Awire with a resistance of 10 ohm is stretched so that its new length is four times its original length. find the resistance of the longer wire, assuming that volume of the materials are unchanged.

Let original length be L. The new length is therefore 4L.

Let original cross sectional surface area of the wire be equal to πr^2.

This means original volume was L x πr^2 = Lπr^2

The volume is the same but the length is different so 4L x new surface area must be equal to Lπr^2. Let new surface area be equal to Y.

4L x Y = Lπr^2
=> Y = (πr^2 )/ 4

Using the resistivity formula,
R = pL/A. p which is resistivity is a constant so it stays the same

But this time, instead of L we have 4L and instead of πr^2 we have (πr^2)/4.

so the new resistance
= (4Lp)/ {(πr^2)/4}
= 16 (pL)/(πr^2)
= 16 (pL)/A. because πr^2 is A
since pL/A is equal to R from the formula, this is equal to
16 R.

R was 10 ohms
therefore new resistance is 16 x 10 = 160 ohms

Answer the above statement is ( false );