2009 Archives

31. The number of times the hands of a watch are at right angle between 4

The number of times the hands of a watch are at right angle between 4 p.m. to 10 p.m. is :

This question was previously asked in

UPSC CAPF – 2009

The correct option is C.

The hands of a clock are at a right angle (90 degrees) 22 times in a 12-hour period. This is slightly less than twice per hour because the minute hand gains on the hour hand. The right angles occur approximately every $12/11 \times 30 \approx 32.7$ minutes relative to the previous right angle position. The times when the hands are exactly at right angles are around H:15 and H:45. The exact times in minutes past H:00 are given by $M = \frac{12}{11}(5H \pm 15)$.
The hours between 4 p.m. to 10 p.m. cover the interval [4:00 p.m., 10:00 p.m.]. This is a 6-hour period.
Let’s list the times when the hands are at right angles in the 12-hour cycle (using H=0 for 12, H=1 for 1, …, H=11 for 11):
$M = \frac{12}{11}(5H \pm 15)$
H=4: $\frac{12}{11}(20 \pm 15) \implies \frac{60}{11} \approx 5.45$ (4:05 p.m.), $\frac{420}{11} \approx 38.18$ (4:38 p.m.). (2 times)
H=5: $\frac{12}{11}(25 \pm 15) \implies \frac{120}{11} \approx 10.91$ (5:10 p.m.), $\frac{480}{11} \approx 43.63$ (5:43 p.m.). (2 times)
H=6: $\frac{12}{11}(30 \pm 15) \implies \frac{180}{11} \approx 16.36$ (6:16 p.m.), $\frac{540}{11} \approx 49.09$ (6:49 p.m.). (2 times)
H=7: $\frac{12}{11}(35 \pm 15) \implies \frac{240}{11} \approx 21.82$ (7:21 p.m.), $\frac{600}{11} \approx 54.54$ (7:54 p.m.). (2 times)
H=8: $\frac{12}{11}(40 \pm 15) \implies \frac{300}{11} \approx 27.27$ (8:27 p.m.), $\frac{660}{11} = 60$ (9:00 p.m.). (2 times)
H=9: $\frac{12}{11}(45 \pm 15) \implies \frac{360}{11} \approx 32.73$ (9:32 p.m.), $\frac{720}{11} \approx 65.45$ (10:05 p.m.). (1 time in the range [9:00, 10:00])

The times in the interval [4:00 p.m., 10:00 p.m.] are:
4:05, 4:38, 5:10, 5:43, 6:16, 6:49, 7:21, 7:54, 8:27, 9:00, 9:32.
All these 11 times are within the specified range.

However, the option C is 10. This suggests a different interpretation. A common interpretation in such problems is to count the number of times the hands form a right angle strictly *between* the listed hour marks.
Times within (4:00, 5:00): 4:05, 4:38 (2)
Times within (5:00, 6:00): 5:10, 5:43 (2)
Times within (6:00, 7:00): 6:16, 6:49 (2)
Times within (7:00, 8:00): 7:21, 7:54 (2)
Times within (8:00, 9:00): 8:27 (1) – 9:00 is a boundary
Times within (9:00, 10:00): 9:32 (1) – 9:00 and 10:00 are boundaries
Summing these gives 2 + 2 + 2 + 2 + 1 + 1 = 10.
This interpretation excludes the instance at 9:00 p.m., which falls exactly on an hour boundary.

In a 12-hour period, the hands are at right angles 22 times. The times 3:00 and 9:00 are special as they occur exactly on the hour. In the intervals that contain 3 or 9 (i.e., 2-3, 3-4, 8-9, 9-10), one of the two right angles for that hour interval falls exactly on the hour mark (3:00 or 9:00). If the question implies counting instances strictly between the full hour points, the 9:00 p.m. instance would be excluded from both the (8,9) and (9,10) intervals when summed this way.

32. Kamla got married 6 years ago. Today her age is $1\frac{1}{4}$ times h

Kamla got married 6 years ago. Today her age is $1\frac{1}{4}$ times her age at the time of marriage. Her son’s age is $\frac{1}{10}$ times her present age. What is her son’s age ?

