Resistance of wire, Resistivity and Conductivity Questions in English

(A) The resistance $R$ of a wire is given by $R = \rho \frac{l}{A}$. Since the volume $V = l \times A$ remains constant during stretching,$A = \frac{V}{l}$.
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 $l' = l + 0.1l = 1.1l$.
The new resistance $R'$ is given by $R' = R \times (\frac{l'}{l})^2$.
$R' = 10 \times (1.1)^2 = 10 \times 1.21 = 12.1\, \Omega$.
Note: For small changes,the approximation $\Delta R/R \approx 2 \Delta l/l$ gives a $20\%$ increase,but the exact calculation yields $12.1\, \Omega$.

(D) Since both wires are made of copper,their resistivities are equal,i.e.,$\rho_A = \rho_B = \rho$.
The resistance of a wire is given by the formula $R = \rho \frac{l}{A}$,where $l$ is the length and $A$ is the cross-sectional area.
For wire $A$: $R_A = \rho \frac{3}{A}$.
For wire $B$: $R_B = \rho \frac{5}{A}$.
Comparing the two,since $3 < 5$ and the material and diameter (and thus area $A$) are the same,it follows that $R_A < R_B$.
Therefore,the correct relation is $R_A < R_B$ with $\rho_A = \rho_B$.

(C) The formula for the resistance of a conductor at temperature $T$ is given by $R_T = R_0 (1 + \alpha \Delta T)$.
Here,$R_T = 3R_0$,$\alpha = 4 \times 10^{-3} / ^oC$,and $\Delta T = T - 0 = T$.
Substituting the values: $3R_0 = R_0 (1 + 4 \times 10^{-3} \times T)$.
Dividing both sides by $R_0$: $3 = 1 + 0.004T$.
$2 = 0.004T$.
$T = 2 / 0.004 = 2000 / 4 = 500 ^oC$.
Therefore,the temperature is $500 ^oC$.

(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.
For $t = 50^{\circ}C$, $R_{50} = R_0(1 + 50\alpha) = 5$ --- $(1)$
For $t = 100^{\circ}C$, $R_{100} = R_0(1 + 100\alpha) = 6$ --- $(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{R_0(1 + 100\alpha)}{R_0(1 + 50\alpha)} = \frac{6}{5}$
$5(1 + 100\alpha) = 6(1 + 50\alpha)$
$5 + 500\alpha = 6 + 300\alpha$
$200\alpha = 1$
$\alpha = \frac{1}{200} = 0.005^{\circ}C^{-1}$
Substituting $\alpha$ into equation $(1)$:
$5 = R_0(1 + 50 \times 0.005)$
$5 = R_0(1 + 0.25)$
$5 = R_0(1.25)$
$R_0 = \frac{5}{1.25} = 4\, \Omega$.

(D) Given that the resistances are equal,$R_A = R_B$.
Using the formula for resistance $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$,we have:
$\rho_A \frac{l_A}{\pi r_A^2} = \rho_B \frac{l_B}{\pi r_B^2}$
Given $\rho_B = 2\rho_A$ and $d_B = 2d_A$ (which implies $r_B = 2r_A$):
$\rho_A \frac{l_A}{r_A^2} = (2\rho_A) \frac{l_B}{(2r_A)^2}$
$\rho_A \frac{l_A}{r_A^2} = 2\rho_A \frac{l_B}{4r_A^2}$
$l_A = \frac{1}{2} l_B$
Therefore,$\frac{l_B}{l_A} = 2$.

(D) Let $R_0$ be the resistance of both conductors at $0\,^{\circ}C$.
For the first conductor at $t_1\,^{\circ}C$,the resistance is $R_1 = R_0(1 + \alpha_1 t_1)$.
For the second conductor at $t_2\,^{\circ}C$,the resistance is $R_2 = R_0(1 + \alpha_2 t_2)$.
Given that $R_1 = R_2$,we have:
$R_0(1 + \alpha_1 t_1) = R_0(1 + \alpha_2 t_2)$
$1 + \alpha_1 t_1 = 1 + \alpha_2 t_2$
$\alpha_1 t_1 = \alpha_2 t_2$
Therefore,the ratio $\alpha_1 / \alpha_2 = t_2 / t_1$.

(D) 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 at operating temperature.
Since the resistance increases with temperature,the resistance at room temperature follows the same trend as the resistance at operating temperature.
For a fixed voltage $V$,$R = \frac{V^2}{P}$.
Thus,$R_{100} = \frac{V^2}{100}$,$R_{60} = \frac{V^2}{60}$,and $R_{40} = \frac{V^2}{40}$.
Comparing these values,we get $R_{40} > R_{60} > R_{100}$.
Taking the reciprocal,we get $\frac{1}{R_{100}} > \frac{1}{R_{60}} > \frac{1}{R_{40}}$.

(B) The formula for resistance at temperature $T$ is $R_T = R_0[1 + \alpha(T - T_0)]$.
Given: $\alpha = 0.00125\,^{\circ}C^{-1}$,$R_1 = 1\,\Omega$ at $T_1 = 300\,K$ $(27\,^{\circ}C)$,and $R_2 = 2\,\Omega$ at $T_2 = ?$.
Using the relation $R_2 = R_1[1 + \alpha(t_2 - t_1)]$,where $t_1 = 27\,^{\circ}C$:
$2 = 1[1 + 0.00125(t_2 - 27)]$
$2 - 1 = 0.00125(t_2 - 27)$
$1 = 0.00125(t_2 - 27)$
$t_2 - 27 = \frac{1}{0.00125} = 800$
$t_2 = 800 + 27 = 827\,^{\circ}C$.
Converting to Kelvin: $T_2 = 827 + 273 = 1100\,K$.

