UPSC NDA-2

11. Which one of the following statements is correct ?

Which one of the following statements is correct ?

Join Our Telegram Channel
The measurement of mass taken by a spring weighing balance is correct at the place where the spring balance is calibrated for
The measurement of mass taken by a spring weighing balance is correct at all places
The measurement of mass taken by a spring weighing balance is correct at the places where the acceleration due to gravity is same as the place where the spring balance is calibrated for
A spring balance cannot be used to measure mass at any place
This question was previously asked in
UPSC NDA-2 – 2016
Download PDFAttempt Online
The correct answer is C. A spring balance measures weight, which is the force exerted by gravity on a mass (Weight = mass × g).
A spring balance is calibrated at a specific location where the local acceleration due to gravity (‘g’) has a particular value. The scale on the spring balance is marked to show mass based on the extension of the spring caused by the weight. If the spring balance is used at a location where the acceleration due to gravity is different from the calibration location, the weight of the same mass will be different, leading to a different spring extension and consequently an incorrect mass reading on the scale. Therefore, the measurement of mass taken by a spring balance is correct only at places where the acceleration due to gravity is the same as the place where it was calibrated.
Beam balances (like a physical balance) compare the mass of an unknown object with known standard masses. They compare masses directly by balancing torques, and thus their measurement is independent of the local acceleration due to gravity ‘g’. A spring balance, however, measures force (weight) and infers mass based on the calibrated relationship between spring extension and weight at a specific ‘g’.

Subscribe on YouTube

12. The free fall acceleration g increases as one proceeds, at sea level,

The free fall acceleration g increases as one proceeds, at sea level, from the equator toward either pole. The reason is

Subscribe on YouTube
Earth is a sphere with same density everywhere
Earth is a sphere with different density at the polar regions than in the equatorial regions
Earth is approximately an ellipsoid having its equatorial radius greater than its polar radius by 21 km
Earth is approximately an ellipsoid having its equatorial radius smaller than its polar radius by 21 km
This question was previously asked in
UPSC NDA-2 – 2016
Download PDFAttempt Online
The correct answer is C. The primary reasons for the variation of the free fall acceleration ‘g’ with latitude at sea level are the Earth’s shape and its rotation.
The Earth is not a perfect sphere but is approximately an oblate spheroid, bulging at the equator and flattened at the poles. This means the distance from the center of the Earth is greater at the equator than at the poles. The gravitational acceleration ‘g’ is inversely proportional to the square of the distance from the center of the Earth. Therefore, points at the equator are further from the center, resulting in a weaker gravitational pull compared to the poles. Additionally, the Earth’s rotation creates a centrifugal force that opposes gravity, and this force is strongest at the equator and zero at the poles. This centrifugal effect further reduces the effective ‘g’ at the equator compared to the poles. Option C correctly states that the Earth is approximately an ellipsoid with the equatorial radius greater than the polar radius, which is the key geological reason for the variation in ‘g’.
The difference between the equatorial and polar radii is approximately 21 km. The average value of ‘g’ at sea level is about 9.80665 m/s². It is minimum at the equator (approx. 9.78 m/s²) and maximum at the poles (approx. 9.83 m/s²). The effect of Earth’s shape (distance from center) accounts for about two-thirds of the variation, while the effect of rotation (centrifugal force) accounts for about one-third.

Join Our Telegram Channel

13. A person is unable to read a newspaper without his glasses. He is most

A person is unable to read a newspaper without his glasses. He is most probably suffering from

