A load of mass M kg is suspended from a steel wire of length 2m and radius 1.0 mm in Searle's apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8. The new value of increase in length of the steel wire is:

In Searle’s apparatus the extension in the steel wire is obtained from Young’s modulus formula

$$\Delta L=\frac{F\,L}{A\,Y}$$

where $$\Delta L$$ is the increase in length, $$F$$ the tensile force in the wire, $$L$$ the original length, $$A$$ the cross-sectional area and $$Y$$ Young’s modulus of steel. For a given wire $$L,\;A$$ and $$Y$$ are fixed, so the extension is directly proportional to the applied force:

$$\Delta L\;\propto\;F$$

Initially the wire supports the load in air only. The force is simply the weight

$$F_1 = M\,g$$

and the corresponding extension is given as

$$\Delta L_1 = 4.0\;\text{mm}$$

Now the load is completely immersed in a liquid. The liquid exerts an upthrust (buoyant force) that reduces the effective weight. We determine this buoyant force step by step.

The relative density (specific gravity) of the load is 8, so its density is

$$\rho_{\text{load}} = 8\,\rho_{\text{water}}$$

The relative density of the liquid is 2, so the liquid’s density is

$$\rho_{\text{liq}} = 2\,\rho_{\text{water}}$$

The volume of the load is

$$V = \frac{M}{\rho_{\text{load}}} = \frac{M}{8\,\rho_{\text{water}}}$$

The buoyant force equals the weight of the displaced liquid:

$$F_b = \rho_{\text{liq}}\,V\,g = (2\,\rho_{\text{water}})\left(\frac{M}{8\,\rho_{\text{water}}}\right)g = \frac{2}{8}\,M\,g = \frac{1}{4}\,M\,g$$

Hence the apparent (effective) weight of the load while immersed is

$$F_2 = F_1 - F_b = M\,g - \frac{1}{4}\,M\,g = \frac{3}{4}\,M\,g$$

Because extension is proportional to force, the new extension $$\Delta L_2$$ satisfies

$$\frac{\Delta L_2}{\Delta L_1} = \frac{F_2}{F_1} = \frac{\frac{3}{4}M\,g}{M\,g} = \frac{3}{4}$$

Substituting $$\Delta L_1 = 4.0\;\text{mm}$$, we obtain

$$\Delta L_2 = \frac{3}{4}\times 4.0\;\text{mm} = 3.0\;\text{mm}$$

Hence, the correct answer is Option D.