UPSC CAPF Archives

1. Which one of the following states has the maximum number of registered

Which one of the following states has the maximum number of registered E-Waste recyclers / dismantlers ?

[amp_mcq option1=”Maharashtra” option2=”Tamil Nadu” option3=”Karnataka” option4=”Uttar Pradesh” correct=”option1″]

This question was previously asked in

UPSC CAPF – 2016

The correct answer is Maharashtra.

– This question pertains to the number of registered E-Waste recyclers and dismantlers, which falls under the E-Waste Management Rules. The Central Pollution Control Board (CPCB) is the regulatory body in India.
– Based on CPCB data available around the time frame from which this question likely originates (e.g., data from 2016-2019), Maharashtra has consistently had the highest number of registered E-Waste dismantling and recycling units among all Indian states.
– States like Tamil Nadu, Karnataka, and Uttar Pradesh also have a significant number of registered units, but Maharashtra has typically led in this count.

The number of registered facilities can change over time as new licenses are granted and others expire or are revoked. However, historical data places Maharashtra at the top for having the maximum number of registered E-Waste recyclers/dismantlers. This is often attributed to the state’s large population, high e-waste generation, and relatively developed industrial infrastructure for processing waste.

2. Perth located on 118° East Longitude will be celebrating New Year even

Perth located on 118° East Longitude will be celebrating New Year event on 1^st of January 2017 at 6:00 AM. At that time, what would be the time at Los Angeles located on 110° West Longitude ?

[amp_mcq option1=”9:12 PM of 1^st January 2017″ option2=”2:48 PM of 31^st December 2016″ option3=”11:40 PM of 31^st December 2016″ option4=”5:28 AM of 1^st January 2017″ correct=”option2″]

This question was previously asked in

UPSC CAPF – 2016

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The correct answer is B) 2:48 PM of 31st December 2016.

– Longitude of Perth = 118° East.
– Longitude of Los Angeles = 110° West.
– The total difference in longitude = 118° (East) + 110° (West) = 228°.
– The Earth rotates 360° in 24 hours.
– The time difference for 1° longitude = 24 hours / 360° = 1/15 hours = 4 minutes.
– The total time difference between Perth and Los Angeles = 228° * 4 minutes/° = 912 minutes.
– Convert minutes to hours: 912 minutes / 60 minutes/hour = 15 hours and 12 minutes (912 = 15 * 60 + 12).
– Perth is located in the East, so its time is ahead of Los Angeles, which is in the West.
– Time at Los Angeles = Time at Perth – Time difference.
– Time at Perth = 6:00 AM on 1st January 2017.
– Subtract 15 hours 12 minutes from 6:00 AM, 1st January 2017.
– Subtracting 6 hours from 6:00 AM Jan 1st gives 0:00 AM (midnight) on Jan 1st.
– We need to subtract another 15h 12m – 6h = 9h 12m.
– Subtracting 9 hours from 0:00 AM Jan 1st gives 15:00 (3:00 PM) on 31st December 2016.
– Subtracting the final 12 minutes from 3:00 PM Dec 31st gives 14:48 PM (2:48 PM) on 31st December 2016.

The International Date Line is located roughly around 180° longitude. Moving west across the International Date Line means going back one day. The total longitudinal difference of 228° is less than 180° in either direction from the prime meridian, but crossing the 0° longitude is involved. Perth (118°E) and Los Angeles (110°W) are on opposite sides of the prime meridian, and their difference is calculated by summing their longitudes. Since Perth is East of Los Angeles, its time is ahead, and the date can be different. In this case, going back 15 hours 12 minutes from Jan 1st in Perth takes us back into Dec 31st in Los Angeles.

3. Which of the following statements are correct ? 1. Kolkata port is t

Which of the following statements are correct ?

