Civil Engineering - Hydraulics - Discussion
Discussion Forum : Hydraulics - Section 1 (Q.No. 48)
48.
A tank 4m x 3m x 2m containing an oil of specific gravity 0.83 is moved with an acceleration g/2 m sec2. The ratio of the pressures at its bottom when it is moving vertically up and down, is
Discussion:
12 comments Page 1 of 2.
Umesh said:
10 years ago
How please explain?
SAK said:
9 years ago
How please explain?
Andhokar said:
9 years ago
Please explain the solution.
ALO said:
9 years ago
Pressure ,P = Rho*g*h.
According to the condition.
When it is moving vertically up, g = ( 9.8 - 9.8/2).
when it is moving vertically down, g = ( 9.8 + 9.8/2).
Then Ratio = .83*( 9.8+9.8/2)*2 : .83*( 9.8-9.8/2)*2.
So it must me 3 : 1.
According to the condition.
When it is moving vertically up, g = ( 9.8 - 9.8/2).
when it is moving vertically down, g = ( 9.8 + 9.8/2).
Then Ratio = .83*( 9.8+9.8/2)*2 : .83*( 9.8-9.8/2)*2.
So it must me 3 : 1.
(2)
Aks said:
9 years ago
Good job @Alo.
R K Budumuru said:
9 years ago
Ratio of the pressures moving vertically up and down should be 1 : 3 not 3 : 1.
Abhishek Anand said:
9 years ago
Effective g = g+a for upward motion.
= g-a for downward motion.
So ans = 3 : 1.
= g-a for downward motion.
So ans = 3 : 1.
Dhanu said:
8 years ago
Correct @Abhishek.
(1)
Garry said:
8 years ago
The given answer is correct. It will be 3 by 1.
When the body will go upward acceleration will be G+a and when the body will go downward acceleration will be G-a.
When the body will go upward acceleration will be G+a and when the body will go downward acceleration will be G-a.
Monu said:
8 years ago
Why height is taken same for both condition? @Alo.
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