Question

if the length of the wire in increase by5% what will be the percentage change in its resistance

Answer 1

Assuming the total volume of the wire remains constant,
A1L1 = A2L2 where A1, A2 and L1, L2 represent the initial and final cross-sectional areas and lengths of the wire respectively.

Hence, when the wire stretched to an increased length (5% more than the original),
L2 = (105*L1)/100
or, L1/L2 = 100/105
or, A2/A1 = 100/105

Now, the resistance of a wire, R = pL/A where p is the resistivity, L and A are the length and area of the wire respectively.

Thus, R2/R1 = (pL2/A2) / (pL1/A1)
or, R2/R1 = (L2/L1) * (A1/A2)
or, R2/R1 = 105/100 * 105/100
or, R2/R1 = (105/100)^2
or, R2 = (11025/10000)*R1
or, R2 = (110.25/100)*R1
or, R2 = 110.25% R1

Hence, the percentage change in resistance due to 5% increase in length is (110.25 - 100)
= 10.25