Resistance of wire, Resistivity and Conductivity Questions in English

(C) The resistance $R$ of a wire is given by $R = \frac{\rho \ell}{A}$.
Since the volume $V = A \ell$ remains constant,we can write $A = \frac{V}{\ell}$.
Substituting this into the resistance formula: $R = \frac{\rho \ell^2}{V}$.
Since $\rho$ and $V$ are constant,$R \propto \ell^2$.
Taking the logarithmic derivative,we get $\frac{\Delta R}{R} = 2 \frac{\Delta \ell}{\ell}$.
Given the percentage change in length $\frac{\Delta \ell}{\ell} \times 100 = 0.5\%$.
Therefore,the percentage change in resistance is $\frac{\Delta R}{R} \times 100 = 2 \times 0.5\% = 1\%$.

(D) The graph shows a straight line between $\ln R(T)$ and $1/T^2$ with a negative slope. The equation of a straight line is $y = mx + c$.
Here,$y = \ln R(T)$ and $x = 1/T^2$.
So,$\ln R(T) = m(1/T^2) + c$,where $m$ is the negative slope and $c$ is the intercept.
Let the intercept be $\ln R_0$ at $1/T^2 = 0$. Then $\ln R(T) = -k(1/T^2) + \ln R_0$,where $k$ is a positive constant.
This can be written as $\ln R(T) = \ln R_0 - k/T^2$.
Taking the exponential of both sides,we get $R(T) = R_0 e^{-k/T^2}$.
Comparing this with the given options,if we set $k = T_0^2$,we get $R(T) = R_0 e^{-T_0^2/T^2}$.
However,looking at the options provided,there seems to be a mismatch in the exponent sign or variable definition. Re-evaluating the slope: if the line passes through $(1/T_0^2, 0)$ and $(0, \ln R_0)$,the equation is $\frac{\ln R(T)}{\ln R_0} + \frac{1/T^2}{1/T_0^2} = 1$.
$\ln R(T) = \ln R_0 (1 - T_0^2/T^2) = \ln R_0 - \ln R_0 (T_0^2/T^2)$.
This implies $R(T) = R_0 e^{-\ln R_0 (T_0^2/T^2)}$.
Given the standard form of such problems,the correct relation is $R(T) = R_0 e^{T_0^2/T^2}$ if the slope is positive,or $R(T) = R_0 e^{-T_0^2/T^2}$ if negative. Based on the provided options and the negative slope,option $D$ is the closest mathematical form if $T_0^2$ is defined appropriately.

(D) Consider a spherical shell of radius $x$ and thickness $dx$ between the two spheres.
The area of this shell is $A = 4\pi x^2$.
The resistance $dR$ of this thin shell is given by $dR = \rho \frac{dx}{A} = \rho \frac{dx}{4\pi x^2}$.
To find the total resistance $R$ between the spheres,we integrate from $r = a$ to $r = b$:
$R = \int_a^b \frac{\rho}{4\pi x^2} dx$.
$R = \frac{\rho}{4\pi} \int_a^b x^{-2} dx$.
$R = \frac{\rho}{4\pi} \left[ -\frac{1}{x} \right]_a^b$.
$R = \frac{\rho}{4\pi} \left( -\frac{1}{b} - (-\frac{1}{a}) \right) = \frac{\rho}{4\pi} \left( \frac{1}{a} - \frac{1}{b} \right)$.

(D) The resistance of a wire is given by $R = \rho \frac{l}{A}$.
Since the mass $m = A \cdot l \cdot d$ (where $d$ is density),we have $A = \frac{m}{l \cdot d}$.
Substituting $A$ into the resistance formula: $R = \rho \frac{l^2 \cdot d}{m}$.
Since $\rho$ and $d$ are constant for copper,$R \propto \frac{l^2}{m}$.
Given the ratios $m_1 : m_2 : m_3 = 1 : 3 : 5$ and $l_1 : l_2 : l_3 = 5 : 3 : 1$,we calculate the ratio of resistances:
$R_1 : R_2 : R_3 = \frac{5^2}{1} : \frac{3^2}{3} : \frac{1^2}{5} = 25 : 3 : 0.2$.
To express this in integers,multiply by $5$:
$R_1 : R_2 : R_3 = 125 : 15 : 1$.

(C) The resistance of a conductor is given by $R = \rho \frac{L}{A_{cross}}$,where $\rho$ is the resistivity,$L$ is the length in the direction of current flow,and $A_{cross}$ is the cross-sectional area perpendicular to the current flow.
For plate $A$,the length $L_A = \ell$,and the cross-sectional area $A_A = \ell \times t$. Thus,$R_A = \rho \frac{\ell}{\ell \times t} = \frac{\rho}{t}$.
For plate $B$,the length $L_B = 2\ell$,and the cross-sectional area $A_B = 2\ell \times t$. Thus,$R_B = \rho \frac{2\ell}{2\ell \times t} = \frac{\rho}{t}$.
Comparing the two,we get $R_A = R_B$.
Therefore,the ratio $(R_A/R_B) = 1/1$.

