
The elongation of a steel wire stretched by a force is 'e'. If a wire of the same material of double the length and half the diameter is subjected to double the force, its elongation will be ?
(A) 16e
(B) 4e
(C) \[\dfrac{e}{4}\]
(D) \[\dfrac{e}{16}\]
Answer
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Hint: The elongation of a material upon application of force can be realized by using the concept of Modulus of solids. Since we are talking about wires, we have to take the concept of Young’s Modulus.
Young’s modulus does not depend upon length or area of the wire but it is the property of the material and depends upon the nature of the material.
Complete step by step solution:
The given question can easily be solved by using the Hooke’s law of elasticity. The formula of Young’s modulus is: \[Y=\dfrac{Fl}{A\Delta l}\]
here F is the force acting and A is the area of the wire given by \[A=\pi {{r}^{2}}\]or in terms of diameter as \[A=\pi {{\dfrac{d}{4}}^{2}}\] and \[\Delta l\]is the extension in the length. For first case the equation becomes: \[Y=\dfrac{Fl}{Ae}\]-(1)
Now for the second case, the wire is the same so the Young modulus remains the same. Now the length is doubled and the diameter is halved.
$l'=2l$
$F'=2F$
$d'=\dfrac{d}{2}$
So, new area can be written as:
$A'=\pi {{\dfrac{d'}{4}}^{2}} \\
\Rightarrow A'=\pi {{\dfrac{d}{4\times 4}}^{2}} \\
\Rightarrow A'=\dfrac{\pi {{d}^{2}}}{4}\times \dfrac{1}{4} \\
\therefore A'=\dfrac{A}{4} \\$
So, Young modulus can be written as:
$ Y=\dfrac{F'l'}{A'e'} \\
\Rightarrow Y=\dfrac{2F\times 2l}{\dfrac{A}{4}\times e'} \\
\Rightarrow Y=16\times \dfrac{Fl}{Ae'} \\$
Putting the value of Y from equation (1)
$\Rightarrow \dfrac{Fl}{Ae}=16\times \dfrac{Fl}{Ae'} \\
\Rightarrow e'=16e \\$
So, the correct option comes out to be (d)
Note: For the wires of the same material young’s modulus is the same as young’s modulus of elasticity depending upon the nature of material of the substance. If the material of the wires are different then their values of young’s modulus will also be different. Young’s modulus is the property of the material which measures the elasticity of the material.
If we are given a problem in which two rods of different materials are attached then we have to use two sets of young’s moduli for each of them.
Young’s modulus does not depend upon length or area of the wire but it is the property of the material and depends upon the nature of the material.
Complete step by step solution:
The given question can easily be solved by using the Hooke’s law of elasticity. The formula of Young’s modulus is: \[Y=\dfrac{Fl}{A\Delta l}\]
here F is the force acting and A is the area of the wire given by \[A=\pi {{r}^{2}}\]or in terms of diameter as \[A=\pi {{\dfrac{d}{4}}^{2}}\] and \[\Delta l\]is the extension in the length. For first case the equation becomes: \[Y=\dfrac{Fl}{Ae}\]-(1)
Now for the second case, the wire is the same so the Young modulus remains the same. Now the length is doubled and the diameter is halved.
$l'=2l$
$F'=2F$
$d'=\dfrac{d}{2}$
So, new area can be written as:
$A'=\pi {{\dfrac{d'}{4}}^{2}} \\
\Rightarrow A'=\pi {{\dfrac{d}{4\times 4}}^{2}} \\
\Rightarrow A'=\dfrac{\pi {{d}^{2}}}{4}\times \dfrac{1}{4} \\
\therefore A'=\dfrac{A}{4} \\$
So, Young modulus can be written as:
$ Y=\dfrac{F'l'}{A'e'} \\
\Rightarrow Y=\dfrac{2F\times 2l}{\dfrac{A}{4}\times e'} \\
\Rightarrow Y=16\times \dfrac{Fl}{Ae'} \\$
Putting the value of Y from equation (1)
$\Rightarrow \dfrac{Fl}{Ae}=16\times \dfrac{Fl}{Ae'} \\
\Rightarrow e'=16e \\$
So, the correct option comes out to be (d)
Note: For the wires of the same material young’s modulus is the same as young’s modulus of elasticity depending upon the nature of material of the substance. If the material of the wires are different then their values of young’s modulus will also be different. Young’s modulus is the property of the material which measures the elasticity of the material.
If we are given a problem in which two rods of different materials are attached then we have to use two sets of young’s moduli for each of them.
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