question_answer

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-      [JEE Main 12-Jan-2019 Evening]

A)  4.0 mm           

B)       zero     

C)  5.0 mm           

D)       3.0 mm.

Correct Answer: D Solution : [d] Let \[\rho \]and \[\sigma \]be the density of the liquid and material of the load respectively.

In first case, the extension in the wire is

\[x=Mg/k=V\rho g/k\]                             .(i)

When the load is immersed in the liquid, up thrust + internal force due to extension in wire = weight of the load

\[\Rightarrow V\sigma g+k{{x}_{1}}=V\rho g\Rightarrow {{x}_{1}}=Vg(\rho -\sigma )/k\]

Using (i) and (ii),

\[{{x}_{1}}=\frac{Vgx}{Vg\rho }(\rho -\sigma )=x\left( 1-\frac{\sigma }{\rho } \right)=4\times \left( 1-\frac{2}{8} \right)=3mm\]