
Two wires are made up of same material and have the same material and have the same volume .However wire 1 has cross sectional area A and wire 2 has cross sectional area 3A .If the length of wire 1 increases by $\Delta $x on applying force F ,how much force is needed to stretch wire 2 by same amount?
1. 4F
2. 6F
3. 9F
4. F
Answer
561k+ views
Hint:The above question is based on Young’s modulus concept.
Young Modulus is given by the formula:
$Y = \dfrac{{longitudinal\ stress}}{{strain}}$
$$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}}$$ , F is the force , A is the area and l is the length and $\Delta L$ is the change in length.
Using the above relation we will solve the problem.
Complete step by step solution:
Young’s Modulus is a measure of the ability of a material to withstand changes in length when under lengthwise tension or compression .Young’s modulus is the ratio of longitudinal stress to strain.SI unit of Young’s modulus is Pascal.
$Y = \dfrac{{longitudinal\ stress}}{{strain}}$
$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}}$
Let’s solve the given problem now:
$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta X}}{L}}}$.............1 (for wire 1 ,when stretched to $\Delta x$)
We can write equation 1 as :
$Y = \dfrac{{FL}}{{A\Delta X}}$
We also have relation:
$
$V = A \times L $
$ \dfrac{V}{A} = L $
$ (V is the volume of the ,L is the length and A is the area )
Therefore we can write equation 1 as :
$Y = \dfrac{{FV}}{{{A^2}\Delta X}}$...............2
Similarly, we will proceed for wire 2.
$ \Rightarrow Y = \dfrac{{\dfrac{{F'}}{{3A}}}}{{\dfrac{{\Delta X}}{L}}}$
$ \Rightarrow Y = \dfrac{{F'V}}{{{{(3A)}^2}\Delta X}}$ (substituting values of L and area of wire 2)
$ \Rightarrow Y = \dfrac{{F'V}}{{9{A^2}\Delta X}}$............3
Equating equation 2&3
$\dfrac{{FV}}{{{A^2}\Delta X}}$$ = \dfrac{{F'V}}{{9{A^2}\Delta X}}$( on cancelling the similar terms )
$ \Rightarrow F' = 9F$
Therefore, option 3 is the correct force of wire 2 is 9F.
Note:
Application of Young’s Modulus in our day to day life is in designing a bridge ,one has to keep in mind the traffic load, weight of the bridge ,force of wind .Bridge should also not bend too much nor break. Another application is heavy load lifted by the crane.
Young Modulus is given by the formula:
$Y = \dfrac{{longitudinal\ stress}}{{strain}}$
$$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}}$$ , F is the force , A is the area and l is the length and $\Delta L$ is the change in length.
Using the above relation we will solve the problem.
Complete step by step solution:
Young’s Modulus is a measure of the ability of a material to withstand changes in length when under lengthwise tension or compression .Young’s modulus is the ratio of longitudinal stress to strain.SI unit of Young’s modulus is Pascal.
$Y = \dfrac{{longitudinal\ stress}}{{strain}}$
$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}}$
Let’s solve the given problem now:
$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta X}}{L}}}$.............1 (for wire 1 ,when stretched to $\Delta x$)
We can write equation 1 as :
$Y = \dfrac{{FL}}{{A\Delta X}}$
We also have relation:
$
$V = A \times L $
$ \dfrac{V}{A} = L $
$ (V is the volume of the ,L is the length and A is the area )
Therefore we can write equation 1 as :
$Y = \dfrac{{FV}}{{{A^2}\Delta X}}$...............2
Similarly, we will proceed for wire 2.
$ \Rightarrow Y = \dfrac{{\dfrac{{F'}}{{3A}}}}{{\dfrac{{\Delta X}}{L}}}$
$ \Rightarrow Y = \dfrac{{F'V}}{{{{(3A)}^2}\Delta X}}$ (substituting values of L and area of wire 2)
$ \Rightarrow Y = \dfrac{{F'V}}{{9{A^2}\Delta X}}$............3
Equating equation 2&3
$\dfrac{{FV}}{{{A^2}\Delta X}}$$ = \dfrac{{F'V}}{{9{A^2}\Delta X}}$( on cancelling the similar terms )
$ \Rightarrow F' = 9F$
Therefore, option 3 is the correct force of wire 2 is 9F.
Note:
Application of Young’s Modulus in our day to day life is in designing a bridge ,one has to keep in mind the traffic load, weight of the bridge ,force of wind .Bridge should also not bend too much nor break. Another application is heavy load lifted by the crane.
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