Ohm's Law and Resistance Questions in English

(B) Wire $Y$ has greater resistance.
Resistance $(R)$ of a conductor is given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.
Since both wires have the same cross-sectional area $(A)$ and are made of the same material (same $\rho$),the resistance is directly proportional to the length $(l)$ of the wire $(R \propto l)$.
As the length of wire $Y$ $(2l)$ is greater than the length of wire $X$ $(l)$,wire $Y$ has greater resistance.

(D) The resistance $R$ of a cylindrical conductor depends on the following factors:
$(i)$ Length $(l)$: Resistance is directly proportional to the length $(R \propto l)$.
$(ii)$ Area of cross-section $(A)$: Resistance is inversely proportional to the area of cross-section $(R \propto 1/A)$.
$(iii)$ Nature of the material (resistivity,$\rho$).
$(iv)$ Temperature.
When a conductor is stretched to double its length,its volume remains constant. If the original length is $l$ and the new length is $l' = 2l$,then the new area of cross-section $A'$ will be $A/2$ because $V = A \times l = A' \times l'$.
The new resistance $R'$ is given by:
$R' = \rho \times (l' / A') = \rho \times (2l / (A/2)) = 4 \times (\rho \times l / A) = 4R$.
Thus,the resistance becomes $4$ times the original resistance.

(B) Ammeter $A_{2}$ will show a higher reading.
According to the formula for resistance,$R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the area of cross-section.
Since both wires have the same material ($\rho$ is constant) and the same area of cross-section,the resistance is directly proportional to the length $(R \propto l)$.
Given that wire $A$ is longer than wire $B$,the resistance of wire $A$ $(R_{A})$ is greater than the resistance of wire $B$ $(R_{B})$.
According to Ohm's law,$I = \frac{V}{R}$. Since both resistors are connected in parallel,the potential difference $(V)$ across them is the same.
Therefore,current is inversely proportional to resistance $(I \propto \frac{1}{R})$.
Since $R_{A} > R_{B}$,the current flowing through wire $A$ $(I_{A})$ will be less than the current flowing through wire $B$ $(I_{B})$.
Thus,ammeter $A_{2}$ (connected in series with wire $B$) will show a higher reading than ammeter $A_{1}$ (connected in series with wire $A$).

(A) We know that $V = IR$,which implies $R = V/I$. The slope of the $V-I$ graph represents the resistance $R$ of the wire.
From the given graph,the slope of line $A$ is greater than the slope of line $B$,therefore,the resistance of wire $A$ $(R_A)$ is greater than the resistance of wire $B$ $(R_B)$.
We know that 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 both wires are made of the same material,their resistivity $\rho$ is the same. Since they are of equal thickness,their cross-sectional area $A$ is also the same.
Thus,$R \propto l$. Since $R_A > R_B$,it follows that the length of wire $A$ is greater than the length of wire $B$.

(N/A) Based on the provided $V-I$ graph,we can observe the relationship between voltage $(V)$ and current $(I)$. The table below shows the values of $I$ corresponding to four different values of $V$:

$V$ (volts)	$2$	$4$	$6$	$8$
$I$ (amperes)	$1$	$2$	$3$	$4$

From the graph,when $V = 10 \text{ volts}$,the corresponding value of current $I$ is $5 \text{ amperes}$.
To determine the resistance $(R)$ of the resistor,we use Ohm's Law,which states $R = \frac{V}{I}$. The resistance is equal to the reciprocal of the slope of the $I-V$ graph,or the slope of the $V-I$ graph itself. Using the values from the graph:
$R = \frac{V}{I} = \frac{10 \text{ V}}{5 \text{ A}} = 2 \ \Omega$.

