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Question: A uniform cylindrical wire is subjected to a longitudinal tensile stress of \(5 \times {10^7}\,N\,{m...

A uniform cylindrical wire is subjected to a longitudinal tensile stress of 5×107 N m−25 \times {10^7}\,N\,{m^{ - 2}} . Young's modulus of the material of the wire is 2×1011 N m−22 \times {10^{11}}\,N\,{m^{ - 2}} . The volume change in the wire is 0.02 %0.02\,\% . The fractional change in the radius is
A. 0.25×10−40.25 \times {10^{ - 4}}
B. 0.5×10−40.5 \times {10^{ - 4}}
C. 0.1×10−40.1 \times {10^{ - 4}}
D. 1.5×10−41.5 \times {10^{ - 4}}

Explanation

Solution

When a wire is under stress it obeys Hooke’s law which states that the stress is directly proportional to the strain. Mathematically it is expressed as σ=YΔLL\sigma = Y\dfrac{{\Delta L}}{L} where ΔLL\dfrac{{\Delta L}}{L} is the longitudinal strain. In this question, we are given the value of stress and the young’s modulus through which we will calculate the fractional change in the length. Then we will use the relation V=πr2LV = \pi {r^2}L since the volume change in the wire is given to be 0.02 %0.02\,\% .

Complete step by step answer:
We know that σ=YΔLL\sigma = Y\dfrac{{\Delta L}}{L} where ΔLL\dfrac{{\Delta L}}{L} is the fractional change in length, Y is the Young’s modulus and σ\sigma is the stress.Given that σ=5×107 N m−2\sigma = 5 \times {10^7}\,N\,{m^{ - 2}} and Y=2×1011 N m−2Y = 2 \times {10^{11}}\,N\,{m^{ - 2}} ,
Substituting in the equation we get,
5×107 N m−2=2×1011 N m−2×ΔLL5 \times {10^7}\,N\,{m^{ - 2}} = 2 \times {10^{11}}\,N\,{m^{ - 2}} \times \dfrac{{\Delta L}}{L}
⇒ΔLL=2.5×10−4\Rightarrow \dfrac{{\Delta L}}{L} = 2.5 \times {10^{ - 4}}

Now we know that V=πr2LV = \pi {r^2}L where V is the volume and r is the radius of the cylindrical wire.
ΔVV=πΔ(r2L)πr2L\dfrac{{\Delta V}}{V} = \dfrac{{\pi \Delta ({r^2}L)}}{{\pi {r^2}L}}
Further solving this, we get
ΔVV=2LrΔr+r2ΔLr2L\dfrac{{\Delta V}}{V} = \dfrac{{2Lr\Delta r + {r^2}\Delta L}}{{{r^2}L}}
⇒ΔVV=ΔLL+2Δrr\Rightarrow \dfrac{{\Delta V}}{V} = \dfrac{{\Delta L}}{L} + 2\dfrac{{\Delta r}}{r}
We are given that ΔVV=0.02 %\dfrac{{\Delta V}}{V} = 0.02\,\%
⇒ΔVV=2×10−4\Rightarrow \dfrac{{\Delta V}}{V} = 2 \times {10^{ - 4}}

Now substituting in the equation, we get,
⇒2×10−4=2.5×10−4+2Δrr\Rightarrow 2 \times {10^{ - 4}} = 2.5 \times {10^{ - 4}} + 2\dfrac{{\Delta r}}{r}
Further solving the equation, we get
2Δrr=−0.5×10−42\dfrac{{\Delta r}}{r} = - 0.5 \times {10^{ - 4}}
∴Δrr=−0.25×10−4\therefore \dfrac{{\Delta r}}{r} = - 0.25 \times {10^{ - 4}}

Hence, the correct option is A.

Note: The negative sign denotes that the radius of the wire is decreasing. As the wire is stretched by the tensile stress, its area of cross section reduces. Also, we must carefully note that here LL and rr both were variables. So, we applied the product rule of differentiation. Had one of these been a constant, we would have simply differentiated the quantity.