UPSC CDS-2 Archives

31. Which one of the following agents does not contribute to propagation o

Which one of the following agents does not contribute to propagation of plants through seed dispersal?

[amp_mcq option1=”Wind” option2=”Fungus” option3=”Animal” option4=”Water” correct=”option2″]

This question was previously asked in

UPSC CDS-2 – 2018

The correct answer is B) Fungus.

Seed dispersal is the movement or transport of seeds away from the parent plant. Common agents of seed dispersal include wind (anemochory), water (hydrochory), animals (zoochory), and mechanical means (autochory). Fungi are typically decomposers or pathogens and do not act as agents for the propagation of plants through seed dispersal; they are usually involved in the decay of organic matter or infecting plants.

Animal dispersal can involve seeds sticking to fur/feathers, being ingested and passed through the digestive tract, or being collected and stored by animals. Water dispersal is common for plants near water bodies, with seeds floating away. Wind dispersal often involves lightweight seeds or seeds with wings/parachutes. Fungi play crucial roles in ecosystems, such as nutrient cycling and symbiotic relationships (like mycorrhizae), but not seed dispersal.

32. If the xylem of a plant is mechanically blocked, which of the followin

If the xylem of a plant is mechanically blocked, which of the following functions of the plant will be affected?

[amp_mcq option1=”Transport of water only” option2=”Transport of water and solutes” option3=”Transport of solutes only” option4=”Transport of gases” correct=”option2″]

This question was previously asked in

UPSC CDS-2 – 2018

Download PDF Attempt Online

The correct answer is B) Transport of water and solutes.

Xylem is the primary tissue in vascular plants responsible for transporting water and dissolved mineral nutrients (solutes) from the roots upwards to the rest of the plant, including the leaves. If the xylem is mechanically blocked, this upward movement of both water and solutes will be significantly impaired.

Phloem is another vascular tissue responsible for transporting sugars (produced during photosynthesis) from the leaves to other parts of the plant. Gases are primarily exchanged through stomata on leaves and lenticels on stems, not transported through xylem or phloem in bulk flow. Therefore, blocking xylem specifically affects water and mineral transport.

33. In which one of the following physiological processes, excess water es

In which one of the following physiological processes, excess water escapes in the form of droplets from a plant?

[amp_mcq option1=”Transpiration” option2=”Guttation” option3=”Secretion” option4=”Excretion” correct=”option2″]

This question was previously asked in

UPSC CDS-2 – 2018

Download PDF Attempt Online

Guttation is the process where plants lose water in the form of liquid droplets from specialized pores called hydathodes, usually located at the tips or margins of leaves. This phenomenon typically occurs when transpiration rates are low (due to high humidity) but water uptake by the roots is high (due to high soil moisture and root pressure). The positive root pressure forces water up the xylem and out through the hydathodes.
Transpiration is the loss of water vapor from the plant surface, primarily through stomata.
Secretion is the process where plants release substances produced by their metabolic activity (e.g., nectar, resins, oils).
Excretion is the removal of metabolic waste products, which plants often store internally or deposit in non-living tissues; they don’t have specialized excretory organs like animals.
The specific loss of water in the form of droplets is called guttation.

– Guttation is the loss of water as liquid droplets.
– It occurs through hydathodes.
– It happens when transpiration is low and root pressure is high.
– Transpiration is the loss of water as vapor.

Guttation is different from dew, which is water that condenses on the leaf surface from atmospheric moisture. Guttation water originates from inside the plant. Guttation droplets may contain dissolved mineral salts and sugars, unlike pure condensed water.

34. Which one of the following denotes a ‘true’ fruit?

Which one of the following denotes a ‘true’ fruit?

