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:

• A. \( 1 : 4 \)
• B. \( 1 : 8 \)
• C. \( 1 : 2 \)
• D. \( 8 : 1 \)

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the ratio of the elongation produced in wire \(X\) to that in wire \(Y\).
We are given that they are made of the same material, subjected to the same stretching force, and have known ratios for their lengths and diameters.
Step 2: Key Formula or Approach:
According to Hooke's Law, the longitudinal elongation \(\Delta L\) under a tension load \(F\) is given by:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
where \(L\) is the length, \(A\) is the cross-sectional area, and \(Y\) is Young's modulus.
Expressing the circular cross-sectional area in terms of diameter \(d\) where \(A = \frac{\pi d^2}{4}\):
\[ \Delta L = \frac{4FL}{\pi d^2 Y} \]
Step 3: Detailed Explanation:
1. Both wires are made of the same material, which means their Young's modulus values are equal (\(Y_X = Y_Y\)).
2. They are subjected to the same stretching force (\(F_X = F_Y\)).
3. From this, we establish that the elongation is directly proportional to length and inversely proportional to the square of the diameter:
\[ \Delta L \propto \frac{L}{d^2} \]
4. The given ratios are:
- Length ratio: \(\frac{L_X}{L_Y} = \frac{1}{2}\)
- Diameter ratio: \(\frac{d_X}{d_Y} = \frac{2}{1} \implies \frac{d_Y}{d_X} = \frac{1}{2}\)
5. Set up the ratio for the elongations:
\[ \frac{\Delta L_X}{\Delta L_Y} = \left(\frac{L_X}{L_Y}\right) \times \left(\frac{d_Y}{d_X}\right)^2 \]
6. Substitute the given ratios into the proportionality equation:
\[ \frac{\Delta L_X}{\Delta L_Y} = \left( \frac{1}{2} \right) \times \left( \frac{1}{2} \right)^2 \]
\[ \frac{\Delta L_X}{\Delta L_Y} = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} \]
This gives the ratio \(1 : 8\), which matches Option (B).
Step 4: Final Answer:
The ratio of the elongation produced in wire \(X\) to that in wire \(Y\) is \(1 : 8\), which corresponds to Option (B).