Answer:
(i) The volume of the cylindrical beaker is given by:
V = A x h = (25 cm^2) x (10 cm) = 250 cm^3
The mass of the oil in the beaker is given by:
m_oil = density x volume = (0.8 g/cm^3) x (250 cm^3) = 200 g
The total mass of the beaker and oil is therefore:
m_total = m_beaker + m_oil = 50 kg + 0.2 kg = 50.2 kg
(ii) The volume of the aluminum is given by:
V_aluminum = m_aluminum / density = 66 g / (2.2 g/cm^3) = 30 cm^3
When the aluminum is lowered into the beaker, it displaces an equal volume of oil. Therefore, the volume of oil that overflows is 30 cm^3.
(iii) The final mass of the beaker and its contents is the sum of the mass of the beaker, the mass of the oil remaining in the beaker, and the mass of the aluminum:
m_final = m_beaker + m_oil + m_aluminum = 50 kg + 0.17 kg + 0.066 kg = 50.24 kg
To calculate the mass of the remaining oil, we need to subtract the volume of aluminum from the volume of the beaker and multiply by the density of the oil:
V_remaining_oil = (A x h) - V_aluminum = (25 cm^2 x 10 cm) - 30 cm^3 = 220 cm^3
m_remaining_oil = density x V_remaining_oil = 0.8 g/cm^3 x 220 cm^3 = 176 g
Therefore, the final mass of the beaker and its contents after the overflow liquid has been wiped off is 50.24 kg, and there is 176 g of oil remaining in the beaker