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

To find the length of wire $B$, we start by using the definition of Young's modulus ($Y$), which relates stress and strain for a material. The formula is:
$$ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L / L} = \frac{F \cdot L}{A \cdot \Delta L} $$
In this setup, two wires $A$ and $B$ are joined in series and subjected to a tension $F$. Since they are joined, the tension $F$ acting on both wires is the same. The problem also states that both wires have an equal cross-sectional area ($A$) and that the resulting elongations ($\Delta L$) are equal.

From the formula, we can see that if $F$, $A$, and $\Delta L$ are constant, then the Young's modulus $Y$ is directly proportional to the original length $L$ of the wire:
$$ Y \propto L \implies \frac{Y_A}{Y_B} = \frac{L_A}{L_B} $$
We are given that the ratio of Young's modulus $Y_A/Y_B = 20/11$ and the length of wire $A$ is $L_A = 2.2\text{ m}$. We substitute these values into our ratio equation:
$$ \frac{20}{11} = \frac{2.2}{L_B} $$
To solve for $L_B$, we rearrange the equation:
$$ L_B = \frac{2.2 \times 11}{20} $$
$$ L_B = 0.11 \times 11 = 1.21\text{ m} $$
So, the length of wire $B$ is $1.21$ meters.