(A) $(i)$ The $SI$ unit of resistance is $Ohm (\Omega)$.
$(ii)$ One $Ohm$ is defined as the resistance of a conductor such that when a potential difference of $1\, V$ is applied across its ends, a current of $1\, A$ flows through it.
$(iii)$ Given:
$I = 200\, mA = 0.2\, A$
$V = 0.8\, V$
Using Ohm's Law, $V = IR$:
$R = \frac{V}{I}$
$R = \frac{0.8\, V}{0.2\, A}$
$R = 4\, \Omega$

(N/A) The flow of electric current in a conductor is due to the motion of electrons. These electrons are not completely free to move; they collide with the atoms and ions of the conductor as they drift. These collisions and the electrostatic attraction exerted by the atoms hinder the motion of electrons,which manifests as resistance.
$(b)$ Conductor: $A$ material that allows electric current to flow through it easily due to the presence of free electrons.
Resistor: $A$ component or conductor that has a specific,significant amount of resistance,used to control or limit the flow of current in a circuit.
Resistance: It is the inherent property of a material by virtue of which it opposes the flow of electric current through it.

(D) Given: Resistance $(R) = 6 \, \Omega$,Potential difference $(V) = 12 \, V$.
Using Ohm's law: $V = IR$.
Therefore,current $(I) = \frac{V}{R} = \frac{12 \, V}{6 \, \Omega} = 2 \, A$.
For the second part,the original resistance is $R = \rho \frac{l}{A}$.
When the wire is stretched to double its length,the new length $l' = 2l$. Since the volume of the wire remains constant,$V = A \times l = A' \times l'$.
Thus,$A' = \frac{A \times l}{l'} = \frac{A \times l}{2l} = \frac{A}{2}$.
The new resistance $R' = \rho \frac{l'}{A'} = \rho \frac{2l}{A/2} = 4 \left( \rho \frac{l}{A} \right) = 4R$.
Substituting the original resistance value: $R' = 4 \times 6 \, \Omega = 24 \, \Omega$.

(A) $R = \rho \frac{L}{A}$
Where $R$ is the resistance and $\rho$ is the resistivity of the material.
Initial length of the wire $= L$. The new length is $L' = 2L$ and the new area of cross-section is $A' = A/2$.
$(a)$ The new resistance $R'$ is given by:
$R' = \rho \frac{L'}{A'} = \rho \frac{2L}{A/2} = 4 \left( \rho \frac{L}{A} \right) = 4R$.
Thus,the resistance becomes $4$ times the original resistance.
$(b)$ Resistivity $(\rho)$ is an intrinsic property of the material and depends only on the nature of the material and temperature. Therefore,it remains unchanged.

(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 area of cross-section.
For the first wire,$R_1 = 10\, \Omega = \rho \frac{l}{A}$.
For the second wire,the length $l' = 3l$ and the area $A' = 4A$.
The resistance of the second wire is $R_2 = \rho \frac{l'}{A'} = \rho \frac{3l}{4A}$.
Substituting the value of $R_1$ into the equation,we get $R_2 = \frac{3}{4} \times R_1$.
$R_2 = \frac{3}{4} \times 10\, \Omega = 7.5\, \Omega$.

(N/A) The relationship is defined by Ohm's Law, which states that: "Provided the physical conditions (like temperature) remain constant, the electric current flowing through a conductor is directly proportional to the potential difference applied across its two ends."
The formula is $V = I \times R$, where $V$ is potential difference, $I$ is current, and $R$ is resistance.
The graph of potential difference $(V)$ versus current $(I)$ is a straight line passing through the origin.
Given:
Potential difference $(V)$ = $1.4 \ V$
Current $(I)$ = $0.35 \ A$
Using Ohm's Law, the resistance $(R)$ is calculated as:
$R = \frac{V}{I}$
$R = \frac{1.4}{0.35}$
$R = 4 \ \Omega$
Thus, the resistance of the conductor is $4 \ \Omega$.

