A wire of uniform cross section and length l has a resistance of `16 Omega`. It is cut into four equal parts.Each part is stretched uniformly to length l and all the four streteched parts are connected in parallel. Calculate the total resistance of the combination so formed. Assume that stretching of wire does not cause any change in the density of its material.

Explanation

Let the cross section of the wire changes from A to `A_1` when it is cut in to four parts.
As density remains constant , so volume also remains constant. Volume of each part before stretching=volume of each part after stretching
or, `(Al)/4=A_1l or, A_1=A/4`
So, resistance of each part, `R_1=(rho l)/(A/4)=(4rhol)/A`
Therefore, equivalent resistance of the parallel combination,
`R_(eq)=R_1/4=(rhol)/4=16 Omega` `[because Initial velocity =(rhol)/A=16Omega]`