UPSC CAPF Archives

21. Which one of the following statements is correct ?

Which one of the following statements is correct ?

Iron sulphate and copper sulphate crystals have same number of water of crystallization

Iron sulphate and zinc sulphate crystals have same number of water of crystallization

Zinc sulphate and copper sulphate crystals have same number of water of crystallization

Iron sulphate, copper sulphate and zinc sulphate crystals each have same number of water of crystallization

This question was previously asked in

UPSC CAPF – 2015

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The statement that is correct is that Iron sulphate and zinc sulphate crystals have the same number of water of crystallization.

Crystals of hydrated salts contain a definite number of water molecules incorporated within their crystal structure. This water is called water of crystallization.
– Iron(II) sulphate (Ferrous sulphate) commonly crystallizes as the heptahydrate, FeSO4.7H2O.
– Copper(II) sulphate commonly crystallizes as the pentahydrate, CuSO4.5H2O.
– Zinc sulphate commonly crystallizes as the heptahydrate, ZnSO4.7H2O.

Comparing the number of water molecules:
– Iron sulphate (7) and Copper sulphate (5): Different number.
– Iron sulphate (7) and Zinc sulphate (7): Same number.
– Zinc sulphate (7) and Copper sulphate (5): Different number.
– Iron sulphate (7), Copper sulphate (5), and Zinc sulphate (7): Not all have the same number.

The number of water molecules in a hydrated crystal is specific to the compound and the conditions under which it crystallized. Common hydrated salts often encountered in chemistry include copper(II) sulphate pentahydrate (blue vitriol), iron(II) sulphate heptahydrate (green vitriol), and zinc sulphate heptahydrate (white vitriol).

22. Arrange the following National Parks of India from North to South dire

Arrange the following National Parks of India from North to South direction :
1. Indravati National Park
2. Nagarhole National Park
3. Corbett National Park
4. Madhav National Park
Select the correct answer using the code given below :

1, 3, 2, 4

2, 1, 4, 3

3, 4, 1, 2

2, 3, 4, 1

This question was previously asked in

UPSC CAPF – 2015

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Arranging the National Parks from North to South direction gives the sequence 3, 4, 1, 2.

The approximate locations of the National Parks are:
1. Indravati National Park: Chhattisgarh (Central India)
2. Nagarhole National Park: Karnataka (South India)
3. Corbett National Park: Uttarakhand (Northern India)
4. Madhav National Park: Madhya Pradesh (Central India)

Ordering by approximate latitude from North to South:
– Corbett National Park (Uttarakhand) is located in the northernmost part among the given options, in the Himalayan foothills.
– Madhav National Park (Madhya Pradesh) is located south of Uttarakhand, in central India.
– Indravati National Park (Chhattisgarh) is located south-east of Madhya Pradesh, further south in central India.
– Nagarhole National Park (Karnataka) is located in the southern part of India.

So the order from North to South is: Corbett (3) -> Madhav (4) -> Indravati (1) -> Nagarhole (2). This corresponds to the sequence 3, 4, 1, 2.

Knowing the geographical location of key places like National Parks is important for questions related to mapping and geography. Uttarakhand is part of the North Indian states. Madhya Pradesh and Chhattisgarh are in Central India, with MP generally slightly more northerly than Chhattisgarh. Karnataka is one of the Southern Indian states.

23. Which of the following statements with regard to the continental shelf

Which of the following statements with regard to the continental shelf is NOT correct ?

Shelves are absent close to plate boundaries

They are highly sedimented

They have abrupt falls towards the continental slopes

They are rich fishing areas

This question was previously asked in

UPSC CAPF – 2015

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The statement that is NOT correct with regard to the continental shelf is “Shelves are absent close to plate boundaries”.

