JEE Main 12 Jan 2019 Evening Question 12
Question: 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 [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) 4.0 mm
B) zero
C) 5.0 mm
D) 3.0 mm.
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ \rho $ and $ \sigma $ be the density of the liquid and material of the load respectively.
In first case, the extension in the wire is $ x=Mg/k=V\rho g/k $ -(i)
When the load is immersed in the liquid,
up thrust + internal force due to extension in wire = weight of the load
$ \Rightarrow V\sigma g+kx _1=V\rho g\Rightarrow x _1=Vg(\rho -\sigma )/k -$(ii)
Using (i) and (ii), $ x _1=\frac{Vgx}{Vg\rho }(\rho -\sigma )=x( 1-\frac{\sigma }{\rho } )=4\times ( 1-\frac{2}{8} )=3mm $
