# A) An object weighs 40 g in air and has a volume of 8.0 cm3. What will be its apparent weight when immersed in water?

Jul 7, 2018

#### Answer:

$\text{The object weighs "32 g" in water}$.

#### Explanation:

To avoid confusion between grams and the acceleration due to gravity, I will assign ${g}_{r}$ to indicate the acceleration due to gravity (where we would normally use $g$).

Weight of an object in air is $m a s s \cdot {g}_{r}$. So in this case, the weight is $40 g \cdot {g}_{r}$. (We will not need to multiply by the value of ${g}_{r}$ because we will be comparing with weight of water and will take the same shortcut there).

When immersed in water, the object will displace $8.0 c {m}^{3}$ of water.

The density of water is $1 \frac{g}{c {m}^{3}}$.
Therefore the mass of the displaced water will be $1 \frac{g}{\cancel{c {m}^{3}}} \cdot 8.0 \cancel{c {m}^{3}} = 8.0 g$

The weight of the displaced water is $8.0 g \cdot {g}_{r}$.
Therefore the object has a upward buoyant force equal to $8.0 g \cdot {g}_{r}$.

The weight of the object in air (a downward force) and the buoyant force (an upward force) are both components of the apparent weight in water. Therefore we will have to subtract the 2 forces.

$\text{The apparent weight in water} = 40 g \cdot {g}_{r} - 8.0 g \cdot {g}_{r} = 32 g \cdot {g}_{r}$.

Or, using the terminology the question was asked with: $\text{the object weighs "32 g" in water}$.

I hope this helps,
Steve