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Consider a U-tube filled with mercury except the 18-cm-high portion at the top, as shown in figure. The diameter of the right arm of theU-tube is D = 2 cm, and the diameter of the left arm is 2D. Oil with a specific gravity of 2.72 is poured into the left arm, forcing some mercury from the left arm into the right one. Determine the maximum amount of oil in liters that can be added into the left arm.
Correct answer is 'Range: 0.23 to 0.24'. Can you explain this answer?
Verified Answer
Consider a U-tube filled with mercury except the 18-cm-high portion a...
When oil is poured in left limb, the level of mercury goes down by h1 and in right limb goes up by h2
After adding the oil, ρoil × g(18 + h1) = ρHg × g(h1 + h2)
2.72 × (18 + h1) = 13.6 (h1 + h2)
Also h1 × (2D)2 = D2 × h2
h1 × 4 = h2
2.72 × (18 + h1) = 13.6(5h1)
48.96 + 2.72 h1 = 68 h1
h1 = 0.75 cm
Quantity of oil added
=π / 4 × (0.04)2 × (0.1875
= 2.35 × 102-4m3
= 0.2356 lts.
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Most Upvoted Answer
Consider a U-tube filled with mercury except the 18-cm-high portion a...
When oil is poured in left limb, the level of mercury goes down by h1 and in right limb goes up by h2
After adding the oil, ρoil × g(18 + h1) = ρHg × g(h1 + h2)
2.72 × (18 + h1) = 13.6 (h1 + h2)
Also h1 × (2D)2 = D2 × h2
h1 × 4 = h2
2.72 × (18 + h1) = 13.6(5h1)
48.96 + 2.72 h1 = 68 h1
h1 = 0.75 cm
Quantity of oil added
=π / 4 × (0.04)2 × (0.1875
= 2.35 × 102-4m3
= 0.2356 lts.
Free Test
Community Answer
Consider a U-tube filled with mercury except the 18-cm-high portion a...
When oil is poured in left limb, the level of mercury goes down by h1 and in right limb goes up by h2
After adding the oil, ρoil × g(18 + h1) = ρHg × g(h1 + h2)
2.72 × (18 + h1) = 13.6 (h1 + h2)
Also h1 × (2D)2 = D2 × h2
h1 × 4 = h2
2.72 × (18 + h1) = 13.6(5h1)
48.96 + 2.72 h1 = 68 h1
h1 = 0.75 cm
Quantity of oil added
=π / 4 × (0.04)2 × (0.1875
= 2.35 × 102-4m3
= 0.2356 lts.
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Consider a U-tube filled with mercury except the 18-cm-high portion at the top, as shown in figure. The diameter of the right arm of theU-tube is D = 2 cm, and the diameter of the left arm is 2D. Oil with a specific gravity of 2.72 is poured into the left arm, forcing some mercury from the left arm into the right one. Determine the maximum amount of oil in liters that can be added into the left arm.Correct answer is 'Range: 0.23 to 0.24'. Can you explain this answer?
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Consider a U-tube filled with mercury except the 18-cm-high portion at the top, as shown in figure. The diameter of the right arm of theU-tube is D = 2 cm, and the diameter of the left arm is 2D. Oil with a specific gravity of 2.72 is poured into the left arm, forcing some mercury from the left arm into the right one. Determine the maximum amount of oil in liters that can be added into the left arm.Correct answer is 'Range: 0.23 to 0.24'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Consider a U-tube filled with mercury except the 18-cm-high portion at the top, as shown in figure. The diameter of the right arm of theU-tube is D = 2 cm, and the diameter of the left arm is 2D. Oil with a specific gravity of 2.72 is poured into the left arm, forcing some mercury from the left arm into the right one. Determine the maximum amount of oil in liters that can be added into the left arm.Correct answer is 'Range: 0.23 to 0.24'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a U-tube filled with mercury except the 18-cm-high portion at the top, as shown in figure. The diameter of the right arm of theU-tube is D = 2 cm, and the diameter of the left arm is 2D. Oil with a specific gravity of 2.72 is poured into the left arm, forcing some mercury from the left arm into the right one. Determine the maximum amount of oil in liters that can be added into the left arm.Correct answer is 'Range: 0.23 to 0.24'. Can you explain this answer?.
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