Resistance of wire, Resistivity and Conductivity Questions in English

(D) The resistance of a wire is given by $R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}$.
Since the volume of the wire remains constant during stretching,$V = A_1 L_1 = A_2 L_2$,which implies $\pi r_1^2 L_1 = \pi r_2^2 L_2$.
Thus,$\frac{L_2}{L_1} = \left( \frac{r_1}{r_2} \right)^2$.
The ratio of resistances is $\frac{R_2}{R_1} = \frac{\rho L_2 / A_2}{\rho L_1 / A_1} = \left( \frac{L_2}{L_1} \right) \left( \frac{A_1}{A_2} \right) = \left( \frac{r_1}{r_2} \right)^2 \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{r_1}{r_2} \right)^4$.
Given $r_1 = r$ and $r_2 = \frac{3r}{4}$,we have $\frac{R_2}{R} = \left( \frac{r}{3r/4} \right)^4 = \left( \frac{4}{3} \right)^4 = \frac{256}{81}$.
Therefore,$R_2 = \frac{256R}{81}$.

(D) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is resistivity,$L$ is length,and $A$ is the cross-sectional area.
$1$. As length $L$ increases,resistance $R$ increases.
$2$. As cross-sectional area $A$ decreases,resistance $R$ increases.
$3$. For a conductor,resistivity $\rho$ increases with an increase in temperature,which leads to an increase in resistance $R$.
Therefore,all the given factors lead to an increase in resistance.

(D) Specific resistance,also known as resistivity $(\rho)$,is an intrinsic property of a material.
It 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 $(L)$ or cross-sectional area $(A)$.
Since both wires are made of the same material,their specific resistance will be equal.
Therefore,the ratio of their specific resistance is $1:1$.

(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 $A$ is the cross-sectional area.
Given: $\rho = 48 \times 10^{-8} \, \Omega \, m$,$R = 4.2 \, \Omega$,and diameter $d = 0.4 \, mm = 0.4 \times 10^{-3} \, m$.
The radius $r = \frac{d}{2} = 0.2 \times 10^{-3} \, m$.
The area $A = \pi r^2 = 3.14 \times (0.2 \times 10^{-3})^2 = 3.14 \times 0.04 \times 10^{-6} = 0.1256 \times 10^{-6} \, m^2$.
Rearranging the formula for length: $l = \frac{R \cdot A}{\rho}$.
Substituting the values: $l = \frac{4.2 \times 3.14 \times (0.2 \times 10^{-3})^2}{48 \times 10^{-8}}$.
$l = \frac{4.2 \times 3.14 \times 0.04 \times 10^{-6}}{48 \times 10^{-8}} = \frac{0.52752 \times 10^{-6}}{48 \times 10^{-8}} = \frac{52.752}{48} = 1.1 \, m$.

(D) Copper is a metal (conductor). For metals,the resistance decreases as the temperature decreases because the collision frequency of electrons with lattice ions decreases.
Germanium is a semiconductor. For semiconductors,the resistance increases as the temperature decreases because the number of charge carriers (electrons and holes) decreases exponentially with a decrease in temperature.
Therefore,when cooled from room temperature to $80\, K$,the resistance of the copper strip decreases and the resistance of the germanium strip increases.

$(A)$ Let the initial length be $l_1 = l$. Since the volume of the wire remains constant, $V = A_1 l_1 = A_2 l_2$.
Given that the length is increased by $100 \%$, the new length $l_2 = l_1 + 1.00 l_1 = 2l_1$.
From the volume conservation, $A_1 l_1 = A_2 (2l_1) \Rightarrow A_2 = \frac{A_1}{2}$.
The resistance of a wire is given by $R = \rho \frac{l}{A}$.
Thus, the initial resistance $R_1 = \rho \frac{l_1}{A_1}$ and the final resistance $R_2 = \rho \frac{l_2}{A_2} = \rho \frac{2l_1}{A_1/2} = 4 \rho \frac{l_1}{A_1} = 4R_1$.
The percentage change in resistance is $\frac{\Delta R}{R_1} \times 100 = \frac{R_2 - R_1}{R_1} \times 100 = \frac{4R_1 - R_1}{R_1} \times 100 = 300 \%$.

