Question

Young's modulus of the material of wires X and Y are in the ratio \(4:1\) and the areas of cross sections of the wires X and Y are in the ratio \(2:1\). If the same amount of load is applied to both the wires, the ratio of elongation produced in the wires X and Y will be: (Assume length of the wires X and Y initially are the same)

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

Hint:

For same force and length, elongation is inversely proportional to area and Young’s modulus: \(\Delta L \propto \frac{1}{AY}\).

Answer 1

The Correct Option is A

Step 1: Formula for elongation.
\[ \Delta L = \frac{FL}{AY} \]

Step 2: Identify given ratios.
\[ Y_X:Y_Y = 4:1 \] \[ A_X:A_Y = 2:1 \]

Step 3: Since load and length are same.
\[ \Delta L \propto \frac{1}{AY} \]

Step 4: Write ratio of elongations.
\[ \frac{\Delta L_X}{\Delta L_Y} = \frac{A_Y Y_Y}{A_X Y_X} \]

Step 5: Substitute values.
\[ \frac{\Delta L_X}{\Delta L_Y} = \frac{1 \times 1}{2 \times 4} \]
\[ \frac{\Delta L_X}{\Delta L_Y} = \frac{1}{8} \]

Step 6: Final conclusion.
\[ \boxed{1:8} \] Hence, correct answer is option (A).