UPSC NDA-1 Archives

1. A deep sea diver may hurt his ear drum during diving because of

A deep sea diver may hurt his ear drum during diving because of

[amp_mcq option1=”lack of oxygen” option2=”high atmospheric pressure” option3=”high water pressure” option4=”All of the above” correct=”option3″]

This question was previously asked in

UPSC NDA-1 – 2015

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Pressure in water increases significantly with depth due to the weight of the water column above. A deep sea diver experiences very high external pressure from the surrounding water. The middle ear is normally filled with air at approximately atmospheric pressure (adjusted through the Eustachian tube). If the diver does not equalize the pressure in their middle ear with the external water pressure, the large pressure difference across the ear drum can cause it to bulge inward and potentially rupture.

Hydrostatic pressure in a fluid increases linearly with depth. The significant pressure difference between the external water pressure and the internal ear pressure is the cause of potential ear drum damage for divers.

Divers are trained in techniques (like Valsalva maneuver) to equalize the pressure in their middle ears with the external pressure, preventing barotrauma (pressure injury).

2. A solid is melted and allowed to cool and solidify again. The temperat

A solid is melted and allowed to cool and solidify again. The temperature is measured at equal intervals of time. The graph below shows the change of temperature with time.
[Graph image is part of the question context]
The part of the curve that is practically horizontal is due to

[amp_mcq option1=”latent heat given away by the liquid” option2=”specific heat given away by the liquid” option3=”thermal capacity changes with time keeping temperature constant” option4=”change in density during transformation” correct=”option1″]

This question was previously asked in

UPSC NDA-1 – 2015

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The horizontal part of the cooling curve represents a phase transition. As the liquid cools and reaches its freezing point, it starts to solidify. During the solidification process, the temperature remains constant because the substance is releasing latent heat of fusion (or solidification). This released latent heat compensates for the heat being lost to the surroundings, preventing the temperature from dropping until all the liquid has solidified.

During a phase change (like melting or freezing), the temperature of a pure substance remains constant as latent heat is absorbed or released.

Latent heat is the heat energy absorbed or released by a substance during a phase change at constant temperature. Specific heat is the heat required to raise the temperature of a unit mass of a substance by one degree within a single phase.

3. A person stands on his two feet over a surface and experiences a press

A person stands on his two feet over a surface and experiences a pressure P. Now the person stands on only one foot. He would experience a pressure of magnitude

[amp_mcq option1=”4 P” option2=”P” option3=”$\frac{1}{2}$ P” option4=”2 P” correct=”option4″]

This question was previously asked in

UPSC NDA-1 – 2015

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Pressure is defined as force per unit area (P = F/A). When a person stands on a surface, the force exerted is their weight (F = W). When standing on two feet, the weight is distributed over the total area of two feet ($A_{two}$). So the pressure is $P = W / A_{two}$. When standing on one foot, the weight is distributed over the area of one foot ($A_{one}$). Assuming the area of one foot is approximately half the area of two feet ($A_{one} \approx A_{two}/2$), the new pressure is $P’ = W / A_{one} \approx W / (A_{two}/2) = 2W/A_{two} = 2P$. Thus, the pressure is doubled.

Pressure is inversely proportional to the area over which the force is applied, assuming the force remains constant.

This is a simple application of the definition of pressure. By reducing the contact area by half (from two feet to one foot), the pressure exerted on the surface doubles for the same force (weight).

4. The seeds of flowering plants are made up of

The seeds of flowering plants are made up of

[amp_mcq option1=”Ovary and ovary wall” option2=”Embryo, food reserves and coat” option3=”Cotyledons” option4=”Zygotes” correct=”option2″]

This question was previously asked in

UPSC NDA-1 – 2015

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A seed is a fundamental part of the reproduction process in flowering plants. It develops from the ovule after fertilization. A typical seed is composed of three main parts: the embryo (developed from the zygote), stored food reserves (often in the form of endosperm or cotyledons), and a protective seed coat (developed from the integuments of the ovule).

