2020

131. Tigers have a shorter small intestine compared to cows. The length of

Tigers have a shorter small intestine compared to cows. The length of the small intestine differs in various animals depending on the

[amp_mcq option1=”availability of water in their habitat.” option2=”size of their mouth cavity.” option3=”kind of habitats where they live.” option4=”kind of food they eat.” correct=”option4″]

This question was previously asked in
UPSC CISF-AC-EXE – 2020
Download PDFAttempt Online
Tigers are carnivores and cows are herbivores. The length of the small intestine in animals is primarily adapted to the type of food they eat. Carnivores eat meat, which is relatively easy to digest and absorb, thus requiring a shorter digestive tract. Herbivores eat plant material, which is rich in cellulose and requires a longer, more complex digestive process, often involving fermentation by symbiotic microorganisms. A longer small intestine provides more surface area and time for the digestion and absorption of nutrients from plant matter.
The digestive system, including the length of the small intestine, is adapted to the diet of the animal. Herbivores have longer intestines than carnivores due to the difficulty in digesting plant cellulose.
The digestion of cellulose in herbivores often occurs in specialized chambers (like the rumen in cows) or in the hindgut, requiring a slower passage rate and a larger surface area for absorption, which is facilitated by a longer small intestine and overall digestive tract.

132. Which one of the following nutrients is *not* available in

Which one of the following nutrients is *not* available in fertilizers?

[amp_mcq option1=”Iron” option2=”Nitrogen” option3=”Phosphorus” option4=”Potassium” correct=”option1″]

This question was previously asked in
UPSC CISF-AC-EXE – 2020
Download PDFAttempt Online
The question asks which nutrient is *not* available in fertilizers among the given options. All listed options (Iron, Nitrogen, Phosphorus, Potassium) are essential plant nutrients and are available in various types of fertilizers. Nitrogen, Phosphorus, and Potassium are primary macronutrients commonly found in NPK fertilizers, which form the bulk of the fertilizer industry. Iron is a micronutrient, required in smaller quantities, but is also available in specific micronutrient fertilizers (e.g., iron chelates) and is often included in complete fertilizer blends.
While all options are plant nutrients and can be found in fertilizers, Nitrogen, Phosphorus, and Potassium are the primary macronutrients (needed in large quantities) and form the basis of most common fertilizers. Iron is a micronutrient (needed in smaller quantities).
The phrasing of the question is potentially misleading, as Iron is indeed available in fertilizers. However, in the context of multiple-choice questions where one option must be selected, and considering the prominence of N, P, and K as the main components of standard fertilizers compared to Iron which is a micronutrient supplement, Iron might be considered the intended answer if the question is implicitly distinguishing between macronutrients and micronutrients in the context of bulk fertilization, despite the poor wording. Given the options, Iron is the least likely to be a major component of a general fertilizer compared to N, P, and K.

133. During photosynthesis, O₂ of the atmosphere is fixed with the help of

During photosynthesis, O₂ of the atmosphere is fixed with the help of H₂O and sunlight to synthesize carbohydrate and O₂ is evolved by the splitting of

[amp_mcq option1=”CO₂” option2=”H₂O” option3=”NO₃⁻” option4=”PO₄³⁻” correct=”option2″]

