{1}{\rho}$, we can write the ratio as $\frac{1/\rho}{\rho} = \frac{1}{\rho^2}$.
For a conductor, as the temperature increases, the resistivity $(\rho)$ increases due to increased collisions of electrons with the lattice ions.
Since $\rho$ is in the denominator and is squared, as $\rho$ increases, the value of $\frac{1}{\rho^2}$ decreases.
Therefore, the ratio of conductivity to resistivity decreases.

(B) When two wires are connected in series,the total resistance $R$ is the sum of individual resistances $R_1$ and $R_2$.
$R = R_1 + R_2$
Using the formula $R = \frac{\rho L}{A}$,where $\rho$ is resistivity,$L$ is length,and $A$ is the cross-sectional area:
$\frac{\rho(2l)}{A} = \frac{\rho_1 l}{A} + \frac{\rho_2 l}{A}$
Dividing both sides by $\frac{l}{A}$:
$2\rho = \rho_1 + \rho_2$
$\therefore \rho = \frac{\rho_1 + \rho_2}{2}$

(B) The resistance $R$ of a wire is given by $R = \frac{\rho l}{A}$.
Since volume $V = A \times l$,we can write $A = \frac{V}{l}$.
Substituting this into the resistance formula,we get $R = \frac{\rho l^2}{V}$.
Since density $d = \frac{m}{V}$,we have $V = \frac{m}{d}$.
Thus,$R = \frac{\rho l^2 d}{m}$.
For wires of the same material,$\rho$ and $d$ are constant,so $R \propto \frac{l^2}{m}$.
Given the ratios $m_1:m_2:m_3 = 1:3:5$ and $l_1:l_2:l_3 = 5:3:1$,the ratio of resistances is:
$R_1: R_2: R_3 = \frac{l_1^2}{m_1} : \frac{l_2^2}{m_2} : \frac{l_3^2}{m_3}$
$R_1: R_2: R_3 = \frac{5^2}{1} : \frac{3^2}{3} : \frac{1^2}{5}$
$R_1: R_2: R_3 = 25 : 3 : 0.2$
Multiplying by $5$ to clear the fraction:
$R_1: R_2: R_3 = 125 : 15 : 1$.

(D) The resistance of a wire is given by $R = \frac{\rho l}{A}$.
Since volume $V = A \times l$ remains constant during stretching,we can write $A = \frac{V}{l}$.
Substituting this into the resistance formula: $R = \frac{\rho l^2}{V}$.
Since $\rho$ and $V$ are constant,$R \propto l^2$.
Given the change in length $\frac{\Delta l}{l} = 2\% = 0.02$.
Using the differential method for small changes: $\frac{\Delta R}{R} = 2 \frac{\Delta l}{l}$.
$\frac{\Delta R}{R} = 2 \times 2\% = 4\%$.
Alternatively,using the ratio method:
$R_1 \propto l^2$ and $R_2 \propto (1.02l)^2 = 1.0404l^2$.
$\frac{R_2 - R_1}{R_1} \times 100\% = (1.0404 - 1) \times 100\% = 4.04\% \approx 4\%$.

(C) The resistance $R$ of a wire is given by the formula: $R = \frac{\rho l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.
Rearranging the formula to solve for resistivity $\rho$: $\rho = \frac{RA}{l}$.
Given values: $R = 5 \ \Omega$,$l = 15 \ m$,and $A = 6 \times 10^{-7} \ m^2$.
Substituting these values into the equation:
$\rho = \frac{5 \times 6 \times 10^{-7}}{15}$
$\rho = \frac{30 \times 10^{-7}}{15}$
$\rho = 2 \times 10^{-7} \ \Omega \ m$.

(C) The correct option is $C$.
Silicon is a semiconductor.
In semiconductors,the number density of charge carriers (electrons and holes) increases exponentially with an increase in temperature.
Although the relaxation time decreases with temperature,the increase in the number density of charge carriers dominates,leading to a decrease in resistivity as temperature increases.
In contrast,metals like Copper,Aluminium,and alloys like Nichrome show an increase in resistivity with an increase in temperature.

(D) The correct option is $D$.
Conductivity $(\sigma)$ is an intrinsic property of a material.
It depends only on the nature of the material, its temperature, and pressure.
It does not depend on the physical dimensions of the conductor, such as its length $(L)$ or cross-sectional area $(A)$.
Therefore, when the wire is stretched, its resistance changes, but its conductivity remains the same.

(A) Given: Initial temperature $T_{1} = 100^{\circ} C$,initial resistance $R_{1} = 100 \ \Omega$,final resistance $R_{2} = 200 \ \Omega$,and temperature coefficient $\alpha = 0.005 \ ^{\circ} C^{-1}$.
Using the formula for temperature dependence of resistance: $R_{2} = R_{1}[1 + \alpha(T_{2} - T_{1})]$.
Rearranging for $T_{2}$: $T_{2} - T_{1} = \frac{R_{2} - R_{1}}{\alpha R_{1}}$.
Substituting the values: $T_{2} - 100 = \frac{200 - 100}{0.005 \times 100}$.
$T_{2} - 100 = \frac{100}{0.5} = 200$.
$T_{2} = 200 + 100 = 300^{\circ} C$.

(A) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length of the conductor in the direction of current flow,and $A$ is the cross-sectional area perpendicular to the current flow.
From the relation $R \propto \frac{L}{A}$,we see that for a fixed volume of the material,the resistance is proportional to the square of the length $(L^2)$ or inversely proportional to the square of the area $(A^2)$.
Alternatively,for a fixed rod,the resistance is maximum when the current flows through the smallest cross-sectional area $A$,which results in the longest effective length $L$.
The three possible cross-sectional areas are:
$1$. $A_1 = 10 \text{ cm} \times 1 \text{ cm} = 10 \text{ cm}^2$
$2$. $A_2 = 10 \text{ cm} \times 0.5 \text{ cm} = 5 \text{ cm}^2$
$3$. $A_3 = 1 \text{ cm} \times 0.5 \text{ cm} = 0.5 \text{ cm}^2$
The resistance is maximum when the cross-sectional area $A$ is minimum. The minimum area is $0.5 \text{ cm}^2$,which corresponds to the faces of dimensions $1 \text{ cm} \times 0.5 \text{ cm}$.

(A) In the given graph:
$1$. For material $Z$,resistivity $\rho$ decreases exponentially with an increase in temperature $T$. This is a characteristic property of semiconductors.
$2$. For material $Y$,resistivity $\rho$ increases linearly with temperature $T$. This is a characteristic property of alloys like nichrome,which have a very small temperature coefficient of resistance.
$3$. For material $X$,resistivity $\rho$ increases non-linearly with temperature $T$. This is a characteristic property of metals like copper.
Therefore,$X$ is copper,$Y$ is nichrome,and $Z$ is a semiconductor.

(C) The resistance $(R)$ of a wire is given by the formula: $R = \frac{\rho \ell}{A}$,where $\rho$ is resistivity,$\ell$ is length,and $A$ is the cross-sectional area.
Since the wire is cylindrical,the cross-sectional area $A = \pi r^2 = \pi (D/2)^2 = \frac{\pi D^2}{4}$.
Substituting this into the resistance formula,we get: $R = \frac{\rho \ell}{\pi D^2 / 4} = \frac{4 \rho \ell}{\pi D^2}$.
Since $\rho$,$\ell$,and $\pi$ are constants,we have $R \propto \frac{1}{D^2}$.
This relationship represents an inverse square law,which corresponds to a curve that decreases as $D$ increases. Among the given options,Graph $C$ represents this behavior.