UPSC CISF-AC-EXE

11. Newton’s law of motion cannot be applicable to the particles moving at

Newton’s law of motion cannot be applicable to the particles moving at a speed comparable to the speed of

Join Our Telegram Channel
light
sound
rocket
bullet train
This question was previously asked in
UPSC CISF-AC-EXE – 2017
Download PDFAttempt Online
Newton’s laws of motion are fundamental principles of classical mechanics. Classical mechanics provides an accurate description of the motion of objects in everyday life. However, these laws break down and are not applicable under certain extreme conditions:
1. When the speed of the object is comparable to the speed of light (approximately 3 x 10^8 m/s). In this regime, motion must be described by Einstein’s theory of special relativity. Relativistic effects like time dilation and length contraction become significant.
2. When the size of the object is very small (atomic or subatomic scales). In this regime, quantum mechanics is required to describe the behavior of particles.

The speed of sound, rockets, and bullet trains are all vastly lower than the speed of light, so classical mechanics (and Newton’s laws) apply accurately to objects moving at these speeds.

– Newton’s laws are part of classical mechanics.
– Classical mechanics is an approximation that works well for macroscopic objects at relatively low speeds.
– It fails when speeds approach the speed of light (requiring relativity) or at very small scales (requiring quantum mechanics).
The speed of light is a fundamental constant in the universe. No object with mass can reach or exceed the speed of light. Particles moving at speeds close to light speed are typically subatomic particles accelerated in particle accelerators.

Subscribe on YouTube

12. Longest day/night is experienced during a year in the

Longest day/night is experienced during a year in the

Join Our Telegram Channel
Equator
Pole
Tropics of Cancer/Capricorn
Arctic/Antarctic Circle
This question was previously asked in
UPSC CISF-AC-EXE – 2017
Download PDFAttempt Online
The length of day and night varies significantly with latitude due to the tilt of the Earth’s axis relative to its orbital plane. The maximum variation occurs at the poles (North and South Poles). At the poles, during summer, there is continuous daylight for approximately six months (polar day), and during winter, there is continuous darkness for approximately six months (polar night). This represents the absolute extreme in terms of the duration of continuous day or night experienced during a year. The question asks for the location where the “Longest day/night is experienced during a year”, which refers to the maximum possible duration of either daylight or darkness.
– The Earth’s axial tilt causes seasons and variation in day/night length with latitude.
– The variation is minimal at the equator (always around 12 hours day/night).
– The variation increases towards the poles.
– The poles experience the most extreme variations, including prolonged periods of continuous daylight or darkness.
At the Arctic and Antarctic Circles (approx 66.5° latitude), the sun does not set on the summer solstice (24 hours of daylight) and does not rise on the winter solstice (24 hours of darkness). However, this is a 24-hour period, whereas at the poles, the period extends to roughly six months. Thus, the pole experiences the *longest* continuous day or night period.

Subscribe on YouTube

13. The following are the major peaks of the Himalayas. Arrange them from

The following are the major peaks of the Himalayas. Arrange them from higher to lower altitudes :

  • 1. Kanchunjunga
  • 2. Nanga Parbat
  • 3. Nanda Devi
  • 4. Dhaulagiri

Select the correct answer using the code given below:

1-4-2-3
1-2-4-3
4-1-2-3
2-3-1-4
This question was previously asked in
UPSC CISF-AC-EXE – 2017
Download PDFAttempt Online
To arrange the given Himalayan peaks from higher to lower altitudes, we need to know their approximate heights:
1. Kanchenjunga: Approximately 8,586 meters (world’s 3rd highest peak)
2. Nanga Parbat: Approximately 8,126 meters (world’s 9th highest peak)
3. Nanda Devi: Approximately 7,816 meters (highest peak entirely within India)
4. Dhaulagiri I: Approximately 8,167 meters (world’s 7th highest peak)

Subscribe on YouTube

Arranging these from highest to lowest altitude:
– Kanchenjunga (8586 m)
– Dhaulagiri I (8167 m)
– Nanga Parbat (8126 m)
– Nanda Devi (7816 m)

Join Our Telegram Channel

The correct order according to the numbers given is 1-4-2-3.

