Question:
Two wires $X$ and $Y$ of the same material have lengths in the ratio $1:2$ and diameters in the ratio $2:1$. If they are subjected to the same stretching force, the ratio of the elongation produced in wire $X$ to that in wire $Y$ ($\Delta L_X : \Delta L_Y$) is:
Two wires $X$ and $Y$ of the same material have lengths in the ratio $1:2$ and diameters in the ratio $2:1$. If they are subjected to the same stretching force, the ratio of the elongation produced in wire $X$ to that in wire $Y$ ($\Delta L_X : \Delta L_Y$) is:
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Always watch out for the squaring effect of thickness dimensions! Since area scales with the square of the diameter, doubling the diameter makes a wire four times more resistant to stretching. Wire $X$ is shorter (wants to stretch half as much) but significantly thicker (wants to stretch a quarter as much), yielding \(\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\) instantly.
Updated On: May 26, 2026
- \( 1 : 4 \)
- \( 1 : 8 \)
- \( 1 : 2 \)
- \( 8 : 1 \)
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The Correct Option is B
Solution and Explanation
Concept:
By Hooke's Law within the elastic limit, the longitudinal elongation (\(\Delta L\)) produced in a metallic structural wire under a tension load is dictated by Young's Modulus (\(Y\)):
\[
Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\left(\frac{F}{A}\right)}{\left(\frac{\Delta L}{L}\right)} \quad \implies \quad \Delta L = \frac{F \cdot L}{A \cdot Y}
\]
Expressing the cross-sectional area in terms of diameter \(d\) where \(A = \frac{\pi d^2}{4}\):
\[
\Delta L = \frac{4FL}{\pi d^2 Y}
\]
Since both wires are made of the same material, their Young's Modulus values are identical (\(Y_X = Y_Y\)). Furthermore, they are subjected to the same stretching force (\(F_X = F_Y\)). Therefore, elongation is directly proportional to length and inversely proportional to the square of the diameter:
\[
\Delta L \propto \frac{L}{d^2} \quad \implies \quad \frac{\Delta L_X}{\Delta L_Y} = \left(\frac{L_X}{L_Y}\right) \times \left(\frac{d_Y}{d_X}\right)^2
\]
Step 1: Substituting the given proportional parameters.
We are given the following explicit scale ratios:
• Length ratio: \(\frac{L_X}{L_Y} = \frac{1}{2}\)
• Diameter ratio: \(\frac{d_X}{d_Y} = \frac{2}{1} \implies \frac{d_Y}{d_X} = \frac{1}{2}\) Plugging these values directly into our proportional equation system: \[ \frac{\Delta L_X}{\Delta L_Y} = \left( \frac{1}{2} \right) \times \left( \frac{1}{2} \right)^2 \]
Step 2: Evaluating the final fractional ratio.
\[ \frac{\Delta L_X}{\Delta L_Y} = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} \] Thus, the tracking extension ratio evaluates to \(1 : 8\).
Step 1: Substituting the given proportional parameters.
We are given the following explicit scale ratios:
• Length ratio: \(\frac{L_X}{L_Y} = \frac{1}{2}\)
• Diameter ratio: \(\frac{d_X}{d_Y} = \frac{2}{1} \implies \frac{d_Y}{d_X} = \frac{1}{2}\) Plugging these values directly into our proportional equation system: \[ \frac{\Delta L_X}{\Delta L_Y} = \left( \frac{1}{2} \right) \times \left( \frac{1}{2} \right)^2 \]
Step 2: Evaluating the final fractional ratio.
\[ \frac{\Delta L_X}{\Delta L_Y} = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} \] Thus, the tracking extension ratio evaluates to \(1 : 8\).
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