(D) The resistance $R$ at temperature $\theta$ is given by the formula: $R = R_0(1 + \alpha \theta)$.
Given:
$R_0 = 5\,\Omega$ (at $0\,^oC$)
$R_{100} = 5.75\,\Omega$ (at $100\,^oC$)
$R_{\theta} = 5.15\,\Omega$ (at unknown temperature $\theta$)
First,calculate the temperature coefficient of resistance $\alpha$:
$R_{100} = R_0(1 + \alpha \times 100)$
$5.75 = 5(1 + 100\alpha)$
$1.15 = 1 + 100\alpha$
$0.15 = 100\alpha$
$\alpha = 0.0015\,\,^oC^{-1} = 1.5 \times 10^{-3}\,\,^oC^{-1}$.
Now,find the unknown temperature $\theta$ for $R_{\theta} = 5.15\,\Omega$:
$R_{\theta} = R_0(1 + \alpha \theta)$
$5.15 = 5(1 + 0.0015 \times \theta)$
$1.03 = 1 + 0.0015 \times \theta$
$0.03 = 0.0015 \times \theta$
$\theta = \frac{0.03}{0.0015} = 20\,^oC$.

(B) Given: Initial temperature $T_1 = 30^{\circ}C$,Initial resistance $R_1 = 10\, \Omega$,Temperature coefficient $\alpha = 0.002\, ^{\circ}C^{-1}$.
We want to increase the resistance by $10\%$,so the new resistance $R_2 = R_1 + 0.10 R_1 = 1.10 R_1 = 11\, \Omega$.
The formula for resistance at temperature $T$ is $R = R_0(1 + \alpha T)$.
For $T_1 = 30^{\circ}C$: $R_1 = R_0(1 + 30\alpha) = 10\, \Omega$.
For $T_2$: $R_2 = R_0(1 + \alpha T_2) = 11\, \Omega$.
Dividing the two equations: $\frac{R_2}{R_1} = \frac{1 + \alpha T_2}{1 + 30\alpha} = \frac{11}{10} = 1.1$.
$1 + \alpha T_2 = 1.1(1 + 30\alpha) = 1.1 + 33\alpha$.
$\alpha T_2 = 0.1 + 33\alpha$.
$T_2 = \frac{0.1}{\alpha} + 33$.
Substituting $\alpha = 0.002$: $T_2 = \frac{0.1}{0.002} + 33 = 50 + 33 = 83^{\circ}C$.

(D) The resistance at temperature $\theta$ is given by $R_{\theta} = R_0(1 + \alpha\theta)$,where $R_0$ is the resistance at $0\,^{\circ}\text{C}$.
At $50\,^{\circ}\text{C}$,$5 = R_0(1 + 50\alpha)$ ... $(1)$
At $100\,^{\circ}\text{C}$,$6 = R_0(1 + 100\alpha)$ ... $(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{5}{6} = \frac{1 + 50\alpha}{1 + 100\alpha}$
$5(1 + 100\alpha) = 6(1 + 50\alpha)$
$5 + 500\alpha = 6 + 300\alpha$
$200\alpha = 1 \implies \alpha = \frac{1}{200}$
Substituting $\alpha$ in equation $(1)$:
$5 = R_0(1 + 50 \times \frac{1}{200})$
$5 = R_0(1 + \frac{1}{4})$
$5 = R_0(\frac{5}{4})$
$R_0 = 4\,\Omega$.

(A) Given: Dimensions $l_1 = 50 \ cm = 0.5 \ m$,$A_1 = 1 \ cm \times 1 \ cm = 10^{-4} \ m^2$. Resistivity $\rho = 3.5 \times 10^{-5} \ \Omega \cdot m$.
Case $1$: Resistance between square ends $(l = 0.5 \ m, A = 10^{-4} \ m^2)$:
$R_1 = \rho \frac{l}{A} = 3.5 \times 10^{-5} \times \frac{0.5}{10^{-4}} = 3.5 \times 0.5 \times 10^{-1} = 1.75 \times 10^{-1} = 17.5 \times 10^{-2} \ \Omega$.
Case $2$: Resistance between rectangular ends $(l = 1 \ cm = 0.01 \ m, A = 1 \ cm \times 50 \ cm = 50 \ cm^2 = 50 \times 10^{-4} \ m^2)$:
$R_2 = \rho \frac{l}{A} = 3.5 \times 10^{-5} \times \frac{0.01}{50 \times 10^{-4}} = 3.5 \times 10^{-5} \times \frac{10^{-2}}{50 \times 10^{-4}} = 3.5 \times 10^{-5} \times \frac{1}{50} \times 10^2 = 0.07 \times 10^{-4} = 7 \times 10^{-6} \ \Omega$.

