Liquid water is denser than ice due to
[amp_mcq option1=”higher surface tension” option2=”hydrogen bonding” option3=”van der Waals forces” option4=”covalent bonding” correct=”option2″]
This question was previously asked in
UPSC CAPF – 2017
UPSC CAPF
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) ?
[amp_mcq option1=”West Bengal – Madhya Pradesh – Uttarakhand – Jammu and Kashmir” option2=”Madhya Pradesh – West Bengal – Jammu and Kashmir – Uttarakhand” option3=”Uttarakhand – West Bengal – Madhya Pradesh – Jammu and Kashmir” option4=”West Bengal – Uttarakhand – Madhya Pradesh – Jammu and Kashmir” correct=”option3″]
This question was previously asked in
UPSC CAPF – 2017
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).
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 :
| List I (Mineral) |
List II (Mine) |
| A. Zinc | 1. Amjhore |
| B. Gold | 2. Sukinda |
| C. Chromite | 3. Zawar |
| D. Pyrite | 4. Hutti |
Code :
[amp_mcq option1=”1 2 4 3″ option2=”3 2 4 1″ option3=”3 4 2 1″ option4=”1 4 2 3″ correct=”option3″]
This question was previously asked in
UPSC CAPF – 2017
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.
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 ?
[amp_mcq option1=”Cyprus : Mediterranean Sea” option2=”Falkland : Atlantic Ocean” option3=”Chagos : North Pacific Ocean” option4=”Islas Cocos : Indian Ocean” correct=”option3″]
This question was previously asked in
UPSC CAPF – 2017
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
[amp_mcq option1=”9 : 2″ option2=”11 : 3″ option3=”12 : 5″ option4=”13 : 4″ correct=”option2″]
This question was previously asked in
UPSC CAPF – 2017
– 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.
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.
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 ?
[amp_mcq option1=”3 ½ min” option2=”4 min” option3=”4 ½ min” option4=”4 ¾ min” correct=”option2″]
This question was previously asked in
UPSC CAPF – 2017
– 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.
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 ?
[amp_mcq option1=”37.5 %” option2=”60%” option3=”75%” option4=”120%” correct=”option1″]
This question was previously asked in
UPSC CAPF – 2017
– 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 $0.375 \times 100\% = 37.5\%$.
38. Bronze is an alloy of copper and
Bronze is an alloy of copper and
[amp_mcq option1=”nickel” option2=”iron” option3=”tin” option4=”aluminium” correct=”option3″]
This question was previously asked in
UPSC CAPF – 2017
– 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%.
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 ?
[amp_mcq option1=”Hydrometer” option2=”Hygrometer” option3=”Hypsometer” option4=”Pycnometer” correct=”option2″]
This question was previously asked in
UPSC CAPF – 2017
– 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.
40. Which one of the following is NOT correct about organic farming ?
Which one of the following is NOT correct about organic farming ?
[amp_mcq option1=”It does not use genetically modified seeds” option2=”Synthetic pesticides or fertilizers are not used” option3=”It uses minimal crop rotation” option4=”It uses ecologically protective practices” correct=”option3″]
This question was previously asked in
UPSC CAPF – 2017
– 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.