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In a stream line steady state flow, two points A and B on a stream line are 1 m apart and the velocity of a particle varies linearly from 2 m/s to 5 m/s as the particle moves from A to B. What is the acceleration of fluid at B?
- a)3 m/s2
- b)6 m/s2
- c)9 m/s2
- d)15 m/s2
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
In a stream line steady state flow, two points A and B on a stream li...
For a flow a = U ∂U / ∂x +∂U /∂t
As flow is steady
∂U / ∂t = 0
Also it is given, that as particle moves from A to B its velocity increases linearly,
therefore U = ax + b And
hence
∂U/∂x =UB − UA / xB − xA = 5 − 2 / 1 − 0 = 3/sec
Thus, a(x) = U(x) × 3 m/sec Hence a at B
a(xB) = U(xB) ∂U /∂x
a(xB) = 5 × 3
a(xB) = 15 m/sec
Most Upvoted Answer
In a stream line steady state flow, two points A and B on a stream li...
Solution:
Given data:
Distance between two points A and B on a stream line = 1 m
Velocity of the particle at point A = 2 m/s
Velocity of the particle at point B = 5 m/s
We know that acceleration (a) of a fluid is given by the formula:
a = dv/dt
where,
dv = change in velocity
dt = time taken for the change
As the flow is steady state, the time taken for the particle to move from point A to point B is constant.
Therefore, the acceleration of fluid at point B can be calculated as follows:
dv = (5 - 2) m/s = 3 m/s
dt = distance between A and B / velocity difference between A and B
= 1 m / (5 - 2) m/s
= 1/3 s
a = dv/dt
= 3 m/s / (1/3) s
= 9 m/s^2
Therefore, the acceleration of fluid at point B is 9 m/s^2, which is option 'D'
Given data:
Distance between two points A and B on a stream line = 1 m
Velocity of the particle at point A = 2 m/s
Velocity of the particle at point B = 5 m/s
We know that acceleration (a) of a fluid is given by the formula:
a = dv/dt
where,
dv = change in velocity
dt = time taken for the change
As the flow is steady state, the time taken for the particle to move from point A to point B is constant.
Therefore, the acceleration of fluid at point B can be calculated as follows:
dv = (5 - 2) m/s = 3 m/s
dt = distance between A and B / velocity difference between A and B
= 1 m / (5 - 2) m/s
= 1/3 s
a = dv/dt
= 3 m/s / (1/3) s
= 9 m/s^2
Therefore, the acceleration of fluid at point B is 9 m/s^2, which is option 'D'
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In a stream line steady state flow, two points A and B on a stream line are 1 m apart and the velocity of a particle varies linearly from 2 m/s to 5 m/s as the particle moves from A to B. What is the acceleration of fluid at B?a) 3 m/s2b) 6 m/s2c) 9 m/s2d) 15 m/s2Correct answer is option 'D'. Can you explain this answer?
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