Ohm's Law Questions in English

(A) Ohm's law states that the current flowing through a conductor is directly proportional to the potential difference across its ends,provided the physical conditions (like temperature) remain constant.
For metallic conductors,Ohm's law holds true as long as the temperature remains constant.
At high temperatures,the resistance of metallic conductors increases,causing the $V-I$ graph to become non-linear,which violates Ohm's law.
Electrolytes and diodes are non-ohmic devices because they do not follow a linear relationship between $V$ and $I$.
Therefore,Ohm's law is primarily applicable to metallic conductors at low or constant temperatures.

(B) According to Ohm's law,the potential difference $V$ across a conductor is directly proportional to the current $I$ flowing through it,provided the physical conditions remain constant.
$V = IR$
In this circuit,the cell provides an electromotive force $(e.m.f.)$ $E$,which acts as the potential difference across the resistance $R$ of the ammeter.
Therefore,substituting $E$ for $V$,we get:
$E = IR$

(B) Ohm's law states that the current flowing through a conductor is directly proportional to the potential difference applied across its ends,provided the physical conditions remain constant. This results in a linear $V-I$ graph. For liquids (electrolytes),the relationship between current and potential difference is generally non-linear due to factors like polarization and chemical changes at the electrodes. Therefore,liquids do not follow Ohm's law strictly,meaning they obey it only partially or under specific,limited conditions. Thus,option $B$ is correct.

(B) According to Ohm's law,the ratio $V/i$ is equal to the resistance $R$ of the conductor.
For a metallic wire,the resistance $R$ is given by the relation $R_t = R_0(1 + \alpha \Delta T)$,where $\alpha$ is the temperature coefficient of resistance.
For metals,$\alpha$ is positive,which means that as the temperature increases,the resistance $R$ of the metallic wire increases.
Therefore,the ratio $V/i$ increases as the temperature rises.

(A) According to Ohm's Law,the resistance $R$ is given by the ratio of potential difference $V$ to the current $I$.
$R = \frac{V}{I}$
Given:
Potential difference $V = 60\, V$
Current $I = 15\, A$
Substituting the values:
$R = \frac{60}{15} = 4\, \Omega$
Therefore,the resistance of the coil is $4\, \Omega$.

(C) The correct answer is $C$. Ohm's law states that the current flowing through a conductor is directly proportional to the potential difference across its ends,provided physical conditions remain constant.
Semiconductors are non-ohmic devices,meaning they do not follow Ohm's law because the relationship between voltage and current is non-linear.
Therefore,the statement that Ohm's law is applicable to semiconductors is false.

(B) discharge tube does not follow Ohm's law,which states that the current is directly proportional to the potential difference for a constant resistance. In a discharge tube,the current-voltage $(I-V)$ relationship is non-linear due to the process of secondary ionization of the gas particles inside the tube. As the potential difference increases,the number of charge carriers increases significantly,causing the resistance to change. Therefore,it is classified as a non-ohmic device.

(B) To verify Ohm's law,we need to measure the current flowing through a resistor and the potential difference across it.
$1$. An ammeter is used to measure current and must be connected in series with the component.
$2$. $A$ voltmeter is used to measure potential difference and must be connected in parallel across the component.
$3$. In the provided options,the correct setup is the one where the ammeter is in series with the resistor and the voltmeter is connected in parallel across the resistor to measure the voltage drop across it.
$4$. Based on standard circuit diagrams for Ohm's law verification,the setup where the ammeter is in series and the voltmeter is in parallel is the correct configuration.

(A) For an ohmic resistance,the potential difference $V$ is directly proportional to the current $I$ flowing through it,according to Ohm's Law.
This is expressed as $V \propto I$,which can be written as $V = RI$,where $R$ is the resistance and is a constant.
The equation $V = RI$ represents a straight line passing through the origin in a $V$-$I$ graph,where the slope of the line is equal to the resistance $R$.
Therefore,the graph that represents an ohmic resistance is a straight line passing through the origin.

