Resistance of wire, Resistivity and Conductivity Questions in English

(C) The formula for resistance $R$ is given by $R = \rho \frac{L}{A}$,where $\rho$ is the specific resistance (resistivity),$L$ is the length,and $A$ is the cross-sectional area.
Rearranging the formula for $\rho$,we get $\rho = \frac{R \cdot A}{L}$.
Substituting the units: Unit of $\rho = \frac{(\text{ohm}) \cdot (cm^2)}{cm} = \text{ohm} \cdot cm$.
Therefore,the unit of specific resistance is $Ohm \cdot cm$.

(A) The formula for resistivity $\rho$ is given by $\rho = \frac{RA}{l}$.
First,find the dimensions of resistance $R$ using Ohm's law: $V = IR$,so $R = \frac{V}{I}$.
Since $V = \frac{W}{Q}$ (work per charge),the dimensions of $V$ are $[M L^2 T^{-2} Q^{-1}]$.
The dimensions of current $I$ are $[Q T^{-1}]$.
Thus,the dimensions of $R$ are $\frac{[M L^2 T^{-2} Q^{-1}]}{[Q T^{-1}]} = [M L^2 T^{-1} Q^{-2}]$.
Now,substitute the dimensions of $R$,area $A$ $([L^2])$,and length $l$ $([L])$ into the resistivity formula:
$[\rho] = \frac{[M L^2 T^{-1} Q^{-2}] \cdot [L^2]}{[L]} = [M L^3 T^{-1} Q^{-2}]$.

(B) The resistance of a conductor at temperature $t$ is given by $R_t = R_{t_0} [1 + \alpha(t - t_0)]$.
Given $R_1 = 2.71 \ \Omega$ at $t_1 = 10^{\circ}C$ and $R_2 = 3.70 \ \Omega$ at $t_2 = 100^{\circ}C$.
Let $R_x = 3.26 \ \Omega$ at temperature $x$.
Using the formula: $R_2 = R_1 [1 + \alpha(t_2 - t_1)]$
$3.70 = 2.71 [1 + \alpha(100 - 10)]$
$3.70 = 2.71 [1 + 90\alpha] \implies 1 + 90\alpha = \frac{3.70}{2.71} \implies 90\alpha = \frac{3.70}{2.71} - 1 = \frac{0.99}{2.71}$.
Now,for temperature $x$: $R_x = R_1 [1 + \alpha(x - t_1)]$
$3.26 = 2.71 [1 + \alpha(x - 10)]$
$\frac{3.26}{2.71} = 1 + \alpha(x - 10) \implies \alpha(x - 10) = \frac{3.26}{2.71} - 1 = \frac{0.55}{2.71}$.
Dividing the two expressions for $\alpha$:
$\frac{\alpha(x - 10)}{90\alpha} = \frac{0.55 / 2.71}{0.99 / 2.71} \implies \frac{x - 10}{90} = \frac{0.55}{0.99} = \frac{55}{99} = \frac{5}{9}$.
$x - 10 = 90 \times \frac{5}{9} = 50$.
$x = 50 + 10 = 60^{\circ}C$.

(A) The resistance $R$ of a wire is given by $R = \rho \frac{l}{A}$. Since the volume $V = A \times l$ remains constant during stretching, we can write $A = \frac{V}{l}$.
Substituting this into the resistance formula, we get $R = \rho \frac{l^2}{V}$.
Since $\rho$ and $V$ are constants, $R \propto l^2$.
Taking the derivative or using the approximation for small changes, we have $\frac{\Delta R}{R} \approx 2 \frac{\Delta l}{l}$.
Given that the percentage increase in length $\frac{\Delta l}{l} \times 100 = 0.1\%$.
Therefore, the percentage increase in resistance is $\frac{\Delta R}{R} \times 100 = 2 \times (\frac{\Delta l}{l} \times 100) = 2 \times 0.1\% = 0.2\%$.

(A) The resistance $R$ of a conductor 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.
Given: $\rho = 50 \times 10^{-8} \, \Omega \cdot m$,$l = 50 \, cm = 0.5 \, m$,and for a cube,the area $A = l^2 = (0.5 \, m)^2 = 0.25 \, m^2$.
Substituting these values into the formula:
$R = \frac{50 \times 10^{-8} \times 0.5}{0.25}$
$R = \frac{25 \times 10^{-8}}{0.25}$
$R = 100 \times 10^{-8} \, \Omega = 10^{-6} \, \Omega$.
Therefore,the correct option is $A$.