(C) The resistance 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.
For $t = 50^{\circ}C$,$R_{50} = R_0 [1 + 50\alpha] = 5 \quad \dots(1)$
For $t = 100^{\circ}C$,$R_{100} = R_0 [1 + 100\alpha] = 6 \quad \dots(2)$
Dividing equation $(1)$ by $(2)$:
$\frac{5}{6} = \frac{1 + 50\alpha}{1 + 100\alpha}$
$5(1 + 100\alpha) = 6(1 + 50\alpha)$
$5 + 500\alpha = 6 + 300\alpha$
$200\alpha = 1 \implies \alpha = \frac{1}{200} \, ^{\circ}C^{-1}$
Substituting $\alpha$ into equation $(1)$:
$5 = R_0 [1 + 50 \times \frac{1}{200}]$
$5 = R_0 [1 + \frac{1}{4}]$
$5 = R_0 [\frac{5}{4}]$
$R_0 = 4 \, \Omega$

(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,$V = A_1 L_1 = A_2 L_2$.
Therefore,$L_1 \pi r_1^2 = L_2 \pi r_2^2$,which implies $\frac{L_2}{L_1} = \left( \frac{r_1}{r_2} \right)^2$.
The ratio of resistances is $\frac{R_2}{R_1} = \frac{L_2}{L_1} \cdot \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{r_1}{r_2} \right)^2 \cdot \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{r_1}{r_2} \right)^4$.
Given $r_2 = \frac{r_1}{2}$,we have $\frac{r_1}{r_2} = 2$.
Substituting this into the ratio,$\frac{R_2}{R} = (2)^4 = 16$.
Thus,the new resistance is $R' = 16\, R$.

(D) The resistance of a conductor at temperature $t$ is given by $R_t = R_0(1 + \alpha t)$.
Given $R_1 = 20\,\Omega$ at $t_1 = 20\,^{\circ}C$ and $R_2 = 60\,\Omega$ at $t_2 = 500\,^{\circ}C$.
Using the ratio: $\frac{R_1}{R_2} = \frac{1 + \alpha t_1}{1 + \alpha t_2} \Rightarrow \frac{20}{60} = \frac{1 + 20\alpha}{1 + 500\alpha}$.
Solving for $\alpha$: $\frac{1}{3} = \frac{1 + 20\alpha}{1 + 500\alpha} \Rightarrow 1 + 500\alpha = 3 + 60\alpha \Rightarrow 440\alpha = 2 \Rightarrow \alpha = \frac{1}{220}\,^{\circ}C^{-1}$.
Now,to find the temperature $t$ where $R = 25\,\Omega$,we use: $\frac{R_1}{R} = \frac{1 + \alpha t_1}{1 + \alpha t}$.
$\frac{20}{25} = \frac{1 + 20(1/220)}{1 + t(1/220)} \Rightarrow 0.8 = \frac{1 + 1/11}{1 + t/220} \Rightarrow 0.8 = \frac{12/11}{1 + t/220}$.
$1 + \frac{t}{220} = \frac{12}{11 \times 0.8} = \frac{12}{8.8} = \frac{120}{88} = \frac{15}{11}$.
$\frac{t}{220} = \frac{15}{11} - 1 = \frac{4}{11} \Rightarrow t = \frac{4}{11} \times 220 = 80\,^{\circ}C$.

(A) The formula for resistance is $R = \rho \frac{l}{A}$.
Given: Resistivity $\rho = 50 \times 10^{-8} \, \Omega m$,length $l = 50 \, cm = 0.5 \, m$.
For a cube,the area of cross-section $A = l^2 = (0.5 \, m)^2 = 0.25 \, m^2$.
Substituting the values into the formula:
$R = (50 \times 10^{-8}) \times \frac{0.5}{0.25} = (50 \times 10^{-8}) \times 2 = 100 \times 10^{-8} \, \Omega = 10^{-6} \, \Omega$.

(B) The resistance of a conductor is given by $R = \rho \frac{l}{A}$.
For the wire: $l_1 = 1.0\, m$,$d_1 = 0.6\, cm = 0.6 \times 10^{-2}\, m$,$r_1 = 0.3 \times 10^{-2}\, m$,$R_1 = 3.0 \times 10^{-3}\, \Omega$.
$R_1 = \rho \frac{l_1}{\pi r_1^2} \Rightarrow \rho = \frac{R_1 \pi r_1^2}{l_1}$.
For the disc: $l_2 = 1.0\, mm = 1.0 \times 10^{-3}\, m$,$d_2 = 2.0\, cm = 2.0 \times 10^{-2}\, m$,$r_2 = 1.0 \times 10^{-2}\, m$.
$R_2 = \rho \frac{l_2}{\pi r_2^2} = \left( \frac{R_1 \pi r_1^2}{l_1} \right) \frac{l_2}{\pi r_2^2} = R_1 \left( \frac{l_2}{l_1} \right) \left( \frac{r_1}{r_2} \right)^2$.
Substituting the values: $R_2 = (3.0 \times 10^{-3}) \times \left( \frac{1.0 \times 10^{-3}}{1.0} \right) \times \left( \frac{0.3 \times 10^{-2}}{1.0 \times 10^{-2}} \right)^2$.
$R_2 = (3.0 \times 10^{-6}) \times (0.3)^2 = (3.0 \times 10^{-6}) \times 0.09 = 0.27 \times 10^{-6} = 2.7 \times 10^{-7}\, \Omega$.

