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Sagot :
Answer:
[tex]$10.97 \ g/cm^3$[/tex]
Explanation:
Given :
Mass of a bar of lead = 115.2 g
Initial water level [tex]$\text{in the graduated cylinder}$[/tex] = 25 mL
Final water level [tex]$\text{in the graduated cylinder}$[/tex] = 35.5 mL
Difference in the water level = 35.5 - 25
= 10.5 mL
= [tex]10.5 \ cm^3[/tex]
We know that when a body is submerged in water, it displaces its own volume of water.
Therefore, the volume of the lead bar = volume of the water displaced = 10.5 mL = [tex]10.5 \ cm^3[/tex]
We know that mathematically, density is the ratio of mass of body to its volume.
Density of the lead bar is given by :
[tex]$\rho =\frac{\text{mass}}{\text{volume}}$[/tex]
[tex]$\rho =\frac{\text{115.2 g}}{\text{10.5 cm}^3}$[/tex]
= [tex]$10.97 \ g/cm^3$[/tex]
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