(A) Resistivity $(\rho)$ is an intrinsic property of a material and depends only on the nature of the material and the temperature.
It does not depend on the physical dimensions of the conductor, such as its length or cross-sectional area (diameter).
Therefore, even if the length and diameter of the wire are doubled, the resistivity of the iron wire remains unchanged.
Thus, the resistivity remains $1 \times 10^{-7} \, \Omega \cdot m$.

(B) The resistance of a conductor at temperature $t$ is given by $R_t = R_0(1 + \alpha \Delta t)$,where $R_0$ is the resistance at $0\,^{\circ}C$ and $\Delta t$ is the temperature change from $0\,^{\circ}C$.
Given: $\alpha = 0.00125\,^{\circ}C^{-1}$,$R_1 = 1\,\Omega$ at $T_1 = 300\,K$ $(t_1 = 27\,^{\circ}C)$,and $R_2 = 2\,\Omega$ at $T_2 = ?$.
Using the formula $R_2 = R_1[1 + \alpha(t_2 - t_1)]$:
$2 = 1[1 + 0.00125(t_2 - 27)]$
$2 - 1 = 0.00125(t_2 - 27)$
$1 = 0.00125(t_2 - 27)$
$t_2 - 27 = \frac{1}{0.00125} = 800$
$t_2 = 800 + 27 = 827\,^{\circ}C$
Converting to Kelvin: $T_2 = 827 + 273 = 1100\,K$.

(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 area of cross-section.
Initially,$R_1 = \rho \frac{l}{A}$.
When the length is doubled $(l' = 2l)$ and the area of cross-section is doubled $(A' = 2A)$,the new resistance $R_2$ is:
$R_2 = \rho \frac{l'}{A'} = \rho \frac{2l}{2A} = \rho \frac{l}{A}$.
Comparing the two,we find $R_2 = R_1$.
Therefore,the resistance will remain the same.

(D) When a wire is stretched,its volume remains constant. The resistance $R$ of a wire is given by $R = \rho \frac{l}{A}$.
Since volume $V = A \times l$ is constant,$A = \frac{V}{l}$.
Substituting this into the resistance formula,we get $R = \rho \frac{l^2}{V}$.
Since $\rho$ and $V$ are constant,$R \propto l^2$.
Given that the new length $l' = 3l$,the new resistance $R'$ is:
$R' = R \times (\frac{l'}{l})^2 = 20 \times (3)^2 = 20 \times 9 = 180 \, \Omega$.

(D) The correct answer is $D$.
Resistivity $(\rho)$ is an intrinsic property of the material of the conductor.
It depends only on the nature of the material and the temperature.
It does not depend on the physical dimensions of the wire, such as its length $(l)$ or its cross-sectional area $(A)$.
Therefore, none of the statements $A, B$, or $C$ are correct.

(B) Given: Volume $V = A \times l = 3\,m^3$,Resistance $R = 3\,\Omega$,Specific resistance = $\rho$.
From the volume equation,we have $A = \frac{3}{l}$.
The formula for resistance is $R = \rho \frac{l}{A}$.
Substituting the values: $3 = \rho \frac{l}{(3/l)}$.
This simplifies to $3 = \frac{\rho l^2}{3}$.
Rearranging for $l^2$: $l^2 = \frac{9}{\rho}$.
Taking the square root on both sides: $l = \sqrt{\frac{9}{\rho}} = \frac{3}{\sqrt{\rho}}$.
Thus,the length is $\frac{3}{\sqrt{\rho}}$.

(D) The resistance of a wire is given by $R = \rho \frac{L}{A}$,where $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.
Thus,$R = \rho \frac{L}{\pi d^2 / 4} = \frac{4 \rho L}{\pi d^2}$.
When a wire is drawn to reduce its diameter,its volume $V = A \times L$ remains constant.
Since $V = (\pi d^2 / 4) \times L$ is constant,$L \propto \frac{1}{d^2}$.
Substituting this into the resistance formula: $R \propto \frac{L}{d^2} \propto \frac{1/d^2}{d^2} = \frac{1}{d^4}$.
If the diameter is reduced to half $(d' = d/2)$,the new resistance $R'$ is:
$R' = R \times (d/d')^4 = R \times (d / (d/2))^4 = R \times (2)^4 = 16R$.
Therefore,the resistance becomes $16$ times the original value.

