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A student determined the viscosity of a liquid using the falling-ball method - Edexcel - A-Level Physics - Question 12 - 2023 - Paper 1

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A student determined the viscosity of a liquid using the falling-ball method. (a) When the ball is falling at terminal velocity the following equation applies drag... show full transcript

Worked Solution & Example Answer:A student determined the viscosity of a liquid using the falling-ball method - Edexcel - A-Level Physics - Question 12 - 2023 - Paper 1

Step 1

Describe how the student could use her measurements to calculate a value for the drag force acting on the ball.

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Answer

To calculate the drag force acting on the ball, the student would carry out the following steps:

  1. Measure the Mass of the Ball: The student would use the balance to weigh the ball and determine its mass (m).

  2. Calculate the Weight of the Ball: Using the mass measured and the acceleration due to gravity (g, approximately 9.81 m/s²), the student can calculate the weight of the ball (W) using the formula:

    W=mimesgW = m imes g

  3. Determine the Upthrust Effect: The student should also know the density (ρ) of the liquid. Using the volume (V) of the ball, which can be calculated from its radius (r) by the formula:

    V=43πr3V = \frac{4}{3} \pi r^3

    The upthrust (U) can then be calculated using:

    U=ρliquid×g×VU = \rho_{liquid} \times g \times V

  4. Calculate the Drag Force: Finally, the drag force (F_d) can be calculated by rearranging the equation given in the question:

    Fd=W−UF_d = W - U

    By substituting the values derived from the measurements, the drag force acting on the ball can be determined.

Step 2

Calculate the viscosity of the liquid.

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Answer

To calculate the viscosity of the liquid, the formula for viscosity (η) relating to terminal velocity (v), radius (r), drag force (F_d), and fluid density (ρ) can be used:

η=Fd6πrvη = \frac{F_d}{6 \pi r v}

Substituting in the values from the question:

  • Fd=1.1×10−2 NF_d = 1.1 \times 10^{-2} \text{ N}
  • r=0.50×10−2 mr = 0.50 \times 10^{-2} \text{ m}
  • v=5.4×10−1 m/sv = 5.4 \times 10^{-1} \text{ m/s}

The formula becomes:

η=1.1×10−26π(0.50×10−2)(5.4×10−1)η = \frac{1.1 \times 10^{-2}}{6 \pi (0.50 \times 10^{-2}) (5.4 \times 10^{-1})}

Calculating this:

  • First, calculate the denominator:
6π(0.50×10−2)(5.4×10−1)≈5.1×10−26 \pi (0.50 \times 10^{-2}) (5.4 \times 10^{-1}) \approx 5.1 \times 10^{-2}
  • Now, substitute back into the viscosity formula:
η=1.1×10−25.1×10−2≈0.216 Pa.sη = \frac{1.1 \times 10^{-2}}{5.1 \times 10^{-2}} \approx 0.216 \text{ Pa.s}

Thus, the viscosity of the liquid is approximately 0.216 Pa.s.

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