(D) The resistance $R$ of a conductor is given by $R = \rho \frac{l}{A}$.
Here,the cross-sectional area $A$ is the area of the hollow part: $A = \pi (r_2^2 - r_1^2)$.
Given: $r_2 = 10 \, cm = 0.1 \, m$,thickness $t = 5 \, mm = 0.005 \, m$.
So,inner radius $r_1 = r_2 - t = 10 \, cm - 0.5 \, cm = 9.5 \, cm = 0.095 \, m$.
$A = \pi [(0.1)^2 - (0.095)^2] = \pi [0.01 - 0.009025] = \pi [0.000975] \, m^2$.
Now,$R = (1.7 \times 10^{-8}) \times \frac{5}{\pi \times 0.000975} \approx 2.77 \times 10^{-5} \, \Omega$.

(A) The resistance of a wire is given by $R = \rho \frac{l}{A}$.
Since the volume $V = A \times l$ remains constant,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,$R \propto l^2$.
Taking the derivative,we get $\frac{\Delta R}{R} = 2 \frac{\Delta l}{l}$.
Given $\frac{\Delta l}{l} = 0.1\%$,the percentage change in resistance is $\frac{\Delta R}{R} \times 100 = 2 \times 0.1\% = 0.2\%$.

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

(A) The resistance of a wire is given by $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.
Since all wires are made of the same material ($\rho$ is constant) and have the same length ($l$ is constant),the resistance depends only on the cross-sectional area $A$.
For wire $A$: The cross-section is a square frame with outer side $\sqrt{3}a$ and inner side $\sqrt{2}a$. Area $A_A = (\sqrt{3}a)^2 - (\sqrt{2}a)^2 = 3a^2 - 2a^2 = a^2$.
For wire $B$: The cross-section is a square frame with outer side $\sqrt{2}a$ and inner side $a$. Area $A_B = (\sqrt{2}a)^2 - (a)^2 = 2a^2 - a^2 = a^2$.
For wire $C$: The cross-section is a solid square with side $a$. Area $A_C = a^2$.
Since $A_A = A_B = A_C = a^2$,it follows that $R_A = R_B = R_C$.

(B) Given: Resistivity $\rho = 3 \times 10^{-7} \, \Omega \cdot m$.
Dimensions of the block are $1 \, cm \times 1 \, cm \times 100 \, cm$.
For the resistance between the rectangular faces (the $1 \, cm \times 100 \, cm$ faces),the length $l$ of the current path is the remaining dimension,$l = 1 \, cm = 10^{-2} \, m$.
The area of cross-section $A$ is $1 \, cm \times 100 \, cm = 100 \, cm^2 = 100 \times 10^{-4} \, m^2 = 10^{-2} \, m^2$.
The formula for resistance is $R = \rho \frac{l}{A}$.
Substituting the values: $R = (3 \times 10^{-7} \, \Omega \cdot m) \times \frac{10^{-2} \, m}{10^{-2} \, m^2}$.
$R = 3 \times 10^{-7} \, \Omega$.

(A) The resistance of a wire is given by $R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}$.
Since the lengths $L$ and resistivity $\rho$ are the same for both segments,the resistance $R$ is inversely proportional to the square of the radius: $R \propto \frac{1}{r^2}$.
Let $r_{AB} = 3r$ and $r_{BC} = r$. Then the ratio of resistances is $\frac{R_{AB}}{R_{BC}} = \frac{r_{BC}^2}{r_{AB}^2} = \frac{r^2}{(3r)^2} = \frac{1}{9}$.
Since the segments $AB$ and $BC$ are in series,the same current $I$ flows through both.
According to Ohm's Law,the potential difference $V = IR$.
Therefore,the ratio of potential differences is $\frac{V_{AB}}{V_{BC}} = \frac{I R_{AB}}{I R_{BC}} = \frac{R_{AB}}{R_{BC}} = \frac{1}{9}$.

(C) Let the initial length be $l$ and area be $A$. The initial resistance is $R = \rho \frac{l}{A}$.
When the wire is stretched by $10\%$,the new length $l' = l + 0.1l = 1.1l$.
Since the volume $V = Al$ remains constant,$A'l' = Al \Rightarrow A' = \frac{Al}{1.1l} = \frac{A}{1.1}$.
The new resistance $R' = \rho \frac{l'}{A'} = \rho \frac{1.1l}{A/1.1} = (1.1)^2 \rho \frac{l}{A} = 1.21 R$.
Specific resistance (resistivity) $\rho$ is a property of the material and depends only on the temperature,not on the dimensions of the wire.
Therefore,the new resistance becomes $1.21$ times the original,and the specific resistance remains the same.

(C) The resistance of a wire is given by $R = \rho \frac{l}{A} = 4 \,\Omega$ .... $(i)$
When the wire is stretched to twice its original length,the new length $l^{\prime} = 2l$.
Since the volume of the wire remains constant during stretching,$V = lA = l^{\prime}A^{\prime}$.
Substituting $l^{\prime} = 2l$,we get $lA = (2l)A^{\prime}$,which implies $A^{\prime} = \frac{A}{2}$.
The new resistance $R^{\prime}$ is given by $R^{\prime} = \rho \frac{l^{\prime}}{A^{\prime}}$.
Substituting the values,$R^{\prime} = \rho \frac{2l}{A/2} = 4 \left( \rho \frac{l}{A} \right)$.
Using equation $(i)$,$R^{\prime} = 4 \times 4 \,\Omega = 16 \,\Omega$.

