A piece of iron weighs 44.6 gf in air. If the density of iron is `8.9xx10^(3)kgm^(-3)`, find the weight of iron piece when immersed in water.

Answer

To find the weight of the iron piece when immersed in water, we can follow these steps:

### Step 1: Convert the weight in air to mass
Given that the weight of the iron piece in air is 44.6 gf, we can convert this to kilograms (kg) using the conversion factor \(1 \text{ gf} = 0.00981 \text{ N}\).

\[ \text{Weight in Newtons} = 44.6 \, \text{gf} \times 0.00981 \, \text{N/gf} = 0.437 \, \text{N} \]

Now, we can find the mass \(m\) using the formula \(W = mg\), where \(g = 9.8 \, \text{m/s}^2\):

\[ m = \frac{W}{g} = \frac{0.437}{9.8} \approx 0.0446 \, \text{kg} \]

### Step 2: Calculate the volume of the iron piece
We know the density of iron is given as \(8.9 \times 10^3 \, \text{kg/m}^3\). We can use the formula for density:

\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \implies \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]

Substituting the values:

\[ V = \frac{0.0446 \, \text{kg}}{8.9 \times 10^3 \, \text{kg/m}^3} \approx 5.01 \times 10^{-6} \, \text{m}^3 \]

### Step 3: Calculate the buoyant force when immersed in water
The buoyant force \(F_b\) can be calculated using Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced. The density of water is \(1000 \, \text{kg/m}^3\).

\[ F_b = \text{Density of water} \times g \times V \]

Substituting the values:

\[ F_b = 1000 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2 \times 5.01 \times 10^{-6} \, \text{m}^3 \approx 0.0492 \, \text{N} \]

### Step 4: Calculate the apparent weight in water
The apparent weight \(W'\) of the iron piece when immersed in water can be calculated as:

\[ W' = W - F_b \]

Substituting the values:

\[ W' = 0.437 \, \text{N} - 0.0492 \, \text{N} \approx 0.3878 \, \text{N} \]

### Step 5: Convert the apparent weight back to gram force
To convert the apparent weight back to gram force:

\[ W'_{\text{gf}} = \frac{W'}{g} \times 1000 \]

Substituting the values:

\[ W'_{\text{gf}} = \frac{0.3878}{9.8} \times 1000 \approx 39.5 \, \text{gf} \]

### Final Answer
The weight of the iron piece when immersed in water is approximately **39.5 gf**.

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