A wire has a resistance of 100 ohms. It is compressed to half of its original length keeping volume same then what will be its new resistance in ohm?

2 Answers
Jun 18, 2018

25 color(white)(l) Omega

Explanation:

The equation

R=rho*l/A

gives the resistance of a conductor of uniform cross-section area A, length l, and resistivity rho (in appropriate units).

It appears that a cylinder of height l properly models the geometric of a section of this wire.

V_"wire section"=V_"right cylinder" = A * l

A= 1/l * V_"right cylinder"

The question states that the volume of the section doesn't change as it is compressed, meaning that the value of 2 V_"wire section""/"pi stays constant. Halving l would double the value of A.

A'=1/(l') * V_"right cylinder"=1/(0.5color(white)(l) l) * V_"right cylinder"=2 color(white)(l) A

Thus

  • l'=0.5 color(white)(l) l
  • A' = 2 color(white)(l) A

The resistivity of the wire rho is independent of the wire's dimensions. Thus

R'=rho*(l')/(A')=rho * (0.5 color(white)(l) l)/(2 color(white)(l) A)=1/4 color(white)(l) R= 25 color(white)(l) Omega

Jun 18, 2018

25 ohm

Explanation:

L=lenght
A=cross section
Omega=resistance
V=volume

Omega propto (L/A)
V=A*L
V is constant
L rarr L/2
A rarr 2A
Omega rarr Omega prime
Omega prime/Omega
=(L/2)/(2A) div L/A
=1/4
Omega = 100 ohm
Omega prime = 25 ohm