– Knowledge of the approximate altitudes of major Himalayan peaks.
– Ability to compare and rank numbers.
All these peaks are located in the Himalayas. Kanchenjunga lies on the border between Nepal and India. Dhaulagiri and Nanga Parbat are in Nepal and Pakistan-administered Kashmir respectively. Nanda Devi is located entirely within the state of Uttarakhand, India.

14. Chinook is

Chinook is

Subscribe on YouTube
A very warm and dry wind on the eastern slopes of the Rockies
A violent and extremely cold wind of the Tundras
An extremely cold wind in central Siberia
A dry and dusty wind off the west coast of Africa blowing from the deserts
This question was previously asked in
UPSC CISF-AC-EXE – 2017
Download PDFAttempt Online
Chinook is a type of foehn wind, which is a warm, dry, downslope wind that occurs on the leeward side (sheltered side) of a mountain range. The Chinook is specifically associated with the eastern slopes of the Rocky Mountains in North America. As moist air rises over the western slopes, it cools, condenses, and releases precipitation. On the eastern slopes, the air descends, warms adiabatically (due to compression), and becomes drier. This warm, dry wind can cause rapid temperature increases.
– Chinook is a warm, dry wind.
– It occurs on the leeward side of mountain ranges, specifically the eastern slopes of the Rockies.
– It is a type of foehn wind.
Foehn winds occur in many parts of the world, such as the Alps (where they are called Foehn), the Andes (Zonda), and others. The Chinook in North America is famous for causing dramatic temperature rises and rapid snowmelt, earning it the nickname “snow eater.”

Join Our Telegram Channel

15. Gir National Park is famous for

Gir National Park is famous for

Join Our Telegram Channel
Tiger
Lion
Migratory birds
One-horn rhinoceros
This question was previously asked in
UPSC CISF-AC-EXE – 2017
Download PDFAttempt Online
Gir National Park and Wildlife Sanctuary in Gujarat, India, is globally renowned as the last and only abode of the Asiatic Lion (*Panthera leo persica*). While it also hosts other wildlife and birds, its primary fame comes from its successful conservation of the Asiatic Lion population.
– Gir National Park is specifically known for conserving the Asiatic Lion.
– Other wildlife parks in India are famous for different species (e.g., Kaziranga for One-horned Rhinoceros, Ranthambore/Corbett for Tigers).
The Asiatic Lion population in Gir had dwindled to very low numbers in the early 20th century but has significantly recovered due to conservation efforts. It is now the only wild population of Asiatic lions in the world.

Subscribe on YouTube

16. 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 (City) List II (River)
A. Lucknow 1. Ganga
B. Jabalpur 2. Tapti
C. Surat 3. Narmada
D. Patna 4. Gomti

Code :

Join Our Telegram Channel
1 2 3 4
4 3 2 1
1 3 2 4
4 2 3 1
This question was previously asked in
UPSC CISF-AC-EXE – 2017
Download PDFAttempt Online
The correct matching of cities with the rivers they are located on is:
– Lucknow is situated on the banks of the Gomti River.
– Jabalpur is located near the Narmada River.
– Surat is located on the Tapi (Tapti) River.
– Patna is located on the banks of the Ganga River.
Matching the List I (City) with List II (River):
A. Lucknow – 4. Gomti
B. Jabalpur – 3. Narmada
C. Surat – 2. Tapti
D. Patna – 1. Ganga
The correct sequence of matches is 4-3-2-1.
– Knowledge of the geographical locations of major Indian cities and the rivers flowing through them.
– Specific knowledge of the location of Lucknow (Gomti), Jabalpur (Narmada), Surat (Tapti), and Patna (Ganga).
Many significant Indian cities are located on river banks, which have historically been important for water supply, transport, and trade. Examples include Delhi (Yamuna), Kolkata (Hooghly/Ganga), Varanasi (Ganga), Ahmedabad (Sabarmati), Chennai (Cooum and Adyar), etc.

