Answer:
25.2 ml
Explanation:
When the aluminium block is inserted in the container, the overall amount of only the water in the beaker can equal V o.
The weight of the water expelled by the plastic container should be equal to the weight of the aluminium block, according to the buoyancy balance relation.
i.e.
[tex]\rho_w V_wg = m_{Al}g = \rho_{Al}V_{Al} g \\ \\ V_w = \dfrac{\rho_{Al}V_{AL}}{\rho_{w}}[/tex]
When the aluminium block is inserted into the plastic container, the initial volume of water = 60 ml
[tex]V_i = V_o + V_w[/tex]
[tex]V_i = V_o + \dfrac{\rho_{Al}V_{Al}}{\rho_{w}}---(1)[/tex]
When the aluminium block is placed outside the container, the volume of the water
[tex]V_f = V_o +V_{Al} ---(2)[/tex]
By subtracting equation (1) and (2)
[tex]V_i -V_f = V_o + \dfrac{\rho_{Al} V_{Al}}{\rho_w}- ( V_o + V_{Al}}) \\ \\ =\dfrac{\rho _{Al}V_{Al}}{\rho_w}-V_{Al} \\ \\ = V_{Al} \Big( \dfrac{\rho_{Al}}{\rho_{w}}-1 \Big)[/tex]
since;
[tex]m_{Al} = 40 g[/tex]
[tex]V _{Al} = \dfrac{40 \ g}{2.7 \ g/cm^3} \\ \\ V_{Al} = 14.815 \ cm^3[/tex]
Similarly;
[tex]\dfrac{\rho_{Al}}{\rho_{w}}= \dfrac{2.7 }{1.0}[/tex]
= 2.7
[tex]V_i -V_f =14.815\Big( 2.7-1 \Big) \\ \\ V_i -V_f = 25.1855 \ ml \\ \\ = \mathbf{25.2 \ ml}[/tex]