(B) Since both metal wires are of identical dimensions,their length and area of cross-section are the same. Let them be $l$ and $A$ respectively.
The resistance of the first wire is $R_1 = \frac{l}{\sigma_1 A}$ ...$(i)$
The resistance of the second wire is $R_2 = \frac{l}{\sigma_2 A}$ ...$(ii)$
Since they are connected in series,their effective resistance is $R_s = R_1 + R_2$.
$R_s = \frac{l}{\sigma_1 A} + \frac{l}{\sigma_2 A} = \frac{l}{A} \left( \frac{1}{\sigma_1} + \frac{1}{\sigma_2} \right)$ ...$(iii)$
If $\sigma_{eff}$ is the effective conductivity of the combination,then the total length is $2l$ and the total resistance is $R_s = \frac{2l}{\sigma_{eff} A}$ ...$(iv)$
Equating equations $(iii)$ and $(iv)$,we get:
$\frac{2l}{\sigma_{eff} A} = \frac{l}{A} \left( \frac{\sigma_1 + \sigma_2}{\sigma_1 \sigma_2} \right)$
$\frac{2}{\sigma_{eff}} = \frac{\sigma_1 + \sigma_2}{\sigma_1 \sigma_2}$
$\sigma_{eff} = \frac{2 \sigma_1 \sigma_2}{\sigma_1 + \sigma_2}$

(B) The resistance of a wire of length $l$,cross-sectional area $A$,and resistivity $\rho$ is given by $R = \rho \frac{l}{A}$.
When the wire is stretched to $n$ times its original length,the new length becomes $l' = nl$.
Since the volume of the wire remains constant during the process,$V = A \cdot l = A' \cdot l'$.
Substituting $l' = nl$,we get $A' = \frac{A \cdot l}{nl} = \frac{A}{n}$.
The new resistance $R'$ is given by $R' = \rho \frac{l'}{A'}$.
Substituting the values of $l'$ and $A'$,we get $R' = \rho \frac{nl}{A/n} = n^2 \left( \rho \frac{l}{A} \right) = n^2 R$.

(C) Let $\rho$ be the resistivity of the material.
The resistance $R$ of a conductor is given by $R = \rho \frac{L}{A}$,where $L$ is the length of the conductor in the direction of current flow and $A$ is the cross-sectional area perpendicular to the current flow.
$1$. For contact $A-A$ (Top and bottom faces):
The length $L = x$ and the area $A = 2x \times 4x = 8x^2$.
$R_{AA} = \rho \frac{x}{8x^2} = \frac{\rho}{8x}$.
$2$. For contact $B-B$ (Left and right faces):
The length $L = 2x$ and the area $A = x \times 4x = 4x^2$.
$R_{BB} = \rho \frac{2x}{4x^2} = \frac{\rho}{2x} = \frac{4\rho}{8x}$.
$3$. For contact $C-C$ (Front and rear faces):
The length $L = 4x$ and the area $A = x \times 2x = 2x^2$.
$R_{CC} = \rho \frac{4x}{2x^2} = \frac{2\rho}{x} = \frac{16\rho}{8x}$.
Comparing the resistances,$R_{CC} > R_{BB} > R_{AA}$.
Thus,the maximum electrical resistance is obtained for contact $C-C$.

(A) We know that resistance $R$ is given by $R = \rho \frac{l}{S}$.
Conductivity $\sigma$ is the reciprocal of resistivity $\rho$,so $\sigma = \frac{1}{\rho}$.
Conductance $G$ is the reciprocal of resistance $R$,so $G = \frac{1}{R}$,which implies $R = \frac{1}{G}$.
Substituting $R = \frac{1}{G}$ into the resistance formula: $\frac{1}{G} = \rho \frac{l}{S}$.
Rearranging for $\rho$: $\rho = \frac{S}{Gl}$.
Since $\sigma = \frac{1}{\rho}$,we have $\sigma = \frac{Gl}{S}$.
Alternatively,using the given options,we check the expression $\sigma = \frac{Gl}{S}$.
Wait,let us re-evaluate the expression $\sigma = \frac{1}{\rho}$.
From $R = \rho \frac{l}{S}$,we have $\rho = \frac{RS}{l}$.
Thus,$\sigma = \frac{1}{\rho} = \frac{l}{RS}$.
Since $G = \frac{1}{R}$,we can write $\sigma = \frac{Gl}{S}$.
Looking at the options provided,if we consider $\sigma = \frac{1}{\rho}$ and $G = \frac{1}{R}$,then $GR = 1$. Therefore,$\sigma = \frac{GR}{\rho}$ is mathematically consistent as $\sigma = \frac{1}{\rho} \times 1 = \frac{1}{\rho} \times (GR) = \frac{GR}{\rho}$.

(C) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{l}{A_{cs}}$,where $\rho$ is the resistivity,$l$ is the length,and $A_{cs}$ is the cross-sectional area.
For a square plate of side length $l$ and thickness $t$,the cross-sectional area through which current flows is $A_{cs} = l \times t$.
Thus,the resistance is $R = \rho \frac{l}{l \times t} = \frac{\rho}{t}$.
Since both plates $A$ and $B$ are made of the same metal (same $\rho$) and have the same thickness $(t)$,the resistance of each plate is independent of its side length.
Therefore,$R_A = \frac{\rho}{t}$ and $R_B = \frac{\rho}{t}$.
The ratio of the resistances is $\frac{R_A}{R_B} = \frac{\rho/t}{\rho/t} = 1:1$.

