UPSC CAPF

41. Which one of the following elements is present in the green pigment of

Which one of the following elements is present in the green pigment of leaves ?

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Magnesium
Iron
Calcium
Copper
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Magnesium is present in the green pigment of leaves (chlorophyll).
– The green pigment in leaves is called chlorophyll.
– Chlorophyll molecules are crucial for photosynthesis, the process by which plants convert light energy into chemical energy.
– The core structure of the chlorophyll molecule is a porphyrin ring, similar to the heme group in hemoglobin, but with a central magnesium atom coordinating the nitrogen atoms, whereas hemoglobin has iron at its center.
While iron is essential for the synthesis of chlorophyll, it is not a structural component of the chlorophyll molecule itself. Magnesium is the central metal ion within the chlorophyll structure. Calcium is important for cell wall structure and signaling, and copper is a cofactor in various enzymes.

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42. Which one of the following tiger reserves of India has ‘Bhoorsingh the

Which one of the following tiger reserves of India has ‘Bhoorsingh the Barasingha’ as its official mascot ?

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Nameri tiger reserve
Ranthambhore tiger reserve
Panna tiger reserve
Kanha tiger reserve
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Kanha tiger reserve of India has ‘Bhoorsingh the Barasingha’ as its official mascot.
– Kanha Tiger Reserve, located in Madhya Pradesh, became the first tiger reserve in India to officially introduce a mascot, ‘Bhoorsingh the Barasingha’ (swamp deer), in 2017.
– The aim was to connect people with wildlife and encourage conservation efforts, particularly focusing on the barasingha, which is the state animal of Madhya Pradesh and a species found in abundance in Kanha.
Barasingha (Rucervus duvaucelii) is a vulnerable species of deer. Kanha National Park is particularly famous for its conservation efforts for the hard ground barasingha, saving it from near extinction. The mascot initiative helps in raising awareness about this unique species and the park’s ecosystem.

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43. 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 :

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List I
(Vegetation)
A. Chaparrals
B. Maquis
C. Fynbos
D. Malle scrubs

List II
(Region)
1. North America
2. Southern Europe
3. South Africa
4. Australia

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Code :

A-1, B-2, C-3, D-4
A-1, B-3, C-2, D-4
A-4, B-3, C-2, D-1
A-4, B-2, C-3, D-1
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The correct match for the vegetation types found in regions with Mediterranean climates is:
A. Chaparrals – 1. North America
B. Maquis – 2. Southern Europe
C. Fynbos – 3. South Africa
D. Malle scrubs – 4. Australia
– Chaparrals are shrubland plant communities found primarily in California, USA, which has a Mediterranean climate.
– Maquis is a dense growth of evergreen shrubs, characteristic of the Mediterranean region of southern Europe.
– Fynbos is a distinct type of shrubland vegetation found in the Cape Floristic Region of South Africa, an area with a Mediterranean climate.
– Mallee scrub is a native vegetation type in southern Australia, characterized by shrubby eucalyptus species, found in regions with a Mediterranean climate.
Mediterranean climate regions (characterized by hot, dry summers and mild, wet winters) are found in different parts of the world. Despite being geographically separated, these regions often develop similar vegetation types due to convergent evolution. The names like Chaparral, Maquis, Fynbos, and Mallee are local terms for these broadly similar vegetation structures.

44. 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
(Hydro power project)
A. Maithon project
B. Salal project
C. Rana Pratap Sagar project
D. Tehri project

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List II
(River)
1. Chambal river
2. Bhagirathi river
3. Barakar river
4. Chenab river

Code :

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A-2, B-1, C-4, D-3
A-2, B-4, C-1, D-3
A-3, B-1, C-4, D-2
A-3, B-4, C-1, D-2
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The correct match is:
A. Maithon project – 3. Barakar river
B. Salal project – 4. Chenab river
C. Rana Pratap Sagar project – 1. Chambal river
D. Tehri project – 2. Bhagirathi river
– The Maithon Dam is built on the Barakar River, a tributary of the Damodar River, in Jharkhand.
– The Salal Dam is a run-of-the-river hydropower project on the Chenab River in Jammu and Kashmir.
– The Rana Pratap Sagar Dam is built on the Chambal River in Rajasthan.
– The Tehri Dam is the highest dam in India, built on the Bhagirathi River, a major tributary of the Ganges, in Uttarakhand.
These projects are significant hydroelectric power generation facilities in India, contributing to the energy needs of the respective regions. Matching dams/projects with their corresponding rivers is a common type of question in geography sections of competitive exams.

