Q132 / 180 · COMEDK 2023 (Memory Based) · Physics

Two wire of same material having radius in ratio $2: 1$ and lengths in ratio $1: 2$. If same force is applied on them, then ratio of their change in length will be

  • A
    1 : 1
  • B
    1 : 2
  • C
    1 : 4
  • Correct answer
    D
    1 : 8

Solution

Given, ratio of radius of two wires, $\frac{r_1}{r_2}=\frac{2}{1}$ and ratio in their lengths, $\frac{l_1}{l_2}=\frac{1}{2}$ When same force is applied on them, then ratio change in their lengths, $\frac{\Delta l_1}{\Delta l_2}=\text { ? }$ We known that, Young's modulus, $\begin{gathered} Y=\frac{F l}{A \Delta l} \Rightarrow \Delta l=\frac{F l}{A Y} \\ \Delta l \propto \frac{l}{A} \\ \therefore \quad \frac{\Delta l_1}{\Delta l_2}=\frac{l_1 / A_1}{l_2 / A_2}=\frac{l_1 A_2}{l_2 A_2} \\ \Rightarrow \quad \frac{\Delta l_1}{\Delta l_2}=\frac{l_1 / A_1}{r_2^2 / r_1^2}=\frac{1}{2} \cdot\left(\frac{1}{2}\right)^2=\frac{1}{8} \end{gathered}$

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

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