(C) The resistance $R$ of a conductor is given by $R = \frac{\rho l}{A}$, where $\rho$ is the resistivity, $l$ is the length, and $A$ is the cross-sectional area.
For conductor $A$ (solid wire), the radius $r_A = 0.5 \, mm$. The cross-sectional area is $A_A = \pi r_A^2$.
Thus, $R_A = \frac{\rho l}{\pi r_A^2}$.
For conductor $B$ (hollow tube), the outside radius $r_{out} = 1.0 \, mm$ and the inside radius $r_{in} = 0.5 \, mm$. The cross-sectional area is $A_B = \pi (r_{out}^2 - r_{in}^2) = \pi (1.0^2 - 0.5^2) = \pi (1 - 0.25) = 0.75 \pi \, mm^2$.
Comparing the areas, $A_B = \pi (1.0^2 - 0.5^2) = 3 \pi (0.5^2) = 3 A_A$.
Since $R \propto \frac{1}{A}$, we have $\frac{R_A}{R_B} = \frac{A_B}{A_A} = \frac{3 A_A}{A_A} = 3$.

(B) Let the original length of the wire be $l$ and its original resistance be $R = \frac{\rho l}{A}$.
Let the fraction of the length stretched be $a$. The length of the part to be stretched is $al$,and the remaining part is $l(1-a)$.
The resistance of the unstretched part is $R_1 = \frac{\rho l(1-a)}{A} = R(1-a)$.
When the part of length $al$ is stretched to a new length $l'$,the total length becomes $l' + l(1-a) = \frac{3}{2}l$,so $l' = \frac{3}{2}l - l + al = l(\frac{1}{2} + a)$.
Since the volume of the stretched part remains constant,$A'l' = A(al)$,so $A' = \frac{A(al)}{l'} = \frac{A(al)}{l(\frac{1}{2} + a)} = \frac{Aa}{\frac{1}{2} + a}$.
The resistance of the stretched part is $R' = \frac{\rho l'}{A'} = \frac{\rho l(\frac{1}{2} + a)}{\frac{Aa}{\frac{1}{2} + a}} = R \frac{(\frac{1}{2} + a)^2}{a}$.
The total resistance is $R_1 + R' = 4R$.
$R(1-a) + R \frac{(\frac{1}{2} + a)^2}{a} = 4R$.
$(1-a) + \frac{\frac{1}{4} + a + a^2}{a} = 4$.
$a - a^2 + \frac{1}{4} + a + a^2 = 4a$.
$2a + \frac{1}{4} = 4a$.
$2a = \frac{1}{4} \Rightarrow a = \frac{1}{8}$.

(A) The resistivity of a conductor is given by $\rho = \frac{m}{n e^{2} \tau}$,where $m$ is the mass of an electron,$e$ is the charge on an electron,$n$ is the number of free electrons per unit volume,and $\tau$ is the relaxation time.
From this,we see that $\rho \propto \frac{1}{\tau}$.
Also,the conductivity $\sigma$ is the reciprocal of resistivity,so $\sigma = \frac{1}{\rho}$.
Therefore,$\sigma \propto \tau$.
The ratio of resistivity to conductivity is $\frac{\rho}{\sigma} = \frac{\rho}{1/\rho} = \rho^{2}$.
Since $\rho \propto \frac{1}{\tau}$,we have $\frac{\rho}{\sigma} \propto \frac{1}{\tau^{2}}$.
When the temperature of a conductor increases,the thermal velocity of electrons increases,which leads to more frequent collisions. Consequently,the relaxation time $\tau$ decreases.
Since $\frac{\rho}{\sigma} \propto \frac{1}{\tau^{2}}$,as $\tau$ decreases,the ratio $\frac{\rho}{\sigma}$ increases.

(C) The resistance at temperature $T$ is given by $R_T = R_0(1 + \alpha \Delta T)$,where $R_0$ is the resistance at $0\,^{\circ}C$.
Given:
$R_1 = 100\,\Omega$ at $T_1 = 100\,^{\circ}C$
$R_2 = 200\,\Omega$ at $T_2 = T$
$\alpha = 0.005\,^{\circ}C^{-1}$
Using the formula $R = R_0(1 + \alpha T)$:
$100 = R_0(1 + 0.005 \times 100) = R_0(1 + 0.5) = 1.5 R_0$
$R_0 = \frac{100}{1.5} = \frac{200}{3}\,\Omega$
Now,for $R_2 = 200\,\Omega$:
$200 = R_0(1 + 0.005 \times T)$
$200 = \frac{200}{3}(1 + 0.005 T)$
$3 = 1 + 0.005 T$
$2 = 0.005 T$
$T = \frac{2}{0.005} = 400\,^{\circ}C$.

