1. A field channel has cultivable commanded area of 2000 hectares. The intensities of irrigation for gram and wheat are 30% and 50% respectively. Gram has a kor period of 18 days, kor depth of 12 cm, while wheat has a kor period of 18 days and a kor depth of 15 cm. The discharge (in m³/s) required in the field channel to supply water to the commanded area during the kor period is ____

1. 1. 0.57 m³/s

   
   
   

   2. 1.427 m³/s

   
   
   

   3. 2 m³/s

   
   
   

   4. 2.5 m³/s

   
   
   

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1. View Hint View Answer Discuss in Forum

   1.427
   

   Q =	A
   	D

   
   

   Duty, D = 8.64	B
   	Δ

   
   B – kor period (days)
   Δ – kor depth (m)
   Gram
   

   QG =	AG	=	2000 × 0.3		= 0.463 m³
   	DG		8.64 ×	18
   				0.12

   
   Wheat
   

   QW =	AW	=	2000 × 0.5		= 0.964 m³
   	DW		8.64 ×	18
   				0.15

   
   Discharge required = Q₁ + Q₂ = 1.427 m³/s

   Correct Option: B

   1.427
   

   Q =	A
   	D

   
   

   Duty, D = 8.64	B
   	Δ

   
   B – kor period (days)
   Δ – kor depth (m)
   Gram
   

   QG =	AG	=	2000 × 0.3		= 0.463 m³
   	DG		8.64 ×	18
   				0.12

   
   Wheat
   

   QW =	AW	=	2000 × 0.5		= 0.964 m³
   	DW		8.64 ×	18
   				0.15

   
   Discharge required = Q₁ + Q₂ = 1.427 m³/s

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2. A rectangular channel of 2.5 m width is carrying a discharge of 4 m³/s. Considering that acceleration due to gravity as 9.81 m/s², the velocity of flow (in m/s) corresponding to the critical depth (at which the specific energy is minimum) is ________.

1. 1. 1.5 m/s

   
   
   

   2. 2 m/s

   
   
   

   3. 2.5 m/s

   
   
   

   4. 3 m/s

   
   
   

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1. View Hint View Answer Discuss in Forum

   2.45 to 2.55
   

   y³c =	q³
   	g

   
   

   q =	Q	=	4	= 1.6m³/s/m
   	B		2.5

   
   

   ∴ yc =	(1.6)²	= 0.639 m
   	9.81

   
   At critical depth,
   

   ∴ yc =	Vc
   	2

   
   

   	V²	=	0.639
   	2g		2

   
   ∴ V = √0.639 × g = 2.5 m/s

   Correct Option: C

   2.45 to 2.55
   

   y³c =	q³
   	g

   
   

   q =	Q	=	4	= 1.6m³/s/m
   	B		2.5

   
   

   ∴ yc =	(1.6)²	= 0.639 m
   	9.81

   
   At critical depth,
   

   ∴ yc =	Vc
   	2

   
   

   	V²	=	0.639
   	2g		2

   
   ∴ V = √0.639 × g = 2.5 m/s

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3. The transplantation of rice requires 10 days and total depth of water required during transplantation is 48 cm. During transplantation, there is an effective rainfall (useful for irrigation) of 8 cm. The duty of irrigation water (in hectares/cumec) is

1. 1. 612

   
   
   

   2. 216

   
   
   

   3. 300

   
   
   

   4. 108

   
   
   

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1. View Hint View Answer Discuss in Forum

   D = 864 B/Δ hectares/cumecs
   

   =	864 × 10	= 216 hectares/cumecs
   	(48 – 8) × 10

   Correct Option: B

   D = 864 B/Δ hectares/cumecs
   

   =	864 × 10	= 216 hectares/cumecs
   	(48 – 8) × 10

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4. Wheat crop requires 55 cm of water during 120 days of base period. The total rainfall during this period is 100 mm. Assume the irrigation efficiency to be 60%. The area (in ha) of the land which can be irrigated with a canal flow of 0.01 m³/s is

1. 1. 13.82

   
   
   

   2. 18.85

   
   
   

   3. 23.04

   
   
   

   4. 230.40

   
   
   

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1. View Hint View Answer Discuss in Forum

   Actual water depth required (D₁) = 55 – total rainfall
   = 55 – 10 = 45 cm
   

   D1 =	864 × B
   	D

   
   

   ⇒ 45 =	864 × 120
   	D

   
   D = 2304 ha/cumec
   1 m³/s discharge → 2304 hectares (irrigation)
   ∴ 0.01 m³/s discharge → 2304 × 0.01 = 23.04 hectares
   

   Actual irrigated area = 23.04 ×	60	= 13.824 hectares.
   	100

   Correct Option: A

   Actual water depth required (D₁) = 55 – total rainfall
   = 55 – 10 = 45 cm
   

   D1 =	864 × B
   	D

   
   

   ⇒ 45 =	864 × 120
   	D

   
   D = 2304 ha/cumec
   1 m³/s discharge → 2304 hectares (irrigation)
   ∴ 0.01 m³/s discharge → 2304 × 0.01 = 23.04 hectares
   

   Actual irrigated area = 23.04 ×	60	= 13.824 hectares.
   	100

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5. An agricultural land of 437 ha is to be irrigated for a particular crop. The base period of the crop is 90 days and the total depth of water required by the crop is 105 cm. If a rainfall of 15 cm occurs during the base period, the duty of irrigation water is

1. 1. 437 ha/cumec

   
   
   

   2. 486 ha/cumec

   
   
   

   3. 741 ha/cumec

   
   
   

   4. 864 ha/cumec

   
   
   

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1. View Hint View Answer Discuss in Forum

   Net water required = 105 – 15 = 90 cm
   D × Delta = Duty × Base period
   

   D =	1 × (90 × 24 × 60 × 60)	× 104 (m² to hectare) = 864 hectares/curve
   	0.9

   
   

   Duty =	864 × B	=	864 × 90	= 864 ha/cumec
   	Δ		90

   Correct Option: D

   Net water required = 105 – 15 = 90 cm
   D × Delta = Duty × Base period
   

   D =	1 × (90 × 24 × 60 × 60)	× 104 (m² to hectare) = 864 hectares/curve
   	0.9

   
   

   Duty =	864 × B	=	864 × 90	= 864 ha/cumec
   	Δ		90

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