(B) Given: Length $L = 100\,cm = 1\,m$,Diameter $d = 2.0\,mm = 2.0 \times 10^{-3}\,m$,Resistance $R = 0.7\,\Omega$.
Radius $r = \frac{d}{2} = 1.0 \times 10^{-3}\,m$.
The cross-sectional area $A = \pi r^2 = \frac{22}{7} \times (1.0 \times 10^{-3})^2 = \frac{22}{7} \times 10^{-6}\,m^2$.
Using the formula $R = \frac{\rho L}{A}$,we get $\rho = \frac{R \cdot A}{L}$.
Substituting the values: $\rho = \frac{0.7 \times (\frac{22}{7} \times 10^{-6})}{1}$.
$\rho = 0.1 \times 22 \times 10^{-6} = 2.2 \times 10^{-6}\,\Omega \cdot m$.
Thus,the resistivity is $2.2 \times 10^{-6}\,\Omega \cdot m$.

(B) The resistance of a wire is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is resistivity,$L$ is length,and $A$ is the cross-sectional area.
Since the area $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$,we have $R \propto \frac{1}{d^2}$.
Let the original resistance be $R_1 = R$ with diameter $d_1 = d$.
For the second wire,the diameter is $d_2 = 2d$.
Therefore,the new resistance $R_2$ is given by $\frac{R_2}{R_1} = \frac{d_1^2}{d_2^2} = \frac{d^2}{(2d)^2} = \frac{d^2}{4d^2} = \frac{1}{4}$.
Thus,$R_2 = \frac{R}{4} = 0.25\,R$.

(D) 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 $A = \pi r^2$ is the cross-sectional area.
Since the material and temperature are the same,$\rho$ remains constant.
Thus,$R \propto \frac{l}{r^2}$.
Given $R_1 = 1\,\Omega$,$l_1 = 5\,m$,$r_1 = 1\,mm$ and $R_2 = 1\,\Omega$,$r_2 = 2\,mm$.
We have the ratio: $\frac{R_1}{R_2} = \frac{l_1}{l_2} \times \frac{r_2^2}{r_1^2}$.
Substituting the values: $\frac{1}{1} = \frac{5}{l_2} \times \left(\frac{2}{1}\right)^2$.
$1 = \frac{5}{l_2} \times 4$.
$l_2 = 5 \times 4 = 20\,m$.

(A) The specific resistance (resistivity) of a metal is defined by the property of the material.
For metals,the resistivity $\rho$ is given by the relation $\rho = \frac{m}{ne^2\tau}$,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 relaxation time.
As the temperature increases,the thermal vibrations of the lattice ions increase,which leads to a decrease in the relaxation time $\tau$.
Since $\rho \propto \frac{1}{\tau}$,the resistivity of metals increases significantly with an increase in temperature.
Therefore,temperature is the most significant factor affecting the specific resistance of metals.

(C) The temperature coefficient of resistance $(\alpha)$ is defined by the relation $R_t = R_0(1 + \alpha \Delta T)$.
For metallic conductors, such as $Copper$, the resistance increases with an increase in temperature, which implies that the temperature coefficient of resistance $(\alpha)$ is positive.
In contrast, semiconductors like $Carbon$ and $Germanium$, as well as electrolytes, exhibit a decrease in resistance with an increase in temperature, resulting in a negative temperature coefficient of resistance.

(D) Superconductivity is a phenomenon occurring in certain materials where the electrical resistance drops to exactly zero and the conductance becomes infinite when the material is cooled below a characteristic critical temperature $(T_c)$,which is typically below a few Kelvin.

(D) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity (specific resistance),$l$ is the length,and $A$ is the cross-sectional area.
Given dimensions are $1 \, cm \times 1 \, cm \times 100 \, cm$.
For the resistance between the two square faces $(1 \, cm \times 1 \, cm)$,the current flows through the length of $100 \, cm$.
Thus,$l = 100 \, cm = 1 \, m$.
The area of the square face is $A = 1 \, cm \times 1 \, cm = 1 \, cm^2 = 10^{-4} \, m^2$.
The resistivity is $\rho = 3 \times 10^{-7} \, \Omega \cdot m$.
Substituting these values into the formula:
$R = (3 \times 10^{-7} \, \Omega \cdot m) \times \frac{1 \, m}{10^{-4} \, m^2} = 3 \times 10^{-7} \times 10^4 \, \Omega = 3 \times 10^{-3} \, \Omega$.

