Civil Engineering - GATE Exam Questions - Discussion

Discussion Forum : GATE Exam Questions - Section 2 (Q.No. 25)
25.
A body moving through still water at 6 m/sec produces a water velocity of 4 m/sec at a point 1 m ahead. The difference in pressure between the nose and the point 1 m ahead would be
2, 000 N/m2
10, 000 N/m2
19, 620 N/m2
98, 100 N/m2
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
10 comments Page 1 of 1.

Citu said:   4 years ago
The right answer is (A)
At point 1, velocity of water =0
At point 2, relative velocity of water = 2 m/sec.

Apply Bernoulli theory, answer will be 2000N/m2.

Butaarts said:   5 years ago
Thanks @Radha.

Swar said:   6 years ago
Thank you @Radha.

Radha said:   6 years ago
(P1/ ρg) + (v1^2/2g) +h= (P2/ρg) + (v2^2/2g)+h.

Apply Bernoulli's equation.

in question asked for pressure difference, i.e, (P1-P2).
Get all pressure head terms one side and velocity head terms another side; datum heads will get cancelled.

Now "g" is both sides of the equation so cancel it.
you will be remaining with;
(P1-P2)/d = (v1^2-v2^2)/2.
=>(P1-P2)= d*(v1^2-v2^2)/2 * assuming d as density of water.

Put all the values the answer will be : 10,000 N/m2

Hope this will help you.
(2)

Dhanu said:   7 years ago
How to solve this. Please, anyone, tell me.

Akki said:   7 years ago
Here, ρ = 1000.

Ranganadh said:   7 years ago
@Abhi.

ρ = 2.

Abhi said:   8 years ago
What is the value of ρ?

Snehal Wankhede said:   9 years ago
(P1/ ρg) + (v1^2/2g) + h= (P2/ρg) + (v2^2/2g).
= (v2^2/2g) - h = (6^2/2g); v2 = 4m/s.
= (36 - 16) * 1000/2 = 10,000 N/m2.

Maanna said:   9 years ago
What's the formula for this?

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