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Jet of water of 2.5 cm diameter, moving with a velocity of 10m / s strikes a hinged square plate of weight 98.1 N at the centre of the plate. The plate is of uniform thickness. At what angle the plate will swing?
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Jet of water of 2.5 cm diameter, moving with a velocity of 10m / s str...
Introduction
To determine the angle at which the plate will swing when struck by a jet of water, we need to analyze the forces and torques involved in this scenario.
Parameters Given:
- Diameter of the water jet: 2.5 cm
- Velocity of the water jet: 10 m/s
- Weight of the plate: 98.1 N
Calculation of Force from Water Jet:
- Cross-sectional Area (A):
- Area = π(d/2)^2 = π(0.0125)^2 ≈ 4.91 x 10^-4 m²
- Mass Flow Rate (ṁ):
- ṁ = ρ * A * v
- Assuming water density (ρ) = 1000 kg/m³,
- ṁ ≈ 1000 * 4.91 x 10^-4 * 10 ≈ 4.91 kg/s
- Force exerted by the jet (F):
- F = ṁ * v = 4.91 * 10 ≈ 49.1 N
Torque Calculation:
- Distance from hinge to the center:
- For a square plate, the distance (r) from the hinge to the center is half the side length (s).
- Assuming the side length is equal to the diameter of the jet, r = 0.025 m / 2 ≈ 0.0125 m.
- Torque (τ):
- τ = F * r = 49.1 N * 0.0125 m ≈ 0.61375 Nm
Equilibrium Condition:
- The plate will reach an equilibrium angle (θ) when the torque due to the water force equals the torque due to the weight of the plate.
- Weight Torque:
- Weight acts downwards at the center of the plate, creating a counteracting torque.
- Calculating the angle:
- Using the relation between torque and angle, we can find θ where the torques balance.
Conclusion:
The angle at which the plate swings can be determined by the ratio of the torques, considering the weight of the plate and the force exerted by the water.
This approach outlines the basic calculations and principles necessary to solve for the angle of swing. Further detailed analysis may require specific dimensions of the plate and additional parameters for precise calculations.
To determine the angle at which the plate will swing when struck by a jet of water, we need to analyze the forces and torques involved in this scenario.
Parameters Given:
- Diameter of the water jet: 2.5 cm
- Velocity of the water jet: 10 m/s
- Weight of the plate: 98.1 N
Calculation of Force from Water Jet:
- Cross-sectional Area (A):
- Area = π(d/2)^2 = π(0.0125)^2 ≈ 4.91 x 10^-4 m²
- Mass Flow Rate (ṁ):
- ṁ = ρ * A * v
- Assuming water density (ρ) = 1000 kg/m³,
- ṁ ≈ 1000 * 4.91 x 10^-4 * 10 ≈ 4.91 kg/s
- Force exerted by the jet (F):
- F = ṁ * v = 4.91 * 10 ≈ 49.1 N
Torque Calculation:
- Distance from hinge to the center:
- For a square plate, the distance (r) from the hinge to the center is half the side length (s).
- Assuming the side length is equal to the diameter of the jet, r = 0.025 m / 2 ≈ 0.0125 m.
- Torque (τ):
- τ = F * r = 49.1 N * 0.0125 m ≈ 0.61375 Nm
Equilibrium Condition:
- The plate will reach an equilibrium angle (θ) when the torque due to the water force equals the torque due to the weight of the plate.
- Weight Torque:
- Weight acts downwards at the center of the plate, creating a counteracting torque.
- Calculating the angle:
- Using the relation between torque and angle, we can find θ where the torques balance.
Conclusion:
The angle at which the plate swings can be determined by the ratio of the torques, considering the weight of the plate and the force exerted by the water.
This approach outlines the basic calculations and principles necessary to solve for the angle of swing. Further detailed analysis may require specific dimensions of the plate and additional parameters for precise calculations.
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Jet of water of 2.5 cm diameter, moving with a velocity of 10m / s strikes a hinged square plate of weight 98.1 N at the centre of the plate. The plate is of uniform thickness. At what angle the plate will swing?
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Jet of water of 2.5 cm diameter, moving with a velocity of 10m / s strikes a hinged square plate of weight 98.1 N at the centre of the plate. The plate is of uniform thickness. At what angle the plate will swing? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Jet of water of 2.5 cm diameter, moving with a velocity of 10m / s strikes a hinged square plate of weight 98.1 N at the centre of the plate. The plate is of uniform thickness. At what angle the plate will swing? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Jet of water of 2.5 cm diameter, moving with a velocity of 10m / s strikes a hinged square plate of weight 98.1 N at the centre of the plate. The plate is of uniform thickness. At what angle the plate will swing?.
Jet of water of 2.5 cm diameter, moving with a velocity of 10m / s strikes a hinged square plate of weight 98.1 N at the centre of the plate. The plate is of uniform thickness. At what angle the plate will swing? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Jet of water of 2.5 cm diameter, moving with a velocity of 10m / s strikes a hinged square plate of weight 98.1 N at the centre of the plate. The plate is of uniform thickness. At what angle the plate will swing? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Jet of water of 2.5 cm diameter, moving with a velocity of 10m / s strikes a hinged square plate of weight 98.1 N at the centre of the plate. The plate is of uniform thickness. At what angle the plate will swing?.
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