
The diameter of a wire decreases by $5\%$ when wire is subjected to a stress of $8\times {{10}^{10}}\,N/{{m}^{2}}$. If $\sigma =0.5$ for material of wire then Young’s modulus of material of wire:
$\begin{align}
& a)\,4\times {{10}^{10}}N/{{m}^{2}} \\
& b)\,4\times {{10}^{11}}N/{{m}^{2}} \\
& c)\,8\times {{10}^{10}}N/{{m}^{2}} \\
& d)\,8\times {{10}^{11}}N/{{m}^{2}} \\
\end{align}$
Answer
576.3k+ views
Hint: Change in diameter of a wire is given, for calculating Young’s modulus we need stress and strain value of a wire, stress is given and strain is change in length, which needs to be calculated first before calculating the final answer.
Formula used:
$Young's\,Modulus(Y)=\dfrac{Stress}{Strain}$
Complete answer:
Young's modulus ($E$), the Young modulus or the modulus of elasticity in tension, is a mechanical property that measures the tensile stiffness of a solid material. It quantifies the relationship between tensile stress $\sigma $ (force per unit area) and axial strain $\varepsilon $ (proportional deformation) in the linear elastic region of a material and is determined using the formula:
$E=\dfrac{\sigma }{\varepsilon }$
First, we need to calculate the stress of the wire, so that we can substitute that value in the Young’s Modulus equation to get the final answer.
$Volume\,of\,the\,wire\,(V)=(\dfrac{\pi {{d}^{2}}}{4})\times l$
Differentiate the above formula, so that we can get:
$\dfrac{\Delta V}{V}=2(\dfrac{\Delta d}{d})+\dfrac{\Delta l}{l}$, where $\dfrac{\Delta V}{V}\approx 0$, as there is no change in the volume of wire.
$\dfrac{\Delta d}{d}=-(0.05)$, as the diameter of the wire decreases by $5\%$, and $\dfrac{\Delta l}{l}$ is strain, which we need to find.
After substituting the value in the equation, we get:
$\begin{align}
& 0=2(-0.05)+\dfrac{\Delta l}{l} \\
& \dfrac{\Delta l}{l}=0.1 \\
\end{align}$
which means, $Strain=0.1$
Now, Putting the value of Stress and Strain in the Young’s Modulus equation, we get:
$\begin{align}
& Y=\dfrac{8\times {{10}^{10}}N/{{m}^{2}}}{0.1} \\
& Y=8\times {{10}^{11}}N/{{m}^{2}} \\
\end{align}$
So, the correct answer is Option (d).
Note:
The value of $\sigma $, given in the question is there to confuse you, as the Young’s modulus formula can also be written in terms of $\sigma $. So, understand the question properly first, then answer it or else you will never get the correct answer.
Formula used:
$Young's\,Modulus(Y)=\dfrac{Stress}{Strain}$
Complete answer:
Young's modulus ($E$), the Young modulus or the modulus of elasticity in tension, is a mechanical property that measures the tensile stiffness of a solid material. It quantifies the relationship between tensile stress $\sigma $ (force per unit area) and axial strain $\varepsilon $ (proportional deformation) in the linear elastic region of a material and is determined using the formula:
$E=\dfrac{\sigma }{\varepsilon }$
First, we need to calculate the stress of the wire, so that we can substitute that value in the Young’s Modulus equation to get the final answer.
$Volume\,of\,the\,wire\,(V)=(\dfrac{\pi {{d}^{2}}}{4})\times l$
Differentiate the above formula, so that we can get:
$\dfrac{\Delta V}{V}=2(\dfrac{\Delta d}{d})+\dfrac{\Delta l}{l}$, where $\dfrac{\Delta V}{V}\approx 0$, as there is no change in the volume of wire.
$\dfrac{\Delta d}{d}=-(0.05)$, as the diameter of the wire decreases by $5\%$, and $\dfrac{\Delta l}{l}$ is strain, which we need to find.
After substituting the value in the equation, we get:
$\begin{align}
& 0=2(-0.05)+\dfrac{\Delta l}{l} \\
& \dfrac{\Delta l}{l}=0.1 \\
\end{align}$
which means, $Strain=0.1$
Now, Putting the value of Stress and Strain in the Young’s Modulus equation, we get:
$\begin{align}
& Y=\dfrac{8\times {{10}^{10}}N/{{m}^{2}}}{0.1} \\
& Y=8\times {{10}^{11}}N/{{m}^{2}} \\
\end{align}$
So, the correct answer is Option (d).
Note:
The value of $\sigma $, given in the question is there to confuse you, as the Young’s modulus formula can also be written in terms of $\sigma $. So, understand the question properly first, then answer it or else you will never get the correct answer.
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