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A cylindrical metallic wire is stretched to increase its length by 5%. Calculate the percentage change in its resistance. |
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Answer» Solution :It is given that l. =l + 5% of `l = l (1 + 5/100)` As on stretching the wire its volume remains constant, HENCE A.l= A.l. `rArr A. = (Al)/(l.) = (Al)/(l (1 + 5/100) ) = (A)/( (1+ 5/100))` ` therefore ` New resistance `R. (rho l.)/(A.) = (rho l (1 + 5/100)^2)/(A ) = R.(1 + 5/100)^2 = R (1 + 10/100)` ` therefore ` PERCENTAGE increase in resistance `Delta R = (R. - R)/(R ) XX 100% = (R (1 + 10/100)-R)/(R ) xx 100%= 10%` |
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