Question

Two wires of cross sectional area \(1 \text{ cm}^2\) and \(2 \text{ cm}^2\) and lengths 20 cm and 30 cm are connected to the same load. If their extensions are same, find the ratio of their Young's modulus :

• A. \(\frac{4}{2}\)
• B. \(\frac{4}{3}\)
• C. \(\frac{1}{2}\)
• D. \(\frac{3}{2}\)

Hint:

When variables like load and extension are "same", isolate the remaining variables in the formula. Here, \(Y \cdot \frac{A}{L} = \text{constant}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Young's Modulus (\(Y\)) is a measure of the stiffness of a solid material. It is the ratio of stress to strain.
Step 2: Key Formula or Approach:
Young's Modulus \(Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\Delta L}\).
Step 3: Detailed Explanation:
Given that the load (\(F\)) and extension (\(\Delta L\)) are the same for both wires:
\[ Y \propto \frac{L}{A} \]
Let the first wire have \(L_1 = 20 \text{ cm}\) and \(A_1 = 1 \text{ cm}^2\).
Let the second wire have \(L_2 = 30 \text{ cm}\) and \(A_2 = 2 \text{ cm}^2\).
The ratio of their Young's moduli is:
\[ \frac{Y_1}{Y_2} = \frac{L_1 / A_1}{L_2 / A_2} = \frac{20 / 1}{30 / 2} \]
\[ \frac{Y_1}{Y_2} = \frac{20}{15} = \frac{4}{3} \]
Step 4: Final Answer:
The ratio of their Young's moduli is \(\frac{4}{3}\).