– Continental shelves are the extended perimeter of each continent, associated with relatively shallow water compared to the open ocean.
– Statement B is correct: Continental shelves receive large amounts of sediment from rivers and coastal erosion, making them highly sedimented environments.
– Statement C is correct: The continental shelf ends at the shelf break, where there is a relatively abrupt increase in slope leading down to the continental slope and then the abyssal plain.
– Statement D is correct: Due to shallow waters, sunlight penetration, and nutrient-rich upwellings, continental shelves are highly productive ecosystems supporting abundant marine life, making them important fishing areas.
– Statement A is incorrect: While continental shelves are often narrower or even largely absent at active plate boundaries (like convergent or transform margins, e.g., along the west coast of South America or parts of California) compared to passive margins (e.g., along the east coast of North America), they are not universally absent at all plate boundaries. Some types of plate boundaries or specific locations on active margins can still have continental shelves, although perhaps limited in width. Saying they are “absent” is too strong a generalization and therefore incorrect.

Passive continental margins are created by rifting and continental breakup, leading to wide, stable shelves. Active continental margins, associated with plate boundaries, are tectonically active and can have narrow or non-existent shelves depending on the specific geological processes occurring (like subduction, collision, or transform faulting).

24. At what time between 2 and 3 will the hour and minute hands of a clock

At what time between 2 and 3 will the hour and minute hands of a clock be 12 minutes divisions apart ?

20 minutes past 2

24 5/11 minutes past 2

24 minutes past 2

24 12/13 minutes past 2

This question was previously asked in

UPSC CAPF – 2015

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The hour and minute hands of a clock will be 12 minutes divisions apart at 24 minutes past 2.

12 minute divisions on a clock correspond to an angle of 12 * 6 = 72 degrees (since there are 60 minute divisions in 360 degrees, so each division is 6 degrees).

At 2:00, the hour hand is at the 2 o’clock mark (10 minute divisions or 60 degrees from 12), and the minute hand is at the 12 o’clock mark (0 degrees). The angle between them is 60 degrees, with the hour hand ahead of the minute hand.

The minute hand moves at 6 degrees per minute, and the hour hand moves at 0.5 degrees per minute. The relative speed of the minute hand with respect to the hour hand is 6 – 0.5 = 5.5 degrees per minute.

Let ‘t’ be the number of minutes past 2 when the hands are 72 degrees apart.
At time ‘t’, the angle of the minute hand from 12 is 6t degrees.
At time ‘t’, the angle of the hour hand from 12 is (60 + 0.5t) degrees.

We want the absolute difference between these angles to be 72 degrees:
|6t – (60 + 0.5t)| = 72
|5.5t – 60| = 72

Two possible cases:
1) 5.5t – 60 = 72
5.5t = 132
t = 132 / 5.5 = 1320 / 55 = 264 / 11 = 24 minutes.
This corresponds to the minute hand being ahead of the hour hand by 72 degrees.

2) 5.5t – 60 = -72
5.5t = -12
t = -12 / 5.5. This is a negative time, which is not applicable for the time between 2 and 3. This situation (hour hand 72 degrees ahead of minute hand) occurs before 2:00.

Therefore, the only time between 2 and 3 when the hands are 12 minute divisions (72 degrees) apart is exactly 24 minutes past 2.

Clock problems often involve the relative speed of the hands. The minute hand completes a full circle (360 degrees) in 60 minutes, while the hour hand completes a full circle in 12 hours (720 minutes). The relative speed of the minute hand is 5.5 degrees/minute (or 11/2 degrees/minute) faster than the hour hand. This relative speed is key to calculating when specific relative positions occur.

25. ‘A’, ‘B’ and ‘C’ started their independent businesses with equal amoun

‘A’, ‘B’ and ‘C’ started their independent businesses with equal amounts of capital. During the first year ‘A’ made 10% profit, ‘B’ incurred 10% loss and ‘C’ made a profit of 5%. In the second year ‘A’ incurred 20% loss, ‘B’ made profit of 20% and ‘C’ made profit of 5%. Which of the following is FALSE at the end of second year ?

'C' is the richest

'A' is the poorest

'B' is the richest

'C' is richer than 'B'

This question was previously asked in

UPSC CAPF – 2015

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Let the initial capital be P.