This question was previously asked in
UPSC CISF-AC-EXE – 2020
Download PDFAttempt Online
During photosynthesis, $O_2$ is evolved by the splitting of H₂O.
– Photosynthesis is the process by which green plants and some other organisms use sunlight to synthesize foods with the help of chlorophyll pigment.
– The overall equation for photosynthesis is $6\text{CO}_2 + 6\text{H}_2\text{O} + \text{Light Energy} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$.
– The process occurs in two main stages: the light-dependent reactions and the light-independent reactions (Calvin cycle).
– The light-dependent reactions take place in the thylakoid membranes of chloroplasts. In this stage, light energy is absorbed by chlorophyll, leading to the splitting of water molecules ($H_2O$). This process is called photolysis.
– Photolysis of water ($2H_2O \rightarrow 4e^- + 4H^+ + O_2$) releases electrons (used in the electron transport chain to produce ATP and NADPH), protons ($H^+$), and oxygen gas ($O_2$) as a byproduct. This oxygen is released into the atmosphere.
– The light-independent reactions (Calvin cycle) take place in the stroma of chloroplasts and use the ATP and NADPH produced during the light reactions to fix carbon dioxide ($CO_2$) from the atmosphere and synthesize carbohydrates.
– The question contains a slight inaccuracy by stating “$O_2$ of the atmosphere is fixed with the help of $H_2O$ and sunlight…”. It should be “$CO_2$ of the atmosphere is fixed…”. However, the second part of the sentence correctly asks what $O_2$ is evolved from.
– Based on the mechanism of photosynthesis, the oxygen gas evolved comes directly from the splitting of water molecules during the light-dependent stage.
The origin of oxygen produced during photosynthesis was a significant scientific question. Experiments using isotopic tracers ($^{18}O$) demonstrated that the oxygen evolved comes from water and not from carbon dioxide.

134. Which of the following organelle(s) in an animal cell would have DNA a

Which of the following organelle(s) in an animal cell would have DNA and RNA?

[amp_mcq option1=”Nucleus only” option2=”Nucleus and mitochondria only” option3=”Nucleus, mitochondria and ribosomes” option4=”Mitochondria only” correct=”option2″]

This question was previously asked in
UPSC CISF-AC-EXE – 2020
Download PDFAttempt Online
In an animal cell, the Nucleus and mitochondria only would have DNA and RNA.
– DNA (Deoxyribonucleic Acid) is the primary genetic material in most organisms. In eukaryotic cells, the majority of DNA is found in the nucleus (nuclear DNA), forming chromosomes.
– Eukaryotic cells also contain DNA in certain organelles: mitochondria (mitochondrial DNA or mtDNA) and, in plant cells and some other eukaryotes, chloroplasts (chloroplast DNA or cpDNA). Animal cells lack chloroplasts.
– RNA (Ribonucleic Acid) is involved in protein synthesis and other cellular processes. Various types of RNA (mRNA, tRNA, rRNA) are transcribed from DNA.
– In animal cells:
– Nucleus: Contains the main genome (DNA) and is where transcription occurs, producing various types of RNA. Thus, the nucleus has both DNA and RNA.
– Mitochondria: Contain their own circular DNA (mtDNA) and also have ribosomes (mitoribosomes) made of ribosomal RNA (mt-rRNA) and proteins, and produce their own mRNA and tRNA. Thus, mitochondria have both DNA and RNA.
– Ribosomes: Are complex molecular machines responsible for protein synthesis. They are composed of ribosomal RNA (rRNA) and proteins. Ribosomes themselves do not contain DNA. They read mRNA to build proteins.
– Therefore, in an animal cell, the nucleus and mitochondria are the organelles that contain both DNA and RNA.
While RNA is widely present in the cytoplasm (as mRNA being translated, tRNA carrying amino acids, and ribosomes containing rRNA), the question asks about *organelles* that contain both DNA and RNA. Ribosomes are technically organelles (or macromolecular machines depending on definition), but they lack DNA. The nucleus and mitochondria fit the criteria.

135. The kingdom Plantae as laid down by R. Whittaker comprises which of th

The kingdom Plantae as laid down by R. Whittaker comprises which of the following group of plants?

