A string $A$ of length $0.314$ m and Young's modulus $2 \times 10^{10}\ \text{N/m}^2$ 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 $= 10\ \text{m/s}^2$) (Mass of both strings is to be neglected as compared to the mass of load)
Answer: (B) $2$
Both strings carry $F = 8$ N and have $A = \pi(0.2 \times 10^{-3})^2 = 1.256 \times 10^{-7}\ \text{m}^2$.
$\Delta L_A = \dfrac{8 \times 0.314}{1.256 \times 10^{-7} \times 2 \times 10^{10}} = 1$ mm.
$B$ has twice the length and twice the Young's modulus, so $\dfrac{L}{Y}$ is the same: $\Delta L_B = 1$ mm.
Total $= 2$ mm.
Solution by Sreeraj P, M.Sc Physics