UPSC CISF-AC-EXE Archives

31. Recently, in which one of the following countries did the military ous

Recently, in which one of the following countries did the military oust the Government and seize power ?

[amp_mcq option1=”Belarus” option2=”Costa Rica” option3=”Sudan” option4=”Turkmenistan” correct=”option3″]

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The correct answer is Sudan.

The question asks about a country where the military recently ousted the government and seized power. Among the options provided, Sudan experienced a significant military coup in recent history (relative to the likely timeframe of the question, possibly late 2021 or early 2022). In October 2021, the Sudanese military, led by Abdel Fattah al-Burhan, dissolved the transitional civilian government and arrested civilian leaders, effectively seizing power. This event led to widespread protests and international condemnation.

While other countries listed might face political challenges, none have experienced a direct military coup ousting the government in the same prominent manner as Sudan did in October 2021. Belarus has an authoritarian regime, Costa Rica is a stable democracy without a standing army, and Turkmenistan is also an authoritarian state, but none saw a recent military coup of the type described. The coup in Sudan was a major international news event at the time.

32. Kaleshwaram Lift Irrigation Project, often discussed in media, is buil

Kaleshwaram Lift Irrigation Project, often discussed in media, is built across which one of the following rivers ?

[amp_mcq option1=”Cauvery” option2=”Godavari” option3=”Krishna” option4=”Penneru” correct=”option2″]

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The correct answer is Godavari.

The Kaleshwaram Lift Irrigation Project (KLIP) is a massive multi-purpose irrigation project located in Telangana, India. It is designed to lift water from the Godavari river and its tributaries and distribute it to various parts of the state for irrigation, drinking water, and industrial use. The project involves a complex network of barrages, reservoirs, tunnels, pumping stations, and canals. It is notable for its scale, particularly the extensive use of lift irrigation technology to raise water to higher elevations.

The Kaleshwaram Project is built at the confluence of the Godavari River and its tributary Pranahita River at Kaleshwaram, Telangana. It is considered one of the world’s largest lift irrigation projects. The project aims to provide irrigation water to lakhs of acres and drinking water to millions of people.

33. What is the state of matter in a glowing fluorescent tube ?

What is the state of matter in a glowing fluorescent tube ?

[amp_mcq option1=”Gas” option2=”Liquid” option3=”Plasma” option4=”Bose-Einstein Condensate” correct=”option3″]

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The correct answer is Plasma.

A fluorescent tube typically contains an inert gas (like argon) and a small amount of mercury vapour at low pressure. When an electric voltage is applied across the electrodes, it accelerates electrons within the tube. These electrons collide with the gas atoms and mercury atoms, causing them to become ionized (losing electrons) and excited. This creates a state of matter consisting of a collection of positively charged ions and negatively charged electrons, along with neutral atoms, that is overall electrically neutral. This state is known as plasma. The excited mercury atoms and ions emit ultraviolet (UV) light. The inside of the tube is coated with a phosphorescent material (phosphor) that absorbs the UV light and re-emits it as visible light, causing the tube to glow.

Plasma is often referred to as the fourth state of matter, distinct from solid, liquid, and gas. It is essentially an ionized gas. Plasma is the most abundant state of matter in the visible universe, found in stars, nebulae, lightning, auroras, and certain human-made applications like fluorescent lamps, neon signs, and plasma televisions.

34. Which one of the following compounds is present in antacid tablets or

Which one of the following compounds is present in antacid tablets or medicines which are used to cure indigestion and acidity in the stomach ?

[amp_mcq option1=”Sodium hydroxide” option2=”Potassium hydroxide” option3=”Lithium hydroxide” option4=”Aluminium hydroxide” correct=”option4″]

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The correct answer is Aluminium hydroxide.

Antacid tablets or medicines are used to neutralize excess stomach acid (hydrochloric acid, HCl) to relieve indigestion and acidity. Antacids are typically bases or basic salts.
Let’s look at the given options:
A) Sodium hydroxide (NaOH): This is a strong base (lye). It is highly corrosive and would cause severe internal burns if ingested. It is not used as an antacid.
B) Potassium hydroxide (KOH): Similar to sodium hydroxide, this is also a strong base (caustic potash) and is highly corrosive. It is not used as an antacid.
C) Lithium hydroxide (LiOH): A strong base. Not used in antacids.
D) Aluminium hydroxide (Al(OH)₃): This is a weak base commonly used in antacid formulations, often combined with magnesium hydroxide (Mg(OH)₂) or calcium carbonate (CaCO₃). Aluminium hydroxide reacts with stomach acid to form aluminium chloride and water: Al(OH)₃(s) + 3HCl(aq) → AlCl₃(aq) + 3H₂O(l). It helps neutralize acid and can also help protect the stomach lining.

