7339362 - = = . . . Y FA∆ll 301 X 9 8π4d2 X 15 20 00894...

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Solution: For a wire, Initial length of wire, = . l 15 2 m Maximum change in length, = . = . ∆l 8 94 mm 0 00894 m Mass of object, = m 301 kg Young’s modulus of wire, = Y 91 GPa Let diameter of wire be d From Hooke’s law: Young’s Modulus = stress/strain
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Unformatted text preview: = = . . . Y FA∆ll 301 X 9 8π4d2 X 15 20 00894 = . . . π4d2 301 X 9 8 X 15 291 X 109 X 0 00894 = . . . d2 301 X 9 8 X 15 2 X 491 X 109 X 0 00894 X π Hence, diameter, = . d 8 378 mm...
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This note was uploaded on 07/22/2011 for the course ME 3 taught by Professor Prof.ramachandran during the Spring '11 term at Indian Institute of Technology, Kharagpur.

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