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
2 rad/sec
3 rad/sec
4 rad/sec
5 rad/sec
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
23 comments Page 2 of 3.

KANYA said:   6 years ago
@Kiran, 12.28÷60 = 0.2 not 2.

Kiran said:   6 years ago
@ Deepak. You are close to the answer. Just covert the units. 125.28 is in rad/min. Divide by 60 to get 2 rad/sec.

12.28/60 ~ 2 rad/sec.

Lalit pathade said:   6 years ago
Vessel height =( angula velocity^2 * radius^2)/2g.
(40) =(angular velocity ^2 * (sqrt(5)) ^2) /(2*9.81).
Angular velocity = 2rad/sec.

Muhammad Saqib said:   6 years ago
Explain how 2rad/sec correct? Please explain me.

Mrutyunjay said:   8 years ago
How? Please explain.

Anurag said:   7 years ago
12.52 rad/sec is the right answer.

Deepak said:   7 years ago
According to me, it is 125.28 radian/sec.

Shubham said:   7 years ago
Formula is.

Vessel height =( angula velocity"2 * radius"2)/2g.

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.

Anshu said:   7 years ago
√ ((0.4*2*9.81)-0.05) = 2.79.

Hence answer is A correct.


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