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Grade: 12
        
The length of a given cylindrical wire is increased by 100%.Due to the conquent decrease in diameter change in the resistance of the wire will be 
 
 
one month ago

Answers : (2)

Arun
22583 Points
							
Dear Akshit
 
As we know R = (rho) L/A
 
Now as length is increased by 100% , hence new length = 2L
 
Vold = Vnew
 
A L = Anew (2L)
 
hence
 
new Resistance = rho * 2L/(A/2) = 4 (rho)L/A
 
hence new resistance = 4R
 
hencec increase = 300%
one month ago
Vikas TU
8810 Points
							
Dear student 
Alternate method
The total volume remains the same before and after stretching.
Therefore A x l = A’ x l’
Here l’ = 2 l
A’ = A x l / l’ = A x l / 2 l = A/2
Percentage change in resistance

= Rf – Ri / Rl x 100 = ?
 l’/A’ - l/A / l ? l/A x 100
= [ (l’/A’ * A/ l) – 1 ] x 100 [(2 l/A/2 x A / l)-1 ] x 100
= 300%
one month ago
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