UPSC CAPF Archives

31. Liquid water is denser than ice due to

Liquid water is denser than ice due to

higher surface tension

hydrogen bonding

van der Waals forces

covalent bonding

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UPSC CAPF – 2017

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Liquid water is denser than ice. This anomalous property is due to the unique structure formed by hydrogen bonding. In ice, water molecules form a rigid, open lattice structure (like a hexagonally ordered network) held together by hydrogen bonds. This structure is less compact than the arrangement of molecules in liquid water. In liquid water, while hydrogen bonds still exist, they are constantly breaking and reforming, allowing molecules to pack closer together, thus increasing the density compared to ice.

Hydrogen bonding in water is responsible for several of its anomalous properties, including the fact that solid water (ice) is less dense than liquid water.

Most substances contract and become denser when they solidify. Water expands and becomes less dense when it freezes. This property is crucial for life on Earth, as ice floats on water bodies, insulating the liquid water below and preventing lakes and rivers from freezing solid from the bottom up.

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32. As per Census 2011, which one of the following is the correct descendi

As per Census 2011, which one of the following is the correct descending order of States in India in respect of sex ratio (female per thousand of males) ?

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West Bengal - Madhya Pradesh - Uttarakhand - Jammu and Kashmir

Madhya Pradesh - West Bengal - Jammu and Kashmir - Uttarakhand

Uttarakhand - West Bengal - Madhya Pradesh - Jammu and Kashmir

West Bengal - Uttarakhand - Madhya Pradesh - Jammu and Kashmir

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According to Census 2011 data, the sex ratio (females per 1000 males) for the listed states is approximately:
Uttarakhand: 963
West Bengal: 950
Madhya Pradesh: 931
Jammu and Kashmir: 889 (for the undivided state)
Arranging these in descending order of sex ratio gives: Uttarakhand (963) > West Bengal (950) > Madhya Pradesh (931) > Jammu and Kashmir (889).

Sex ratio data for selected Indian states as per Census 2011. Kerala had the highest sex ratio (1084) and Haryana the lowest (879) among major states in 2011.

Sex ratio is an important demographic indicator reflecting the balance between the number of females and males in a population. Regional variations in sex ratio can be influenced by factors like birth rates, mortality rates, migration, and societal practices (e.g., sex-selective abortion, neglect of female children).

33. 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 (Mineral)	List II (Mine)
A. Zinc	1. Amjhore
B. Gold	2. Sukinda
C. Chromite	3. Zawar
D. Pyrite	4. Hutti

Code :

1 2 4 3

3 2 4 1

3 4 2 1

1 4 2 3

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UPSC CAPF – 2017

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Matching List I (Mineral) with List II (Mine):
A. Zinc: Zawar mines are a major site for zinc mining in Rajasthan. (A-3)
B. Gold: Hutti mines in Karnataka are well-known gold mines. (B-4)
C. Chromite: Sukinda Valley in Odisha is the largest chromite producing area in India. (C-2)
D. Pyrite: Amjhore in Bihar (now Jharkhand) was historically known for pyrite deposits and mining. (D-1)
The correct match is A-3, B-4, C-2, D-1.

Knowledge of important mineral deposits and associated mining locations in India.

Other important mineral mining areas in India include Kolar Gold Fields (KGF – historically significant for gold, but production is minimal now), Khetri for Copper, Singareni for Coal, Iron ore belts in Odisha, Jharkhand, Chhattisgarh, Karnataka.

34. Which one of the following pairs of Island and Sea/Ocean is NOT correc

Which one of the following pairs of Island and Sea/Ocean is NOT correctly matched ?

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Cyprus : Mediterranean Sea

Falkland : Atlantic Ocean

Chagos : North Pacific Ocean

Islas Cocos : Indian Ocean

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The pair ‘Chagos : North Pacific Ocean’ is NOT correctly matched. The Chagos Archipelago is located in the central Indian Ocean, approximately 500 km south of the Maldives archipelago.

Geographical location of major island groups and the seas/oceans they are situated in.

A) Cyprus is correctly located in the Mediterranean Sea.
B) The Falkland Islands are correctly located in the South Atlantic Ocean.
D) Islas Cocos can refer to multiple islands. If it refers to Cocos (Keeling) Islands, they are in the Indian Ocean. If it refers to Cocos Island off Costa Rica, it’s in the Pacific Ocean. However, given the clear incorrectness of option C (Chagos in North Pacific), C is the intended answer.