(C) Given: Both wires have the same mass $(m)$ and same material (same density $\rho_d$ and resistivity $\rho$).
Since mass = density $\times$ volume,and density is the same,the volume $(V = A \cdot l)$ must be the same for both wires.
$V_A = V_B \Rightarrow A_A l_A = A_B l_B \Rightarrow \frac{l_A}{l_B} = \frac{A_B}{A_A}$.
Since $A = \pi r^2 = \pi (d/2)^2$,we have $\frac{A_B}{A_A} = \frac{d_B^2}{d_A^2}$.
Given $d_A = \frac{d_B}{2}$,so $\frac{d_B}{d_A} = 2$. Thus,$\frac{A_B}{A_A} = (2)^2 = 4$.
Therefore,$\frac{l_A}{l_B} = 4$.
The resistance is given by $R = \rho \frac{l}{A}$.
Taking the ratio: $\frac{R_A}{R_B} = \frac{l_A}{l_B} \times \frac{A_B}{A_A} = 4 \times 4 = 16$.
Given $R_A = 24 \, \Omega$,then $R_B = \frac{R_A}{16} = \frac{24}{16} = 1.5 \, \Omega$.

(B) The resistance $R$ 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.
Conductance $C$ is the reciprocal of resistance,so $C = \frac{1}{R} = \frac{A}{\rho l}$.
From this expression,it is clear that $C \propto \frac{1}{l}$.
If the length $l$ is doubled $(l' = 2l)$,the new conductance $C'$ becomes $C' = \frac{A}{\rho (2l)} = \frac{1}{2} \left( \frac{A}{\rho l} \right) = \frac{1}{2} C$.
Therefore,the conductance is halved.

(A) The resistance of a wire is given by $R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}$.
Since the volume of the wire remains constant during stretching,$V = A_1 L_1 = A_2 L_2$,which implies $L_1 r_1^2 = L_2 r_2^2$.
Therefore,$\frac{L_2}{L_1} = \left( \frac{r_1}{r_2} \right)^2$.
The ratio of the new resistance $R_2$ to the original resistance $R_1$ is $\frac{R_2}{R_1} = \frac{\rho L_2 / A_2}{\rho L_1 / A_1} = \left( \frac{L_2}{L_1} \right) \left( \frac{A_1}{A_2} \right) = \left( \frac{r_1}{r_2} \right)^2 \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{r_1}{r_2} \right)^4$.
Given that the radius is reduced to half,$r_2 = \frac{r_1}{2}$,so $\frac{r_1}{r_2} = 2$.
Substituting this into the formula,$\frac{R_2}{R} = (2)^4 = 16$.
Thus,the new resistance is $R_2 = 16 \, R$.

(D) When a wire of resistance $R$ is stretched to $n$ times its original length,its volume remains constant.
Since $V = A \cdot L$,if the length becomes $nL$,the new area $A'$ must be $A/n$ to keep the volume constant.
The resistance is given by $R = \rho \frac{L}{A}$.
The new resistance $R'$ is given by $R' = \rho \frac{nL}{A/n} = n^2 \rho \frac{L}{A} = n^2 R$.
Given that the length is stretched to $n = 4$ times,the new resistance is $R' = (4)^2 R = 16R$.

(A) Lead wires are used to connect a power source to a load. To minimize power loss $(P = I^2R)$ and voltage drop $(V = IR)$ across the wires, the resistance $(R)$ of the wires should be as low as possible.
Since resistance is given by the formula $R = \rho \frac{L}{A}$, where $\rho$ is resistivity, $L$ is length, and $A$ is the cross-sectional area $(A = \pi r^2)$, to decrease resistance, we need to increase the cross-sectional area $(A)$.
Increasing the area means increasing the diameter of the wire.
Therefore, lead wires should have a larger diameter to ensure low resistance.

(D) Standard resistors are designed to maintain a constant resistance value regardless of changes in temperature.
Constantan and manganin are alloys that exhibit two key properties:
$1$. They have high resistivity,which allows for the construction of compact resistors.
$2$. They have a very low temperature coefficient of resistance,meaning their resistance changes negligibly with temperature fluctuations.
Therefore,both properties make them ideal for standard resistance.
Thus,the correct option is $(D)$.

