Question:
A wire is stretched to double its length. New resistance becomes:
A wire is stretched to double its length. New resistance becomes:
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For any wire stretched by a factor of $n$ (where $l' = n \cdot l$), the new resistance is given by the shortcut formula:
\[ R' = n^2 \cdot R \] Since the length is doubled ($n = 2$), the new resistance is $2^2 \cdot R = 4R$.
\[ R' = n^2 \cdot R \] Since the length is doubled ($n = 2$), the new resistance is $2^2 \cdot R = 4R$.
Updated On: Jun 3, 2026
- $R/2$
- $R$
- $2R$
- $4R$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
This problem requires determining the change in the electrical resistance of a conductive wire when it is mechanically stretched such that its final length becomes twice its original length.
Step 2: Key Formula or Approach:
The resistance of a uniform conductor is given by:
\[ R = \rho \frac{l}{A} \] where $\rho$ is the resistivity of the material, $l$ is the length, and $A$ is the cross-sectional area. Since the material volume remains constant during stretching, we use $V = A \cdot l = \text{constant}$.
Step 3: Detailed Explanation:
• The initial resistance of the wire is represented as $R_1 = \rho \frac{l_1}{A_1}$, where $l_1$ is the initial length and $A_1$ is the initial cross-sectional area.
• When a wire is physically stretched, its total volume ($V$) remains constant because no material is added or removed. The volume of a cylinder is $V = A \cdot l$.
• Thus, the relationship between initial and final parameters is: $V_1 = V_2 \implies A_1 l_1 = A_2 l_2$.
• We are given that the wire is stretched to double its initial length, so the new length is $l_2 = 2l_1$.
• Using the conservation of volume, we can solve for the new cross-sectional area $A_2$:
\[ A_2 = A_1 \left( \frac{l_1}{l_2} \right) = A_1 \left( \frac{l_1}{2l_1} \right) = \frac{A_1}{2} \]
• The new resistance $R_2$ of the stretched wire is given by:
\[ R_2 = \rho \frac{l_2}{A_2} \]
• Substituting the values of $l_2 = 2l_1$ and $A_2 = \frac{A_1}{2}$ into the formula:
\[ R_2 = \rho \frac{2l_1}{\left(\frac{A_1}{2}\right)} = 4 \left( \rho \frac{l_1}{A_1} \right) = 4R_1 \]
• This calculation shows that as the length increases and the cross-sectional area simultaneously decreases, both factors work together to increase the total resistance of the wire.
• Specifically, doubling the length doubles the resistance, and halving the area doubles it again, resulting in a net increase of four times the original value.
Step 4: Final Answer:
The new resistance of the stretched wire is $4R$.
This problem requires determining the change in the electrical resistance of a conductive wire when it is mechanically stretched such that its final length becomes twice its original length.
Step 2: Key Formula or Approach:
The resistance of a uniform conductor is given by:
\[ R = \rho \frac{l}{A} \] where $\rho$ is the resistivity of the material, $l$ is the length, and $A$ is the cross-sectional area. Since the material volume remains constant during stretching, we use $V = A \cdot l = \text{constant}$.
Step 3: Detailed Explanation:
• The initial resistance of the wire is represented as $R_1 = \rho \frac{l_1}{A_1}$, where $l_1$ is the initial length and $A_1$ is the initial cross-sectional area.
• When a wire is physically stretched, its total volume ($V$) remains constant because no material is added or removed. The volume of a cylinder is $V = A \cdot l$.
• Thus, the relationship between initial and final parameters is: $V_1 = V_2 \implies A_1 l_1 = A_2 l_2$.
• We are given that the wire is stretched to double its initial length, so the new length is $l_2 = 2l_1$.
• Using the conservation of volume, we can solve for the new cross-sectional area $A_2$:
\[ A_2 = A_1 \left( \frac{l_1}{l_2} \right) = A_1 \left( \frac{l_1}{2l_1} \right) = \frac{A_1}{2} \]
• The new resistance $R_2$ of the stretched wire is given by:
\[ R_2 = \rho \frac{l_2}{A_2} \]
• Substituting the values of $l_2 = 2l_1$ and $A_2 = \frac{A_1}{2}$ into the formula:
\[ R_2 = \rho \frac{2l_1}{\left(\frac{A_1}{2}\right)} = 4 \left( \rho \frac{l_1}{A_1} \right) = 4R_1 \]
• This calculation shows that as the length increases and the cross-sectional area simultaneously decreases, both factors work together to increase the total resistance of the wire.
• Specifically, doubling the length doubles the resistance, and halving the area doubles it again, resulting in a net increase of four times the original value.
Step 4: Final Answer:
The new resistance of the stretched wire is $4R$.
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