(D) According to Ohm's Law,$V = iR$,which implies $R = V/i$.
In the given $V-i$ graph,the resistance $(R)$ at any point is equal to the slope of the tangent at that point,which is given by $\tan \theta = \frac{dV}{di}$.
As we move from point $A$ to $D$ along the curve,the angle $\theta$ that the tangent makes with the current axis $(i)$ decreases.
Since the slope of the curve decreases as we move from $A$ to $D$,the resistance at these points follows the order $R_A > R_B > R_C > R_D$.
Comparing this with the given options:
$A)$ $R_C = R_D$ is incorrect.
$B)$ $R_B > R_A$ is incorrect as $R_A > R_B$.
$C)$ $R_C > R_B$ is incorrect as $R_B > R_C$.
Therefore,the correct option is $D$.

(A) The slope of the $V-I$ graph represents the resistance $R$ of the conductor,as $R = V/I$.
From the given figure,the slope of the line corresponding to temperature $T_1$ is greater than the slope of the line corresponding to temperature $T_2$.
Therefore,$R_{T_1} > R_{T_2}$.
For a metallic conductor,the resistance increases with an increase in temperature.
Since the resistance at $T_1$ is greater than the resistance at $T_2$,it follows that $T_1 > T_2$.

(C) The resistance $R$ is defined by the relation $R = V/I$. In an $I-V$ graph,the slope of the curve is given by $dI/dV$,which is equal to $1/R$.
For negative resistance,the slope $dI/dV$ must be negative.
Looking at the graph,in the portion $CD$,as the voltage $V$ increases,the current $I$ decreases.
Therefore,the slope $dI/dV$ is negative in the region $CD$,which corresponds to negative resistance.

(D) According to Ohm's Law,$V = iR$,which implies $R = \frac{V}{i}$.
In the given $V-i$ graph,the slope of the straight line represents the resistance $R$.
Taking a point on the graph,for $i = 4 \text{ A}$,the corresponding voltage is $V = 4 \text{ V}$.
Therefore,$R = \frac{V}{i} = \frac{4}{4} = 1 \text{ } \Omega$.

(A) The resistance $R$ is defined by the relation $R = \frac{V}{I}$.
In a $V-I$ graph, the resistance is given by the slope of the graph, but here the graph is plotted as $I$ versus $V$.
Therefore, the conductance $G = \frac{I}{V}$ is represented by the slope of the $I-V$ graph.
Resistance $R = \frac{1}{\text{slope of } I-V \text{ graph}}$.
At point $A$, the slope of the tangent to the curve is negative (as the current $I$ decreases with increasing voltage $V$).
Since the slope is negative, the resistance $R$ is negative at point $A$.

(C) For a conductor,the resistance $R$ is directly proportional to the temperature $T$,i.e.,$R \propto T$.
In a $V-i$ graph,the slope represents the resistance $R = \frac{V}{i}$.
From the figure,for temperature $T_1$,the slope is $\tan \theta$. Thus,$R_1 = \tan \theta \propto T_1$,which gives $T_1 = k \tan \theta$ for some constant $k$.
For temperature $T_2$,the angle with the $V$-axis is $\theta$,so the angle with the $i$-axis is $(90^\circ - \theta)$. Thus,$R_2 = \tan(90^\circ - \theta) = \cot \theta \propto T_2$,which gives $T_2 = k \cot \theta$.
Now,$(T_2 - T_1) = k(\cot \theta - \tan \theta)$.
Using trigonometric identities:
$(T_2 - T_1) = k \left( \frac{\cos \theta}{\sin \theta} - \frac{\sin \theta}{\cos \theta} \right) = k \left( \frac{\cos^2 \theta - \sin^2 \theta}{\sin \theta \cos \theta} \right) = k \left( \frac{\cos 2\theta}{\frac{1}{2} \sin 2\theta} \right) = 2k \cot 2\theta$.
Therefore,$(T_2 - T_1) \propto \cot 2\theta$.

