UPSC NDA-2 Archives

31. In which one among the following situations, the bulb would glow the m

In which one among the following situations, the bulb would glow the most ? (Consider all batteries are the same)

Diagram (a)

Diagram (b)

Diagram (c)

Diagram (d)

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Assuming Diagram (b) shows two identical batteries connected in series to a single bulb, this configuration would result in the highest voltage across the bulb, making it glow the brightest.

– The brightness of a bulb (assumed to be resistive) is proportional to the power dissipated, P. Power can be calculated as P = V²/R or P = I²R, where V is the voltage across the bulb, I is the current through the bulb, and R is its resistance. For a given bulb (fixed R), maximum brightness corresponds to maximum voltage or maximum current.
– Assuming each battery provides a voltage V:
– Diagram (a) (one battery, one bulb): Voltage across bulb = V.
– Diagram (b) (two batteries in series, one bulb): The voltages of batteries in series add up. Total voltage = 2V. This voltage is across the bulb.
– Diagram (c) (two batteries in parallel, one bulb): Batteries in parallel maintain the same voltage as a single battery (assuming identical ideal batteries). Total voltage = V. This voltage is across the bulb. (Parallel connection increases total current capacity, but not voltage).
– Diagram (d) (one battery, two bulbs in series): The voltage V from the battery is divided between the two bulbs. Assuming identical bulbs, voltage across each bulb = V/2.
– Comparing the voltage across the bulb in each case: (a) V, (b) 2V, (c) V, (d) V/2.
– Since the power dissipated (and thus brightness) is proportional to V², the bulb will glow brightest when the voltage across it is highest, which is in case (b) with two batteries in series (2V).

Connecting batteries in series increases the total voltage supplied, while connecting identical batteries in parallel increases the total charge capacity and allows the supply of a larger current for a longer duration, without increasing the voltage (ideally). Connecting bulbs in series increases the total resistance of the circuit, reducing the current and the voltage across each individual bulb (and thus brightness, assuming they are identical).

32. Lightning is due to

Lightning is due to

The flow of charges between different parts of the cloud

The short-circuiting of charges between the upper and lower surfaces of the cloud

The collection of positively charged particles on the base and collection of negatively charged particles at the top of the cloud

The induction of positive charge on the ground below the negative charge at the base of the cloud

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The correct answer is B. Lightning is fundamentally a rapid, large-scale electrical discharge, often described as a “short-circuiting” of charges, which frequently occurs between different charged regions within a thundercloud, typically between the upper and lower parts.

– Lightning is an electrical discharge phenomena occurring in the atmosphere.
– It is caused by the build-up of electric charge differences within cumulonimbus clouds (thunderclouds) or between the cloud and the ground.
– Charge separation within a cloud typically results in positive charges accumulating at the top and negative charges accumulating at the bottom, although complex charge distributions exist.
– When the potential difference between these regions (or between the cloud and ground) exceeds the dielectric strength of the air, an electrical breakdown occurs, creating a conductive channel.
– The lightning strike is the rapid flow of electric charge (electrons and ions) through this channel. This discharge can happen within the cloud (intra-cloud lightning), between clouds (inter-cloud lightning), or between the cloud and the ground (cloud-to-ground lightning).
– Option B describes the rapid discharge (“short-circuiting”) between upper and lower charged regions within a cloud, which is a common type of lightning (intra-cloud lightning). While option A is also a flow of charges, B is more specific about the location and uses the term “short-circuiting” which captures the rapid discharge aspect. Option C describes the cause (charge separation) but gets the common polarity wrong (usually positive at top, negative at bottom). Option D describes ground charge induction, a condition for ground strikes, not the general cause of lightning.

Cloud-to-ground lightning is the most dangerous type. The rapid heating of the air along the lightning channel causes it to expand explosively, creating the sound waves we perceive as thunder.