1. Kolkata port is the only riverine major port of India
2. The port of Cochin is located on the Willington Island
3. Maharashtra has three major ports
4. Mundra port is India’s largest private sector port

Select the correct answer using the code given below :

[amp_mcq option1=”1 and 2 only” option2=”3 and 4 only” option3=”2, 3 and 4″ option4=”1, 2 and 4″ correct=”option4″]

This question was previously asked in

UPSC CAPF – 2016

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The correct answer is D) 1, 2 and 4.

– Statement 1: Kolkata port is the only riverine major port of India. This statement is correct. Kolkata Port (Syama Prasad Mookerjee Port) is located on the Hooghly River, making it a riverine port. It is India’s only major port situated on a river.
– Statement 2: The port of Cochin is located on the Willington Island. This statement is correct. Cochin Port is built on Willingdon Island, a man-made island in the Vembanad Lake, Kerala.
– Statement 3: Maharashtra has three major ports. This statement is incorrect. Maharashtra has two major ports: Mumbai Port and Jawaharlal Nehru Port Trust (JNPT). There are other ports in Maharashtra, but only these two are classified as major ports.
– Statement 4: Mundra port is India’s largest private sector port. This statement is correct. Mundra Port, located in Gujarat and operated by Adani Ports and SEZ Limited, is India’s largest private port by cargo handling capacity and volume.

As per the classification by the Ministry of Shipping, Government of India, there are 12 major ports in India. Statement 3 is factually incorrect regarding the number of major ports in Maharashtra. Statements 1, 2, and 4 provide accurate information about the respective ports.

4. Which one of the following inequalities is always true for positive re

Which one of the following inequalities is always true for positive real numbers x, y ?

[amp_mcq option1=”xy > x + y” option2=”(x + y) < (x + y)²" option3="x + y < x² + y²" option4="1 + x + y < (1 + x + y)²" correct="option4"]

This question was previously asked in

UPSC CAPF – 2016

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The correct answer is D) 1 + x + y < (1 + x + y)².

– The question asks for the inequality that is *always* true for positive real numbers x and y.
– A) xy > x + y: If x=1, y=1, 1 > 2 is false. Not always true.
– B) (x + y) < (x + y)²: Let z = x + y. Since x, y > 0, z > 0. The inequality is z < z². This is equivalent to z² - z > 0, or z(z-1) > 0. Since z > 0, this is true only if z-1 > 0, i.e., z > 1. If x=0.1, y=0.1, x+y=0.2, which is not greater than 1. 0.2 < (0.2)² = 0.04 is false. Not always true. - C) x + y < x² + y²: If x=1, y=1, 1+1 < 1²+1² is 2 < 2, which is false. If x=0.5, y=0.5, 0.5+0.5 < 0.5²+0.5² is 1 < 0.25+0.25=0.5, which is false. Not always true. - D) 1 + x + y < (1 + x + y)²: Let w = 1 + x + y. Since x and y are positive real numbers (x>0, y>0), 1 + x + y must be greater than 1 (w > 1). The inequality becomes w < w². This is equivalent to w² - w > 0, or w(w-1) > 0. Since w > 1, both w and (w-1) are positive. Therefore, their product w(w-1) is always positive. Thus, w < w² is always true when w > 1. As 1 + x + y > 1 for positive x, y, the inequality 1 + x + y < (1 + x + y)² is always true.

The inequality z < z² is true for z < 0 or z > 1. Since x, y are positive, 1 + x + y is always greater than 1, falling into the w > 1 range.

5. Which one of the following is different from the remaining three ?

Which one of the following is different from the remaining three ?

[amp_mcq option1=”Triangle” option2=”Square” option3=”Circle” option4=”Ellipse” correct=”option3″]

This question was previously asked in

UPSC CAPF – 2016

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The correct answer is Circle.

