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A water sprinkler has an 8 mm diameter nozzle at either end of the rotating arm. It is discharging water at right angle to the rotating arm in opposite direction at a velocity of 10 m/s. If the axis of rotation is at a distance of 0.2 m from one end and 0.24 m from other end, then the angular velocity of arm is (neglect frictional effect)
- a)8.33 rad/s
- b)45.082 rad/s
- c)91.67 rad/s
- d)2500 rad/s
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A water sprinkler has an 8 mm diameter nozzle at either end of the ro...
Given, R1 = 0.2 m, R2 = 0.24 m
Suppose the angular velocity of sprinkle is ω rad/s then the absolute velocity of flow through nozzle at both ends (discharge water velocity and sprinkle velocity will be in opposite direction)
V1 = 10 – 0.2 ω
V2 = 10 – 0.24 ω
There is no external torque, so total angular momentum will be conserved,
ρ Q [V1R1 + V2R2] = 0
ρ Q ≠ 0
V1R1 + V2R2 = 0
(10 – 0.2 ω ) × 0.2 + (10 – 0.24 ω) × 0.24 = 0
2 – 0.04 ω + 2.4 – 0.0576 ω = 0
ω = 45.082 rad/s
Most Upvoted Answer
A water sprinkler has an 8 mm diameter nozzle at either end of the ro...
Given data:
- Diameter of the nozzle = 8 mm
- Velocity of water discharge = 10 m/s
- Distance of the axis of rotation from one end = 0.2 m
- Distance of the axis of rotation from the other end = 0.24 m
To find:
- Angular velocity of the arm
Let's assume:
- The length of the arm = L
- The angular velocity of the arm = ω
The linear velocity of the water coming out of the nozzle is given by the equation:
v = ω * r
where v is the linear velocity, ω is the angular velocity, and r is the distance of the point from the axis of rotation.
The diameter of the nozzle is 8 mm, so the radius of the nozzle is 4 mm or 0.004 m.
The linear velocity of the water coming out of the nozzle is given as 10 m/s. We can use this information to find the angular velocity of the arm.
At one end of the arm:
v1 = ω * 0.004 m
At the other end of the arm:
v2 = ω * 0.004 m
The distance of the axis of rotation from one end is 0.2 m, so the linear velocity at this point is given as:
v1 = ω * 0.004 m = 0.2 m/s
Similarly, the distance of the axis of rotation from the other end is 0.24 m, so the linear velocity at this point is given as:
v2 = ω * 0.004 m = 0.24 m/s
From the above equations, we can write:
ω * 0.004 m = 0.2 m/s
ω = 0.2 m/s / 0.004 m
ω = 50 rad/s
Similarly,
ω * 0.004 m = 0.24 m/s
ω = 0.24 m/s / 0.004 m
ω = 60 rad/s
Since the angular velocity of the arm is the same at both ends, we can take the average of the two values:
Average angular velocity = (50 rad/s + 60 rad/s) / 2
Average angular velocity = 55 rad/s
Therefore, the angular velocity of the arm is 55 rad/s, which is closest to option B (45.082 rad/s).
- Diameter of the nozzle = 8 mm
- Velocity of water discharge = 10 m/s
- Distance of the axis of rotation from one end = 0.2 m
- Distance of the axis of rotation from the other end = 0.24 m
To find:
- Angular velocity of the arm
Let's assume:
- The length of the arm = L
- The angular velocity of the arm = ω
The linear velocity of the water coming out of the nozzle is given by the equation:
v = ω * r
where v is the linear velocity, ω is the angular velocity, and r is the distance of the point from the axis of rotation.
The diameter of the nozzle is 8 mm, so the radius of the nozzle is 4 mm or 0.004 m.
The linear velocity of the water coming out of the nozzle is given as 10 m/s. We can use this information to find the angular velocity of the arm.