35. One year ago, a father was four times as old as his son. After six yea

One year ago, a father was four times as old as his son. After six years his age exceeds twice his son’s age by 9 years. The ratio of their present age is

9 : 2

11 : 3

12 : 5

13 : 4

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UPSC CAPF – 2017

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The ratio of their present age is 11 : 3.

– Let the present age of the son be $S$ years and the present age of the father be $F$ years.
– One year ago, the son’s age was $S-1$ and the father’s age was $F-1$.
– According to the first condition: $F-1 = 4(S-1)$. This simplifies to $F-1 = 4S – 4$, so $F = 4S – 3$ (Equation 1).
– After six years, the son’s age will be $S+6$ and the father’s age will be $F+6$.
– According to the second condition: The father’s age ($F+6$) exceeds twice his son’s age ($2(S+6)$) by 9 years. So, $F+6 = 2(S+6) + 9$.
– This simplifies to $F+6 = 2S + 12 + 9$, which is $F+6 = 2S + 21$. So, $F = 2S + 15$ (Equation 2).
– Now we have a system of two linear equations for $F$ and $S$:
1) $F = 4S – 3$
2) $F = 2S + 15$
– Equating the expressions for $F$: $4S – 3 = 2S + 15$.
– Subtract $2S$ from both sides: $2S – 3 = 15$.
– Add 3 to both sides: $2S = 18$.
– Divide by 2: $S = 9$. The son’s present age is 9 years.
– Substitute $S=9$ into Equation 1 (or Equation 2) to find $F$:
$F = 4(9) – 3 = 36 – 3 = 33$. The father’s present age is 33 years.
– The ratio of their present age (Father : Son) is $F : S = 33 : 9$.
– This ratio can be simplified by dividing both numbers by their greatest common divisor, which is 3.
$33 \div 3 = 11$
$9 \div 3 = 3$
– The simplified ratio is 11 : 3.

It’s always a good idea to check the answer with the original conditions.
Present ages: Father = 33, Son = 9.
One year ago: Father = 32, Son = 8. Is 32 four times 8? Yes, $32 = 4 \times 8$. (Condition 1 satisfied).
After six years: Father = $33+6 = 39$, Son = $9+6 = 15$. Is 39 equal to twice the son’s age plus 9? $2 \times 15 + 9 = 30 + 9 = 39$. Yes. (Condition 2 satisfied).
The calculated ages satisfy both conditions.

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36. Two pipes A and B can fill a tank in 12 minutes and 16 minutes respect

Two pipes A and B can fill a tank in 12 minutes and 16 minutes respectively. If both the pipes are opened together, then after how much time, B should be closed so that the tank is full in 9 minutes ?

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3 ½ min

4 min

4 ½ min

4 ¾ min

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UPSC CAPF – 2017

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Pipe B should be closed after 4 minutes so that the tank is full in 9 minutes.

– Pipe A fills the tank in 12 minutes, so its filling rate is $1/12$ of the tank per minute.
– Pipe B fills the tank in 16 minutes, so its filling rate is $1/16$ of the tank per minute.
– Let $t$ be the time (in minutes) for which both pipes are open together.
– After time $t$, pipe B is closed, and only pipe A continues to fill the tank for the remaining time.
– The total time to fill the tank is 9 minutes. So, pipe A works for the entire 9 minutes. Pipe B works only for the first $t$ minutes.
– Amount filled by A in 9 minutes = Rate of A $\times$ Time A worked = $(1/12) \times 9 = 9/12 = 3/4$ of the tank.
– Amount filled by B in $t$ minutes = Rate of B $\times$ Time B worked = $(1/16) \times t = t/16$ of the tank.
– The total amount filled is the sum of the amounts filled by A and B, which is 1 full tank.
– So, $(3/4) + (t/16) = 1$.
– To solve for $t$, multiply the entire equation by the least common multiple of 4 and 16, which is 16:
$16 \times (3/4) + 16 \times (t/16) = 16 \times 1$
$4 \times 3 + t = 16$
$12 + t = 16$
$t = 16 – 12 = 4$.
– Therefore, pipe B should be closed after 4 minutes.

Alternatively, we can set up the equation based on the duration both pipes work together and the duration only A works. Let B be closed after $t$ minutes. Both A and B work for $t$ minutes, and A works alone for $(9-t)$ minutes.
Work done by A and B together in $t$ minutes = $(1/12 + 1/16) \times t = (4/48 + 3/48) \times t = (7/48)t$.
Work done by A alone in $(9-t)$ minutes = $(1/12) \times (9-t)$.
Total work = $(7/48)t + (1/12)(9-t) = 1$.
Multiply by 48: $7t + 4(9-t) = 48 \Rightarrow 7t + 36 – 4t = 48 \Rightarrow 3t + 36 = 48 \Rightarrow 3t = 12 \Rightarrow t = 4$.
Both methods yield the same result.

