The correct answer is 21%.Concept:Resistance (R): Resistance is a measure of the opposition to current flow in an electrical circuit.It is measured in ohms, symbolized by the Greek letter omega (?).Ohms is named after George Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance. \(R = \frac{? l}{A}\) where ? is resistivity, l is length, and A is the area of the cross-section of the wire.Explanation: Suppose original length = L, Area of cross-section = A, radius of wire = r and Resistance = RSince the wire is the same, the resistivity ? will be the same.After the decrease in length, let us take new length = L', radius of wire = r' and new resistance = R'L' = 110 L/ 100 = 11 L/ 10( since there is 10% increase)The volume of the wire remains the same ( given).So, Volume of Original Wire = volume of new wire ? r2 L = ? r ' 2 L'r2L = r' 2 L'r' 2 = r2 L / L'r' 2 = 10 r2 L / 11 Lr' 2 = 10 r2 / 11Decrease in resistance = \(\frac{R'-R}{R} À” 100\)\(\rm \frac{\left(\frac{\rho L'}{A'}-\frac{\rho L{}}{A}\right)}{\frac{\rho L}{A}}\times100\)\(\rm \frac{\left(\frac{\rho L'}{\pi r'^2}-\frac{\rho L{}}{\pi r^2}\right)}{\frac{\rho L}{\pi r^2}}\times100\)The resistance increases by 21%.