At the end of the first year:
A’s capital = P + 10% of P = 1.1P
B’s capital = P – 10% of P = 0.9P
C’s capital = P + 5% of P = 1.05P

At the end of the second year:
A’s capital = 1.1P – 20% of 1.1P = 1.1P * (1 – 0.20) = 1.1P * 0.80 = 0.88P
B’s capital = 0.9P + 20% of 0.9P = 0.9P * (1 + 0.20) = 0.9P * 1.20 = 1.08P
C’s capital = 1.05P + 5% of 1.05P = 1.05P * (1 + 0.05) = 1.05P * 1.05 = 1.1025P

Comparing the capitals at the end of the second year:
C (1.1025P) > B (1.08P) > A (0.88P).
So, C is the richest, B is the second richest, and A is the poorest.

Let’s evaluate the statements:
A) ‘C’ is the richest: C’s capital (1.1025P) is the highest. This is TRUE.
B) ‘A’ is the poorest: A’s capital (0.88P) is the lowest. This is TRUE.
C) ‘B’ is the richest: B’s capital (1.08P) is not the highest (C is higher). This is FALSE.
D) ‘C’ is richer than ‘B’: C’s capital (1.1025P) is greater than B’s capital (1.08P). This is TRUE.

The statement which is FALSE at the end of the second year is ‘B’ is the richest.

This problem involves calculating successive percentage changes. When calculating a percentage change on a previous amount, the new amount becomes the base for the next percentage change. It’s important to calculate the value at the end of the first year before applying the percentage changes for the second year.

26. Hemocyanin is an oxygen-transport metalloprotein present in some inver

Hemocyanin is an oxygen-transport metalloprotein present in some invertebrate animals. This protein contains :

one copper atom

two copper atoms

one iron atom

one magnesium atom

This question was previously asked in

UPSC CAPF – 2015

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Hemocyanin is an oxygen-transport protein found in many invertebrates. It is a metalloprotein that contains copper atoms.

Hemocyanin utilizes copper atoms to bind and transport oxygen. Each oxygen binding site in hemocyanin contains two copper atoms.

In contrast, hemoglobin, the oxygen-transport protein in vertebrates, uses iron atoms (specifically, heme groups containing iron) for oxygen binding. Hemocyanin is responsible for the blue colour of the blood (hemolymph) in organisms like mollusks (e.g., snails, squids) and some arthropods (e.g., spiders, crustaceans) when oxygenated.

27. Match List-I with List-II and select the correct answer using the code

Match List-I with List-II and select the correct answer using the code given below the Lists :

List-I (Element)	List-II (Application)
A. Isotope of Uranium	1. Treatment of cancer
B. Isotope of Cobalt	2. Treatment of goitre
C. Isotope of Iodine	3. Treatment of secondary cancer
D. Isotope of Radium	4. Nuclear fuel

Code :

	A	B	C	D
(a)	3	2	1	4
(b)	4	2	1	3
(c)	4	1	2	3
(d)	3	1	2	4

A-3, B-2, C-1, D-4

A-4, B-2, C-1, D-3

A-4, B-1, C-2, D-3

A-3, B-1, C-2, D-4

This question was previously asked in

UPSC CAPF – 2015

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The correct matching is A-4, B-1, C-2, D-3.

– Isotopes of Uranium, specifically Uranium-235, are widely used as fuel in nuclear power plants because they are fissile.
– Cobalt-60 is a radioactive isotope of cobalt used in radiotherapy for the treatment of cancer.
– Iodine-131 is a radioactive isotope of iodine used in the treatment of goitre and thyroid cancer due to its accumulation in the thyroid gland.
– Radium isotopes, particularly Radium-226, were historically used in brachytherapy (a type of radiotherapy) for treating various cancers, including secondary cancers (metastases), although its use has largely been replaced by other isotopes like Cobalt-60 or Iridium-192 due to safety concerns and availability.

Radioisotopes have numerous applications in medicine, industry, and research, utilizing their radioactive properties or as tracers. Their specific application depends on the element’s chemical properties, the isotope’s half-life, and the type and energy of radiation emitted.

