A load of mass $M \mathrm{~kg}$ is suspended from a steel wire of length 2 m and radius 1.0 mm in Se

Question

A load of mass $M \mathrm{~kg}$ is suspended from a steel wire of length 2 m 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\

Professor PC Thomas Classes · Answer

(d) : Let $ho$ and $\sigma$ be the density of the liquid and material of the load respectively.\ In first case, the extension in the wire is $$\begin{equation*} x=M g / k=V ho g / k 	ag{i} \end{equation*}$$ When the load is immersed in the liquid, upthrust + internal force due to extension in wire $=$ weight of the load\ $\Rightarrow \quad V \sigma g+k x_{1}=V ho g$ $$\begin{equation*} \Rightarrow \quad x_{1}=V g(ho-\sigma) / k 	ag{ii} \end{equation*}$$ Using (i) and (ii),\ $x_{1}=\frac{V g x}{V g ho}(ho-\sigma)=x\left(1-\frac{\sigma}{ho}ight)$\ $=4 	imes\left(1-\frac{2}{8}ight)=3 \mathrm{~mm}$\