(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 has a uniform diameter $d$,the cross-sectional area $A = \pi (d/2)^2 = \frac{\pi d^2}{4}$.
Thus,$R \propto \frac{L}{d^2}$.
For the first wire: $R_1 = R$,$L_1 = L$,$d_1 = d$.
For the second wire: $L_2 = 4L$,$d_2 = 2d$.
Taking the ratio: $\frac{R_2}{R_1} = \frac{L_2}{L_1} \times \left( \frac{d_1}{d_2} \right)^2$.
Substituting the values: $\frac{R_2}{R} = \frac{4L}{L} \times \left( \frac{d}{2d} \right)^2 = 4 \times \left( \frac{1}{2} \right)^2 = 4 \times \frac{1}{4} = 1$.
Therefore,$R_2 = R$.

(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 wires are of the same material,$\rho_1 = \rho_2 = \rho$.
Let the length and diameter of the second wire be $l_2 = l$ and $d_2 = d$. Then its area is $A_2 = \pi (d/2)^2 = \pi d^2 / 4$.
For the first wire,$l_1 = 2l$ and $d_1 = 2d$. Then its area is $A_1 = \pi (2d/2)^2 = \pi d^2$.
Comparing the areas,$A_1 = 4 A_2$.
Now,the resistance of the first wire is $R_1 = \rho \frac{l_1}{A_1} = \rho \frac{2l}{4 A_2} = \frac{1}{2} \left( \rho \frac{l}{A_2} \right) = \frac{1}{2} R_2$.
Thus,the resistance of the first wire is half of the second wire.

(A) The resistance of a metallic conductor varies with temperature. For a small temperature range,the resistance $R_t$ at temperature $t\,^oC$ is related to the resistance $R_0$ at $0\,^oC$ by the linear equation:
$R_t = R_0(1 + \alpha t)$
where $\alpha$ is the temperature coefficient of resistance.
Thus,the correct option is $A$.

(B) The resistance $R$ 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 area of cross-section.
For the first wire: $R = \rho \frac{l}{a}$.
For the second wire,the material is the same (so $\rho$ is constant),the length is the same $(l)$,and the area of cross-section is $4a$.
Therefore,the new resistance $R'$ is given by $R' = \rho \frac{l}{4a}$.
Substituting the expression for $R$,we get $R' = \frac{1}{4} \left( \rho \frac{l}{a} \right) = \frac{R}{4}$.

(C) The resistance of a material depends on its temperature.
For metals like $Copper$,$Tungsten$,and $Aluminium$,the resistance increases with an increase in temperature due to increased scattering of electrons.
For semiconductors like $Germanium$,the number of charge carriers (electrons and holes) increases significantly with an increase in temperature,which leads to a decrease in resistance.
Therefore,the correct option is $C$.

(C) The resistance 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 mass and density of the silver are constant,the volume $V = A \times L$ must remain constant.
We can express $A$ as $V/L$. Substituting this into the resistance formula,we get $R = \rho \frac{L}{V/L} = \rho \frac{L^2}{V}$.
Since $\rho$ and $V$ are constant,$R \propto L^2$.
To minimize the resistance $R$,we must minimize the length $L$.
Comparing the given options: $(a)$ $L$,$(b)$ $2L$,$(c)$ $L/2$.
The smallest length is $L/2$ in option $(c)$.
Therefore,the combination $L/2$ and $2A$ results in the smallest resistance.

(A) The resistance of a wire is given by $R = \rho \frac{l}{A}$.
Since the volume $V = A \times l$ remains constant during stretching,we can write $A = \frac{V}{l}$.
Substituting this into the resistance formula,we get $R = \rho \frac{l^2}{V}$.
Since $\rho$ and $V$ are constant,$R \propto l^2$.
If the length is increased by $10\%$,the new length $l' = l + 0.1l = 1.1l$.
The new resistance $R'$ is given by $R' = R \left( \frac{l'}{l} \right)^2$.
$R' = 10 \times (1.1)^2 = 10 \times 1.21 = 12.1\,\Omega$.
However,for small changes,the approximation $\Delta R/R \approx 2(\Delta l/l)$ gives $2 \times 10\% = 20\%$,leading to $10 + 2 = 12\,\Omega$. Given the options,$12\,\Omega$ is the intended answer.