– Triangle and Square are polygons, which are closed figures formed by straight line segments.
– Circle and Ellipse are closed curved shapes, not polygons. They are also both conic sections (formed by intersecting a cone with a plane).
– This grouping places Triangle and Square in one category and Circle and Ellipse in another, which doesn’t allow one to be different from the other three.
– Let’s consider other properties:
– Triangle: Can have varying angles and side lengths (scalene, isosceles, equilateral). Curvature is concentrated at vertices.
– Square: All angles are 90 degrees, all sides equal. Constant zero curvature along sides, infinite curvature at vertices.
– Ellipse: Varying curvature along the curve.
– Circle: Constant curvature along the curve. A circle is a special case of an ellipse where the two foci coincide and the eccentricity is zero.
– The property that distinguishes the Circle from the other three is its constant curvature. Triangle, Square, and Ellipse all have curvature that varies or is concentrated at points (infinite curvature at vertices for polygons).

Other possible but less compelling distinctions could be made (e.g., minimum sides for polygon – Triangle, regularity – Square and equilateral Triangle are regular polygons, Circle is ‘most regular’ curve), but constant curvature provides a clear mathematical distinction that groups Triangle, Square, and Ellipse (non-constant/infinite curvature) against Circle (constant curvature).

6. A circular coin of radius 1 cm is allowed to roll freely on the periph

A circular coin of radius 1 cm is allowed to roll freely on the periphery over a circular disc of radius 10 cm. If the disc has no movement and the coin completes one revolution rolling on the periphery over the disc and without slipping, then what is the number of times the coin rotated about its centre ?

[amp_mcq option1=”10″ option2=”10.5″ option3=”11″ option4=”12″ correct=”option3″]

This question was previously asked in

UPSC CAPF – 2016

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The correct answer is 11.

– Let R be the radius of the large disc (10 cm) and r be the radius of the coin (1 cm).
– The coin rolls on the periphery of the disc without slipping.
– The path traced by the center of the coin is a circle with radius R + r = 10 + 1 = 11 cm.
– When a smaller circle (radius r) rolls on the outside of a larger circle (radius R), the number of rotations the smaller circle makes about its own center for one complete revolution around the larger circle is given by the formula (R/r) + 1.
– In this case, R=10 cm and r=1 cm.
– Number of rotations = (10 cm / 1 cm) + 1 = 10 + 1 = 11.
– This formula accounts for the rotation due to the rolling action (distance rolled / circumference of coin) and the additional rotation due to the coin revolving around the large disc (one full revolution).
– The distance rolled by the coin along the edge of the disc is the circumference of the disc, which is 2 * pi * R = 2 * pi * 10 = 20 pi cm. The rotation due to this rolling is (20 pi) / (2 * pi * r) = 20 pi / (2 * pi * 1) = 10 rotations (R/r).
– As the coin’s center completes one revolution around the large disc’s center (path circumference 2*pi*(R+r)), the coin itself also makes one full turn due to the change in orientation as it circles the origin.
– The total number of rotations relative to a fixed direction is the sum of rotations from rolling and rotation from revolution.
– Total rotations = (R/r) + 1 = 10 + 1 = 11.

This result is related to the study of epicycloids, the path traced by a point on the circumference of the smaller circle. The number of rotations of the smaller circle about its center relative to a fixed external frame during one revolution around the larger circle is (R/r) + 1 when rolling on the outside, and (R/r) – 1 when rolling on the inside (hypocycloid).

7. A device can write 100 digits in 1 minute. It starts writing natural n

A device can write 100 digits in 1 minute. It starts writing natural numbers. The device is stopped after running it for half an hour. It is found that the last number it was writing is incomplete. The number is :

[amp_mcq option1=”3000″ option2=”3001″ option3=”1026″ option4=”1027″ correct=”option4″]

This question was previously asked in

UPSC CAPF – 2016

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The correct answer is 1027.

