Meteorology and Dispersion Questions and Answers

Q: The height of tropopause over the equator is about..........km.

Over the equator the tropopause is about 16 km high.The tropopause, the boundary between the troposphere and the stratosphere, is highest over the equator, about 16-18 km, because strong heating and convection there lift the troposphere. It is lowest over the poles, about 8 km.8 km is the polar height, and 10 km and 12 km are typical mid-latitude values, so none of them is the equatorial figure.

Q: Dry adiabatic lapse rate is about ...............°C/km.

The dry adiabatic lapse rate is about 9.8°C/km.A parcel of unsaturated air that rises expands and cools without exchanging heat with its surroundings. The rate of cooling is\[ \begin{aligned} \Gamma_{d} &= \frac{g}{c_{p}} = \frac{9.81}{1004}\ \text{K/m} \\ &\approx 9.8\ ^{\circ}\text{C/km} \end{aligned} \]Saturated air cools more slowly, about 4.5-6°C/km, because condensation releases latent heat, so 4.5 is the wet (saturated) rate. About 6.5°C/km is the average environmental lapse rate. 7 and 12 are not the dry adiabatic value.

Q: The dispersal of pollutants in air is dependent on :

Dispersal depends on wind speed and thermal stability both.Wind speed sets how fast a plume is carried away and diluted horizontally: the ground-level concentration from a source falls roughly in proportion to the wind speed. Thermal stability, set by the vertical temperature profile (the lapse rate), decides how much the plume can mix vertically. An unstable atmosphere with a superadiabatic lapse rate gives strong vertical mixing, while a stable layer or an inversion traps pollutants near the ground.That is why dispersion models such as the Gaussian plume use both a wind speed and a stability class (Pasquill A-F). 'Wind speed only' and 'thermal stability only' each leave out half of the picture, and humidity is not a primary control on dispersion.

Q: Hydrostatic equation giving the relation between height and pressure at a given temperature :

The hydrostatic equation is \(\frac{dp}{dz} = -\rho g\).Take a thin horizontal slab of air of thickness \(dz\) and unit area. Its weight is \(\rho g\,dz\), and in hydrostatic balance this weight is supported by the difference in pressure between its lower and upper faces, so the pressure falls with height by exactly that weight:\[ \begin{aligned} p(z) - p(z+dz) &= \rho g\,dz \\ \frac{dp}{dz} &= -\rho g \end{aligned} \]The minus sign says pressure decreases upward. The temperature enters only through the density, since \(\rho = p/RT\) for air. The options containing T in place of g are dimensionally wrong (pressure per metre must equal density times an acceleration), and \(-\rho/g\) divides where it should multiply.

Q: In an adiabatic process the ................ remains unchanged.

In an adiabatic process the entropy remains unchanged.Adiabatic means no heat crosses the boundary of the system: \(dQ = 0\). For a reversible process the change in entropy is \(dS = dQ/T\), so \(dS = 0\) and the process is isentropic. In meteorology a rising dry air parcel is treated this way, and its conserved quantity is the potential temperature, which is a measure of entropy.Temperature, density and pressure all change in an adiabatic process: a rising parcel expands as pressure falls, its density drops, and it cools at the dry adiabatic lapse rate because it does work on its surroundings. Keeping temperature constant would describe an isothermal process instead.

Q: If 10 calories of heat is added to a system which does not change its specific volume, then as per the first law of thermodynamics the change in internal energy of the system will be :

The change in internal energy is positive.The first law of thermodynamics is \(dQ = dU + p\,dV\). If the specific volume does not change, \(dV = 0\), so no work is done and all the heat added goes into internal energy:\[ \begin{aligned} dU &= dQ - p\,dV \\ &= 10 - 0 = +10\ \text{cal} \end{aligned} \]The internal energy rises by 10 calories, which shows up as a rise in temperature. It cannot be zero, since heat was added and none left as work, and it cannot be negative, since that would need the system to lose energy. So 'either zero or negative' is also wrong.

Q: An inversion layer is seen between 1000 m to 1500 m. If the temperature at 1000 m is 280 K, what will be temperature at the top of the layer (negative lapse rate 1.0°C/100 m) ?

The temperature at the top of the layer is 285 K.In an inversion the temperature rises with height, which is what a negative lapse rate means. Here it rises by 1.0 °C for every 100 m through a layer 500 m thick:\[ \begin{aligned} \Delta T &= \frac{1.0}{100} \times (1500 - 1000) \\ &= 5\ \text{K} \\ T_{top} &= 280 + 5 = 285\ \text{K} \end{aligned} \]A change of 1 °C equals a change of 1 K, so no conversion is needed. 278 K would mean the air cooled with height, the opposite of an inversion; 290 K uses twice the gradient, and 284 K does not follow from the figures given.