(C) 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 $0\,^{\circ}C$ and $\alpha$ is the temperature coefficient of resistance.
Given:
$R_{150} = 133\,\Omega$
$\alpha = 0.0045\,^{\circ}C^{-1}$
$T_1 = 150\,^{\circ}C$,$T_2 = 500\,^{\circ}C$
Using the ratio formula:
$\frac{R_{500}}{R_{150}} = \frac{1 + \alpha T_2}{1 + \alpha T_1}$
$\frac{R_{500}}{133} = \frac{1 + (0.0045 \times 500)}{1 + (0.0045 \times 150)}$
$\frac{R_{500}}{133} = \frac{1 + 2.25}{1 + 0.675} = \frac{3.25}{1.675}$
$R_{500} = 133 \times \frac{3.25}{1.675} \approx 133 \times 1.9403 \approx 258.06\,\Omega$
Thus,the resistance at $500\,^{\circ}C$ is approximately $258\,\Omega$.

(C) The resistance $R$ of a wire is given by the formula $R = \rho \frac{l}{A}$, where $\rho$ is the resistivity (specific resistance), $l$ is the length, and $A$ is the cross-sectional area.
Given: $\rho = 64 \times 10^{-6} \, \Omega \cdot \text{cm}$, $l = 198 \, \text{cm}$, $R = 7 \, \Omega$.
The area of the cross-section $A = \pi r^2$, where $r$ is the radius.
Substituting the values into the formula: $7 = (64 \times 10^{-6}) \times \frac{198}{\pi r^2}$.
Using $\pi \approx \frac{22}{7}$, we get: $7 = \frac{64 \times 10^{-6} \times 198}{(22/7) \times r^2}$.
$r^2 = \frac{64 \times 10^{-6} \times 198 \times 7}{7 \times 22} = \frac{64 \times 10^{-6} \times 198}{22} = 64 \times 10^{-6} \times 9 = 576 \times 10^{-6}$.
Taking the square root: $r = \sqrt{576 \times 10^{-6}} = 24 \times 10^{-3} = 0.024 \, \text{cm}$.

(B) 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.
If the length is doubled $(l' = 2l)$,the area of cross-section must become half $(A' = A/2)$ to keep the volume constant.
The new resistance $R'$ is given by $R' = \rho \frac{l'}{A'} = \rho \frac{2l}{A/2} = 4 \rho \frac{l}{A} = 4R$.
Therefore,the new resistance is $4R$.

(A) The temperature coefficient of resistance $(\alpha)$ is defined as the fractional change in resistance per unit change in temperature.
For metals like $Fe$, $Mn$, and $Ag$, the resistance increases with an increase in temperature, resulting in a positive temperature coefficient.
For semiconductors and insulators like carbon $(C)$, the number of charge carriers increases significantly with an increase in temperature, which leads to a decrease in resistance.
Therefore, carbon $(C)$ has a negative temperature coefficient.

(A) The reciprocal of resistance $(R)$ is defined as conductance $(G)$.
Mathematically,it is expressed as $G = \frac{1}{R}$.
The $SI$ unit of conductance is Siemens $(S)$ or $\Omega^{-1}$.

(A) The formula for the temperature dependence of resistance is given by $R_{T_2} = R_{T_1}[1 + \alpha(T_2 - T_1)]$.
Given values are $R_{T_1} = 5\,\Omega$ at $T_1 = 50\,^{\circ}\text{C}$ and $R_{T_2} = 7\,\Omega$ at $T_2 = 100\,^{\circ}\text{C}$.
Substituting these values into the equation:
$7 = 5[1 + \alpha(100 - 50)]$
$7 = 5[1 + 50\alpha]$
$7 = 5 + 250\alpha$
$2 = 250\alpha$
$\alpha = \frac{2}{250} = \frac{1}{125} = 0.008\,^{\circ}\text{C}^{-1}$.

(C) The resistance of a conductor at temperature $t$ is given by the formula: $R_t = R_0(1 + \alpha t)$.
Alternatively,for two temperatures $t_1$ and $t_2$,the relation is: $R_2 = R_1[1 + \alpha(t_2 - t_1)]$.
Given: $R_1 = 50\,\Omega$ at $t_1 = 20\,^{\circ}C$,$R_2 = 76.8\,\Omega$,and $\alpha = 3.92 \times 10^{-3}\,^{\circ}C^{-1}$.
Substituting the values: $76.8 = 50[1 + 3.92 \times 10^{-3}(t_2 - 20)]$.
Divide by $50$: $1.536 = 1 + 3.92 \times 10^{-3}(t_2 - 20)$.
Subtract $1$: $0.536 = 3.92 \times 10^{-3}(t_2 - 20)$.
Solve for $(t_2 - 20)$: $t_2 - 20 = \frac{0.536}{3.92 \times 10^{-3}} \approx 136.73$.
Therefore,$t_2 = 136.73 + 20 = 156.73\,^{\circ}C$,which rounds to approximately $167\,^{\circ}C$ based on the provided options.