28. Match List-I with List-II and select the correct answer using the code

Match List-I with List-II and select the correct answer using the code given below the Lists :

List-I (Tropical Cyclone)	List-II (Location)
A. Cyclones	1. USA
B. Hurricanes	2. East Asia
C. Typhoons	3. Australia
D. Willy-willies	4. India

Code :

	A	B	C	D
(a)	4	2	1	3
(b)	4	1	2	3
(c)	3	1	2	4
(d)	3	2	1	4

A-4, B-2, C-1, D-3

A-4, B-1, C-2, D-3

A-3, B-1, C-2, D-4

A-3, B-2, C-1, D-4

This question was previously asked in

UPSC CAPF – 2015

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The correct matching is A-4, B-1, C-2, D-3.

– Tropical cyclones are known by different regional names.
– Cyclones is the term commonly used for tropical cyclones in the Indian Ocean region, including areas around India.
– Hurricanes is the term used in the Atlantic Ocean and the northeastern Pacific Ocean, affecting areas like the USA.
– Typhoons is the term used in the northwestern Pacific Ocean, affecting East Asia.
– Willy-willies is an informal term sometimes used for tropical cyclones in the Australian region.

These tropical storms are essentially the same phenomenon – intense low-pressure systems that form over warm tropical waters – but their nomenclature varies depending on the geographical basin where they originate. Other regional names include Bagyo (Philippines) and Severe Cyclonic Storm (India Meteorological Department classification).

29. Match List-I with List-II and select the correct answer using the code

Match List-I with List-II and select the correct answer using the code given below the Lists :

List-I (Landform feature)	List-II (Location)
A. Alluvial fans	1. Mountainous Areas
B. ‘V’-shaped valleys	2. Coasts
C. Deltas	3. Lower Reaches of the river
D. Ox-bow lakes	4. Mountain Foothills

Code :

	A	B	C	D
(a)	4	1	2	3
(b)	4	2	1	3
(c)	3	2	1	4
(d)	3	1	2	4

A-4, B-1, C-2, D-3

A-4, B-2, C-1, D-3

A-3, B-2, C-1, D-4

A-3, B-1, C-2, D-4

This question was previously asked in

UPSC CAPF – 2015

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The correct matching is A-4, B-1, C-2, D-3.

– Alluvial fans are fan-shaped deposits of sediment formed at the base of a mountain or at the mouth of a narrow valley where a stream flows onto a flatter surface, characteristic of mountain foothills.
– ‘V’-shaped valleys are formed by the erosional action of rivers in their upper course, typically found in mountainous areas where downcutting is dominant.
– Deltas are landforms created by deposition of sediment that is carried by a river as the flow leaves its mouth and enters slower-moving or standing water, usually the sea or a lake. They are characteristic features of the lower reaches of rivers, often found at coasts.
– Ox-bow lakes are crescent-shaped lakes formed in abandoned meander loops of a river, typically found in the flat, alluvial plains of the river’s lower course.

The lower reaches of a river are often characterized by meandering, the formation of ox-bow lakes, and ultimately the development of a delta at its mouth on the coast. V-shaped valleys are associated with the youthful stage and mountainous terrain, while alluvial fans form where mountain streams transition to flatter land.

30. What is the respective ratio of students studying in Standard III of s

What is the respective ratio of students studying in Standard III of schools A and B together to those studying in Standard VI of schools C and D together ?

53 : 52

43 : 47

25 : 27

39 : 38

This question was previously asked in

UPSC CAPF – 2015

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The question asks for the ratio of students in specific standards across different schools. This requires a data source listing the number of students per standard for each school. Assuming such data was available, the steps to solve this are outlined below. The provided option A is the correct answer based on the original source data not included here.

To determine the required ratio:
1. Calculate the total number of students studying in Standard III from schools A and B together (Sum of students in Standard III of School A and Standard III of School B).
2. Calculate the total number of students studying in Standard VI from schools C and D together (Sum of students in Standard VI of School C and Standard VI of School D).
3. Express the result from step 1 to the result from step 2 as a ratio and simplify it to its lowest terms.
The necessary student count data per standard per school is missing from the prompt.

Ratios are used to compare the relative size of two quantities. They are often expressed in the form A:B. Simplifying a ratio involves dividing both parts by their greatest common divisor.