One end of a metal wire is fixed to a ceiling and a load of $2\ \text{kg}$ hangs from the other end. A similar wire is attached to the bottom of the load and another load of $1\ \text{kg}$ hangs from this lower wire. Then the ratio of the longitudinal strain of the upper wire to that of the lower wire will be ______.
[Area of cross section of wire $= 0.005\ \text{cm}^2$, $Y = 2\times10^{11}\ \text{N m}^{-2}$ and $g = 10\ \text{m s}^{-2}$]
Numerical value type. Enter your answer.
Answer: 3
The upper wire carries both loads, $T_1 = 3g = 30\ \text{N}$; the lower wire carries only $T_2 = 1g = 10\ \text{N}$.
Strain $= \dfrac{T}{AY}$ and the wires are identical, so
$$\frac{\text{strain}_1}{\text{strain}_2} = \frac{30}{10} = 3$$
Solution by Sreeraj P, M.Sc Physics