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The standard sea level atmospheric pressure is equivalent to
- a)10.2 m of freshwater of density (ρ) = 998 kg/m3
- b)10.1 m of saline water of density (ρ) = 1025 kg/m3
- c)12.5 m of kerosene of density (ρ) = 800 kg/m3
- d)6.4 m of carbon tetrachloride of density (ρ) = 1590 kg/m3
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The standard sea level atmospheric pressure is equivalent toa)10.2 m ...
Pressure at sea level = 101.325 kPa
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Checking according to given options.
P = hρg
a) ⇒ 10.2 m of fresh water (ρ = 998 kg/m3)
P = 10.2 × 998 × 9.81 = 99861.87 Pa = 99.861 kPa
b) ⇒ 10.1 m of saline water (ρ = 1025 kg/m3)
P = 10.1 × 1025 × 9.81 = 101.558 kPa
c) ⇒ 12.5 m of kerosene of ρ = 800 kg/m3
P = 12.5 × 800 × 9.81 = 98.1 kPa
d) ⇒ 6.4 m of carbon tetrachloride of ρ = 1590 kg/m3
⇒ 6.4 × 1590 × 9.81 = 99.826 kPa
So option b is nearest, so correct
Most Upvoted Answer
The standard sea level atmospheric pressure is equivalent toa)10.2 m ...
Explanation:
The standard sea level atmospheric pressure is 101325 Pa. This pressure can be expressed in terms of height of a fluid column of a particular density. Let's calculate the height of different fluids that would result in the same pressure as the standard sea level atmospheric pressure.
a) 10.2 m of freshwater of density (ρ) = 998 kg/m3
The pressure at a depth of h meters in a fluid of density ρ is given by P = ρgh, where g is the acceleration due to gravity. Therefore, the height of fresh water required to generate a pressure of 101325 Pa is h = P/(ρg) = 101325/(998*9.81) = 10.2 m.
b) 10.1 m of saline water of density (ρ) = 1025 kg/m3
Saline water is denser than fresh water, so a lower column height is required to generate the same pressure. Using the same formula as above, the height of saline water required is h = P/(ρg) = 101325/(1025*9.81) = 10.1 m.
c) 12.5 m of kerosene of density (ρ) = 800 kg/m3
Kerosene is less dense than water, so a higher column height is required to generate the same pressure. Using the same formula as above, the height of kerosene required is h = P/(ρg) = 101325/(800*9.81) = 12.5 m.
d) 6.4 m of carbon tetrachloride of density (ρ) = 1590 kg/m3
Carbon tetrachloride is much denser than water, so a much lower column height is required to generate the same pressure. Using the same formula as above, the height of carbon tetrachloride required is h = P/(ρg) = 101325/(1590*9.81) = 6.4 m.
Therefore, the correct answer is option B, 10.1 m of saline water of density (ρ) = 1025 kg/m3.
The standard sea level atmospheric pressure is 101325 Pa. This pressure can be expressed in terms of height of a fluid column of a particular density. Let's calculate the height of different fluids that would result in the same pressure as the standard sea level atmospheric pressure.
a) 10.2 m of freshwater of density (ρ) = 998 kg/m3
The pressure at a depth of h meters in a fluid of density ρ is given by P = ρgh, where g is the acceleration due to gravity. Therefore, the height of fresh water required to generate a pressure of 101325 Pa is h = P/(ρg) = 101325/(998*9.81) = 10.2 m.
b) 10.1 m of saline water of density (ρ) = 1025 kg/m3
Saline water is denser than fresh water, so a lower column height is required to generate the same pressure. Using the same formula as above, the height of saline water required is h = P/(ρg) = 101325/(1025*9.81) = 10.1 m.
c) 12.5 m of kerosene of density (ρ) = 800 kg/m3
Kerosene is less dense than water, so a higher column height is required to generate the same pressure. Using the same formula as above, the height of kerosene required is h = P/(ρg) = 101325/(800*9.81) = 12.5 m.
d) 6.4 m of carbon tetrachloride of density (ρ) = 1590 kg/m3
Carbon tetrachloride is much denser than water, so a much lower column height is required to generate the same pressure. Using the same formula as above, the height of carbon tetrachloride required is h = P/(ρg) = 101325/(1590*9.81) = 6.4 m.
Therefore, the correct answer is option B, 10.1 m of saline water of density (ρ) = 1025 kg/m3.
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The standard sea level atmospheric pressure is equivalent toa)10.2 m of freshwater of density (ρ) = 998 kg/m3b)10.1 m of saline water of density (ρ) = 1025 kg/m3c)12.5 m of kerosene of density (ρ) = 800 kg/m3d)6.4 m of carbon tetrachloride of density (ρ) = 1590 kg/m3Correct answer is option 'B'. Can you explain this answer?
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The standard sea level atmospheric pressure is equivalent toa)10.2 m of freshwater of density (ρ) = 998 kg/m3b)10.1 m of saline water of density (ρ) = 1025 kg/m3c)12.5 m of kerosene of density (ρ) = 800 kg/m3d)6.4 m of carbon tetrachloride of density (ρ) = 1590 kg/m3Correct answer is option 'B'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about The standard sea level atmospheric pressure is equivalent toa)10.2 m of freshwater of density (ρ) = 998 kg/m3b)10.1 m of saline water of density (ρ) = 1025 kg/m3c)12.5 m of kerosene of density (ρ) = 800 kg/m3d)6.4 m of carbon tetrachloride of density (ρ) = 1590 kg/m3Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The standard sea level atmospheric pressure is equivalent toa)10.2 m of freshwater of density (ρ) = 998 kg/m3b)10.1 m of saline water of density (ρ) = 1025 kg/m3c)12.5 m of kerosene of density (ρ) = 800 kg/m3d)6.4 m of carbon tetrachloride of density (ρ) = 1590 kg/m3Correct answer is option 'B'. Can you explain this answer?.
The standard sea level atmospheric pressure is equivalent toa)10.2 m of freshwater of density (ρ) = 998 kg/m3b)10.1 m of saline water of density (ρ) = 1025 kg/m3c)12.5 m of kerosene of density (ρ) = 800 kg/m3d)6.4 m of carbon tetrachloride of density (ρ) = 1590 kg/m3Correct answer is option 'B'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about The standard sea level atmospheric pressure is equivalent toa)10.2 m of freshwater of density (ρ) = 998 kg/m3b)10.1 m of saline water of density (ρ) = 1025 kg/m3c)12.5 m of kerosene of density (ρ) = 800 kg/m3d)6.4 m of carbon tetrachloride of density (ρ) = 1590 kg/m3Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The standard sea level atmospheric pressure is equivalent toa)10.2 m of freshwater of density (ρ) = 998 kg/m3b)10.1 m of saline water of density (ρ) = 1025 kg/m3c)12.5 m of kerosene of density (ρ) = 800 kg/m3d)6.4 m of carbon tetrachloride of density (ρ) = 1590 kg/m3Correct answer is option 'B'. Can you explain this answer?.
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