(D) According to Ohm's Law,$V = iR$,which implies $R = V/i$.
In a $V-i$ graph,the resistance $R$ is represented by the slope of the line with respect to the $i$-axis.
The slope of a line is defined as $\tan(\phi)$,where $\phi$ is the angle the line makes with the $i$-axis.
Given that the graph makes an angle $\theta$ with the $V$-axis,the angle it makes with the $i$-axis is $(90^\circ - \theta)$.
Therefore,the resistance $R = \tan(90^\circ - \theta) = \cot \theta$.

(A) $Cu$ voltameter with soluble electrodes obeys Ohm's law,resulting in a linear $V-I$ graph passing through the origin,as shown in graph $B$.
In a water voltameter,there is a back $e.m.f.$ due to polarization. When the applied voltage $V$ is small $(V < 1.7 \, V)$,very little current flows,and the device does not obey Ohm's law.
As soon as $V$ exceeds $1.7 \, V$ (the decomposition voltage),the current increases steadily according to Ohm's law,as shown in graph $A$.

(A) The slope of the $V-I$ graph represents the resistance $R$ of the conductor,as $R = \frac{V}{I}$.
From the given graph,the slope for temperature $T_1$ is greater than the slope for temperature $T_2$.
Therefore,$R_1 > R_2$.
For a metallic conductor,the resistance increases with an increase in temperature.
Since $R_1 > R_2$,it follows that $T_1 > T_2$.

(A) According to Ohm's law,the potential difference $V$ across a conductor is directly proportional to the current $I$ flowing through it,provided the temperature and other physical conditions remain constant.
$V \propto I$
$V = IR$
Where $R$ is the resistance,which is a constant for an ohmic resistor.
This equation represents a straight line passing through the origin in a $V-I$ graph.
Therefore,the graph showing a linear relationship between $V$ and $I$ represents an ohmic resistor.

(A) According to Ohm's law,$V = IR$.
Taking the logarithm on both sides,we get $\log\ V = \log\ (IR) = \log\ I + \log\ R$.
This equation is of the form $y = mx + c$,where $y = \log\ V$,$x = \log\ I$,slope $m = 1$,and intercept $c = \log\ R$.
Since this is a linear equation,the graph between $\log\ V$ and $\log\ I$ is a straight line.

(D) According to Ohm's law,$V = iR$,so $R = V/i$.
In the given graph,the $V$-axis is vertical and the $i$-axis is horizontal.
The angle $\theta$ is given with the $V$-axis.
Therefore,the slope of the graph with respect to the $i$-axis is $\tan \phi$,where $\phi$ is the angle with the $i$-axis.
Since $\phi = 90^\circ - \theta$,the slope is $\tan(90^\circ - \theta) = \cot \theta$.
Since the slope of a $V-i$ graph represents resistance $R$,we have $R \propto \cot \theta$.

(B) According to Ohm's law,$R = \frac{V}{I}$.
In the given $I-V$ graph,the slope of the line is $\frac{I}{V} = \frac{1}{R}$.
Since the slope of the line for $T_1$ is greater than the slope of the line for $T_2$,we have $\frac{1}{R_1} > \frac{1}{R_2}$,which implies $R_1 < R_2$.
For a metallic conductor,resistance $R$ increases with an increase in temperature $T$.
Since $R_1 < R_2$,it follows that $T_1 < T_2$.

(C) The slope of the $I-V$ graph is given by $\text{Slope} = \frac{I}{V} = \frac{1}{R}$.
From the graph,the slope of the line corresponding to $T_1$ is greater than the slope of the line corresponding to $T_2$.
Therefore,$\frac{1}{R_1} > \frac{1}{R_2}$,which implies $R_1 < R_2$.
For a metallic wire,resistance increases with an increase in temperature.
Since $R_1 < R_2$,it follows that $T_1 < T_2$.

