Respuesta :

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]

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