
Calculate the force required to increase the length of a steel wire of cross-sectional area ${{10}^{-6}}{{m}^{2}}$ by 0.5% . Given, ${{Y}_{steel}}=2\times {{10}^{11}}N{{m}^{-2}}$ .
Answer
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Hint: We will assume the cross-sectional area of the rod remains constant throughout the process and then apply Hooke’s law. Hooke’s law states that the stress applied on an inelastic body is directly proportional to the strain generated in the body. And the constant of proportionality is called Young’s Modulus of Elasticity.
Complete step-by-step answer:
We will first assign some useful terms that we are going to use later in our solution. Let the length of the steel wire be given by ‘l’. Then, we can write the change in length of the steel wire as ‘$\vartriangle l$ ’.
Now, the percentage strain in the wire has been given to us in the problem as:
$\begin{align}
& \Rightarrow \dfrac{\vartriangle l}{l}\%=0.5 \\
& \therefore \dfrac{\vartriangle l}{l}=0.5\times {{10}^{-2}} \\
\end{align}$
Also, let the area of cross-section of the steel wire be denoted by ‘A’. Then, it has been given to us in the problem that:
$\Rightarrow A={{10}^{-6}}{{m}^{2}}$
Now, the Hooke’s law for an inelastic material is given by the following equation:
$\begin{align}
& \Rightarrow Stress\propto Strain \\
& or,\dfrac{F}{A}=Y\dfrac{\vartriangle l}{l} \\
\end{align}$
Where, Y is the constant of proportionality called the Young’s modulus of elasticity and has been given to us in the problem as:
$\Rightarrow Y=2\times {{10}^{-11}}N{{m}^{-2}}$
Thus, putting the values of all the respective terms in the equation for Hooke’s law, we get:
$\begin{align}
& \Rightarrow \dfrac{F}{{{10}^{-6}}}=2\times {{10}^{11}}\left( 0.5\times {{10}^{-2}} \right) \\
& \Rightarrow F={{10}^{3}}N \\
& \therefore F=1000N \\
\end{align}$
Hence, a force of 1000N is required to increase the length of a steel wire of cross-sectional area ${{10}^{-6}}{{m}^{2}}$ by 0.5% .
Note: It should be noted that we were able to calculate this force without knowing the length of the wire, but the fractional change in length only. Thus, we can conclude with a statement that for a wire of certain cross-sectional area and fractional change in length, the amount of force required to bring about this change is independent of the length of the wire.
Complete step-by-step answer:
We will first assign some useful terms that we are going to use later in our solution. Let the length of the steel wire be given by ‘l’. Then, we can write the change in length of the steel wire as ‘$\vartriangle l$ ’.
Now, the percentage strain in the wire has been given to us in the problem as:
$\begin{align}
& \Rightarrow \dfrac{\vartriangle l}{l}\%=0.5 \\
& \therefore \dfrac{\vartriangle l}{l}=0.5\times {{10}^{-2}} \\
\end{align}$
Also, let the area of cross-section of the steel wire be denoted by ‘A’. Then, it has been given to us in the problem that:
$\Rightarrow A={{10}^{-6}}{{m}^{2}}$
Now, the Hooke’s law for an inelastic material is given by the following equation:
$\begin{align}
& \Rightarrow Stress\propto Strain \\
& or,\dfrac{F}{A}=Y\dfrac{\vartriangle l}{l} \\
\end{align}$
Where, Y is the constant of proportionality called the Young’s modulus of elasticity and has been given to us in the problem as:
$\Rightarrow Y=2\times {{10}^{-11}}N{{m}^{-2}}$
Thus, putting the values of all the respective terms in the equation for Hooke’s law, we get:
$\begin{align}
& \Rightarrow \dfrac{F}{{{10}^{-6}}}=2\times {{10}^{11}}\left( 0.5\times {{10}^{-2}} \right) \\
& \Rightarrow F={{10}^{3}}N \\
& \therefore F=1000N \\
\end{align}$
Hence, a force of 1000N is required to increase the length of a steel wire of cross-sectional area ${{10}^{-6}}{{m}^{2}}$ by 0.5% .
Note: It should be noted that we were able to calculate this force without knowing the length of the wire, but the fractional change in length only. Thus, we can conclude with a statement that for a wire of certain cross-sectional area and fractional change in length, the amount of force required to bring about this change is independent of the length of the wire.
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