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Two wire are made of the same material and same volume .Cross-sectional area of wire 1 is A and wire 2 is 3A. If the length of the wire 1 increase by dx on applying force F, how much force needed to stretch wire 2 by the same amount? |
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Answer» Let the length of first wire be Land that of 2nd wire be l. As the volumes of two wires are same then L×A=l×3A So L/l=3 Let Young's modulus of the material be Y Then for elongation of dx length of first wire we can write Y=(F1/A)/(dx/L), where F1 is the force required. For similar elongation of second wire if force required be F2 then Y=(F2/3A)/(dx/l) So (F2/3A)/(dx/l)=(F1/A)/(dx/L) =>F2/F1=L/l×3=3×3=9 =>F2=9×F1 Hence 9 times force is required in 2nd case to cause similar elongation. |
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