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Question: A steel rod of length 1 m and radius 10 mm is stretched by a force 100 KN along its length. The stre...

A steel rod of length 1 m and radius 10 mm is stretched by a force 100 KN along its length. The stress produced in the rod is YSteel=2×1011Nm−2Y_{Steel} = 2 \times 10^{11}Nm^{- 2}

A

3.18×106Nm−23.18 \times 10^{6}Nm^{- 2}

B

3.18×107Nm−23.18 \times 10^{7}Nm^{- 2}

C

3.18×108Nm−23.18 \times 10^{8}Nm^{- 2}

D

3.18×109Nm−23.18 \times 10^{9}Nm^{- 2}

Answer

3.18×108Nm−23.18 \times 10^{8}Nm^{- 2}

Explanation

Solution

: Here, r=10mm=10×10−3m=10−2mr = 10mm = 10 \times 10^{- 3}m = 10^{- 2}m

L = 1m, F =100 KN = 100×103N=105N100 \times 10^{3}N = 10^{5}N

Stress produced in the rod is

Stress=FA=Fπr2=100×103N3.14×(10−2m)2=3.18×108Nm−2Stress = \frac{F}{A} = \frac{F}{\pi r^{2}} = \frac{100 \times 10^{3}N}{3.14 \times (10^{- 2}m)^{2}} = 3.18 \times 10^{8}Nm^{- 2}