(A) Given: $\rho_B = 2\rho_A$ and $d_B = 2d_A$.
Since the wires are circular,the area of cross-section is $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.
For the resistances to be equal,$R_B = R_A$.
Using the formula $R = \frac{\rho l}{A}$,we have $\frac{\rho_B l_B}{A_B} = \frac{\rho_A l_A}{A_A}$.
Substituting the values: $\frac{2\rho_A l_B}{\frac{\pi (2d_A)^2}{4}} = \frac{\rho_A l_A}{\frac{\pi d_A^2}{4}}$.
Simplifying the equation: $\frac{2\rho_A l_B}{4 A_A} = \frac{\rho_A l_A}{A_A} \Rightarrow \frac{2 l_B}{4} = l_A$.
Therefore,$\frac{l_B}{l_A} = \frac{4}{2} = 2$.

(D) We know that the resistance at temperature $t$ is given by the formula:
$R_{t} = R_{0}(1 + \alpha t)$
where $R_{t}$ is the resistance at $t\, ^\circ C$,$R_{0}$ is the resistance at $0\, ^\circ C$,and $\alpha$ is the temperature coefficient of resistance.
For $t = 50\, ^\circ C$:
$5 = R_{0}(1 + 50\alpha)$ ......$(i)$
For $t = 100\, ^\circ C$:
$6 = R_{0}(1 + 100\alpha)$ ......$(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{5}{6} = \frac{R_{0}(1 + 50\alpha)}{R_{0}(1 + 100\alpha)}$
$\frac{5}{6} = \frac{1 + 50\alpha}{1 + 100\alpha}$
Cross-multiplying:
$5(1 + 100\alpha) = 6(1 + 50\alpha)$
$5 + 500\alpha = 6 + 300\alpha$
$200\alpha = 1$
$\alpha = \frac{1}{200} = 0.005\, ^\circ C^{-1}$
Now,substitute $\alpha$ back into equation $(i)$ to find $R_{0}$:
$5 = R_{0}(1 + 50 \times 0.005)$
$5 = R_{0}(1 + 0.25)$
$5 = R_{0}(1.25)$
$R_{0} = \frac{5}{1.25} = 4\, \Omega$
Thus,the resistance at $0\, ^\circ C$ is $4\, \Omega$.

(D) Let $j$ be the current density.
Since the current $I$ spreads over a hemispherical surface of area $2 \pi r^2$,the current density is $j = \frac{I}{2 \pi r^2}$.
Using Ohm's law in the form $E = \rho j$,we get:
$E = \rho \left( \frac{I}{2 \pi r^2} \right) = \frac{\rho I}{2 \pi r^2}$.

(B) The resistance of a wire is given by $R = \frac{\rho l}{A}$.
Since the volume $V = A \times l$ remains constant when the wire is stretched,we can write $A = \frac{V}{l}$.
Substituting this into the resistance formula,we get $R = \frac{\rho l^2}{V}$.
Since $\rho$ and $V$ are constants,$R \propto l^2$.
Taking the derivative or using the approximation for small changes,the fractional change in resistance is given by $\frac{\Delta R}{R} = 2 \frac{\Delta l}{l}$.
Given that the wire is stretched by $0.1 \%$,we have $\frac{\Delta l}{l} = 0.1 \% = 0.001$.
Therefore,$\frac{\Delta R}{R} = 2 \times 0.1 \% = 0.2 \%$.
Since the change is positive,the resistance will increase by $0.2 \%$.

(C) The resistance $R$ of a metallic wire is given by the formula $R = \rho \frac{\ell}{A} = \rho \frac{\ell}{\pi r^2}$,where $\rho$ is resistivity,$\ell$ is length,and $r$ is the radius.
From this,we see that $R \propto \frac{\ell}{r^2}$.
Let the initial resistance be $R = k \frac{\ell}{r^2}$.
For option $C$,the new length $\ell' = 2\ell$ and the new radius $r' = \frac{r}{2}$.
The new resistance $R'$ is given by $R' = k \frac{\ell'}{(r')^2} = k \frac{2\ell}{(r/2)^2} = k \frac{2\ell}{r^2/4} = 8 \times k \frac{\ell}{r^2} = 8R$.
Therefore,the resistance becomes $8$ times when the length is doubled and the radius is halved.

(D) The wire consists of two parallel conductors: the inner core and the outer shell.
$1$. Resistance of the inner core $(R_1)$: The radius is $R$,resistivity is $\rho$,and length is $l$. Thus,$R_1 = \frac{\rho l}{\pi R^2}$.
$2$. Resistance of the outer shell $(R_2)$: The inner radius is $R$ and the outer radius is $2R$ (since thickness is $R$). The area of cross-section is $A_2 = \pi((2R)^2 - R^2) = 3\pi R^2$. The resistivity is $2\rho$. Thus,$R_2 = \frac{2\rho l}{3\pi R^2}$.
$3$. Equivalent resistance $(R_{eq})$: Since they are in parallel,$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{\pi R^2}{\rho l} + \frac{3\pi R^2}{2\rho l} = \frac{\pi R^2}{\rho l} (1 + 1.5) = \frac{2.5 \pi R^2}{\rho l} = \frac{5 \pi R^2}{2 \rho l}$.
Therefore,$R_{eq} = \frac{2 \rho l}{5 \pi R^2}$.

(C) The resistance of a conductor increases with an increase in temperature because the collision frequency of electrons increases,leading to a positive temperature coefficient of resistance.
Conversely,the resistance of a semiconductor decreases with an increase in temperature because more charge carriers (electrons and holes) are generated due to thermal excitation,leading to 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 conclusion is that $P$ is a conductor and $Q$ is a semiconductor.