At one end of the arm:
v1 = ω * 0.004 m
At the other end of the arm:
v2 = ω * 0.004 m
The distance of the axis of rotation from one end is 0.2 m, so the linear velocity at this point is given as:
v1 = ω * 0.004 m = 0.2 m/s
Similarly, the distance of the axis of rotation from the other end is 0.24 m, so the linear velocity at this point is given as:
v2 = ω * 0.004 m = 0.24 m/s
From the above equations, we can write:
ω * 0.004 m = 0.2 m/s
ω = 0.2 m/s / 0.004 m
ω = 50 rad/s
Similarly,
ω * 0.004 m = 0.24 m/s
ω = 0.24 m/s / 0.004 m
ω = 60 rad/s
Since the angular velocity of the arm is the same at both ends, we can take the average of the two values:
Average angular velocity = (50 rad/s + 60 rad/s) / 2
Average angular velocity = 55 rad/s
Therefore, the angular velocity of the arm is 55 rad/s, which is closest to option B (45.082 rad/s).
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A water sprinkler has an 8 mm diameter nozzle at either end of the rotating arm. It is discharging water at right angle to the rotating arm in opposite direction at a velocity of 10 m/s. If the axis of rotation is at a distance of 0.2 m from one end and 0.24 m from other end, then the angular velocity of arm is (neglect frictional effect)a)8.33 rad/sb)45.082 rad/sc)91.67 rad/sd)2500 rad/sCorrect answer is option 'B'. Can you explain this answer?
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A water sprinkler has an 8 mm diameter nozzle at either end of the rotating arm. It is discharging water at right angle to the rotating arm in opposite direction at a velocity of 10 m/s. If the axis of rotation is at a distance of 0.2 m from one end and 0.24 m from other end, then the angular velocity of arm is (neglect frictional effect)a)8.33 rad/sb)45.082 rad/sc)91.67 rad/sd)2500 rad/sCorrect answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A water sprinkler has an 8 mm diameter nozzle at either end of the rotating arm. It is discharging water at right angle to the rotating arm in opposite direction at a velocity of 10 m/s. If the axis of rotation is at a distance of 0.2 m from one end and 0.24 m from other end, then the angular velocity of arm is (neglect frictional effect)a)8.33 rad/sb)45.082 rad/sc)91.67 rad/sd)2500 rad/sCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A water sprinkler has an 8 mm diameter nozzle at either end of the rotating arm. It is discharging water at right angle to the rotating arm in opposite direction at a velocity of 10 m/s. If the axis of rotation is at a distance of 0.2 m from one end and 0.24 m from other end, then the angular velocity of arm is (neglect frictional effect)a)8.33 rad/sb)45.082 rad/sc)91.67 rad/sd)2500 rad/sCorrect answer is option 'B'. Can you explain this answer?.
A water sprinkler has an 8 mm diameter nozzle at either end of the rotating arm. It is discharging water at right angle to the rotating arm in opposite direction at a velocity of 10 m/s. If the axis of rotation is at a distance of 0.2 m from one end and 0.24 m from other end, then the angular velocity of arm is (neglect frictional effect)a)8.33 rad/sb)45.082 rad/sc)91.67 rad/sd)2500 rad/sCorrect answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A water sprinkler has an 8 mm diameter nozzle at either end of the rotating arm. It is discharging water at right angle to the rotating arm in opposite direction at a velocity of 10 m/s. If the axis of rotation is at a distance of 0.2 m from one end and 0.24 m from other end, then the angular velocity of arm is (neglect frictional effect)a)8.33 rad/sb)45.082 rad/sc)91.67 rad/sd)2500 rad/sCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A water sprinkler has an 8 mm diameter nozzle at either end of the rotating arm. It is discharging water at right angle to the rotating arm in opposite direction at a velocity of 10 m/s. If the axis of rotation is at a distance of 0.2 m from one end and 0.24 m from other end, then the angular velocity of arm is (neglect frictional effect)a)8.33 rad/sb)45.082 rad/sc)91.67 rad/sd)2500 rad/sCorrect answer is option 'B'. Can you explain this answer?.
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