UPSC CAPF Archives

31. Which one of the following layers of atmosphere has high concentration

Which one of the following layers of atmosphere has high concentration of ions ?

Stratosphere

Exosphere

Thermosphere

Troposphere

This question was previously asked in

UPSC CAPF – 2016

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The Thermosphere is the layer of the atmosphere that has a high concentration of ions.

The Thermosphere contains a region called the ionosphere, where solar radiation causes atoms and molecules to become ionized (gain or lose electrons), creating a high concentration of ions and free electrons. This ionization is responsible for phenomena like the aurora borealis/australis and facilitates the reflection of radio waves used for long-distance communication.

The atmosphere is divided into several layers based on temperature profiles: Troposphere, Stratosphere, Mesosphere, Thermosphere, and Exosphere.
– Troposphere: Lowest layer, where weather occurs.
– Stratosphere: Contains the ozone layer. Temperature increases with altitude.
– Mesosphere: Temperature decreases with altitude. Meteors burn up here.
– Thermosphere: Temperature increases significantly with altitude due to absorption of high-energy solar radiation. It includes the ionosphere.
– Exosphere: Outermost layer, where the atmosphere thins out into space.

32. Consider the following Statements and Conclusions: Statements : Moha

Consider the following Statements and Conclusions:
Statements :

Mohan is a good sportsman.
Sportsmen are healthy.

Conclusions :

I. All healthy persons are sportsmen.
II. Mohan is healthy.

Which one of the following is correct?

Only Conclusion I follows

Only Conclusion II follows

Both Conclusion I and Conclusion II follow

Neither Conclusion I nor Conclusion II follows

This question was previously asked in

UPSC CAPF – 2016

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Only Conclusion II follows from the given statements.

The statements are: (1) Mohan is a good sportsman. (2) Sportsmen are healthy.
From statement (2), we know that the property of being ‘healthy’ applies to all ‘sportsmen’. Statement (1) tells us that Mohan is a ‘sportsman’. Since Mohan belongs to the category ‘sportsmen’, and all members of this category are ‘healthy’ according to statement (2), it logically follows that Mohan is ‘healthy’. Thus, Conclusion II is valid.
Conclusion I states: “All healthy persons are sportsmen.” This is the converse of statement (2). Statement (2) “All Sportsmen are Healthy” (S -> H) does not imply “All Healthy persons are Sportsmen” (H -> S). There can be healthy persons who are not sportsmen (e.g., a healthy artist or student). Therefore, Conclusion I does not follow logically.

This is another example of deductive reasoning or syllogism.
Statement 1: Mohan ∈ Sportsmen
Statement 2: Sportsmen ⊆ Healthy Persons (All S are H)
From these, it follows that Mohan ∈ Healthy Persons. (Conclusion II)
The converse of “All S are H” is “All H are S”, which is not necessarily true.

33. Which one of the following figures correctly represents the relations

Which one of the following figures correctly represents the relations between Jupiter, Mars and Planets?

This question was previously asked in

UPSC CAPF – 2016

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The figure representing two separate circles within a larger circle correctly represents the relationship between Jupiter, Mars, and Planets. Although the description mentions “overlapping circles”, option B typically represents this relationship in Venn diagrams. Assuming the diagram for option B visually shows two distinct, non-overlapping circles inside a larger one, then it is the correct representation.

Jupiter is a planet, and Mars is a planet. Jupiter and Mars are distinct celestial bodies and do not overlap. Therefore, the set of Jupiter is a subset of the set of Planets, and the set of Mars is a subset of the set of Planets. The sets of Jupiter and Mars are mutually exclusive (they do not share members). This is best represented by a large circle encompassing two smaller, separate circles.

Venn diagrams are used to visually represent the relationships between different sets. The largest circle represents the most inclusive set (Planets). The smaller circles represent the subsets (Jupiter, Mars). Since Jupiter and Mars are individual, distinct planets, their circles should be separate from each other within the larger ‘Planets’ circle. Based on typical Venn diagram options, option B’s *visual* representation, despite the potentially inaccurate textual description “overlapping circles”, is likely intended to show this correct relationship.