Common active ingredients in antacids include Aluminium hydroxide (Al(OH)₃), Magnesium hydroxide (Mg(OH)₂), Calcium carbonate (CaCO₃), and Sodium bicarbonate (NaHCO₃). Aluminium and magnesium based antacids can have side effects; aluminium salts can cause constipation, while magnesium salts can cause diarrhoea. Combination products are often used to balance these effects. Sodium bicarbonate is fast-acting but can cause belching and flatulence due to CO₂ production, and can be a concern for people on sodium-restricted diets. Calcium carbonate is effective but can also cause belching and constipation.

35. Which one of the following substances shows sublimation behaviour or p

Which one of the following substances shows sublimation behaviour or property at standard temperature and pressure ?

[amp_mcq option1=”Solid carbon dioxide” option2=”Water” option3=”Liquid nitrogen” option4=”Iron” correct=”option1″]

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The correct answer is Solid carbon dioxide.

Sublimation is a phase transition where a substance changes directly from a solid to a gas without passing through a liquid phase. The question asks which substance exhibits this behaviour at standard temperature and pressure (STP). STP is typically defined as 0°C (273.15 K) and 1 atmosphere (101.325 kPa).
Let’s examine the options:
A) Solid carbon dioxide (Dry ice): At atmospheric pressure (around 1 atm), solid CO₂ sublimes directly into gaseous CO₂ at -78.5°C. While this temperature is below the typical definition of standard temperature (0°C or 25°C), it readily sublimes at standard atmospheric pressure. Compared to other options, it’s the classic example of sublimation under common conditions. Its triple point is -56.6°C at 5.18 atm; below the triple point pressure, it only exists as solid and gas.
B) Water: At 1 atm, ice melts at 0°C and water boils at 100°C. Water exists as a liquid between 0°C and 100°C at 1 atm. Sublimation of ice can occur below 0°C, but water does not typically sublime from solid to gas at standard temperature (0°C or 25°C) and pressure (1 atm) as its primary transition.
C) Liquid nitrogen: Nitrogen boils at -196°C at 1 atm. It is a gas at standard temperature. It doesn’t sublime (solid to gas) at standard temperature and pressure; it transitions from liquid to gas at a much lower temperature.
D) Iron: Iron is a solid at standard temperature and pressure. It melts at 1538°C and boils at 2862°C at 1 atm. It does not sublime under standard conditions.
Solid carbon dioxide is the substance among the options that is well-known for undergoing sublimation at standard atmospheric pressure, even if the temperature is not strictly “standard room temperature”. Given the context of such questions, dry ice is the intended answer for a substance showing sublimation behaviour at standard pressure.

Other substances that sublime at standard atmospheric pressure include iodine, naphthalene (mothballs), camphor, and arsenic. The triple point of a substance is the temperature and pressure at which the three phases (solid, liquid, and gas) of that substance coexist in thermodynamic equilibrium. If the pressure is below the triple point pressure, heating the solid at that pressure will result in sublimation rather than melting. For CO₂, the triple point pressure (5.18 atm) is above standard atmospheric pressure (1 atm), which is why it sublimes instead of melting at 1 atm.

36. If α and β are the roots of the equation x² – 7x + 11 = 0, then the va

If α and β are the roots of the equation x² – 7x + 11 = 0, then the value of α³ + β³ is equal to :

[amp_mcq option1=”112″ option2=”77″ option3=”49″ option4=”224″ correct=”option1″]

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The correct answer is 112.

