Question 65: Properties of Matter

Question

Young's moduli of the material of wires A and B are in the ratio of $$1: 4$$, while its area of cross sections are in the ratio of $$1: 3$$. If the same amount of load is applied to both the wires, the amount of elongation produced in the wires $$\mathrm{A}$$ and $$\mathrm{B}$$ will be in the ratio of
[Assume length of wires A and B are same]

Accepted Answer

Given the formula for elongation in a material due to a force:
$$
\Delta L = \frac{FL}{AY}
$$
where:
F is the force applied,
L is the original length,
A is the cross-sectional area of the material, and
Y is Young's modulus of the material.
The ratio of the elongations in the two wires A and B is given by:
$$
\frac{\Delta L_1}{\Delta L_2} = \frac{F_1}{F_2} 	imes \frac{A_2}{A_1} 	imes \frac{Y_2}{Y_1}
$$
Since the same force is applied on both wires (i.e., ($F_1/F_2$ = 1)), the areas are in the ratio 1:3 (i.e., ($A_2/A_1$ = 3)), and the Young's moduli are in the ratio 1:4 (i.e., ($Y_2/Y_1$ = 4)), substituting these values into the equation gives:
$$
\frac{\Delta L_1}{\Delta L_2} = 1 	imes 3 	imes 4 = 12
$$
So, the ratio of the elongations is 12:1, which indicates that wire A will elongate 12 times more than wire B when the same force is applied.

Answer

1 : 12

Answer

1 : 36

Answer

12 : 1

Answer

36 : 1

Properties of Matter

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