A seed is the structure containing the embryonic plant, along with food, enclosed in a protective outer covering.

The embryo consists of the radicle (embryonic root), plumule (embryonic shoot), and one or two cotyledons (seed leaves). The stored food provides nourishment for the embryo during germination before it can photosynthesize.

5. Gametophytes of sexually reproducing flowering plants are A) Haploid B

Gametophytes of sexually reproducing flowering plants are
A) Haploid
B) Diploid
C) Tetraploid
D) Polyploid

[amp_mcq option1=”Haploid” option2=”Diploid” option3=”Tetraploid” option4=”Polyploid” correct=”option1″]

This question was previously asked in

UPSC NDA-1 – 2015

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In the life cycle of sexually reproducing flowering plants (angiosperms), the dominant generation is the sporophyte, which is diploid (2n). This sporophyte produces spores through meiosis. These spores develop into the gametophytes (male gametophyte or pollen grain, and female gametophyte or embryo sac). The gametophytes are haploid (n). They produce gametes (sperm and egg) through mitosis.

The gametophyte generation in plants is haploid, producing gametes by mitosis. The sporophyte generation is diploid, producing spores by meiosis.

In flowering plants, the gametophyte generation is significantly reduced and dependent on the sporophyte. The male gametophyte is the pollen grain, and the female gametophyte is the embryo sac within the ovule.

6. To a perpendicular to the plane of ecliptic, Earth’s axis of rotation

To a perpendicular to the plane of ecliptic, Earth’s axis of rotation makes an angle of 23½ degrees. Had this angle been zero degree, which one among the following would result ?
A) There would have been no season
B) The length of day and night would have been the same throughout the year
C) The length of the day and night would have been the same all over the earth
D) All of the above

[amp_mcq option1=”There would have been no season” option2=”The length of day and night would have been the same throughout the year” option3=”The length of the day and night would have been the same all over the earth” option4=”All of the above” correct=”option4″]

This question was previously asked in

UPSC NDA-1 – 2015

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If Earth’s axis of rotation made an angle of zero degrees with the perpendicular to the plane of the ecliptic, it would mean the axis is perpendicular to the orbital plane. This configuration would eliminate the tilt that causes seasons. With no tilt, the sun’s rays would be directly overhead at the equator throughout the year. Consequently, every point on Earth would experience approximately 12 hours of daylight and 12 hours of darkness every day of the year, and there would be no significant seasonal variations in temperature or day length at any given latitude.

The tilt of Earth’s axis of rotation (currently 23.5 degrees) is the primary cause of seasons, varying day lengths throughout the year at latitudes away from the equator, and varying angles of incidence of solar radiation.

A zero-degree tilt would mean the terminator (the line separating day from night) would always pass through the poles. This would result in uniform day/night lengths of 12 hours everywhere on Earth throughout the year and the absence of distinct seasons.

7. 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
(National Park/Wildlife Sanctuary)
A. Chandra Prabha
B. Silent Valley
C. Valley of flowers
D. Indravati

List II
(State)
1. Uttarakhand
2. Chhattisgarh
3. Uttar Pradesh
4. Kerala

Code :
A) 2 1 4 3
B) 3 4 1 2
C) 2 4 1 3
D) 3 1 4 2

[amp_mcq option1=”2 1 4 3″ option2=”3 4 1 2″ option3=”2 4 1 3″ option4=”3 1 4 2″ correct=”option2″]

This question was previously asked in

UPSC NDA-1 – 2015

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The correct matching is Chandra Prabha (Uttar Pradesh), Silent Valley (Kerala), Valley of Flowers (Uttarakhand), and Indravati (Chhattisgarh). This corresponds to the code 3, 4, 1, 2.

This question tests knowledge of the geographical locations of prominent National Parks and Wildlife Sanctuaries in India.

Chandra Prabha Wildlife Sanctuary is located in Chandauli district of Uttar Pradesh. Silent Valley National Park is in the Nilgiri Hills of Kerala. Valley of Flowers National Park is in the Chamoli district of Uttarakhand, known for its endemic alpine flowers. Indravati National Park is in the Bijapur district of Chhattisgarh.