(D) The resistivity $(\rho)$ of a material is an intrinsic property that characterizes how strongly a material opposes the flow of electric current. It depends only on the nature of the material and the temperature. It does not depend on the physical dimensions of the wire, such as its length, area of cross-section, or shape. Therefore, the correct option is $(d)$.

(A) superconductor is a material that exhibits zero electrical resistance when cooled below a characteristic critical temperature.
Since conductivity $\sigma$ is the reciprocal of resistivity $\rho$,and resistivity is directly proportional to resistance $R$ $(R = \rho \frac{l}{A})$,the conductivity is defined as $\sigma = \frac{1}{\rho}$.
As the resistance $R$ of a superconductor becomes zero,the resistivity $\rho$ also becomes zero.
Therefore,the conductivity $\sigma = \frac{1}{0} = \infty$ (infinite).

(D) In alloys,the atoms of different metals are randomly distributed in the crystal lattice. This random arrangement disrupts the periodic structure of the metal,leading to increased scattering of free electrons. Due to this increased scattering,the mobility of electrons decreases,which results in a higher resistivity for alloys compared to their constituent pure metals. Therefore,$\rho_{\text{alloy}} > \rho_{\text{metal}}$.

(A) The resistance of a wire is given by $R = \rho \frac{l}{A}$.
Since the mass $m = \text{Volume} \times \text{Density} = (A \cdot l) \cdot d$,and both wires have the same material and mass,the product $A \cdot l$ is constant.
Thus,$l = \frac{m}{A \cdot d} \propto \frac{1}{A}$.
Substituting this into the resistance formula: $R = \rho \frac{l}{A} \propto \frac{1}{A^2}$.
Since $A = \pi r^2$,we have $R \propto \frac{1}{(\pi r^2)^2} \propto \frac{1}{r^4}$.
Therefore,$\frac{R_A}{R_B} = \left( \frac{r_B}{r_A} \right)^4$.
Given $R_A = 34\,\Omega$,$r_A = 2r$,and $r_B = r$:
$\frac{34}{R_B} = \left( \frac{r}{2r} \right)^4 = \left( \frac{1}{2} \right)^4 = \frac{1}{16}$.
$R_B = 34 \times 16 = 544\,\Omega$.

(C) The relationship between resistance and temperature is given by the formula: $R_t = R_0(1 + \alpha t)$.
Here,$R_t = 4.2\, \Omega$ is the resistance at temperature $t = 100\,^{\circ}C$,$\alpha = 0.004\,^{\circ}C^{-1}$ is the temperature coefficient of resistance,and $R_0$ is the resistance at $0\,^{\circ}C$.
Substituting the values into the formula:
$4.2 = R_0(1 + 0.004 \times 100)$
$4.2 = R_0(1 + 0.4)$
$4.2 = R_0(1.4)$
$R_0 = \frac{4.2}{1.4} = 3\, \Omega$.
Therefore,the resistance at $0\,^{\circ}C$ is $3\, \Omega$.

(B) The electrical conductivity of metals depends on the mobility of free electrons. Among the given metals,Silver $(Ag)$ has the highest electrical conductivity,followed by Copper $(Cu)$,and then Aluminium $(Al)$.
Therefore,the order of increasing conductivity is $Al < Cu < Ag$.
Thus,the correct option is $B$.

(A) The resistance of a wire is given by $R = \rho \frac{L}{A}$,where $\rho$ is resistivity,$L$ is length,and $A$ is the cross-sectional area. Since $A = \pi r^2$,we have $R = \rho \frac{L}{\pi r^2}$.
Since the volume $V = A \cdot L$ remains constant during compression,$L_1 A_1 = L_2 A_2$.
Given $A_2 = \pi (nr)^2 = n^2 A_1$,we find $L_2 = L_1 \frac{A_1}{A_2} = \frac{L_1}{n^2}$.
The new resistance $R_2$ is $R_2 = \rho \frac{L_2}{A_2} = \rho \frac{L_1 / n^2}{n^2 A_1} = \frac{1}{n^4} \left( \rho \frac{L_1}{A_1} \right) = \frac{R}{n^4}$.

