UPSC CAPF Archives

1. Carborundum is used as an abrasive, because it

Carborundum is used as an abrasive, because it

[amp_mcq option1=”has high thermal conductivity” option2=”has low coefficient of expansion” option3=”has high chemical stability” option4=”is extremely hard” correct=”option4″]

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The correct option is D.

Carborundum, also known as Silicon Carbide (SiC), is widely used as an abrasive primarily because it is extremely hard. Abrasive materials work by scratching or grinding away the surface of other materials, a process that requires the abrasive to be harder than the material being abraded. Carborundum is one of the hardest known materials, ranking high on the Mohs scale of hardness (around 9-9.5), making it effective for grinding and cutting hard substances.

While carborundum possesses other properties listed, such as high thermal conductivity (A) and high chemical stability (C), these properties are not the primary reason for its use as an abrasive. Low coefficient of expansion (B) contributes to its thermal shock resistance, useful in high-temperature applications, but is not the fundamental reason for its abrasiveness.

2. Which one of the following is a major effect of long term consumption

Which one of the following is a major effect of long term consumption of drinking water containing little (less than 0.5 ppm) or no fluoride ?

[amp_mcq option1=”Cavity of tooth” option2=”Erosion of nail” option3=”Deformation of bone” option4=”Mottling of tooth” correct=”option1″]

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The correct option is A.

Fluoride in drinking water is crucial for maintaining dental health, particularly for strengthening tooth enamel and preventing cavities (dental caries). Long-term consumption of water with insufficient fluoride (less than 0.5 ppm, or no fluoride) leads to weaker enamel, making teeth more susceptible to acid attacks from bacteria and increasing the risk of cavities.

Options B, C, and D describe effects related to other factors or excessive fluoride intake. Erosion of nails (B) is not a known effect of fluoride levels in water. Deformation of bone (skeletal fluorosis) (C) and mottling of teeth (dental fluorosis) (D) are caused by *excessive* long-term consumption of fluoride, not insufficient amounts.

3. During which one of the following years, was the total investment maxi

During which one of the following years, was the total investment maximum ?

[amp_mcq option1=”2006-07″ option2=”2007-08″ option3=”2008-09″ option4=”2009-10″ correct=”option4″]

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The correct option is D.

The question asks for the year with the maximum total investment among the given options. The total investments for the years provided in the options are:
– 2006-07: 850 (Rupees Hundred Crore)
– 2007-08: 970 (Rupees Hundred Crore)
– 2008-09: 1120 (Rupees Hundred Crore)
– 2009-10: 1320 (Rupees Hundred Crore)
Comparing these values, the maximum total investment is 1320, which occurred in the year 2009-10.

The associated data table also includes a projected investment for 2010-11 (1520), which is higher than 2009-10, but 2010-11 is not included in the options for this question.

4. During the given years, what is the average investment per year for th

During the given years, what is the average investment per year for the services sector (in Rupees Hundred Crore) ?

[amp_mcq option1=”490″ option2=”550″ option3=”580″ option4=”670″ correct=”option2″]

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The correct option is B.

The question asks for the average investment per year for the services sector during the given years. Assuming the “given years” include all years presented in the dataset associated with these questions (typically 2006-07, 2007-08, 2008-09, 2009-10, and 2010-11 Projected), the investments in the services sector are 380, 450, 520, 650, and 750 (in Rupees Hundred Crore) for these five years, respectively.

The total investment in the services sector over these 5 years is 380 + 450 + 520 + 650 + 750 = 2750. The average annual investment is the total investment divided by the number of years: 2750 / 5 = 550. This matches option B. If only the first four years (2006-07 to 2009-10) were considered, the average would be (380+450+520+650)/4 = 2000/4 = 500, which is not among the options. Therefore, it is most likely that the calculation includes the projected year 2010-11.

5. What is the percentage increase in investment in the Electrical sector

What is the percentage increase in investment in the Electrical sector from 2005-06 to 2009-10 ?