(N/A) Electric current is defined as the rate of flow of electric charge through a conductor.
The $SI$ unit of electric current is the ampere $(A)$.
According to $Ohm's$ law,$V = I \times R$,where $V$ is the potential difference,$I$ is the current,and $R$ is the resistance.
From this,$R = V / I$.
If the potential difference across a conductor is $1$ volt and the current flowing through it is $1$ ampere,then the resistance $R$ is given by $R = 1 \text{ volt} / 1 \text{ ampere} = 1 \text{ ohm } (\Omega)$.
Thus,$1$ ohm is defined as the resistance of a conductor when a potential difference of $1$ volt applied across its ends causes a current of $1$ ampere to flow through it.

(B) Copper and aluminium are used because they have low electrical resistivity,which results in low resistance and minimal energy loss during transmission. Additionally,they are highly ductile,allowing them to be easily drawn into thin wires.
$(b)$ The resistance of a wire is given by the formula $R = \frac{\rho L}{A}$,where $\rho$ is resistivity,$L$ is length,and $A$ is the cross-sectional area.
Since $R$ and $L$ are equal for both wires,we have $A = \frac{\rho L}{R}$,which implies $A \propto \rho$.
Given that the resistivity of manganin is higher than that of copper $(\rho_{\text{manganin}} > \rho_{\text{copper}})$,it follows that the cross-sectional area of the manganin wire must be greater than that of the copper wire $(A_{\text{manganin}} > A_{\text{copper}})$.
Therefore,the manganin wire is thicker.

(N/A) $(i)$ The resistance $R$ of a wire is directly proportional to its length $L$ $(R \propto L)$. When the length is doubled,the resistance also becomes double. Therefore,the new resistance is $2 \times 10 \, \Omega = 20 \, \Omega$.
$(ii)$ The resistance $R$ is inversely proportional to the square of the diameter $D$ $(R \propto 1/D^2)$. When the diameter is doubled,the resistance becomes $1/4$ of its original value. Therefore,the new resistance is $10/4 \, \Omega = 2.5 \, \Omega$.
$(b)$ The heating elements of electrical devices are made of alloys because alloys have a much higher resistivity than their constituent metals. Furthermore,alloys do not oxidize (burn) readily even at high temperatures,which increases the durability and efficiency of the heating element.

(A-B, B-A) $(i)$ Element $B$: It is used for the filament of an electric bulb because it has a high resistivity $\left(520 \times 10^{-8} \, \Omega m\right)$. High resistivity allows the filament to produce a large amount of heat when current passes through it,enabling it to glow.
$(ii)$ Element $A$: It is used for electrical transmission lines because it has a very low resistivity $\left(1.62 \times 10^{-8} \, \Omega m\right)$. Low resistivity ensures minimal energy loss in the form of heat during the transmission of electric power over long distances.

(A) $Ohm's$ law states that the electric current flowing through a metallic conductor is directly proportional to the potential difference across its ends,provided that the physical conditions (like temperature) remain constant.
Given:
Current $(I)$ = $0.2 \, A$
Potential difference $(V)$ = $0.8 \, V$
Using the formula $R = V / I$:
$R = 0.8 \, V / 0.2 \, A$
$R = 4 \, \Omega$
Therefore,the resistance of the conductor is $4 \, \Omega$.

(N/A) The $V-I$ graph for a nichrome wire is a straight line passing through the origin,as nichrome follows Ohm's law.
Given:
Resistivity $\rho = 4.8 \times 10^{-7} \, \Omega \cdot m$
Length $L = 625 \, mm = 0.625 \, m$
Resistance $R = 4 \, \Omega$
Using the formula for resistance: $R = \frac{\rho L}{A}$
Rearranging for area $A$: $A = \frac{\rho L}{R}$
Substituting the values:
$A = \frac{4.8 \times 10^{-7} \times 0.625}{4}$
$A = 1.2 \times 10^{-7} \times 0.625$
$A = 0.75 \times 10^{-7} \, m^2$ or $7.5 \times 10^{-8} \, m^2$

