A Perspex box has a 10cm x 12cm rectangular base and contains water to a height of 10 cm. A piece of rock of mass 623 g is lowered into the water and the level rises to 12 cm. Calculate the density of the rock.
Answers
Answer:
(a) The volume of water is 100 cm³
(b) The volume of the rock is 20 cm³
(c) The density of the rock is 30 g/cm³
Explanation:
The given parameters of the perspex box are;
The area of the base of the box, A = 10 cm²
The initial level of water in the box, h₁ = 10 cm
The mass of the rock placed in the box, m = 600 g
The final level of water in the box, h₂ = 12 cm
(a) The volume of water in the box, 'V', is given as follows;
V = A × h₁
∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³
The volume of water in the box, V = 100 cm³
(b) When the rock is placed in the box the total volume, V_TV
T
, is given by the sum of the rock, V_rV
r
, and the water, V, is given as follows;
V_TV
T
= V_rV
r
+ V
V_TV
T
= A × h₂
∴ V_TV
T
= 10 cm² × 12 cm = 120 cm³
The total volume, V_TV
T
= 120 cm³
The volume of the rock, V_rV
r
= V_TV
T
- V
∴ V_rV
r
= 120 cm³ - 100 cm³ = 20 cm³
The volume of the rock, V_rV
r
= 20 cm³
(c) The density of the rock, ρ = (Mass of the rock, m)/(The volume of the rock)
∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³ .