QJEEM24Phc862: JEE Main 2024 Physics - Class 11 JEEM24Phc862 Solution: One end of a metal wire is fixed to a ceiling and a load of…

To solve this problem, we use the fact that the longitudinal strain in a wire is given by the formula:

where $\Delta L$ is the change in length, $L$ is the original length, $F$ is the force applied, $A$ is the area of cross-section of the wire, and $Y$ is the Young's modulus of the material of the wire.

The force applied by each load due to gravity is calculated using $F = mg$ , where $m$ is the mass of the load and $g$ is the acceleration due to gravity ( $10 \text{ m/s}^2$ in this case).

For the upper wire, the total force applied is the weight of both masses (2 kg and 1 kg):

For the lower wire, the force applied is just the weight of the 1 kg mass:

Since the area $A$ and Young's modulus $Y$ are the same for both wires, these values cancel out when we calculate the ratio of the strains. Therefore, the ratio of the strains is directly proportional to the ratio of the forces:

Hence, the ratio of longitudinal strain of the upper wire to that of the lower wire is 3.