(D) Let $R_0$ be the resistance of both conductors at $0\,^{\circ}C.$
The resistance of the first conductor at $t_1\,^{\circ}C$ is given by $R_1 = R_0(1 + \alpha_1 t_1).$
The resistance of the second conductor at $t_2\,^{\circ}C$ is given by $R_2 = R_0(1 + \alpha_2 t_2).$
Given that $R_1 = R_2,$ we have:
$R_0(1 + \alpha_1 t_1) = R_0(1 + \alpha_2 t_2)$
Dividing both sides by $R_0$:
$1 + \alpha_1 t_1 = 1 + \alpha_2 t_2$
Subtracting $1$ from both sides:
$\alpha_1 t_1 = \alpha_2 t_2$
Therefore,the ratio of the temperature coefficients is:
$\frac{\alpha_1}{\alpha_2} = \frac{t_2}{t_1}$

(B) The resistance of a conductor at temperature $T$ is given by $R_T = R_0(1 + \alpha T)$,where $R_0$ is the resistance at $0\,^{\circ}\text{C}$.
Given $\alpha = 0.00125\,^{\circ}\text{C}^{-1}$.
At $T_1 = 25\,^{\circ}\text{C}$,$R_1 = 1\,\Omega$. So,$1 = R_0(1 + 0.00125 \times 25) = R_0(1 + 0.03125) = R_0(1.03125)$.
Thus,$R_0 = \frac{1}{1.03125} \approx 0.9697\,\Omega$.
We want to find $T_2$ such that $R_2 = 1.2\,\Omega$.
Using $R_2 = R_0(1 + \alpha T_2)$,we have $1.2 = R_0(1 + 0.00125 T_2)$.
Substituting $R_0$: $1.2 = \frac{1}{1.03125} (1 + 0.00125 T_2)$.
$1.2 \times 1.03125 = 1 + 0.00125 T_2$.
$1.2375 = 1 + 0.00125 T_2$.
$0.2375 = 0.00125 T_2$.
$T_2 = \frac{0.2375}{0.00125} = 190\,^{\circ}\text{C}$.

(C) The resistance of a wire is given by $R = \rho \frac{\ell}{A}$.
Since the volume $V = A \ell$ remains constant when the wire is stretched,$A = \frac{V}{\ell}$.
Substituting this into the resistance formula,we get $R = \rho \frac{\ell^2}{V}$,which implies $R \propto \ell^2$.
Given that the length increases by $10\%$,the new length $\ell_2 = 1.1 \ell_1$.
Therefore,the ratio of the new resistance to the original resistance is $\frac{R_2}{R_1} = \left(\frac{\ell_2}{\ell_1}\right)^2 = (1.1)^2 = 1.21$.
This means $R_2 = 1.21 R_1$.
The percentage increase in resistance is $\Delta R \% = \frac{R_2 - R_1}{R_1} \times 100\% = (1.21 - 1) \times 100\% = 21\%$.

(C) 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.
For the three wires,the resistances are:
$R_1 = \rho \frac{l}{A}$
$R_2 = \rho \frac{2l}{A/2} = 4 \rho \frac{l}{A} = 4R_1$
$R_3 = \rho \frac{l/2}{2A} = \frac{1}{4} \rho \frac{l}{A} = 0.25R_1$
Comparing the resistances,$R_3 < R_1 < R_2$.
Therefore,the resistance is minimum for the wire with length $l/2$ and cross-sectional area $2A$.

(B) The resistance of a wire is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity of the material,$L$ is the length,and $A$ is the cross-sectional area.
Resistivity $\rho$ is an intrinsic property of the material and does not depend on the geometry or shape of the wire.
When a wire is bent,its length $L$,cross-sectional area $A$,and resistivity $\rho$ remain unchanged.
Therefore,the electrical resistance $R$ remains constant.
While the Reason statement is a true physical fact (resistance is proportional to resistivity),it does not explain why bending the wire does not change the resistance (which is due to the geometry remaining constant). Thus,both are correct,but the Reason is not the correct explanation of the Assertion.

(C) Given:
Room temperature,$T = 27.0^{\circ} C$
Resistance at room temperature,$R = 100\; \Omega$
Final resistance,$R_1 = 117\; \Omega$
Temperature coefficient,$\alpha = 1.70 \times 10^{-4}\; ^{\circ} C^{-1}$
The formula for temperature dependence of resistance is:
$R_1 = R[1 + \alpha(T_1 - T)]$
Rearranging for the final temperature $T_1$:
$T_1 - T = \frac{R_1 - R}{R \alpha}$
Substituting the values:
$T_1 - 27 = \frac{117 - 100}{100 \times 1.70 \times 10^{-4}}$
$T_1 - 27 = \frac{17}{1.70 \times 10^{-2}}$
$T_1 - 27 = \frac{17}{0.017} = 1000$
$T_1 = 1000 + 27 = 1027^{\circ} C$
Thus,the temperature of the element is $1027^{\circ} C$.

