Reasoning: Resistance depends on length $\mathbf{(L)}$, cross-sectional area $\mathbf{(A)}$, and resistivity $\mathbf{\rho}$ (nature of material).

Reasoning: $\mathbf{R \propto L}$ $\mathbf{\Longrightarrow}$ Doubling length doubles resistance.

Reasoning: Since $\mathbf{R \propto L}$, tripling length triples resistance if area remains constant.

Reasoning: Larger area provides more paths for charge flow, reducing resistance.

Reasoning: $\mathbf{R \propto}$ $\frac{\mathbf{1}}{\mathbf{A}}$

Reasoning: $\mathbf{\rho}$ represents the intrinsic opposition offered by a material to current flow.

Reasoning: Resistivity is a property of the material and is independent of dimensions.

Reasoning: From $\mathbf{\rho}\mathbf{=}$ $\frac{\mathbf{RA}}{\mathbf{L}}$, unit is ohm-meter $\mathbf{(\Omega \cdot m)}$.

Reasoning: Resistance is directly proportional to length.

Reasoning: Greater cross-sectional area reduces resistance.

Reasoning: $\mathbf{R =}$ $\frac{\mathbf{(1\ \times}\mathbf{\ 10}^{\mathbf{- 6}}\mathbf{)\ \times \ (4)}}{\mathbf{2\ \times}\mathbf{\ 10}^{\mathbf{- 6}}}\mathbf{= 2\ \Omega}$

Reasoning: Lower resistivity means better conductivity.

Reasoning: $\mathbf{R = \rho}$ $\frac{\mathbf{2L}}{\mathbf{2A}}\mathbf{=}\mathbf{\rho}\frac{\mathbf{L}}{\mathbf{A}}$

Reasoning: Greater length and smaller area both increase resistance.

Reasoning: All three factors appear in $\mathbf{R}\mathbf{= \rho}$ $\frac{\mathbf{L}}{\mathbf{A}}$.

Reasoning: $\mathbf{A =}\mathbf{\pi r}^{\mathbf{2}}$ $\mathbf{\ \Longrightarrow \ }$ Doubling radius makes area four times larger, reducing resistance to one-fourth.
⮚ Step by Step Solution:
● Step 1: The cross-sectional area $\mathbf{A}$ of a wire with radius $\mathbf{r}$ is given by $\mathbf{A = \pi r}^{\mathbf{2}}$.
● Step 2: The resistance $\mathbf{R}$ of the wire is given by $\mathbf{R = \rho}$ $\frac{\mathbf{L}}{\mathbf{A}}\mathbf{= \rho}\frac{\mathbf{L}}{\mathbf{\pi r}^{\mathbf{2}}}\mathbf{\ }$.
● Step 3: If the radius $\mathbf{r}$ is doubled, the new radius is $\mathbf{2r}$. The new cross-sectional area $\mathbf{A}^{\mathbf{'}}\mathbf{= \pi}\mathbf{(2r)}^{\mathbf{2}}\mathbf{=}\mathbf{4\pi r}^{\mathbf{2}}$.
The new resistance $\mathbf{R}^{\mathbf{'}}\mathbf{= \rho}$ $\frac{\mathbf{L}}{\mathbf{4\pi r}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\left( \mathbf{\rho}\frac{\mathbf{L}}{\mathbf{\pi r}^{\mathbf{2}}} \right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{(R) =}\frac{\mathbf{R}}{\mathbf{4}}$. Thus, the resistance is reduced to one-fourth of its initial value.

Reasoning: Material difference means different resistivity.

Reasoning: Lower resistivity means lower resistance for identical dimensions.

Reasoning: High resistivity strongly opposes current flow.

Reasoning: $\mathbf{R \propto}$ $\frac{\mathbf{1}}{\mathbf{A}}$ $\mathbf{\Longrightarrow}$ Its resistance becomes twice the original value $\mathbf{(2R)}$.
⮚ Why the Resistance Changes:
● The Formula: Resistance $\mathbf{(R)}$ depends on the material’s resistivity $\mathbf{(\rho)}$ length $\mathbf{(L)}$, and cross-sectional area $\mathbf{(A)}$ through the equation $\mathbf{R = \rho}$ $\frac{\mathbf{L}}{\mathbf{A}}\mathbf{\ }$.
● Length $\mathbf{(L)}$: Stays the same
● Area $\mathbf{(A)}$: Reduced to half $\mathbf{(A\ /\ 2)}$
● Calculation: $\mathbf{R}^{\mathbf{'}}\mathbf{= \rho}$ $\frac{\mathbf{L}}{\mathbf{A/2}}\mathbf{= 2}\left( \mathbf{\rho}\frac{\mathbf{L}}{\mathbf{A}} \right)\mathbf{= 2R}$
● Inverse Relation: Because resistance is inversely proportional to the cross-sectional area, making the area half makes the resistance twice as large.

