Question

Two wires are joined together and elongated with force as shown in the figure. If \( \dfrac{Y_1}{Y_2} = \dfrac{20}{11} \), find \( \dfrac{\ell_1}{\ell_2} \) so that they have same elongation.

• A. \( \frac{11}{20} \)
• B. \( \frac{20}{11} \)
• C. \( \frac{11}{10} \)
• D. \( \frac{10}{11} \)

Answer 1

The Correct Option is B

Concept: Elongation of a wire under tension is given by \[ \Delta \ell = \frac{T\ell}{YA} \] where

• \(T\) = applied force
• \(\ell\) = length of wire
• \(Y\) = Young's modulus
• \(A\) = cross-sectional area

Step 1: Condition for equal elongation.} Since both wires experience the same force and have same area, \[ \Delta \ell_1 = \Delta \ell_2 \] \[ \frac{T\ell_1}{Y_1A} = \frac{T\ell_2}{Y_2A} \]
Step 2: Simplify equation.} Cancel \(T\) and \(A\): \[ \frac{\ell_1}{Y_1} = \frac{\ell_2}{Y_2} \] \[ \frac{\ell_1}{\ell_2} = \frac{Y_1}{Y_2} \]
Step 3: Substitute given ratio.} \[ \frac{Y_1}{Y_2} = \frac{20}{11} \] Thus \[ \frac{\ell_1}{\ell_2} = \frac{20}{11} \] Final Result \[ \frac{\ell_1}{\ell_2} = \frac{20}{11} \]