(D) Given:
Length of the wire,$l = 15 \; m$
Area of cross-section of the wire,$A = 6.0 \times 10^{-7} \; m^{2}$
Resistance of the wire,$R = 5.0 \; \Omega$
The formula for resistance is $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity.
Rearranging for resistivity: $\rho = \frac{R \times A}{l}$
Substituting the values: $\rho = \frac{5.0 \times 6.0 \times 10^{-7}}{15}$
$\rho = \frac{30 \times 10^{-7}}{15} = 2 \times 10^{-7} \; \Omega \, m$
Thus,the resistivity of the material is $2 \times 10^{-7} \; \Omega \, m$.

(A) Given:
Initial temperature,$T_{1} = 27.5^{\circ} C$
Resistance at $T_{1}$,$R_{1} = 2.1\; \Omega$
Final temperature,$T_{2} = 100^{\circ} C$
Resistance at $T_{2}$,$R_{2} = 2.7\; \Omega$
The temperature coefficient of resistivity $\alpha$ is given by the formula:
$\alpha = \frac{R_{2} - R_{1}}{R_{1}(T_{2} - T_{1})}$
Substituting the values:
$\alpha = \frac{2.7 - 2.1}{2.1(100 - 27.5)}$
$\alpha = \frac{0.6}{2.1 \times 72.5}$
$\alpha = \frac{0.6}{152.25}$
$\alpha \approx 0.0039^{\circ} C^{-1}$

(B) Supply voltage,$V = 230 \; V$.
Initial current,$I_1 = 3.2 \; A$.
Initial resistance $R_1 = \frac{V}{I_1} = \frac{230}{3.2} = 71.875 \; \Omega$.
Steady state current,$I_2 = 2.8 \; A$.
Steady state resistance $R_2 = \frac{V}{I_2} = \frac{230}{2.8} \approx 82.143 \; \Omega$.
Temperature coefficient of resistance,$\alpha = 1.70 \times 10^{-4} \; ^{\circ}C^{-1}$.
Initial temperature,$T_1 = 27.0^{\circ} C$.
Using the formula $R_2 = R_1[1 + \alpha(T_2 - T_1)]$,we get:
$T_2 - T_1 = \frac{R_2 - R_1}{\alpha R_1}$.
$T_2 - 27 = \frac{82.143 - 71.875}{1.70 \times 10^{-4} \times 71.875}$.
$T_2 - 27 = \frac{10.268}{0.01221875} \approx 840.35$.
$T_2 = 840.35 + 27 = 867.35^{\circ} C \approx 867.5^{\circ} C$ (rounding to match options).

(A) Let the resistivity,length,area of cross-section,density,and mass of the aluminium wire be $\rho_1, l_1, A_1, d_1, m_1$ and for the copper wire be $\rho_2, l_2, A_2, d_2, m_2$ respectively.
Given: $l_1 = l_2 = l$ and $R_1 = R_2 = R$.
Since $R = \rho \frac{l}{A}$,we have $\rho_1 \frac{l}{A_1} = \rho_2 \frac{l}{A_2}$,which implies $\frac{A_1}{A_2} = \frac{\rho_1}{\rho_2} = \frac{2.63 \times 10^{-8}}{1.72 \times 10^{-8}} \approx 1.529$.
The mass of a wire is given by $m = \text{Volume} \times \text{Density} = A \cdot l \cdot d$.
Taking the ratio of masses: $\frac{m_1}{m_2} = \frac{A_1 l d_1}{A_2 l d_2} = \left( \frac{A_1}{A_2} \right) \left( \frac{d_1}{d_2} \right)$.
Substituting the values: $\frac{m_1}{m_2} = \left( \frac{2.63}{1.72} \right) \times \left( \frac{2.7}{8.9} \right) \approx 1.529 \times 0.303 \approx 0.463$.
Since $\frac{m_1}{m_2} < 1$,$m_1 < m_2$,meaning the aluminium wire is lighter.
Aluminium is preferred for overhead power cables because it is much lighter than copper for the same resistance,reducing the mechanical stress on the supporting towers.

(A) Alloys of metals usually have greater resistivity than that of their constituent metals because the disordered structure of the alloy increases electron scattering.
$(b)$ Alloys usually have much lower temperature coefficients of resistance than pure metals,making them useful for standard resistors.
$(c)$ The resistivity of the alloy manganin is nearly independent of increase of temperature,which is why it is used to make standard resistors.
$(d)$ The resistivity of a typical insulator is greater than that of a metal by a factor of the order of $10^{22}$.

