A stress of $3.18 imes 10^{8}Nm^{- 2}$ is applied to a steel rod of length $1m$ along its length. Its Young’s modulus is $2 imes 10^{11}Nm^{- 2}$. Then the elongation produced in the rod in $mm$ is

Question

A stress of $3.18 	imes 10^{8}Nm^{- 2}$ is applied to a steel rod of length $1m$ along its length. Its Young’s modulus is $2 	imes 10^{11}Nm^{- 2}$. Then the elongation produced in the rod in $mm$ is

Dr. Abubacker · Accepted Answer

(d) $Y = \frac{F/A}{\mathrm{\Delta}l/l}$
Given, $F/A =$ stress $= 3.18 	imes 10^{8}Nm^{- 2}$
$l = 1m,\ Y = 2 	imes 10^{11}Nm^{- 2}$
$\mathrm{\Delta}l = \frac{lF/A}{Y} = \frac{1 	imes 3.18 	imes 10^{8}}{2 	imes 10^{11}} = 1.59 	imes 10^{- 3}m = 1.59mm$

Q. A stress of $3.18 \times 10^{8}Nm^{- 2}$ is applied to a steel rod of length $1m$ along its length. Its Young’s modulus is $2 \times 10^{11}Nm^{- 2}$. Then the elongation produced in the rod in $mm$ is

Explanation

Reviewed by Dr. Abubacker

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