A beaker when partly filled with water has total mass `20.00g`. If a piece of metal with density `3.00g//cm^(3)` and volume `1.00cm^(3)` is suspended by a thin string, so that it is submerged in the water but does not rest on the bottom of the beaker, how much dies the beaker then apper to weigh if it is resting on a scale?

Explanation

To solve the problem, we need to determine the apparent weight of the beaker when the metal piece is submerged in the water. Here’s the step-by-step solution:

### Step 1: Understand the forces acting on the beaker
The beaker experiences two main forces:
- The weight of the beaker and the water inside it.
- The buoyant force acting on the submerged metal piece.

### Step 2: Calculate the weight of the beaker and water
The total mass of the beaker when partly filled with water is given as 20.00 g. To convert this mass into weight (force), we use the formula:
\[ \text{Weight} = \text{mass} \times g \]
where \( g \approx 9.8 \, \text{m/s}^2 \).

First, convert the mass from grams to kilograms:
\[ 20.00 \, \text{g} = 0.020 \, \text{kg} \]

Now, calculate the weight:
\[ \text{Weight}_{\text{beaker}} = 0.020 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 0.196 \, \text{N} \]

### Step 3: Calculate the buoyant force acting on the metal piece
The buoyant force can be calculated using Archimedes' principle:
\[ \text{Buoyant Force} = \text{density of fluid} \times \text{volume submerged} \times g \]

Given:
- Density of water \( \rho = 1000 \, \text{kg/m}^3 \)
- Volume of the metal piece \( V = 1.00 \, \text{cm}^3 = 1.00 \times 10^{-6} \, \text{m}^3 \)

Now, calculate the buoyant force:
\[ \text{Buoyant Force} = 1000 \, \text{kg/m}^3 \times 1.00 \times 10^{-6} \, \text{m}^3 \times 9.8 \, \text{m/s}^2 \]
\[ = 0.0098 \, \text{N} \]

### Step 4: Calculate the apparent weight of the beaker
The apparent weight of the beaker when the metal piece is submerged is the sum of the weight of the beaker and the buoyant force:
\[ \text{Apparent Weight} = \text{Weight}_{\text{beaker}} + \text{Buoyant Force} \]
\[ = 0.196 \, \text{N} + 0.0098 \, \text{N} \]
\[ = 0.2058 \, \text{N} \]

### Step 5: Convert the apparent weight back to grams
To express this weight in grams, we can use the conversion:
\[ \text{Weight in grams} = \frac{\text{Apparent Weight}}{g} \]
\[ = \frac{0.2058 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 0.02096 \, \text{kg} = 20.96 \, \text{g} \]

### Final Answer
Thus, the beaker appears to weigh approximately **20.96 g** when the metal piece is submerged.

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