Ratio of Elongations in Suspended Wires Supporting a Block

To find the ratio of elongations in the copper and steel wires, $\frac{\Delta l_c}{\Delta l_s}$, we analyze the system in equilibrium and apply Hooke's Law.

Step 1: Equilibrium of Tensions in the Wires

Let $T_s$ and $T_c$ be the tensions in the steel wire and copper wire, respectively.

Resolving forces horizontally for the suspended block: $T_s \cos 60^\circ = T_c \cos 30^\circ$

Substituting the values of the trigonometric ratios: $T_s \cdot \left(\frac{1}{2}\right) = T_c \cdot \left(\frac{\sqrt{3}}{2}\right)$

$T_s = \sqrt{3} T_c \implies \frac{T_c}{T_s} = \frac{1}{\sqrt{3}}$

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Step 2: Formula for Elongation

From Young's Modulus definition, the elongation $\Delta l$ in a wire of length $l$, cross-sectional area $A$, and Young's modulus $Y$ under tension $T$ is given by: $\Delta l = \frac{T \cdot l}{A \cdot Y}$

Thus, the elongations in the copper and steel wires are: $\Delta l_c = \frac{T_c \cdot l_c}{A_c \cdot Y_c}$

$\Delta l_s = \frac{T_s \cdot l_s}{A_s \cdot Y_s}$

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Step 3: Calculating the Ratio $\frac{\Delta l_c}{\Delta l_s}$

Taking the ratio of the two elongations: $\frac{\Delta l_c}{\Delta l_s} = \left( \frac{T_c}{T_s} \right) \times \left( \frac{l_c}{l_s} \right) \times \left( \frac{A_s}{A_c} \right) \times \left( \frac{Y_s}{Y_c} \right)$

Given data:

• $\frac{T_c}{T_s} = \frac{1}{\sqrt{3}}$
• Length ratio: $\frac{l_c}{l_s} = \frac{\sqrt{3}\text{ m}}{1\text{ m}} = \sqrt{3}$
• Area ratio: $A_c = A_s \implies \frac{A_s}{A_c} = 1$
• Young's moduli ratio: $\frac{Y_s}{Y_c} = \frac{2 \times 10^{11}\text{ N/m}^2}{1 \times 10^{11}\text{ N/m}^2} = 2$

Substituting these values into the ratio equation: $\frac{\Delta l_c}{\Delta l_s} = \left( \frac{1}{\sqrt{3}} \right) \times \left( \sqrt{3} \right) \times (1) \times (2) = 2$

Thus, the ratio $\frac{\Delta l_c}{\Delta l_s}$ is $2$.