Q: Subtropical jet stream with its core of strong winds (exceeding 60 kt) at nearly 200 hPa ...................... .

The subtropical jet stream moves equatorward in winter.The subtropical westerly jet sits at the poleward edge of the Hadley cell, near 200 hPa. Its position follows the sun: in the northern winter, when the Hadley circulation and the thermal gradient shift south, the jet moves towards the equator and lies over northern India at about 25-30°N, south of the Himalaya. In summer it shifts poleward to about 40°N, north of the Tibetan plateau, and its withdrawal from India is linked to the onset of the monsoon.So it does not stay in the same place through the year, it does not move equatorward in summer, and it does not move poleward in winter; each of those reverses the seasonal shift.

Q: In a dry atmosphere under adiabatic condition the lapse rate is ..............

The dry adiabatic lapse rate is − g/cp, that is \(dT/dz = -g/c_{p}\).For a rising parcel of dry air the first law with \(dQ = 0\) gives \(c_{p}\,dT = \alpha\,dp\), and the hydrostatic equation gives \(\alpha\,dp = -g\,dz\). Together:\[ \begin{aligned} c_{p}\,dT &= -g\,dz \\ \frac{dT}{dz} &= -\frac{g}{c_{p}} \approx -9.8\ \text{K/km} \end{aligned} \]The specific heat at constant pressure appears because the parcel expands freely as it rises. Using the specific heat at constant volume, g/cv, would ignore the work of expansion. The forms involving R are not dimensionally a temperature gradient.

The height of tropopause over the equator is about..........km.

A.10
B.8
C.16
D.12

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

Over the equator the tropopause is about 16 km high.

The tropopause, the boundary between the troposphere and the stratosphere, is highest over the equator, about 16-18 km, because strong heating and convection there lift the troposphere. It is lowest over the poles, about 8 km.

8 km is the polar height, and 10 km and 12 km are typical mid-latitude values, so none of them is the equatorial figure.

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Dry adiabatic lapse rate is about ...............°C/km.

A.7
B.4.5
C.9.8
D.12

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The dry adiabatic lapse rate is about 9.8°C/km.

A parcel of unsaturated air that rises expands and cools without exchanging heat with its surroundings. The rate of cooling is

\[ \begin{aligned} \Gamma_{d} &= \frac{g}{c_{p}} = \frac{9.81}{1004}\ \text{K/m} \\ &\approx 9.8\ ^{\circ}\text{C/km} \end{aligned} \]

Saturated air cools more slowly, about 4.5-6°C/km, because condensation releases latent heat, so 4.5 is the wet (saturated) rate. About 6.5°C/km is the average environmental lapse rate. 7 and 12 are not the dry adiabatic value.

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Tropical cyclones are most frequently observed in the Bay of Bengal during :

A.July - August
B.Jan. - Feb.
C.March - May
D.Oct. - Dec.

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

Tropical cyclones form most often over the Bay of Bengal in October to December, the post-monsoon season.

The bay has two cyclone seasons, April-May before the monsoon and October-December after it, and the post-monsoon season is the more active: the sea surface is still warm after the summer, the monsoon trough has moved south over the bay, and vertical wind shear is low once the strong monsoon upper-level easterlies weaken. Cyclones such as Phailin (October 2013) and Hudhud (October 2014) are typical of this season.

July-August is the peak of the south-west monsoon, when strong shear favours monsoon depressions rather than cyclones. January-February is too cool and dry. March-May is the secondary season, active mainly in May, but it produces fewer storms than October-December.

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The dispersal of pollutants in air is dependent on :

A.Wind speed only
B.Wind speed and thermal stability both
C.Thermal stability only
D.Wind speed and humidity

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

Dispersal depends on wind speed and thermal stability both.

Wind speed sets how fast a plume is carried away and diluted horizontally: the ground-level concentration from a source falls roughly in proportion to the wind speed. Thermal stability, set by the vertical temperature profile (the lapse rate), decides how much the plume can mix vertically. An unstable atmosphere with a superadiabatic lapse rate gives strong vertical mixing, while a stable layer or an inversion traps pollutants near the ground.

That is why dispersion models such as the Gaussian plume use both a wind speed and a stability class (Pasquill A-F). 'Wind speed only' and 'thermal stability only' each leave out half of the picture, and humidity is not a primary control on dispersion.

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Hydrostatic equation giving the relation between height and pressure at a given temperature :

A.\(\frac{dp}{dz} = -\rho T\)
B.\(\frac{dp}{dz} = \frac{-\rho}{T}\)
C.\(\frac{dp}{dz} = -\rho g\)
D.\(\frac{dp}{dz} = \frac{-\rho}{g}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The hydrostatic equation is \(\frac{dp}{dz} = -\rho g\).