(B) To verify Ohm's law,we use the relation $V = IR$,where $I$ is the current flowing through a resistor and $V$ is the potential difference across that specific resistor.
$1$. An ammeter must be connected in series with the resistor to measure the current $I$ flowing through it.
$2$. $A$ voltmeter must be connected in parallel with the resistor to measure the potential difference $V$ across it.
Looking at the options,the setup in option $(b)$ correctly places the ammeter in series with the resistor and the voltmeter in parallel across the resistor. Therefore,option $(b)$ is the correct setup.

(B) From Ohm's law,$V = IR$,which implies $I = \frac{1}{R} V$.
Comparing this with the equation of a straight line $y = mx$,the slope of the $I-V$ graph is given by $\text{slope} = \frac{dI}{dV} = \frac{1}{R}$.
From the given graph,the slope for temperature $T_1$ is greater than the slope for temperature $T_2$.
Therefore,$\frac{1}{R_1} > \frac{1}{R_2}$,which implies $R_1 < R_2$.
For a metallic wire,resistance increases with an increase in temperature $(R \propto T)$.
Since $R_1 < R_2$,it follows that $T_1 < T_2$.

(B) $1$. Option $A$ is incorrect because alloys with low temperature coefficients of resistance (like Manganin or Constantan) are used,not necessarily those with very low resistivity.
$2$. Option $B$ is correct. The microscopic form of Ohm's law is given by $\vec{J} = \sigma \vec{E}$,which can be written as $\vec{E} = \rho \vec{J}$,where $\rho$ is resistivity.
$3$. Option $C$ is incorrect because $V = IR$ is only valid for Ohmic conductors,not for non-Ohmic materials like semiconductors or diodes.
$4$. Therefore,only statement $B$ is true.

(D) The Assertion is incorrect because Ohm's law is not applicable to all conducting elements. Materials that obey Ohm's law are called ohmic conductors (e.g.,metallic conductors),while those that do not are called non-ohmic conductors (e.g.,junction diodes,transistors,electrolytes).
The Reason is also incorrect because Ohm's law is not a fundamental law of nature like Newton's laws or Maxwell's equations. It is an empirical relationship that holds true only for certain materials under specific conditions.
Therefore,both the Assertion and the Reason are incorrect.

(N/A) To determine the nature of the conductor,we calculate the ratio $R = V/I$ for several data points:
For $I = 0.2 \, A, V = 3.94 \, V \implies R = 3.94 / 0.2 = 19.7 \, \Omega$.
For $I = 1.0 \, A, V = 19.7 \, V \implies R = 19.7 / 1.0 = 19.7 \, \Omega$.
For $I = 8.0 \, A, V = 158.0 \, V \implies R = 158.0 / 8.0 = 19.7 \, \Omega$.
Since the ratio $V/I$ remains constant at $19.7 \, \Omega$ for all given values,the resistor obeys Ohm's law.
Therefore,we conclude that manganin is an ohmic conductor with a constant resistance of $19.7 \, \Omega$.

(N/A) In $1828$, German scientist George Simon Ohm formulated the law as follows:
"Under constant physical conditions, the ratio of the potential difference $(V)$ between two points of a conductor to the current $(I)$ flowing through it is constant."
$\therefore V \propto I$
$\therefore V = RI$, where $R$ is the constant of proportionality.
$\therefore \frac{V}{I} = R$
Here, $R$ is called the resistance of the conductor. The unit of resistance is $\frac{\text{Volt}}{\text{Ampere}} = \text{Ohm} (\Omega)$. The dimensional formula for resistance is $[M^1 L^2 T^{-3} A^{-2}]$.