[amp_mcq option1=”When only the thalamus of the flower grows and develops into a fruit” option2=”When only the receptacle of the flower develops into a fruit” option3=”When fruit originates only from the calyx of a flower” option4=”When only the ovary of the flower grows into a fruit” correct=”option4″]

This question was previously asked in

UPSC CDS-2 – 2018

Download PDF Attempt Online

In botany, a ‘true’ fruit develops exclusively from the ovary of a flower after fertilization. The ovary wall matures into the pericarp (fruit wall), and the ovules inside the ovary develop into seeds.
A) When only the thalamus…: If the thalamus develops into the edible part of the fruit, it is considered a false fruit or accessory fruit (e.g., apple, pear).
B) When only the receptacle…: Similar to the thalamus, if the receptacle contributes significantly to the fruit structure, it’s an accessory fruit (e.g., strawberry, where the receptacle becomes the fleshy part).
C) When fruit originates only from the calyx…: The calyx (sepals) can sometimes persist or even enlarge in the fruit (e.g., in strawberry where persistent calyx is present, or ground cherry where it encloses the fruit), but the fruit itself develops from the ovary. If the calyx formed the primary fruit structure, it would not be a true fruit originating *only* from the ovary.
D) When only the ovary of the flower grows into a fruit: This is the definition of a true fruit. The fruit wall (pericarp) is derived from the ovary wall.

– A true fruit develops solely from the ovary.
– False fruits (accessory fruits) involve other floral parts besides the ovary in their formation (e.g., thalamus, receptacle, calyx).
– The ovary contains ovules, which develop into seeds after fertilization.

Examples of true fruits include tomato, mango, peach, cherry, and grape. Examples of false fruits include apple, pear, strawberry, fig, and pineapple. Parthenocarpic fruits develop from an unfertilized ovary and are typically seedless (e.g., some varieties of bananas and grapes), and are still considered true fruits as they originate from the ovary.

35. Which of the following is not a primary function of a green leaf?

Which of the following is not a primary function of a green leaf?

[amp_mcq option1=”Manufacture of food” option2=”Interchange of gases” option3=”Evaporation of water” option4=”Conduction of food and water” correct=”option4″]

This question was previously asked in

UPSC CDS-2 – 2018

Download PDF Attempt Online

The primary functions of a green leaf are:
A) Manufacture of food (Photosynthesis): Green leaves contain chlorophyll and are the main sites for converting light energy, CO₂, and water into glucose (food). This is a primary function.
B) Interchange of gases: Leaves have stomata, pores that regulate the exchange of gases (CO₂ intake and O₂ release) with the atmosphere. This is a primary function essential for photosynthesis and respiration.
C) Evaporation of water (Transpiration): Water is lost from the plant mainly through stomata on the leaves in the form of water vapor. This process helps in the ascent of sap and cooling. This is considered a primary function related to water balance and transport.
D) Conduction of food and water: While vascular tissues (xylem for water, phloem for food/sugars) are present in the leaf veins and transport substances within the leaf and to/from the rest of the plant, the *conduction* itself is a transport process performed by these tissues, not a primary metabolic or exchange *function* of the leaf tissue as a whole. The primary functions are the synthesis (photosynthesis) and exchange processes.
Therefore, conduction of food and water is not considered a primary function of the leaf organ itself, unlike photosynthesis, gas exchange, and transpiration.

– Primary functions of leaves include photosynthesis, gas exchange (CO2, O2), and transpiration.
– Conduction is a transport process occurring within the vascular tissues (xylem and phloem) found within the leaf veins.
– The question asks for a primary *function* of the leaf, distinguishing between synthesis/exchange processes and transport.

The vascular bundles (veins) provide structural support and facilitate the transport of water to the photosynthetic cells and the transport of sugars away from the leaf. While essential for the leaf’s operation, this transport is a supportive function rather than a primary function like energy conversion or gas exchange.

36. Which of the following is/are the main absorbing organ/organs of

Which of the following is/are the main absorbing organ/organs of plants?

[amp_mcq option1=”Root only” option2=”Leaf only” option3=”Root and leaf only” option4=”Root, leaf and bark” correct=”option3″]

This question was previously asked in

UPSC CDS-2 – 2018

Download PDF Attempt Online

Plants absorb substances from their environment through specialized organs.
Roots are the primary organs for absorbing water and mineral nutrients from the soil. They are well-adapted for this function with root hairs increasing surface area.
Leaves are the primary organs for absorbing carbon dioxide (CO₂) from the atmosphere, which is essential for photosynthesis. They also play a role in absorbing water vapor or even liquid water directly from the surface (e.g., dew), though this is usually not the main source of water compared to root absorption.
Bark, particularly older bark, is generally not a primary absorbing organ for water or nutrients from the substrate or air, although some gas exchange might occur through lenticels.
Considering the main substances absorbed by plants for their metabolism (water, minerals, CO₂), both roots (from soil) and leaves (from air) function as primary absorbing organs. Thus, “Root and leaf only” best describes the main absorbing organs among the given options.