34. Premises: All apples are golden in colour. No golden coloured things a

Premises: All apples are golden in colour. No golden coloured things are cheap.
Conclusions:

I. All apples are cheap.
II. Golden coloured apples are not cheap.

Which of the following conclusions follow logically?

Only I

Only II

Both I and II

Neither I nor II

This question was previously asked in

UPSC CAPF – 2015

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The correct conclusion that follows logically is “Golden coloured apples are not cheap.”

The premises are: (1) All apples are golden in colour. (2) No golden coloured things are cheap.
From premise (1), we know that the set of apples is a subset of golden coloured things. From premise (2), we know that the set of golden coloured things and the set of cheap things have no overlap.
Combining these, since all apples are golden coloured, and no golden coloured things are cheap, it follows that no apples are cheap.
Conclusion I states: “All apples are cheap.” This is directly contradicted by the logical consequence of the premises (“No apples are cheap”). Thus, Conclusion I does not follow.
Conclusion II states: “Golden coloured apples are not cheap.” Apples that are golden coloured are, by definition, golden coloured things. According to premise (2), no golden coloured things are cheap. Therefore, golden coloured apples are not cheap. This conclusion follows directly from the premises.

This is a syllogistic reasoning problem. Representing the sets: A = Apples, G = Golden coloured things, C = Cheap things.
Premise 1: All A are G.
Premise 2: No G are C. This implies No C are G, and if something is G, it is not C.
Conclusion I: All A are C. (False)
Conclusion II: Apples that are G are not C. Since all A are G, this is equivalent to saying A are not C. From “All A are G” and “No G are C”, we deduce “No A are C” (which means A are not C). Therefore, Conclusion II follows.

35. Premises: All film stars are playback singers. All film directors are

Premises: All film stars are playback singers. All film directors are film stars.
Conclusions:

I. All film directors are playback singers.
II. Some film stars are film directors.

Which of the following conclusions follow logically?

Only I

Only II

Both I and II

Neither I nor II

This question was previously asked in

UPSC CAPF – 2015

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Both Conclusion I and Conclusion II follow logically from the given premises.

Premise 1: All film stars are playback singers. (FS ⊆ PS)
Premise 2: All film directors are film stars. (FD ⊆ FS)
Conclusion I: All film directors are playback singers.
From Premise 2, the set of film directors (FD) is a subset of the set of film stars (FS). From Premise 1, the set of film stars (FS) is a subset of the set of playback singers (PS). If A ⊆ B and B ⊆ C, then A ⊆ C. Therefore, FD ⊆ PS. This means all film directors are playback singers. Conclusion I is true.
Conclusion II: Some film stars are film directors.
Premise 2 states that all film directors are film stars (FD ⊆ FS). This is a universal affirmative statement (‘All A are B’). In traditional logic, a universal affirmative statement ‘All A are B’ implies the particular affirmative statement ‘Some B are A’, provided that the set A is not empty. Assuming there is at least one film director (which is standard in such problems unless specified otherwise), then there is at least one member in the set FD. Since every member of FD is also in FS, there must be at least one member in FS that is also in FD. Thus, some film stars are film directors. Conclusion II is true.

This problem demonstrates the transitivity of the subset relationship (for Conclusion I) and the conversion of a universal affirmative statement (for Conclusion II, assuming existential import). If we strictly adhere to modern formal logic without existential import for universal statements, ‘All A are B’ does not necessarily imply ‘Some B are A’ if the set A is empty. However, in typical syllogism questions, especially in the context of competitive exams like UPSC, the assumption of existential import for the subject class of universal statements is generally followed unless explicitly negated. Therefore, ‘All film directors are film stars’ implies ‘Some film stars are film directors’ as long as there exists at least one film director.

36. The next three (03) items have two premises and two conclusions. If th

The next three (03) items have two premises and two conclusions. If the premises are assumed to be true (irrespective of factuality), then, in respect of each of the items given below, which of the following conclusions follow logically ?
Premises: All metals are grey in colour. Some metals are heavy.
Conclusions:

I. All heavy metals are grey in colour.
II. All light metals are not grey in colour.

Only I

Only II

Both I and II

Neither I nor II

This question was previously asked in

UPSC CAPF – 2015

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Only Conclusion I follows logically from the given premises.