(N/A) Resistance is the property of a conductor to oppose the flow of electric current through it.
The $SI$ unit of resistance is $Ohm$ ($\Omega$). One $Ohm$ is defined as the resistance of a conductor through which a current of $1$ $Ampere$ flows when a potential difference of $1$ $Volt$ is applied across its ends.
The factors on which the resistance of a conductor depends are:
$1$. Length of the conductor $(l)$
$2$. Area of cross-section $(A)$
$3$. Nature of the material (resistivity, $\rho$)
$4$. Temperature of the conductor
$(i)$ Since resistance $R \propto l$, if the length is doubled, the resistance becomes $2$ times the original value.
$(ii)$ Since resistance $R \propto \frac{1}{A}$ and $A = \pi r^2$, if the radius $r$ is doubled, the area $A$ becomes $4$ times $(A' = \pi(2r)^2 = 4\pi r^2)$. Consequently, the resistance becomes $\frac{1}{4}$th of its original value.

(B) The circuit diagram consists of an electric bulb,a voltmeter connected in parallel to the bulb,an ammeter connected in series,a cell,and a key,as shown in the provided image.
$(b)$ Given:
Current $(I) = 0.4 \ A$
Potential difference $(V) = 2 \ V$
According to Ohm's law,$V = IR$
Therefore,$R = \frac{V}{I} = \frac{2 \ V}{0.4 \ A} = 5 \ \Omega$
So,the resistance of the electric bulb is $5 \ \Omega$.
$(c)$ The law applied is Ohm's law. The $V-I$ graph for an ohmic conductor is a straight line passing through the origin,as shown in the provided image.

(N/A) Ohm's law states that the current flowing through a metallic conductor is directly proportional to the potential difference applied across its ends,provided its physical conditions (like temperature) remain constant.
Mathematically,this is expressed as:
$V \propto I$
$V = IR$
Here,$V$ is the potential difference,$I$ is the current,and $R$ is the constant of proportionality known as the resistance of the conductor.
Verification:
To verify Ohm's law,set up a circuit as shown in the figure,consisting of a nichrome wire $R$,an ammeter,a voltmeter,a battery,and a rheostat $(Rh)$.
$1$. Close the key $K$ and note the current $I$ in the ammeter and potential difference $V$ in the voltmeter for a specific position of the rheostat.
$2$. Change the position of the rheostat to vary the current in the circuit and record the corresponding values of $V$ and $I$.
$3$. Calculate the ratio $V/I$ for each set of readings. You will observe that the ratio $V/I$ is constant.
$4$. Plot a graph of $V$ (on the y-axis) against $I$ (on the x-axis). The graph will be a straight line passing through the origin,which confirms that $V$ is directly proportional to $I$,thus verifying Ohm's law.

(N/A) An ammeter is used to measure electric current in an electric circuit.
One ampere is defined as the flow of one coulomb of electric charge through a circuit in one second.
$(b)$ $(i)$ Variable resistor (Rheostat) and $(ii)$ Closed plug key (switch).
$(c)$ $(i)$ The circuit diagram consists of a battery of four cells,an ammeter in series,a nichrome wire $XY$ in series,and a voltmeter connected in parallel across $XY$.
$(ii)$ Since the graph is a straight line passing through the origin,the ratio $V/I$ is constant,which represents the resistance $R$ of the nichrome wire.
From the graph,at $V = 1.6 \ V$,$I = 0.6 \ A$.
$R = V/I = 1.6 / 0.6 = 2.67 \ \Omega$.
Since the ratio $V/I$ is constant for all values $(0.8 \ V, 1.2 \ V, 1.6 \ V)$,it confirms Ohm's Law.

