Question

Two wires \(A\) and \(B\) made of different materials have lengths \(6.0\) cm and \(5.4\) cm, and areas of cross-sections \(3.0\times10^{-5}\,\text{m}^2\) and \(4.5\times10^{-5}\,\text{m}^2\), respectively. They are stretched by the same magnitude under the same load. If the ratio of Young’s modulus of \(A\) to that of \(B\) is \(x:3\), find the value of \(x\).

• A. \(5\)
• B. \(4\)
• C. \(2\)
• D. \(1\)

Hint:

When load and extension are same, compare Young’s modulus directly using \( Y\propto \frac{L}{A} \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Young's Modulus (\(Y\)) is a property of the material defined as the ratio of tensile stress to tensile strain.
Step 2: Key Formula or Approach:
\[ Y = \frac{F/A}{\Delta L / L} = \frac{F L}{A \Delta L} \]
Step 3: Detailed Explanation:
Given: \(F\) and \(\Delta L\) are the same for both wires.
Thus, \(Y \propto \frac{L}{A}\).
The ratio of Young's Moduli is:
\[ \frac{Y_A}{Y_B} = \frac{L_A}{A_A} \times \frac{A_B}{L_B} \]
\[ \frac{Y_A}{Y_B} = \frac{6.0}{3.0 \times 10^{-5}} \times \frac{4.5 \times 10^{-5}}{5.4} \]
\[ \frac{Y_A}{Y_B} = \frac{6.0}{5.4} \times \frac{4.5}{3.0} \]
\[ \frac{Y_A}{Y_B} = \frac{10}{9} \times \frac{1.5}{1} = \frac{15}{9} = \frac{5}{3} \]
Given \(\frac{Y_A}{Y_B} = \frac{x}{3}\), comparing gives \(x = 5\).
Step 4: Final Answer:
The value of \(x\) is \(5\).