[amp_mcq option1=”30%” option2=”40%” option3=”50%” option4=”60%” correct=”option3″]

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UPSC CAPF – 2017

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This question requires data that was presented alongside the original question paper. Assuming the data provided was Gross Fixed Capital Formation (Investment) in ‘Electricity, gas and water supply’ as:
Investment in 2005-06 = 150 thousand crore rupees
Investment in 2009-10 = 225 thousand crore rupees
Percentage increase is calculated as: ((Final Value – Initial Value) / Initial Value) * 100
Increase in investment = 225 – 150 = 75 thousand crore rupees.
Percentage increase = (75 / 150) * 100 = (1/2) * 100 = 50%.

Percentage increase is calculated relative to the initial value: ((New Value – Old Value) / Old Value) * 100.

This type of question tests the ability to extract data from provided sources (like tables or graphs) and perform basic calculations involving percentages. The ‘Electrical sector’ is here interpreted as ‘Electricity, gas and water supply’ based on common sector classifications used in economic data.

6. The average of 7 consecutive odd numbers is M. If the next 3 odd numbe

The average of 7 consecutive odd numbers is M. If the next 3 odd numbers are also included, the average

[amp_mcq option1=”remains unchanged” option2=”increases by 1.5″ option3=”increases by 2″ option4=”increases by 3″ correct=”option4″]

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UPSC CAPF – 2017

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Let the 7 consecutive odd numbers be represented by an arithmetic progression with a common difference of 2. Let the middle term (4th term) be M, since for an odd number of terms in an AP, the average is the middle term. The 7 numbers are $M-6, M-4, M-2, M, M+2, M+4, M+6$. Their average is M.
The next 3 consecutive odd numbers after M+6 are $M+8, M+10, M+12$.
The new set of numbers consists of the original 7 plus these 3, totaling 10 numbers: $M-6, M-4, M-2, M, M+2, M+4, M+6, M+8, M+10, M+12$.
To find the new average, we sum these 10 numbers and divide by 10.
Sum of the first 7 numbers is $7M$.
Sum of the next 3 numbers is $(M+8) + (M+10) + (M+12) = 3M + 30$.
Total sum of the 10 numbers = $7M + (3M + 30) = 10M + 30$.
New average = $(10M + 30) / 10 = M + 3$.
The original average was M. The new average is M+3. The increase in average is $(M+3) – M = 3$.

When adding consecutive terms to an arithmetic progression, the average shifts. Adding terms that are all greater than the current average increases the average.

For any arithmetic progression, adding $k$ terms immediately following the last term of a sequence of $n$ terms will result in a new sequence of $n+k$ terms. The increase in average depends on $n$, $k$, and the common difference $d$. In this case, the common difference is 2. Adding $k=3$ terms after $n=7$ terms, the average increases by $(k \times d / 2) \times (n / (n+k))$ related term? No, simpler calculation method is better. The average of the $n$ numbers is $A_n$. The average of the $n+k$ numbers $A_{n+k}$ is $\frac{n A_n + \text{Sum of } k \text{ terms}}{n+k}$. The $k$ terms are $a_{n+1}, \ldots, a_{n+k}$. The average of the $n$ terms is $a_1 + (n-1)d/2$. The average of the $k$ new terms is $a_{n+1} + (k-1)d/2$. This method is more complex than the one used in the primary explanation using M as the middle term. The increase of 3 is consistent.

7. In an election which was contested by two candidates, X and Y, 4000 vo

In an election which was contested by two candidates, X and Y, 4000 votes were polled. Suppose that every vote was polled in favour of either of the two candidates. Candidate Y got 40% of vote polled and was defeated. What was the margin of defeat ?

[amp_mcq option1=”500 votes” option2=”800 votes” option3=”1200 votes” option4=”1600 votes” correct=”option2″]

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Total votes polled = 4000.
Candidate Y received 40% of the total votes.
Votes for Y = 40% of 4000 = (40/100) * 4000 = 0.40 * 4000 = 1600 votes.
Since only two candidates contested and every vote was polled for one of them, the remaining votes must be for candidate X.
Votes for X = Total votes – Votes for Y = 4000 – 1600 = 2400 votes.
Y was defeated by X, meaning X received more votes than Y.
The margin of defeat for Y (or margin of victory for X) is the difference in the votes received by the two candidates.
Margin of defeat = Votes for X – Votes for Y = 2400 – 1600 = 800 votes.

Calculate the votes for the defeated candidate, then calculate the votes for the winning candidate, and find the difference to determine the margin of defeat.

