A merchant in Kathmandu sells you a solid gold 1-kg statue for a very reasonable price. When you get home, you wonder whether you got taken, so you lower the statue into a container of water and measure the volume of displaced water. Show that for 1 kg of pur gold, the volume of water displaced is 51.8cm3 (Hint: Archimedes' Principal)

51.81ml

To determine the volume of water displaced by the solid gold statue, we can use Archimedes' Principle.

Archimedes' Principle states that the buoyant force exerted by a fluid on an object submerged in it is equal to the weight of the fluid displaced by the object.

Here's how you can calculate the volume of water displaced by the 1 kg solid gold statue:

Step 1: Determine the density of gold
The density of gold is known to be approximately 19.3 grams per cubic centimeter (g/cm^3). Since we need the density in kilograms per cubic centimeter (kg/cm^3) for the rest of the calculation, we need to convert the density to kg/cm^3.
Density of gold = 19.3 g/cm^3 = 0.0193 kg/cm^3

Step 2: Calculate the weight of the gold statue
The weight of the gold statue is given as 1 kg.

Step 3: Calculate the volume of water displaced
According to Archimedes' Principle, the weight of the water displaced is equal to the weight of the gold statue.
Weight of water displaced = Weight of gold statue = 1 kg

Since weight = mass x gravitational acceleration, and gravitational acceleration is approximately 9.8 m/s^2:
Weight = mass x gravitational acceleration
1 kg = mass x 9.8 m/s^2
mass = 1 kg / (9.8 m/s^2)
mass = 0.102 kg

To solve for volume, we can use the equation:
Density = mass / volume

Plugging in the known values:
0.0193 kg/cm^3 = 0.102 kg / volume

Solving for volume:
volume = 0.102 kg / 0.0193 kg/cm^3
volume ≈ 5.28 cm^3

Therefore, the volume of water displaced by a 1 kg solid gold statue is approximately 5.28 cm^3.

To determine the volume of water displaced by the solid gold statue, we can use Archimedes' Principle. According to this principle, when an object is submerged in a fluid, it experiences an upward buoyant force equal to the weight of the fluid it displaces.

Now, let's break down the steps to calculate the volume of the water displaced by the gold statue:

1. Start by calculating the weight of the gold statue. Since it is mentioned that the statue weighs 1 kg, we can multiply the mass by the acceleration due to gravity (9.8 m/s²) to find the weight. Therefore, the weight of the gold statue is 1 kg × 9.8 m/s² = 9.8 N (Newtons).

2. The weight of the water displaced by the statue will be equal to the weight of the statue itself (as per Archimedes' Principle). Therefore, the weight of the water displaced is also 9.8 N.

3. Now, use the density of water to find the volume of the water displaced. The density of water is approximately 1 g/cm³ or 1000 kg/m³. Since the weight of the water displaced is 9.8 N and the density of water is 1000 kg/m³, we can use the formula:

Weight of water displaced = Density of water × Volume of water displaced × Acceleration due to gravity

Substituting the given values, we get:

9.8 N = 1000 kg/m³ × Volume of water displaced × 9.8 m/s²

Canceling out the factors of acceleration due to gravity, we are left with:

1 = Volume of water displaced × 1000

Rearranging the equation, we find:

Volume of water displaced = 1 / 1000 = 0.001 m³

4. The volume we obtained in the previous step is in cubic meters, but the question specifies the volume in cubic centimeters. To convert from cubic meters to cubic centimeters, we need to multiply by 1,000,000. Therefore:

Volume of water displaced = 0.001 m³ × 1,000,000 cm³/m³ = 1000 cm³

Hence, the volume of water displaced by a 1 kg solid gold statue is 1000 cm³.

Therefore, based on the given information, the calculated volume of water displaced by a 1 kg solid gold statue is 1000 cm³.

mass = volume x density.

mass = 1000 g
density of gold is 19.3 g/cc
Solve for volume.