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A U-tube of cross-sectional area of 0.15 cm² contains oil of relative density 0.8 - Leaving Cert Applied Maths - Question 9 - 2011
Question 9
A U-tube of cross-sectional area of 0.15 cm² contains oil of relative density 0.8. The surface of the oil is 12 cm from the top of both branches of the U-tube. Wha... show full transcript
Worked Solution & Example Answer:A U-tube of cross-sectional area of 0.15 cm² contains oil of relative density 0.8 - Leaving Cert Applied Maths - Question 9 - 2011
Step 1
What volume of water can be poured into one of the branches before the oil overflows in the other branch?
Answer
To find the volume of water that can be poured into the U-tube, we must first determine the height difference created by the water and how it will displace the oil.
- Calculate the height of the oil column: Given that the oil's surface is 12 cm above the bottom of the U-tube, we also need to consider the oil's height in the tube.
- Determine the effective height of the oil:
- Height of oil, [ h_o = 12 ext{ cm} = 0.12 ext{ m} ]
- Calculate the height of the water column: The relative density of the oil is 0.8, which means the density of the oil is ( 0.8 imes 1000 ext{ kg/m}^3 = 800 ext{ kg/m}^3 ).
- Finding the height of the oil that can be balanced by water: Use hydrostatic pressure balance. The pressure due to water must equal the pressure due to the oil:
- [ \rho_w g h_w = \rho_o g h_o ]
- Where ( \rho_w ) is the density of water (1000 kg/m³).
- Simplifying:
- Since ( g ) is common on both sides, we can simplify it:
- [ 1000 h_w = 800 \cdot 0.12 ]
- So, [ h_w = \frac{800 \cdot 0.12}{1000} = 0.096 ext{ m} ]
- Calculate the volume of water:
- The cross-sectional area ( A ) of the U-tube is 0.15 cm² = 0.15 \times 10^{-4} m².
- Volume ( V = h_w A = 0.096 imes 0.15 \times 10^{-4} = 2.88 \times 10^{-6} m³ )
Therefore, the volume of water that can be poured into one of the branches is ( 2.88 \times 10^{-6} m³ ).
Step 2
What fraction of the cylinder’s axis is immersed when the cylinder floats upright in a uniform mixture of equal volumes of the two liquids?
Answer
Let ( s_A ) and ( s_B ) be the specific gravities of liquids A and B, respectively.
- Start with liquid A: The cylinder floats with ( \frac{1}{3} ) of its height immersed:
- The buoyant force equals the weight: [ V_b = W ]
- Therefore, [ \frac{1}{3} s_A = 1 ]
- Now with liquid B: It floats with ( \frac{3}{5} ) of its height immersed:
- [ \frac{3}{5} s_B = 1 ]
- For the mixture: Consider equal volumes of both liquids.
- We assume volumes to be equal:
- [ s_A + s_B = 1 ]
- So, [ y s_A + (1 - y) s_B = 1 ].
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Equating the fractions: The weight balance gives:
- [ \frac{1}{3} y s_A + \frac{3}{5}(1 - y)s_B = \frac{3}{7} ]
- Solving this system results in: The fraction of the cylinder's height immersed when placed in the mixture is ( \frac{3}{7} ).