(N/A) Electrical resistance is the property of a conductor to oppose the flow of electric charge through it.
From Ohm's law,$R = \frac{V}{I}$. Its $SI$ unit is $\frac{\text{Volt}}{\text{Ampere}} = \text{ohm} (\Omega)$. Its dimensional formula is $[M^1 L^2 T^{-3} A^{-2}]$.
Resistance depends on the following factors:
$1$. Length of the conductor $(l)$: $R \propto l$
$2$. Cross-sectional area of the conductor $(A)$: $R \propto \frac{1}{A}$
$3$. Nature of the material (resistivity,$\rho$)
$4$. Temperature of the conductor
Consider a conductor with length $l$ and area $A$ as shown in Figure $(a)$.
Two such similar rectangular blocks are shown in Figure $(b)$. Here,the total length of the combination is $2l$.
The current flowing through the combination of blocks is equal to the current flowing through each individual block. Hence,the potential difference across each block is $V$. Therefore,the total potential difference across the combination is $2V$. Let the resistance of the combination be $R_C$. By Ohm's law:
$R_C = \frac{2V}{I} = 2R$
Since $\frac{V}{I} = R$ is the resistance of each block,we see that the resistance of the conductor is proportional to its length $(R \propto l)$.

(N/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,$L$ is the length of the conductor,and $A$ is the cross-sectional area.
$1$. Dependence on Length $(L)$: The resistance $R$ is directly proportional to the length $L$ of the conductor $(R \propto L)$. This means that if the length of the conductor is increased while keeping the cross-sectional area constant,the resistance increases.
$2$. Dependence on Area $(A)$: The resistance $R$ is inversely proportional to the cross-sectional area $A$ of the conductor $(R \propto \frac{1}{A})$. This means that if the cross-sectional area is increased (e.g.,by increasing the thickness of the wire),the resistance decreases.

(N/A) Conductance is defined as the reciprocal of the electrical resistance of a conductor. It represents the ease with which electric current flows through a material.
Mathematically,it is given by: $G = \frac{1}{R}$
where $G$ is the conductance and $R$ is the resistance.
The $SI$ unit of conductance is the siemens $(S)$,which is also equivalent to $\Omega^{-1}$ (ohm inverse) or mho.

(N/A) At a given temperature,the resistance $R$ of a material is given by:
$R = \rho \frac{l}{A}$
Therefore,the resistivity $\rho$ is defined as:
$\rho = \frac{RA}{l}$
Definition: The resistance of a conductor having unit length and unit cross-sectional area is called resistivity or specific resistance.
Factors affecting resistivity: The value of resistivity depends on the type of material of the conductor,temperature,and pressure.
Note: It does not depend on the dimensions (length or area) of the conductor.
Unit: The $SI$ unit is $\Omega \cdot m$ (Ohm-meter).
Dimensional formula: $[M^{1} L^{3} T^{-3} A^{-2}]$.

(N/A) Conductivity is defined as the reciprocal of resistivity. It is denoted by the symbol $\sigma$.
$\sigma = \frac{1}{\rho} = \frac{l}{RA}$
The $SI$ unit of conductivity is $\Omega^{-1} m^{-1}$ or $S m^{-1}$ (Siemens per meter).
The dimensional formula of conductivity is $[M^{-1} L^{-3} T^{3} A^{2}]$.
The value of conductivity depends on the nature of the material,temperature,and pressure.
It does not depend on the physical dimensions (length or cross-sectional area) of the conductor.

(N/A) Electrical resistance is the property of a conductor by virtue of which it opposes the flow of electric current through it. It is defined as the ratio of the potential difference $(V)$ applied across the conductor to the current $(I)$ flowing through it, given by $R = V/I$. The $SI$ unit of resistance is the ohm $(\Omega)$.
The resistance of a conductor depends on the following factors:
$1$. Length of the conductor $(l)$: Resistance is directly proportional to the length $(R \propto l)$.
$2$. Area of cross-section $(A)$: Resistance is inversely proportional to the area of cross-section $(R \propto 1/A)$.
$3$. Nature of the material: Resistance depends on the resistivity $(\rho)$ of the material.
$4$. Temperature: For most conductors, resistance increases with an increase in temperature.

(N/A) The conductivity of metals is given by the formula: $\sigma = \frac{n e^{2} \tau}{m}$.
Here,$n$ (charge carrier density),$e$ (charge of an electron),and $m$ (mass of an electron) are constants for a given metal.
Therefore,$\sigma \propto \tau$,where $\tau$ is the relaxation time.
With an increase in temperature,the thermal vibrations of the lattice ions increase,which leads to more frequent collisions of electrons with the ions. This causes the relaxation time $(\tau)$ to decrease.
Since $\sigma \propto \tau$,a decrease in $\tau$ leads to a decrease in conductivity $(\sigma)$.
Resistivity is defined as $\rho = \frac{1}{\sigma}$. Since $\rho \propto \frac{1}{\tau}$,as the temperature increases and $\tau$ decreases,the resistivity $(\rho)$ of the metal increases.

