UPSC CAPF Archives

21. A test consists of 25 MCQs. Each correct answer gives +4 marks and inc

A test consists of 25 MCQs. Each correct answer gives +4 marks and incorrect answer gives -1 mark. If a candidate scores 74 marks, then how many questions were left unattempted?

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UPSC CAPF – 2022

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Let c be the number of correct answers, i be the number of incorrect answers, and u be the number of unattempted questions.
Total questions: c + i + u = 25
Total score: 4c – i = 74 (unattempted questions contribute 0 marks).
We need to find the value of u. We can test the given options for u.
If u = 4, then c + i = 25 – 4 = 21.
We have the system of equations:
1) c + i = 21
2) 4c – i = 74
Add the two equations: (c + i) + (4c – i) = 21 + 74 => 5c = 95 => c = 19.
Substitute c = 19 into equation 1: 19 + i = 21 => i = 21 – 19 = 2.
Check the score: 4c – i = 4(19) – 2 = 76 – 2 = 74. This matches the given score.
Therefore, the number of unattempted questions is 4.

– Set up equations based on the given information: total number of questions and the scoring system.
– Use variables to represent the number of correct, incorrect, and unattempted questions.
– The total number of questions is the sum of correct, incorrect, and unattempted questions.
– The total score is calculated based on the marks per correct and incorrect answer.

Trying other options for ‘u’ would not yield integer values for ‘c’ and ‘i’ (as shown in the thought process), confirming that u=4 is the unique solution. This type of problem involves solving a system of linear equations, possibly with the added constraint that the variables must be non-negative integers.

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22. A coin is tossed 3 times. The probability of getting exactly 2 heads

A coin is tossed 3 times. The probability of getting exactly 2 heads is

$rac{1}{3}$

$rac{3}{8}$

$rac{1}{2}$

$rac{5}{8}$

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UPSC CAPF – 2022

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When a fair coin is tossed, there are two possible outcomes: Heads (H) or Tails (T). When tossed 3 times, the total number of possible outcomes is 2³ = 8. The sample space is {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. We are interested in the outcomes with exactly 2 heads. These are {HHT, HTH, THH}. There are 3 favorable outcomes. The probability of getting exactly 2 heads is the number of favorable outcomes divided by the total number of outcomes: 3/8.

– List all possible outcomes (sample space) for the given number of trials.
– Identify the outcomes that satisfy the condition (favorable outcomes).
– Probability = (Number of favorable outcomes) / (Total number of outcomes).
– For independent events like coin tosses, the total number of outcomes for ‘n’ trials is 2^n.

This type of problem can also be solved using the binomial probability formula: P(X=k) = C(n, k) * p^k * (1-p)^(n-k), where n is the number of trials (3), k is the number of successes (2 heads), p is the probability of success on a single trial (0.5 for heads), and C(n, k) is the binomial coefficient “n choose k”.
C(3, 2) = 3! / (2! * 1!) = 3.
P(exactly 2 heads) = C(3, 2) * (0.5)² * (0.5)¹ = 3 * 0.25 * 0.5 = 3 * 0.125 = 0.375 = 3/8.

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23. Two friends 10 km apart start running towards each other at speeds of

Two friends 10 km apart start running towards each other at speeds of 10 km/hr and 14 km/hr respectively. After how much time will they meet each other?

20 minutes

25 minutes

28 minutes

30 minutes

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UPSC CAPF – 2022

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When two objects move towards each other, their relative speed is the sum of their individual speeds. The distance between the two friends is 10 km. Their speeds are 10 km/hr and 14 km/hr.
Relative speed = 10 km/hr + 14 km/hr = 24 km/hr.
The time taken to meet is the distance divided by the relative speed.
Time = Distance / Relative Speed = 10 km / 24 km/hr = 10/24 hours.
To convert hours to minutes, multiply by 60:
Time in minutes = (10/24) * 60 = (5/12) * 60 = 5 * 5 = 25 minutes.

