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Question: Data for a steel wire on an electric guitar are listed. Diameter =\(5.0\times {{10}^{-4}}m\) Yo...

Data for a steel wire on an electric guitar are listed.
Diameter =5.0×10−4m5.0\times {{10}^{-4}}m
Young’s modulus =2.0×1011Pa2.0\times {{10}^{11}}Pa
Tension = 20N20N
The wire snaps and contracts elastically. Assume the wire obeys Hooke’s law.
By what percentage does the length ll of a piece of the wire contract?
A 1.3×10−41.3\times {{10}^{-4}} %
B 5.1×10−45.1\times {{10}^{-4}} %
C 1.3×10−21.3\times {{10}^{-2}} %
D 5.1×10−25.1\times {{10}^{-2}} %

Explanation

Solution

Recall that the young’s modulus of a wire is given by the ratio of tensile stress to longitudinal strain. In the question, we are actually asked the percentage longitudinal strain. You could rearrange the above expression and then get the longitudinal strain. You are directly given all the required values. Substituting all that and then multiplying with 100 will give you required percentage contraction.
Formula used:
Expression for young’s modulus,
Y=σε=FAΔllY=\dfrac{\sigma }{\varepsilon }=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}

Complete answer:
In the question, we are given some data on an electric guitar. We are given its diameter, Young’s modulus and tension. In the question the wire snaps and contracts elastically and also obeys Hooke’s law. If the length of the wire was given byll then, we are asked to find what percentage of that length is contracting.
From the definition of Young’s modulus (Y), we know that it is the ratio of tensile (or compressive) stress(σ)\left( \sigma \right) to the longitudinal strain(ε)\left( \varepsilon \right). That is,
Y=σεY=\dfrac{\sigma }{\varepsilon } …………………… (1)
Tensile stress just like any other stress is the force acting per unit cross-sectional area, that is,
σ=FA\sigma =\dfrac{F}{A} ………………………………. (2)
Longitudinal strain is the ratio of change in length to the original length, that is,
ε=Δll\varepsilon =\dfrac{\Delta l}{l} ……………………………… (3)
In the question we are actually asked to find the percentage strain.
Substituting (2) and (3) in (1), we get,
⇒Y=FAΔll\Rightarrow Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}
⇒Δll=FA×Y\Rightarrow \dfrac{\Delta l}{l}=\dfrac{F}{A\times Y} …………………………………. (4)
Given values:
F=20NF=20N
Y=2.0×1011PaY=2.0\times {{10}^{11}}Pa
Diameter of the wire is given as,
d=5.0×10−4md=5.0\times {{10}^{-4}}m
So, radius of the wire will be,
r=2.5×10−4mr=2.5\times {{10}^{-4}}m
Then, area of cross-section is given by,
A=π×r2A=\pi \times {{r}^{2}}
A=π×(2.5×10−4)2=19.625×10−8m2A=\pi \times {{\left( 2.5\times {{10}^{-4}} \right)}^{2}}=19.625\times {{10}^{-8}}{{m}^{2}}
Substituting all these values in (4) we get,
Δll=20N19.625×10−8m2×2.0×1011Pa\dfrac{\Delta l}{l}=\dfrac{20N}{19.625\times {{10}^{-8}}{{m}^{2}}\times 2.0\times {{10}^{11}}Pa}
⇒Δll=2039.25×103\Rightarrow \dfrac{\Delta l}{l}=\dfrac{20}{39.25\times {{10}^{3}}}
⇒Δll=0.51×10−3\Rightarrow \dfrac{\Delta l}{l}=0.51\times {{10}^{-3}}
We are asked to find the percentage contraction, that is,Δll×100\dfrac{\Delta l}{l}\times 100
⇒Δll×100=0.51×10−3×100\Rightarrow \dfrac{\Delta l}{l}\times 100=0.51\times {{10}^{-3}}\times 100
⇒5.1×10−2\Rightarrow 5.1\times {{10}^{-2}}
Therefore, the percentage contraction in length is
5.1×10−25.1\times {{10}^{-2}} %

So, the correct answer is “Option D”.

Note:
You may note that we have directly substituted the value of young’s modulus in Pascal. This is because we know that,
1Pa=1Nm−21Pa=1N{{m}^{-{{2}^{{}}}}}
Also, you should be careful while dealing with the powers of ten.