Question

The length of the metallic wire is $l_1$ when the tension in it is $T_1$ It is $l_2$ when the tension is $T_2$. The original length of the wire will be:

• A. $\frac{\left(l_1+l_2\right)}{2}$
• B. $\frac{\left(T_1{l}_1-T_2{l}_2\right)}{\left(T_2-T_1\right)}$
• C. $\frac{\left(T_2{l}_1+T_1{l}_2\right)}{\left(T_2+T_1\right)}$
• D. $\frac{\left(T_2{l}_1-T_1{l}_2\right)}{\left(T_2-T_1\right)}$

Answer 1

The correct option is D

$\frac{\left(T_2{l}_1-T_1{l}_2\right)}{\left(T_2-T_1\right)}$

Step 1: Given that:

The length of the first metallic wire is $l_1$

The tension of the first metallic wire$T_1$

The length of the second metallic wire is $l_2$

The tension of the second metallic wire is $T_2$

Step 3: Draw the required diagram:

Step 2: Formula used:

The general formula of the young modulus in terms of stress and strain$=\frac{Stress}{Strain}$

Stress can be calculated as -$stress=\frac{F}{A}$

where $F$is the force exerted by the object under tension, $A$ is the cross-sectional area

Strain will be calculated as- $Strain=\frac{∆l}{l}$

$∆l$ is the change in length and $l$ is the actual length.

Thus putting values we get young's modulus as-

$y=\frac{{\displaystyle \frac{F}{A}}}{{\displaystyle \frac{∆l}{l}}}$

Step 3: Calculating the original length:

Therefore for the first wire's tension $T_1$:

Using the formula of young's modulus-

For the first metallic wire-

Substituting the known values-

$y=\frac{{\displaystyle \frac{T_1}{A}}}{{\displaystyle \frac{\left(l-l_1\right)}{l}}}\dots \dots \dots \dots \dots \dots \dots \dots \left(1\right)$

Now for the second wire's tension $T_2$

Substituting the known values-

$y=\frac{{\displaystyle \frac{T_2}{A}}}{{\displaystyle \frac{\left(l-l_2\right)}{l}}}\dots \dots \dots \dots \dots \dots \dots \dots \dots \left(2\right)$

Step 4: Now equating the equations:

Therefore, equating equation (1) and equation (2) we get-

$$\begin{gathered} \Rightarrow \frac{{\displaystyle \frac{T_1}{A}}}{{\displaystyle \frac{\left(l-l_1\right)}{l}}}=\frac{{\displaystyle \frac{T_2}{A}}}{{\displaystyle \frac{\left(l-l_2\right)}{l}}} \\ \Rightarrow \frac{T_1}{\left(l-l_1\right)}=\frac{T_2}{\left(l-l_2\right)} \\ \Rightarrow T_{1}l-T_1{l}_2=T_{2}l-T_2{l}_1 \\ \Rightarrow l(T_1-T_2)=T_1{l}_2-T_2{l}_1 \\ \Rightarrow l=\frac{T_2{l}_1-T_1{l}_2}{T_2-T_1} \end{gathered}$$

Therefore the original length of the wire will be $\frac{\left(T_2{l}_1-T_1{l}_2\right)}{\left(T_2-T_1\right)}$.

Hence, option D is the correct answer.