Fluid Mechanics MCQs on “Pressure Distribution in a Liquid Subjected to Horizontal / Vertical Acceleration”.
1. A rectangular tank is moving horizontally in the direction of its length with a constant acceleration of 3.6 m/s2. If tank is open at the top then calculate the angle of water surface to the horizontal.
a) 20.15
b) 69.84
c) 40.30
d) None of the mentioned
Answer: a
Clarification: tanθ=a/g
tanθ=3.6/9.8
θ=20.15⁰.
2. A rectangular tank is moving horizontally in the direction of its length with a constant acceleration of 4.8 m/s2. The length of tank is 7 m and depth is 1.5 m. If tank is open at the top then calculate the maximum pressure intensity at the bottom.
a) 6.3 N/cm2
b) 3.15 N/cm2
c) 12.6 N/cm2
d) 1.6 N/cm2
Answer: b
Clarification: tanθ=a/g
tanθ=4.8/9.8
θ=26.07⁰
h= d+(L/2)tanθ
= 1.5+3.5tan26.07
= 3.21 m
p=ρ*g*h
=3.15 N/cm2.
3. A rectangular tank is moving horizontally in the direction of its length with a constant acceleration of 5.5 m/s2. The length of tank is 5.5 m and depth is 2 m. If tank is open at the top then calculate the minimum pressure intensity at the bottom.
a) 3.8 N/cm2
b) 1.9 N/cm2
c) 5.7 N/cm2
d) 2.6 N/cm2
Answer: b
Clarification: tanθ=a/g
tanθ=5.5/9.8
θ=29.28⁰
h= d-(L/2)tanθ
= 2-2.75 tan29.28
= 3.21 m
p=ρ*g*h
=1.9 N/cm2.
4. A rectangular tank is moving horizontally in the direction of its length with a constant acceleration of 4.5 m/s2.The length, width and depth of tank are 7 m, 3m, 2.5m respectively. If tank is open at the top then calculate the total force due to water acting on higher pressure end of the tank.
a) 1.07 MN
b) 2.14 MN
c) 4.28 MN
d) 4.35 MN
Answer: a
Clarification: tanθ=a/g
tanθ=4.5/9.8
θ=24.64⁰
h= d+(L/2)tanθ
= 2.5+3.5tan24.64
= 4.1 m
F=wAĥ
=9810*2.68*4.1
= 1.07 MN.
5. A tank containing water upto a depth of 500 mm is moving vertically upward with a constant acceleration of 2.45 m/s2. Find the force exerted by fluid of specific gravity .65 on the side of tank,width of tank is 1m.
a) 996.1 N
b) 1992.2 N
c) 498.06 N
d) 124.5 N
Answer: a
Clarification: p=ρ*g*h*(1+a/g)
=650*9.81*0.5*(1+2.45/9.81)
=3984.5 N/m2
F=wAĥ
= 650*9.81*0.5*0.5*3984.5
= 996.1N.
6. A tank containing water upto a depth of 750 mm is moving vertically downward with a constant acceleration of 3.45 m/s2. Find the force exerted by fluid of specific gravity .85 on the side of tank,width of tank is 2m
a) 2682.75 N
b) 5365.5 N
c) 1341.25 N
d) 4024.5 N
Answer: a
Clarification: p=ρ*g*h*(1-a/g)
=750*9.81*0.75*(1-3.45/9.81)
=3577 N/m2
F=wAĥ
= 0.5*0.75*3984.5*2
= 2682.75 N.
7. A tank containing water upto a depth of 650 mm is stationary. Find the force exerted by fluid of specific gravity .55 on the side of tank,width of tank is 1.5m
a) 1709.9 N
b) 3419.4N
c) 6838.8 N
d) 1367.75 N
Answer: a
Clarification: p=ρ*g*h
=550*9.81*0.65
=3507 N/m2
F=wAĥ
= 0.5*0.65*3507*1.5
= 1709.7N.
8. The pressure intensity at the bottom remains same, even if the tank moves with constant horizontal acceleration.
a) True
b) False
Answer: b
Clarification: The pressure intensity at the bottom differs due to variation in height as tank moves with constant acceleration.
9. There will be development of shear stress due to the dynamic motion of tank or container.
a) True
b) False
Answer: b
Clarification: The water in tank is at rest even if the tank is moving.
10. If the tank is moving vertically, which of its component is subjected to maximum total pressure?
a) Lower part of vertical walls
b) Higher part of vertical walls
c) Base
d) None of the mentioned
Answer: c
Clarification: Base bores the direct pressure.