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lio A tennis ball has a radius of 3.5 cm . Three tennis balls fit exactly into a cylindrical tube. Using \( \pi=3 \cdot 14 \), calculate: (i) The volume of the three tennis balls (ii) The height of the tube (iii) The radius of the tube (iv) The volume of the tube, correct to two decimal places (v) The fraction of the volume of the tube taken up by the three tennis balls

Ask by Macdonald Hart. in Ireland
Mar 09,2025

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(i) Volume of three tennis balls: ≈538.5 cm³ (ii) Height of the tube: 21 cm (iii) Radius of the tube: 3.5 cm (iv) Volume of the tube: ≈807.77 cm³ (v) Fraction of the tube’s volume occupied by the tennis balls: ≈66.7%

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The Deep Dive

To find the volume of three tennis balls, we start with the formula for the volume of a sphere: \[ V = \frac{4}{3} \pi r^3 \] Given the radius \( r = 3.5 \) cm and \( \pi = 3.14 \): (i) The volume of one tennis ball is: \[ V = \frac{4}{3} \cdot 3.14 \cdot (3.5)^3 = \frac{4}{3} \cdot 3.14 \cdot 42.875 \approx 179.59 \text{ cm}^3 \] Thus, the volume of three tennis balls is approximately: \[ 3 \cdot 179.59 \approx 538.77 \text{ cm}^3 \] (ii) Since three tennis balls fit exactly in the tube, the height of the tube is three times the diameter of one tennis ball. The diameter \( d = 2 \cdot 3.5 = 7 \) cm, so the height is: \[ 3 \cdot 7 \text{ cm} = 21 \text{ cm} \] (iii) The radius of the tube is equal to the radius of one tennis ball, which is: \[ 3.5 \text{ cm} \] (iv) The volume of the cylindrical tube is given by the formula: \[ V_{\text{tube}} = \pi r^2 h \] Substituting the values in: \[ \pi = 3.14, r = 3.5 \text{ cm}, h = 21 \text{ cm} \] \[ V_{\text{tube}} = 3.14 \cdot (3.5)^2 \cdot 21 = 3.14 \cdot 12.25 \cdot 21 \approx 831.47 \text{ cm}^3 \] (v) The fraction of the volume of the tube taken up by the three tennis balls is: \[ \text{Fraction} = \frac{\text{Volume of balls}}{\text{Volume of tube}} = \frac{538.77}{831.47} \approx 0.65 \] In conclusion: (i) Volume of three tennis balls: \(\approx 538.77 \text{ cm}^3\) (ii) Height of the tube: \(21 \text{ cm}\) (iii) Radius of the tube: \(3.5 \text{ cm}\) (iv) Volume of the tube: \(\approx 831.47 \text{ cm}^3\) (v) Fraction of volume occupied: \(\approx 0.65\) or \(65\%\).

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