33. Which one among the following figures correctly represents the ray dia

Which one among the following figures correctly represents the ray diagram ? (Consider the lens to be thin)

Figure (a)

Figure (b)

Figure (c)

Figure (d)

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Assuming Figure (a) correctly depicts a standard ray diagram for a thin lens following the rules of refraction, option A is the correct answer.

– Ray diagrams for lenses use standard principal rays to trace the path of light and locate the image formed by the lens. The rules for these rays are based on the lens’s focal points and optical center.
– For a convex (converging) lens: (1) A ray parallel to the principal axis passes through the focal point on the other side after refraction. (2) A ray passing through the optical center goes straight through without deviation. (3) A ray passing through the focal point on the object side becomes parallel to the principal axis after refraction.
– For a concave (diverging) lens: (1) A ray parallel to the principal axis appears to diverge from the focal point on the same side after refraction. (2) A ray passing through the optical center goes straight through without deviation. (3) A ray directed towards the focal point on the other side becomes parallel to the principal axis after refraction.
– A correct ray diagram will accurately show the refraction of light rays according to these rules and the formation of the image where the refracted rays (or their extensions) intersect. Without the actual figures, it is impossible to verify which diagram is correct; this explanation assumes diagram (a) follows the correct physics principles for the lens depicted.

Ray diagrams are a useful tool in geometric optics for visualizing the formation of images by lenses and mirrors and understanding the properties (real/virtual, inverted/erect, magnified/diminished) of the image. Accuracy in drawing the rays according to the rules is crucial.

34. Which of the following statements give characteristics of contact forc

Which of the following statements give characteristics of contact forces ?

1. It appears between an object when it is in contact with some other object
2. It satisfies the third law of motion
3. It may appear between a pair of solid and fluid

Select the answer using the code given below :

1 and 3 only

2 and 3 only

1 and 2 only

1, 2 and 3

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The correct answer is D, as all three statements are correct characteristics of contact forces.

– Contact forces are forces that arise when two objects are in physical contact with each other. Examples include normal force, friction, tension, and air resistance. Statement 1 accurately describes this.
– Newton’s Third Law of Motion states that for every action, there is an equal and opposite reaction. All forces in nature, including contact forces, obey this law. For instance, if object A exerts a contact force on object B, then object B simultaneously exerts an equal and opposite contact force on object A. Statement 2 is correct.
– Contact forces can exist between solids, between a solid and a fluid (liquid or gas), or between different fluids. For example, buoyancy and drag force are contact forces between a solid and a fluid, or within a fluid. Statement 3 is correct.

Forces are broadly classified into contact forces and non-contact forces (like gravitational, electromagnetic, and nuclear forces) which act over a distance. The statements correctly describe the nature and properties of contact forces.

35. An astronaut whose weight on the Earth is 600 N experiences weightless

An astronaut whose weight on the Earth is 600 N experiences weightlessness on International Space Station orbiting around the Earth. It means that

acceleration of the astronaut is zero

normal reaction of the space station floor on the astronaut is zero

gravitational pull of earth on the astronaut is zero

space station applies a centrifugal force on the astronaut

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The correct answer is B. An astronaut experiences weightlessness on the International Space Station (ISS) because the normal reaction force from the space station floor on the astronaut is zero.

– Weight is the force exerted on an object due to gravity. On Earth, we feel our weight because of the normal force exerted by the surface supporting us, which balances gravity.
– The ISS and everything in it, including the astronaut, are constantly in freefall around the Earth. They are orbiting because they have a high tangential velocity while simultaneously accelerating towards the Earth due to gravity.
– In freefall, there is no supporting surface providing a normal reaction force to counteract gravity. The feeling of weight comes from this reaction force. Since this support force is absent, the astronaut feels weightless.
– The gravitational pull of Earth on the astronaut is not zero in orbit; it is still significant (around 90% of Earth’s surface gravity at ISS altitude) and is what keeps the ISS in orbit.
– The astronaut is accelerating towards the Earth (centripetal acceleration required for circular motion), so their acceleration is not zero.
– Centrifugal force is a fictitious force experienced in a rotating frame of reference; it’s not a real force causing weightlessness.

