JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Elasticity & SHM

The stretch of the wire is proportional to the tension in it. In air the tension is the full weight and the extension is $4.0$ mm. Immersing the load adds an upthrust equal to the weight of displaced liquid; with the load's relative density $8$ and the liquid's $2$, this buoyant force is $\frac{2}{8}Mg=\frac{Mg}{4}$. The effective tension becomes $Mg-\frac{Mg}{4}=\frac{3Mg}{4}$, and since extension scales with tension the new extension is $\frac{3}{4}\times 4.0=3.0$ mm.