Question : In order to quadruple the resistance of a wire of uniform cross section, a fraction of its length was uniformly streched till the final length of the entire wire was 1.5 times the initial length. The fraction is
a) 1/2
b) 1/4
c) 1/6
d) 1/8
Doubt by Khuspreet
Solution :
R = Total Resistance of the Original Wire
L = Original Length of the wire
A = Original Area of cross section of the wire
X = Length of the wire which is actually streched.
X/L = Fraction of the wire which is actually streched
L = Original Length of the wire
A = Original Area of cross section of the wire
X = Length of the wire which is actually streched.
X/L = Fraction of the wire which is actually streched
R' = Total resistance of the Streched Wire
L' =Final Length of the total wire
A' = Final Area of cross section of the streched wire
L'=(3/2)L
L'=1.5L
We know, the volume of the streched part will remain constant.
X×A = (0.5L+X)×A'
A'=XA/(0.5L+X) — (1)
Also
R' = 4R (Given)
A'=XA/(0.5L+X) — (1)
Also
R' = 4R (Given)
4R = R'
4R = R1+R2
[R1 and R2 are connected in Series]
4R = R1+R2
[R1 and R2 are connected in Series]
4ρL/A = 4[ρ(L-X)/A + ρ(0.5L+X)/A']
4ρL/A = ρ(L-X)/A + ρ(0.5L+X)/[XA/(0.5L+X)]
[Using eq (1)]
4ρL/A = ρ(L-X)/A + ρ(0.5L+X)/[XA/(0.5L+X)]
[Using eq (1)]
4ρL/A = ρ(L-X)/A + ρ(0.5L+X)2/LA
4ρL/A = ρ/A[(L-X) + ρ(0.5L+X)2/X]
4L = (L-X) + (0.5L+X)2/X
4LX =X(L-X)+ (0.5L+X)2
4LX-X(L-X) = (0.5L+X)2
X(4L-L+X) = (0.5L)2+X2+2(0.5L)(X)
X(3L+X) = 0.25L2+X2+LX
3LX+X2 = 0.25L2+X2+LX
3LX-LX = 0.25L2
2LX=0.25L2
2X=0.25L
X/L = 0.25/2
X/L = 25/200
X/L = 1/8
4ρL/A = ρ/A[(L-X) + ρ(0.5L+X)2/X]
4L = (L-X) + (0.5L+X)2/X
4LX =X(L-X)+ (0.5L+X)2
4LX-X(L-X) = (0.5L+X)2
X(4L-L+X) = (0.5L)2+X2+2(0.5L)(X)
X(3L+X) = 0.25L2+X2+LX
3LX+X2 = 0.25L2+X2+LX
3LX-LX = 0.25L2
2LX=0.25L2
2X=0.25L
X/L = 0.25/2
X/L = 25/200
X/L = 1/8
So, The required fraction is 1/8
Hence, d) would be the correct option.