(N/A) $(i)$ Ohm's law.
It states that,provided the physical conditions remain the same,the current flowing through a conductor is directly proportional to the potential difference across its two ends.
Mathematically,we have $V \propto I$ or $V = IR$,where $R$ is the resistance of the conductor.
$(ii)$ The $V-I$ graph is a straight line passing through the origin. This justifies the law because a straight-line graph indicates that the potential difference $V$ is directly proportional to the current $I$ $(V \propto I)$.
$(iii)$ $(a)$ Variable resistor (Rheostat): It is used to vary the resistance in the circuit to control the current.
$(b)$ Closed switch (or closed key): It allows the current to flow through the circuit by completing the path.

(D) Resistance is defined as the opposition offered by a conductor to the flow of electric current.
It depends on the following factors:
$(i)$ Nature of the material of the conductor.
$(ii)$ Length of the conductor $(L)$.
$(iii)$ Area of cross-section of the conductor $(A)$.
$A$ rheostat or variable resistor is the device used to change the resistance in an electric circuit without changing the voltage source.
Calculation:
Given: $L = 50 \, m$,$A = 0.01 \, mm^2 = 0.01 \times 10^{-6} \, m^2$,$\rho = 5 \times 10^{-8} \, \Omega \, m$.
Using the formula $R = \frac{\rho L}{A}$:
$R = \frac{5 \times 10^{-8} \times 50}{0.01 \times 10^{-6}}$
$R = \frac{250 \times 10^{-8}}{10^{-8}}$
$R = 250 \, \Omega$.

(D) Let the length of the original wire be $l$ and its radius be $r$.
The original resistance is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$.
The original volume is $V_1 = \pi r^2 l$.
When the radius is halved,the new radius becomes $r' = \frac{r}{2}$. Let the new length be $l'$.
Since the volume remains constant,$V_1 = V_2$,so $\pi r^2 l = \pi (\frac{r}{2})^2 l'$.
$\pi r^2 l = \pi \frac{r^2}{4} l'$,which simplifies to $l' = 4l$.
The new resistance is $R' = \rho \frac{l'}{\pi (r')^2} = \rho \frac{4l}{\pi (r/2)^2}$.
$R' = \rho \frac{4l}{\pi (r^2/4)} = 16 \left( \rho \frac{l}{\pi r^2} \right)$.
Therefore,$R' = 16R$.
The new resistance is $16$ times the original resistance.

(A) Given: Potential difference $(V)$ = $1.5 \ V$,Resistance $(R)$ = $30 \ \Omega$.
According to Ohm's Law,$V = IR$.
Therefore,the current $(I)$ is given by the formula: $I = \frac{V}{R}$.
Substituting the values: $I = \frac{1.5}{30} = 0.05 \ A$.
Thus,the current flowing through the circuit is $0.05 \ A$.

(N/A) Given: Potential difference $V = 220 \ V$,Resistance $R = 1200 \ \Omega$.
According to Ohm's law,$V = IR$,so $I = V / R$.
$I = 220 \ V / 1200 \ \Omega = 0.1833 \ A \approx 0.18 \ A$.
$(b)$ Given: Potential difference $V = 220 \ V$,Resistance $R = 100 \ \Omega$.
According to Ohm's law,$I = V / R$.
$I = 220 \ V / 100 \ \Omega = 2.2 \ A$.

(B) Given: Initial potential difference $V_1 = 60\, V$,Initial current $I_1 = 4\, A$.
According to Ohm's law,$V = IR$,where $R$ is the resistance of the heater. Since the heater remains the same,its resistance $R$ is constant.
First,calculate the resistance: $R = V_1 / I_1 = 60\, V / 4\, A = 15\, \Omega$.
Now,for the new potential difference $V_2 = 120\, V$,the new current $I_2$ is calculated as:
$I_2 = V_2 / R = 120\, V / 15\, \Omega = 8\, A$.
Alternatively,since $I \propto V$,doubling the potential difference from $60\, V$ to $120\, V$ will double the current from $4\, A$ to $8\, A$.

