Question

Two wires A and B made of same material and areas of cross-section in the ratio $1:2$ are stretched by same force. If the masses of the wires A and B are in the ratio $2:3$, then the ratio of the elongations of the wires A and B is

• A. $1:2$
• B. $8:3$
• C. $1:3$
• D. $4:3$

Hint:

When length is not explicitly given but mass is, substitute $L = \text{Mass}/(\text{Density} \times \text{Area})$. This modifies the dependency on Area from $A^{-1}$ to $A^{-2}$.

Answer 1

The Correct Option is B

Step 1: Formula for Elongation:
Elongation $\Delta L = \frac{FL}{AY}$, where $F$ is force, $L$ is length, $A$ is area, $Y$ is Young's modulus.
Step 2: Express Length in terms of Mass:
Mass $m = \text{Density}(\rho) \times \text{Volume} = \rho \cdot A \cdot L$. So, $L = \frac{m}{\rho A}$. Substitute $L$ into the elongation formula: \[ \Delta L = \frac{F}{AY} \left( \frac{m}{\rho A} \right) = \frac{Fm}{A^2 Y \rho} \]
Step 3: Apply Ratios:
Given: Same material $\implies Y, \rho$ are constant. Same force $\implies F$ is constant. Ratio of Areas: $A_A : A_B = 1:2$. Ratio of Masses: $m_A : m_B = 2:3$. \[ \frac{\Delta L_A}{\Delta L_B} = \frac{m_A / A_A^2}{m_B / A_B^2} = \left( \frac{m_A}{m_B} \right) \left( \frac{A_B}{A_A} \right)^2 \]
Step 4: Calculate:
\[ \frac{\Delta L_A}{\Delta L_B} = \left( \frac{2}{3} \right) \left( \frac{2}{1} \right)^2 = \frac{2}{3} \times 4 = \frac{8}{3} \]