Question

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?

Answer 1

`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`

Answer 2

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`