(A) Given:
Resistance $R = 26 \, \Omega$
Length $L = 1 \, m$
Diameter $D = 0.3 \, mm = 3 \times 10^{-4} \, m$
Radius $r = D/2 = 1.5 \times 10^{-4} \, m$
The formula for resistance is $R = \rho \frac{L}{A}$,where $A = \pi r^2 = \pi (D/2)^2 = \frac{\pi D^2}{4}$.
Rearranging for resistivity $\rho$:
$\rho = \frac{R A}{L} = \frac{R \pi D^2}{4 L}$
Substituting the values:
$\rho = \frac{26 \times 3.14 \times (3 \times 10^{-4})^2}{4 \times 1}$
$\rho = \frac{26 \times 3.14 \times 9 \times 10^{-8}}{4}$
$\rho = 1.8369 \times 10^{-6} \, \Omega \, m \approx 1.84 \times 10^{-6} \, \Omega \, m$.

(A) Given initial resistance $R = 4 \, \Omega$.
When a wire is doubled on itself,its length $L$ becomes half $(L_n = L/2)$ and its area of cross-section $A$ becomes double $(A_n = 2A)$.
The resistance of a wire is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity of the material.
The new resistance $R_n$ is given by:
$R_n = \rho \frac{L_n}{A_n} = \rho \frac{L/2}{2A} = \frac{1}{4} \left( \rho \frac{L}{A} \right) = \frac{R}{4}$.
Substituting the value of $R = 4 \, \Omega$:
$R_n = \frac{4 \, \Omega}{4} = 1 \, \Omega$.
Thus,the new resistance of the wire is $1 \, \Omega$.

(A) Given: Length $L = 1.0 \, m$,Diameter $D = 0.2 \, mm = 0.2 \times 10^{-3} \, m$.
Radius $r = D/2 = 0.1 \times 10^{-3} \, m = 10^{-4} \, m$.
Resistance $R = 10 \, \Omega$.
Formula for resistivity: $\rho = \frac{R \cdot A}{L}$,where $A = \pi r^2$.
Substituting the values: $A = 3.14 \times (10^{-4})^2 = 3.14 \times 10^{-8} \, m^2$.
$\rho = \frac{10 \times 3.14 \times 10^{-8}}{1.0} = 3.14 \times 10^{-7} \, \Omega \cdot m$.

(A) Given: Current $I = 10 \ A$,Potential difference $V = 220 \ V$.
According to Ohm's law,the relationship between voltage,current,and resistance is given by $V = I \times R$.
To find the resistance $R$,we rearrange the formula: $R = V / I$.
Substituting the given values: $R = 220 \ V / 10 \ A$.
Therefore,$R = 22 \ \Omega$.

(N/A) The circuit diagram is as shown in the image.
Activity to find the resistance $R$:
$1$. Set up a circuit using the wire of resistance $R$,an ammeter,a voltmeter,a plug key,and four cells of $1.5 \, V$ each.
$2$. First,use one cell as the source and note the readings of the ammeter $(I)$ and voltmeter $(V)$.
$3$. Increase the number of cells one by one and note the corresponding readings of $I$ and $V$.
$4$. Calculate the ratio $V/I$ for each set of readings. The average of these ratios gives the resistance $R$ of the wire.
Numerical calculation:
Given: Current $(I) = 0.2 \, A$,Resistance $(R) = 20 \, \Omega$.
Using Ohm's Law: $V = I \times R$
$V = 0.2 \, A \times 20 \, \Omega = 4 \, V$.
Thus,the potential difference across the ends of the resistor is $4 \, V$.

(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 area of cross-section.
Since both wires are of the same metal,their resistivity $\rho$ is the same. Given that $A$ is the same for both,$R \propto L$.
Therefore,the ratio of resistances is $\frac{R_A}{R_B} = \frac{L_A}{L_B} = \frac{2}{1}$.
According to Ohm's Law,$I = \frac{V}{R}$. Since the potential difference $V$ is the same for both,$I \propto \frac{1}{R}$.
Thus,the ratio of currents is $\frac{I_A}{I_B} = \frac{R_B}{R_A} = \frac{1}{2}$.

