A piece of wood weighs 2 N in air and 1 N when fully submerged in water. Calculate its relative density.

From the diagrams:

• The weight of the wood in air ($W_1$) is indicated by the pointer on the left spring balance.
• The weight of the wood in water ($W_2$) is indicated by the pointer on the right spring balance.

$$W_1 = 2\, \text{N}$$ $$W_2 = 1\, \text{N}$$

The apparent loss of weight is the difference between the weight in air and the weight in water.

$$\text{Apparent loss of weight} = W_1 - W_2$$ $$\text{Apparent loss of weight} = 2\, \text{N} - 1\, \text{N}$$ $$\text{Apparent loss of weight} = 1\, \text{N}$$

Relative density is calculated as the ratio of the weight of the substance in air to the apparent loss of weight in water (which is equal to the weight of the displaced water).

$$\text{Relative density} = \frac{\text{Weight of wood in air}}{\text{Apparent loss of weight in water}}$$ $$\text{Relative density} = \frac{W_1}{W_1 - W_2}$$ $$\text{Relative density} = \frac{2\, \text{N}}{1\, \text{N}}$$ $$\text{Relative density} = 2$$

The apparent loss of weight of the wood when submerged in water is due to the upthrust (or buoyant force) exerted by the water. According to Archimedes' principle, this upthrust is equal to the weight of the water displaced by the submerged part of the wood.

Here are the answers:

a)

i) Weight ($W_1$) of the wood in air: $\boxed{\text{2 N}}$

ii) Weight ($W_2$) of the wood in water: $\boxed{\text{1 N}}$

b)

i) Apparent loss of weight of the wood in water: $\boxed{\text{1 N}}$

ii) Relative density of the wood: $\boxed{\text{2}}$

c) The apparent loss of weight of the wood in water is caused by the upthrust (or buoyant force) exerted by the water on the submerged wood. This upthrust is equal to the weight of the water displaced by the wood, as stated by Archimedes' principle.