Finding the Actual Mass of a Submerged Object

This problem involves understanding how the apparent mass of an object changes when it is fully immersed in a fluid like water. The reduction in mass is due to the buoyant force exerted by the fluid. We can use the concepts of apparent mass, actual mass, relative density, and buoyant force to solve this.

Understanding the Key Concepts

• Actual Mass: The true mass of the object measured in vacuum or air (ignoring buoyancy of air, which is usually negligible).
• Apparent Mass: The mass of the object when measured while submerged in a fluid. It is the actual mass minus the buoyant force divided by the acceleration due to gravity.
• Buoyant Force: The upward force exerted by a fluid that opposes the weight of a partially or fully immersed object. According to Archimedes' principle, this force is equal to the weight of the fluid displaced by the object.
• Relative Density (Specific Gravity): The ratio of the density of a substance to the density of a reference substance, usually water for liquids and solids. \[ \text{Relative Density (RD)} = \frac{\text{Density of Substance}}{\text{Density of Water}} \]

Relating Apparent Mass, Actual Mass, and Relative Density

When an object is fully immersed in water, the buoyant force acts upwards, reducing its effective weight.

Let \(m\) be the actual mass of the metal piece and \(m_{apparent}\) be its apparent mass when immersed in water.

The relationship between apparent mass, actual mass, and buoyant force is given by: \[ m_{apparent} \times g = m \times g - F_B \] where \(g\) is the acceleration due to gravity and \(F_B\) is the buoyant force.

Dividing by \(g\), we get: \[ m_{apparent} = m - \frac{F_B}{g} \]

The buoyant force \(F_B\) is equal to the weight of the water displaced. If \(V\) is the volume of the metal piece (and thus the volume of water displaced) and \(\rho_{water}\) is the density of water, then: \[ F_B = V \times \rho_{water} \times g \]

So, the buoyant force divided by \(g\) is: \[ \frac{F_B}{g} = V \times \rho_{water} \]

Substituting this into the apparent mass equation: \[ m_{apparent} = m - V \times \rho_{water} \]

The actual mass \(m\) of the metal piece can also be expressed in terms of its volume \(V\) and density \(\rho_{metal}\): \[ m = \rho_{metal} \times V \] From this, we can write the volume as: \[ V = \frac{m}{\rho_{metal}} \]

Substitute this expression for \(V\) into the apparent mass equation: \[ m_{apparent} = m - \left( \frac{m}{\rho_{metal}} \right) \times \rho_{water} \] \[ m_{apparent} = m \left( 1 - \frac{\rho_{water}}{\rho_{metal}} \right) \]

We know that relative density (RD) is the ratio of the density of the metal to the density of water: \[ \text{RD} = \frac{\rho_{metal}}{\rho_{water}} \] Therefore, the ratio \(\frac{\rho_{water}}{\rho_{metal}}\) is equal to \(\frac{1}{\text{RD}}\).

Substitute this into the equation for apparent mass: \[ m_{apparent} = m \left( 1 - \frac{1}{\text{RD}} \right) \]

Now, we can rearrange this equation to solve for the actual mass \(m\): \[ m_{apparent} = m \left( \frac{\text{RD} - 1}{\text{RD}} \right) \] \[ m = m_{apparent} \times \frac{\text{RD}}{\text{RD} - 1} \]

Step-by-Step Calculation

Given:

• Apparent mass (\(m_{apparent}\)) = 60 gm
• Relative Density (RD) = 2.5

Using the formula derived: \[ m = m_{apparent} \times \frac{\text{RD}}{\text{RD} - 1} \]

Substitute the given values: \[ m = 60 \times \frac{2.5}{2.5 - 1} \] \[ m = 60 \times \frac{2.5}{1.5} \]

Now, perform the calculation: \[ m = 60 \times \frac{2.5}{1.5} \] \[ m = 60 \times \frac{25}{15} \] \[ m = 60 \times \frac{5}{3} \] \[ m = 20 \times 5 \] \[ m = 100 \text{ gm} \]

Result and Conclusion

The actual mass of the piece of metal is 100 gm.

When the metal is immersed in water, it displaces a volume of water that weighs \(100 \text{ gm} - 60 \text{ gm} = 40 \text{ gm}\). The density of water is approximately 1 gm/cm³. So, the volume of the metal is about 40 cm³.

The density of the metal is relative density \( \times \) density of water = \(2.5 \times 1 \text{ gm/cm}^3 = 2.5 \text{ gm/cm}^3\).

The actual mass is density \( \times \) volume = \(2.5 \text{ gm/cm}^3 \times 40 \text{ cm}^3 = 100 \text{ gm}\). This confirms our result.

Revision Table: Mass, Density, and Buoyancy

Additional Information: Archimedes' Principle and Applications

Archimedes' principle is fundamental to understanding buoyancy. It states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

This principle explains why ships float (they displace a weight of water equal to their own weight), why hot air balloons rise (hot air is less dense than surrounding air), and how hydrometers work (they sink to a depth where their weight equals the weight of the displaced fluid).

The concept of apparent weight (or mass) is crucial in various applications, such as measuring the density of irregular solids or determining the composition of alloys.

The formula \(m = m_{apparent} \times \frac{\text{RD}}{\text{RD} - 1}\) is a useful shortcut for problems involving apparent mass in water and relative density.