Q125 / 180 · COMEDK 2023 (Morning Shift) · 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
- Correct answerA$1: 8$
- B$1: 2$
- C$1: 4$
- D$1: 1$
Solution
The change in length $\Delta L$ of a wire is given by the formula $\Delta L = \dfrac{FL}{AY}$, where $F$ is the applied force, $L$ is the original length, $A$ is the cross-sectional area, and $Y$ is Young's modulus of the material.
Since both wires are made of the same material, $Y$ is the same for both. The force $F$ applied is also the same. Thus, $\Delta L \propto \dfrac{L}{A}$.
The area $A$ is given by $\pi r^2$, so $\Delta L \propto \dfrac{L}{r^2}$.
Given the ratio of radii $r_1 : r_2 = 2 : 1$ and the ratio of lengths $L_1 : L_2 = 1 : 2$, we have:
$\dfrac{\Delta L_1}{\Delta L_2} = \left( \dfrac{L_1}{L_2} \right) \times \left( \dfrac{r_2}{r_1} \right)^2$
Substituting the given values:
$\dfrac{\Delta L_1}{\Delta L_2} = \left( \dfrac{1}{2} \right) \times \left( \dfrac{1}{2} \right)^2 = \dfrac{1}{2} \times \dfrac{1}{4} = \dfrac{1}{8}$.
Answer: $1: 8$
Since both wires are made of the same material, $Y$ is the same for both. The force $F$ applied is also the same. Thus, $\Delta L \propto \dfrac{L}{A}$.
The area $A$ is given by $\pi r^2$, so $\Delta L \propto \dfrac{L}{r^2}$.
Given the ratio of radii $r_1 : r_2 = 2 : 1$ and the ratio of lengths $L_1 : L_2 = 1 : 2$, we have:
$\dfrac{\Delta L_1}{\Delta L_2} = \left( \dfrac{L_1}{L_2} \right) \times \left( \dfrac{r_2}{r_1} \right)^2$
Substituting the given values:
$\dfrac{\Delta L_1}{\Delta L_2} = \left( \dfrac{1}{2} \right) \times \left( \dfrac{1}{2} \right)^2 = \dfrac{1}{2} \times \dfrac{1}{4} = \dfrac{1}{8}$.
Answer: $1: 8$
Hence, the correct answer is option: 'A'.