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.
Step 1: Understanding the Concept:
When two wires are connected end-to-end (in series) and pulled, the same force (tension) acts on both. By applying Hooke's Law for elasticity, we can relate their elongations to their original lengths and Young's moduli.
Step 2: Key Formula or Approach:
The elongation \(\Delta \ell\) of a wire under tension is:
\[ \Delta \ell = \frac{F \ell}{A Y} \]
Step 3: Detailed Explanation:
The problem states that the elongation is the same for both wires:
\[ \Delta \ell_1 = \Delta \ell_2 \]
Assuming they have identical cross-sectional areas (\(A_1 = A_2 = A\)) and share the same force \(F\):
\[ \frac{F \ell_1}{A Y_1} = \frac{F \ell_2}{A Y_2} \]
Cancel out the common terms \(F\) and \(A\):
\[ \frac{\ell_1}{Y_1} = \frac{\ell_2}{Y_2} \]
Rearrange the equation to find the ratio of their lengths:
\[ \frac{\ell_1}{\ell_2} = \frac{Y_1}{Y_2} \]
We are given that \(\frac{Y_1}{Y_2} = \frac{20}{11}\). Therefore:
\[ \frac{\ell_1}{\ell_2} = \frac{20}{11} \]
Step 4: Final Answer:
The ratio \(\frac{\ell_1}{\ell_2}\) is \(20/11\).
