Q135 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Physics

Two metal wires $A$ and $B$ have length $L$ and $3 L$ respectively. The radius of cross-sectional circular area of wire $A$ and $B$ are $R$ and $2 R$, respectively. These wires are joined end to end along their axis. When one end of the combined system is fixed and other end is pulled with a constant force $F$, the elongation in both the wires is equal. If $Y_A$ and $Y_B$ are Young's modulus of wire $A$ and $B$, then the $Y_B / Y_A$ is

  • Correct answer
    A
    $\frac{3}{4}$
  • B
    $\frac{4}{3}$
  • C
    $\frac{2}{3}$
  • D
    $\frac{3}{2}$

Solution

For metal wire $A, L_A=L, R_A=R$ For metal wire $B, L_B=3 L, R_B=2 R$ When the ends of the of metal wire $A$ and $B$ are joined and pulled with a constant force $F$ after fixing the one end of combined system, then change in length of wire $A$ and $B$ is same. $\begin{array}{ll} \text { i.e., } & \Delta L_A=\Delta L_B \\ \Rightarrow & \frac{F L_A}{A_A Y_A}=\frac{F L_B}{A_B Y_B} \Rightarrow \frac{F L}{\pi R_A^2 Y_1}=\frac{F \times 3 L}{\pi R_B^2 Y_2} \\ \Rightarrow & \frac{1}{\pi R^2 Y_A}=\frac{3}{\pi(2 R)^2 Y_B} \Rightarrow \frac{1}{\pi R^2 Y_A}=\frac{3}{4 \pi R^2 Y_B} \\ \Rightarrow & \frac{1}{Y_A}=\frac{3}{4 Y_B} \Rightarrow \frac{Y_B}{Y_A}=\frac{3}{4} \end{array}$

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

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