(A) For a metallic wire,the slope of the $I-V$ graph represents the conductance,which is the reciprocal of resistance $(R = V/I)$.
From the graph,the slope is given by $\Delta I / \Delta V = 1/R$.
As the temperature of a metallic conductor increases,its resistance $(R)$ increases.
Since $R$ increases with temperature,the slope $(1/R)$ decreases as the temperature increases.
Therefore,the line with the smallest slope corresponds to the highest temperature.
In the given graph,line $1$ has the smallest slope,so it represents the highest temperature.

(A) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area. The area $A = \pi r^2 = \pi (D/2)^2 = \frac{\pi D^2}{4}$.
$(i)$ When $V$ is halved: The resistance $R$ depends only on the physical dimensions and material properties of the conductor. Therefore,$R$ remains unchanged.
$(ii)$ When $L$ is halved: Since $R \propto L$,if the length $L$ is halved,the resistance $R$ also becomes halved.
$(iii)$ When $D$ is doubled: Since $R \propto \frac{1}{A}$ and $A \propto D^2$,we have $R \propto \frac{1}{D^2}$. If $D$ is doubled,$R$ becomes $\frac{1}{(2)^2} = \frac{1}{4}$ of its original value.

(B) The graph shows the variation of current $(I)$ on the $y$-axis with respect to voltage $(V)$ on the $x$-axis.
According to Ohm's law, $V = IR$, which implies $I = (1/R)V$.
The slope of the $I-V$ graph is given by $\text{slope} = I/V = 1/R$.
From the graph, the slope of line $X$ is greater than the slope of line $Y$ $(\text{slope}_X > \text{slope}_Y)$.
Since $\text{slope} = 1/R$, this means $1/R_X > 1/R_Y$, which implies $R_X < R_Y$.
Resistance $(R)$ is related to resistivity $(\rho)$ by the formula $R = \rho(L/A)$.
Given that both wires have the same length $(L)$ and the same diameter (and thus the same cross-sectional area $A$), the resistance is directly proportional to the resistivity $(R \propto \rho)$.
Since $R_Y > R_X$, it follows that the resistivity of wire $Y$ is greater than the resistivity of wire $X$ $(\rho_Y > \rho_X)$.

(B) 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.
Since both resistors are made of the same material ($\rho$ is constant) and have the same radii ($A$ is constant), the resistance $R$ is directly proportional to the length $L$ $(R \propto L)$.
Given $L_{1} > L_{2}$, it follows that $R_{1} > R_{2}$.
In a $V-I$ graph, the slope represents the resistance $(R = \frac{V}{I} = \text{slope})$. A steeper slope indicates a higher resistance.
Comparing the two graphs, graph $B$ has a greater slope than graph $A$, meaning the resistance of graph $B$ is greater than that of graph $A$.
Since $R_{1} > R_{2}$, graph $B$ corresponds to the longer length $L_{1}$.

(A) 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.
When the wire is melted and recast to half its length $(L' = L / 2)$,its volume remains constant.
Since $V = A \times L$,we have $A \times L = A' \times L'$.
Substituting $L' = L / 2$,we get $A \times L = A' \times (L / 2)$,which implies $A' = 2A$.
The new resistance $R'$ is given by $R' = \rho \frac{L'}{A'}$.
Substituting the new values: $R' = \rho \frac{L / 2}{2A} = \frac{1}{4} \times \rho \frac{L}{A}$.
Therefore,$R' = R / 4$.

(A) The power rating of a bulb is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.
From this formula,we can express resistance as $R = \frac{V^2}{P}$.
Since the voltage $V$ is constant $(220 \, V)$ for all three 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 lowest resistance.
Comparing the given power ratings: $60 \, W > 40 \, W > 25 \, W$.
Thus,the $60 \, W$ bulb has the lowest resistance.