8. Which one of the following Indian states has the highest proportion of

Which one of the following Indian states has the highest proportion of area under forest cover ?
A) Madhya Pradesh
B) Sikkim
C) Meghalaya
D) Mizoram

[amp_mcq option1=”Madhya Pradesh” option2=”Sikkim” option3=”Meghalaya” option4=”Mizoram” correct=”option4″]

This question was previously asked in

UPSC NDA-1 – 2015

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Among the given options, Mizoram has the highest proportion (percentage) of its geographical area under forest cover. According to the India State of Forest Report (ISFR) 2021, Mizoram has 84.93% of its geographical area under forest cover, which is the highest percentage among all Indian states. While Madhya Pradesh has the largest total area under forest cover, its percentage is much lower (25.14% as per ISFR 2021). Sikkim and Meghalaya also have high percentages but typically lower than Mizoram.

Distinguishing between total forest area and the proportion of forest cover when comparing states, and knowing which state holds the highest percentage.

The states in the North-Eastern region of India are generally characterized by high forest cover percentages due to their topography and climate. Understanding the data from the India State of Forest Report is crucial for answering such questions.

9. Washing Soda is the common name for A) Calcium Carbonate B) Magnesium

Washing Soda is the common name for
A) Calcium Carbonate
B) Magnesium Carbonate
C) Sodium Carbonate
D) Potassium Carbonate

[amp_mcq option1=”Calcium Carbonate” option2=”Magnesium Carbonate” option3=”Sodium Carbonate” option4=”Potassium Carbonate” correct=”option3″]

This question was previously asked in

UPSC NDA-1 – 2015

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Washing Soda is the common name for Sodium Carbonate ($\text{Na}_2\text{CO}_3$). It is typically sold as the decahydrate, $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$. It is widely used as a cleaning agent, in laundry, and in the manufacture of glass and paper.

Knowing the common names of important chemical compounds.

Calcium Carbonate ($\text{CaCO}_3$) is found in limestone, marble, and chalk. Magnesium Carbonate ($\text{MgCO}_3$) is a component of minerals like magnesite. Potassium Carbonate ($\text{K}_2\text{CO}_3$) is also known as potash and is used in various industrial applications, including soap and glass making.

10. Why is Graphite used in electrolytic cells ? A) Graphite is soft and c

Why is Graphite used in electrolytic cells ?
A) Graphite is soft and can be easily moulded into electrodes
B) Graphite is made of layers of carbon atoms which can slide
C) Graphite is inert to most of the chemicals and remains intact in electrolytic cells
D) Graphite is a good conductor of electricity

[amp_mcq option1=”Graphite is soft and can be easily moulded into electrodes” option2=”Graphite is made of layers of carbon atoms which can slide” option3=”Graphite is inert to most of the chemicals and remains intact in electrolytic cells” option4=”Graphite is a good conductor of electricity” correct=”option4″]

This question was previously asked in

UPSC NDA-1 – 2015

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Graphite is used in electrolytic cells primarily because it is a good conductor of electricity. Electrodes serve as points of contact between the external electrical circuit and the electrolyte, facilitating the flow of current needed for electrolysis. Being a good conductor is a fundamental requirement for this function. While its relative inertness to many chemicals (option C) and ability to withstand high temperatures are also beneficial properties that make it a suitable material for electrodes in specific processes, the primary reason for its *use as an electrode* is its ability to conduct electricity. Options A and B describe physical properties that are not the main reasons for its use in this context.

Identifying the essential property of a material required for it to function as an electrode in an electrolytic cell.

Electrodes in electrolytic cells are either anodes or cathodes, where oxidation and reduction reactions occur, respectively. These reactions require the transfer of electrons between the external circuit and the ions in the electrolyte, which is facilitated by the electrical conductivity of the electrode material. Graphite’s conductivity stems from the delocalized electrons within its layered structure.