Question

A wire is stretched 1 mm by a force F. If a second wire of same material, same length and 4 times the diameter of the first wire is stretched by the same force F, then the elongation of the second wire is

• A. 1/8 mm
• B. 8 mm
• C. 16 mm
• D. 1/16 mm

Hint:

Young's modulus: $Y = \frac{FL}{A\Delta L}$. Area is proportional to the square of the diameter.

Answer 1

The Correct Option is D

Young's modulus $Y = \frac{FL}{A\Delta L}$, where $F$ is the force, $L$ is the length, $A$ is the cross-sectional area, and $\Delta L$ is the elongation. Since the material and length are the same for both wires, we have: $Y = \frac{FL}{A_1 \Delta L_1} = \frac{FL}{A_2 \Delta L_2}$ $\frac{1}{A_1 \Delta L_1} = \frac{1}{A_2 \Delta L_2}$ $A_1\Delta L_1 = A_2\Delta L_2$ Given that the diameter of the second wire is 4 times the diameter of the first wire, the area of the second wire is $A_2 = 16A_1$. $A_1(1) = 16A_1 \Delta L_2$ $\Delta L_2 = \frac{1}{16}$ mm.