(B) The unit of resistance is the $ohm$ ($\Omega$), not the $ampere$.
The $ampere$ $(A)$ is the $SI$ unit of electric current.
Therefore, the statement is $False$.

(A) According to Ohm's Law,$V = IR$,where $V$ is potential difference,$I$ is current,and $R$ is resistance.
Rearranging for current,we get $I = V/R$.
If the resistance $R$ is doubled $(R' = 2R)$ while the potential difference $V$ remains constant,the new current $I'$ becomes $I' = V/(2R) = (1/2) \times (V/R) = I/2$.
Therefore,the current is halved.

(A) The resistance of a metallic conductor is given by the formula $R = \frac{ml}{ne^2 \tau A}$.
As the temperature increases,the thermal velocity of the free electrons increases,which leads to more frequent collisions between electrons and ions.
This results in a decrease in the relaxation time $(\tau)$.
Since $R \propto \frac{1}{\tau}$,a decrease in relaxation time leads to an increase in the resistance of the conductor.
Therefore,the statement is true.

(B) The statement is $False$.
Insulators are materials that do not allow electric current to flow through them easily.
This is because they possess a very high electrical resistance,which prevents the movement of electrons.

(A) The resistance of a material depends on its resistivity $(\rho)$.
Rubber is an insulator, meaning it has a very high resistivity, which prevents the flow of electric current.
Copper is a conductor, meaning it has a very low resistivity, which allows electric current to flow easily.
Since resistance $R = \rho \frac{l}{A}$, for pieces of the same dimensions (length $l$ and area $A$), the material with higher resistivity will have higher resistance.
Therefore, the resistance of rubber is significantly higher than that of copper.

(A) The resistivity of pure metals is given by the relation $\rho_T = \rho_0 [1 + \alpha(T - T_0)]$,where $\alpha$ is the temperature coefficient of resistivity.
For pure metals,$\alpha$ is positive,which means that as the temperature $T$ increases,the resistivity $\rho_T$ also increases.
This happens because,with an increase in temperature,the frequency of collisions between free electrons and the vibrating ions of the metal lattice increases,leading to greater resistance to the flow of current.

(B) $Ohm's$ law states that the current flowing through a conductor is directly proportional to the potential difference across its ends,provided the temperature and other physical conditions remain constant. The mathematical expression is $V = IR$. Power is a separate concept defined as $P = VI$ or $P = I^2R$. Therefore,the statement is False.

(B) The statement is False.
Resistance in a conducting wire is primarily caused by the collisions of electrons with the atoms and ions of the conductor's lattice structure as they drift through it.
While electrons do exert repulsive forces on each other,this is not the primary cause of electrical resistance in a conductor.

(C) 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 temperature and other physical conditions remain constant.
Mathematically,this is expressed as $V = IR$ or $I = V/R$.
Here,$R$ is the resistance,which is a constant for a given conductor at a constant temperature.
Since $I \propto V$,as the voltage $(V)$ increases,the current $(I)$ also increases,assuming the resistance $(R)$ remains constant.
Therefore,option $C$ is the correct statement.

(A) According to Ohm's Law,the relationship between potential difference $(V)$,current $(I)$,and resistance $(R)$ is given by $V = I \times R$.
Given:
Resistance $(R)$ = $10\, \Omega$
Potential difference $(V)$ = $1.5\, V$
To find the current $(I)$,we rearrange the formula: $I = V / R$.
Substituting the values: $I = 1.5\, V / 10\, \Omega = 0.15\, A$.
To convert the current from Amperes $(A)$ to milliamperes $(mA)$,we multiply by $1000$: $0.15\, A \times 1000 = 150\, mA$.
Therefore,the electric current flowing through the wire is $150\, mA$.

(D) The resistivity or specific resistance of a material is an intrinsic property of the material.
It does not depend on the physical dimensions of the conductor,such as its length or area of cross-section.
It depends only on the nature of the material and the temperature of the conductor.
Therefore,the correct factor is the material of the wire.