Given: let radius of the wire initially : r1 radius of the wire after it decreases by 0.1% : r2= r1- (1/1000 )r1 = r1- (1/1000)r1 = (999/1000)r1= 0.999 r1 Area of the wire initially : A1 Area of the wire after it is stretched becomes : A2 Resistance initially of the wire : R1 Resistance after it is stretched : R2 lenght of the wire initially,l1lenght after it is stretched ,l2 Now, the most important thing to be kept in mind while solving this problem is that volume of the wire remains constant.Hence, l1A1= l2A2 l1×π r12= l2×π r22 l2= l1× r12/ r22= l1/0.99801We know that resistance( R) is given by the following equation: R = ρ(l/A)Where ‘ρ’ is the resistivity of the wire which will remain the same throughout as it is the property of the material of which the wire is made up.Taking the ratio of the resistances: R1/R2= (l1/A1)/(l2/A2) R1/R2= (l1A2)/(A1l2)Substituting the values of l2and A1in the above equation we get:R1/R2= (l1π r22)/( l2×π r22)R1/R2= l1/l2× ( r2/ r1)2R1= 0.9960 R2Or the percentage increase in the resistance of the wire is =(R2- R1)/R1×100 = 0.004×100 = 0.4 = 0.4% increase in resistance