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 loads 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 :

Answer

3

Explanation

F = Mg
`rArr` stress `sigma = (F)/(A) = (Mg)/(A)`
Since `(sigma)/(in) = gamma rarr in = (sigma)/(gamma)`
In first case, `in = (Delta l)/(l) = (4 xx 10^(-3))/(2) = 2 xx 10^(-3)`
`rarr gamma = (Mg//A)/((2 xx 10^(-3)))`
After dipping in liquid,
Consider equilibrium of block of mass M.
Where F is the force applied by steel wire and
B = Buoyant force `= rho_(1)Vg`
Since `M = rho V` and `(rhol)/(rho) = (2)/(8) = 0.25` (given)
`B = (Mg)/(4)`
`F = Mg - (Mg)/(4) = (3Mg)/(4) , " " in = (F//A)/(Y) = (3 xx (Mg)/(4A))/(Y) = (3Mg)/(4AY)`
`E = (3Mg)/(4A) xx (2 xx 10^(-3))/(Mg//A) = 1.5 xx 10^(-3) , (Delta l)/(l) = 1.5 xx 10^(-3)`
`Delta l = 2 xx 1.5 xx 10^(-3) m , Delta l = 3mm`