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A cylindrical rod of circular cross section is subjected to a tensile force and the longitudinal strain produced in the rod is 2%. If the radius of the rod is doubled and the applied force is also doubled, the longitudinal strain produced will be

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Detailed Solution

$e = \frac{∆ l}{l} = \frac{F}{A . Y} = \frac{F}{{πr}^{2} Y} \Rightarrow \frac{e_{2}}{e_{1}} = {(\frac{r_{1}}{r_{2}})}^{2} (\frac{F_{2}}{F_{1}}) = {(\frac{1}{2})}^{2} x 2 = \frac{1}{2}$

$\Rightarrow e_{2} = \frac{1}{2} x 2 % = 1 %$

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