[amp_mcq option1=”Bryophyta, Pteridophyta, Protista, Gymnosperms and Angiosperms” option2=”Thallophyta, Pteridophyta, Fungi, Gymnosperms and Angiosperms” option3=”Thallophyta, Bryophyta, Monera, Pteridophyta and Gymnosperms” option4=”Thallophyta, Bryophyta, Pteridophyta, Gymnosperms and Angiosperms” correct=”option4″]

This question was previously asked in
UPSC CISF-AC-EXE – 2020
Download PDFAttempt Online
The kingdom Plantae as laid down by R. Whittaker comprises Thallophyta, Bryophyta, Pteridophyta, Gymnosperms and Angiosperms.
– R.H. Whittaker proposed the five-kingdom classification system: Monera, Protista, Fungi, Plantae, and Animalia.
– According to Whittaker’s system, the kingdom Plantae includes multicellular eukaryotic organisms that are autotrophic (perform photosynthesis) and have cell walls primarily made of cellulose.
– The major groups traditionally placed under Plantae in Whittaker’s system are Algae (specifically multicellular forms), Bryophytes, Pteridophytes, Gymnosperms, and Angiosperms.
– The term ‘Thallophyta’ is an older classification group (or a descriptive term) that included organisms with an undifferentiated body structure (thallus), such as algae, fungi, and lichens. In Whittaker’s system, Fungi were placed in a separate kingdom, and Algae were distributed among Protista (unicellular eukaryotes) and Plantae (multicellular eukaryotes).
– Option D lists Thallophyta, Bryophyta, Pteridophyta, Gymnosperms, and Angiosperms. Given the other options include members of Monera, Protista, and Fungi (separate kingdoms), option D is the most accurate representation of the groups typically included in Whittaker’s Plantae. ‘Thallophyta’ in this context likely refers to the multicellular algae forms included in Plantae by Whittaker, such as some green algae (e.g., Spirogyra, Ulothrix) and brown/red algae.
Whittaker’s five-kingdom system was a significant step in biological classification, separating organisms based on cell structure, mode of nutrition, and ecological role. While ‘Thallophyta’ is somewhat an outdated term as a formal taxon in modern cladistics, its inclusion in Option D reflects the historical context and how these groups were often discussed under the umbrella of Plantae in earlier classification frameworks like Whittaker’s, distinguishing them from Embryophytes (Bryophytes, Pteridophytes, Gymnosperms, Angiosperms).

136. A diameter PQ is drawn to a circle whose diameter length is 1 m. A squ

A diameter PQ is drawn to a circle whose diameter length is 1 m. A square is drawn using the diameter PQ as one of its sides. Assuming that $\pi$ is 22/7, what is the area of the part of the square lying outside the circle ?

[amp_mcq option1=”3/28 sq. m” option2=”11/28 sq. m” option3=”15/28 sq. m” option4=”17/28 sq. m” correct=”option4″]

This question was previously asked in
UPSC CISF-AC-EXE – 2020
Download PDFAttempt Online
The area of the part of the square lying outside the circle is 17/28 sq. m.
– The diameter PQ of the circle has length 1 m. The radius of the circle is $1/2$ m.
– A square is drawn using the diameter PQ as one of its sides. This means the side length of the square is equal to the length of the diameter, which is 1 m.
– The area of the square is side * side = $1^2 = 1$ sq. m.
– The area of the circle is $\pi r^2 = \pi (1/2)^2 = \pi/4$ sq. m.
– Using $\pi = 22/7$, the area of the circle is $(22/7) / 4 = 22/28 = 11/14$ sq. m.
– Let’s interpret “using the diameter PQ as one of its sides” to mean that the line segment PQ forms one boundary of the square. Let PQ lie on the x-axis from (0,0) to (1,0). The square would then occupy the region $0 \le x \le 1$ and $0 \le y \le 1$ (assuming it’s drawn above PQ). The circle with diameter PQ is centered at the midpoint of PQ, which is (0.5, 0), and has radius 0.5. Its equation is $(x-0.5)^2 + y^2 = 0.5^2 = 0.25$.
– The area of the part of the square lying outside the circle is the Area of the Square minus the Area of the region common to both the square and the circle.
– The square is defined by $0 \le x \le 1$ and $0 \le y \le 1$. The circle is defined by $(x-0.5)^2 + y^2 \le 0.25$.
– The part of the circle within the square is where $y \ge 0$ and $(x-0.5)^2 + y^2 \le 0.25$. This describes the upper semi-circle bounded by the diameter PQ (the base of the square).
– The area of this upper semi-circle is half the area of the full circle = (Area of circle) / 2 = $(\pi/4) / 2 = \pi/8$.
– Using $\pi = 22/7$, the area of the semi-circle is $(22/7) / 8 = 22/56 = 11/28$ sq. m.
– The area of the part of the square lying outside this semi-circular region is Area of Square – Area of the semi-circle.
– Area outside = 1 sq. m – 11/28 sq. m = $(28/28) – (11/28) = (28-11)/28 = 17/28$ sq. m.
The crucial part of this problem is correctly interpreting the geometric setup based on the phrase “A square is drawn using the diameter PQ as one of its sides”. The standard interpretation in such geometry problems is that the diameter forms one edge of the square, and the region considered is the area within the square but outside the shape (or part of it) defined by the diameter.

