UPSC NDA-2 Archives

11. The part of the human eye on which the image is formed is

The part of the human eye on which the image is formed is

[amp_mcq option1=”pupil” option2=”cornea” option3=”retina” option4=”iris” correct=”option3″]

This question was previously asked in

UPSC NDA-2 – 2020

Light enters the human eye through the cornea, passes through the pupil (regulated by the iris), the lens, and then is focused on the retina. The retina is the light-sensitive tissue located at the back of the eye containing photoreceptor cells (rods and cones), where the image is formed and converted into electrical signals that are sent to the brain via the optic nerve.

The retina is the screen at the back of the eye where the image is formed by the lens system.

The cornea and lens are responsible for refracting (bending) light to focus it on the retina. The iris controls the size of the pupil, regulating the amount of light entering the eye. The image formed on the retina is real and inverted.

12. Light of uniform intensity impinges perpendicularly on a totally refle

Light of uniform intensity impinges perpendicularly on a totally reflecting surface. If the area of the surface is halved, the radiation force on it will become

[amp_mcq option1=”double” option2=”half” option3=”four times” option4=”one fourth” correct=”option2″]

This question was previously asked in

UPSC NDA-2 – 2020

Download PDF Attempt Online

The radiation force (F) on a totally reflecting surface is given by F = 2IA/c, where I is the intensity of light, A is the area of the surface, and c is the speed of light. The question states that the light has uniform intensity (I is constant) and impinges perpendicularly. When the area of the surface is halved, the new area A₂ = A₁/2. The new radiation force F₂ will be F₂ = 2IA₂/c = 2I(A₁/2)/c = (2IA₁/c) / 2 = F₁/2. Thus, the radiation force on it will become half.

Radiation force on a surface is proportional to the intensity of light and the area of the surface. For a totally reflecting surface, the force is twice that on a totally absorbing surface for the same intensity and area.

The radiation pressure is defined as the force per unit area (P = F/A). For a totally reflecting surface, the radiation pressure is P = 2I/c. The force is the product of pressure and area: F = P * A = (2I/c) * A.

13. A metallic wire having resistance of 20 Ω is cut into two equal parts

A metallic wire having resistance of 20 Ω is cut into two equal parts in length. These parts are then connected in parallel. The resistance of this parallel combination is equal to

[amp_mcq option1=”20 Ω” option2=”10 Ω” option3=”5 Ω” option4=”15 Ω” correct=”option3″]

This question was previously asked in

UPSC NDA-2 – 2020

Download PDF Attempt Online

The original metallic wire has a resistance R = 20 Ω. When it is cut into two equal parts in length, the resistance of each part is halved, assuming the material and cross-sectional area remain uniform. So, each part has a resistance R₁ = R₂ = R/2 = 20 Ω / 2 = 10 Ω. When these two parts are connected in parallel, the equivalent resistance (Rₚ) is given by the formula 1/Rₚ = 1/R₁ + 1/R₂. Substituting the values, 1/Rₚ = 1/10 Ω + 1/10 Ω = 2/10 Ω = 1/5 Ω. Therefore, the equivalent resistance Rₚ = 5 Ω.

The resistance of a wire is directly proportional to its length. For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances.

For two resistors R₁ and R₂ in parallel, the equivalent resistance can also be calculated as Rₚ = (R₁ * R₂) / (R₁ + R₂). In this case, Rₚ = (10 Ω * 10 Ω) / (10 Ω + 10 Ω) = 100 / 20 = 5 Ω.

14. Two planets orbit the Sun in circular orbits, with their radius of orb

Two planets orbit the Sun in circular orbits, with their radius of orbit as R₁ = R and R₂ = 4R. Ratio of their periods (T₁/T₂) around the Sun will be

[amp_mcq option1=”1/16″ option2=”1/8″ option3=”1/4″ option4=”1/2″ correct=”option2″]

This question was previously asked in

UPSC NDA-2 – 2020

Download PDF Attempt Online

This problem can be solved using Kepler’s Third Law of planetary motion, which states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (R) of its orbit: T² ∝ R³. For circular orbits, the semi-major axis is simply the radius (R). Thus, (T₁/T₂)² = (R₁/R₂)³. Given R₁ = R and R₂ = 4R, we have (T₁/T₂)² = (R / 4R)³ = (1/4)³ = 1/64. Taking the square root of both sides, T₁/T₂ = √(1/64) = 1/8.

