JEE MAIN - Physics (2019 - 12th January Evening Slot - No. 1)
A load of mass M kg is suspended from a steel wire of length 2m 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:
The new value of increase in length of the steel wire is:
5.0 mm
zero
3.0 mm
4.0 mm
Explanation
_12th_January_Evening_Slot_en_1_1.png)
$${F \over A} = y.{{\Delta \ell } \over \ell }$$
$$\Delta \ell \propto F$$ . . .. (i)
T $$=$$ mg
T $$=$$ mg $$-$$ fB $$=$$ mg $$-$$ $${m \over {{\rho _b}}}.{\rho _\ell }.$$g
$$ = \left( {1 - {{{\rho _\ell }} \over {{\rho _b}}}} \right)$$ mg
$$ = \left( {1 - {2 \over 8}} \right)$$ mg
T' $$=$$ $${3 \over 4}$$ mg
From (i)
$${{\Delta \ell '} \over {\Delta \ell }} = {{T'} \over T} = {3 \over 4}$$
$$\Delta \ell ' = {3 \over 4}.\Delta \ell = 3$$ mm
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