137. R walks a long distance every Sunday. He walks 2 km towards the north

R walks a long distance every Sunday. He walks 2 km towards the north from his house and then turns right; he walks another 2 km and again turns right; next he walks 5 km and turns left; he further walks 2 km and stops. He rests for some time and returns home following a straight route without any turning point. What is the distance R walks after he has rested ?

[amp_mcq option1=”11 km” option2=”7 km” option3=”6 km” option4=”5 km” correct=”option4″]

This question was previously asked in
UPSC CISF-AC-EXE – 2020
Download PDFAttempt Online
The distance R walks after he has rested is 5 km.
– Let R’s house be the starting point (0,0).
– 1. Walks 2 km towards North: Reaches point (0, 2).
– 2. Turns right (East) and walks 2 km: Reaches point (0+2, 2) = (2, 2).
– 3. Turns right (South) and walks 5 km: Reaches point (2, 2-5) = (2, -3).
– 4. Turns left (East) and walks 2 km: Reaches point (2+2, -3) = (4, -3).
– R stops at the point (4, -3). This point is 4 km East and 3 km South of his house (0,0).
– He returns home following a straight route from (4, -3) to (0,0).
– This is the distance between these two points, which can be calculated using the distance formula or by recognizing it as the hypotenuse of a right triangle with legs of length 4 km and 3 km.
– Distance = $\sqrt{(4-0)^2 + (-3-0)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ km.
– The distance R walks after he has rested is the length of this straight return journey, which is 5 km.
Direction and distance problems can be solved by representing the movements on a coordinate plane. The final position relative to the starting point can then be determined, and the straight-line distance calculated using the Pythagorean theorem.

138. Consider the following number : $3^5 \times 5^5 \times 6^{10} \times 1

Consider the following number : $3^5 \times 5^5 \times 6^{10} \times 10^6 \times 15^{12} \times 12^{15} \times 25^7$
What is the number of consecutive zeros at the end of the number given above ?

[amp_mcq option1=”50″ option2=”46″ option3=”37″ option4=”35″ correct=”option3″]

This question was previously asked in
UPSC CISF-AC-EXE – 2020
Download PDFAttempt Online
The number of consecutive zeros at the end of the number is 37.
– The number of consecutive zeros at the end of an integer is determined by the number of times 10 is a factor in its prime factorization. Since $10 = 2 \times 5$, we need to count the number of pairs of factors (2, 5). The number of zeros is equal to the minimum of the total number of factors of 2 and the total number of factors of 5 in the prime factorization of the given number.
– The given number is $3^5 \times 5^5 \times 6^{10} \times 10^6 \times 15^{12} \times 12^{15} \times 25^7$.
– Prime factorize each term:
– $3^5 = 3^5$
– $5^5 = 5^5$
– $6^{10} = (2 \times 3)^{10} = 2^{10} \times 3^{10}$
– $10^6 = (2 \times 5)^6 = 2^6 \times 5^6$
– $15^{12} = (3 \times 5)^{12} = 3^{12} \times 5^{12}$
– $12^{15} = (2^2 \times 3)^{15} = (2^2)^{15} \times 3^{15} = 2^{30} \times 3^{15}$
– $25^7 = (5^2)^7 = 5^{14}$
– Combine the prime factors:
– Factors of 2: $2^{10} \times 2^6 \times 2^{30} = 2^{10+6+30} = 2^{46}$. Total power of 2 is 46.
– Factors of 5: $5^5 \times 5^6 \times 5^{12} \times 5^{14} = 5^{5+6+12+14} = 5^{37}$. Total power of 5 is 37.
– (Factors of 3 are $3^5 \times 3^{10} \times 3^{12} \times 3^{15}$, but these do not contribute to zeros).
– The number of factors of 2 is 46. The number of factors of 5 is 37.
– The number of pairs of (2, 5) is $\min(46, 37) = 37$.
– Therefore, there are 37 consecutive zeros at the end of the number.
Counting trailing zeros involves finding the highest power of 10 that divides the number. Since $10 = 2 \times 5$, this is equivalent to finding the highest power of 5 in the prime factorization of the number (as there are always more factors of 2 than 5 in typical integers).

