Respuesta :
Answer:
[tex]S_i=\frac{9}{10} =0.9[/tex]
Explanation:
Given:
- volume of oil in the cylinder, [tex]V_o=60\ cm^2[/tex]
- volume of the oil level when the ice is immersed, [tex]V=90\ cm^3[/tex]
- the volume level of oil when the ice melted, [tex]V'=87\ cm^3[/tex]
Now, therefore the volume of ice:
[tex]V_i=V-V_o[/tex]
[tex]V_i=90-60[/tex]
[tex]V_i=30\ cm^3[/tex]
Now the volume of water:
[tex]V_w=V'-V_o[/tex]
[tex]V_w=87-60[/tex]
[tex]V_w=27\ cm^3[/tex]
As we know that the relative density is the ratio of density of the substance to the density of water.
So, the relative density of ice:
[tex]S_i=\frac{\rho_i}{\rho_w}[/tex] .....................(1)
as we know that density is given as:
[tex]\rm \rho=\frac{mass}{volume}[/tex]
now eq. (1)
[tex]S_i=\frac{m}{V_{i}}\div \frac{m}{V_w}[/tex]
where, m = mass of the water or the ice which remains constant in any phase
[tex]S_i=\frac{V_w}{V_i}[/tex]
[tex]S_i=\frac{27}{30}[/tex]
[tex]S_i=\frac{9}{10} =0.9[/tex]