45. Two men set out at the same time to walk towards each other from point

Two men set out at the same time to walk towards each other from points A and B, 72 km apart. The first man walks at the speed of 4 kmph while the second walks 2 km in the first hour, 2½ km in the second hour, 3 km in the third hour, and so on. The two men will meet

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in 8 hours
nearer to A than B
nearer to B than A
midway between A and B
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The two men will meet midway between A and B.
Let the time taken for them to meet be T hours. The distance between A and B is 72 km.
Man 1 starts from A at a constant speed of 4 kmph. In T hours, Man 1 covers a distance of 4T km.
Man 2 starts from B. In the first hour, he covers 2 km. In the second hour, 2.5 km. In the third, 3 km, and so on. This is an arithmetic progression of distances covered per hour with first term a = 2 and common difference d = 0.5.
The distance covered by Man 2 in T hours is the sum of the first T terms of this AP:
Sum = (T/2) * [2a + (T-1)d] = (T/2) * [2*2 + (T-1)*0.5] = (T/2) * [4 + 0.5T – 0.5] = (T/2) * [3.5 + 0.5T].
When they meet, the sum of the distances covered by both men equals the total distance:
Distance by Man 1 + Distance by Man 2 = 72
4T + (T/2) * (3.5 + 0.5T) = 72
4T + 1.75T + 0.25T^2 = 72
0.25T^2 + 5.75T – 72 = 0
Multiplying by 4 to clear decimals:
T^2 + 23T – 288 = 0
Using the quadratic formula T = [-b ± sqrt(b^2 – 4ac)] / 2a:
T = [-23 ± sqrt(23^2 – 4*1*(-288))] / 2*1
T = [-23 ± sqrt(529 + 1152)] / 2
T = [-23 ± sqrt(1681)] / 2
Since sqrt(1681) = 41 (as 40^2=1600, 41^2=1681), and time must be positive:
T = (-23 + 41) / 2 = 18 / 2 = 9 hours.
They meet after 9 hours.
Distance covered by Man 1 in 9 hours = 4 kmph * 9 hours = 36 km from A.
Distance covered by Man 2 in 9 hours = Sum of 9 terms of AP (a=2, d=0.5) = (9/2) * [2*2 + (9-1)*0.5] = 4.5 * [4 + 8*0.5] = 4.5 * [4 + 4] = 4.5 * 8 = 36 km from B.
Since they meet after covering 36 km from A and 36 km from B, which adds up to the total distance of 72 km, they meet exactly midway between A and B.
Option A states they meet in 8 hours, which was disproven by the calculation (at 8 hours, the total distance covered is 32km + 30km = 62km, less than 72km). Options B and C are incorrect because meeting midway means they are equidistant from A and B. Therefore, only option D accurately describes the meeting point based on the given data. The problem likely includes option A as a distractor based on a potential miscalculation or a similar problem with different parameters.

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46. A, B, C, D, E and F compared their marks in an examination and found t

A, B, C, D, E and F compared their marks in an examination and found that A obtained the highest marks, B obtained more marks than D, C obtained more than at least two others and E had not obtained the lowest marks.
Consider the following statements :

  • Statement 1 : At least two members obtained less marks than C
  • Statement 2 : E and F obtained the same marks

Which of the above statement(s) is/are sufficient to identify the one with the lowest marks ?

Both 1 and 2
Neither 1 nor 2
1 only
2 only
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Statement 2 alone is sufficient to identify the one with the lowest marks.
Let the people be ranked from 1 (highest) to 6 (lowest). We are given:
1. A is 1st.
2. B > D.
3. C > at least two others (C is among ranks 2, 3, 4).
4. E is not 6th (E is among ranks 1, 2, 3, 4, 5). Since A is 1st, E is among 2, 3, 4, 5.
From these facts, the person with the lowest marks (rank 6) can be B, D, or F.

Statement 1: At least two members obtained less marks than C. This is already stated in the problem (C > at least two others). This provides no new information to distinguish between D and F as the lowest. So, Statement 1 alone is insufficient.

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Statement 2: E and F obtained the same marks. Also, E is not 6th (from initial facts). If E and F are tied and E is not the lowest, then F also cannot be the lowest. Since the lowest person must be from {B, D, F} (after eliminating A, C, E as candidates for lowest), and F is eliminated by Statement 2, the lowest person must be from {B, D}. Given B > D, D must have the lowest marks among B and D. Therefore, D has the lowest marks. Statement 2 alone is sufficient.