– When objects move towards each other, use relative speed, which is the sum of their speeds.
– Use the formula Time = Distance / Speed.
– Ensure consistent units (convert hours to minutes if necessary).

This is a classic relative motion problem. The concept of relative speed simplifies calculating the time until they meet or the distance covered by one relative to the other. If they were moving away from each other, the relative speed would be the difference between their speeds.

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24. There is a group of 5 people among which there is one couple. In how m

There is a group of 5 people among which there is one couple. In how many ways can these 5 people be seated in a row having 5 chairs if the couple is to be seated next to each other?

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120

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To solve this problem, we treat the couple as a single unit. This means we are arranging 4 entities: the couple (as one unit) and the remaining 3 individuals. These 4 entities can be arranged in 4! ways. Since the two members of the couple can swap positions within their unit, there are 2! ways for them to be seated relative to each other. The total number of ways is the product of the ways to arrange the units and the ways to arrange within the unit: 4! × 2! = 24 × 2 = 48.

– Treat the constrained group (the couple) as a single unit.
– Calculate the number of permutations of the resulting units (the couple unit + the other individuals).
– Calculate the number of permutations within the constrained group (the couple).
– The total number of arrangements is the product of these two numbers.

If there were no constraints, 5 people could be seated in 5! = 120 ways. The constraint that the couple sits together reduces the number of possibilities significantly. This method of treating a group that must stay together as a single unit is a standard technique in permutation problems.

25. The atomic radius of hydrogen atom is

The atomic radius of hydrogen atom is

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37 nanometer

37 picometer

17 picometer

57 picometer

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UPSC CAPF – 2022

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The atomic radius of a hydrogen atom is approximately 37 picometers (pm) when considered as a covalent radius, or about 52.9 pm for the Bohr radius. Given the options, 37 picometer is the closest and a commonly cited value for the covalent radius of hydrogen.

– The size of an atom is often described by its radius.
– Different definitions of atomic radius exist (e.g., covalent radius, van der Waals radius, Bohr radius) which yield different values.
– 1 picometer (pm) = 10⁻¹² meters.
– 1 nanometer (nm) = 10⁻⁹ meters = 1000 picometers.
– A radius of 37 nanometers would be extremely large, equivalent to 37,000 picometers, which is not the size of a hydrogen atom.

The Bohr radius (a₀) for the ground state of hydrogen is approximately 52.9 pm. The covalent radius of hydrogen, often determined from the H-H bond length in H₂ (about 74 pm), is taken as half of this length, i.e., 37 pm. The van der Waals radius of hydrogen is significantly larger, around 120 pm. The question does not specify which type of radius, but 37 pm is a valid and commonly used value for its covalent radius.

26. Calcium oxide reacts with water to produce slaked lime. It is an examp

Calcium oxide reacts with water to produce slaked lime. It is an example of

combination reaction

decomposition reaction

oxidation reaction

addition reaction

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UPSC CAPF – 2022

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The reaction described is between calcium oxide ($\text{CaO}$) and water ($\text{H}_2\text{O}$) to produce slaked lime, which is calcium hydroxide ($\text{Ca(OH)}_2$).
The balanced chemical equation for this reaction is:
$\text{CaO(s)} + \text{H}_2\text{O(l)} \rightarrow \text{Ca(OH)}_2\text{(aq)}$

Let’s analyze the types of reactions given in the options:
A) Combination reaction: A reaction in which two or more reactants combine to form a single product. In this reaction, calcium oxide and water (two reactants) combine to form calcium hydroxide (a single product). This matches the definition of a combination reaction.
B) Decomposition reaction: A reaction in which a single compound breaks down into two or more simpler substances. This is the opposite of the given reaction.
C) Oxidation reaction: A reaction involving the loss of electrons or increase in oxidation state. While redox aspects might be present in the formation of bonds, the primary classification based on the change in the number of substances is combination. The oxidation states of Ca (+2), O (-2), and H (+1) do not change overall.
D) Addition reaction: A reaction in which atoms are added to a molecule across a multiple bond (like a double or triple bond). This term is mainly used in organic chemistry for reactions involving unsaturated hydrocarbons. While water is added to CaO, the term ‘combination reaction’ is the standard classification for this type of inorganic reaction.