Premise 1: All metals are grey in colour. (M ⊆ G)
Premise 2: Some metals are heavy. (Some M are H)
Conclusion I: All heavy metals are grey in colour.
“Heavy metals” refers to the set of things that are both heavy and metal (H ∩ M). According to Premise 1, anything that is a metal is grey (M ⊆ G). Therefore, if something is a heavy metal, it must first be a metal, and thus it must be grey. So, H ∩ M ⊆ G. Conclusion I logically follows.
Conclusion II: All light metals are not grey in colour.
“Light metals” refers to the set of things that are metals but not heavy (M ∩ not-H). According to Premise 1, all metals are grey (M ⊆ G). This means *any* metal, regardless of whether it is heavy or light, must be grey. Therefore, light metals are grey. Conclusion II, stating that light metals are *not* grey, contradicts Premise 1 and is false.

This is a standard problem involving categorical syllogisms. Using Venn diagrams can also help visualize the relationships. Draw three overlapping circles for Metals (M), Grey (G), and Heavy (H). Premise 1 (All M are G) means the part of M outside G is empty. Premise 2 (Some M are H) means there is at least one element in the intersection of M and H. Conclusion I (All H∩M are G) means the part of (H∩M) outside G is empty, which is true because M is entirely within G. Conclusion II (All M∩not-H are not-G) means the part of (M∩not-H) inside G is empty, which is false because M is entirely within G, meaning M∩not-H must also be within G.

37. In a radioactive decay of a nucleus, an electron is also emitted. This

In a radioactive decay of a nucleus, an electron is also emitted. This may happen due to the fact that :

electrons are present inside a nucleus

an electron is created at the time of conversion of a neutron into proton

an electron is created at the time of conversion of a proton into a neutron

electrons need to be emitted for conservation of momentum

This question was previously asked in

UPSC CAPF – 2015

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In radioactive beta minus (β⁻) decay, an electron is emitted because a neutron is converted into a proton within the nucleus.

Beta minus (β⁻) decay is a type of radioactive decay in which a neutron (n) within an atomic nucleus is converted into a proton (p). In this process, an electron (e⁻) and an electron antineutrino (ν̄e) are emitted from the nucleus. The reaction is typically written as: n → p + e⁻ + ν̄e. The electron is not pre-existing within the nucleus; it is created during this transformation. The atomic number of the nucleus increases by one, while the mass number remains unchanged.

Electrons are fundamental particles and are not constituents of the nucleus; protons and neutrons are the nucleons. The electron emitted in beta decay originates from the conversion of a neutron. Another type of beta decay is beta plus (β⁺) decay, where a proton converts into a neutron, emitting a positron (e⁺) and an electron neutrino (νe): p → n + e⁺ + νe. Electron capture is an alternative process where an electron from an inner atomic shell is captured by a proton in the nucleus, leading to the conversion of a proton into a neutron and emission of a neutrino. Momentum and energy conservation rules are followed in all radioactive decay processes, and the emission of the neutrino/antineutrino is necessary for conserving energy, momentum, and angular momentum, but the *reason* for electron emission in β⁻ decay is the fundamental weak interaction process of neutron decay.

38. Heavy water of an atomic reactor is :

Heavy water of an atomic reactor is :

deionised water

an oxide of heavier isotope of oxygen

a mixture of ice and water

an oxide of heavier isotope of hydrogen

This question was previously asked in

UPSC CAPF – 2015

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Heavy water is an oxide of a heavier isotope of hydrogen.

Heavy water has the chemical formula D₂O, where D represents Deuterium. Deuterium is an isotope of hydrogen that contains one proton and one neutron in its nucleus (whereas common hydrogen, Protium, has only one proton). Thus, deuterium is a heavier isotope of hydrogen. Heavy water (D₂O) is water composed of two deuterium atoms and one oxygen atom.

