Q. 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 :

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Solution:

$\cfrac{F}{A} \, = \, y. \cfrac{\Delta \ell}{\ell}$ $\Delta \ell \, \, ∝ \, F \, \, \, \, \, \, \, \, \, \, \, \,$...(1) $T \, = \, mg$ $T \, = \, mg \, - \, f_B \, = \, mg \, - \, \cfrac{m}{\rho_b}.\rho \ell.g$ $\, \, \, \, \, \, \, \, \, =\bigg(1 \, - \cfrac{\rho_\ell}{\rho_b}\bigg) mg$ $\, \, \, \, \, \, \, \, \, \, \, \, =\bigg(1 - \cfrac{2}{8} \bigg) mg$ $\, \, \, \, \, \, \, \, \, \, \, \, T' \, = \, \cfrac{3}{4}mg$ $From \, (1)$ $\cfrac{\Delta \ell'}{\Delta \ell} = \cfrac{T'}{T} \, = \, \cfrac{3}{4}$ $\Delta \ell' \, \, = \cfrac{3}{4} \, .\Delta \ell \, \, =3 \, mm$

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