Subscribe on YouTube

17. The sum of two numbers is 143. If the greater number is divided by the

The sum of two numbers is 143. If the greater number is divided by the difference of the numbers, the quotient is 7. What is the difference of the two numbers ?

9
11
14
15
This question was previously asked in
UPSC CISF-AC-EXE – 2017
Download PDFAttempt Online
Let the two numbers be \(x\) and \(y\), with \(x > y\).
According to the problem:
1. The sum of the two numbers is 143: \(x + y = 143\)
2. The greater number (\(x\)) divided by the difference of the numbers (\(x – y\)) is 7: \(\frac{x}{x – y} = 7\)

From the second equation, we get:
\(x = 7(x – y)\)
\(x = 7x – 7y\)
\(7y = 7x – x\)
\(7y = 6x\)
\(x = \frac{7y}{6}\)

Now substitute this expression for \(x\) into the first equation:
\(\frac{7y}{6} + y = 143\)
To combine the terms on the left side, find a common denominator:
\(\frac{7y + 6y}{6} = 143\)
\(\frac{13y}{6} = 143\)
\(13y = 143 \times 6\)
\(y = \frac{143 \times 6}{13}\)
Since \(143 = 13 \times 11\), we have:
\(y = \frac{13 \times 11 \times 6}{13}\)
\(y = 11 \times 6\)
\(y = 66\)

Now find the value of \(x\) using \(x = \frac{7y}{6}\):
\(x = \frac{7 \times 66}{6}\)
\(x = 7 \times 11\)
\(x = 77\)

Join Our Telegram Channel

The two numbers are 77 and 66.
The question asks for the difference of the two numbers, which is \(x – y\):
Difference = \(77 – 66 = 11\)

– Translate the word problem into a system of two linear equations with two variables.
– Solve the system of equations to find the values of the two numbers.
– Calculate the required difference between the numbers.
Verifying the solution:
Sum: \(77 + 66 = 143\) (Correct)
Difference: \(77 – 66 = 11\)
Greater number divided by difference: \(77 / 11 = 7\) (Correct)
The difference of the two numbers is indeed 11.

Subscribe on YouTube

18. Given that 1st November 2019 is a Friday. Then 1st November 2022 is a

Given that 1st November 2019 is a Friday. Then 1st November 2022 is a

Join Our Telegram Channel
Monday
Tuesday
Wednesday
Thursday
This question was previously asked in
UPSC CISF-AC-EXE – 2017
Download PDFAttempt Online
If 1st November 2019 is a Friday, then 1st November 2022 is a Tuesday.
To find the day of the week after a certain number of years, we need to calculate the number of “odd days” between the two dates. An odd day is the remainder when the total number of days is divided by 7.
A normal year (365 days) has 365 mod 7 = 1 odd day.
A leap year (366 days) has 366 mod 7 = 2 odd days.
The period from 1st November 2019 to 1st November 2022 spans three full years: the period covering 2020, the period covering 2021, and the period covering 2022 up to Nov 1.
– From Nov 1, 2019 to Nov 1, 2020: This period includes the entire year 2020, which is a leap year (divisible by 4). Thus, this period adds 2 odd days.
– From Nov 1, 2020 to Nov 1, 2021: This period includes the entire year 2021, which is a normal year. Thus, this period adds 1 odd day.
– From Nov 1, 2021 to Nov 1, 2022: This period includes the entire year 2022, which is a normal year. Thus, this period adds 1 odd day.
Total odd days from 1st November 2019 to 1st November 2022 = 2 (from 2020) + 1 (from 2021) + 1 (from 2022) = 4 odd days.
Starting from Friday, we move forward by 4 days:
Friday + 1 day = Saturday
Saturday + 1 day = Sunday
Sunday + 1 day = Monday
Monday + 1 day = Tuesday
So, 1st November 2022 is a Tuesday.
Leap years occur every 4 years, except for years divisible by 100 but not by 400. 2020 is a leap year because it is divisible by 4. Years like 1900 are not leap years, while 2000 is a leap year. This rule helps determine the number of odd days accurately over longer periods.