Take a thin horizontal slab of air of thickness \(dz\) and unit area. Its weight is \(\rho g\,dz\), and in hydrostatic balance this weight is supported by the difference in pressure between its lower and upper faces, so the pressure falls with height by exactly that weight:

\[ \begin{aligned} p(z) - p(z+dz) &= \rho g\,dz \\ \frac{dp}{dz} &= -\rho g \end{aligned} \]

The minus sign says pressure decreases upward. The temperature enters only through the density, since \(\rho = p/RT\) for air. The options containing T in place of g are dimensionally wrong (pressure per metre must equal density times an acceleration), and \(-\rho/g\) divides where it should multiply.

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In an adiabatic process the ................ remains unchanged.

A.Temperature
B.Entropy
C.Density
D.Pressure

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

In an adiabatic process the entropy remains unchanged.

Adiabatic means no heat crosses the boundary of the system: \(dQ = 0\). For a reversible process the change in entropy is \(dS = dQ/T\), so \(dS = 0\) and the process is isentropic. In meteorology a rising dry air parcel is treated this way, and its conserved quantity is the potential temperature, which is a measure of entropy.

Temperature, density and pressure all change in an adiabatic process: a rising parcel expands as pressure falls, its density drops, and it cools at the dry adiabatic lapse rate because it does work on its surroundings. Keeping temperature constant would describe an isothermal process instead.

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If 10 calories of heat is added to a system which does not change its specific volume, then as per the first law of thermodynamics the change in internal energy of the system will be :

A.Zero
B.Positive
C.Negative
D.Either (A) or (C)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The change in internal energy is positive.

The first law of thermodynamics is \(dQ = dU + p\,dV\). If the specific volume does not change, \(dV = 0\), so no work is done and all the heat added goes into internal energy:

\[ \begin{aligned} dU &= dQ - p\,dV \\ &= 10 - 0 = +10\ \text{cal} \end{aligned} \]

The internal energy rises by 10 calories, which shows up as a rise in temperature. It cannot be zero, since heat was added and none left as work, and it cannot be negative, since that would need the system to lose energy. So 'either zero or negative' is also wrong.

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An inversion layer is seen between 1000 m to 1500 m. If the temperature at 1000 m is 280 K, what will be temperature at the top of the layer (negative lapse rate 1.0°C/100 m) ?

A.290 K
B.278 K
C.284 K
D.285 K

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The temperature at the top of the layer is 285 K.

In an inversion the temperature rises with height, which is what a negative lapse rate means. Here it rises by 1.0 °C for every 100 m through a layer 500 m thick:

\[ \begin{aligned} \Delta T &= \frac{1.0}{100} \times (1500 - 1000) \\ &= 5\ \text{K} \\ T_{top} &= 280 + 5 = 285\ \text{K} \end{aligned} \]

A change of 1 °C equals a change of 1 K, so no conversion is needed. 278 K would mean the air cooled with height, the opposite of an inversion; 290 K uses twice the gradient, and 284 K does not follow from the figures given.

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Subtropical jet stream with its core of strong winds (exceeding 60 kt) at nearly 200 hPa ...................... .

A.Moves equatorward in summer
B.Remains at same location in winter and summer
C.Moves equatorward in winter
D.Moves poleward in winter

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The subtropical jet stream moves equatorward in winter.

The subtropical westerly jet sits at the poleward edge of the Hadley cell, near 200 hPa. Its position follows the sun: in the northern winter, when the Hadley circulation and the thermal gradient shift south, the jet moves towards the equator and lies over northern India at about 25-30°N, south of the Himalaya. In summer it shifts poleward to about 40°N, north of the Tibetan plateau, and its withdrawal from India is linked to the onset of the monsoon.

So it does not stay in the same place through the year, it does not move equatorward in summer, and it does not move poleward in winter; each of those reverses the seasonal shift.

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10.

In a dry atmosphere under adiabatic condition the lapse rate is ..............

A.− g/cv
B.− g/cp
C.− g/cpR
D.− gcp/R

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The dry adiabatic lapse rate is − g/c_p, that is \(dT/dz = -g/c_{p}\).

For a rising parcel of dry air the first law with \(dQ = 0\) gives \(c_{p}\,dT = \alpha\,dp\), and the hydrostatic equation gives \(\alpha\,dp = -g\,dz\). Together:

\[ \begin{aligned} c_{p}\,dT &= -g\,dz \\ \frac{dT}{dz} &= -\frac{g}{c_{p}} \approx -9.8\ \text{K/km} \end{aligned} \]

The specific heat at constant pressure appears because the parcel expands freely as it rises. Using the specific heat at constant volume, g/c_v, would ignore the work of expansion. The forms involving R are not dimensionally a temperature gradient.

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