(A) 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 $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$,we have $R \propto \frac{l}{d^2}$.
For option $A$: $R_A \propto \frac{50}{(0.5)^2} = \frac{50}{0.25} = 200$.
For option $B$: $R_B \propto \frac{100}{(1)^2} = 100$.
For option $C$: $R_C \propto \frac{200}{(2)^2} = \frac{200}{4} = 50$.
For option $D$: $R_D \propto \frac{300}{(3)^2} = \frac{300}{9} \approx 33.3$.
Comparing these values,the ratio $\frac{l}{d^2}$ is maximum for option $A$. Therefore,the wire in option $A$ has the highest electrical resistance.

(C) When a wire is stretched,its volume remains constant. Since $Volume = Area \times Length$,if the length becomes $n$ times,the area becomes $1/n$ times.
Resistance $R = \rho \frac{l}{A}$. Therefore,$R \propto l^2$.
Given that the length becomes $2$ times $(l_2 = 2l_1)$,the new resistance $R_2$ is:
$R_2 = R_1 \times (2)^2 = 4R_1$.
The change in resistance is $\Delta R = R_2 - R_1 = 4R_1 - R_1 = 3R_1$.
The ratio of the change in resistance to the initial resistance is $\frac{\Delta R}{R_1} = \frac{3R_1}{R_1} = 3:1$.

(C) The resistance of a wire is given by $R = \rho \frac{l}{A}$. Since the volume $V = A \times l$ remains constant, $A = \frac{V}{l}$.
Substituting this, we get $R = \rho \frac{l^2}{V}$, which implies $R \propto l^2$.
If the length increases by $10\%$, the new length $l_2 = l_1 + 0.10 l_1 = 1.1 l_1$.
The ratio of resistances is $\frac{R_2}{R_1} = \left( \frac{l_2}{l_1} \right)^2 = (1.1)^2 = 1.21$.
Thus, $R_2 = 1.21 R_1$.
The percentage change in resistance is $\frac{R_2 - R_1}{R_1} \times 100 = (1.21 - 1) \times 100 = 21\%$.

(C) The resistance $R$ of a wire is given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the specific resistance (resistivity),$l$ is the length,and $A$ is the cross-sectional area.
Given the ratios:
$\frac{\rho_1}{\rho_2} = \frac{2}{3}$
$\frac{l_1}{l_2} = \frac{3}{4}$
$\frac{A_1}{A_2} = \frac{4}{5}$
The ratio of their resistances is $\frac{R_1}{R_2} = \frac{\rho_1}{\rho_2} \times \frac{l_1}{l_2} \times \frac{A_2}{A_1}$.
Substituting the given values:
$\frac{R_1}{R_2} = \frac{2}{3} \times \frac{3}{4} \times \frac{5}{4} = \frac{2 \times 3 \times 5}{3 \times 4 \times 4} = \frac{30}{48} = \frac{5}{8}$.
Thus,the ratio of their resistances is $5:8$.

(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 the diameters are equal,the cross-sectional areas $A$ are equal for both wires.
The resistances are $R_1 = \frac{\rho_1 l_1}{A}$ and $R_2 = \frac{\rho_2 l_2}{A}$.
When joined in series,the equivalent resistance is $R_{eq} = R_1 + R_2$.
The equivalent wire has total length $(l_1 + l_2)$ and the same cross-sectional area $A$,so $R_{eq} = \frac{\rho_{eq}(l_1 + l_2)}{A}$.
Equating the expressions: $\frac{\rho_{eq}(l_1 + l_2)}{A} = \frac{\rho_1 l_1}{A} + \frac{\rho_2 l_2}{A}$.
Canceling $A$ from both sides,we get $\rho_{eq}(l_1 + l_2) = \rho_1 l_1 + \rho_2 l_2$.
Therefore,the equivalent resistivity is $\rho_{eq} = \frac{\rho_1 l_1 + \rho_2 l_2}{l_1 + l_2}$.