37. The length of a rectangle is increased by 60%. By what per cent would

The length of a rectangle is increased by 60%. By what per cent would the width have to be decreased to maintain the same area ?

37.5 %

60%

75%

120%

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The width would have to be decreased by 37.5% to maintain the same area.

– Let the original length be $L$ and original width be $W$. The original area is $A = L \times W$.
– The length is increased by 60%, so the new length $L’$ is $L + 0.60L = 1.60L$.
– Let the new width be $W’$. The new area $A’ = L’ \times W’ = 1.60L \times W’$.
– To maintain the same area, $A’ = A$, so $1.60L \times W’ = L \times W$.
– We can solve for $W’$: $W’ = \frac{L \times W}{1.60L} = \frac{W}{1.60} = \frac{W}{8/5} = \frac{5}{8}W$.
– The decrease in width is $W – W’ = W – \frac{5}{8}W = \frac{3}{8}W$.
– The percentage decrease in width is $\frac{\text{Decrease in width}}{\text{Original width}} \times 100\% = \frac{(3/8)W}{W} \times 100\% = \frac{3}{8} \times 100\%$.
– $\frac{3}{8} = 0.375$, so the percentage decrease is $0.375 \times 100\% = 37.5\%$.

This problem demonstrates the inverse relationship between dimensions when the area is kept constant. If one dimension is increased, the other must be decreased proportionally to maintain the same area. The percentage change in one dimension results in a different percentage change in the other dimension for a constant area, especially when the changes are expressed relative to the original values.

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38. Bronze is an alloy of copper and

Bronze is an alloy of copper and

nickel

iron

tin

aluminium

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Bronze is an alloy of copper and tin.

– An alloy is a mixture of metals, or a mixture of a metal and another element.
– Bronze is historically one of the most important alloys, known for its hardness and durability compared to pure copper.
– Typically, bronze consists primarily of copper, usually with tin as the main additive. The amount of tin varies but is commonly between 5% and 20%.

Brass is another common copper alloy, primarily consisting of copper and zinc. Other elements like aluminum, manganese, phosphorus, or silicon may be added to bronze to achieve specific properties, but tin is the defining secondary element for classical bronze.

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39. Which one of the following instruments is used for measuring moisture

Which one of the following instruments is used for measuring moisture content of air ?

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Hydrometer

Hygrometer

Hypsometer

Pycnometer

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A Hygrometer is used for measuring moisture content of air.

– A hygrometer is an instrument used to measure the amount of water vapor in the air, soil, or confined spaces. This is commonly known as humidity.
– A hydrometer is used to measure the specific gravity (or relative density) of liquids, i.e., the ratio of the density of the liquid to the density of water.
– A hypsometer is used to measure elevations or the boiling point of a liquid.
– A pycnometer is a device used to measure the density of solids or liquids.

Different types of hygrometers exist, including mechanical hygrometers, electrical hygrometers (using resistance or capacitance), and dew-point hygrometers. They are essential tools in meteorology, HVAC systems, and various industrial and scientific applications where humidity control is important.

40. Which one of the following is NOT correct about organic farming ?

Which one of the following is NOT correct about organic farming ?

It does not use genetically modified seeds

Synthetic pesticides or fertilizers are not used

It uses minimal crop rotation

It uses ecologically protective practices

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UPSC CAPF – 2017

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The statement “It uses minimal crop rotation” is NOT correct about organic farming.

– Organic farming is an agricultural system that avoids the use of synthetic fertilizers, pesticides, herbicides, genetically modified organisms (GMOs), and growth hormones. (Statements A and B are correct).
– It focuses on ecological balance, biodiversity conservation, and sustainable practices to maintain soil fertility and manage pests and diseases. (Statement D is correct).
– Crop rotation is a fundamental practice in organic farming. It is used extensively to improve soil health, break pest and disease cycles, manage weeds, and diversify nutrient uptake from the soil, thereby reducing reliance on external inputs. Organic farming uses *extensive*, not minimal, crop rotation.

Other common practices in organic farming include using compost and manure for soil fertility, biological pest control, cover cropping, and conservation tillage. The goal is to build healthy soil and a resilient ecosystem.

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