Reasoning: $\mathbf{R = \rho}$ $\frac{\mathbf{2L}}{\mathbf{A\ /\ 2}}\mathbf{= 4\ \rho}\frac{\mathbf{L}}{\mathbf{A}}$
In a wire’s length is doubled and its area is halved, its resistance becomes four times its original value $\mathbf{(4R)}$.
⮚ Resistance Formula:
● Original resistance: $\mathbf{R = \rho}$ $\frac{\mathbf{L}}{\mathbf{A}}$ (where $\mathbf{\rho}$ is resistivity, $\mathbf{L}$ is length, and $\mathbf{A}$ is area).
● New length: $\mathbf{L' = 2L}$
● New area: $\mathbf{A}^{\mathbf{'}}\mathbf{=}$ $\frac{\mathbf{A}}{\mathbf{2}}$
⮚ Calculation Steps:
● Substitute new values into the formula $\mathbf{\Longrightarrow}$ $\mathbf{R}^{\mathbf{'}}\mathbf{= \rho}$ $\frac{\mathbf{2L}}{\mathbf{A/2}}$
● Simplify the fraction $\mathbf{\Longrightarrow}$ $\mathbf{R}^{\mathbf{'}}\mathbf{= \rho}$ $\left( \mathbf{2L \times}\frac{\mathbf{2}}{\mathbf{A}} \right)\mathbf{= 4}\left( \mathbf{\rho}\frac{\mathbf{L}}{\mathbf{A}} \right)$
● Result $\mathbf{\Longrightarrow}$ $\mathbf{R}^{\mathbf{'}}\mathbf{= 4R}$

Reasoning: Lower resistivity directly lowers resistance.

Reasoning: Halving the length halves the resistance.
⮚ Step by Solution:
● Step 1: Given the original resistance of the wire is $\mathbf{R}$ and the wire is cut into two equal parts.
● Step 2: Each part will have half the length of the original wire. $\mathbf{i.e}$, $\mathbf{L\ /\ 2}$.
● Step 3: The resistance of each part will be $\mathbf{R\ /\ 2}$.
⮚ Final Answer: The resistance of each part is $\mathbf{R\ /\ 2}$.

Reasoning: Resistivity is an intrinsic property of the material.

Reasoning: For equal resistance, $\frac{\mathbf{L}}{\mathbf{A}}$ must remain constant.
If one wire is twice as long and has equal resistance from the same material, its area must be twice as large (doubled).
⮚ Resistance Formula: $\mathbf{R = \rho}$ $\frac{\mathbf{L}}{\mathbf{A}}$
● $\mathbf{R =}$ resistance
● $\mathbf{\rho =}$ material resistivity ● $\mathbf{L =}$ length ● $\mathbf{A =}$ cross-sectional area
⮚ Calculation Details:
● Let the first wire have length $\mathbf{L}_{\mathbf{1}}$ an area $\mathbf{A}_{\mathbf{1}}$, with resistance $\mathbf{R}_{\mathbf{1}}\mathbf{= \rho}$ $\frac{\mathbf{L}_{\mathbf{1}}}{\mathbf{A}_{\mathbf{1}}}$.
● Let the second wire have length $\mathbf{L}_{\mathbf{2}}\mathbf{= 2}\mathbf{L}_{\mathbf{1}}$ and equal resistance $\mathbf{R}_{\mathbf{2}}\mathbf{=}\mathbf{R}_{\mathbf{1}}$.
● Setting the resistances equal $\mathbf{\Longrightarrow}$ $\mathbf{\rho}$ $\frac{\mathbf{L}_{\mathbf{1}}}{\mathbf{A}_{\mathbf{1}}}\mathbf{=}\mathbf{\rho}\frac{\mathbf{2L}_{\mathbf{1}}}{\mathbf{A}_{\mathbf{2}}}$
● Canceling $\mathbf{\rho}$ and $\mathbf{L}_{\mathbf{1}}$ $\mathbf{\Longrightarrow}$ $\frac{\mathbf{1}}{\mathbf{A}_{\mathbf{1}}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{A}_{\mathbf{2}}}$
● Solving for $\mathbf{A}_{\mathbf{2}}$ $\mathbf{\Longrightarrow}$ $\mathbf{A}_{\mathbf{2}}\mathbf{= 2\ }\mathbf{A}_{\mathbf{1}}$

Reasoning: Rearranging gives $\mathbf{\rho =}$ $\frac{\mathbf{RA}}{\mathbf{L}}$.