2 years

3 years

4 years

5 years

This question was previously asked in

UPSC CAPF – 2009

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The correct option is B.

Let Kamla’s present age be $A$ years and her age at the time of marriage be $M$ years.
She got married 6 years ago, so $A = M + 6$.
Her present age is $1\frac{1}{4}$ times her age at marriage, so $A = \frac{5}{4} M$.
Equating the two expressions for $A$: $M + 6 = \frac{5}{4} M$.
Multiplying by 4: $4M + 24 = 5M$.
Subtracting $4M$: $24 = M$.
So, Kamla’s age at marriage was 24 years.
Her present age is $A = M + 6 = 24 + 6 = 30$ years.
Her son’s age is $\frac{1}{10}$ times her present age.
Son’s age = $\frac{1}{10} \times 30 = 3$ years.

This problem involves setting up equations based on given information about ages and solving them simultaneously. The fraction $1\frac{1}{4}$ is equivalent to $\frac{5}{4}$ or $1.25$.

33. ‘A’ completes a work in 10 days, while ‘B’ and ‘C’ complete it in 12 a

‘A’ completes a work in 10 days, while ‘B’ and ‘C’ complete it in 12 and 15 days respectively. In how many days can ‘A’, ‘B’ and ‘C’ together complete the work ?

This question was previously asked in

UPSC CAPF – 2009

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‘A’, ‘B’, and ‘C’ together can complete the work in 4 days.

To find the time taken when working together, we calculate the individual daily work rates, sum them up to find the combined daily rate, and then find the reciprocal of the combined rate.

A’s daily work rate = 1/10. B’s daily work rate = 1/12. C’s daily work rate = 1/15.
Combined daily work rate = 1/10 + 1/12 + 1/15.
Finding a common denominator (LCM of 10, 12, 15 is 60):
Combined rate = (6/60) + (5/60) + (4/60) = 15/60 = 1/4.
They complete 1/4 of the work per day.
Time taken to complete the whole work = 1 / (Combined rate) = 1 / (1/4) = 4 days.

34. Which one of the following types of forests cover the largest area in

Which one of the following types of forests cover the largest area in India ?

Tropical dry deciduous

Tropical moist deciduous

Tropical evergreen

Himalayan moist temperate

This question was previously asked in

UPSC CAPF – 2009

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Tropical dry deciduous forests cover the largest area in India.

Tropical dry deciduous forests are found in vast areas of the Indian subcontinent where rainfall is moderate (70-100 cm annually), and they shed their leaves during the dry season. This type of forest is widespread across the central parts of India.

Tropical moist deciduous forests require higher rainfall (100-200 cm) and are also significant but cover a smaller area than dry deciduous forests. Tropical evergreen forests are limited to high rainfall regions (over 200 cm), such as parts of the Western Ghats and Northeast India. Himalayan moist temperate forests are specific to the Himalayan region.

35. Which one of the following does not control climate ?

Which one of the following does not control climate ?

Latitude

Altitude

Longitude

Forest cover

This question was previously asked in

UPSC CAPF – 2009

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Longitude does not directly control climate.

Climate is primarily controlled by factors that affect the distribution of solar radiation and heat transfer in the atmosphere and oceans. These include Latitude (influences angle of sun’s rays), Altitude (affects temperature and pressure), Proximity to water bodies, Ocean Currents, Prevailing Winds, Topography, and Vegetation Cover. Longitude primarily relates to position east-west and time zones and does not have a direct causal influence on climate.

While locations at similar longitudes might share similar climates if other factors are constant (e.g., coastal regions at the same latitude), longitude itself is not a climate-controlling factor like latitude or altitude are. Forest cover is considered a significant factor influencing local and regional climate, primarily through its effects on albedo, evapotranspiration, and carbon cycling.

36. Which one of the following is a planetary wind ?

Which one of the following is a planetary wind ?

Monsoon

Trade

Chinook

Mistral

This question was previously asked in

UPSC CAPF – 2009

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Trade winds are a type of planetary wind.

