1. Given data

• Weight of block: $W=100\,\text{N}$
• Cross-sectional area of both wires: $A=0.5\,\text{cm}^2=0.5\times 10^{-4}\,\text{m}^2$
• Copper wire:
  • Length $L_c=\sqrt{3}\,\text{m}$
  • Angle with ceiling $30^\circ$
  • Young's modulus $Y_c=1\times 10^{11}\,\text{N/m}^2$
• Steel wire:
  • Length $L_s=1\,\text{m}$
  • Angle with ceiling $60^\circ$
  • Young's modulus $Y_s=2\times 10^{11}\,\text{N/m}^2$

We need to find: $\frac{\Delta l_c}{\Delta l_s}$

2. Find tensions in the two wires using equilibrium

Let tension in copper wire be $T_c$ and in steel wire be $T_s$.

Since the block is in equilibrium, horizontal components must cancel: $T_c\cos 30^\circ = T_s\cos 60^\circ$ $T_c\cdot \frac{\sqrt{3}}{2} = T_s\cdot \frac{1}{2}$ $T_s = \sqrt{3}\,T_c$

Now vertical components balance the weight: $T_c\sin 30^\circ + T_s\sin 60^\circ = 100$ $T_c\cdot \frac{1}{2} + T_s\cdot \frac{\sqrt{3}}{2} = 100$

Substitute $T_s=\sqrt{3}T_c$: $\frac{T_c}{2} + \frac{\sqrt{3}}{2}(\sqrt{3}T_c)=100$ $\frac{T_c}{2} + \frac{3T_c}{2}=100$ $2T_c=100$ $T_c=50\,\text{N}$

Hence, $T_s=\sqrt{3}\times 50=50\sqrt{3}\,\text{N}$

3. Use formula for elongation

For a wire, $\Delta l = \frac{TL}{AY}$

So, $\Delta l_c = \frac{T_cL_c}{AY_c}, \qquad \Delta l_s = \frac{T_sL_s}{AY_s}$

Therefore, \frac{\Delta l_c}{\Delta l_s}=rac{T_cL_c}{AY_c}\cdot \frac{AY_s}{T_sL_s}

Area cancels: \frac{\Delta l_c}{\Delta l_s}=rac{T_cL_cY_s}{T_sL_sY_c}

Substitute values: \frac{\Delta l_c}{\Delta l_s}=rac{50\cdot \sqrt{3}\cdot 2\times 10^{11}}{50\sqrt{3}\cdot 1\cdot 1\times 10^{11}}

Everything cancels except 2: $\frac{\Delta l_c}{\Delta l_s}=2$

4. Final answer

$\boxed{2}$