Question

The Young’s modulus of three materials are in the ratio 2 : 2 : 1. Three wires made of these materials have their cross-sectional areas in the ratio 1 : 2 : 3. For a given stretching force the elongation's in the three wires are in the ratio

• A. 1 : 2 : 3
• B. 3 : 2 : 1
• C. 5 : 4 : 3
• D. 6 : 3 : 4

Answer 1

Answer: (D)

6 : 3 : 4

Answer 2

$l = \frac{FL}{AY}$ and for a given stretching force $l \propto \frac{1}{AY}$

Let three wires have young's modulus 2Y, 2Y and Y and their cross sectional areas are A, 2A and $3A$ respectively.

$l_{1}:l_{2}:l_{3} = \frac{1}{A_{1}Y_{1}}:\frac{1}{A_{2}Y_{2}}:\frac{1}{A_{3}Y_{3}} = \frac{1}{A \times 2Y}:\frac{1}{2A \times 2Y}:\frac{1}{3A \times Y} = \frac{1}{2}:\frac{1}{4}:\frac{1}{3} = 6:3:4$.