One end of a metal wire is fixed to a ceiling and a load of 2 kg hangs from the other end. A similar wire is attached to the bottom of the load and another load of 1 kg hangs from this lower wire. Then the ratio of longitudinal strain of upper wire to that of the lower wire will be [Area of cross section of wire = $$0.005 \text{ cm}^2$$, $$Y = 2 \times 10^{11} \text{ N m}^{-2}$$ and $$g = 10 \text{ m s}^{-2}$$]

Find the ratio of longitudinal strain of the upper wire to the lower wire.

The upper wire supports both loads (2 kg + 1 kg), while the lower wire supports only the bottom load (1 kg).

Force on upper wire: $$F_1 = (2 + 1) \times g = 3 \times 10 = 30$$ N.

Force on lower wire: $$F_2 = 1 \times g = 1 \times 10 = 10$$ N.

Strain $$\epsilon = \frac{\Delta L}{L} = \frac{F}{YA}$$ (from Young's modulus $$Y = \frac{F/A}{\Delta L/L}$$).

Since both wires are similar (same $$Y$$ and $$A$$): $$\epsilon \propto F$$.

$$ \frac{\epsilon_1}{\epsilon_2} = \frac{F_1}{F_2} = \frac{30}{10} = 3 $$

The answer is 3.