(B) For two wires connected in series,the total resistance $R_{eq}$ is the sum of individual resistances: $R_{eq} = R_1 + R_2$.
Since the wires have the same dimensions (length $l$ and cross-sectional area $A$),the resistance of each wire is given by $R = \rho \frac{l}{A}$.
Substituting this into the series combination formula: $\rho_{eq} \frac{2l}{A} = \rho_1 \frac{l}{A} + \rho_2 \frac{l}{A}$.
Dividing both sides by $\frac{l}{A}$,we get: $2\rho_{eq} = \rho_1 + \rho_2$.
Therefore,the equivalent resistivity is $\rho_{eq} = \frac{\rho_1 + \rho_2}{2}$.

(B) Given: $R = 150 \; \Omega$,$R_{0} = 100 \; \Omega$,and $\Delta t = 227^{\circ} C - 27^{\circ} C = 200^{\circ} C$.
Using the formula $R = R_{0}(1 + \alpha \Delta t)$:
$150 = 100(1 + \alpha \times 200)$
$1.5 = 1 + 200\alpha$
$0.5 = 200\alpha$
$\alpha = 0.5 / 200 = 2.5 \times 10^{-3} /^{\circ} C$.
Thus,Statement-$1$ is true.
Regarding Statement-$2$: The linear approximation $R = R_{0}(1 + \alpha \Delta t)$ is derived from the Taylor expansion $R = R_{0} e^{\alpha \Delta t} \approx R_{0}(1 + \alpha \Delta t + \dots)$,which is valid only when $\alpha \Delta t << 1$. In this problem,$\Delta R = 50 \; \Omega$,which is $50\%$ of $R_{0}$. Since $\Delta R$ is not much smaller than $R_{0}$,the linear approximation is technically inaccurate for large temperature changes. However,the statement claims the formula is *only* valid for small changes,which is a standard physical constraint for linear approximations. Therefore,Statement-$2$ is also true and provides the theoretical basis for the limitations of the formula used in Statement-$1$.

(C) Constantan is an alloy of copper and nickel.
It is widely used in making standard resistances because it has a very low (negligible) temperature coefficient of resistance.
This property ensures that the resistance value of the wire remains almost constant even when there are fluctuations in the surrounding temperature,making it ideal for precision instruments.

(A) 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.
For a hollow tube,the cross-sectional area $A$ is given by $A = \pi (r_2^2 - r_1^2)$,where $r_2$ is the outer radius and $r_1$ is the inner radius.
Given:
Length $l = 5\,m$
Outer diameter $D_2 = 10\,cm$,so outer radius $r_2 = 5\,cm = 0.05\,m$.
Wall thickness $t = 5\,mm = 0.5\,cm = 0.005\,m$.
Inner radius $r_1 = r_2 - t = 5\,cm - 0.5\,cm = 4.5\,cm = 0.045\,m$.
Resistivity $\rho = 1.7 \times 10^{-8} \,\Omega m$.
Calculating the area $A$:
$A = \pi [(0.05)^2 - (0.045)^2] = \pi [0.0025 - 0.002025] = \pi [0.000475] \,m^2$.
Now,calculating the resistance $R$:
$R = (1.7 \times 10^{-8}) \times \frac{5}{\pi \times 0.000475} \approx \frac{8.5 \times 10^{-8}}{0.001492} \approx 5.69 \times 10^{-5} \,\Omega$.
Rounding to the nearest provided option,the resistance is approximately $5.6 \times 10^{-5} \,\Omega$.

(A) Let $l$ be the original length of the wire and $A$ be its original cross-sectional area. The original resistance is $R = \rho \frac{l}{A}$.
Let a length $x$ of the wire be stretched to a new length $x' = x + 0.5l$. Since the volume of this part remains constant,$A x = A' x'$,where $A'$ is the new cross-sectional area.
Thus,$A' = A \frac{x}{x + 0.5l}$.
The new resistance $R'$ is the sum of the resistance of the unstretched part $(l-x)$ and the stretched part $(x+0.5l)$:
$R' = \rho \frac{l-x}{A} + \rho \frac{x+0.5l}{A'} = \rho \frac{l-x}{A} + \rho \frac{(x+0.5l)^2}{Ax}$.
Given $R' = 4R$,we have $4 \frac{l}{A} = \frac{l-x}{A} + \frac{(x+0.5l)^2}{Ax}$.
Multiplying by $Ax$,we get $4lx = x(l-x) + (x+0.5l)^2$.
$4lx = lx - x^2 + x^2 + lx + 0.25l^2$.
$4lx = 2lx + 0.25l^2$.
$2lx = 0.25l^2$.
$x/l = 0.25/2 = 1/8$.