(C) The volume of the wire remains constant during stretching. Therefore,$V = A_1 l_1 = A_2 l_2$.
Since $A = \pi r^2$,we have $\pi r^2 l_1 = \pi (r/2)^2 l_2$.
This simplifies to $r^2 l_1 = \frac{r^2}{4} l_2$,which gives $l_2 = 4 l_1$.
The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$.
Thus,$R \propto \frac{l}{r^2}$.
Let $R_1 = R$ and $R_2$ be the new resistance.
$\frac{R_2}{R_1} = \frac{l_2}{l_1} \times (\frac{r_1}{r_2})^2$.
Substituting the values: $\frac{R_2}{R} = (4) \times (\frac{r}{r/2})^2 = 4 \times 2^2 = 4 \times 4 = 16$.
Therefore,the new resistance is $16R$.

(B) The resistivity or specific resistance of a material is an intrinsic property of the 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,radius,or cross-sectional area.
Since both wires are made of the same material,their specific resistance (resistivity) will be the same.
Therefore,the ratio of their specific resistance is $1 : 1$.
Hence,option $B$ is correct.

(C) The volume $V$ of the wire is given by $V = \pi r^2 l$,where $r$ is the radius and $l$ is the length.
Since the volume remains constant during stretching,$\frac{\Delta V}{V} = 2\frac{\Delta r}{r} + \frac{\Delta l}{l} = 0$.
This implies $\frac{\Delta l}{l} = -2\frac{\Delta r}{r} \dots (1)$.
The resistance $R$ is given by $R = \frac{\rho l}{\pi r^2}$.
Taking the relative error,we get $\frac{\Delta R}{R} = \frac{\Delta l}{l} - 2\frac{\Delta r}{r}$.
Substituting equation $(1)$ into this expression:
$\frac{\Delta R}{R} = (-2\frac{\Delta r}{r}) - 2\frac{\Delta r}{r} = -4\frac{\Delta r}{r}$.
Given the radius decreases by $0.1\%$,we have $\frac{\Delta r}{r} = -0.1\%$.
Therefore,the percentage increase in resistance is $\frac{\Delta R}{R} = -4 \times (-0.1\%) = 0.4\%$.

(C) Given: Initial resistance $R_0 = 100\,\Omega$,Voltage $V = 220\,V$,Current $I = 2\,A$,and temperature change $\Delta t = 500\,^{\circ}C$.
First,calculate the resistance $R_t$ at the higher temperature using Ohm's Law: $R_t = \frac{V}{I} = \frac{220}{2} = 110\,\Omega$.
The formula for temperature dependence of resistance is $R_t = R_0(1 + \alpha \Delta t)$.
Substituting the values: $110 = 100(1 + \alpha \times 500)$.
$1.1 = 1 + 500\alpha$.
$0.1 = 500\alpha$.
$\alpha = \frac{0.1}{500} = \frac{1}{5000} = 0.0002\,^{\circ}C^{-1}$.
Therefore,$\alpha = 2 \times 10^{-4}\,^{\circ}C^{-1}$.

(A) Given: Initial resistance $R_1 = 100\, \Omega$,initial radius $r_1 = r$,final radius $r_2 = r/2$.
The resistance of a wire is given by $R = \frac{\rho l}{A}$.
Since the volume $V$ of the wire remains constant during recasting,$V = A \cdot l = \text{constant}$.
Substituting $l = V/A$ into the resistance formula,we get $R = \frac{\rho V}{A^2}$.
Since $\rho$ and $V$ are constant,$R \propto \frac{1}{A^2}$.
Given $A = \pi r^2$,we have $R \propto \frac{1}{(\pi r^2)^2} \propto \frac{1}{r^4}$.
Therefore,$\frac{R_2}{R_1} = \left( \frac{r_1}{r_2} \right)^4$.
Substituting the values: $\frac{R_2}{100} = \left( \frac{r}{r/2} \right)^4 = (2)^4 = 16$.
$R_2 = 16 \times 100 = 1600\, \Omega$.

(B) We know that resistivity is given by $\rho = \frac{R A}{\ell}$.
Conductivity $\sigma = \frac{1}{\rho} = \frac{\ell}{R A}$.
Since $V = I R$,we have $R = \frac{V}{I}$. Substituting this,$\sigma = \frac{\ell I}{V A}$.
The dimensions are:
$[\ell] = [L]$
$[I] = [I]$
$[A] = [L^2]$
$[V] = [M L^2 T^{-3} I^{-1}]$
Substituting these into the formula:
$\sigma = \frac{[L][I]}{[M L^2 T^{-3} I^{-1}] [L^2]} = \frac{[L][I]}{[M L^3 T^{-3} I^{-1}]}$
$\sigma = [M^{-1} L^{-3} T^3 I^2]$.

(C) The initial resistance of the wire is given by $R = \rho \frac{l}{A}$,where $l$ is the length and $A$ is the cross-sectional area.
When the wire is bent at the middle by $180^o$ and the ends are twisted,the new length becomes $l' = \frac{l}{2}$ and the new cross-sectional area becomes $A' = 2A$.
The new resistance $R'$ is given by $R' = \rho \frac{l'}{A'}$.
Substituting the new values: $R' = \rho \frac{l/2}{2A} = \frac{1}{4} \left( \rho \frac{l}{A} \right) = \frac{R}{4}$.