A load of mass \(M\,kg\) 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 ........ \(mm\)

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Correct Answer

(A) \(3\)

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Detailed explanation

In first case:

\(\frac{{\frac{{mg}}{A}}}{{\left( {\frac{{\Delta {\ell _1}}}{\ell }} \right)}} = Y\,\,\,\,\, \Rightarrow \,\,\,\Delta {\ell _1} = \frac{{mg}}{{\left( {\frac{{YA}}{\ell }} \right)}}\)

In second case:

\(\frac{{\frac{{mg - B}}{A}}}{{\left( {\frac{{\Delta {\ell _1}}}{\ell }} \right)}} = Y\,\,\,\, \Rightarrow \,\,\Delta {\ell _2} = \frac{{mg - B}}{{\left( {\frac{{YA}}{\ell }} \right)}} = \frac{{\left( {\frac{{3mg}}{4}} \right)}}{{\left( {\frac{{YA}}{\ell }} \right)}}\)

\( \Rightarrow \,\,\Delta {\ell _2} = \frac{3}{4}\Delta {\ell _1} = 3\,mm\)

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