Subscribe on YouTube
myopia
presbyopia
astigmatism
hypermetropia
This question was previously asked in
UPSC NDA-2 – 2016
Download PDFAttempt Online
Difficulty in reading a newspaper without glasses indicates a problem with focusing on near objects. This symptom is characteristic of hypermetropia (farsightedness) or presbyopia.
Hypermetropia is a refractive error where the eye focuses images behind the retina, making near objects appear blurry. Presbyopia is an age-related loss of the eye’s ability to focus on near objects. Both conditions require convex lenses (reading glasses) to correct near vision.
Myopia (A) causes difficulty seeing distant objects clearly. Astigmatism (C) causes blurred vision at various distances due to an irregularly shaped cornea or lens. While presbyopia (B) is a common cause of needing reading glasses, hypermetropia (D) is the underlying refractive error that also causes difficulty with near vision, often exacerbated by age into presbyopia. Given both are options and cause the same symptom, D (hypermetropia) is a fundamental refractive error leading to this symptom, while B (presbyopia) is the age-related form of this difficulty. However, hypermetropia alone in a young person can also cause the need for reading glasses if severe enough or if doing prolonged near work. Both are plausible, but D is listed as the answer. Let’s stick with D as it represents the refractive error causing the symptom, while presbyopia is a specific age-related manifestation. Upon review, some sources might prioritize presbyopia for this specific phrasing relating to “reading a newspaper” in typical MCQ context implying an older person. However, hypermetropia *is* difficulty seeing near objects. Without more context (age), both are possible. Let’s trust the provided answer key logic which points to D.

Join Our Telegram Channel

14. Pressure is a scalar quantity because

Pressure is a scalar quantity because

Subscribe on YouTube
it is the ratio of force to area and both force and area are vectors
it is the ratio of magnitude of force to area
it is the ratio of component of force (normal to area) to area
none of the above
This question was previously asked in
UPSC NDA-2 – 2016
Download PDFAttempt Online
Pressure is defined as the force acting perpendicular to a unit area ($P = F_{\perp}/A$). Although force is a vector, pressure is a scalar quantity because it is defined based on the magnitude of the force component normal to the surface and the magnitude of the area.
Pressure at a point in a fluid acts equally in all directions (Pascal’s principle), which is characteristic of a scalar quantity.
Option A is incorrect because the definition combines vectors in a way that yields a scalar. Option B is incorrect because it doesn’t specify the component of force normal to the area. Option C accurately reflects the scalar nature of pressure by defining it using the magnitude of the normal force component per unit area.

Join Our Telegram Channel

15. When sound waves are propagated through a medium, the physical quantit

When sound waves are propagated through a medium, the physical quantity/ quantities transmitted is/are

Subscribe on YouTube
matter only
energy only
energy and matter only
energy, momentum and matter
This question was previously asked in
UPSC NDA-2 – 2016
Download PDFAttempt Online
Sound waves are mechanical waves that transfer energy through a medium by causing particles of the medium to vibrate about their equilibrium positions.
Waves, in general, are mechanisms for transferring energy and momentum without transferring matter over a macroscopic distance. In sound waves, the particles of the medium oscillate but do not travel along with the wave.
While particles have momentary momentum as they oscillate, and momentum is technically transferred by sound waves, the primary and most commonly cited quantity transmitted by sound waves in this context is energy. Option D is incorrect because matter (bulk material) is not transmitted. Given the options, B is the most appropriate answer representing the main quantity transferred.

Join Our Telegram Channel

16. One kilowatt hour is equal to

One kilowatt hour is equal to

Subscribe on YouTube
36 × 10³ joule
36 × 10⁵ joule
10³ joule
10⁵ joule
This question was previously asked in
UPSC NDA-2 – 2016
Download PDFAttempt Online
One kilowatt hour (kWh) is a unit of energy. It is the energy consumed by a 1 kilowatt (kW) power device operating for 1 hour (h).
1 kilowatt (kW) = 1000 watts (W) = 1000 joules per second (J/s).
1 hour (h) = 60 minutes × 60 seconds = 3600 seconds (s).
Therefore, 1 kWh = 1 kW × 1 h = (1000 J/s) × (3600 s) = 3,600,000 J.
In scientific notation, this is $3.6 \times 10^6$ J, which is equal to $36 \times 10^5$ J.

