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Physics Question on mechanical properties of solids

A structural steel rod has a radius of 10 mm and a length of 1 m. A 100 kN force stretches it along its length. The strain on the rod is (Ysteel=200?109 N m−2)(Y_{steel} = 200 ? 10^9 \,N \,m^{-2})

A

1.6 mm

B

2.6 mm

C

3.6 mm

D

4.6 mm

Answer

1.6 mm

Explanation

Solution

Here, r=10 mm=10?10−3 mr = 10\, mm = 10 ? 10^{-3}\, m L=1 mL = 1\, m F=100 kN=100?103 N=105 NF = 100\, kN = 100 ? 10^3\, N = 10^5\, N Stress on the rod is given by Stress =FA=Fπr2=\frac{F}{A}=\frac{F}{\pi r^{2}} =100×103 N3.14×(10−2 m)2=3.18×108 N m−2=\frac{100\times10^{3}\,N}{3.14\times\left(10^{-2}\,m\right)^{2}}=3.18\times10^{8}\,N\,m^{-2} Elongation, ΔL=(F/A)LY=(3.18×108 N m−2)(1 m)2×1011 N m−2\Delta L=\frac{\left(F / A\right)L}{Y}=\frac{\left(3.18\times10^{8}\,N\,m^{-2}\right)\left(1\,m\right)}{2\times10^{11}\,N\,m^{-2}} =1.59×10−3 m=1.59 mm=1.59\times10^{-3}\,m=1.59\,mm The strain on the rod is given by Strain =ΔLL= \frac{\Delta L}{L} =1.59×10−3 m1 m=1.59×10−3 m=1.59 mm=\frac{1.59\times10^{-3}\,m}{1\,m}=1.59\times10^{-3}\,m=1.59\,mm =1.6 mm=1.6\,mm