Gradually Varied MCQ Quiz - Objective Question with Answer for Gradually Varied - Download Free PDF

Last updated on Apr 21, 2025

Latest Gradually Varied MCQ Objective Questions

Gradually Varied Question 1:

Find the slope of the free water surface of a rectangular stream 15 m wide and 3 m deep. The slope of the stream bed is 1 in 5000. Total discharge is 29 m3/s. Assume the slope of the energy line = 1 in 21,700 and that the depth is increasing in the direction of the flow.

  1. 1 in 9300
  2. 1 in 6400
  3. 1 in 8200
  4. 1 in 4500

Answer (Detailed Solution Below)

Option 2 : 1 in 6400

Gradually Varied Question 1 Detailed Solution

Concept:

The gradually varied flow (GVF) equation derived from energy slope and channel bed slope is:

\(\frac{dy}{dx} = \frac{S_0 - S_f}{1 - F_r^2} \)

Given:

Width of stream (b) = 15 m

Depth of stream (y) = 3 m

Discharge (Q) = 29 m³/s

Bed slope S0=15000=0.0002" id="MathJax-Element-7-Frame" role="presentation" style="position: relative;" tabindex="0">S0=15000=0.0002

Energy slope Sf=121700=0.000046" id="MathJax-Element-8-Frame" role="presentation" style="position: relative;" tabindex="0">Sf=121700=0.000046

Step 1: Area and velocity

\( A = b \cdot y = 15 \cdot 3 = 45 \, m^2 \)

\( V = \frac{Q}{A} = \frac{29}{45} = 0.644 \, m/s \)

Step 2: Froude number

\( F_r = \frac{V}{\sqrt{g y}} = \frac{0.644}{\sqrt{9.81 \cdot 3}} = \frac{0.644}{5.427} \approx 0.1186 \)

Step 3: Apply GVF equation

\(\frac{dy}{dx} = \frac{0.0002 - 0.000046}{1 - (0.1186)^2} = \frac{0.000154}{1 - 0.0141} = \frac{0.000154}{0.9859} \approx 0.0001562\)

Step 4: Find slope ratio

\( \frac{1}{0.0001562} \approx 6400 \)

Gradually Varied Question 2:

The rectangular channel will be most economical when:

  1. the width of the channel is half the depth of flow
  2. the width of the channel is two times the depth of flow
  3. the depth of flow is two times the width of the channe
  4. the depth of flow is three times the width of the channel 

Answer (Detailed Solution Below)

Option 2 : the width of the channel is two times the depth of flow

Gradually Varied Question 2 Detailed Solution

Explanation:

Most economical section is the one whose wetted perimeter is minimum for the given value of discharge.

Full Test 2 (31-80) images Q.55

For minimum wetted perimeter,

d = b/2

∴ d/b = 1 : 2

Note:

For most economical Channel

S.No.

Shape

Hydraulic Radius

1

GATE CE FT 5 (SLOT 1) images Q14

Rectangular Channel

\({y_m} = \frac{y_0}{2}\)

2

GATE CE FT 5 (SLOT 1) images Q14b

Trapezoidal channel

\({y_m} = \frac{y_0}{2}\)

3

Triangular Channel

\({y_m} = \frac{y_0}{{2\sqrt 2 }}\)

4

896

Circular channel

\({y_m} = 0.29d\)

Gradually Varied Question 3:

A channel of rectangular section, 5 m wide, carries water at the rate of 20 m3 /s at a depth of 1 m. Find the value of Manning’s constant, if the bed slope is 1 in 900.

  1. \(\frac{1}{120}\left(\frac{5}{7}\right)^{\frac{2}3{}}\)
  2. \(\frac{1}{100}\left(\frac{5}{7}\right)^{\frac{2}3{}}\)
  3. \(\frac{1}{150}\left(\frac{5}{7}\right)^{\frac{2}3{}}\)
  4. \(\frac{1}{180}\left(\frac{5}{7}\right)^{\frac{2}3{}}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{1}{120}\left(\frac{5}{7}\right)^{\frac{2}3{}}\)

Gradually Varied Question 3 Detailed Solution

Explanation:

Given Data:

  • Channel Width (B) = 5 m
  • Depth (y) = 1 m
  • Discharge (Q) = 20 m³/s
  • Bed Slope (S) = 1/900

Step 1: Calculate Flow Area (A)

\( A = B \times y = 5 \times 1 = 5 m^2 \)

Step 2: Calculate Wetted Perimeter (P)

\( P = B + 2y = 5 + 2(1) = 7 m \)

Step 3: Calculate Hydraulic Radius (R)

\( R = \frac{A}{P} = \frac{5}{7} m \)

Step 4: Find Velocity (V) using Q = AV

\( V = \frac{Q}{A} = \frac{20}{5} = 4 \text{ m/s} \)

Step 5: Solve for Manning’s Constant (n)

Using Manning’s equation:

\( V = \frac{1}{n} R^{2/3} S^{1/2} \)

Substituting values:

\( 4 = \frac{1}{n} \times \left(\frac{5}{7}\right)^{2/3} \times \left(\frac{1}{900}\right)^{1/2} \)

Solving for \( n \):

\( n = \frac{1}{120} \times \left(\frac{5}{7}\right) \)

 

Gradually Varied Question 4:

In the context of open channel flow, torrential flow is also known as _________.