Kepler’s Third Law relates the orbital period and orbital radius of planets orbiting the same central body: T² ∝ R³.

Kepler’s Laws are empirical laws describing the motion of planets around the Sun. Newton’s Law of Universal Gravitation provides the theoretical basis for Kepler’s Laws. For circular orbits, the speed v is constant, and the period T = 2πR/v. The gravitational force provides the centripetal force: GMm/R² = mv²/R. Substituting v = 2πR/T gives GMm/R² = m(2πR/T)²/R, which simplifies to T² = (4π²/GM) R³, confirming T² ∝ R³.

15. Cartilage is NOT found in

Cartilage is NOT found in

[amp_mcq option1=”larynx” option2=”nose” option3=”ear” option4=”urinary bladder” correct=”option4″]

This question was previously asked in

UPSC NDA-2 – 2020

Download PDF Attempt Online

Cartilage is a type of flexible connective tissue found in various parts of the body that require support and flexibility. The larynx (voice box), nose, and ear all contain cartilage as a major structural component. The urinary bladder is a hollow muscular organ that stores urine; its wall is primarily composed of smooth muscle tissue, not cartilage.

Cartilage is a structural component of the skeletal system and certain organs, providing support and flexibility. It is different from muscle tissue, which forms the walls of organs like the urinary bladder.

Examples of locations where cartilage is found include joints (articular cartilage), ribs, trachea, bronchi, intervertebral discs, and the structures mentioned in options A, B, and C. The urinary bladder wall consists of the detrusor muscle, a type of smooth muscle.

16. Which one of the following plant tissues has dead cells ?

Which one of the following plant tissues has dead cells ?

[amp_mcq option1=”Epidermis” option2=”Parenchyma” option3=”Collenchyma” option4=”Sclerenchyma” correct=”option4″]

This question was previously asked in

UPSC NDA-2 – 2020

Download PDF Attempt Online

Sclerenchyma is a plant tissue that provides mechanical support and strength to the plant. Sclerenchyma cells typically have thick, lignified secondary cell walls and are often dead at maturity, their function being purely mechanical support provided by their rigid walls. Examples include fibers (long, slender cells) and sclereids (variously shaped cells found in fruits, seeds, etc.).

Plant tissues can be broadly classified into meristematic (dividing cells) and permanent (differentiated cells). Permanent tissues can be simple (made of one type of cell) or complex (made of more than one type of cell). Parenchyma, Collenchyma, and Sclerenchyma are simple permanent tissues. Parenchyma and Collenchyma are composed of living cells.

– Epidermis is the outermost protective layer of cells, usually living.
– Parenchyma cells are living, thin-walled cells involved in photosynthesis, storage, secretion, etc.
– Collenchyma cells are living cells with unevenly thickened primary cell walls, providing mechanical support to growing stems and petioles.

17. In prokaryotic organisms, nuclear region is not surrounded by a membra

In prokaryotic organisms, nuclear region is not surrounded by a membrane. This undefined nuclear region is known as

[amp_mcq option1=”Nucleic acid” option2=”Nucleoid” option3=”Nucleolus” option4=”Nucleosome” correct=”option2″]

This question was previously asked in

UPSC NDA-2 – 2020

Download PDF Attempt Online

In prokaryotic organisms (like bacteria and archaea), the genetic material (DNA) is located in a region of the cytoplasm that is not enclosed by a nuclear membrane. This ill-defined, irregularly shaped region where the genetic material is concentrated is called the nucleoid.

Prokaryotes lack membrane-bound organelles, including a true nucleus. Eukaryotic cells, on the other hand, have a membrane-bound nucleus containing their genetic material.

– Nucleic acid refers to the genetic material itself (DNA or RNA).
– Nucleolus is a dense structure found within the nucleus of eukaryotic cells, involved in ribosome synthesis.
– Nucleosome is a structural unit of a eukaryotic chromosome, consisting of a length of DNA coiled around a core of histone proteins.

18. Which one of the following statements is NOT correct ?

Which one of the following statements is NOT correct ?