– Roots absorb water and minerals from the soil.
– Leaves absorb carbon dioxide from the atmosphere.
– Both are essential absorptive processes for plant survival.
– Bark is generally not a primary absorbing organ.

While roots absorb water and minerals, leaves are the main sites of CO₂ absorption, which is crucial for photosynthesis. Some specialized structures like epiphytic roots or insectivorous leaves might have unique absorption functions, but in general terms for typical vascular plants, roots and leaves are the main absorbing interfaces with the environment (soil and atmosphere respectively).

37. Which one of the following does not convert electrical energy into l

Which one of the following does not convert electrical energy into light energy?

[amp_mcq option1=”A candle” option2=”A light-emitting diode” option3=”A laser” option4=”A television set” correct=”option1″]

This question was previously asked in

UPSC CDS-2 – 2018

Download PDF Attempt Online

The question asks which option does *not* convert electrical energy into light energy.
A) A candle: A candle produces light through the process of combustion, where chemical energy stored in the wax is converted into heat and light energy. It does not use electrical energy.
B) A light-emitting diode (LED): LEDs are semiconductor devices that convert electrical energy directly into light through electroluminescence.
C) A laser: Lasers produce light through stimulated emission. Many types of lasers, particularly semiconductor lasers, are powered by electrical energy. Other lasers may be powered by optical energy, but the principle is often related to electrical excitation or pumping.
D) A television set: A television set’s display screen (whether CRT, LCD, LED, or Plasma) converts electrical energy into light to form images.
Therefore, a candle is the only option that does not convert electrical energy into light energy.

– Identify the energy conversion process in each device.
– Electrical energy to light energy conversion is the key process to look for.
– Understand the working principle of each listed item.

Incandescent bulbs, fluorescent lamps, and gas discharge lamps are other common examples of devices that convert electrical energy into light energy. Photosynthesis in plants converts light energy into chemical energy.

38. Two persons are holding a rope of negligible mass horizontally. A 20 k

Two persons are holding a rope of negligible mass horizontally. A 20 kg mass is attached to the rope at the midpoint; as a result the rope deviates from the horizontal direction. The tension required to completely straighten the rope is (g = 10 m/s²)

[amp_mcq option1=”200 N” option2=”20 N” option3=”10 N” option4=”infinitely large” correct=”option4″]

This question was previously asked in

UPSC CDS-2 – 2018

Download PDF Attempt Online

When a mass is attached to the midpoint of a horizontally held rope, the rope sags downwards due to the weight of the mass (W = mg). The tension in the rope acts along the direction of the rope segments on either side of the mass. For the rope to be in equilibrium, the vector sum of the two tensions and the weight must be zero. If the rope is perfectly horizontal, the tension vectors would have only horizontal components (neglecting the mass of the rope itself). However, the weight of the attached mass acts vertically downwards. To balance this downward force, there must be an equal and opposite upward force provided by the vertical components of the tension in the rope. As the rope approaches a perfectly horizontal state, the angle (θ) it makes with the horizontal approaches zero. The vertical component of the tension in each half of the rope is T * sin(θ), where T is the tension. The total upward vertical force is 2 * T * sin(θ). To balance the weight W, 2 * T * sin(θ) = W. If the rope is to be perfectly straight and horizontal (θ = 0), then sin(θ) = sin(0) = 0. For 2 * T * sin(θ) to equal the non-zero weight W (20 kg * 10 m/s² = 200 N), the tension T must tend towards infinity as sin(θ) tends towards zero. Thus, an infinitely large tension is required to completely straighten the rope.

– Static equilibrium requires the net force in all directions to be zero.
– When a mass hangs from a rope, the weight acts vertically downwards.
– Tension in the rope must have a vertical component to balance the weight.
– For a nearly horizontal rope, the angle with the horizontal is very small.
– As the angle approaches zero, the sine of the angle approaches zero, requiring tension to approach infinity to provide a non-zero vertical force.

This scenario represents a classic physics problem illustrating the vector resolution of forces and the limit as an angle approaches zero. In any real-world scenario, the rope will always sag slightly, having a non-zero angle, because no rope can withstand infinite tension.