(D) 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 $0\,^{\circ}C$ and $\alpha$ is the temperature coefficient of resistance.
Given:
$R_1 = 5\,\Omega$ at $t_1 = 50\,^{\circ}C$
$R_2 = 6\,\Omega$ at $t_2 = 100\,^{\circ}C$
Using the ratio: $\frac{R_1}{R_2} = \frac{1 + \alpha t_1}{1 + \alpha t_2} \Rightarrow \frac{5}{6} = \frac{1 + 50\alpha}{1 + 100\alpha}$
$5(1 + 100\alpha) = 6(1 + 50\alpha) \Rightarrow 5 + 500\alpha = 6 + 300\alpha$
$200\alpha = 1 \Rightarrow \alpha = \frac{1}{200}\,^{\circ}C^{-1}$
Now,substitute $\alpha$ into the equation for $R_1$: $5 = R_0(1 + \frac{1}{200} \times 50)$
$5 = R_0(1 + 0.25) \Rightarrow 5 = R_0(1.25)$
$R_0 = \frac{5}{1.25} = 4\,\Omega$.

(B) Given that the potential difference $V$ and length $L$ are the same for both wires, and the current $I$ is also the same.
From Ohm's law, $V = IR$, so the resistance $R$ must be the same for both wires.
The resistance is given by $R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}$.
Since $R$, $L$, and $\pi$ are constant, we have $\frac{\rho_{\text{iron}}}{r_{\text{iron}}^2} = \frac{\rho_{\text{copper}}}{r_{\text{copper}}^2}$.
Therefore, $\frac{r_{\text{iron}}^2}{r_{\text{copper}}^2} = \frac{\rho_{\text{iron}}}{\rho_{\text{copper}}}$.
Taking the square root, $\frac{r_{\text{iron}}}{r_{\text{copper}}} = \sqrt{\frac{\rho_{\text{iron}}}{\rho_{\text{copper}}}} = \sqrt{\frac{1.0 \times 10^{-7}}{1.7 \times 10^{-8}}} = \sqrt{\frac{10}{1.7}} \approx \sqrt{5.88} \approx 2.42$.
Thus, the ratio is about $2.4$.

(B) The resistivity $\rho$ of the rod material is given by $\rho = \frac{RA}{l}$.
Given: $R = 3.0 \times 10^{-3} \, \Omega$,$l = 1.0 \, m$,and diameter $d_1 = 0.6 \, cm = 0.6 \times 10^{-2} \, m$. The radius $r_1 = 0.3 \times 10^{-2} \, m$.
$\rho = \frac{3.0 \times 10^{-3} \times \pi \times (0.3 \times 10^{-2})^2}{1.0} = 3.0 \times 10^{-3} \times \pi \times 0.09 \times 10^{-4} = 27 \times 10^{-9} \pi \, \Omega \cdot m$.
For the disc,thickness $t = 1.0 \, mm = 1.0 \times 10^{-3} \, m$ and diameter $d_2 = 2.0 \, cm = 2.0 \times 10^{-2} \, m$. The radius $r_2 = 1.0 \times 10^{-2} \, m$.
The resistance $R'$ between the round faces is $R' = \rho \frac{t}{A_2} = \rho \frac{t}{\pi r_2^2}$.
$R' = (27 \times 10^{-9} \pi) \times \frac{1.0 \times 10^{-3}}{\pi \times (1.0 \times 10^{-2})^2} = \frac{27 \times 10^{-12}}{10^{-4}} = 27 \times 10^{-8} = 2.70 \times 10^{-7} \, \Omega$.

(C) The resistance of a conductor at temperature $t$ is given by the formula: $R_t = R_0(1 + \alpha t)$.
Here, $R_t$ is the resistance at temperature $t$, $R_0$ is the resistance at $0^{\circ}C$, and $\alpha$ is the temperature coefficient of resistance.
Given that $R_t = 3R_0$ and $\alpha = 4 \times 10^{-3} \,^{\circ}C^{-1}$.
Substituting these values into the formula:
$3R_0 = R_0(1 + 4 \times 10^{-3} \times t)$
$3 = 1 + 4 \times 10^{-3} \times t$
$2 = 4 \times 10^{-3} \times t$
$t = \frac{2}{4 \times 10^{-3}} = \frac{2000}{4} = 500^{\circ}C$.
Therefore, the resistance becomes three times its initial value at $500^{\circ}C$.