139. There are four large urns numbered 1 to 4. The number of different way

There are four large urns numbered 1 to 4. The number of different ways all the three balls numbered 1 to 3 can be kept inside the four urns is

[amp_mcq option1=”7″ option2=”12″ option3=”27″ option4=”64″ correct=”option4″]

This question was previously asked in
UPSC CISF-AC-EXE – 2020
Download PDFAttempt Online
The number of different ways all the three balls numbered 1 to 3 can be kept inside the four urns is 64.
– We have 3 distinct balls (numbered 1, 2, and 3).
– We have 4 distinct urns (numbered 1, 2, 3, and 4).
– Each ball can be placed into any one of the four urns, and the placement of one ball does not affect the choices for the other balls.
– For Ball 1, there are 4 possible urns it can be placed in.
– For Ball 2, there are also 4 possible urns it can be placed in, independently of where Ball 1 was placed.
– For Ball 3, there are similarly 4 possible urns it can be placed in, independently of the placement of the other balls.
– The total number of ways to place all three balls is the product of the number of choices for each ball.
– Total ways = (Choices for Ball 1) * (Choices for Ball 2) * (Choices for Ball 3) = 4 * 4 * 4 = $4^3$.
– $4^3 = 64$.
This is a problem involving permutations with repetition, specifically distributing distinct items into distinct bins. If there are $k$ distinct items and $n$ distinct bins, and each item can go into any bin, the total number of ways is $n^k$. Here, items are balls (k=3) and bins are urns (n=4).

140. The first ten letters of the English Alphabet are used in place of the

The first ten letters of the English Alphabet are used in place of the integers 0, 1, 2, …, 9 respectively. Then what is the value of BAG – ADD + FIG ?

[amp_mcq option1=”GFI” option2=”GAD” option3=”GJF” option4=”GFJ” correct=”option4″]

This question was previously asked in
UPSC CISF-AC-EXE – 2020
Download PDFAttempt Online
The value of BAG – ADD + FIG in the given code is GFJ.
– The first ten letters of the English Alphabet (A, B, C, D, E, F, G, H, I, J) are used in place of integers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.
– A=0, B=1, C=2, D=3, E=4, F=5, G=6, H=7, I=8, J=9.
– The question asks for the value of BAG – ADD + FIG. Assuming the letters represent digits in a number based on their position:
– BAG corresponds to the number formed by B(1), A(0), G(6), which is 106.
– ADD corresponds to the number formed by A(0), D(3), D(3), which is 033 or 33.
– FIG corresponds to the number formed by F(5), I(8), G(6), which is 586.
– Perform the calculation: 106 – 33 + 586.
– 106 – 33 = 73.
– 73 + 586 = 659.
– Now convert the result 659 back into letters using the given mapping:
– 6 corresponds to G.
– 5 corresponds to F.
– 9 corresponds to J.
– The resulting sequence of letters is GFJ.
This type of problem requires interpreting letters as numerical digits based on a defined code and then performing standard arithmetic operations. The result is then converted back into the letter code.
Page 1Page 2Page 3Page 4Page 5