– The device writes 100 digits per minute for 30 minutes, so a total of 100 * 30 = 3000 digits are written.
– Natural numbers start from 1.
– Digits used for 1-digit numbers (1-9): 9 numbers * 1 digit/number = 9 digits. (Numbers 1 to 9 completed)
– Digits used for 2-digit numbers (10-99): 90 numbers * 2 digits/number = 180 digits. (Numbers 10 to 99 completed)
– Total digits used for 1-digit and 2-digit numbers = 9 + 180 = 189 digits. (Numbers 1 to 99 completed)
– Remaining digits to be written = 3000 – 189 = 2811 digits.
– These remaining digits are used for 3-digit numbers (100-999) and then 4-digit numbers (1000-…).
– Digits used for all 3-digit numbers (100-999): 900 numbers * 3 digits/number = 2700 digits. (Numbers 100 to 999 completed)
– Total digits used for 1-digit, 2-digit, and 3-digit numbers = 189 + 2700 = 2889 digits. (Numbers 1 to 999 completed)
– Remaining digits = 3000 – 2889 = 111 digits.
– These 111 digits are used for writing 4-digit numbers (1000, 1001, …). Each 4-digit number uses 4 digits.
– The digits come from the sequence 1000, 1001, 1002, …
– Number of full 4-digit numbers whose digits are included in the 111 digits = floor(111 / 4) = 27 numbers.
– These 27 numbers are 1000, 1001, …, 1000 + (27 – 1) = 1026.
– Digits used for these 27 full 4-digit numbers = 27 * 4 = 108 digits.
– Total digits used so far = 2889 (up to 999) + 108 (for 1000 to 1026) = 2997 digits.
– The numbers written completely are 1, 2, …, 999, 1000, …, 1026.
– Remaining digits to reach 3000 = 3000 – 2997 = 3 digits.
– These 3 digits are the first three digits of the next number in the sequence, which is 1027.
– The digits of 1027 are 1, 0, 2, 7.
– The device writes the 2998th digit (‘1’ of 1027), the 2999th digit (‘0’ of 1027), and the 3000th digit (‘2’ of 1027).
– The device stops after writing the digit ‘2’ of the number 1027.
– The last number it was writing is 1027, and it is incomplete (only ‘102’ has been written).

The calculation steps carefully account for the digits used by numbers of increasing length (1-digit, 2-digits, 3-digits) until the total number of digits approaches 3000, at which point the next number in the sequence is partially written.

8. 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 (Isotope)	List II (Application)
A. Carbon	1. Treatment of goitre
B. Cobalt	2. Calculation of age of the earth
C. Iodine	3. Treatment of cancer
D. Uranium	4. Fuel in nuclear reactors

Code :

[amp_mcq option1=”2 3 1 4″ option2=”2 1 3 4″ option3=”4 1 3 2″ option4=”4 3 1 2″ correct=”option1″]

This question was previously asked in

UPSC CAPF – 2016

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The correct answer is A) 2 3 1 4.

– B. Cobalt (specifically Cobalt-60) is widely used in radiotherapy for the treatment of cancer due to its emission of gamma rays. This matches with application 3.
– C. Iodine (specifically Iodine-131) is used in the diagnosis and treatment of thyroid conditions, including goitre, as the thyroid gland absorbs iodine. This matches with application 1.
– D. Uranium (specifically Uranium-235) is the primary fuel used in nuclear reactors for generating energy through nuclear fission. Uranium isotopes (like Uranium-238) are also used in radiometric dating (Uranium-Lead dating) to calculate the age of rocks and the Earth. Application 4 (Fuel in nuclear reactors) and Application 2 (Calculation of age of the earth) are both valid uses of Uranium isotopes.
– A. Carbon: Carbon-14 is used for radiocarbon dating, but this is for dating organic materials up to around 50,000 years old, not for the age of the Earth itself (which is billions of years, dated using isotopes like Uranium). Carbon (graphite) is used as a moderator in some nuclear reactors, but not as fuel. Neither application 2 nor 4 is a direct or primary application of Carbon isotopes compared to other options.

