In order to quadruple the resistance of a wire of uniform cross-section; a fraction of its length was stretched uniformly till the final length of the entire wire was 1.5 times the initial length. Prove that the value of the fraction is 1/8
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It is not an 'easy sum'. I guess you are being sarcastic. But it is quite interesting to work out.
Call the initial length L, the initial resistance R. the initial cross-sectional area A and the resistivity ρ.
R = ρL/A
Call the fraction of L that is stretched, f. This means a length of fL is stretched.
The final system behaves as 2 wires in series: unstretched section (1) and stretched section (2).
___________________________
For the unstretched section (1):
L₁ = L-fL = L(1-f)
R₁ = ρL₁/A
= ρL(1-f)/A
___________________________
For the stretched section (2):
L₁ + L₂ = 1.5L
L₂ = 1.5L-L₁
= 1.5L - L(1-f)
= 1.5L - L + Lf
= 0.5L+ Lf
= L(0.5 + f)
___________________________
A section has been stretched from length fL to L₂. So (to conserve volume) its cross-sectional area has changed by a factor fL/L₂.
A₂ = A * fL/L₂
Since L₂ = L(0.5+f)
A₂ = A * fL/(L(0.5+f))
= Af/(0.5+f)
___________________________
R₂ = ρL₂/A₂
= ρL(0.5+f)/(Af/(0.5+f))
= ρL(0.5+f)²/(Af)
___________________________
Since the 'resistors' are in series and the total new resistance is 4R:
R₁ + R₂ = 4R
ρL(1-f)/A + ρL(0.5+f)²/(Af) = 4ρL/A
1 - f + (0.5+f)²/f = 4
f - f² + (0.5+f)² = 4f
f - f² + 0.25+f +f² = 4f
2f + 0.25 = 4f
2f = 0.25
f = 0.25/2 = 1/8
Call the initial length L, the initial resistance R. the initial cross-sectional area A and the resistivity ρ.
R = ρL/A
Call the fraction of L that is stretched, f. This means a length of fL is stretched.
The final system behaves as 2 wires in series: unstretched section (1) and stretched section (2).
___________________________
For the unstretched section (1):
L₁ = L-fL = L(1-f)
R₁ = ρL₁/A
= ρL(1-f)/A
___________________________
For the stretched section (2):
L₁ + L₂ = 1.5L
L₂ = 1.5L-L₁
= 1.5L - L(1-f)
= 1.5L - L + Lf
= 0.5L+ Lf
= L(0.5 + f)
___________________________
A section has been stretched from length fL to L₂. So (to conserve volume) its cross-sectional area has changed by a factor fL/L₂.
A₂ = A * fL/L₂
Since L₂ = L(0.5+f)
A₂ = A * fL/(L(0.5+f))
= Af/(0.5+f)
___________________________
R₂ = ρL₂/A₂
= ρL(0.5+f)/(Af/(0.5+f))
= ρL(0.5+f)²/(Af)
___________________________
Since the 'resistors' are in series and the total new resistance is 4R:
R₁ + R₂ = 4R
ρL(1-f)/A + ρL(0.5+f)²/(Af) = 4ρL/A
1 - f + (0.5+f)²/f = 4
f - f² + (0.5+f)² = 4f
f - f² + 0.25+f +f² = 4f
2f + 0.25 = 4f
2f = 0.25
f = 0.25/2 = 1/8