Answer:
[tex]\rho_{s} = 4 [/tex]
[tex]\rho_{l} = 0.6 [/tex]
[tex]\rho{liq} = 600 kg/m^{3}[/tex]
Given:
Weight of solid in air, [tex]w_{sa} = 200 N[/tex]
Weight of solid in water, [tex]w_{sw} = 150 N[/tex]
Weight of solid in liquid, [tex]w_{sl} = 170 N[/tex]
Solution:
Calculation of:
1. Relative density of solid, [tex]\rho_{s}[/tex]
[tex]\rho_{s} = \frac{w_{sa}}{w_{sa} - w_{sw}}[/tex]
[tex]\rho_{s} = \frac{200}{200 - 150} = 4 [/tex]
2. Relative density of liquid, [tex]\rho_{l}[/tex]
[tex]\rho_{l} = \frac{w_{sa} - w_{sl}}{w_{sa} - w_{sw}}[/tex]
[tex]\rho_{l} = \frac{200 - 170}{200 - 150} = 0.6[/tex]
3. Density of liquid in S.I units:
Also, we know:
[tex]\rho{l} = \frac{\rho_{liq}}{\rho_{w}}[/tex]
where
[tex] = {\rho_{liq}}[/tex] = density of liquid
[tex] = {\rho_{w}} = 1000 kg/m^{3}[/tex] = density of water
Now, from the above formula:
[tex]0.6 = \frac{\rho_{liq}}{1000}[/tex]
[tex]\rho{liq} = 600 kg/m^{3}[/tex]