Answer:
[tex]\rho _{liquid}=1995.07kg/m^{3}[/tex]
Explanation:
When the rock is immersed in unknown liquid the forces that act on it are shown as under
1) Tension T by the string
2) Weight W of the rock
3) Force of buoyancy due to displaced liquid B
For equilibrium we have [tex]T_{3}+B = W_{rock}[/tex]
[tex]T_{3}+\rho _{Liquid}V_{rock}g[/tex]=[tex]W_{rock}.....(\alpha)[/tex]
When the rock is suspended in air for equilibrium we have
[tex]T_{1}=W_{rock}....(\beta)[/tex]
When the rock is suspended in water for equilibrium we have
[tex]T_{2}[/tex] + [tex]\rho _{water}V_{rock}g[/tex]=[tex]W_{rock}.....(\gamma)[/tex]
Using the given values of tension and solving α,β,γ simultaneously for [tex]\rho _{Liquid}[/tex] we get
[tex]W_{rock}=51.9N\\31.6+1000\times V_{rock}\times g=51.9N\\\\11.4+\rho _{liquid}V_{rock}g=51.9N\\\\[/tex]
Solving for density of liquid we get
[tex]\rho _{liquid}=\frac{51.9-11.4}{51.9-31.6}\times 1000[/tex]
[tex]\rho _{liquid}=1995.07kg/m^{3}[/tex]
