Respuesta :
Answer:
Explanation:
Given
mass of aluminium ball [tex]m_{object}=4\ kg[/tex]
apparent mass of aluminium ball in liquid [tex]m_{apparent}=2.1\ kg[/tex]
If [tex]\rho _{object}[/tex] and [tex]\rho _{liquid}[/tex] is the density of object and liquid then
mass of object can be written as
[tex]\rho _{object}\times V=4---1[/tex]
[tex]\rho _{object}\times V-\rho _{apparent}\times V=2.1----2[/tex]
where V is the volume of object
divide 1 and 2 we get
[tex]\frac{\rho _{object}\times V}{\rho _{object}\times V-\rho _{liquid}\times V}=\frac{4}{2.1}[/tex]
[tex]\frac{\rho _{object}}{\rho _{object}-\rho _{liquid}}=\frac{4}{2.1}[/tex]
[tex]\frac{\rho _{liquid}}{\rho _{object}}=\frac{4-2.1}{4}\ \ \left [ \frac{\rho _{liquid}}{\rho _{object}}=\frac{m_{object}-m_{apparent}}{m_{object}}\right ][/tex]
[tex]\rho _{liquid}=\rho _{object}\times 0.475[/tex]
Density of aluminium is [tex]2710\ kg/m^3[/tex]
[tex]\rho_{liquid}=2710\times 0.475=1287.25\ kg/m^3[/tex]
