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Home  >>  EAMCET  >>  Physics
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A 4.0 m long copper wire of cross sectional area $ 1.2 \: cm^2$ is stretched by a force of $ 4.8 \times 10^3N$ If Young's modulus for copper is $ Y = 1.2 \times 10^{11}\: N/m^2$ , the increase in length of wire and strain energy stored per unit volume are :

$\begin {array} {1 1} (a)\;1.32 \times 10^{-4} m, \: 66 \times 10^3J \\ (b)\;132 \times 10^{-4} m, \: 6.6 \times 10^2J \\ (c)\;13.2 \times 10^{-4} m, \: 6.6 \times 10^3J \\ (d)\;0.132 \times 10^{-4} m, \: 66 \times 10^4J \end {array}$

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Ans : (3)
$ 13.2 \times 10^{-4} m, \: 6.6 \times 10^3J$
answered Feb 27, 2014 by thanvigandhi_1
edited Mar 4, 2014 by thanvigandhi_1
 

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