Question

There are two wires of same material and same length while the diameter of second wire is two times the diameter of the first wire. Then the ratio of extensions produced in the wires by applying same load will be

• A. 1:1
• B. 1:2
• C. 2:1
• D. 4:1

Hint:

When force, length, and material are the same, elongation is inversely proportional to the square of the diameter ($\Delta L \propto \frac{1}{D^2}$). If diameter doubles, elongation becomes $1/4$. Thus, the ratio of elongations will be $4:1$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the ratio of extensions (elongations) in two wires. The wires are made of the same material (same Young's modulus), have the same length, but one has twice the diameter of the other. They are subjected to the same load.
Step 2: Key Formula or Approach:
Young's modulus ($Y$) is defined as: \[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\Delta L} \] Rearranging for elongation ($\Delta L$): \[ \Delta L = \frac{FL}{AY} \] The cross-sectional area of a wire is \( A = \pi r^2 = \pi \left( \frac{D}{2} \right)^2 = \frac{\pi D^2}{4} \), where \( D \) is the diameter.
So, \( \Delta L = \frac{FL}{(\frac{\pi D^2}{4})Y} = \frac{4FL}{\pi D^2 Y} \).
Step 3: Detailed Explanation:
Given:
- Same material: \( Y_1 = Y_2 = Y \)
- Same length: \( L_1 = L_2 = L \)
- Same load: \( F_1 = F_2 = F \)
- Diameter of second wire is two times the first: \( D_2 = 2D_1 \)
Let \( \Delta L_1 \) and \( \Delta L_2 \) be the extensions of the first and second wires, respectively.
\[ \Delta L_1 = \frac{4FL}{\pi D_1^2 Y} \] \[ \Delta L_2 = \frac{4FL}{\pi D_2^2 Y} \] Form the ratio \( \frac{\Delta L_1}{\Delta L_2} \): \[ \frac{\Delta L_1}{\Delta L_2} = \frac{\frac{4FL}{\pi D_1^2 Y}}{\frac{4FL}{\pi D_2^2 Y}} = \frac{D_2^2}{D_1^2} \] Substitute \( D_2 = 2D_1 \): \[ \frac{\Delta L_1}{\Delta L_2} = \frac{(2D_1)^2}{D_1^2} = \frac{4D_1^2}{D_1^2} = \frac{4}{1} \] Step 4: Final Answer:
The ratio of extensions produced in the wires is 4:1.