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A solid sphere floats at rest in water - Leaving Cert Applied Maths - Question 9 - 2007

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A solid sphere floats at rest in water. The radius of the sphere is 7 cm. Half of the sphere lies below the surface of the water. Find, correct to one place of decim... show full transcript

Worked Solution & Example Answer:A solid sphere floats at rest in water - Leaving Cert Applied Maths - Question 9 - 2007

Step 1

Find, correct to one place of decimals, the weight of the sphere.

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Answer

To find the weight of the sphere that floats in water, we need to use the principle of buoyancy. The volume of the submerged part of the sphere is given by:

  1. Calculate the volume of the sphere that is submerged:

    Since half of the sphere lies below the water, the submerged volume can be calculated as:

    Vsubmerged=12×43πr3V_{submerged} = \frac{1}{2} \times \frac{4}{3} \pi r^3 where ( r = 0.07 \text{ m} )

    Vsubmerged=12×43π(0.07)3=46π(0.000343)≈0.0007163 m3V_{submerged} = \frac{1}{2} \times \frac{4}{3} \pi (0.07)^3 = \frac{4}{6} \pi (0.000343) \approx 0.0007163 \text{ m}^3

  2. Calculate the buoyant force (B):

    Using the density of water ( \rho = 1000 \text{ kg/m}^3 ), the buoyant force can be expressed as:

    B=ρgVsubmergedB = \rho g V_{submerged}

    Approximating ( g \approx 10 \text{ m/s}^2 ):

    B=1000×10×(0.0007163)≈7.2 NB = 1000 \times 10 \times (0.0007163) \approx 7.2 \text{ N}

The weight of the sphere (W) is equal to the buoyant force:

Thus, the weight of the sphere is 7.2 N.

Step 2

Find the tension in the string.

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Answer

For the solid cylinder, we need to determine the tension in the string by considering the forces acting on it:

  1. Calculate the weight of the cylinder (W):

    The density of the cylinder can be found from its relative density:

    Relative Density=Density of CylinderDensity of Water  ⟹  Density of Cylinder=5×1000=5000 kg/m3\text{Relative Density} = \frac{\text{Density of Cylinder}}{\text{Density of Water}} \implies \text{Density of Cylinder} = 5 \times 1000 = 5000 \text{ kg/m}^3

    The volume of the cylinder is:

    V=πr2h=π(0.03)2(0.14)=π(0.0009)(0.14)≈0.000395extm3V = \pi r^2 h = \pi (0.03)^2 (0.14) = \pi (0.0009)(0.14) \approx 0.000395 ext{ m}^3

    The weight of the cylinder is:

    W=ρgV=5000×10×0.000395≈19.75 NW = \rho g V = 5000 \times 10 \times 0.000395 \approx 19.75 \text{ N}

  2. Apply the equilibrium of forces:

    The forces acting on the cylinder are the tension (T), buoyant force (B), and weight (W):

    T+B=W  ⟹  T+(0.9⋅Wwater)=WT + B = W \implies T + (0.9 \cdot W_{water}) = W

    Replacing ( B ) with ( 0.9 \cdot \frac{W}{5} = \frac{41 W}{50} $$ so the equation becomes:

    T+41W50=WT + \frac{41 W}{50} = W

    Therefore:

    T=W−41W50  ⟹  T=9W50T = W - \frac{41 W}{50} \implies T = \frac{9 W}{50}

    Substituting the value of W found previously:

    T=4150⋅5000⋅10⋅(0.03)2(0.14)≈16.236 NT = \frac{41}{50} \cdot 5000 \cdot 10 \cdot (0.03)^2(0.14) \approx 16.236 \text{ N}

Thus, the tension in the string is approximately 16.24 N.

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