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How to solve this problem- 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 relat

 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)
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A admin

 

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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