[amp_mcq option1=”Biomass is a renewable source of energy” option2=”Gobar gas is produced when cow-dung, crop residues, vegetable waste and sewage are allowed to decompose in the absence of oxygen” option3=”Biogas generation reduces soil and water pollution” option4=”Heating capacity of biogas is very low” correct=”option4″]

This question was previously asked in

UPSC NDA-2 – 2020

Download PDF Attempt Online

Statement D is NOT correct. Biogas primarily consists of methane (about 50-75%) and carbon dioxide (25-50%), with trace amounts of other gases. Methane is a flammable gas with a significant calorific value (heat produced per unit volume/mass when burned). While its calorific value is lower than that of natural gas (which is mostly pure methane), it is comparable to or higher than many other common fuels used in rural areas (like wood or dung cakes) and is sufficient for cooking, heating, and generating electricity, indicating a reasonable heating capacity, not a ‘very low’ one.

– Statement A is correct: Biomass (organic matter) is renewable because it can be replenished relatively quickly through natural processes like plant growth.
– Statement B is correct: Biogas is produced through anaerobic digestion, the decomposition of organic matter by microorganisms in the absence of oxygen. This process is used in biogas plants.
– Statement C is correct: Using organic waste to produce biogas reduces the amount of waste that would otherwise pollute soil and water bodies. The digested slurry (digestate) can also be used as a nutrient-rich fertilizer, replacing chemical fertilizers and improving soil health.

Biogas technology provides a sustainable way to manage organic waste, produce renewable energy, and create organic fertilizer, contributing to a circular economy and reducing greenhouse gas emissions, particularly methane emissions from uncontrolled decomposition.

19. Which one of the following is NOT the unit of energy ?

Which one of the following is NOT the unit of energy ?

[amp_mcq option1=”Joule” option2=”Watt-hr” option3=”Newton-metre” option4=”kg-metre/sec²” correct=”option4″]

This question was previously asked in

UPSC NDA-2 – 2020

Download PDF Attempt Online

Let’s analyze the given units:
A) Joule (J) is the standard SI unit of energy and work.
B) Watt-hour (Wh) is a unit of energy, calculated as power (Watt) multiplied by time (hour). 1 Wh = 1 J/s * 3600 s = 3600 J.
C) Newton-metre (Nm) is the unit of work (Force x Distance). Work is a form of energy transfer, so Nm is equivalent to a Joule (1 Nm = 1 J). Note: Nm is also the unit of torque, but in the context of options alongside Joule and Watt-hour, it represents work/energy.
D) kg-metre/sec² is the unit of force (mass x acceleration, F = ma). 1 kg⋅m/s² = 1 Newton (N). It is not a unit of energy.

Energy can be expressed in various units depending on the context (e.g., Joules, kilowatt-hours, calories, foot-pounds). Force and Energy are distinct physical quantities with different units.

Energy can be defined as the capacity to do work. Work done by a force is the product of the force and the distance moved in the direction of the force (W = Fd). The unit of force is Newton (N), and the unit of distance is metre (m), so the unit of work (and energy) is N*m, which is defined as the Joule (J).

20. Two equal resistors R are connected in parallel, and a battery of 12 V

Two equal resistors R are connected in parallel, and a battery of 12 V is connected across this combination. A dc current of 100 mA flows through the circuit as shown below :
The value of R is

[amp_mcq option1=”120 Ω” option2=”240 Ω” option3=”60 Ω” option4=”100 Ω” correct=”option2″]

This question was previously asked in

UPSC NDA-2 – 2020

Download PDF Attempt Online

Two equal resistors R are connected in parallel. The equivalent resistance (Req) of two equal resistors in parallel is given by Req = R/2. According to Ohm’s Law, the voltage (V) across the circuit is equal to the current (I) flowing through it multiplied by the equivalent resistance (Req): V = I * Req.
Given V = 12 V and I = 100 mA = 0.1 A.
So, 12 V = 0.1 A * (R/2).
Rearranging the equation to solve for R:
12 = 0.1 * R / 2
12 = 0.05 * R
R = 12 / 0.05
R = 12 / (5/100)
R = 12 * (100/5)
R = 12 * 20
R = 240 Ω.

For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances: 1/Req = 1/R1 + 1/R2 + … For two equal resistors R, 1/Req = 1/R + 1/R = 2/R, so Req = R/2.

Ohm’s Law (V = IR) is fundamental to analyzing simple electric circuits. Units must be consistent (Volts, Amperes, Ohms). Milliamperes (mA) must be converted to Amperes (A) by dividing by 1000 (100 mA = 0.1 A).