A metal wire of resistance 6 ohm is stretched so that its length is increased to twice its original length.Calculate its new resistance.

Resistance ‘R = 6ohm’ is related to length ‘L’, area of cross-section ‘A’ and resistivity ‘ρ’ as,

R = ρL/A

When L is doubled ,

As we know that the volume of the wire remains same.

old volume = new volume

AL = A'L' ....(i) ( as volume = area x length )

L' = 2L

So, from (i)

A' = A/2

So the new resistance is ,

R' =  ρL'/A'

or

R'  = ρ(2L)/(A/2) = 4 (ρL/A)

=> R' = 4R = 4 x 6ohm = 24 ohm .