Question

A piece of wire is redrawn without change in volume so that it's radius is halved. Compare the new resistance with the original resistance...

Answer 1

Answer:

The new Radius is half, which makes the new cross sectional area {pi*radius²} equals pi*(R/2)², which becomes (pi/4)*R². This is 1/4 of the original area {not one half}, and the length is mulitplied by 4 to maintain same volume.  

So we have 4 divided by (1/4) equals 16. The new resistance is 16 times the original resistance

Explanation:

Answer 2

New resistance is 16 times of old resistance value.

Given:

• radius is halved.
• piece of wire is redrawn without change in volume

To Find:

Compare the new resistance with the original resistance...

Solution:

The new resistance is given by:

R' = (ρL) / A' = (ρL) / [(π/4)R^2] = (4ρL) / πR^2,

where R is the radius of the original wire, and L is the length of the new wire (which is four times the length of the original wire, as you have pointed out).

Substituting R/2 for R in the above equation, we get:

R' = (4ρL) / π(R/2)^2 = (16ρL) / πR^2,

which is 16 times the original resistance.

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