  1. tranquil flow
  2. shooting flow
  3. critical flow
  4. alternate depth

Answer (Detailed Solution Below)

Option 2 : shooting flow

Gradually Varied Question 4 Detailed Solution

In open channel flow, torrential flow, also known as shooting flow, occurs when the flow velocity is greater than the wave velocity. This condition indicates a supercritical flow state where the Froude number is greater than one (Fr > 1). In this state, disturbances cannot propagate upstream, making the flow highly energetic and rapid. This is in contrast to subcritical flow (tranquil flow), where the Froude number is less than one (Fr < 1), and disturbances can travel upstream. Torrential flow is characterized by its high velocity and is typically found in steep channels or during high-flow conditions.

Gradually Varied Question 5:

Flow in an open channel is classified as turbulent if the Reynold's number of flow is ____ .

  1. less than 2000
  2. more than 4000
  3. less than 500
  4. more than 2000

Answer (Detailed Solution Below)

Option 4 : more than 2000

Gradually Varied Question 5 Detailed Solution

Turbulent Flow in Open Channels:

  • Flow in an open channel is classified based on the Reynolds number (Re), which is a dimensionless quantity used to predict flow patterns in different fluid flow situations.
  • The Reynolds number is given by the formula:  \(R_e = {\rho VD \over \mu}\), where ρ is the fluid density, V is the flow velocity, D is the characteristic length (such as hydraulic diameter), and μ is the dynamic viscosity of the fluid.
  • If the Reynolds number of the flow in an open channel is more than 2000, the flow is considered to be turbulent. Turbulent flow is characterized by chaotic changes in pressure and velocity.
  • In contrast, flow with a Reynolds number less than 500 is typically classified as laminar flow, where the fluid flows in parallel layers with minimal disruption between them.
  • Transitional flow occurs between Reynolds numbers of approximately 500 and 2000, where the flow can shift between laminar and turbulent states.

Top Gradually Varied MCQ Objective Questions

A rectangular channel of bed width 2 m is to be laid at a bed slope of 1 in 1000. Find the hydraulic radius of the canal cross-section for the maximum discharge condition? Take Chezy’s constant as 50

  1. 0.5 m
  2. 2 m
  3. 1 m
  4. 0.25 m

Answer (Detailed Solution Below)

Option 1 : 0.5 m

Gradually Varied Question 6 Detailed Solution

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Concept:

Most efficient channel: A channel is said to be efficient if it carries the maximum discharge for the given cross-section which is achieved when the wetted perimeter is kept a minimum.

Rectangular Section:

Full Test 2 (31-80) images Q.55

Area of the flow, A = b × d

Wetted Perimeter, P = b + 2 × d  

For the most efficient Rectangular channel, the two important conditions are

  1. b = 2 × d
  2.  \(R = \frac{A}{P} = \frac{{b\times d}}{{b + 2 \times d}} = \frac{{2{d^2}}}{{4d}} = \frac{d}{2}= \frac{b}{4}\)

Calculation

Given: b = 2 m

 \(R = \frac{2}{4}\)

R = 0.5

quesImage111

Rectangular channel section Trapezoidal channel section
  1. R = y / 2
  2. A = 2y
  3. T = 2y
  4. P = 4y
  5. D = y
  1. R = y / 2
  2. \(A = \sqrt 3 {y^2}\)
  3. \(T = \frac{{4y}}{{\sqrt 3 }}\)
  4. \(P = {{2y\sqrt 3}}{{ }}\)
  5.  D = 3y / 4

Where R = hydraulic radius, A = Area of flow, P = wetted perimeter, y = depth of flow, T = Top width

For obtaining the most economical trapezoidal channel section with depth of flow = 3 m, what is the hydraulic mean radius ?

  1. 1.5 m
  2. 3.0 m
  3. 2.0 m
  4. 1.0 m

Answer (Detailed Solution Below)

Option 1 : 1.5 m

Gradually Varied Question 7 Detailed Solution

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Concept:

Most economical section is the one whose wetted perimeter is minimum for the given value of discharge.

GATE CE FT 5 (SLOT 1) images Q14b

Hydraulic mean radius,

\({y_m} = \frac{y_0}{2} = \frac{3}{2} = 1.5\; m\)

Important Points

For most economical Channel

S.No.