The given quadratic equation is x² – 7x + 11 = 0.
Let the roots be α and β.
According to Vieta’s formulas, for a quadratic equation ax² + bx + c = 0, the sum of the roots is α + β = -b/a and the product of the roots is αβ = c/a.
For the given equation x² – 7x + 11 = 0 (where a=1, b=-7, c=11):
Sum of roots: α + β = -(-7)/1 = 7.
Product of roots: αβ = 11/1 = 11.
We need to find the value of α³ + β³.
We can use the algebraic identity for the sum of cubes: α³ + β³ = (α + β)(α² – αβ + β²).
We can express α² + β² in terms of (α + β) and αβ:
α² + β² = (α + β)² – 2αβ.
Substituting this into the identity:
α³ + β³ = (α + β)[((α + β)² – 2αβ) – αβ]
α³ + β³ = (α + β)((α + β)² – 3αβ).
Now, substitute the values we found for (α + β) and αβ:
α + β = 7
αβ = 11
α³ + β³ = (7)((7)² – 3 × 11)
α³ + β³ = 7(49 – 33)
α³ + β³ = 7(16)
α³ + β³ = 112.

Vieta’s formulas provide a relationship between the coefficients of a polynomial and the sums and products of its roots. For a quadratic equation ax² + bx + c = 0, the formulas are: sum of roots = -b/a, product of roots = c/a. These are very useful for solving problems involving symmetric expressions of roots without explicitly finding the roots themselves.

37. From a pack of 5 green balls and 4 red balls, 2 balls are drawn at ran

From a pack of 5 green balls and 4 red balls, 2 balls are drawn at random. What is the probability that both the balls are of the same colour ?

[amp_mcq option1=”5/9″ option2=”4/9″ option3=”2/9″ option4=”None of the above” correct=”option2″]

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The correct answer is 4/9.

Total number of balls in the pack is 5 green + 4 red = 9 balls.
We are drawing 2 balls at random.
The total number of ways to choose 2 balls out of 9 is given by the combination formula C(n, k) = n! / (k! * (n-k)!):
Total outcomes = C(9, 2) = 9! / (2! * 7!) = (9 × 8) / (2 × 1) = 36.
We want the probability that both balls are of the same colour. This can happen in two mutually exclusive ways:
Case 1: Both balls are green.
Number of ways to choose 2 green balls out of 5 = C(5, 2) = 5! / (2! * 3!) = (5 × 4) / (2 × 1) = 10.
Case 2: Both balls are red.
Number of ways to choose 2 red balls out of 4 = C(4, 2) = 4! / (2! * 2!) = (4 × 3) / (2 × 1) = 6.
The number of favourable outcomes (both balls of the same colour) = Number of ways (both green) + Number of ways (both red) = 10 + 6 = 16.
The probability that both balls are of the same colour is (Favourable outcomes) / (Total outcomes) = 16 / 36.
Simplifying the fraction, 16/36 = (4 × 4) / (9 × 4) = 4/9.

This problem involves basic concepts of probability and combinations. The calculation for combinations C(n, k) represents the number of ways to choose k items from a set of n distinct items without regard to the order of selection. When dealing with probabilities of multiple events, we sum the probabilities of mutually exclusive events (like drawing two green OR two red).

38. Which one of the following is the largest 3-digit number which when di

Which one of the following is the largest 3-digit number which when divided by 12, 15 and 18 respectively, gives a remainder 5 in each case ?

[amp_mcq option1=”955″ option2=”905″ option3=”995″ option4=”755″ correct=”option2″]

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The correct answer is 905.

Let the number be N. The condition states that when N is divided by 12, 15, and 18, the remainder is always 5. This means N – 5 is divisible by 12, 15, and 18. Therefore, N – 5 must be a multiple of the Least Common Multiple (LCM) of 12, 15, and 18.
The prime factorization of 12 is 2² × 3.
The prime factorization of 15 is 3 × 5.
The prime factorization of 18 is 2 × 3².
The LCM (12, 15, 18) = 2² × 3² × 5 = 4 × 9 × 5 = 180.
So, N – 5 = 180k for some integer k.
N = 180k + 5.
We are looking for the largest 3-digit number of this form. The largest 3-digit number is 999.
We need 180k + 5 ≤ 999.
180k ≤ 994.
k ≤ 994 / 180 ≈ 5.52.
The largest integer value for k is 5.
Substituting k = 5 into the formula for N:
N = 180 × 5 + 5 = 900 + 5 = 905.
905 is a 3-digit number. Let’s check if it gives a remainder of 5 when divided by 12, 15, and 18:
905 ÷ 12 = 75 with remainder 5. (900 = 12 * 75)
905 ÷ 15 = 60 with remainder 5. (900 = 15 * 60)
905 ÷ 18 = 50 with remainder 5. (900 = 18 * 50)
Since k=5 gives 905, and any larger integer k would result in a number greater than 999 (e.g., k=6 gives 180*6+5 = 1085), 905 is the largest 3-digit number satisfying the condition.