(A) All the conductors have equal lengths $(l)$ and are made of the same material (same resistivity $\rho$).
The area of cross-section of conductor $A$ is $A_A = (\sqrt{3}a)^2 - (\sqrt{2}a)^2 = 3a^2 - 2a^2 = a^2$.
The area of cross-section of conductor $B$ is $A_B = (\sqrt{2}a)^2 - (a)^2 = 2a^2 - a^2 = a^2$.
The area of cross-section of conductor $C$ is $A_C = (a)^2 = a^2$.
Since the resistance is given by $R = \rho \frac{l}{A}$,and all conductors have the same $\rho$,$l$,and $A$,their resistances are equal:
$R_A = R_B = R_C$.

(D) The slope of the $I-V$ curve is given by $\text{slope} = \frac{I}{V} = \frac{1}{R}$.
Since resistance $R = \frac{\rho L}{A}$, we have $\text{slope} = \frac{1}{R} = \frac{A}{\rho L}$.
From this expression, it is clear that the slope is inversely proportional to the length $L$ of the wire $(\text{slope} \propto \frac{1}{L})$.
Therefore, if the length of the wire is increased, the slope of the curve will decrease.
Thus, option $(d)$ is correct.

(B) The given equation is ${R_t} = {R_0} + {R_0}\alpha t + {R_0}\beta {t^2}$.
This is a quadratic equation of the form $y = ax^2 + bx + c$,which represents a parabola.
From the graph,we observe that the resistance ${R_t}$ increases with temperature $t$,and the slope of the curve $\frac{dR_t}{dt} = {R_0}\alpha + 2{R_0}\beta t$ is positive and increasing.
Since the curve is concave upwards,the second derivative $\frac{d^2R_t}{dt^2} = 2{R_0}\beta$ must be positive,which implies $\beta > 0$.
Also,at $t = 0$,the slope is ${R_0}\alpha$. Since the curve starts with a positive slope,$\alpha$ must be positive.
Therefore,both $\alpha$ and $\beta$ are positive. The correct option is $(b)$.

(B) Let the resistivity at a distance $x$ from the left end be $\rho = (\rho_0 + ax)$,where $\rho_0$ is the resistivity at $x = 0$ and $a$ is a positive constant.
According to Ohm's law,the current density $J$ is given by $J = \frac{i}{A} = \frac{E}{\rho}$,where $i$ is the constant current and $A$ is the uniform cross-sectional area.
Therefore,the electric field intensity $E$ at a distance $x$ is $E = \frac{i \rho}{A} = \frac{i(\rho_0 + ax)}{A}$.
This equation shows that $E$ is a linear function of $x$ of the form $E = mx + c$,where $m = \frac{ia}{A}$ and $c = \frac{i\rho_0}{A}$.
Since $\rho_0 > 0$,the intercept $c$ is non-zero,and the slope $m$ is positive.
Thus,the graph of $E$ versus $x$ is a straight line with a positive slope and a positive $y$-intercept. This corresponds to the graph shown in option $(B)$.

(B) Tungsten is a metal (conductor). For metals,the resistance increases with an increase in temperature because the collision frequency of electrons increases.
Carbon is a non-metal (semiconductor). For semiconductors,the resistance decreases with an increase in temperature because more charge carriers are released as the temperature rises.
Therefore,when hot,the resistance of the tungsten filament increases,and the resistance of the carbon filament decreases.

(A) The power rating of a bulb is given by $P = \frac{V^2}{R}$,where $V$ is the rated voltage and $R$ is the resistance of the filament.
Since $V$ is constant for both bulbs,$P \propto \frac{1}{R}$.
The resistance of a wire is given by $R = \rho \frac{L}{A}$,where $L$ is the length and $A$ is the cross-sectional area (thickness).
Given that the lengths $L$ are the same and the material (resistivity $\rho$) is the same,$R \propto \frac{1}{A}$.
Substituting this into the power relation,we get $P \propto \frac{1}{1/A} \implies P \propto A$.
Therefore,the bulb with higher power $(100 \, W)$ must have a larger cross-sectional area (thicker filament) compared to the bulb with lower power $(60 \, W)$.

