The length of a metal wire is ${l_1}$when the tension in it is \[{F_1}\] and ${l_2}$ when the tension in it is${F_2}$. The natural length of the wire is
A. $\dfrac{{{l_1}{F_1} + {l_2}{F_2}}}{{{F_1} + {F_2}}}$
B. \[\dfrac{{{l_2} - {l_1}}}{{{F_2} - {F_1}}}\]
C. $\dfrac{{{l_1}{F_2} - {l_2}{F_1}}}{{{F_2} - {F_1}}}$
D. $\dfrac{{{l_1}{F_1} - {l_2}{F_2}}}{{{F_2} - {F_1}}}$
Answer
569.4k+ views
Hint: Tension force: The tension force is defined as the force that is transmitted through rope, string or wire when pulled by forces acting from opposite sides. Tension force is directed over the length of wire. Tension is opposite of compression. Tension doesn’t work on its own but only transfers.
Complete step by step solution:
Hooke’s law: It states that the strain of the material is proportional to the applied stress within the plastic limit of that material.
Let be natural length of wire be ${l_0}$
Using Hooke’s law
$ \Rightarrow Y = \dfrac{{F{l_0}}}{{A\Delta l}}$
$ \Rightarrow \Delta l = l - {l_0}$
So the following equation will become,
$ \Rightarrow \Delta l = \dfrac{{F{l_0}}}{{AY}}$
$ \Rightarrow l - {l_0} = \dfrac{{F{l_0}}}{{AY}}$
Case 1: Tension in T1 and length of the wire $l = {l_1}$
$ \Rightarrow {l_1} - {l_0} = \dfrac{{{F_1}{l_0}}}{{AY}}$ …(1)
Case 2: Tension in T2 and length of wire $l = {l_2}$
$ \Rightarrow {l_2} - {l_0} = \dfrac{{{F_2}{l_0}}}{{AY}}$ …(2)
Dividing the equation 2 from 1
$ \Rightarrow \dfrac{{{l_1} - {l_0}}}{{{l_2} - {l_0}}} = \dfrac{{{F_1}}}{{{F_2}}}$
\[ \Rightarrow {F_2}({l_1} - {l_0}) = {F_1}({l_2} - {l_0})\]
\[ \Rightarrow {F_2}{l_1} - {F_2}{l_0} = {F_1}{l_2} - {F_1}{l_0}\]
$ \Rightarrow {l_0} = \dfrac{{{l_1}{F_2} - {l_2}{F_1}}}{{{F_2} - {F_1}}}$
So the answer is (C)$\dfrac{{{l_1}{F_2} - {l_2}{F_1}}}{{{F_2} - {F_1}}}$.
Note: Young modulus is the ratio of stress (force per unit area) to strain (change in length to the original length). Young’s modulus of any material will never change. It depends on the temperature and pressure. Young’s modulus is used to calculate the change in dimension of the object. It also describes the relative stiffness of the material.
Complete step by step solution:
Hooke’s law: It states that the strain of the material is proportional to the applied stress within the plastic limit of that material.
Let be natural length of wire be ${l_0}$
Using Hooke’s law
$ \Rightarrow Y = \dfrac{{F{l_0}}}{{A\Delta l}}$
$ \Rightarrow \Delta l = l - {l_0}$
So the following equation will become,
$ \Rightarrow \Delta l = \dfrac{{F{l_0}}}{{AY}}$
$ \Rightarrow l - {l_0} = \dfrac{{F{l_0}}}{{AY}}$
Case 1: Tension in T1 and length of the wire $l = {l_1}$
$ \Rightarrow {l_1} - {l_0} = \dfrac{{{F_1}{l_0}}}{{AY}}$ …(1)
Case 2: Tension in T2 and length of wire $l = {l_2}$
$ \Rightarrow {l_2} - {l_0} = \dfrac{{{F_2}{l_0}}}{{AY}}$ …(2)
Dividing the equation 2 from 1
$ \Rightarrow \dfrac{{{l_1} - {l_0}}}{{{l_2} - {l_0}}} = \dfrac{{{F_1}}}{{{F_2}}}$
\[ \Rightarrow {F_2}({l_1} - {l_0}) = {F_1}({l_2} - {l_0})\]
\[ \Rightarrow {F_2}{l_1} - {F_2}{l_0} = {F_1}{l_2} - {F_1}{l_0}\]
$ \Rightarrow {l_0} = \dfrac{{{l_1}{F_2} - {l_2}{F_1}}}{{{F_2} - {F_1}}}$
So the answer is (C)$\dfrac{{{l_1}{F_2} - {l_2}{F_1}}}{{{F_2} - {F_1}}}$.
Note: Young modulus is the ratio of stress (force per unit area) to strain (change in length to the original length). Young’s modulus of any material will never change. It depends on the temperature and pressure. Young’s modulus is used to calculate the change in dimension of the object. It also describes the relative stiffness of the material.
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