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Set up the apparatus as shown, then fill the burette with water to the brim. Drain this water into a 100ml measuring cylinder and record its volume, V2. Calculate the volume of excess water above zero mark, V3.

Physics

ScanSolve AI Answer

Answer

Volume of oil

45.36 cm345.36~\text{cm}^3

Volume of water

68.05 cm368.05~\text{cm}^3

Step-by-step solution

5 steps
  1. Find the radius of the tube.

    Diameter d=3.8d = 3.8 cm

    r=d2=3.82=1.9 cmr = \frac{d}{2} = \frac{3.8}{2} = 1.9~\text{cm}
  2. Compute r2r^2.

    r2=(1.9)2=3.61 cm2r^2 = (1.9)^2 = 3.61~\text{cm}^2
  3. Heights of liquids (from diagram

    total height 10 cm, oil layer 4 cm).
    Height of oil hoil=4h_{\text{oil}} = 4 cm
    Height of water hwater=10−4=6h_{\text{water}} = 10 - 4 = 6 cm

  4. Volume of oil.

    Formula: Voil=πr2hoilV_{\text{oil}} = \pi r^2 h_{\text{oil}} cm³
    Substitute:

    Voil=π×3.61 cm2×4 cmV_{\text{oil}} = \pi \times 3.61~\text{cm}^2 \times 4~\text{cm}

    First, 3.61×4=14.443.61 \times 4 = 14.44

    Voil=π×14.44=45.36 cm3V_{\text{oil}} = \pi \times 14.44 = 45.36~\text{cm}^3

    (using π≈3.14\pi \approx 3.14)

  5. Volume of water.

    Formula: Vwater=πr2hwaterV_{\text{water}} = \pi r^2 h_{\text{water}} cm³
    Substitute:

    Vwater=π×3.61 cm2×6 cmV_{\text{water}} = \pi \times 3.61~\text{cm}^2 \times 6~\text{cm}

    First, 3.61×6=21.663.61 \times 6 = 21.66

    Vwater=π×21.66=68.05 cm3V_{\text{water}} = \pi \times 21.66 = 68.05~\text{cm}^3

    (using π≈3.14\pi \approx 3.14)

    Final answers:
    Volume of oil: 45.36 cm3\boxed{45.36~\text{cm}^3}
    Volume of water: 68.05 cm3\boxed{68.05~\text{cm}^3}

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