AP Physics 1 · Topic 8.1

Internal Structure and Density Practice

Part of Fluids.(TOP-8.A)

Practice questions

3

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Sample questions

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  1. Sample 1difficulty 1/5

    A block of aluminum has mass 540 g540~\text{g} and volume 200 cm3200~\text{cm}^3. What is its density?

    • A

      1.4 g/cm31.4~\text{g/cm}^3

    • B

      5.4 g/cm35.4~\text{g/cm}^3

    • C

      2.7 g/cm32.7~\text{g/cm}^3

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    • D

      27 g/cm327~\text{g/cm}^3

    Why

    ρ=m/V=540/200=2.7 g/cm3\rho = m/V = 540/200 = 2.7~\text{g/cm}^3 (matches aluminum).

  2. Sample 2difficulty 1/5

    A liquid of mass 400 g400~\text{g} fills 0.50 L0.50~\text{L}. Its density is

    • A

      0.4 g/cm30.4~\text{g/cm}^3

    • B

      2.0 g/cm32.0~\text{g/cm}^3

    • C

      1.0 g/cm31.0~\text{g/cm}^3

    • D

      0.8 g/cm30.8~\text{g/cm}^3

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    Why

    ρ=m/V=400/500=0.80 g/cm3\rho = m/V = 400/500 = 0.80~\text{g/cm}^3 (1 L=1000 cm31~\text{L} = 1000~\text{cm}^3).

  3. Sample 3difficulty 3/5

    A solid block weighs 20 N20~\text{N} in air and 15 N15~\text{N} when fully submerged in water. Its volume (using ρw=1000 kg/m3\rho_w = 1000~\text{kg/m}^3, g10 m/s2g \approx 10~\text{m/s}^2) is

    • A

      103 m310^{-3}~\text{m}^3

    • B

      2×103 m32 \times 10^{-3}~\text{m}^3

    • C

      1.5×103 m31.5 \times 10^{-3}~\text{m}^3

    • D

      5×104 m35 \times 10^{-4}~\text{m}^3

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    Why

    Buoyancy: FB=2015=5 NF_B = 20 - 15 = 5~\text{N}. V=FB/(ρg)=5/10000=5×104 m3V = F_B/(\rho g) = 5/10000 = 5 \times 10^{-4}~\text{m}^3.