Understanding Weight Differences: Water vs. Mercury in a Container

This problem asks us to compare the weight of water and the weight of mercury when they fill the same container. The key concept here is density, which relates mass to volume, and how mass relates to weight.

Relating Weight, Mass, Volume, and Density

The weight of an object or substance is determined by its mass and the acceleration due to gravity. The mass of a substance, in turn, depends on its density and the volume it occupies.

• Weight (W) = Mass (m) × Acceleration due to gravity (g)
• Mass (m) = Density (\(\rho\)) × Volume (V)

Combining these, we get: Weight (W) = Density (\(\rho\)) × Volume (V) × Acceleration due to gravity (g)

Applying Concepts to the Problem

We have a container that is filled first with water and then with mercury. Since it's the same container, the volume of the substance filling it is the same in both cases. Let's denote this volume by \(V\).

Let \(\rho_{water}\) be the density of water and \(\rho_{mercury}\) be the density of mercury. We are given that \(\rho_{mercury} = 13.6 \times 10^3 \text{ kg/m}^3\). The standard density of water is approximately \(\rho_{water} = 1.0 \times 10^3 \text{ kg/m}^3\).

Weight of Water (X)

The weight of the water filling the container is given as X. Using the formula:

\[ X = \rho_{water} \times V \times g \]

Substituting the density of water:

\[ X = (1.0 \times 10^3 \text{ kg/m}^3) \times V \times g \quad \text{(Equation 1)} \]

Weight of Mercury (Y)

The weight of the mercury filling the same container is given as Y. Using the formula:

\[ Y = \rho_{mercury} \times V \times g \]

Substituting the density of mercury:

\[ Y = (13.6 \times 10^3 \text{ kg/m}^3) \times V \times g \quad \text{(Equation 2)} \]

Comparing the Weights

Now we need to find the relationship between X and Y. We can divide Equation 2 by Equation 1:

\[ \frac{Y}{X} = \frac{(13.6 \times 10^3 \text{ kg/m}^3) \times V \times g}{(1.0 \times 10^3 \text{ kg/m}^3) \times V \times g} \]

Notice that the volume \(V\) and the acceleration due to gravity \(g\) are the same in both the numerator and the denominator, so they cancel out:

\[ \frac{Y}{X} = \frac{13.6 \times 10^3}{1.0 \times 10^3} \]

\[ \frac{Y}{X} = 13.6 \]

To find the relationship between Y and X, we can rearrange this equation:

\[ Y = 13.6 \times X \]

This shows that the weight of the mercury is 13.6 times the weight of the water when they occupy the same volume.

Step-by-Step Solution

1. Identify that the container has a fixed volume, \(V\), which is the same for both water and mercury.
2. Recall the formula for weight: Weight = Density × Volume × Acceleration due to gravity (\(W = \rho V g\)).
3. Write down the weight of water: \(X = \rho_{water} V g\).
4. Write down the weight of mercury: \(Y = \rho_{mercury} V g\).
5. Use the given density of mercury (\(\rho_{mercury} = 13.6 \times 10^3 \text{ kg/m}^3\)) and the known density of water (\(\rho_{water} = 1.0 \times 10^3 \text{ kg/m}^3\)).
6. Substitute the densities into the equations for X and Y: \[ X = (1.0 \times 10^3) V g \] \[ Y = (13.6 \times 10^3) V g \]
7. Divide the equation for Y by the equation for X: \[ \frac{Y}{X} = \frac{(13.6 \times 10^3) V g}{(1.0 \times 10^3) V g} \]
8. Cancel out the common terms (\(V\) and \(g\)): \[ \frac{Y}{X} = \frac{13.6 \times 10^3}{1.0 \times 10^3} = 13.6 \]
9. Rearrange the equation to find the relationship: \(Y = 13.6 X\).
10. Compare this result with the given options.

Key Concepts Revisited

• Density: A measure of how much mass is contained in a given volume (\(\rho = m/V\)). Denser substances have more mass for the same volume.
• Weight: The force of gravity on an object's mass (\(W = mg\)). Since \(g\) is constant at a given location, weight is directly proportional to mass.
• Volume: The amount of space a substance occupies. In this problem, the volume of the container is constant.

Because mercury is much denser than water, the same volume of mercury will have much more mass than the same volume of water. Since weight is proportional to mass, the mercury will weigh significantly more.

Revision Table: Comparing Water and Mercury

Additional Information: Specific Gravity

The ratio of the density of a substance to the density of a reference substance (usually water for liquids and solids) is called specific gravity or relative density. It is a dimensionless quantity.

Specific Gravity of Mercury = \(\frac{\text{Density of Mercury}}{\text{Density of Water}}\)

Specific Gravity of Mercury = \(\frac{13.6 \times 10^3 \text{ kg/m}^3}{1.0 \times 10^3 \text{ kg/m}^3} = 13.6\)

This specific gravity value directly tells us how many times denser mercury is than water. Consequently, for the same volume, mercury is 13.6 times heavier than water.