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Ice Nucleation and Ice Crystal Growth Questions and Answers

81.

Assertion (A): Falling snow drifts much further horizontally than rain from the same cloud.
Reason (R): Its low fall speed leaves it exposed to the wind for far longer.

Answer: D

Both are true and the reason gives the cause, and the displacement can reach many kilometres in strong wind. Radar estimates of snowfall must account for this drift explicitly. The measured snowfall may originate far from where it is recorded.

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Answer: A

Such probes image particles as they pass through a laser beam and size them directly. Shattering of large particles on the probe inlet has been a persistent source of overcounting in older datasets. Reanalysis of historical flights continues on this basis.

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83.

Assertion (A): Older aircraft measurements of ice crystal concentration may have been overestimates.
Reason (R): Large particles shattering on the probe inlet were counted as many small ones.

Answer: B

Both are true and the reason gives the artefact, and modified probe tips have since reduced it substantially. Some historical reports of extreme concentrations are now regarded with caution. Instrumental artefacts can mimic real secondary production.

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Answer: C

The lattice match allows it to nucleate ice at relatively warm supercooled temperatures. Whether seeding produces a measurable increase in precipitation remains contested. Randomised trials have produced mixed and contested results.

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85.

Consider the following statements about glaciogenic seeding.
1. It aims to add ice nuclei to a supercooled cloud.
2. It can only work where supercooled liquid is present.
3. It is effective in clouds that are already fully glaciated.

Answer: D

The first two are correct and define the technique. The third is precisely where seeding cannot help, since there is no supercooled liquid left for added nuclei to act upon. Seeding can only redistribute water that is already liquid.

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86.

Three of the following are required for the Wegener Bergeron Findeisen process to operate. Which is the odd one out?

Answer: A

The process is thermodynamic and needs no updraught, though one helps sustain the mixed state. The other three are genuine requirements without which no vapour transfer can occur at all. An updraught sustains the process but does not drive it.

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Answer: B

With no liquid remaining there is no reservoir to draw from and the process stops. Further crystal growth then depends on the ambient ice supersaturation alone. Growth then proceeds far more slowly than during the mixed stage.

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88.

Saturation vapour pressure over ice at minus twelve degrees is lower than over water by roughly

Answer: C

The difference peaks near a tenth of the water value at that temperature. That modest looking excess is nonetheless enough to transfer the cloud's liquid water into ice within minutes. Small vapour pressure differences act over very short timescales.

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89.

Arrange these temperatures in the order a rising parcel encounters them in a deep convective cloud.
1. minus forty degrees Celsius
2. zero degrees Celsius
3. the tropopause
4. minus fifteen degrees Celsius

Answer: D

The parcel passes from all liquid through the mixed phase region into complete glaciation. Each regime contributes a different microphysical process to the precipitation the storm produces. Each layer contributes something different to the outcome.

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Answer: A

Correct answer: zero and minus forty degrees Celsius

Above zero everything is liquid; below about minus 38 to minus 40 homogeneous freezing removes the last supercooled droplets. Between those two limits ice and liquid can coexist, and that is the layer where riming, the Wegener–Bergeron–Findeisen process, secondary ice production and aircraft icing all occur.

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