Question

The two wires $A$ and $B$ of equal cross-section but of different materials are joined together. The ratio of Young's modulus of wire $A$ and wire $B$ is $20/11$. When the joined wire is kept under certain tension the elongations in the wires $A$ and $B$ are equal. If the length of wire $A$ is $2.2\text{ m}$, then the length of wire $B$ is _______ $\text{m}$.

• A. 1.1
• B. 2.22
• C. 1.21
• D. 4.44

Hint:

Use the Young's modulus formula $Y = \frac{FL}{A\Delta L}$ and notice that since tension, area, and elongation are the same for both wires, $L$ is directly proportional to $Y$.

Answer 1

The Correct Option is C

This problem requires understanding how the physical properties of a material determine its elongation under stress. Young's modulus is defined as the ratio of tensile stress to tensile strain:
$$ Y = \frac{F/A}{\Delta L/L} $$
Rearranging this to solve for the elongation ($\Delta L$), we get:
$$ \Delta L = \frac{F \cdot L}{A \cdot Y} $$
We are told that for wire $A$ and wire $B$, the elongations are equal ($\Delta L_A = \Delta L_B$). Since the wires are connected together and under tension, the force $F$ is same for both. They also share the same cross-sectional area $A$. Thus:
$$ \frac{F \cdot L_A}{A \cdot Y_A} = \frac{F \cdot L_B}{A \cdot Y_B} $$
Canceling the common terms $F$ and $A$ from both sides of the equation, we are left with:
$$ \frac{L_A}{Y_A} = \frac{L_B}{Y_B} \implies L_B = L_A \cdot \frac{Y_B}{Y_A} $$
Given that $Y_A/Y_B = 20/11$, the reciprocal ratio $Y_B/Y_A$ is $11/20$. The original length of wire $A$ is $L_A = 2.2\text{ m}$. Plugging these values in:
$$ L_B = 2.2 \times \frac{11}{20} $$
$$ L_B = 0.11 \times 11 = 1.21\text{ m} $$
Therefore, the calculated length of wire $B$ is $1.21\text{ m}$.