Civil Engineering - Hydraulics - Discussion
Discussion Forum : Hydraulics - Section 3 (Q.No. 2)
2.
A cylindrical vessel 40 cm high is revolved about its vertical axis so that the water touches the bottom when it just spills out. If the radius of the cylinder is 5 cm, the angular velocity of rotation, is
Discussion:
23 comments Page 3 of 3.
Deepankar said:
4 years ago
According to my knowledge, the answer is 125.28 rad/sec.
(1)
BILAL KHAN Zimri said:
3 years ago
Correct ans is 125.28 rad/sec.
Equate all in meters for ease.
The equation is:
H - y(min) = (w^2 x R^2)/(2g).
So, here y(min) =0.
Therefore 0.4 = (w^2 x .0005)/(2 x 9.81).
Solve it, you directly get an answer in rad/sec.
After solving. w= 125.28 rad/sec.
And Just covert the units. 125.28 is in rad/min. Divide by 60 to get 2 rad/sec, we will get the answer.
Equate all in meters for ease.
The equation is:
H - y(min) = (w^2 x R^2)/(2g).
So, here y(min) =0.
Therefore 0.4 = (w^2 x .0005)/(2 x 9.81).
Solve it, you directly get an answer in rad/sec.
After solving. w= 125.28 rad/sec.
And Just covert the units. 125.28 is in rad/min. Divide by 60 to get 2 rad/sec, we will get the answer.
(3)
Inayat Ullah Kakar said:
9 months ago
Formula:
H = w^2× r^2/2g
H = 40cm=0.4m
r = √5,
r^2 = 0.0005m.
Rearrange the above equation:
w^2 = H×2×g/r^2
= 0.4×2×10/0.0005
= 16000
w = √16000.
w = 126 rad/ minutes.
Now rad/sec;
w = 126/60
w= 2 rad/sec
So, the given answer is correct.
H = w^2× r^2/2g
H = 40cm=0.4m
r = √5,
r^2 = 0.0005m.
Rearrange the above equation:
w^2 = H×2×g/r^2
= 0.4×2×10/0.0005
= 16000
w = √16000.
w = 126 rad/ minutes.
Now rad/sec;
w = 126/60
w= 2 rad/sec
So, the given answer is correct.
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