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Calculate the increase in energy of steel rod of 2m and crosssectional area $4cm^2$,when a 500N force is applied on it.$Y_{steel}=2\times 10^{11}N/m^2$

$\begin{array}{1 1}(A)\;2.04\times 10^{-5}J\\(B)\;3.14\times 10^{-5}J\\(C)\;4.35\times 10^{-5}J\\(D)\;\text{None}\end{array} $

1 Answer

Energy density =$\large\frac{1}{2}$$\times$Stress$\times$strain
$\Rightarrow \large\frac{1}{2}$$\times \large\frac{(stress)^2}{Y}$
Energy=$\large\frac{1}{2} \large\frac{(stress)^2}{Y}$$\times$ volume
$\Rightarrow \large\frac{1}{2}\times \frac{500/4\times 10^{-4}}{2\times 10^{11}}$$\times 2\times 4\times 10^{-4}$
$\Rightarrow \large\frac{2500}{2\times 10^{11}}\times \frac{1}{4\times 10^{-4}}$
$\Rightarrow \large\frac{314.5}{10^7}$
$3.145\times 10^{-5}J$
Hence (B) is the correct answer.
answered May 8, 2014 by sreemathi.v

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