(C) The power rating of a bulb is given by the formula $P = \frac{V^2}{R}$,where $V$ is the rated voltage and $R$ is the resistance of the bulb.
From this relation,we can see that $R = \frac{V^2}{P}$.
Since the rated voltage $V$ is the same for all bulbs,the resistance $R$ is inversely proportional to the power $P$ $(R \propto \frac{1}{P})$.
Therefore,the bulb with the highest power rating will have the minimum resistance.
Comparing the given power ratings ($40\,W$,$60\,W$,and $100\,W$),the $100\,W$ bulb has the highest power.
Thus,the $100\,W$ bulb has the minimum resistance.

(A) The resistance of a conductor increases with an increase in temperature because the collisions of electrons with the lattice ions increase.
For a conductor,the relationship is given by $R_t = R_0(1 + \alpha \Delta T)$,where $\alpha$ is the temperature coefficient of resistance.
Since $R_t > R_0$ for an increase in temperature,the value of $\alpha$ must be positive.
Therefore,the temperature coefficient of resistance for a conductor is always positive.

(D) The resistance of a conductor increases with an increase in temperature because the collision frequency of electrons with lattice ions increases. Thus,conductors have a positive temperature coefficient of resistance.
In contrast,the resistance of a semiconductor decreases with an increase in temperature because more charge carriers (electrons and holes) are generated due to thermal excitation,which increases the conductivity. Thus,semiconductors have a negative temperature coefficient of resistance.
Since the resistance of $P$ increases with temperature,$P$ is a conductor.
Since the resistance of $Q$ decreases with temperature,$Q$ is a semiconductor.
Therefore,the correct option is $(d)$.

(C) In metals,the number of free charge carriers (electrons) is essentially constant with temperature. As temperature increases,the scattering of electrons by lattice vibrations increases,leading to an increase in resistance.
In semiconductors,the number of charge carriers (electrons and holes) increases exponentially with temperature due to the thermal excitation of electrons from the valence band to the conduction band. This increase in the number of charge carriers dominates over the scattering effect,causing the resistance to decrease as temperature increases.
Therefore,the fundamental difference arises due to the variation of the number of charge carriers with temperature.

(D) The temperature coefficient of resistance $(\alpha)$ is defined as the fractional change in resistance per unit change in temperature.
For metals like Copper, Aluminium, and Iron, the resistance increases with an increase in temperature, resulting in a positive temperature coefficient of resistance.
For semiconductors like Germanium, the number of charge carriers increases significantly with an increase in temperature, leading to a decrease in resistance.
Therefore, semiconductors exhibit a negative temperature coefficient of resistance.

(C) The resistivity of a material depends on the temperature coefficient of resistance,denoted by $\alpha$.
For conductors,the value of $\alpha$ is positive. As temperature increases,the collision frequency of electrons with the lattice increases,leading to an increase in resistivity.
For semiconductors,the value of $\alpha$ is negative. As temperature increases,more charge carriers (electrons and holes) are generated due to the breaking of covalent bonds,which significantly increases conductivity and thus decreases resistivity.
Therefore,with an increase in temperature,the resistivity of a conductor increases and that of a semiconductor decreases.

(C) In metallic conductors,the number of charge carriers (free electrons) is constant,but as temperature increases,the amplitude of vibration of lattice ions increases,leading to more frequent collisions (scattering),which increases resistance.
In semiconductors,the number of charge carriers (electrons and holes) increases exponentially with temperature due to the breaking of covalent bonds,which significantly decreases resistance.
Therefore,the primary reason for the difference in temperature dependence is the change in the number of charge carriers with temperature.

(A) 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 cylindrical,the area $A = \pi r^2$,where $r$ is the radius.
Therefore,$R = \rho \frac{L}{\pi r^2}$.
When the length $L$ is doubled $(L' = 2L)$ and the radius $r$ is doubled $(r' = 2r)$,the new resistance $R'$ is:
$R' = \rho \frac{2L}{\pi (2r)^2} = \rho \frac{2L}{\pi (4r^2)} = \frac{1}{2} \left( \rho \frac{L}{\pi r^2} \right) = \frac{R}{2}$.
Resistivity $(\rho)$ is a material property that depends only on the nature of the material and temperature,not on the dimensions of the wire. Thus,it remains unchanged.
Therefore,the resistance is halved and the resistivity remains unchanged.

