In order to quadrapule the resistance of a wire of uniform cross section 

a fraction of its length was stretched uniformly till the final length of the wire was 1.5 timed of the original length

 

The value of the fraction elongated of the wire in comparison to original length of the wire is

 

Ans 1/8

 

Please give detailed solution tayes assured 

let length of wire = L

let length of fraction = x

let fraction x be stretched to 't' times its length

 

R = L/A

 

4R = (L-x)/A + tx/(A/t)

4(L/A) = L/A  - x/A + t²x/A

3L/A + x/A = t²x/A

3L + x = t²x

x/L = 3/ t² - 1                    ----------------------(1)

 

 

also given,

new length of wire/original length of wire = 1.5

L - x + tx  / L  = 1.5

0.5L = tx - x

x/L = 0.5 / t-1                    ----------------------(2)

 

comparing (1) and (2)

0.5 / t-1   =  3 /  t²-1

0.5(t+1) = 3

0.5t = 2.5

t = 5

 

now put this value of t in (1)

x/L = 3/ t² - 1

x/L = 3 / 5^2 - 1

x/L = 3/24

x/L = 1/8

 

 

 

*PLZ RATE* cheers!