This type of problem is a classic example of finding a number that satisfies multiple congruence relations (N ≡ 5 mod 12, N ≡ 5 mod 15, N ≡ 5 mod 18), which simplifies to N ≡ 5 mod(LCM(12, 15, 18)).

39. The main constituents of atmospheric air are Oxygen (O₂) and Nitrogen

The main constituents of atmospheric air are Oxygen (O₂) and Nitrogen (N₂). The composition of O₂ and N₂ is approximately :

[amp_mcq option1=”29% and 70% respectively.” option2=”19% and 80% respectively.” option3=”21% and 78% respectively.” option4=”23% and 76% respectively.” correct=”option3″]

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The Earth’s atmosphere is primarily a mixture of gases. The two most abundant gases in dry atmospheric air are Nitrogen (N₂) and Oxygen (O₂).
The approximate composition of dry atmospheric air by volume is:
Nitrogen (N₂) ≈ 78.09%
Oxygen (O₂) ≈ 20.95%
Argon (Ar) ≈ 0.93%
Carbon Dioxide (CO₂) ≈ 0.04%
Trace gases (Neon, Helium, Methane, Krypton, Hydrogen, etc.) make up the rest. Water vapor is also present, but its concentration varies significantly depending on location and weather. The question likely refers to dry air composition.

Looking at the options for the composition of O₂ and N₂:
A) 29% O₂, 70% N₂
B) 19% O₂, 80% N₂
C) 21% O₂, 78% N₂
D) 23% O₂, 76% N₂

Comparing these options to the actual percentages (≈20.95% O₂ and ≈78.09% N₂), option C (21% and 78%) is the closest approximation.

– Atmospheric air is a mixture of gases.
– The main constituents are Nitrogen (N₂) and Oxygen (O₂).
– Their approximate percentages by volume are about 78% Nitrogen and 21% Oxygen.

While Nitrogen and Oxygen make up the vast majority of the atmosphere (about 99%), the remaining 1% includes important gases like Argon, which is the third most abundant gas, and Carbon Dioxide, which plays a significant role in the greenhouse effect, despite its small concentration. These percentages can be taught as roughly 4/5th Nitrogen and 1/5th Oxygen for simplification, but 78% and 21% are more accurate approximate figures often cited.

40. Dyes in black ink can be separated by :

Dyes in black ink can be separated by :

[amp_mcq option1=”evaporation.” option2=”centrifugation.” option3=”sublimation.” option4=”chromatography.” correct=”option4″]

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Black ink is typically a mixture of several coloured dyes. To separate the different components (dyes) of a mixture, various separation techniques can be used. The technique suitable for separating different coloured dyes in ink is chromatography.
Chromatography is a technique used to separate mixtures based on the differential distribution of components between a stationary phase and a mobile phase. For separating dyes in ink, paper chromatography is a common method. The paper acts as the stationary phase, and a solvent (like water or alcohol) acts as the mobile phase. As the solvent moves up the paper by capillary action, it carries the different dyes with it. Dyes that are more soluble in the solvent and have weaker interactions with the paper move faster and further up the paper, resulting in their separation into distinct spots at different heights.

Let’s consider the other options:
A) Evaporation separates a soluble solid from a liquid solvent by heating the solution to evaporate the solvent. This would only leave the mixture of dyes behind, not separate them.
B) Centrifugation separates components of a mixture based on their density by spinning at high speed. It is used for separating solids from liquids or liquids of different densities (e.g., separating cream from milk, separating blood cells from plasma). It is not suitable for separating dissolved dyes.
C) Sublimation is the process where a substance changes directly from a solid to a gas state upon heating (e.g., dry ice, iodine). It is used to separate sublimable solids from non-sublimable ones. Dyes in ink are not typically sublimable.

– Black ink is a mixture of different coloured dyes.
– Chromatography is a separation technique used to separate components of a mixture based on their different affinities for a stationary phase and a mobile phase.
– Paper chromatography is specifically effective for separating different dyes in ink.

Paper chromatography is a simple and effective type of chromatography often demonstrated in schools. Other types include thin-layer chromatography (TLC), column chromatography, gas chromatography (GC), and high-performance liquid chromatography (HPLC), all based on the same principle of differential partitioning between phases.