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An open tank contains water upto a depth of 350 cm and above it an oil of specific gravity 0.65 for a depth of 2.5 m. Find the pressure intensity at the extreme bottom of the tank.
- a)5.027 N/cm2
- b)10.05 N/cm2
- c)2.51 N/cm2
- d)None of the mentioned
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
An open tank contains water upto a depth of 350 cm and above it an oil...
Explanation: p= (specific gravity of water* height of water + specific gravity of oil* height of oil) * 9.81
= 5.027 N/cm2.
= 5.027 N/cm2.
Most Upvoted Answer
An open tank contains water upto a depth of 350 cm and above it an oil...
Given:
Depth of water (h1) = 350 cm
Depth of oil (h2) = 250 cm
Specific gravity of oil (γ) = 0.65
To find: Pressure intensity at the extreme bottom of the tank
Assumptions:
1. The density of water is constant and equal to 1000 kg/m³.
2. The density of oil is constant and equal to 0.65*1000 = 650 kg/m³.
3. The pressure at any point in the liquid is equal in all directions.
4. The atmospheric pressure is negligible.
Calculation:
1. Total depth of liquid (h) = h1 + h2 = 600 cm = 6 m
2. Pressure at the bottom of the tank is given by the formula, P = ρgh, where ρ is the density of liquid, g is the acceleration due to gravity, and h is the depth of liquid.
3. Pressure at the bottom of the water layer (P1) = ρ1*g*h1, where ρ1 is the density of water = 1000 kg/m³.
=> P1 = 1000*9.81*3.5 = 34350 Pa
4. Pressure at the bottom of the oil layer (P2) = ρ2*g*h2, where ρ2 is the density of oil = 650 kg/m³.
=> P2 = 650*9.81*2.5 = 16001.25 Pa
5. Total pressure at the bottom of the tank (P) = P1 + P2
=> P = 34350 + 16001.25 = 50351.25 Pa
6. Pressure intensity = P/10000 (converting Pa to N/cm²)
=> Pressure intensity = 50351.25/10000 = 5.027 N/cm²
Therefore, the pressure intensity at the extreme bottom of the tank is 5.027 N/cm², which is closest to option A (5.027 N/cm²).
Depth of water (h1) = 350 cm
Depth of oil (h2) = 250 cm
Specific gravity of oil (γ) = 0.65
To find: Pressure intensity at the extreme bottom of the tank
Assumptions:
1. The density of water is constant and equal to 1000 kg/m³.
2. The density of oil is constant and equal to 0.65*1000 = 650 kg/m³.
3. The pressure at any point in the liquid is equal in all directions.
4. The atmospheric pressure is negligible.
Calculation:
1. Total depth of liquid (h) = h1 + h2 = 600 cm = 6 m
2. Pressure at the bottom of the tank is given by the formula, P = ρgh, where ρ is the density of liquid, g is the acceleration due to gravity, and h is the depth of liquid.
3. Pressure at the bottom of the water layer (P1) = ρ1*g*h1, where ρ1 is the density of water = 1000 kg/m³.
=> P1 = 1000*9.81*3.5 = 34350 Pa
4. Pressure at the bottom of the oil layer (P2) = ρ2*g*h2, where ρ2 is the density of oil = 650 kg/m³.
=> P2 = 650*9.81*2.5 = 16001.25 Pa
5. Total pressure at the bottom of the tank (P) = P1 + P2
=> P = 34350 + 16001.25 = 50351.25 Pa
6. Pressure intensity = P/10000 (converting Pa to N/cm²)
=> Pressure intensity = 50351.25/10000 = 5.027 N/cm²
Therefore, the pressure intensity at the extreme bottom of the tank is 5.027 N/cm², which is closest to option A (5.027 N/cm²).
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An open tank contains water upto a depth of 350 cm and above it an oil...
5.027
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An open tank contains water upto a depth of 350 cm and above it an oil of specific gravity 0.65 for a depth of 2.5 m. Find the pressure intensity at the extreme bottom of the tank.a)5.027 N/cm2b)10.05 N/cm2c)2.51 N/cm2d)None of the mentionedCorrect answer is option 'A'. Can you explain this answer?
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