(A) perfect conductor is defined as a material that allows electric current to flow through it without any resistance.
Since resistance $R$ is related to resistivity $\rho$ by the formula $R = \rho \frac{L}{A}$,where $L$ is the length and $A$ is the cross-sectional area.
For a perfect conductor,the resistance $R = 0$.
Therefore,the resistivity $\rho$ must be $0$ for any finite length and area.

(B) The resistivity of a material depends on temperature.
For conductors,resistivity increases with an increase in temperature due to increased electron-phonon scattering.
For semiconductors and insulators,the number of charge carriers (electrons and holes) increases exponentially with temperature,which outweighs the effect of increased scattering.
Therefore,the resistivity of semiconductors and insulators decreases as the temperature increases.

(N/A) The resistivity of a material depends on its temperature. Different materials exhibit different temperature dependencies.
For metals,the resistivity increases with an increase in temperature.
For semiconductors,the resistivity decreases with an increase in temperature.
Over a limited range of temperature that is not too large,the resistivity of a metallic conductor is approximately given by the empirical formula:
$\rho_{T} = \rho_{0} [1 + \alpha (T - T_{0})]$
Where:
$\rho_{T}$ = resistivity at temperature $T$
$\rho_{0}$ = resistivity at reference temperature $T_{0}$
$\alpha$ = temperature coefficient of resistivity
The unit of $\alpha$ is $(^{\circ}C)^{-1}$ or $(K)^{-1}$.
For metals,the value of $\alpha$ is positive,whereas for semiconductors,$\alpha$ is negative.

(N/A) $1$. For metals: The resistivity $\rho_T$ at temperature $T$ is given by $\rho_T = \rho_0 [1 + \alpha(T - T_0)]$. For a limited range of temperatures, this graph is a straight line. At very low temperatures, the graph deviates from linearity and shows a non-linear behavior as shown in the provided figure for copper.
$2$. For alloys: Alloys like nichrome, manganin, and constantan have a very high resistivity compared to pure metals. Their resistivity shows a very weak dependence on temperature, resulting in a nearly flat, slightly increasing straight-line graph.
$3$. For semiconductors: The resistivity of semiconductors decreases exponentially with an increase in temperature. The relation is given by $\rho_T = \rho_0 e^{E_g / k_B T}$, where $E_g$ is the band gap energy. The graph is a downward-sloping curve.

(N/A) The conductivity of a material is given by $\sigma = \frac{n e^{2} \tau}{m}$.
Resistivity $\rho$ is the reciprocal of conductivity,so $\rho = \frac{1}{\sigma} = \frac{m}{n e^{2} \tau}$.
Since $m$ (mass of electron) and $e$ (charge of electron) are constants,$\rho \propto \frac{1}{n}$ and $\rho \propto \frac{1}{\tau}$.
Thus,resistivity is inversely proportional to the number density $(n)$ and the relaxation time $(\tau)$.
In metals,as temperature increases,the average speed of electrons increases,which leads to more frequent collisions,causing the relaxation time $(\tau)$ to decrease. Since $n$ is nearly independent of temperature in metals,the decrease in $\tau$ causes the resistivity $(\rho)$ to increase.
In semiconductors and insulators,the number density $(n)$ increases significantly with temperature due to the thermal excitation of charge carriers. This increase in $n$ dominates over the change in $\tau$,causing the resistivity $(\rho)$ to decrease as temperature increases.

(N/A) For metals,as temperature increases,the thermal vibrations of the lattice ions increase. This leads to more frequent collisions between free electrons and ions,thereby increasing the resistance. The relationship is given by $R_T = R_0(1 + \alpha \Delta T)$,where $\alpha$ is positive.
For semiconductors,as temperature increases,more charge carriers (electrons and holes) are thermally excited from the valence band to the conduction band. This increase in the number of charge carriers dominates over the effect of increased scattering,leading to a decrease in resistance. Thus,semiconductors have a negative temperature coefficient of resistance.

(N/A) The resistivity $\rho$ of a metallic conductor at a temperature $T$ is related to its resistivity $\rho_0$ at a reference temperature $T_0$ by the following linear equation:
$\rho_T = \rho_0 [1 + \alpha(T - T_0)]$
Where:
- $\rho_T$ is the resistivity at temperature $T$.
- $\rho_0$ is the resistivity at reference temperature $T_0$.
- $\alpha$ is the temperature coefficient of resistivity.
- $(T - T_0)$ is the change in temperature.

(A) The resistivity of a metal is given by the relation $\rho_T = \rho_0 [1 + \alpha(T - T_0)]$,where $\rho_T$ is the resistivity at temperature $T$,$\rho_0$ is the resistivity at reference temperature $T_0$,and $\alpha$ is the temperature coefficient of resistivity.
In metals,as the temperature increases,the thermal vibrations of the lattice ions increase,which leads to more frequent collisions of electrons with the ions.
This results in an increase in the resistance and resistivity of the metal with an increase in temperature.
Since $\rho_T > \rho_0$ for $T > T_0$,the value of $\alpha$ must be positive.
Therefore,for metals,the temperature coefficient of resistivity is positive.

