UPPSC Lec'21 ME: Daily Practices Quiz. 23-Nov-2021 |_00.1
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UPPSC Lec’21 ME: Daily Practices Quiz. 23-Nov-2021

Quiz: Mechanical Engineering
Exam: UPPSC AE/Lecturer
Topic: Fluid Mechanics

Each question carries 3 marks
Negative marking: 1 mark
Time: 8 Minutes

Q1. Determine the total pressure on a circular plate of diameter 1.5 m which is placed vertically in water in such a way that the center of plate is 3 m below the surface of water.
(a) 52.002 N
(b) 52002.81 N
(c) 5200.281 N
(d) 520.028 N

Q2. A piece of metal of specific gravity 7 Floats in mercury of specific gravity 13.6 what fraction of its volume is under mercury?
(a) 0.5
(b) 0.4
(c) 0.515
(d) 0.415

Q3. Meta-centric height is the distance between-
(a) Meta Centre and water surface
(b) Meta Centre and Centre of buoyancy
(c) Meta Centre and Centre of gravity
(d) Meta-Centre and centroid

Q4. An object weighs 60 gm in air, 50 gm in water and 40 gm in oil. Then the specific gravity of the oil will be __________.
(a) 0.25
(b) 1
(c) 1.5
(d) 2

Q5. A block of ice floating over water in a vessel slowly melts in it. The water level in the vessel will
(a) start rising
(b) start falling
(c) will remain constant
(d) will depend on temperature of water

Q6. When a body floating in liquid is displaced slightly, it oscillates about ________.
(a) Center of gravity of body
(b) Center of pressure
(c) Center of buoyancy
(d) Meta Centre

S1. Ans.(b)
Sol. f = ρgAx̅
= 1000 ×9.81×π/4 (1.5)²×3
= 52002.8 N

S2. Ans.(c)
Sol. Let v is the volume of metal block, and v’
is the volume of metal block part which is submerged
weight of metal block = weight of fluid displaced
ρ_metal×g×v=ρ_mercury×g×v^’
7000×g×v=13600×g×v^’
v^’=7000/13600 v
v^’=0.515 V

S3. Ans.(c)
Sol. Meta-centric height is the distance between- Meta Centre and Centre of gravity

S4. Ans.(d)
Sol. Weight of body in different medium = (weight of body in air – buoyancy force due to displaced fluid)
weight of body in water = 50 gm
∵ 50= 60-F_y
F_y=10 { ∵Bouyeny force (ρ_y )=ρ.gv}
ρ.g.V=10×10^(-3)
V=(10×10^(-3))/(1000 g)
V=(10×10^(-6))/( g)
When it is submerged in oil then its buoyant force.
Fy^’=60-40=20 gm
fy^’=20×10^(-3)=ρ_oil×g×V
20×10^(-3)=ρ_oil×g×(10×10^(-6))/g
ρ_oil=2000 kg\/m^3
Specific grawity =ρ_oil/ρ_water =2

S5. Ans.(c)
Sol. Assume density of ice and water are same i.e.
(ρ_ω=ρ_ice )
Weight of the ice block = weight of fluid displaced
ρ_ice.g.V_ice=ρ_ω.g.V_fluid
∵ρ_ω=ρ_ice
∵V_ice=V_fluid
So, level of water remains constant even after the melting of ice.

S6. Ans.(d)
Sol. Metacenter is the point about a body will oscillate after it will displace slightly.

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Published by VINAYAK KUMAR

Graduated from BIT Sindri, Postgraduated from IIT BHU Varanasi

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