Subscribe on YouTube

19. In a certain year, a school had 60% boys and 40% girls as students. In

In a certain year, a school had 60% boys and 40% girls as students. In the next five years the number of boys decreased by 10% and the number of girls increased by 10%. What is the change in total roll strength of the school in the five years ?

Subscribe on YouTube
3% increase
2% decrease
No change
5% decrease
This question was previously asked in
UPSC CISF-AC-EXE – 2017
Download PDFAttempt Online
The change in total roll strength of the school in the five years is a 2% decrease.
Assume the initial total roll strength is 100 units.
Initial Boys = 60% of 100 = 60 units.
Initial Girls = 40% of 100 = 40 units.
In the next five years:
Number of boys decreased by 10%. New number of boys = 60 * (1 – 0.10) = 60 * 0.90 = 54 units.
Number of girls increased by 10%. New number of girls = 40 * (1 + 0.10) = 40 * 1.10 = 44 units.
New total roll strength = New Boys + New Girls = 54 + 44 = 98 units.
The change in total roll strength = New Total – Initial Total = 98 – 100 = -2 units.
The percentage change = (Change / Initial Total) * 100 = (-2 / 100) * 100 = -2%.
A -2% change represents a 2% decrease.
This calculation shows how percentage changes applied to different subgroups affect the overall total, depending on the initial proportions of the subgroups.

Join Our Telegram Channel

20. X paid ₹ 47 for certain cups of tea and coffee. If tea costs ₹ 5 per c

X paid ₹ 47 for certain cups of tea and coffee. If tea costs ₹ 5 per cup and coffee costs ₹ 8 per cup, which one of the following statements is correct ?

Subscribe on YouTube
He drank 8 cups of tea and coffee.
He drank the same number of cups of tea and coffee.
He drank more tea than coffee.
He drank more coffee than tea.
This question was previously asked in
UPSC CISF-AC-EXE – 2017
Download PDFAttempt Online
He drank more coffee than tea.
Let ‘t’ be the number of cups of tea and ‘c’ be the number of cups of coffee. The cost equation is 5t + 8c = 47. We need to find non-negative integer solutions for t and c.
We can try possible values for c:
If c=0, 5t = 47 (not possible for integer t)
If c=1, 5t = 47 – 8 = 39 (not possible for integer t)
If c=2, 5t = 47 – 16 = 31 (not possible for integer t)
If c=3, 5t = 47 – 24 = 23 (not possible for integer t)
If c=4, 5t = 47 – 32 = 15 => t = 3. This is a valid integer solution (t=3, c=4).
If c=5, 5t = 47 – 40 = 7 (not possible for integer t)
If c=6, 5t = 47 – 48 = -1 (not possible for non-negative t)
The only valid solution is t=3 cups of tea and c=4 cups of coffee.
Based on this solution:
A) Total cups = 3 + 4 = 7, not 8. (Incorrect)
B) Number of tea cups (3) is not the same as coffee cups (4). (Incorrect)
C) He drank 3 cups of tea and 4 cups of coffee. 3 is not more than 4. (Incorrect)
D) He drank 4 cups of coffee and 3 cups of tea. 4 is more than 3. (Correct)
This is a simple linear Diophantine equation where we are looking for non-negative integer solutions. Since the coefficients are relatively small, trial and error is an efficient method to find the solution.

Join Our Telegram Channel
Page 1Page 2Page 3Page 4Page 5
Test 1Test 2Test 3Test 4Test 5Test 6Test 7Test 8Test 9Test 10Test 11Test 12Test 13Test 14Test 15Test 16Test 17Test 18Test 19Test 20Test 21Test 22Test 23Test 24