The state of apparent weightlessness in orbit is often referred to as microgravity. It is not due to the absence of gravity, but rather the state of continuous freefall. The ISS is continuously falling towards Earth, but its high orbital speed causes it to miss the Earth, resulting in an orbit.

36. Shown in the figure are two hollow cubes C₁ and C₂ of negligible mass

Shown in the figure are two hollow cubes C₁ and C₂ of negligible mass partially filled (depicted by darkened area) with liquids of densities ρ₁ and ρ₂, respectively, floating in water (density ρw). The relationship between ρ₁, ρ₂ and ρw is

[amp_mcq option1=”ρ₂ < ρw < ρ₁" option2="ρ₂ < ρ₁ < ρw" option3="ρ₁ < ρ₂ < ρw" option4="ρ₁ < ρw < ρ₂" correct="option4"]

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The correct answer is D) ρ₁ < ρw < ρ₂.

Both cubes are floating in water. According to Archimedes’ principle, a floating object displaces a volume of fluid whose weight is equal to the weight of the object. The weight of each hollow cube is solely due to the liquid inside, as the cube itself has negligible mass. Let A be the base area and H be the total height of the cubes (assuming they are identical). Let h₁ and h₂ be the submerged heights in water, and H₁_liquid and H₂_liquid be the heights of the liquid inside.
For Cube C₁:
Weight of liquid inside = (Volume of liquid inside) * ρ₁ * g = (A * H₁_liquid) * ρ₁ * g.
Buoyant force = (Volume submerged) * ρw * g = (A * h₁) * ρw * g.
Since it’s floating, (A * H₁_liquid) * ρ₁ * g = (A * h₁) * ρw * g, which simplifies to H₁_liquid * ρ₁ = h₁ * ρw, or ρ₁ = (h₁ / H₁_liquid) * ρw.
From the figure, the submerged height h₁ is significantly less than the height of the liquid inside H₁_liquid. Therefore, (h₁ / H₁_liquid) < 1, which implies ρ₁ < ρw. For Cube C₂: Weight of liquid inside = (A * H₂_liquid) * ρ₂ * g. Buoyant force = (A * h₂) * ρw * g. Since it's floating, (A * H₂_liquid) * ρ₂ * g = (A * h₂) * ρw * g, which simplifies to H₂_liquid * ρ₂ = h₂ * ρw, or ρ₂ = (h₂ / H₂_liquid) * ρw. From the figure, the submerged height h₂ is significantly greater than the height of the liquid inside H₂_liquid. Therefore, (h₂ / H₂_liquid) > 1, which implies ρ₂ > ρw.

Combining the results, we have ρ₁ < ρw and ρ₂ > ρw. This means ρ₁ is less than the density of water, while ρ₂ is greater than the density of water. Therefore, the relationship between the densities is ρ₁ < ρw < ρ₂.

37. Two resistances of 5.0 Ω and 7.0 Ω are connected in series and the com

Two resistances of 5.0 Ω and 7.0 Ω are connected in series and the combi- nation is connected in parallel with a resistance of 36.0 Ω. The equivalent resistance of the combination of three resistors is

24.0 Ω

12.0 Ω

9.0 Ω

6.0 Ω

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The correct answer is C) 9.0 Ω.

First, calculate the equivalent resistance of the two resistors connected in series. For resistors in series, the equivalent resistance (R_series) is the sum of individual resistances:
R_series = R₁ + R₂ = 5.0 Ω + 7.0 Ω = 12.0 Ω.
Next, this series combination (with R_series = 12.0 Ω) is connected in parallel with a third resistance (R₃ = 36.0 Ω). For resistors in parallel, the reciprocal of the equivalent resistance (R_eq) is the sum of the reciprocals of the individual resistances:
1/R_eq = 1/R_series + 1/R₃
1/R_eq = 1/12.0 Ω + 1/36.0 Ω
To add the fractions, find a common denominator, which is 36.
1/R_eq = (3/36) + (1/36) = 4/36
1/R_eq = 1/9
R_eq = 9.0 Ω.

