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 :

Options

A. zero
B. 4.0 mm
C. 3.0 mm
D. 5.0 mm

Answer

C

Explanation

F=Mg
`rarr` stne `sssigma =(F)/(A)=(Mg)/(A)`
Since `(sigma)/(in)=gammararrin=(sigma)/(gamma)`
IN first case, `in=(Deltal)/(l)=(4xx10^(-3))/(2)=2xx10^(-3)`
`rarrgamma=(Mg//A)/((2xx10^(-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_(l)Vg`
Since `MrhoVandrho_(l)/(rho)=(2)/(8)=0.25`(given)

`B=(Mg)/(4)`
`F=Mg-(Mg)/(4)=(3Mg)/(4),in=(F//A)/(Y)=(3xx(Mg)/(4A))/(Y)=(3Mg)/(4AY)`
`E=(3Mg)/(4A)xx(2xx10^(-3))/(Mg//A)=1.5xx10^(-3),(Deltal)/(l)=1.5xx10^(-3)`
`Deltal=2xx1.5xx10^(-3)m,Deltal=3mm`