1.

A resistive wire is stretched till its length is increased by 100%. Due to the consequent decrease in diameter, the change in the resistance of a stretched wire will be

Answer»

3
2
1
0.5

Solution :300 %
LET length of wire is l and area is A.
R = `(rho l)/(A)`
New length of wire l. = l + 100% l = 21 and area is A..
`THEREFORE (R.)/(R) = (rho l. )/(A.) (A)/(rho1)= (A)/(A.) (l.)/(l) = ((l.)/(l) )^(2)= ((2l.)/(l) )^(2) = 4`
`R. = (rho l.)/(A.)`
`therefore R. = 4R = R + 3R = ((4R - R)/(R) XX 100 ) % `
`= ((3R)/(R) xx 100 )%` = 300 %
Percentage change in resistance = 300%


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