Understanding how to combine resistances in series and parallel is fundamental in circuit analysis. Resistances in series add directly, increasing the total resistance. Resistances in parallel combine in a way that the reciprocal of the total resistance is the sum of the reciprocals, resulting in a lower total resistance than the smallest individual resistance.

38. Which one among the following is commonly used as an ‘anti-skinning ag

Which one among the following is commonly used as an ‘anti-skinning agent’ in paints ?

Gelatin

N-methyl pyrrolidone

Pyridine

Polyhydroxy phenol

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The correct answer is D) Polyhydroxy phenol.

‘Skinning’ in paints refers to the formation of a solid or semi-solid film on the surface of the liquid paint when exposed to air. This is primarily due to the oxidation and subsequent polymerization of the drying oils or alkyd resins used as binders, catalyzed by metal driers. Anti-skinning agents are added to paints to prevent this surface oxidation and polymerization. Many anti-skinning agents are volatile antioxidants or complexing agents that temporarily deactivate the metal driers. Phenolic compounds, particularly polyhydroxy phenols and hindered phenols, are effective antioxidants that can scavenge free radicals and prevent the oxidation chain reactions that lead to skinning. Oximes (like methyl ethyl ketoxime) are also commonly used as anti-skinning agents by complexing with metal driers. Polyhydroxy phenol represents a class of compounds known for their antioxidant properties, making them suitable for use as anti-skinning agents.

Gelatin is a protein used as a gelling agent, thickener, or binder. N-methyl pyrrolidone (NMP) is a strong polar solvent. Pyridine is a basic organic compound often used as a solvent or catalyst. None of these are primarily used as anti-skinning agents in paints, unlike phenolic compounds or oximes.

39. The chemical reaction: 2AgCl (s) → 2Ag (s) + Cl₂ (g) takes place

The chemical reaction:
2AgCl (s) → 2Ag (s) + Cl₂ (g)
takes place

in dark

in sunlight

on heating

under high pressure

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The correct answer is B) in sunlight.

The chemical reaction shown, 2AgCl (s) → 2Ag (s) + Cl₂ (g), is the decomposition of silver chloride into silver and chlorine gas. Silver halides like silver chloride (AgCl) and silver bromide (AgBr) are known to be sensitive to light. This decomposition reaction is catalyzed by light, particularly sunlight or bright artificial light. This property is fundamental to traditional black and white photography where silver halides in photographic film undergo decomposition upon exposure to light, forming a latent image.

This type of decomposition reaction, which occurs in the presence of light, is called a photolytic decomposition reaction. While some decomposition reactions occur on heating (thermolysis), the decomposition of silver chloride is prominently triggered by light energy. Dark conditions prevent this reaction. High pressure would likely favour the solid reactant over the gaseous product, not cause decomposition.

40. The mass of 0.5 mole of N₂ gas is

The mass of 0.5 mole of N₂ gas is

7 g

14 g

21 g

28 g

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The correct answer is B) 14 g.

The mass of a substance can be calculated using the number of moles and its molar mass. The formula is: Mass = Number of moles × Molar mass.
Nitrogen gas is diatomic, represented as N₂. The atomic mass of Nitrogen (N) is approximately 14 g/mol. Therefore, the molar mass of N₂ gas is 2 × 14 g/mol = 28 g/mol.
Given the number of moles is 0.5 mole, the mass of 0.5 mole of N₂ gas is 0.5 mol × 28 g/mol = 14 g.

A mole is a unit of amount of substance, defined as containing Avogadro’s number (approximately 6.022 × 10²³) of elementary entities (atoms, molecules, ions, etc.). The molar mass of a substance is the mass of one mole of that substance.