(N/A) Ohm's law states that the current flowing through a conductor is directly proportional to the potential difference applied across its ends,provided physical conditions remain constant. However,it has certain limitations:
$(a)$ $V$ and $I$ are not proportional to each other for all materials.
In the provided graph,the dotted line represents the region where Ohm's law is obeyed (linear behavior). The continuous curve indicates where Ohm's law is not obeyed (non-ohmic behavior).
When current flows through a conductor,Joule heating occurs. As the temperature of the conductor increases,its resistance changes. Consequently,for such materials,the $V-I$ graph is a non-linear curve.
Examples: Diodes,Transistors.
$(b)$ The relationship between $I$ and $V$ depends on the polarity of the applied voltage $V$. Thus,for the same magnitude of positive or negative voltage,different current values are obtained.
Example: $PN$ junction diode.

(A) Ohm's law states that the current $I$ flowing through a conductor is directly proportional to the potential difference $V$ applied across its ends,provided the physical conditions (like temperature) remain constant.
Mathematically,this is expressed as $V = IR$,where $R$ is the resistance of the conductor.
Since $R$ is constant for an ohmic material,the equation $V = IR$ represents a linear relationship.
Therefore,the graph of $V$ versus $I$ is a straight line passing through the origin $(0,0)$.

(A) function is one-one if every input value maps to a unique output value. In the context of electrical circuits, the $V \to I$ relation represents the current $I$ as a function of voltage $V$.
For an Ohmic conductor, the relationship is linear $(V = IR)$, meaning for every $V$, there is a unique $I$, making it a one-one function.
However, for non-Ohmic devices (like $p-n$ junction diodes or thermistors), the $V-I$ characteristic curve is non-linear.
Specifically, in devices exhibiting negative differential resistance (like a tunnel diode), the current $I$ may decrease as voltage $V$ increases over a certain range.
Furthermore, in devices with hysteresis or complex $V-I$ characteristics, multiple values of $V$ might correspond to the same value of $I$, or the relationship may not be strictly monotonic.
Therefore, because the current does not always increase uniquely and monotonically with voltage across all types of materials, the $V \to I$ relation is not universally a one-one function.

(A) To verify $Ohm's$ law,we need to measure the current flowing through the resistor and the potential difference across it.
$1$. Ammeter: An ammeter is designed to measure the current flowing through a circuit. To ensure the same current passes through the ammeter as the circuit component,it must be connected in series with the resistor.
$2$. Voltmeter: $A$ voltmeter is designed to measure the potential difference (voltage) between two points. To measure the voltage across a specific resistor,it must be connected in parallel with that resistor.
Therefore,the correct configuration is that the ammeter is connected in series and the voltmeter is connected in parallel.

(A) Given: $\frac{R}{l} = \frac{1}{2} \, \Omega/cm$ and $l = 5 \, cm$.
First,calculate the resistance $R$ of the wire:
$R = \frac{1}{2} \times l = \frac{1}{2} \times 5 = 2.5 \, \Omega$.
Using Ohm's law,the current $i$ is given by:
$i = \frac{V}{R}$.
Given $V = 1 \, V$,we have:
$i = \frac{1}{2.5} = \frac{10}{25} = 0.4 \, A$.
Therefore,the correct current flowing through the wire is $0.4 \, A$.

(D) Given: Potential difference $V = 10 \,V$,Resistance $R = 1000 \,\Omega$,Time $t = 300 \,s$.
First,calculate the current $I$ using Ohm's law: $I = \frac{V}{R} = \frac{10}{1000} = 0.01 \,A$.
The total charge $Q$ flowing in time $t$ is given by $Q = I \times t = 0.01 \times 300 = 3 \,C$.
The number of electrons $n$ is given by the relation $Q = n \times e$,where $e = 1.6 \times 10^{-19} \,C$.
Therefore,$n = \frac{Q}{e} = \frac{3}{1.6 \times 10^{-19}} = 1.875 \times 10^{19}$ electrons.