39. The position vector of a particle is r⃗ = 2t 2 x̂ + 3t ŷ + 4 ẑ. The

The position vector of a particle is r⃗ = 2t² x̂ + 3t ŷ + 4 ẑ. Then the instantaneous velocity v⃗ and acceleration a⃗ respectively lie

[amp_mcq option1=”on xy-plane and along z-direction” option2=”on yz-plane and along x-direction” option3=”on yz-plane and along y-direction” option4=”on xy-plane and along x-direction” correct=”option4″]

This question was previously asked in

UPSC CDS-2 – 2018

Download PDF Attempt Online

The position vector is given by r⃗ = 2t² x̂ + 3t ŷ + 4 ẑ.
The instantaneous velocity vector v⃗ is the first derivative of the position vector with respect to time:
v⃗ = dr⃗/dt = d/dt (2t² x̂ + 3t ŷ + 4 ẑ) = (d/dt 2t²) x̂ + (d/dt 3t) ŷ + (d/dt 4) ẑ = 4t x̂ + 3 ŷ + 0 ẑ = 4t x̂ + 3 ŷ.
A vector of the form Ax̂ + Bŷ has components only in the x and y directions. Such a vector lies in the xy-plane.
The instantaneous acceleration vector a⃗ is the first derivative of the velocity vector with respect to time:
a⃗ = dv⃗/dt = d/dt (4t x̂ + 3 ŷ) = (d/dt 4t) x̂ + (d/dt 3) ŷ = 4 x̂ + 0 ŷ = 4 x̂.
A vector of the form Ax̂ has a component only in the x direction. Such a vector lies along the x-direction.
Therefore, the instantaneous velocity v⃗ lies on the xy-plane, and the instantaneous acceleration a⃗ lies along the x-direction.

– Velocity is the time derivative of the position vector.
– Acceleration is the time derivative of the velocity vector.
– A vector is in a plane if it has non-zero components only in the directions defining that plane.
– A vector is along an axis if it has a non-zero component only in the direction of that axis.

In this case, the position vector has a constant z-component (4). This means the particle is always located on the plane z=4, which is parallel to the xy-plane. The motion is confined to this plane z=4. The velocity vector (4t x̂ + 3 ŷ) correctly reflects motion in the x and y directions within this plane. The acceleration vector (4 x̂) indicates that the acceleration is constant and directed purely in the positive x-direction.

40. If two miscible liquids of same volume but different densities P 1 an

If two miscible liquids of same volume but different densities P₁ and P₂ are mixed, then the density of the mixture is given by

[amp_mcq option1=”(P₁ + P₂) / 2″ option2=”2P₁P₂ / (P₁ + P₂)” option3=”2P₁P₂ / (P₁ – P₂)” option4=”P₁P₂ / (P₁ + P₂)” correct=”option1″]

This question was previously asked in

UPSC CDS-2 – 2018

Download PDF Attempt Online

Let V be the volume of each liquid. The total volume of the mixture is V_total = V + V = 2V.
The mass of the first liquid is m₁ = density × volume = P₁V.
The mass of the second liquid is m₂ = density × volume = P₂V.
The total mass of the mixture is m_total = m₁ + m₂ = P₁V + P₂V = (P₁ + P₂)V.
The density of the mixture is ρ_mixture = m_total / V_total = (P₁ + P₂)V / (2V) = (P₁ + P₂) / 2.
This formula is valid when equal volumes of two miscible liquids are mixed.

– Density is mass per unit volume.
– When mixing, total mass is the sum of individual masses, and total volume is the sum of individual volumes (assuming no volume change upon mixing, which is typical for miscible liquids unless otherwise specified).
– The calculation is based on the given condition that the liquids have the ‘same volume’.

If two miscible liquids of *same mass* m but different densities P₁ and P₂ are mixed, the volume of the first liquid is V₁ = m/P₁, and the volume of the second liquid is V₂ = m/P₂. The total mass is 2m, and the total volume is V₁ + V₂ = m/P₁ + m/P₂ = m(P₁ + P₂)/(P₁P₂). The density of the mixture would be (2m) / [m(P₁ + P₂)/(P₁P₂)] = 2P₁P₂ / (P₁ + P₂), which is option B. The question specifies ‘same volume’, not ‘same mass’.