A piece of gold weighs 10g in air and 9g in water. What is the volume of cavity? (Density of gold =`19.3 g cm^(-3)`)

Options

A. 0.282 cc
B. 0.482 cc
C. 0.682cc
D. None of these

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the weights
- The true weight of the gold piece in air is 10 grams.
- The apparent weight of the gold piece in water is 9 grams.

### Step 2: Calculate the weight of the liquid displaced
Using the formula for apparent weight:
\[ \text{Apparent Weight} = \text{True Weight} - \text{Weight of Liquid Displaced} \]
Substituting the given values:
\[ 9 \text{ g} = 10 \text{ g} - W_{\text{liquid}} \]
Solving for \(W_{\text{liquid}}\):
\[ W_{\text{liquid}} = 10 \text{ g} - 9 \text{ g} = 1 \text{ g} \]

### Step 3: Calculate the volume of liquid displaced
Using the formula:
\[ \text{Volume of Liquid Displaced} = \frac{\text{Weight of Liquid Displaced}}{\text{Density of Liquid}} \]
Since the density of water is \(1 \text{ g/cm}^3\):
\[ \text{Volume of Liquid Displaced} = \frac{1 \text{ g}}{1 \text{ g/cm}^3} = 1 \text{ cm}^3 \]

### Step 4: Calculate the actual volume of gold
Using the formula:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
For gold:
\[ \text{Volume of Gold} = \frac{10 \text{ g}}{19.3 \text{ g/cm}^3} \approx 0.518 \text{ cm}^3 \]

### Step 5: Calculate the volume of the cavity
The volume of the cavity can be found by subtracting the actual volume of gold from the volume of liquid displaced:
\[ \text{Volume of Cavity} = \text{Volume of Liquid Displaced} - \text{Volume of Gold} \]
\[ \text{Volume of Cavity} = 1 \text{ cm}^3 - 0.518 \text{ cm}^3 = 0.482 \text{ cm}^3 \]

### Final Answer
The volume of the cavity in the gold piece is approximately \(0.482 \text{ cm}^3\).

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