(B) material is defined as a superconductor when its electrical resistance drops to zero.
This phenomenon occurs when the material is cooled below or at a specific temperature known as the critical temperature $(T_c)$.
Therefore,a conductor behaves as a superconductor at or below its critical temperature.

(C) The correct statement is that resistance is the opposition to the flow of current.
$A$. Electric current is a scalar quantity,not a vector,because it does not follow vector addition laws.
$B$. For a conductor,resistivity increases with an increase in temperature.
$C$. Resistance is defined as the property of a material to oppose the flow of electric current through it. This is correct.
$D$. Current density is a vector quantity.

(C) We know that Ohm's law is given by $V = I R$.
Substituting $V = E \cdot l$ and $R = \rho \cdot \frac{l}{A}$,where $l$ is the length and $A$ is the cross-sectional area of the conductor,we get:
$E \cdot l = I \cdot \rho \cdot \frac{l}{A}$
Dividing both sides by $l$,we get:
$E = \left( \frac{I}{A} \right) \rho$
Since current density $J = \frac{I}{A}$,we substitute this into the equation:
$E = J \rho$ or $E = \rho J$.

(D) The $I-V$ characteristic of a copper wire of length $L$ and area of cross-section $A$ is shown in the figure.
Since the $I-V$ characteristic is a straight line,it obeys Ohm's law.
The slope of the $I-V$ graph is given by $\text{slope} = \tan \theta = \frac{I}{V}$.
From Ohm's law,$V = IR$,so $\frac{I}{V} = \frac{1}{R}$.
Since $R = \rho \frac{L}{A}$,we have $\text{slope} = \frac{1}{\rho \frac{L}{A}} = \frac{A}{\rho L}$.
$(i)$ If the length $L$ is increased,the slope $\frac{A}{\rho L}$ decreases.
(ii) If the area $A$ is increased,the slope $\frac{A}{\rho L}$ increases.
(iii) If a steel wire of the same dimensions is used,since $\rho_{\text{steel}} > \rho_{\text{copper}}$,the slope $\frac{A}{\rho L}$ decreases.
(iv) If the experiment is performed at a higher temperature,the resistivity $\rho$ of copper increases,so the slope $\frac{A}{\rho L}$ decreases.
Therefore,the correct statement is that the slope becomes less if the length of the wire is increased.

(B) According to Ohm's law,$V = IR$,which can be written as $I = \frac{1}{R}V$.
Comparing this with the equation of a straight line $y = mx$,the slope of the $I-V$ graph is $\frac{I}{V} = \frac{1}{R}$.
Thus,the slope of the $I-V$ graph is inversely proportional to the resistance $(Slope \propto \frac{1}{R})$.
In the given diagram,the slope of line $Q$ is greater than the slope of line $P$ (i.e.,$Slope_Q > Slope_P$).
Therefore,$\frac{1}{R_Q} > \frac{1}{R_P}$.
This implies that $R_P > R_Q$.

(D) Ohm's law is applicable to conductors under constant physical conditions.
Ohm's law states that the current flowing through a conductor is directly proportional to the potential difference applied across its ends,provided the temperature and other physical conditions remain constant.
Mathematically,this is expressed as $V = IR$,where $V$ is the potential difference,$I$ is the current,and $R$ is the resistance.
Since diodes,transistors,and electrolytes are non-ohmic devices,they do not follow this linear relationship.

(A) For a conductor,the resistance $R$ is given by the slope of the $V-I$ graph,where $R = \frac{V}{I}$.
From the figure,the slope of the line corresponding to $T_{1}$ is greater than the slope of the line corresponding to $T_{2}$.
Therefore,$R_{1} > R_{2}$.
For a metallic conductor,the resistance increases with an increase in temperature $(R \propto T)$.
Since $R_{1} > R_{2}$,it follows that $T_{1} > T_{2}$.

