Two wires $A$ and $B$ of same cross-section are connected end to end. When same tension is created in both wires, the elognation in $B$ wire is twice the elongation in $A$ wire. If $L_A$ and $L_B$ are the initial lengths of the wires $A$ and $B$ respectively, then (Young's modulus of material of wire $A=2 \times 10^{11} \mathrm{Nm}^{-2}$ and Young's modulus of material of wire $B=1.1 \times 10^{11} \mathrm{Nm}^{-2}$ ).

• Correct answer
  A

  $\frac{L_A}{L_B}=\frac{10}{11}$
• B

  $\frac{L_A}{L_B}=\frac{4}{5}$
• C

  $\frac{L_A}{L_B}=\frac{9}{11}$
• D

  $\frac{L_A}{L_B}=\frac{3}{7}$

Solution

The given situation is shown below.
Here, elongation in wire $B$ is twice of elongation in wire $A$ on the application of same tension $T$. i.e $\Delta L_B=2 \Delta L_A$ Young's modulus of wire $A$ and $B$ are given as $Y_A=\frac{T L_A}{A \cdot \Delta L_A}$ and $Y_B=\frac{T L_B}{A \Delta L_B}$ $\therefore \quad \frac{Y_A}{Y_B}=\frac{L_A}{L_B} \times \frac{\Delta L_B}{\Delta L_A} \Rightarrow \frac{2 \times 10^{11}}{1.1 \times 10^{11}}=\frac{L_A}{L_B} \times \frac{2 \Delta L_A}{\Delta L_A}$ $\Rightarrow \quad \frac{2}{11}=\frac{L_A}{L_B} \times 2 \Rightarrow \frac{L_A}{L_B}=\frac{1}{11}=\frac{10}{11}$

Hence, the correct answer is option: 'A'.