(C) The resistance $R$ of a wire is given by the formula $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$,where $\rho$ is the resistivity,$l$ is the length,and $r$ is the radius.
Since all wires are made of the same material,$\rho$ is constant.
For option $A$: $R_A \propto \frac{40}{1^2} = 40$.
For option $B$: $R_B \propto \frac{40}{2^2} = 10$.
For option $C$: $R_C \propto \frac{80}{1^2} = 80$.
For option $D$: $R_D \propto \frac{80}{2^2} = 20$.
Comparing the values,$R_C$ is the highest. Therefore,the wire with radius $1 \, mm$ and length $80 \, m$ has the highest resistance.

(A) The resistance $R$ of a wire is given by $R = \frac{\rho l}{A}$.
Since the volume $V = A \times l$ is constant,we can write $A = \frac{V}{l}$.
Substituting this into the resistance formula: $R = \frac{\rho l}{(V/l)} = \frac{\rho l^2}{V}$.
Since $\rho$ (resistivity) and $V$ (volume) are constant,we have $R \propto l^2$.
When the length $l$ is doubled $(l' = 2l)$,the new resistance $R'$ becomes $R' \propto (2l)^2 = 4l^2$.
Therefore,$R' = 4R$,which means the resistance becomes four times the original value.

(A) superconductor is a material that exhibits zero electrical resistance when cooled below a characteristic critical temperature.
Since electrical conductivity $(\sigma)$ is the reciprocal of electrical resistivity $(\rho)$, i.e., $\sigma = 1 / \rho$.
As the resistivity $(\rho)$ of a superconductor is zero, its conductivity $(\sigma)$ becomes $1 / 0$, which is infinite.
Therefore, the conductivity of a superconductor is infinite.

(A) Copper is a metal (conductor). For metals,the resistivity $\rho$ decreases as temperature $T$ decreases because the relaxation time increases.
Silicon is a semiconductor. For semiconductors,the resistivity $\rho$ increases as temperature $T$ decreases because the number of free charge carriers decreases exponentially with temperature.
Therefore,when cooled from $300 \, K$ to $60 \, K$,the resistivity of copper decreases and the resistivity of silicon increases.

(C) Standard resistors are designed to maintain a constant resistance value over a wide range of temperatures.
For a material to be suitable for a standard resistor, its resistance should not change significantly with temperature.
The temperature coefficient of resistance $(\alpha)$ for such materials must be very low or negligible.
Tungsten is often used in specific applications, but for standard resistors, alloys like Manganin or Constantan are typically preferred because they have a very low temperature coefficient of resistance.
However, in the context of this specific question, the property that makes a material suitable for a standard resistor is that its resistance remains stable, which implies a negligible temperature coefficient of resistance.

(A) 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 area $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$,we can write $R = \rho \frac{4L}{\pi d^2}$.
Let the first wire have length $L_1 = L$ and diameter $d_1 = d$. Its resistance is $R_1 = \rho \frac{4L}{\pi d^2}$.
For the second wire,$L_2 = 2L$ and $d_2 = 2d$. Its resistance is $R_2 = \rho \frac{4(2L)}{\pi (2d)^2} = \rho \frac{8L}{4\pi d^2} = \rho \frac{2L}{\pi d^2}$.
Comparing $R_1$ and $R_2$: $R_1 = 2 \times (\rho \frac{2L}{\pi d^2}) = 2 R_2$.
Thus,the resistance of the first wire is double the resistance of the second wire.

(D) The specific resistance, also known as resistivity $(\rho)$, is an intrinsic property of the material of the wire.
It depends only on the nature of the material and the temperature of the conductor.
It does not depend on the physical dimensions of the wire, such as its length $(l)$ or cross-sectional area $(A)$.
Therefore, if the length of the wire is doubled, the resistivity remains unchanged.

