Final answer:
Using the density formula, the volume of the brass cylinder is calculated to be 383.05 cm³, and the tension in the string when submerged is 27.72 N, which is derived from the apparent weight loss due to the water displacement.
Step-by-step explanation:
To calculate the volume of the brass cylinder, we use the formula for density, \(\rho = \frac{m}{V}\), where \(\rho\) is the density, \(m\) is the mass, and \(V\) is the volume. Given that the mass \(m\) is 3.21 kg (or 3210 g, since density is given in g/cm³) and the density \(\rho\) is 8.38 g/cm³, the volume \(V\) of the cylinder can be calculated as follows:
\(V = \frac{m}{\rho} = \frac{3210 \, g}{8.38 \, g/cm^3} = 383.05 \, cm^3\).
To calculate the tension in the string when the cylinder is submerged in water, we first need to determine the apparent weight loss of the cylinder, which equals the weight of the water displaced. Since the volume of the cylinder is 383.05 cm³, it will displace the same volume of water when fully submerged. The mass of the water displaced is equal to its volume multiplied by its density (which is 1.00 g/cm³). Thus, the tension in the string is equal to the actual weight of the brass cylinder in air minus the weight of the water displaced.
The weight of the brass cylinder in air is \(m_{brass} \cdot g = 3210 \, g \cdot 9.81 \, m/s^2 = 31.48 \, N\) (converting grams to kilograms by dividing by 1000).
The weight of the water displaced is \(m_{water} \cdot g = V_{water} \cdot \rho_{water} \cdot g = 383.05 \, cm^3 \cdot 1.00 \, g/cm^3 \cdot 9.81 \, m/s^2 = 3.76 \, N\) (again, converting grams to kilograms).
The tension in the string, which is the apparent weight of the submerged cylinder, can now be calculated as \(T = m_{brass} \cdot g - m_{water} \cdot g = 31.48 \, N - 3.76 \, N = 27.72 \, N\).
So the volume of the cylinder is 383.05 cm³ and the tension in the string when submerged is 27.72 N.