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A wire of length 2.5 m has a percentage strain of 0.012% under a tensile force. The extension produce in the wire will beA. 0.03 mmB. 0.3 mmC. 0.3 mmD. 0.03 m |
Answer» Correct Answer - B (b) Here, original length, L=2.5 m ltrbgt Strain `=(DeltaL)/(L)=0.012% =(0.012)/(100)` `therefore DeltaL`= Strain `xx` L or `DeltaL`= Extension `=(0.012)/(100)xxL` `=(0.012xx2.5)/(100)=3xx10^(-4) m =0.3 mm` |
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