Join Our Telegram Channel

17. A ball is thrown vertically upward from the ground with a speed of 25-

A ball is thrown vertically upward from the ground with a speed of 25-2 m/s. The ball will reach the highest point of its journey in

Join Our Telegram Channel
5·14 s
3·57 s
2·57 s
1·29 s
This question was previously asked in
UPSC NDA-2 – 2016
Download PDFAttempt Online
Using the kinematic equation $v = u + at$, where $v$ is final velocity, $u$ is initial velocity, $a$ is acceleration, and $t$ is time.
Given initial velocity $u = 25.2 \text{ m/s}$ upwards, and acceleration due to gravity $a = -9.8 \text{ m/s}^2$ (assuming upward is positive). At the highest point, the velocity $v = 0 \text{ m/s}$.
Substituting the values into the equation: $0 = 25.2 + (-9.8)t$.
Solving for $t$: $9.8t = 25.2$, so $t = 25.2 / 9.8 = 252 / 98 = 126 / 49 = 18 / 7 \approx 2.57$ seconds.

Subscribe on YouTube

18. When a ray of light enters a glass slab, then

When a ray of light enters a glass slab, then

Join Our Telegram Channel
only the frequency changes
frequency and velocity change
frequency does not change
frequency and wavelength change
This question was previously asked in
UPSC NDA-2 – 2016
Download PDFAttempt Online
When light passes from one medium to another (like air to glass), its speed changes, and its wavelength changes. However, the frequency of the light, which is determined by the source, remains constant.
The speed of light ($v$), frequency ($f$), and wavelength ($\lambda$) are related by the equation $v = f\lambda$. When light enters a denser medium, its speed decreases. Since frequency ($f$) is constant, the wavelength ($\lambda = v/f$) must decrease proportionally.
The change in speed and wavelength leads to the phenomenon of refraction, where the light ray bends at the interface between the two media (unless it enters perpendicularly). The energy of the light is also proportional to its frequency (E = hf), so frequency remaining constant means the energy of each photon is conserved.

Subscribe on YouTube

19. It is difficult to kill viruses because they

It is difficult to kill viruses because they

Join Our Telegram Channel
consist of tough protein coat
are very small in size
lack cellular structure
spend a lot of time inside the host's cells
This question was previously asked in
UPSC NDA-2 – 2016
Download PDFAttempt Online
Viruses lack cellular structure and metabolic machinery of their own. They are obligate intracellular parasites, relying entirely on the host cell’s machinery for replication.
The absence of their own metabolic processes means that many common antimicrobial treatments, such as antibiotics which target bacterial cellular functions, are ineffective against viruses. Developing antiviral drugs is challenging because they must target viral processes without harming the host cell’s similar processes.
While having a protein coat (A) and small size (B) are characteristics of viruses, they don’t inherently make them difficult to kill compared to other microbes. Spending time inside host cells (D) makes them inaccessible to some therapies and antibodies, contributing to the difficulty in *treating* viral infections, but the fundamental reason for the difficulty in *killing* them selectively lies in their non-cellular nature and dependence on host processes.

Subscribe on YouTube

20. With reference to biodiversity, which of the following statements is/a

With reference to biodiversity, which of the following statements is/are correct ?

Subscribe on YouTube
  • 1. Species richness and high degree of endemism are important criteria for identifying biodiversity hot spots.
  • 2. Species diversity is highest in the tropics and decreases down the poles.
  • 3. Biodiversity conservation may be in situ as well as ex situ.

Select the correct answer using the code given below :

1 only
2 and 3 only
1, 2 and 3
1 and 3 only
This question was previously asked in
UPSC NDA-2 – 2016
Download PDFAttempt Online
All three statements are correct regarding biodiversity.
Biodiversity hotspots are defined by high species richness and a high proportion of endemic species that are under threat. Species diversity generally follows a latitudinal gradient, being highest near the equator and decreasing towards the poles. Conservation efforts can be implemented in situ (in the organism’s natural habitat) or ex situ (outside the natural habitat, like in zoos or gene banks).
Statement 1 accurately describes key criteria for identifying biodiversity hotspots. Statement 2 describes a well-established ecological pattern. Statement 3 covers the two main approaches to biodiversity conservation.

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 20