Related Formula
Y = StressStrain = (F / A)/(Δ / ) = (F )/(A Δ )$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta \ell / \ell} = \frac{F \ell}{A \Delta \ell}$$
Core Logic
Since the wires are stretched by the same magnitude (Δ A = Δ B$\Delta \ell_A = \Delta \ell_B$) under the same load (FA = FB$F_A = F_B$), the Young's Modulus Y$Y$ is proportional to ( )/(A)$\frac{\ell}{A}$.
Step 1: Setting up the Ratio
(YA)/(YB) = (( A)/( B)) ((AB)/(AA))$$\frac{Y_A}{Y_B} = \left(\frac{\ell_A}{\ell_B}\right) \left(\frac{A_B}{A_A}\right)$$
Substitute the given values:
A = 6.0 ~cm$\ell_A = 6.0 \mathrm{~cm}$
B = 5.4 ~cm$\ell_B = 5.4 \mathrm{~cm}$
AA = 3.0 × 10⁻⁵ ~m²$A_A = 3.0 \times 10^{-5} \mathrm{~m}^2$
AB = 4.5 × 10⁻⁵ ~m²$A_B = 4.5 \times 10^{-5} \mathrm{~m}^2$
Step 2: Calculation
(YA)/(YB) = ((6.0)/(5.4)) ( 4.5 × 10⁻⁵3.0 × 10⁻⁵)$$\frac{Y_A}{Y_B} = \left(\frac{6.0}{5.4}\right) \left(\frac{4.5 \times 10^{-5}}{3.0 \times 10^{-5}}\right)$$
= ((60)/(54)) ((4.5)/(3.0))$$= \left(\frac{60}{54}\right) \left(\frac{4.5}{3.0}\right)$$
= ((10)/(9)) ((1.5)/(1)) = (15)/(9) = (5)/(3)$$= \left(\frac{10}{9}\right) \left(\frac{1.5}{1}\right) = \frac{15}{9} = \frac{5}{3}$$
Step 3: Finding x
Given ratio is x : 3$x : 3$.
(x)/(3) = (5)/(3) ⇒ x = 5$$\frac{x}{3} = \frac{5}{3} \Rightarrow x = 5$$
Pattern Recognition
Whenever "stretched by same magnitude under same load" appears, strip out F$F$ and Δ$\Delta \ell$ to leave Y ∝ / A$Y \propto \ell / A$.
Chapter Mix
Class 11 Physics: Mechanical Properties of Solids