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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 :
Option 1)
zero
Option 2)
5.0 mm
Option 3)
3.0 mm
Option 4)
4.0 mm

Answers (1)

best_answer

Elongation -

$\Delta l= \frac{FL}{AY}$

- wherein

$\Delta l= Elongation$

$T=mg$ - (i) $T^{'}-mg +F_{B} = 0$ - (ii)

We known that

$\frac{F}{A} = Y.\frac{\Delta l}{l}$

$\Rightarrow \Delta l\propto F$ - (iii)

$T^{'}=mg - F_{B}$

$T^{'}=mg - \frac{m}{\rho _{b}}.g$

$T^{'}=\left ( 1-\frac{\rho _{l}}{\rho_{b}} \right )mg = \left ( 1-\frac{2}{8} \right )mg = \frac{3}{4}mg$ ........iv

From equation i, iii and iv

$\frac{\Delta l^{'}}{\Delta l} = \frac{T^{'}}{T} = \frac{\frac{3}{4}mg}{mg}$

$\Delta l^{'} = \frac{3}{4}\Delta l=\frac{3}{4}\times 4$

$\Delta l^{'} = 3mm$

Option 1)

zero

Option 2)

5.0 mm

Option 3)
3.0 mm
Option 4)

4.0 mm

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