The ranking from highest to lowest would be A > C > … > … > … > D if Statement 2 is true. For example, a possible ranking could be A > C > B > E=F > D, or A > B > C > E=F > D, or A > B > E=F > C > D, etc., as long as A is first, C is in the top 4, E=F are not last, and B>D, with D being the last.

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47. The difference between the compound interest and the simple interest f

The difference between the compound interest and the simple interest for 2 years on a sum of money is Rs. 60. If the simple interest for 2 years is Rs. 1440, what is the rate of interest ?

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$4 rac{1}{6}$ %
$6 rac{1}{4}$ %
8%
$8 rac{1}{3}$ %
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The rate of interest is $8\frac{1}{3}$ %.
Let the principal be P and the rate of interest be R% per annum.
Simple Interest (SI) for 2 years = Rs. 1440.
SI for 1 year = 1440 / 2 = Rs. 720.
We know that SI for 1 year = (P * R * 1) / 100. So, PR/100 = 720.
The difference between Compound Interest (CI) and Simple Interest (SI) for 2 years is the interest earned on the first year’s simple interest.
Difference (CI – SI) for 2 years = Interest on SI of 1st year.
Given difference = Rs. 60.
Interest on SI of 1st year = (SI for 1st year * R * 1) / 100.
So, 60 = (720 * R) / 100.
6000 = 720 * R.
R = 6000 / 720 = 600 / 72 = 100 / 12 = 25 / 3.
25/3 % = $8\frac{1}{3}$ %.
The formula for the difference between CI and SI for 2 years is P * (R/100)^2. We could also solve this using this formula if P was known or easily derivable. Since SI for 2 years is 1440, P * R * 2 / 100 = 1440, so PR/100 = 720. Substitute this into the difference formula: Difference = (PR/100) * (R/100) = 720 * (R/100).
Given difference = 60. So, 60 = 720 * (R/100).
R/100 = 60/720 = 1/12.
R = 100/12 = 25/3 = $8\frac{1}{3}$ %. Both methods yield the same result.

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48. Leakage of which one of the following gases had caused Bhopal Gas Trag

Leakage of which one of the following gases had caused Bhopal Gas Tragedy in the year 1984 ?

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Methyl isocyanate
Hexamethylene diisocyanate
Isophorone diisocyanate
Isothiocyanate
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The leakage of Methyl Isocyanate (MIC) gas caused the Bhopal Gas Tragedy in 1984.
The Bhopal Gas Tragedy occurred on the night of December 2–3, 1984, at the Union Carbide India Limited (UCIL) pesticide plant in Bhopal, Madhya Pradesh. A leak of methyl isocyanate gas and other chemicals resulted in thousands of deaths and long-term health issues for many more.
Methyl Isocyanate (MIC) is a highly toxic organic compound used in the production of pesticides. It is a volatile, flammable liquid. The incident is considered one of the world’s worst industrial disasters.

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49. Which of the following diseases are caused by the consumption of water

Which of the following diseases are caused by the consumption of water contaminated by mercury and nitrate ?

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Minamata disease and Osteoporosis
Osteoporosis and Blue Baby Syndrome
Minamata disease and Blue Baby Syndrome
Osteoporosis and Minamata disease
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Consumption of water contaminated by mercury can cause Minamata disease, and contamination by nitrate can cause Blue Baby Syndrome.
Minamata disease is a neurological syndrome caused by severe mercury poisoning. It was first discovered in Minamata, Japan, caused by the release of methylmercury into industrial wastewater. Blue Baby Syndrome (Methemoglobinemia) is caused by the consumption of nitrates, especially in infants. Nitrates are converted to nitrites, which interfere with the oxygen-carrying capacity of hemoglobin, causing a bluish discoloration of the skin.
Osteoporosis is a disease where bones become weak and brittle, typically linked to calcium and Vitamin D deficiency, hormonal changes, and aging, not directly caused by mercury or nitrate water contamination.

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50. Which one of the following polymers is made of protein ?

Which one of the following polymers is made of protein ?

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Rubber
Cotton
Wool
Jute
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Wool is a polymer made of protein.
Wool fiber is primarily composed of a protein called keratin. Proteins are natural polymers made up of repeating units of amino acids linked by peptide bonds.
Rubber (natural rubber) is a polymer of isoprene, a hydrocarbon. Cotton and Jute are natural fibers composed primarily of cellulose, which is a polysaccharide (a polymer of glucose units). These are carbohydrate polymers, not protein polymers.

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