(D) The resistance $R$ of a wire is given by $R = \rho \frac{l}{A}$.
Since the volume $V = lA$ 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 constant,$R \propto l^2$.
For small changes,the percentage change in resistance is given by $\frac{\Delta R}{R} \times 100 \approx 2 \times (\frac{\Delta l}{l} \times 100)$.
Given that the length increases by $0.1 \%$,we have $\frac{\Delta l}{l} \times 100 = 0.1 \%$.
Therefore,the percentage change in resistance is $2 \times 0.1 \% = 0.2 \%$.
Since the length increases,the resistance increases by $0.2 \%$.

(B) 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,we have $A = \frac{V}{\ell}$.
Substituting this into the resistance formula,we get $R = \rho \frac{\ell^2}{V}$.
Since $\rho$ and $V$ are constant,$R \propto \ell^2$.
Taking the derivative or using the approximation for small changes: $\frac{\Delta R}{R} \approx 2 \frac{\Delta \ell}{\ell}$.
Given $\frac{\Delta \ell}{\ell} = 0.1\%$,the percentage change in resistance is $\frac{\Delta R}{R} \times 100 = 2 \times 0.1\% = 0.2\%$.

(A) The power of a heater is given by $P = \frac{V^2}{R}$. Assuming the voltage $V$ remains constant,$P \propto \frac{1}{R}$.
The resistance at temperature $T$ is given by $R_T = R_0(1 + \alpha T)$.
Therefore,$\frac{P_{200}}{P_{800}} = \frac{R_{800}}{R_{200}} = \frac{R_0(1 + \alpha \times 800)}{R_0(1 + \alpha \times 200)}$.
Substituting the given values: $\alpha = 4 \times 10^{-4} \ ^oC^{-1}$,$P_{800} = 500 \ W$.
$\frac{P_{200}}{500} = \frac{1 + (4 \times 10^{-4} \times 800)}{1 + (4 \times 10^{-4} \times 200)} = \frac{1 + 0.32}{1 + 0.08} = \frac{1.32}{1.08}$.
$P_{200} = 500 \times \frac{1.32}{1.08} \approx 611.11 \ W$.
Rounding to the nearest integer,the power is $611 \ W$.

(B) Specific resistance, also known as resistivity $(\rho)$, is an intrinsic property of a material.
It depends on the nature of the material and the temperature.
For metallic conductors, resistivity increases with an increase in temperature due to increased scattering of electrons by vibrating lattice ions.
Resistivity does not depend on the physical dimensions of the conductor, such as its length or cross-sectional area.

(D) Let the initial length be $\ell$ and the initial resistance be $R = \rho \frac{\ell}{A}$.
Since the length is increased by $100\,\%$,the new length $\ell' = \ell + 1.00\ell = 2\ell$.
Since the volume $V = A \times \ell$ remains constant,$A' \ell' = A \ell \implies A' (2\ell) = A \ell \implies A' = \frac{A}{2}$.
The new resistance $R' = \rho \frac{\ell'}{A'} = \rho \frac{2\ell}{A/2} = 4 \left( \rho \frac{\ell}{A} \right) = 4R$.
The percentage change in resistance is given by $\frac{R' - R}{R} \times 100\,\% = \frac{4R - R}{R} \times 100\,\% = 3 \times 100\,\% = 300\,\%$.

(A) The specific resistance or resistivity $\rho$ of a material depends on temperature.
For metals like copper,the resistivity decreases as the temperature decreases because the scattering of electrons by lattice vibrations reduces.
For semiconductors like silicon,the resistivity increases as the temperature decreases because the number of charge carriers (electrons and holes) decreases significantly due to the freezing out of carriers.
Therefore,when cooling from $300 \ K$ to $60 \ K$,the specific resistance of copper decreases,while the specific resistance of silicon increases.

(C) The resistance $R_t$ at temperature $t$ is given by the formula: $R_t = R_0(1 + \alpha t)$.
Given that $R_t = 3R_0$ and $\alpha = 4 \times 10^{-3}{}^{\circ} C^{-1}$.
Substituting these values into the equation:
$3R_0 = R_0(1 + 4 \times 10^{-3} \times t)$
$3 = 1 + 4 \times 10^{-3} \times t$
$2 = 4 \times 10^{-3} \times t$
$t = \frac{2}{4 \times 10^{-3}} = \frac{2000}{4} = 500 ^{\circ} C$.
Therefore, the temperature is $500 ^{\circ} C$.