1.

If length of a given wire increases by 10% due to stretching, the resistance of the wire increasesby

Answer»

0.21
0.2
0.1
0.05

Solution :Here l.=l+ 10% of l=1.11. As VOLUME of wire REMAINS unchanged, so the new cross-section area
`A. = (Al)/(l.) = (Al)/(1.1l) = (A)/(1.1)`
` THEREFORE ` New resistance `R. = (rho l.)/(A.) = (rho (1.1l))/((A//1.1)) = 1.21 (rho l)/(A) = 1.21 R`
` therefore (DELTA R)/(R ) = (R.- R)/(R ) = 0.21 = 21%`


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