Precipitation MCQ Quiz in मल्याळम - Objective Question with Answer for Precipitation - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

Interception is the process by which precipitation is temporarily stored on the "surface" of vegetation, soil, or other objects before it reaches the ground and evaporates directly.

Transpiration, on the other hand, is the process by which water is lost from the leaves of plants "through tiny pores called stomata".

Additional Information

Infiltration:

(i) When water is applied to the surface of the soil, a part of it seeps into the soil. This movement of water through soil is called infiltration.

(ii) This infiltration water used in increasing the moisture of soil up to field capacity and extra water goes down and becomes ground water.

Transpiration:

(i) Transpiration is the process by which water leaves the body of a living plant and reaches the atmosphere as water vapour. The transpiration occurs when the process of photosynthesis is run. 

Evaporation:

(i) Evaporation is the process in which a liquid changes to the gaseous state at the free surface, below the boiling point through the transfer of heat energy.

Concept

The optimum number of rain gauges is given by

\(N = {\left( {\frac{{{C_v}}}{\epsilon}} \right)^2} \)

Where,

Cv is the coefficient of variation

ϵ is error percentage

The coefficient of variation is given by

\({C_v} = \frac{{{\sigma _{n - 1}}}}{{\bar P}}\)

Where,

\(\bar P\) is average rainfall

σ is the standard deviation

Calculation:

Given,

Number of Rain gauge = 9, error = 10 %, Cv = ?

The optimum number of rain gauges

\(N = {\left( {\frac{{{C_v}}}{\epsilon}} \right)^2} \)

\(9= {\left( {\frac{{{C_v}}}{10}} \right)^2} \)

C_V = 30

Concept:

Water budget method:

By applying mass balance around water body and by writing water budget equation, evaporation is estimated.

Water budget equation around water body is

Σinflow - Σoutflow = ± ΔS

Σinflow includes rainfall and runoff

Σoutflow includes water used, evaporation and seepage

ΔS = change in storage 

ΔS is positive when water level rises.

ΔS is negative when water level falls.    

Calculation:

Area of Lake = 15 km2, Precipitation = 700 mm/year = 0.7 m

Average inflow Qin = 1.4 m3/s = 1.4 × 60 × 60 × 24 × 365 / (15 × 10⁶) = 2.943 m

Average outflow Qout = 1.6 m3/s = 1.6 × 60 × 60 × 24 × 365 / (15 × 106) = 3.364 m

As mentioned there is no net water exchange between the lake and the groundwater (ΔS = 0)

⇒ Inflow = Outflow

⇒ 2.943 m + 0.7 m = Evaporation + 3.364

∴ Evaporation = 0.279 m = 279 mm

Concept:

The diameter of the receiver tipping bucket rain gauge is 30 cm (300 mm).

A tipping bucket rain gauge consists of a circular collector funnel with a brass or gunmetal rim that directs the rain into a pair of tipping buckets. After rainwater has filled one side of the bucket with a small amount of rain, the assembly tips. An electrical pulse is generated on each tilt and is recorded to provide data of rain fallen with time.

It has 30 cm (300 mm) diameter sharp-edged receiver and at the end of the receiver is provided with a funnel. Pair of buckets are pivoted under this funnel in such a way that when one bucket receives precipitation (rainfall) of 0.25 mm, it tips discharging its rainfall into the container, bringing the other bucket under the funnel.

Forms of precipitation 

Explanation:

The mean annual PET (in cm) over various parts of the country is shown in the form of isopleths. lsopleths are the line on a map through places having equal depth of “evapotranspiration”.

Additional Information

Explanation:

Symons rain gauge:

• The Symons rain gauge is a non-recording rain gauge used by the meteorological department of India.
• In Symons rain gauge, rainwater is collected in a cylindrical bottle which is measured manually on daily basis.
• It mainly consists of a funnel and a receiver.
• It consists of a circular receiving area that is 127mm in diameter and connected to a funnel.
• The rim of the collector is set at a height of 30.5 cm above ground level.
• The funnel discharges the rainfall into receiving funnel.
• Funnels and receiving vessels are placed in metal containers.
• The receiving bottle normally holds a maximum of 10 cm of rain and in case of heavy rainfall, the measurement must be done frequently.
• The last reading should be taken at 8:30 am and the sum of the previous readings in the last 24 hours should be recorded as the total rainfall for the day.

The correct answer is Frontal rainfall.

Key Points

• Frontal rainfall occurs when: Two air masses meet, one a warm air mass and one a cold air mass. The lighter, less dense, warm air is forced to rise over the denser, cold air.
  • This causes the warm air to cool and begin to condense. As the warm air is forced to rise further condensation occurs and rain is formed. Cyclonic rainfall is another term for frontal rainfall.
• Orographic or relief rainfall is produced when moist air is lifted as it moves over a mountain range. As the air rises and cools, orographic clouds form and serve as the source of the precipitation, most of which falls upwind of the mountain ridge.
• When the land warms up, it heats the air above it.
  • This causes the air to expand and rise. As the air rises it cools and condenses. If this process continues then rain will fall. This type of rainfall is called convectional rainfall.

Concept:

Estimation of missing rainfall data:

i) Arithmetic Mean Method:

It is used if the normal annual precipitation of the nearby stations is 10% of within the normal annual precipitation at station X.

\({{\rm{P}}_{\rm{x}}} = \frac{{{{\rm{P}}_1} + {{\rm{P}}_2} + {{\rm{P}}_2} + \ldots {{\rm{P}}_{\rm{n}}}}}{{\rm{n}}}\)

ii) Normal Ratio Method:

It is used if normal precipitation at any of the selected stations differ by more than 10% of selected station.

\({{\rm{P}}_{\rm{x}}} = \frac{{{{\rm{N}}_{\rm{x}}}}}{{\rm{N}}}\left[ {\frac{{{{\rm{P}}_1}}}{{{{\rm{N}}_1}}} + \frac{{{{\rm{P}}_2}}}{{{{\rm{N}}_2}}} + \ldots \frac{{{{\rm{P}}_{\rm{n}}}}}{{{{\rm{N}}_{\rm{n}}}}}} \right]\)

Where,

P1, P2, …Pn are respective annual precipitation for different stations.

N1, N2, …Nn are respective average annual/normal precipitation for different stations.

Concept:

Pan coefficient (CP) = Pond evaporation (E) / Pan evaporation (EP)

Calculation:

Given,

Pan coefficient (C_P) = 0.70

Pan evaporation (E_P) = 45 mm

Pond evaporation (E) = C_P x E_P

E = 0.70 x 45 = 31.5 mm