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The reaction fits perfectly the definition of a combination reaction. It is also a highly exothermic reaction, releasing a significant amount of heat, which is why it is also called the slaking of lime.

– Identifying the reactants and products of the reaction: Calcium oxide + Water -> Calcium hydroxide.
– Understanding the definition of different types of chemical reactions.
– Classifying the reaction based on the change in the number of reactants and products. Two reactants form one product, characteristic of a combination reaction.

This reaction is a classic example of an inorganic combination reaction and the process of slaking quicklime ($\text{CaO}$) to produce slaked lime ($\text{Ca(OH)}_2$). Slaked lime has many uses, including in mortars, plasters, and water treatment. The reaction is also an example of a hydrolysis reaction because water is a reactant. However, among the given choices, “combination reaction” is the most appropriate primary classification.

27. Hydrogenation of alkenes can be carried out in the presence of

Hydrogenation of alkenes can be carried out in the presence of

copper

zinc

aluminium

nickel

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Hydrogenation is a chemical reaction between molecular hydrogen ($\text{H}_2$) and another compound, usually in the presence of a catalyst. Hydrogenation of alkenes involves adding hydrogen across the carbon-carbon double bond, converting the alkene into a saturated alkane. This reaction typically requires a catalyst to lower the activation energy, as the $\text{H-H}$ bond in hydrogen is strong.
Common catalysts used for the hydrogenation of alkenes are transition metals, particularly from Group 8, 9, and 10 of the periodic table. These metals, such as platinum ($\text{Pt}$), palladium ($\text{Pd}$), rhodium ($\text{Rh}$), and nickel ($\text{Ni}$), adsorb both the alkene and hydrogen molecules onto their surface, facilitating the reaction.

Looking at the options provided:
A) Copper: Copper can catalyze some hydrogenation reactions, but it is generally less active than Ni, Pt, or Pd for alkene hydrogenation.
B) Zinc: Zinc is not a typical catalyst for the hydrogenation of alkenes.
C) Aluminium: Aluminium is not used as a catalyst for the hydrogenation of alkenes; it is a reactive metal itself.
D) Nickel: Nickel is a very common and widely used catalyst for the hydrogenation of alkenes, especially in industrial processes (e.g., hydrogenation of vegetable oils to margarine – the Sabatier-Senderens process often uses Ni).

Therefore, nickel is a standard catalyst for the hydrogenation of alkenes.

– Hydrogenation of alkenes is the addition of H₂ to the double bond.
– This reaction requires a catalyst.
– Transition metals like Ni, Pt, Pd, and Rh are common hydrogenation catalysts.
– Nickel is a common and cost-effective catalyst for alkene hydrogenation.

The catalytic hydrogenation of alkenes is a heterogeneous catalytic process where the reactants and catalyst are in different phases (alkenes/hydrogen are gases or liquids, catalyst is a solid metal). The reaction proceeds via a mechanism where both the alkene and hydrogen are adsorbed onto the catalyst surface. Raney nickel is a porous form of nickel alloy commonly used for hydrogenation.

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28. White gold is an alloy of

White gold is an alloy of

gold, nickel and palladium

gold, cobalt and palladium

gold, titanium and platinum

gold, magnesium and palladium

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White gold is an alloy of gold with at least one white metal. The purpose of alloying gold with white metals is to produce a white appearance and often increase its hardness and durability. Common white metals used in white gold alloys include nickel, palladium, silver, and platinum.
– Nickel is a common and inexpensive white metal used in white gold alloys, especially in the United States. It provides hardness but can cause allergic reactions in some individuals.
– Palladium is a more expensive white metal used in white gold alloys, particularly in Europe. It produces a whiter alloy and is hypoallergenic.
– Platinum is also a white metal and can be used, often as part of a platinum group metal mix including palladium.
– Silver is a white metal sometimes used in smaller quantities to modify color and workability.

