Question

A uniform rod of Young's modulus Y is stretched by two tension forces \[{T_1}\] and \[{T_2}\] such that the rods get expanded to length \[{L_1}\] and \[{L_2}\] respectively. Find the initial length of the rod.
A. $\dfrac{L_1 T_1 - L_2 T_2}{T_1 - T_2}$
B. $\dfrac{L_2 T_1 - L_1 T_2}{T_2 - T_1}$
C. $\dfrac{L_1 T_2 - L_2 T_1}{T_2 - T_1}$
D. \[({L_1}/{T_1}) \times ({T_2}/{L_2})\]

Answer 1

Hint: The Young’s modulus (E) is a property of the material that tells us how easily the material can stretch and deform. It is defined as the ratio of tensile stress (\[\sigma \]) to tensile strain (\[\varepsilon \]). Where strain is extension per unit length \[(\varepsilon = dl/l)\] and stress is the amount of force applied per unit area \[(\sigma = F/A)\] . The measurement of the tensile strength of an object is a very important test to determine when an object or material will bend or break.

Formula Used:
The expression of Young’s modulus is,
\[\dfrac{F}{A} = Y\dfrac{{\Delta l}}{l}\]
Here, F is the force, A is the area of cross-section, $\Delta l$ is the change in length and $l$ is the original length of the rod.

Complete step by step solution:
the initial length of the rod to be \[{L_0}\] and area A. Since tension is the external force acting on the rod in this instance, Hooke's law applies,
\[\dfrac{F}{A} = Y\dfrac{{\Delta l}}{l} \\
\Rightarrow \dfrac{T}{A} = Y\dfrac{{\Delta l}}{l} \\ \]
We may compare the tension and elongation created in the rod in both cases as,
\[\dfrac{{{T_1}}}{A} = \dfrac{{Y({L_1} - {L_0})}}{{{L_0}}}\] …… (i)
Here \[({L_1} - {L_0})\] is the elongation in rod when \[{T_1}\] is applied.
Similarly,
\[\dfrac{{{T_2}}}{A} = \dfrac{{Y({L_2} - {L_0})}}{{{L_0}}}\] …… (ii)
Here \[({L_2} - {L_0})\] is the elongation in rod when \[{T_2}\] is applied.
On dividing Eq.(i) and (ii) we get,
\[\dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{{L_1} - {L_0}}}{{{L_2} - {L_0}}} \\
\Rightarrow {T_1}{L_2} - {T_1}{L_0} = {T_2}{L_1} - {T_2}{L_0} \\ \]
\[\therefore {L_0} = \dfrac{{{L_1}{T_2} - {L_2}{T_1}}}{{{T_2} - {T_1}}}\]

Hence, the correct answer is, option(C).

Note: The stress-strain charts might seem substantially different for several types of materials. Materials that are brittle tend to be particularly strong since they can endure great pressure, don't extend much, and break rapidly. Although ductile materials have a greater elastic zone where the stress-strain relationship is linear, this linearity breaks down at the first turnover (the elastic limit), and the material is no longer able to return to its former shape.