Mechanical Properties of Solids

JEE Main level, single correct, medium

33 of 118

Question 33 of 118

Young’s moduli of the material of wires A and B are in the ratio of 1:41:4, while its area of cross sections are in the ratio of 1:31:3. If the same amount of load is applied to both the wires, the amount of elongation produced in the wires A and B will be in the ratio of [Assume length of wires A and B are same]

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Answer: (A) 12:112:1

The formula to calculate the elongation of a metallic wire because of the application of a load can be written as
ΔL=(FLAY)...(1)\begin{matrix}\Delta L=\left(\dfrac{FL}{AY}\right)\end{matrix}...\left(1\right)
The ratio of the elongations for both the wires is, then, calculated as follows-
∆L1∆L2=FLA1Y1FLA2Y2=A2Y2A1Y1=31×41=121\begin{matrix}\dfrac{∆{L}_{1}}{∆{L}_{2}} & = & \dfrac{\dfrac{FL}{{A}_{1}{Y}_{1}}}{\dfrac{FL}{{A}_{2}{Y}_{2}}} \\ & = & \dfrac{{A}_{2}{Y}_{2}}{{A}_{1}{Y}_{1}} \\ & = & \dfrac{3}{1}\times \dfrac{4}{1} \\ & = & \dfrac{12}{1}\end{matrix}

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