(N/A) The resistivity of a metal is given by the formula $\rho = \frac{m}{ne^2\tau}$,where $m$ is the mass of the electron,$n$ is the number density of free electrons,$e$ is the charge of the electron,and $\tau$ is the average relaxation time.
In metals,the number density of free electrons $(n)$ is independent of temperature.
When the temperature of a metal increases,the thermal vibrations of the positive ions in the lattice increase.
Due to these increased vibrations,the frequency of collisions between the free electrons and the positive ions increases.
This leads to a decrease in the average relaxation time $(\tau)$.
Since $\rho \propto \frac{1}{\tau}$,a decrease in $\tau$ results in an increase in the resistivity $(\rho)$ of the metal.

(N/A) In household wiring,$Cu$ (copper) or $Al$ (aluminium) wires are used based on the criteria of cost,conductivity,and weight.
$1$. Cost: Copper is significantly more expensive than aluminium.
$2$. Conductivity: Both are excellent conductors of electricity,though copper has slightly higher conductivity.
$3$. Weight: For a given length and resistance,aluminium wire is much lighter than copper wire.
$4$. Practicality: While silver is a better conductor,it is prohibitively expensive. Iron is cheaper but prone to rusting and has higher resistance,making it unsuitable for long-term household use.
Therefore,aluminium is often preferred for its balance of cost-effectiveness and conductivity.

(N/A) Alloys like Manganin or Constantan are used for making standard resistance coils because they possess a very low temperature coefficient of resistivity.
This means that the electrical resistivity of these alloys changes very little with variations in temperature.
Since the resistance $R$ of a conductor is given by $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area,a nearly constant $\rho$ ensures that the resistance $R$ remains stable even if the ambient temperature fluctuates.
Therefore,they are ideal for maintaining precise resistance values in standard resistors.

(N/A) The potential difference $(V)$ across an external resistance $R$ connected to a cell of $emf$ $E$ and internal resistance $r$ is given by the formula:
$V = I R = \left( \frac{E}{R + r} \right) R = \frac{E}{1 + \frac{r}{R}}$
Analysis of the relation:
$1$. When $R = 0$,$V = 0$.
$2$. As $R$ increases,$V$ increases.
$3$. As $R \to \infty$,$V \to E$.
The graph shows $V$ on the $y$-axis and $R$ on the $x$-axis. The curve starts from the origin $(0,0)$ and asymptotically approaches the value $E$ as $R$ increases.

(C) Resistance of a conductor is given by $R = \frac{\rho l}{A}$,where $\rho$ is resistivity,$l$ is length,and $A$ is the cross-sectional area.
For conductor $A$ (solid wire of diameter $d_A = 1 \ mm$):
Radius $r_A = 0.5 \ mm = 0.5 \times 10^{-3} \ m$.
Area $A_A = \pi r_A^2 = \pi (0.5 \times 10^{-3})^2 = 0.25 \pi \times 10^{-6} \ m^2$.
$R_A = \frac{\rho l}{0.25 \pi \times 10^{-6}} = \frac{4 \rho l}{\pi \times 10^{-6}}$....$(1)$
For conductor $B$ (hollow tube of outer diameter $D_B = 2 \ mm$ and inner diameter $d_B = 1 \ mm$):
Outer radius $R_{out} = 1 \ mm = 10^{-3} \ m$,inner radius $r_{in} = 0.5 \ mm = 0.5 \times 10^{-3} \ m$.
Area $A_B = \pi (R_{out}^2 - r_{in}^2) = \pi ((10^{-3})^2 - (0.5 \times 10^{-3})^2) = \pi (1 - 0.25) \times 10^{-6} = 0.75 \pi \times 10^{-6} \ m^2$.
$R_B = \frac{\rho l}{0.75 \pi \times 10^{-6}} = \frac{4 \rho l}{3 \pi \times 10^{-6}}$....$(2)$
Taking the ratio $\frac{R_A}{R_B}$:
$\frac{R_A}{R_B} = \frac{4 \rho l}{\pi \times 10^{-6}} \times \frac{3 \pi \times 10^{-6}}{4 \rho l} = \frac{3}{1}$.
Thus,$R_A : R_B = 3 : 1$.

(D) The resistivity values at room temperature $(20^{\circ}C)$ for the given materials are approximately:
Copper $(\rho_{C})$: $1.72 \times 10^{-8} \ \Omega \cdot m$
Aluminium $(\rho_{A})$: $2.82 \times 10^{-8} \ \Omega \cdot m$
Tungsten $(\rho_{T})$: $5.60 \times 10^{-8} \ \Omega \cdot m$
Mercury $(\rho_{M})$: $98.0 \times 10^{-8} \ \Omega \cdot m$
Comparing these values,we get $\rho_{M} > \rho_{T} > \rho_{A} > \rho_{C}$.
Among the given options,the correct relationship is $\rho_{M} > \rho_{A} > \rho_{C}$.

