# Topical Questions/O level Physics/Density

## Question 1

A rectangular block of metal is 50 mm long, 35 mm wide and has a thickness of 3.0 mm. It weighs 0.15 N.

Calculate

(a) The volume of the piece of metal,

(b) The density of the metal. 

[Take the weight of 1 kg to be 10 N].

## Question 2

A measuring cylinder contains 30 cm3 of liquid. When a stone of weight 0.92 N is dropped into the liquid, it sinks to the bottom and the liquid level rises to the 70 cm3 graduation.

Taking the weight of 1 kg to be 10 N, calculate

(i) the mass of the stone, 

(ii) the density of the stone. 

Explain why it would not be possible to use this method to determine the density of cork which would float in the liquid. 

## Question 3

(a) The mass of 600 spherical lead pellets is found to be 66 g and the total volume of the pellets is found to be 5.7 cm3.

Calculate

(i) the total weight of the pellets.

(ii) the volume of one pellet,

(iii) the density of one pellets in kg/m3,

(iv) the number of pellets which has a mass of 1.00 kg. [The force of gravity acting on a mass of 1.00 kg is 10.0N.] 

## Question 4

A block of stone measures 2 m x 2 m x 1 m. It has a mass of 8000 kg.

(a) What is its density?

(b) When it is standing on a bench what is the maximum pressure it can exert on the bench?

Part (a)

$$density = \frac{mass}{volume}$$

$$= \frac{8\ 000\ kg}{2\ m \times 2\ m \times 1\ m}$$

$$= \frac{8\ 000\ kg}{4\ m^3}$$

$$= 2\ 000\ kg/m^3$$

Part (b)

Pressure is inversely proportional to the area in contact, which means the smaller the area in contact the higher the pressure. This means that the maximum pressure occurs when the area of contact is a minimum and in this case the side with the smallest area is the 2 m × 1 m side. So the minimum area in contact with bench = 2 m^^2^^.

The block exerts a force on the bench that is equal to its weight. Therefore the force exerted on the bench = W = mg.

Force = 8 000 kg × 10 N/kg

= 80 000 N.

Using the area of the smallest side:

$$Pressure = \frac{force}{area}$$

$$= \frac{80\ 000\ N}{2\ m^2}$$

= 40 000 Pa.