Shape

Hydraulic Radius

1

GATE CE FT 5 (SLOT 1) images Q14

Rectangular Channel

\({y_m} = \frac{y_0}{2}\)

2

GATE CE FT 5 (SLOT 1) images Q14b

Trapezoidal channel

\({y_m} = \frac{y_0}{2}\)

3

Triangular Channel

\({y_m} = \frac{y_0}{{2\sqrt 2 }}\)

4

896

Circular channel

\({y_m} = 0.29d\)

A rectangular channel with Gradually Varied Flow (GVF) has a changing bed slope. If the change is from a steeper slope to a steep slope, the resulting GVF profile is

  1. S3
  2. S1
  3. S2
  4. either S1 or S2, depending on the magnitude of the slopes

Answer (Detailed Solution Below)

Option 1 : S3

Gradually Varied Question 8 Detailed Solution

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Explanation-

  • When two-channel sections have different bed slopes the condition is called a break in grade.
  • Under this situation following conditions must be remembered for drawing the flow profile.
    • CDL is independent of the bed slope.
    • The steeper the slope lesser is the normal depth of flow.
    • Flow always starts from NDL and tries to meet NDL.
    • Subcritical flow has downstream control and supercritical flow has upstream control.
  • Steep slope-
    • F7 Savita Engineering 06-4-22 D21

 

Given data and Analysis-

  • Flow is changed from steeper to sleep.
  • So the Normal depth of flow will increase in a steep slope, as the slope is decreased.
  • CDL will remain the same for both the slope.
  • So from the figure shown below it can be concluded that flow will be S3 profile.

F7 Savita Engineering 06-4-22 D22

Specific energy of flowing water through a rectangular channel of width 5 m when discharge is 10 m/ s and depth of water is 2 m is:

  1. 1.06 m
  2. 1.02 m
  3. 2.05 m
  4. 2.60 m

Answer (Detailed Solution Below)

Option 3 : 2.05 m

Gradually Varied Question 9 Detailed Solution

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Concept:

The specific energy may be given as

\(E = y + \frac{{V_1^2}}{{2g}}\)

Calculation:

We know that

Q = AV

⇒ 10 = 5 × 2 × V

⇒ V = 1 m/s

\(\begin{array}{l} E = y + \frac{{V_1^2}}{{2g}}\\ E= 2 + \frac{{{{\left( {1} \right)}^2}}}{{2 \times 9.81}} = 2.05\;m \end{array}\)

If the channel cut shown in the figure is an economical cut, then what will be its area?

F1  Ram S 20-08-21 Savita D1

  1. A = 1.414 y2
  2. A = 0.5 y2
  3. A = 2 y2
  4. A = 1.914 y2

Answer (Detailed Solution Below)

Option 4 : A = 1.914 y2

Gradually Varied Question 10 Detailed Solution

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Explanation:

For the economical section, its perimeter should be minimum.

\(A = b \times y + \frac{1}{2} \times y \times y\)

\(A = by + \frac{{{y^2}}}{2}\)

\(b = \frac{A}{y} - \frac{y}{2}\)

Perimeter

\(P = y + b + \sqrt {{y^2} + {y^2}}\)

\(P = y + y\sqrt 2 + \frac{A}{y} - \frac{y}{2}\)

\(\frac{{dP}}{{dy}} = 0\)

\(\frac{{dP}}{{dy}} = 0 = 1 + \sqrt 2 - \frac{1}{2} - \frac{A}{{{y^2}}}\)

A = 1.914 y2

For a hydraulically efficient rectangular channel of bed width 5 m, the hydraulic radius is equal to 

  1. 1.25 m
  2. 2 m
  3. 2.25 m
  4. 1.75 m

Answer (Detailed Solution Below)

Option 1 : 1.25 m

Gradually Varied Question 11 Detailed Solution

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Concept

For a hydraulic efficient rectangular channel

\(depth\;of\;water\;level = \frac{{width\;of\;channel}}{2}\)

\(y = \frac{B}{2}\;\)

\(Hydraullic\;radius\;\left( R \right) = \frac{{wetted\;area}}{{wetted\;perimeter}} = \frac{{B \times y}}{{B + 2 \times y}}\)

\(R = \frac{{B \times \frac{B}{2}}}{{B + 2 \times \frac{B}{2}}} = \frac{B}{4}\)

Calculation

Given, Width of channel (B) = 5 m

\(y = \frac{B}{2} = \frac{5}{2} = 2.5\;m\)

\(R = \frac{B}{4} = \frac{5}{4} = 1.25\;m\)

∴ Hydraulic radius (R) = 1.25 m 

What is the rate of flow in a rectangular channel 4 m wide and 2 m deep with a bed slope of 1 in 900 when it is running full if Chezy’s constant is 50?

