Question

Young's moduli of two wires A and B are in the ratio 7 : 4. Wire A is 2 m long and has radius R. Wire B is 1.5 m long and has radius 2 mm. If the two wires stretch by the same length for a given load, then the value of R is close to:
A. 1.9 mm
B. 1.7 mm
C. 1.3 mm
D. 1.5 mm

Answer 1

This question is based on the concept of Young’s modulus of a material. As the wires are stretched by the same length, so, we will consider it to be a constant value. We will consider two equations, that is, for wires A and B and equate them considering the forces applied to be the same. Then, we can find the value of the radius by using the area of the circle.
Formula used:
$Y=\dfrac{FL}{A\Delta L}$

Complete answer:
The formula for calculating Young’s modulus of the material is given as follows.
$Y=\dfrac{FL}{A\Delta L}$
Where F is the force exerted on the material under tension, L is the original length of the material, A is the cross-sectional area of the material and$\Delta L$is the amount by which the material changes.
Rearrange the terms in order to obtain the equation in terms of the force exerted on the material under tension. As the wires are stretched by the same length, so, we will consider it to be a constant value.

& F=\dfrac{YA}{L}\Delta L \\\ & \Rightarrow F\propto \dfrac{YA}{L} \\\ \end{aligned}$$ The different equations for the wires A and B are as follows. $$\begin{aligned} & F\propto \dfrac{{{Y}_{1}}{{A}_{1}}}{{{L}_{1}}} \\\ & F\propto \dfrac{{{Y}_{2}}{{A}_{2}}}{{{L}_{2}}} \\\ \end{aligned}$$ Equate the above equations. $$\dfrac{{{Y}_{1}}{{A}_{1}}}{{{L}_{1}}}=\dfrac{{{Y}_{2}}{{A}_{2}}}{{{L}_{2}}}$$ Represent the area of the wire as the area of the circle in terms of the radius. So, we have, $$\begin{aligned} & \dfrac{{{Y}_{1}}\pi R_{1}^{2}}{{{L}_{1}}}=\dfrac{{{Y}_{2}}\pi R_{2}^{2}}{{{L}_{2}}} \\\ & \Rightarrow \dfrac{{{Y}_{1}}R_{1}^{2}}{{{L}_{1}}}=\dfrac{{{Y}_{2}}R_{2}^{2}}{{{L}_{2}}} \\\ \end{aligned}$$ From given, we have the values of the ratio of the Young’s modulus, the lengths of the wires and the radius of one of the wires. So, substitute the given values in the above equation. $$\begin{aligned} & \dfrac{(7){{R}^{2}}}{2}=\dfrac{(4)\times {{2}^{2}}}{1.5} \\\ & \Rightarrow {{R}^{2}}=\dfrac{16}{1.5}\times \dfrac{2}{7} \\\ & \Rightarrow R=1.74\,mm \\\ \end{aligned}$$ Therefore, the radius of the wire A can be rounded off to the nearest digit given by, 1.7 mm **So, the correct answer is “Option B”.** **Note:** When the lengths of the wires are stretched have the same value, the force exerted under the tension will be considered to be the same in most of the cases. Even sometimes, they mention that the force exerted is different. In this case, the radius is asked, even they can ask for the diameter of the wire.