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 1 of 3.
Vaish said:
10 years ago
(4*g*h/3).
Abhi said:
9 years ago
y = yo + [(ω^2 x r^2)/(2 x g)].
0.4 = 0 + [(ω^2 x r^2)/(2 x g)].
0.4*2*9.81/0.05 = ω^2.
ω = 12.52 rad/sec.
0.4 = 0 + [(ω^2 x r^2)/(2 x g)].
0.4*2*9.81/0.05 = ω^2.
ω = 12.52 rad/sec.
(1)
M.s.bainsla said:
9 years ago
2 rad/sec is correct. I agree.
Priya said:
8 years ago
Please explain it clearly.
Mrutyunjay said:
8 years ago
How? Please explain.
Arun said:
8 years ago
How 2rad/sec is correct? Explain.
Anurag said:
8 years ago
Explain how.
Priya said:
7 years ago
It should be (4/3*g*h).
Anshu said:
7 years ago
√ ((0.4*2*9.81)-0.05) = 2.79.
Hence answer is A correct.
Hence answer is A correct.
Rohit said:
7 years ago
(4/3 * r * h) * 3.14/180.
Putting the values;
4/3 * 40 * 2.23 * 3.14/180.
= 2.08 rad/sec.
Putting the values;
4/3 * 40 * 2.23 * 3.14/180.
= 2.08 rad/sec.
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