  1. 10.56 m3/s
  2. 13.33 m3/s
  3. 14.38 m3/s
  4. 12.25 m3/s

Answer (Detailed Solution Below)

Option 2 : 13.33 m3/s

Gradually Varied Question 12 Detailed Solution

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Concept:

The Chezy equation can be used to calculate the discharge in the open channel.

\(\rm Q= A × C\sqrt {mi} \)

A - Area of the channel

C - Chezy's coefficient

m - Hydraulic radius (or) Hydraulic mean depth = A/P

P- Perimeter of the channel

i - Bed slope

Calculation:

Given:

b = 4 m, d = 2 m,C = 50

\({\rm{i}} = \frac{{\rm{1}}}{{\rm{900}}} = 0.0011\)

A = 4 × 2 = 8 m2

\({\rm{m}} = \frac{{\rm{A}}}{{\rm{P}}} = \frac{{4\; \times\;2}}{{4 \;+ \;2\; + \;2}} = 1\)

\({\rm{Q}} = \rm A \times {\rm{C}}\sqrt {{\rm{mi}}} = 8 \times50 \sqrt {1\times 0.0011} \)

∴ Q = 13.3 m3/s

An open channel is of isosceles triangle shape, with side slopes 1 vertical and n horizontal. The ratio of the critical depth to specific energy at critical depth will be 

  1. 2/3
  2. 3/4
  3. 4/5
  4. 5/6

Answer (Detailed Solution Below)

Option 3 : 4/5

Gradually Varied Question 13 Detailed Solution

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Concept:

For critical flow, y = yc

Specific energy at critical flow, Ec 

\({E_c} = {y_c} + \frac{{V_c^2}}{{2g}}\)

\({E_c} = {y_c} + \frac{{Q_c^2}}{{2gA_c^2}}\)

For critical flow:

\(\frac{{{Q^2}T}}{{gA_c^3}} = 1 \Rightarrow \frac{{{Q^2}}}{{gA_c^2}} = \frac{{{A_c}}}{T}\)

Where T is the top width

( T = B) 

Put, \(A = \frac{1}{2} \times Base \times Height\)

\({E_c} = {y_c} + \frac{{{A_c}}}{{2T}} = {y_c} + \frac{{0.5 \times B \times {y_c}}}{{2 \times B}} = {y_c} + \frac{{{y_c}}}{4}\)

∴ \(\frac{{{y_c}}}{{{E_c}}} = \frac{4}{5}\)

For subcritical flow in an open channel, the control section for gradually varied flow profiles is

  1. at the downstream end
  2. at the upstream end
  3. at both upstream and downstream ends
  4. at any intermediate section

Answer (Detailed Solution Below)

Option 1 : at the downstream end

Gradually Varied Question 14 Detailed Solution

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Concept:

In an open channel flow control section is a section where for a given discharge the depth of flow is known or it can be controlled.

The Froude number (FR) is given by

\({F_R} = \frac{V}{{\sqrt {gH} }}, \sqrt {gh} - \rm Celerity\)

V - Velocity of flow

H - Hydraulic depth

Celerity is the velocity of a wave when the flow is disturbed.

Subcritical flow:

In Subcritical flow FR < 1, 

\(∴ \sqrt {gh} > V\)

Hence velocity of the wave is more than the velocity of flow, hence it can travel upstream. Here the downstream conditions are controlled.

For subcritical flow downstream is the control section.

Supercritical flow:

In Supercritical flow FR > 1, 

\(∴ \sqrt {gh} < V\)

Hence velocity of the wave is less than the velocity of flow, hence it cannot travel upstream. Here the upstream conditions are controlled.

∴ For supercritical flow upstream is the control section

Key Points

If the wave can travel upstream, downstream is the control section.

If the wave can travel downstream, upstream is the control section.

The flow in a rectangular channel is subcritical. If the width of the channel is reduced at a certain section, the water surface under no-choke condition will

  1. Drop at a downstream section
  2. Rise at a downstream section
  3. Rise at a upstream section
  4. Not undergo any change

Answer (Detailed Solution Below)

Option 1 : Drop at a downstream section

Gradually Varied Question 15 Detailed Solution

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Concept:

When the width of the channel is reduced the term q (Discharge per unit width) increases. The no choke condition means critical depth of flow is not reached.

Case 1: Width of channel is B

The specific energy curve is drawn for the channel

F1 Killi 28.6.21 Pallavi D2

Case 2: Width of channel is B1 (B1 ≤ B)

The specific energy curve corresponding to the width B1 is drawn.

F1 Killi 28.6.21 Pallavi D3

When we consider the subcritical portion the graph,

We find that for the same specific energy the depth decreases when there is reduction in width of the channel. Hence there is a drop at a downstream section.
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