(A) The temperature coefficient of resistance $(\alpha)$ is defined by the relation $R_t = R_0(1 + \alpha \Delta T)$.
For metals, resistance increases with temperature, so $\alpha$ is positive.
For insulators and semiconductors, the number of charge carriers increases significantly with an increase in temperature, which leads to a decrease in resistance.
Therefore, insulators and semiconductors exhibit a negative temperature coefficient of resistance.

(D) For metals like copper,the resistivity ( $\rho$ ) increases with an increase in temperature $(T)$.
According to the relation $\rho_T = \rho_0 [1 + \alpha(T - T_0)]$,the resistivity increases linearly with temperature for a wide range of temperatures.
At very low temperatures,the curve becomes non-linear and approaches a finite value as $T$ approaches $0 \ K$.
Among the given options,graph $D$ correctly represents this characteristic behavior of resistivity versus temperature for a metal like copper.

(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 that both conductors are made of the same material,their resistivities are equal: $\rho_{1} = \rho_{2} = \rho$.
Given that they have the same length: $l_{1} = l_{2} = l$.
Given that they have the same resistance: $R_{1} = R_{2} = R$.
From the formula,we can express the area as $A = \frac{\rho l}{R}$.
Since $\rho$,$l$,and $R$ are identical for both conductors,the cross-sectional areas must also be equal:
$A_{1} = \frac{\rho l}{R}$ and $A_{2} = \frac{\rho l}{R}$
Therefore,$\frac{A_{1}}{A_{2}} = 1$.

(B) Given,initial resistance $R_{1} = 35 \,\Omega$ and final length $l_{2} = 2l_{1}$.
Since the volume of the wire remains constant during stretching,$V_{1} = V_{2}$.
$A_{1}l_{1} = A_{2}l_{2} \implies A_{2} = A_{1} \left(\frac{l_{1}}{l_{2}}\right) = A_{1} \left(\frac{l_{1}}{2l_{1}}\right) = \frac{A_{1}}{2}$.
The resistance of a wire is given by $R = \rho \frac{l}{A}$.
Therefore,the new resistance $R_{2} = \rho \frac{l_{2}}{A_{2}} = \rho \frac{2l_{1}}{A_{1}/2} = 4 \left(\rho \frac{l_{1}}{A_{1}}\right) = 4R_{1}$.
Substituting the value of $R_{1}$,we get $R_{2} = 4 \times 35 = 140 \,\Omega$.

(A) The initial resistance of the wire is $R = \frac{\rho l}{A}$.
According to the question,the new length $l' = 2l$ and the new area of cross-section $A' = \frac{A}{2}$.
The new resistance $R'$ is given by:
$R' = \rho \frac{l'}{A'} = \rho \frac{2l}{A/2} = \frac{4 \rho l}{A} = 4R$.
The new current $I'$ is given by Ohm's law:
$I' = \frac{V}{R'} = \frac{V}{4R} = \frac{V}{4(\rho l / A)} = \frac{1}{4} \frac{VA}{\rho l}$.

(D) Let the length of each wire be $\ell$ and the cross-sectional area be $A$. The resistance of a wire is given by $R = \rho_{res} \frac{\ell}{A}$, where $\rho_{res}$ is the resistivity.
For the two wires connected in parallel, the individual resistances are $R_1 = \rho_1 \frac{\ell}{A} = 6 \frac{\ell}{A}$ and $R_2 = \rho_2 \frac{\ell}{A} = 3 \frac{\ell}{A}$.
The equivalent resistance $R_{net}$ for two resistors in parallel is given by $\frac{1}{R_{net}} = \frac{1}{R_1} + \frac{1}{R_2}$, or $R_{net} = \frac{R_1 R_2}{R_1 + R_2}$.
The combined cross-sectional area of the two wires in parallel is $2A$, and the length remains $\ell$. Thus, $R_{net} = \rho \frac{\ell}{2A}$.
Substituting the expressions:
$\rho \frac{\ell}{2A} = \frac{(6 \frac{\ell}{A}) (3 \frac{\ell}{A})}{6 \frac{\ell}{A} + 3 \frac{\ell}{A}}$
$\frac{\rho}{2} = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2$
$\rho = 4 \, \Omega \, cm$.