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Option A lists gold, nickel, and palladium, which are all standard components found in various white gold alloys.
Option B lists cobalt, which is less commonly used than nickel or palladium in standard white gold jewelry alloys.
Option C lists titanium, which is not typically used in gold alloys for jewelry purposes.
Option D lists magnesium, which is also not typically used in gold alloys for jewelry purposes.

Therefore, gold, nickel, and palladium represent a typical composition of white gold alloys.

– White gold is an alloy of gold with white metals.
– Common white metals alloyed with gold include nickel, palladium, silver, and platinum.
– Option A correctly lists common alloying metals for white gold.

The actual composition of white gold varies. Nickel-white gold alloys are typically 10-20% nickel, with possibly some copper or zinc. Palladium-white gold alloys are typically 10-20% palladium, which is more expensive but produces a softer, more easily worked, and hypoallergenic alloy. Higher karat white gold alloys often use palladium. White gold jewelry is usually plated with rhodium to give it a brighter white finish.

29. Cinnabar is an ore of

Cinnabar is an ore of

mercury

zinc

copper

lead

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UPSC CAPF – 2022

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Cinnabar is a mineral with the chemical formula $\text{HgS}$ (mercury(II) sulfide). It is the most common ore from which the element mercury ($\text{Hg}$) is extracted. Historically, it was mined extensively, and its red color made it a valuable pigment (vermilion). Roasting cinnabar in air converts the mercury(II) sulfide to elemental mercury and sulfur dioxide:
$\text{HgS(s)} + \text{O}_2\text{(g)} \rightarrow \text{Hg(g)} + \text{SO}_2\text{(g)}$

– Cinnabar is a mineral composed of mercury sulfide ($\text{HgS}$).
– An ore is a natural occurrence of a rock or sediment which contains sufficient minerals with economically important elements, metals or gems.
– Cinnabar is the primary ore for the extraction of mercury.

Other common ores of the metals listed are:
– Zinc: Sphalerite ($\text{ZnS}$), Calamine ($\text{ZnCO}_3$).
– Copper: Chalcopyrite ($\text{CuFeS}_2$), Malachite ($\text{Cu}_2(\text{CO}_3)(\text{OH})_2$), Azurite ($\text{Cu}_3(\text{CO}_3)_2(\text{OH})_2$).
– Lead: Galena ($\text{PbS}$).

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30. Borax is prepared from

Borax is prepared from

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calcium carbonate

magnesium carbonate

potassium carbonate

sodium carbonate

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UPSC CAPF – 2022

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Borax is sodium tetraborate decahydrate ($\text{Na}_2\text{B}_4\text{O}_7\cdot10\text{H}_2\text{O}$). It is a compound of sodium and boron. While industrially it is often extracted from naturally occurring borate minerals like kernite or colemanite, it can also be synthesized. One method of synthesizing borax in the lab involves reacting boric acid ($\text{H}_3\text{BO}_3$) with a sodium compound. Among the given options, sodium carbonate ($\text{Na}_2\text{CO}_3$) is a common reactant used in the preparation of borax from boric acid. The reaction is:
$4\text{H}_3\text{BO}_3 + \text{Na}_2\text{CO}_3 \rightarrow \text{Na}_2\text{B}_4\text{O}_7 + 6\text{H}_2\text{O} + \text{CO}_2$
This yields anhydrous sodium tetraborate, which can then be hydrated to form borax. Given the options, sodium carbonate is the most suitable precursor listed for the preparation of borax.

– Borax is a sodium borate compound.
– Its preparation often involves a sodium source and a boron source.
– Sodium carbonate is a common sodium source used in synthesizing borax from boric acid.

Other sodium compounds like sodium hydroxide ($\text{NaOH}$) can also be used to react with boric acid to produce borax, but $\text{Na}_2\text{CO}_3$ is explicitly listed as an option. Borax is an important boron compound used in various applications, including detergents, cosmetics, and as a flux in metallurgy.