A block of weight $100\,\text{N}$ is suspended by copper and steel wires of same cross sectional area $0.5\,\text{cm}^{2}$ and, length $\sqrt{3}\,\text{m}$ and $1\,\text{m}$, respectively. Their other ends are fixed on a ceiling as shown in figure. The angles subtended by copper and steel wires with ceiling are $30^\circ$ and $60^\circ$, respectively. If elongation in copper wire is $(\Delta l_c)$ and elongation in steel wire is $(\Delta l_s)$, then the ratio $\frac{\Delta l_c}{\Delta l_s}$ is ______ [Young's modulus for copper and steel are $1\times10^{11}\,\text{N/m}^{2}$ and $2\times10^{11}\,\text{N/m}^{2}$, respectively.]

Step 1: Identify the given parameters.

The parameters for the copper and steel wires are given as follows:

$$W = 100\text{ N}$$ $$A = 0.5\text{ cm}^2$$ $$l_c = \sqrt{3}\text{ m}$$ $$l_s = 1\text{ m}$$ $$Y_c = 1 \times 10^{11}\text{ N/m}^2$$ $$Y_s = 2 \times 10^{11}\text{ N/m}^2$$

Step 2: Set up the equations for static equilibrium.

Resolve the forces acting on the suspended block in the horizontal and vertical directions.

$$T_s \cos 60^\circ = T_c \cos 30^\circ$$ $$T_s \sin 60^\circ + T_c \sin 30^\circ = W$$

Step 3: Solve for the tensions in the wires.

Express $T_s$ in terms of $T_c$ using the horizontal equilibrium equation.

$$T_s = \sqrt{3} T_c$$

Substitute this into the vertical equilibrium equation.

$$\frac{\sqrt{3}}{2}(\sqrt{3} T_c) + \frac{1}{2} T_c = W$$ $$2 T_c = W$$

Using $W = 100\text{ N}$, solve for both tensions.

$$T_c = 50\text{ N}$$ $$T_s = 50\sqrt{3}\text{ N}$$

Step 4: Formulate the ratio of elongations.

The elongation of a wire under tension is given by Hooke's law.

$$\Delta l = \frac{T l}{A Y}$$

Write the ratio of the elongation of the copper wire to that of the steel wire.

$$\frac{\Delta l_c}{\Delta l_s} = \frac{T_c l_c}{T_s l_s} \cdot \frac{Y_s}{Y_c}$$

Step 5: Calculate the final ratio.

Substitute the known values into the ratio expression.

$$T_c l_c = 50\sqrt{3}\text{ N}\cdot\text{m}$$ $$T_s l_s = 50\sqrt{3}\text{ N}\cdot\text{m}$$

Since $T_c l_c = T_s l_s$, the ratio simplifies to the ratio of the Young's moduli.

$$\frac{\Delta l_c}{\Delta l_s} = \frac{Y_s}{Y_c}$$ $$\frac{\Delta l_c}{\Delta l_s} = \frac{2 \times 10^{11}}{1 \times 10^{11}}$$ $$\frac{\Delta l_c}{\Delta l_s} = 2$$

Answer: 2