1. Ordinate of a 3-hour unit hydrograph for the catchment at t = 3 hours is

1. 1. 2.0 m³/s

   
   
   

   2. 3.0 m³/s

   
   
   

   3. 4.0 m³/s

   
   
   

   4. 5.0 m³/s

   
   
   

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1. View Hint View Answer Discuss in Forum

   Ordinate of 3 h unit UH =	Sum of ordinate of 1 hour UH
   	3

   
   

   Correct Option: C

   Ordinate of 3 h unit UH =	Sum of ordinate of 1 hour UH
   	3

   
   

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2. An isolated 4-hour storm occurred over a catchment as follows:
   
   The φ-index for the catchment is 10 mm/h. The estimated runoff depth from the catchment due to the above storm is

1. 1. 10 mm

   
   
   

   2. 16 mm

   
   
   

   3. 20 mm

   
   
   

   4. 23 mm

   
   
   

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1. View Hint View Answer Discuss in Forum

   Infiltration (i)
   If rainfall > φ index, i = φ index
   If rainfall < φ index, i = rainfall
   ∴ i_{1st hour} = 9 mm,
   i_{2nd hour} = 10 mm
   i_{3rd hour} = 10 mm
   i_{4th hour} = 9 mm.
   Runoff = rainfall – i
   = 0 + (28 – 10) + (12 – 10) + 0
   = 20 mm.

   Correct Option: C

   Infiltration (i)
   If rainfall > φ index, i = φ index
   If rainfall < φ index, i = rainfall
   ∴ i_{1st hour} = 9 mm,
   i_{2nd hour} = 10 mm
   i_{3rd hour} = 10 mm
   i_{4th hour} = 9 mm.
   Runoff = rainfall – i
   = 0 + (28 – 10) + (12 – 10) + 0
   = 20 mm.

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3. The rainfall on five successive days in a catchment were measured as 3, 8, 12, 6 and 2 cms. If the total runoff at the outlet from the catchment was 15 cm, the value of the findex (in mm/hour) is

1. 1. 0.0

   
   
   

   2. 1.04

   
   
   

   3. 1.53

   
   
   

   4. sufficient information not available

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   Windex =	P - R
   	t

   
   

   =	(3 + 8 + 12 + 6 + 2) - 15	= 3.2 cm/d
   	5

   
   

   φindex =	Pe - R	=	(8 + 12 + 6) - 15
   	te		3

   
   = 3.66 cm/d
   = 1.53 mm/h

   Correct Option: B

   
   

   Windex =	P - R
   	t

   
   

   =	(3 + 8 + 12 + 6 + 2) - 15	= 3.2 cm/d
   	5

   
   

   φindex =	Pe - R	=	(8 + 12 + 6) - 15
   	te		3

   
   = 3.66 cm/d
   = 1.53 mm/h

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4. A 6-hour Unit Hydrograph (UH) of a catchment is triangular in shape with a total time base of 36 hours and a peak discharge of 18 m³ /s. The area of the catchment (in sq. km) is

1. 1. 233

   
   
   

   2. 117

   
   
   

   3. 1.2

   
   
   

   4. Sufficient information not available

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Volume of run off = Area of hydrograph
   

   =	1	× 18 × (36 × 60 × 60) = 1166400 m²
   	2

   
   

   Depth of run off = 1 cm =	1	m
   	100

   
   Area of catchment
   

   =	Volume of run off	×	1
   	Depth of run off		(1000)²

   
   

   =	1166400	×	1	≈ 117 km²
   	1/100		(1000)²

   Correct Option: B

   
   Volume of run off = Area of hydrograph
   

   =	1	× 18 × (36 × 60 × 60) = 1166400 m²
   	2

   
   

   Depth of run off = 1 cm =	1	m
   	100

   
   Area of catchment
   

   =	Volume of run off	×	1
   	Depth of run off		(1000)²

   
   

   =	1166400	×	1	≈ 117 km²
   	1/100		(1000)²

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5. Group I lists a few devices while Group II provides information about their uses. Match the devices with their corresponding use.
   

   Group I	Group II
   P. Anemometer	1. Capillary potential of soil water
   Q. Hygrometer	2. Fluid velocity at a specific point in the flow stream
   R. Pitot Tube	3. Water vapour content of air
   S. Tensiometer	4. Wind speed

1. 1. P - 1; Q - 2; R - 3; S - 4

   
   
   

   2. P - 2; Q - 1; R - 4; S - 3

   
   
   

   3. P - 4; Q - 2; R - 1; S - 3

   
   
   

   4. P - 4; Q - 3; R - 2; S - 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

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