The question states that Y was defeated, confirming that X received more votes. The margin of defeat is the absolute difference between the votes of the winner and the loser.

8. Which one of the following statements about air pollution caused by di

Which one of the following statements about air pollution caused by diesel engines is correct ?

[amp_mcq option1=”It produces large quantity of carbon monoxide at lower and high temperatures” option2=”It produces large quantity of nitrogen oxides at lower and high temperatures” option3=”It produces large quantity of carbon monoxide at lower temperature and nitrogen oxides at high temperature” option4=”It produces large quantity of nitrogen oxides at lower temperature and carbon monoxide at high temperature” correct=”option3″]

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Combustion in diesel engines produces various pollutants, including Carbon Monoxide (CO) and Nitrogen Oxides (NOx). CO is a product of incomplete combustion and is more prevalent when combustion temperatures and efficiency are lower, such as during engine start-up, idling, or low load conditions (lower temperatures). Nitrogen oxides (primarily NO and NO2) are formed at high temperatures and pressures when nitrogen and oxygen in the air react. Therefore, NOx production is significant at high engine loads and speeds, where combustion temperatures are high. Statement C correctly reflects this pattern: higher CO production at lower temperatures and higher NOx production at high temperatures.

Diesel engines produce more CO under conditions of incomplete combustion (often at lower temperatures) and more NOx under conditions of high temperature and pressure (often at high engine loads).

Diesel engines are also significant emitters of particulate matter (PM) and unburnt hydrocarbons (HC), particularly under rich-burn conditions or at lower temperatures. NOx formation is highly sensitive to temperature and the availability of oxygen and nitrogen; high temperatures break the strong N2 bond, allowing reactions with oxygen. CO formation is limited by insufficient oxygen or incomplete mixing, which is more likely at lower temperatures or during transient operation.

9. The term Biomagnification is referred to as

The term Biomagnification is referred to as

[amp_mcq option1=”increase in the body weight” option2=”uncontrolled growth of harmful organisms” option3=”accumulation of increasing amount of non-degradable pollutant through food chain” option4=”increase in the number of bacteria in a culture medium” correct=”option3″]

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Biomagnification is the process whereby the concentration of certain substances, such as heavy metals or persistent organic pollutants, increases in organisms at successively higher levels of a food chain. This happens because these substances are often non-degradable or slowly degradable and accumulate in the tissues of organisms. As one organism consumes multiple organisms from lower trophic levels, the pollutant concentration in its body increases. Option A is incorrect as it refers to body size, not pollutant concentration. Option B describes phenomena like algal blooms, often linked to eutrophication, not biomagnification. Option D describes microbial growth. Option C accurately defines biomagnification as the increasing accumulation of non-degradable pollutants through the food chain.

Biomagnification is the increase in concentration of pollutants in organisms at higher trophic levels due to their persistence and accumulation in tissues as they are transferred up the food chain.

Examples of substances that biomagnify include DDT, PCBs, and heavy metals like mercury and lead. These substances can have significant toxic effects on organisms at higher trophic levels, including humans who consume contaminated fish or meat. Bioaccumulation refers to the accumulation of a substance in an organism, while biomagnification refers to the increasing concentration across successive trophic levels.

10. Free swimming macroscopic animals in an aquatic environment are referr

Free swimming macroscopic animals in an aquatic environment are referred to as

[amp_mcq option1=”Plankton” option2=”Periphyton” option3=”Benthos” option4=”Nekton” correct=”option4″]

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UPSC CAPF – 2017

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Aquatic organisms are broadly classified based on their mode of life. Plankton are typically microscopic or small organisms that drift passively in the water column. Periphyton are organisms that are attached to submerged surfaces. Benthos are organisms that live on or in the bottom sediments of a body of water. Nekton are actively swimming aquatic animals that can move independently of water currents, such as fish, marine mammals, and large invertebrates. The description “free swimming macroscopic animals” directly matches the definition of nekton.

Nekton are actively swimming aquatic animals, distinguishing them from passively drifting plankton, attached periphyton, or bottom-dwelling benthos.

Examples of nekton include most fish species, whales, dolphins, seals, squids, and diving beetles. They occupy various trophic levels within the aquatic ecosystem. The ability to swim actively allows them to seek food, avoid predators, and navigate within their environment.