Planetary winds, also known as prevailing winds, are large-scale, relatively constant wind patterns that blow across the globe due to differences in air pressure and the Earth’s rotation. Trade winds (blowing from subtropical high-pressure belts towards the equatorial low-pressure belt), Westerlies (blowing from subtropical highs towards subpolar lows), and Polar Easterlies (blowing from polar highs towards subpolar lows) are primary examples.

Monsoons are seasonal winds driven by differential heating of land and sea, not constant global patterns. Chinook and Mistral are examples of local winds influenced by topography and regional pressure systems.

37. Consider the following statements: 1. The Earth receives the Sun’s e

Consider the following statements:

1. The Earth receives the Sun’s energy at the infrared end of the spectrum.
2. The Earth re-radiates the Sun’s heat as ultraviolet energy.

Which of the above statements is/are correct ?

1 only

2 only

Both 1 and 2

Neither 1 nor 2

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UPSC CAPF – 2009

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Neither statement 1 nor statement 2 is correct.

The Earth receives energy from the Sun primarily in the form of visible light, with significant contributions from infrared and ultraviolet radiation. The peak emission of the Sun’s radiation is in the visible spectrum. The Earth, being much cooler than the Sun, re-radiates absorbed energy primarily as longwave infrared radiation.

Statement 1 is incorrect because the Earth receives energy across the solar spectrum, not just the infrared end. Statement 2 is incorrect because the Earth re-radiates energy as infrared (heat) radiation, not ultraviolet radiation. The greenhouse effect is caused by atmospheric gases absorbing this outgoing infrared radiation.

38. What happens to the water depth as the Tsunami waves approach the coas

What happens to the water depth as the Tsunami waves approach the coast ?

Decreases

Increases

Increases and decreases alternatively

Neither increases nor decreases

This question was previously asked in

UPSC CAPF – 2009

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As Tsunami waves approach the coast, the water depth decreases.

Tsunami waves are long-period waves. As they move from the deep ocean into shallower coastal waters, their speed decreases due to friction with the seabed, but their height increases significantly due to the conservation of energy. This phenomenon is known as shoaling.

In the deep ocean, tsunamis have a long wavelength and small amplitude (height), making them often unnoticed by ships. As they enter shallow water, the wavelength shortens, the speed slows dramatically, and the wave height can build up to several meters or even tens of meters upon reaching the coast.

39. What happens to the soil, where soil water freezes and it tends to for

What happens to the soil, where soil water freezes and it tends to form ice layers parallel with the ground surface ?

It heaves the soil upward in an uneven manner

It heaves the soil downward in an uneven manner

It heaves the soil both upward and downward in an even manner

It remains unaffected

This question was previously asked in

UPSC CAPF – 2009

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When soil water freezes and forms ice layers parallel to the ground surface, this process, known as frost heave, pushes the overlying soil upwards in an uneven manner.

Frost heave occurs when water in the soil pores freezes and expands, and further water is drawn towards the freezing front forming ice lenses. The formation and growth of these ice lenses cause the ground surface to rise.

The unevenness results from variations in soil type, moisture content, and temperature across the area. Frost heave can cause significant damage to roads, buildings, and infrastructure in areas with freezing temperatures and susceptible soils.

40. The laboratory from where carbon dioxide measurements have been taken

The laboratory from where carbon dioxide measurements have been taken since 1958 A.D., which form the basis of present global warming data base is :

Maragheh observatory

Mauna Loa observatory

Griffith observatory

Sydney observatory

This question was previously asked in

UPSC CAPF – 2009

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The Mauna Loa Observatory in Hawaii is the site where continuous measurements of atmospheric carbon dioxide concentration began in 1958, forming the basis of the Keeling Curve and a key dataset for understanding global warming.

The Keeling Curve, named after Charles David Keeling, shows the steady increase in atmospheric CO2 concentration since the late 1950s, along with seasonal fluctuations. This is one of the most important pieces of evidence for anthropogenic climate change.

The Mauna Loa Observatory is operated by the National Oceanic and Atmospheric Administration (NOAA). Its remote location on a volcanic mountain top minimizes local pollution influences, making it an ideal site for monitoring global atmospheric composition.