Heavy water is used in some types of nuclear reactors (like CANDU reactors) as a neutron moderator and coolant. Its function as a moderator is to slow down neutrons, making them more likely to cause nuclear fission in Uranium fuel. D₂O is more effective as a moderator than ordinary water (H₂O) because deuterium absorbs fewer neutrons than protium. Ordinary water also contains a very small natural abundance of D₂O (about 1 part in 6400). Isotopes of oxygen exist (e.g., ¹⁶O, ¹⁷O, ¹⁸O), and water made with ¹⁸O (H₂¹⁸O) is sometimes called “heavy-oxygen water”, but “heavy water” typically refers to D₂O.

39. Two racing cars of masses m₁ and m₂ are moving in circles of radii r₁

Two racing cars of masses m₁ and m₂ are moving in circles of radii r₁ and r₂ respectively. Their speeds are such that each car makes a complete circle in the same time ‘t’. The ratio of angular speed of the first to that of the second car is :

m₁ : m₂

1 : 1

r₁ : r₂

1 : 2

This question was previously asked in

UPSC CAPF – 2015

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The correct answer is 1 : 1.

Angular speed (ω) is defined as the rate of change of angular displacement. For an object moving in a circle, the angular speed is given by ω = 2π / T, where T is the time period (time taken to complete one revolution). The problem states that both cars make a complete circle in the same time ‘t’. Therefore, the time period T is the same for both cars.
For the first car, ω₁ = 2π / t.
For the second car, ω₂ = 2π / t.
The ratio of their angular speeds is ω₁ : ω₂ = (2π / t) : (2π / t) = 1 : 1.
The masses (m₁ and m₂) and radii (r₁ and r₂) of the cars are irrelevant for determining the ratio of angular speeds when the time period is given to be the same.

Linear speed (v) is related to angular speed and radius by v = rω. If the angular speeds are the same but the radii are different, the linear speeds will be different. Centripetal acceleration (a_c = rω²) and centripetal force (F_c = m a_c = m rω²) would also depend on the mass and radius even if the angular speed is constant. This question specifically asks for the ratio of angular speed, which is determined solely by the time period in this case.

40. The age of ‘E’ is twice the age of ‘S’. To find out the difference in

The age of ‘E’ is twice the age of ‘S’. To find out the difference in their ages, which of the following information is/are sufficient ?
I. After five years, the ratio of their ages would be 9 : 5
II. Before ten years, the ratio of their ages was 3 : 1

Only I

Only II

Either I or II

Both I and II

This question was previously asked in

UPSC CAPF – 2015

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Let E be the current age of E and S be the current age of S.
Given: E = 2S. We need to find the difference E – S.
Statement I: After five years, the ratio of their ages is 9:5. So, (E+5)/(S+5) = 9/5. Substituting E=2S, we get (2S+5)/(S+5) = 9/5. This is a linear equation in S, which can be solved to find a unique value for S. Since S can be determined, E can also be determined, and thus the difference E-S can be found. So, Statement I alone is sufficient.
Statement II: Before ten years, the ratio of their ages was 3:1. So, (E-10)/(S-10) = 3/1. Substituting E=2S, we get (2S-10)/(S-10) = 3/1. This is a linear equation in S, which can be solved to find a unique value for S. Since S can be determined, E can also be determined, and thus the difference E-S can be found. So, Statement II alone is sufficient.
Since either statement alone provides enough information to find the difference in their ages, the answer is C.

The question asks whether the given information is sufficient to find the difference in ages. We are given a relationship between their current ages. Each statement provides an additional relationship at a different point in time. We need to check if this system of equations allows us to find the specific ages or their difference.

From Statement I: 5(2S+5) = 9(S+5) => 10S + 25 = 9S + 45 => S = 20. E = 2*20 = 40. Difference = 20.
From Statement II: (2S-10) = 3(S-10) => 2S – 10 = 3S – 30 => S = 20. E = 2*20 = 40. Difference = 20.
Both statements lead to the same ages and difference, confirming that each is independently sufficient.