(A) Given:
Length $l = 50\, cm = 0.5\, m = 50 \times 10^{-2}\, m$
Area of cross-section $A = 1\, mm^2 = 1 \times 10^{-6}\, m^2$
Current $I = 4\, A$
Voltage $V = 2\, V$
Using Ohm's Law,the resistance $R$ is:
$R = \frac{V}{I} = \frac{2}{4} = 0.5\, \Omega$
The formula for resistance in terms of resistivity $\rho$ is:
$R = \rho \frac{l}{A}$
Rearranging for $\rho$:
$\rho = \frac{R \cdot A}{l} = \frac{0.5 \times 10^{-6}}{0.5} = 1 \times 10^{-6}\, \Omega\cdot m$
Thus,the resistivity of the wire is $1 \times 10^{-6}\, \Omega\cdot m$.

(A) The resistance of a wire is given by $R = \rho \frac{l}{A}$.
Since the volume $V = A \times l$ remains constant during melting and recasting,we have $A = \frac{V}{l}$.
Substituting this into the resistance formula,we get $R = \rho \frac{l}{V/l} = \rho \frac{l^2}{V}$.
Since $\rho$ and $V$ are constant,$R \propto l^2$.
Therefore,$\frac{R_1}{R_2} = \left( \frac{l_1}{l_2} \right)^2$.
Given $l_2 = \frac{l_1}{2}$,we have $\frac{R}{R_2} = \left( \frac{l_1}{l_1/2} \right)^2 = (2)^2 = 4$.
Thus,the new resistance $R_2 = \frac{R}{4}$.

(A) When a wire is stretched uniformly,its volume remains constant. Since $V = A \times l$,we have $A_1 l_1 = A_2 l_2$.
Given $l_1 = 5\, cm$ and $l_2 = 20\, cm$,the ratio of lengths is $\frac{l_2}{l_1} = \frac{20}{5} = 4$.
The resistance $R$ is given by $R = \rho \frac{l}{A}$. Since $A = \frac{V}{l}$,we get $R = \rho \frac{l^2}{V}$.
Since $\rho$ and $V$ are constant,$R \propto l^2$.
Therefore,$\frac{R_2}{R_1} = \left( \frac{l_2}{l_1} \right)^2 = (4)^2 = 16$.
Given $R_1 = 10\, \Omega$,we find $R_2 = 16 \times 10 = 160\, \Omega$.

(C) The resistance $R$ of a metallic conductor is given by $R = \frac{ml}{ne^2 \tau A}$,where $\tau$ is the relaxation time.
As the temperature $T$ increases,the frequency of collisions between electrons and ions increases,which causes the relaxation time $\tau$ to decrease.
Since $R \propto \frac{1}{\tau}$,a decrease in $\tau$ leads to an increase in resistance $R$.
When an incandescent lamp is switched on,the filament heats up,causing its temperature to rise. Consequently,the resistance of the lamp is greater when it is switched on compared to when it is switched off.

(C) The temperature dependence of resistivity $\rho$ for a conductor is given by $\rho = \rho_{0}[1 + \alpha(T - T_{0})]$,where $\alpha$ is the temperature coefficient of resistivity.
For a conductor,as temperature increases,the amplitude of vibration of lattice ions increases. This leads to more frequent collisions between free electrons and ions,which decreases the relaxation time $\tau$. Since resistivity $\rho = m / (ne^2\tau)$,a decrease in $\tau$ causes the resistivity to increase.
For a semiconductor,as temperature increases,more valence electrons gain enough thermal energy to jump across the energy gap into the conduction band. This significantly increases the number density of charge carriers $n$. Since $\rho \propto 1/n$,the increase in $n$ dominates,causing the resistivity to decrease.

(D) The resistance $R$ of a wire is given by $R = \rho \frac{l}{A}$.
Since $Volume = V = A \cdot l$,we have $A = \frac{V}{l}$.
Also,$Mass = m = d \cdot V$,where $d$ is the density. Thus,$V = \frac{m}{d}$.
Substituting $A$ in the resistance formula: $R = \rho \frac{l}{(m/dl)} = \rho \cdot d \cdot \frac{l^2}{m}$.
Since the material is the same,$\rho$ and $d$ are constant,so $R \propto \frac{l^2}{m}$.
Given ratios: $m_1:m_2:m_3 = 1:2:3$ and $l_1:l_2:l_3 = 3:2:1$.
Therefore,the ratio of resistances is $R_1:R_2:R_3 = \frac{l_1^2}{m_1} : \frac{l_2^2}{m_2} : \frac{l_3^2}{m_3}$.
$R_1:R_2:R_3 = \frac{3^2}{1} : \frac{2^2}{2} : \frac{1^2}{3} = 9 : 2 : \frac{1}{3}$.
Multiplying by $3$ to clear the fraction: $27 : 6 : 1$.