CIE iGCSE Co-Ordinated Science P1.4 Density Exam Style Questions Paper 1
▶️ Answer/Explanation
Density is defined as mass per unit volume. The equation for density is:
\[ \rho = \frac{m}{V} \]
where \(\rho\) is density, \(m\) is mass, and \(V\) is volume.
Substituting the given values:
\[ \rho = \frac{20 \text{ g}}{5.0 \text{ cm}^3} = 4.0 \text{ g/cm}^3 \]
Therefore, the density of the object is 4.0 g/cm\(^3\).
✅ Answer: (A)
▶️ Answer/Explanation
The volume of the rectangular block is calculated as \( V = 3.0 \times 2.0 \times 2.0 = 12.0 \, \text{cm}^3 \). Using the density formula \( \rho = \frac{m}{V} \), we rearrange to find mass: \( m = \rho \times V = 2.0 \times 12.0 = 24 \, \text{g} \).
✅ Answer: (D)
✅ Answer: (D)
▶️ Answer/Explanation
From the diagram: beaker 1 has volume 40 cm³ and total mass 85 g. Beaker 2 has volume 90 cm³ and total mass 125 g. The difference gives the mass of 50 cm³ of liquid: 125 − 85 = 40 g. Density = mass/volume = 40 g / 50 cm³ = 0.80 g/cm³.
✅ Answer: (B)
✅ Answer: (B)
▶️ Answer/Explanation
Density is calculated using \(\text{density} = \frac{\text{mass}}{\text{volume}}\).
The volume of the block is found by multiplying its three dimensions shown in the diagram, giving \(24\ \text{cm}^3\).
Dividing the mass by this volume, \(\frac{72}{24} = 3.0\), gives the density of the metal.
✅ Answer: (A)
The volume of the block is found by multiplying its three dimensions shown in the diagram, giving \(24\ \text{cm}^3\).
Dividing the mass by this volume, \(\frac{72}{24} = 3.0\), gives the density of the metal.
✅ Answer: (A)
▶️ Answer/Explanation
The volume of the block is \(V = 8.0 \times 1.0 \times 0.10 = 0.80 \text{ m}^3\).
Density is calculated using \(\rho = \dfrac{m}{V} = \dfrac{440}{0.80} = 550 \text{ kg/m}^3\).
This matches option C exactly.
✅ Answer: (C)
Density is calculated using \(\rho = \dfrac{m}{V} = \dfrac{440}{0.80} = 550 \text{ kg/m}^3\).
This matches option C exactly.
✅ Answer: (C)