(C) Ohmic devices obey Ohm's law,which states that the current $I$ flowing through a conductor is directly proportional to the potential difference $V$ applied across its ends,provided the physical conditions (like temperature) remain constant.
This is expressed as $V = IR$,where $R$ is the resistance of the device.
Since $R$ is constant for an Ohmic device,we have $I = (1/R)V$.
This equation represents a straight line passing through the origin $(0,0)$ with a slope of $1/R$.
Among the given options,the graph that shows a straight line passing through the origin is Graph $C$.

(D) The charge passing through the resistor is given by $q = 3t^2 - 2t + 6$.
To find the current $I$, we differentiate the charge with respect to time $t$:
$I = \frac{dq}{dt} = \frac{d}{dt}(3t^2 - 2t + 6) = 6t - 2$.
At time $t = 5 \, s$, the current is:
$I = 6(5) - 2 = 30 - 2 = 28 \, A$.
The resistance is given as $R = 10 \, \Omega$.
Using Ohm's law, the potential difference $V$ is:
$V = I \times R = 28 \, A \times 10 \, \Omega = 280 \, V$.

(C) The $V-I$ graph for a conductor at temperatures $T_1$ and $T_2$ is shown in the figure.
We know that the resistance $R$ of a conductor is directly proportional to its temperature $T$ (i.e.,$R \propto T$).
For the $V-I$ graph,the slope represents the resistance $R = \frac{V}{I}$.
From the figure,the slope of the line for $T_1$ with respect to the $I$-axis is $\tan \theta$. Thus,$R_1 \propto \tan \theta \Rightarrow R_1 = K \tan \theta$,where $K$ is a constant.
Similarly,the slope of the line for $T_2$ with respect to the $I$-axis is $\tan(90^{\circ}-\theta) = \cot \theta$. Thus,$R_2 \propto \cot \theta \Rightarrow R_2 = K \cot \theta$.
Therefore,$T_2 - T_1 \propto R_2 - R_1 = K(\cot \theta - \tan \theta)$.
$T_2 - T_1 \propto K \left( \frac{\cos \theta}{\sin \theta} - \frac{\sin \theta}{\cos \theta} \right) = K \left( \frac{\cos^2 \theta - \sin^2 \theta}{\sin \theta \cos \theta} \right)$.
Using trigonometric identities $\cos^2 \theta - \sin^2 \theta = \cos 2 \theta$ and $\sin \theta \cos \theta = \frac{1}{2} \sin 2 \theta$,we get:
$T_2 - T_1 \propto K \left( \frac{\cos 2 \theta}{\frac{1}{2} \sin 2 \theta} \right) = 2K \cot 2 \theta$.
Thus,$T_2 - T_1 \propto \cot 2 \theta$.

(B) According to Ohm's Law,$V = IR$,which implies $I = \frac{1}{R} V$.
Comparing this with the equation of a straight line $y = mx$,where $y = I$ and $x = V$,the slope $m$ of the $I-V$ graph is given by $m = \tan \theta = \frac{1}{R}$.
From the given graph,the angle $\theta = 60^{\circ}$.
Therefore,$\tan 60^{\circ} = \frac{1}{R}$.
Since $\tan 60^{\circ} = \sqrt{3}$,we have $\sqrt{3} = \frac{1}{R}$.
Thus,the resistance $R = \frac{1}{\sqrt{3}} \Omega$.

(B) For a metal wire,the resistance $R$ increases with an increase in temperature.
The slope of the $I-V$ graph is given by $\frac{I}{V} = \frac{1}{R}$.
From the figure,the slope of the line corresponding to $T_{1}$ is greater than the slope of the line corresponding to $T_{2}$.
Therefore,$\frac{1}{R_{1}} > \frac{1}{R_{2}}$,which implies $R_{1} < R_{2}$.
Since resistance increases with temperature,$R_{1} < R_{2}$ implies $T_{1} < T_{2}$.