A piece of gold `(rho=19.3 g//cm^(3))` has a cavity in it. It weights `38.2g` in air and `36.2g` in water. The volume of the cavity in glold is

Options

A. `0.2cm^(3)`
B. `0.04cm^(3)`
C. `0.02cm^(3)`
D. `0.01cm^(3)`

Explanation

To find the volume of the cavity in the piece of gold, we can follow these steps:

### Step 1: Understand the forces acting on the gold piece
When the gold piece is weighed in air, it has a weight of 38.2 g. When it is weighed in water, it weighs 36.2 g. The difference in weight is due to the buoyant force acting on the gold piece when submerged in water.

### Step 2: Calculate the buoyant force
The buoyant force can be calculated as the difference in weight when the gold piece is in air and when it is in water:
\[ \text{Buoyant Force} = \text{Weight in air} - \text{Weight in water} = 38.2 \, \text{g} - 36.2 \, \text{g} = 2.0 \, \text{g} \]

### Step 3: Use Archimedes' principle
According to Archimedes' principle, the buoyant force is equal to the weight of the water displaced by the submerged object. This can be expressed as:
\[ \text{Buoyant Force} = V_{\text{displaced}} \cdot \rho_{\text{water}} \cdot g \]
where \( V_{\text{displaced}} \) is the volume of the gold piece plus the volume of the cavity, \( \rho_{\text{water}} \) is the density of water (approximately 1 g/cm³), and \( g \) is the acceleration due to gravity (which cancels out in this case).

### Step 4: Set up the equation
Since the buoyant force is 2.0 g, we can set up the equation:
\[ 2.0 \, \text{g} = V_{\text{gold}} \cdot \rho_{\text{water}} + V_{\text{cavity}} \cdot \rho_{\text{water}} \]
This simplifies to:
\[ 2.0 = V_{\text{gold}} + V_{\text{cavity}} \]

### Step 5: Calculate the volume of the gold piece
The volume of the gold piece can be calculated using its mass and density:
\[ V_{\text{gold}} = \frac{\text{mass}}{\text{density}} = \frac{38.2 \, \text{g}}{19.3 \, \text{g/cm}^3} \approx 1.979 \, \text{cm}^3 \]

### Step 6: Substitute back to find the volume of the cavity
Now we can substitute \( V_{\text{gold}} \) back into the equation:
\[ 2.0 = 1.979 + V_{\text{cavity}} \]
Solving for \( V_{\text{cavity}} \):
\[ V_{\text{cavity}} = 2.0 - 1.979 \approx 0.021 \, \text{cm}^3 \]

### Conclusion
The volume of the cavity in the gold piece is approximately \( 0.021 \, \text{cm}^3 \).

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