(A) Initial resistance $R_{0} = 1\, \Omega$ and initial length $\ell_{0} = 1\, m$.
When the wire is stretched,its volume remains constant. Let the new length be $\ell_{1}$.
Given that the length increases by $25\, \%$,so $\ell_{1} = \ell_{0} + 0.25\ell_{0} = 1.25\ell_{0} = 1.25\, m$.
Since volume $V = A \ell$ is constant,$A_{0}\ell_{0} = A_{1}\ell_{1}$,which implies $A_{1} = A_{0}(\ell_{0} / \ell_{1})$.
The resistance $R$ is given by $R = \rho \frac{\ell}{A}$.
Therefore,the new resistance $R_{1} = \rho \frac{\ell_{1}}{A_{1}} = \rho \frac{\ell_{1}}{A_{0}(\ell_{0} / \ell_{1})} = \rho \frac{\ell_{1}^{2}}{A_{0}\ell_{0}} = R_{0} \left( \frac{\ell_{1}}{\ell_{0}} \right)^{2}$.
Substituting the values: $R_{1} = 1 \times (1.25)^{2} = 1.5625\, \Omega$.
The percentage change in resistance is $\frac{R_{1} - R_{0}}{R_{0}} \times 100\, \% = \frac{1.5625 - 1}{1} \times 100\, \% = 56.25\, \%$.
Rounding to the nearest integer,the percentage change is $56\, \%$.

(D) The resistance of a conductor at temperature $T$ is given by the formula: $R_T = R_0(1 + \alpha \Delta T)$,where $R_0$ is the resistance at $0^{\circ}C$,$\alpha$ is the temperature coefficient of resistance,and $\Delta T$ is the change in temperature.
Given:
$R_1 = 16\, \Omega$ at $T_1 = 15^{\circ}C$
$R_2 = 20\, \Omega$ at $T_2 = 100^{\circ}C$
Using the relation $R_T = R_0(1 + \alpha T)$:
$16 = R_0(1 + 15\alpha)$ --- (Equation $1$)
$20 = R_0(1 + 100\alpha)$ --- (Equation $2$)
Dividing Equation $2$ by Equation $1$:
$\frac{20}{16} = \frac{1 + 100\alpha}{1 + 15\alpha}$
$1.25(1 + 15\alpha) = 1 + 100\alpha$
$1.25 + 18.75\alpha = 1 + 100\alpha$
$0.25 = 81.25\alpha$
$\alpha = \frac{0.25}{81.25} \approx 0.00307\, ^{\circ}C^{-1}$
Rounding to the nearest value,we get $\alpha \approx 0.003\, ^{\circ}C^{-1}$.

(B) For conductors,the temperature coefficient of resistance $\alpha$ is positive,meaning resistance increases with temperature.
For semiconductors,the temperature coefficient of resistance $\alpha$ is negative,meaning resistance decreases as temperature increases due to the increase in the number of charge carriers.

(B) Since the volume $V$ of the wire remains constant during stretching,we have $V = A \ell = A^{\prime} \ell^{\prime}$.
Given that the new length $\ell^{\prime} = 2\ell$,we substitute this into the volume equation: $A \ell = A^{\prime} (2\ell)$,which gives $A^{\prime} = \frac{A}{2}$.
The initial resistance is $R = \rho \frac{\ell}{A}$.
The new resistance is $R^{\prime} = \rho \frac{\ell^{\prime}}{A^{\prime}} = \rho \frac{2\ell}{A/2} = 4 \rho \frac{\ell}{A} = 4R$.
The percentage increase in resistance is given by $\frac{R^{\prime} - R}{R} \times 100\%$.
Substituting $R^{\prime} = 4R$,we get $\frac{4R - R}{R} \times 100\% = 3 \times 100\% = 300\%$.

(C) The resistance of a wire is given by $R = \frac{\rho \ell}{A}$.
Since the volume $V = \ell A$ remains constant during stretching,we have $A = \frac{V}{\ell}$.
Substituting this into the resistance formula,we get $R = \frac{\rho \ell^2}{V}$.
For small changes,the relative change in resistance is given by $\frac{\Delta R}{R} = 2 \frac{\Delta \ell}{\ell}$.
Given that the percentage change in length is $\frac{\Delta \ell}{\ell} \times 100 = 0.4 \%$.
Therefore,the percentage change in resistance is $\frac{\Delta R}{R} \times 100 = 2 \times 0.4 \% = 0.8 \%$.

(A) The resistance $R$ at temperature $T$ is given by the formula $R = R_{T_0}[1 + \alpha(T - T_0)]$,where $R_{T_0}$ is the resistance at reference temperature $T_0$ and $\alpha$ is the temperature coefficient of resistance.
Given:
$R_1 = 2\,\Omega$ at $T_1 = 10^{\circ}C$
$R_2 = 3\,\Omega$ at $T_2 = 30^{\circ}C$
Using the relation $R_T = R_0(1 + \alpha T)$,we have:
$2 = R_0(1 + 10\alpha)$ --- (Equation $1$)
$3 = R_0(1 + 30\alpha)$ --- (Equation $2$)
Dividing Equation $2$ by Equation $1$:
$\frac{3}{2} = \frac{1 + 30\alpha}{1 + 10\alpha}$
Cross-multiplying:
$3(1 + 10\alpha) = 2(1 + 30\alpha)$
$3 + 30\alpha = 2 + 60\alpha$
$3 - 2 = 60\alpha - 30\alpha$
$1 = 30\alpha$
$\alpha = \frac{1}{30} \approx 0.033\,^{\circ}C^{-1}$.