Question

Two wires of same length and material are stretched by the same force. If their masses are in the ratio \( 3 : 4 \), then the ratio of their elongations is

• A. \( \dfrac{4}{3} \)
• B. \( \dfrac{5}{3} \)
• C. \( \dfrac{1}{3} \)
• D. \( \dfrac{2}{3} \)

Hint:

For wires of same material and length, elongation is inversely proportional to mass.

Answer 1

The Correct Option is A

Step 1: Write the formula for elongation of a wire.
The elongation of a wire is given by \[ \Delta L = \frac{FL}{AY} \] where \( F \) is force, \( L \) is original length, \( A \) is cross-sectional area and \( Y \) is Young’s modulus.
Step 2: Identify constant quantities.
Since both wires are of the same material and same length, and are stretched by the same force, the quantities \( F \), \( L \), and \( Y \) remain constant.
Thus, \[ \Delta L \propto \frac{1}{A} \]
Step 3: Relate area with mass.
Mass of a wire is given by \[ m = \rho A L \] Since \( \rho \) and \( L \) are constant, \[ m \propto A \] \[ \Rightarrow \Delta L \propto \frac{1}{m} \]
Step 4: Calculate the ratio of elongations.
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{m_2}{m_1} = \frac{4}{3} \]
Step 5: Conclusion.
The ratio of elongations of the two wires is \( \dfrac{4}{3} \).