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9. (a) A uniform cylindrical piece of wood 12 cm long floats in water with its axis vertical and 10 cm of its length immersed - Leaving Cert Applied Maths - Question 9 - 2009
Question 9
9. (a) A uniform cylindrical piece of wood 12 cm long floats in water with its axis vertical and 10 cm of its length immersed. Oil of relative density 0.75 is poure... show full transcript
Worked Solution & Example Answer:9. (a) A uniform cylindrical piece of wood 12 cm long floats in water with its axis vertical and 10 cm of its length immersed - Leaving Cert Applied Maths - Question 9 - 2009
Step 1
Find the depth of the layer of oil.
Answer
To find the depth of the layer of oil, we can use the principle of buoyancy which states that the weight of the fluid displaced is equal to the weight of the object.
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Let the depth of the oil be represented by ( h ). The total length of the cylinder is 12 cm, so the portion submerged in water is 10 cm.
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The volume of the submerged part of the cylinder in water is given as: [ V_{water} = 10 , \text{cm} \times \text{cross-sectional area} ]
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The weight of the wood ( W ) can be expressed as: [ W = \frac{1}{2} \cdot \text{density of wood} \cdot g \cdot V_{cylinder} ]
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The buoyant force from the water is given by: [ B_{water} = V_{water} \cdot \text{density of water} \cdot g ]
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Set up the equation of forces: [ B_{water} + B_{oil} = W ] Where ( B_{oil} = h \cdot \text{cross-sectional area} \cdot 0.75 \cdot g )
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Thus, we have: [ \frac{W}{s} + (h \cdot 0.75) = W ]
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Rearranging gives us: [ h = 8 , \text{cm} ]
Step 2
Find in terms of s the fraction of the length of the rod that is immersed.
Answer
Let the length of the immersed part of the rod be ( x ).
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Take moments about point ( q ): [ B \cdot \sin(\theta) = W \cdot \frac{x}{s} \cdot \sin(\theta) ]
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The buoyant force from water is: [ B = \frac{xW}{s} ]
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Setting up the equilibrium equation results in: [ x^2 - 2a^2 + t^2 = 0 ]
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Solving for the fraction of the immersed length: [ \text{fraction} = \frac{x}{l} = \frac{x}{1 - \sqrt{1 - s}} ]
Step 3