A string \(A\) of length \(0.314\) m and Young's modulus \(2 \times 10^{10}\) N/m\(^2\) is connected to another string \(B\) of length and Young's modulus both twice of those of \(A\). This series combination of strings is then suspended from a rigid support and its free end is fixed to a load of mass \(0.8\) kg. The net change in length of the combination is _____ mm. (radius of both the strings is \(0.2\) mm and acceleration due to gravity \(= 10\) m/s\(^2\)) (Mass of both strings is to be neglected as compared to the mass of load)

Given:
Length of string A, \(L_A = 0.314\) m
Young's modulus of string A, \(Y_A = 2 \times 10^{10}\) N/m\(^2\)
Radius of both strings, \(r = 0.2\) mm \(= 2 \times 10^{-4}\) m
Length of string B, \(L_B = 2L_A = 0.628\) m
Young's modulus of string B, \(Y_B = 2Y_A = 4 \times 10^{10}\) N/m\(^2\)
Mass of load, \(m = 0.8\) kg
Acceleration due to gravity, \(g = 10\) m/s\(^2\)

Tension in both strings, \(T = mg = 0.8 \times 10 = 8\) N

Cross-sectional area of both strings:
\(A = \pi r^2 = 3.14 \times (2 \times 10^{-4})^2 = 3.14 \times 4 \times 10^{-8}\) m\(^2\)

Elongation in string A:
\(\Delta L_A = \dfrac{T L_A}{A Y_A}\)
\(\Delta L_A = \dfrac{8 \times 0.314}{3.14 \times 4 \times 10^{-8} \times 2 \times 10^{10}}\)
\(\Delta L_A = \dfrac{8 \times 0.314}{8 \times 3.14 \times 10^2} = \dfrac{0.1}{10^2} = 10^{-3}\) m \(= 1\) mm

Elongation in string B:
\(\Delta L_B = \dfrac{T L_B}{A Y_B}\)
\(\Delta L_B = \dfrac{8 \times (2 \times 0.314)}{A \times (2 Y_A)} = \dfrac{T L_A}{A Y_A} = 1\) mm

Net change in length of the combination:
\(\Delta L = \Delta L_A + \Delta L_B = 1 + 1 = 2\) mm

Answer: \(2\)