QUIZ: CIVIL (16-JUNE-2020)

Quiz: CIVIL ENGINEERING
Exam: TELANGANA-PSCM
Topic: Hydraulics & Fluid Mechanics

Each question carries 1 mark
Negative marking: No negative marking
Time: 10 Minutes

Q1. Pressure in terms of metres of oil (specific gravity = 0.9) equivalent to 4.5 m of water is
(a) 4.05
(b) 5.0
(c) 3.6
(d) 0.298

Q2. The total pressure on the surface of a vertical sluice gate 2m × 1m with its top 2 m surface beings 0.5 m below the water level will be………….
(a) 500 kg
(b) 1000 kg
(c) 1500 kg
(d) 2000 kg

Q3. Mercury is used in barometers because
(a) it is the best liquid
(b) the height of barometer will be less
(c) its vapour pressure is so low that it may be neglected
(d) both (b) and (c)

Q4. Manometer is used for measuring
(a) Velocity at a point in a fluid
(b) Pressure at a point in a fluid
(c) Difference of pressure between two points
(d) Both (b) and (c)

Q5. When various fluid particles move in zig-zag paths, flow is called:
(a) Laminar flow
(b) Turbulent flow
(c) Uniform flow
(d) None of above

Q6. If a pipe of dia. 30 cm running full with water with velocity 100 m/sec is changed by a pipe of dia. 15cm, then the velocity of water flowing through the pipe will be
(a) 50m/sec
(b) 25m/sec
(c) 200m/sec
(d) 400m/sec

Q7. The hydraulic gradient line
(a) Is always below the total energy line
(b) Is always parallel to bottom
(c) Is always above the total energy line
(d) Is always along the total energy line

Q8. If water enters into the pipe from a reservoir or a tank, then at the entrance into the pipe the head loss will be……………….
(a) V²/2g
(b) 1.2V²/2g
(c) 0.5 V²/2g
(d) ((V_1-V_2)² )/2g

Q9. The discharge through a trapezoidal channel is maximum when:
(a) width of the channel at top is twice the width at bottom
(b) depth of channel is equal to the width at bottom
(c) the sloping side is equal to the half of the width at the top
(d) the sloping side is equal to the width at the bottom

Q10. If the depth and width of a canal is 2 m and 3m respectively, longitudinal slope is 1 in 1000 and Chezy’s constant is 60. The discharge shall be:
(a) 12.6m³/sec.
(b) 9.62m³/sec.
(c) 10.54m³/sec.
(d) 8.50 m³/sec.

SOLUTION

S1. Ans. (b)
Sol. (ρgh) oil = (ρgh) water
1000 × 0.9 × 9.81× h = 1000 × 9.81 × 4.5

h = 5 m of oil

S2. Ans. (d)
Sol.

P =ρgh ̅A
= 1000 × 9.81 × (0.5 + 1/2) × 2 × 1

P = 19620 N.

P = 19620/9.81 kg.

P = 2000 km

S3. Ans.(d)
Sol. Mercury is used in barometers because
→ Its vapour pressure is practically zero.
→ the height of barometer will be less.
→ Its high density.

S4. Ans. (d)
Sol. U-tube manometer ⇒ It is used to measure Low, medium, High, Positive & Negative gauge pressure.
Differential manometer – used for measure pressure difference between two Point
S5. Ans. (b)
Sol. When fluid particle moves in zig – zag path then flow is turbulent flow

S6. Ans. (d)
Sol.

By continuity equation
A₁ V₁ = A₂ V₂
π/4 (30)² × 100 = π/4 × (15)² × V₂

V₂ = 400 m/sec
S7. Ans. (a)
Sol. Hydraulic grade line
(i) Joining the point of piezometric head
(ii) Line may go up or down in the fluid flow
(iii) Always lies below the total energy line.

S8. Ans.(c)
Sol. Loss of energy at the exit of pipe =
(0.5 V²)/2g
S9. Ans. (c)
Sol. In most efficient trapezoidal channel the sloping side in equal to the half of the width at top.

S10. Ans. (c)
Sol.
Depth (y) = 2 m
Width (b) = 3 m
Slope (S) = 1 in 1000
Constant (c) = 60
Q = AC √RS
Q = 3 × 2 × 60 √(A/P×1/100)
Q = 3 × 2 × 60 √((3×2)/(3+4)×1/1000)
Q = 10. 539 m³/sec

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