Reasoning: Low resistivity means less opposition to current.

Reasoning: Resistivity is independent of dimensions.

Reasoning: Since $\mathbf{R = \rho L/A}$, doubling resistivity requires half the length to keep resistance constant.
The Length of Wire $\mathbf{X}$ is twice the length of Wire $\mathbf{Y\ }\left( \mathbf{L}_{\mathbf{X}}\mathbf{= 2}\mathbf{L}_{\mathbf{Y}} \right)$, meaning it is longer because resistance equals resistivity times length divided by area $\left( \mathbf{R =}\frac{\mathbf{\rho L}}{\mathbf{A}} \right)$. Since both wires have equal resistance $\left( \mathbf{R}_{\mathbf{X}}\mathbf{=}\mathbf{R}_{\mathbf{Y}} \right)$ and the same area $\left( \mathbf{A}_{\mathbf{X}}\mathbf{=}\mathbf{A}_{\mathbf{Y}} \right)$ double the resistivity $\mathbf{(}\mathbf{\rho}_{\mathbf{X}}\mathbf{=}\mathbf{2\rho}_{\mathbf{Y}}\mathbf{)}$ requires double the length to keep the resistance balanced.
⮚ Resistance Formula and Setup:
● Resistance equation: $\mathbf{R =}$ $\frac{\mathbf{\rho\ \ .\ \ L}}{\mathbf{A}}$
● Wire X: $\mathbf{R}_{\mathbf{X}}\mathbf{=}$ $\frac{\mathbf{\rho}_{\mathbf{X}}\mathbf{\ \ .\ \ }\mathbf{L}_{\mathbf{X}}}{\mathbf{A}}$
● Wire Y: $\mathbf{R}_{\mathbf{Y}}\mathbf{=}$ $\frac{\mathbf{\rho}_{\mathbf{Y}}\mathbf{\ \ .\ \ }\mathbf{L}_{\mathbf{Y}}}{\mathbf{A}}$
⮚ Step-by-Step Comparison:
● Set resistances equal $\mathbf{\Longrightarrow}$ $\frac{\mathbf{\rho}_{\mathbf{X}}\mathbf{\ \ .\ \ }\mathbf{L}_{\mathbf{X}}}{\mathbf{A}}\mathbf{=}\frac{\mathbf{\rho}_{\mathbf{Y}}\mathbf{\ \ .\ \ }\mathbf{L}_{\mathbf{Y}}}{\mathbf{A}}$
● Cancel out the area $\mathbf{(A)}$ on both sides $\mathbf{\Longrightarrow}$ $\mathbf{\rho}_{\mathbf{X}}\mathbf{\ \ .\ \ }\mathbf{L}_{\mathbf{X}}\mathbf{\ = \ }\mathbf{\rho}_{\mathbf{Y}}\mathbf{\ \ .\ \ }\mathbf{L}_{\mathbf{X}}$
● Substitute $\mathbf{\rho}_{\mathbf{X}}\mathbf{= 2}\mathbf{\rho}_{\mathbf{Y}}$ $\mathbf{\Longrightarrow}$ $\mathbf{2}\mathbf{\rho}_{\mathbf{Y}}\mathbf{\ \ .\ \ }\mathbf{L}_{\mathbf{X}}\mathbf{=}\mathbf{\rho}_{\mathbf{Y}}\mathbf{\ \ .\ \ }\mathbf{L}_{\mathbf{Y}}$
● Cancel out $\mathbf{\rho}_{\mathbf{Y}}$ $\mathbf{\Longrightarrow}$ $\mathbf{2}\mathbf{L}_{\mathbf{X}}\mathbf{=}\mathbf{L}_{\mathbf{Y}}$ or $\mathbf{L}_{\mathbf{X}}\mathbf{= 2}\mathbf{L}_{\mathbf{Y}}$

Reasoning: The formula shows dependence on length, area, and material (resistivity).

Reasoning: Area is proportional to ${\mathbf{(}\mathbf{diameter}\mathbf{)}}^{\mathbf{2}}$. Doubling diameter increases area four times, so resistance becomes $\left( \frac{\mathbf{12}}{\mathbf{4}}\mathbf{= 3\Omega} \right)$.

Reasoning: Resistivity reflects the inherent electrical nature of a material, independent of size and shape.