Let’s examine the pairings in option A: A-2 (Carbon for Age of Earth), B-3 (Cobalt for Cancer), C-1 (Iodine for Goitre), D-4 (Uranium for Fuel). B-3, C-1, and D-4 are correct pairings. A-2 is incorrect as Carbon-14 dating is for organic matter, not the age of the Earth.
Let’s examine the pairings in option D: A-4 (Carbon for Fuel), B-3 (Cobalt for Cancer), C-1 (Iodine for Goitre), D-2 (Uranium for Age of Earth). B-3, C-1, and D-2 are correct pairings. A-4 is incorrect as Carbon is a moderator, not fuel, in reactors.
Both option A and D contain correct pairings for B and C. Both contain a correct pairing for D (either 2 or 4, as both are valid uses). Both contain an incorrect pairing for A. Comparing the incorrect pairings for A, linking Carbon-14 dating to ‘Age of Earth’ might be considered slightly less incorrect in the context of general dating methods related to Earth’s history than linking Carbon (as moderator) to ‘Fuel’. Therefore, option A (A-2, B-3, C-1, D-4) is the most likely intended answer, despite the flaw in the A-2 pairing.

9. Which one of the following is not a colloid ?

Which one of the following is not a colloid ?

[amp_mcq option1=”Milk” option2=”Mud” option3=”Butter” option4=”Boric acid” correct=”option4″]

This question was previously asked in

UPSC CAPF – 2016

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The correct answer is Boric acid.

– Colloids are heterogeneous mixtures where the solute particles are dispersed evenly throughout the solvent as particles that are intermediate in size between those in true solutions and those in suspensions. Colloidal particles typically have a size between 1 nanometer and 1 micrometer.
– True solutions have particle sizes less than 1 nm. Suspensions have particle sizes greater than 1 micrometer, and the particles tend to settle out over time.
– Milk is an emulsion, which is a type of colloid where liquid droplets are dispersed in another liquid.
– Butter is also an emulsion, a type of colloid.
– Mud is primarily a suspension, where solid particles (soil, clay, silt) are dispersed in water. While mud can contain some clay particles of colloidal size, it is generally classified as a suspension because the larger particles settle out.
– Boric acid (H₃BO₃) is a molecular substance that dissolves in water to form a true solution. The particles in a boric acid solution are hydrated boric acid molecules, which are of molecular size, much smaller than colloidal particles.

The Tyndall effect (scattering of light) is a characteristic property of colloids, but not of true solutions. Suspensions may also scatter light, but they are unstable and settle. Boric acid solution will not show the Tyndall effect, unlike milk or muddy water (though settling occurs in mud).

10. Which one of the following statements is not correct ?

Which one of the following statements is not correct ?

[amp_mcq option1=”The rate of evaporation depends on temperature” option2=”The rate of evaporation does not depend on surface area exposed to the atmosphere but on volume of the liquid” option3=”The rate of evaporation depends on humidity of the surroundings” option4=”The rate of evaporation depends on the wind speed” correct=”option2″]

This question was previously asked in

UPSC CAPF – 2016

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The statement that the rate of evaporation does not depend on surface area exposed to the atmosphere but on volume of the liquid is not correct.

Evaporation is the process where liquid turns into gas below its boiling point. It occurs at the surface of the liquid. The rate of evaporation is influenced by several factors:
A) Temperature: Higher temperature increases the kinetic energy of liquid molecules, making it easier for them to escape the surface. (Correct statement)
B) Surface area exposed: A larger surface area allows more molecules to be at the surface and escape into the atmosphere per unit time, thus increasing the rate of evaporation. Volume of the liquid affects the total amount available for evaporation but not the instantaneous rate of evaporation at a given surface area. (Incorrect statement)
C) Humidity of the surroundings: Lower humidity means the air has a lower concentration of water vapor, creating a larger concentration gradient between the liquid surface and the air, leading to faster evaporation. (Correct statement)
D) Wind speed: Wind blows away the saturated air layer above the liquid surface, replacing it with drier air, which increases the rate of evaporation. (Correct statement)

Evaporation is a crucial part of the water cycle. Other factors influencing evaporation include air pressure and the presence of dissolved substances in the liquid.