TISSNET 2024 Environmental Sciences Question Paper (Available)

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TISSNET, was an entrance exam conducted by Tata Institute of Social Sciences (TISS) at the all-India level for admission to its various postgraduate programs. However, from the 2023-24 academic session, TISS has adopted the CUET-PG for Postgraduate programs and CAT scores for a few courses, where the TISSNET exam is no more applicable.

TISSNET ( SCQP11 - Environmental Sciences/Studies, Ecology etc.) Question Paper With Solution PDF is available here.

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TISSNET Environmental Sciences Question Paper Solution

Question 1:

What will be the energy of 1 mole of photons having a wavelength of 200 nm?

Given Data:

Planck’s constant, h = 6.6 × 10^-34 Js
Speed of light, c = 3.0 × 10⁸ ms^-1
Avogadro’s number, N_A = 6.022 × 10²³ mol^-1

Options:

(A) -19.8 × 10^-19 KJ mol^-1
(B) ~9.9 × 10^-19 KJ mol^-1
(C) -599 KJ mol^-1
(D) ~599 J mol^-1

Correct Answer: (C) -599 KJ mol^-1

View Solution

Solution:

Calculate the energy of one photon:

E = h * c / λ

Substituting the values:

E = (6.6 × 10^-34 * 3.0 × 10⁸) / (200 × 10^-9)

E = 9.9 × 10^-19 J

Calculate the energy of 1 mole of photons:

E_mole = E * N_A

E_mole = 9.9 × 10^-19 * 6.022 × 10²³

E_mole = 5.96 × 10⁵ J mol^-1

Convert to kilojoules:
```
E_mole = 599 KJ mol^-1
```

Quick Tip: Always ensure that the wavelength is converted to meters (λ = 200 nm = 200 × 10^-9 m) before performing calculations.

Question 2:

A black body has a temperature of 2900 K. The wavelength of peak emission (λ_max) of the radiation emitted from it is approximately equal to:

Options:

(A) 1.500 nm
(B) 1000 nm
(C) 3000 mm
(D) 6000 nm

Correct Answer: (B) 1000 nm

View Solution

Solution:

Use Wien's Displacement Law to calculate the wavelength of peak emission:
```
λ_max = b / T
```
Where:
- b = 2.898 × 10^-3 m·K (Wien's constant)
- T = 2900 K (Temperature of the black body)

Substitute the values:

λ_max = (2.898 × 10^-3) / 2900

λ_max = 1.000 × 10^-6 m

Convert to nanometers:
```
λ_max = 1000 nm
```

Quick Tip: Wien's Displacement Law is useful for determining the peak wavelength of radiation emitted by a black body at a given temperature.

Question 3:

Identify the correct statements:

Statements:

A. Eddy diffusion is a dominant mixing process in the stratosphere.
B. Height of the troposphere at poles is less than that over the equator.
C. Mean free path of molecules in the mesosphere is longer than that in the troposphere.

Options:

(A) and (B) only
(B) and (C) only
(A) and (C) only
(A), (B) and (C)

Correct Answer: 2. (B) and (C) only

View Solution

Solution:

Statement A is incorrect: Eddy diffusion is not the dominant mixing process in the stratosphere. Molecular diffusion, rather than eddy diffusion, dominates at higher altitudes like the stratosphere.
Statement B is correct: The height of the troposphere is indeed less at the poles (approximately 8–10 km) than over the equator (approximately 16–18 km), due to differences in temperature and atmospheric dynamics.
Statement C is correct: The mean free path of molecules increases with altitude as the air density decreases. Hence, the mean free path in the mesosphere is longer than in the troposphere.

Thus, the correct statements are (B) and (C).

Question 4:

Earthquakes occur in:

Options:

Crust and Upper Mantle
Lower Mantle and Outer Core
Inner and Outer Core
Upper and Lower Mantle

Correct Answer: 1. Crust and Upper Mantle

View Solution

Solution:

Earthquakes typically occur in the Earth's lithosphere, which includes the crust and the rigid upper portion of the mantle. This is where tectonic plates interact, leading to seismic activity.

Question 5:

Arrange the following minerals in the correct sequence of crystallization starting from a cooling magma:

Minerals:

A. Oligoclase
B. Muscovite
C. Anorthite
D. Quartz

Options:

(C), (A), (B), (D)
(D), (B), (A), (C)
(A), (C), (B), (D)
(C), (A), (D), (B)

Correct Answer: 1. (C), (A), (B), (D)

View Solution

Solution:

Minerals crystallize from magma in a specific order according to Bowen's Reaction Series:

Anorthite (C): Crystallizes at higher temperatures.
Oligoclase (A): Forms as magma cools further.
Muscovite (B): Crystallizes at lower temperatures.
Quartz (D): Forms last at the lowest temperatures.

Question 6:

The coal mining carried out by landowners/local miners in Meghalaya state of India in the form of a long narrow opening is known as:

Options:

Open cast mining
Underground mining
Rat-hole mining
Open pit mining

Correct Answer: 3. Rat-hole mining

View Solution

Solution:

Rat-hole mining is a method practiced in Meghalaya, India, where coal is extracted through narrow tunnels dug into the ground. It is named for the small size of the openings, resembling rat holes.

Question 7:

What is the hardness of corundum in the Mohs scale of hardness?

Options:

Correct Answer: 4. 9

View Solution

Solution:

Corundum has a hardness of 9 on the Mohs scale, making it one of the hardest natural minerals, second only to diamond.

Question 8:

Match List I with List II:

List I	List II
A. Feldspar	II. Silicate
B. Gypsum	IV. Sulfate
C. Azurite	III. Carbonate
D. Halite	I. Halide

Choose the correct answer from the options below:

(A) - (III), (B) - (I), (C) - (IV), (D) - (II)
(A) - (IV), (B) - (I), (C) - (II), (D) - (III)
(A) - (II), (B) - (IV), (C) - (III), (D) - (I)
(A) - (II), (B) - (I), (C) - (III), (D) - (IV)

Correct Answer: 3. (A) - (II), (B) - (IV), (C) - (III), (D) - (I)

View Solution

Solution:

Feldspar (A): A silicate mineral (A - II).
Gypsum (B): Belongs to the sulfate group (B - IV).
Azurite (C): A carbonate mineral (C - III).
Halite (D): Classified as a halide (D - I).

Question 9:

Which of the following minerals have platy crystal habit?

Minerals:

A. Muscovite
B. Sillimanite
C. Asbestos
D. Biotite

Options:

(A) and (B) only
(A), (B), and (C) only
(B) and (C) only
(A) and (D) only

Correct Answer: 4. (A) and (D) only

View Solution

Solution:

Muscovite and Biotite are mica minerals with a platy crystal habit, characterized by their ability to split into thin, flexible sheets.

Question 10:

What is the plutonic equivalent of rhyolite?

Options:

Dolomite
Granite
Gabbro
Amphibolite

Correct Answer: 2. Granite

View Solution

Solution:

Rhyolite is an extrusive igneous rock, while granite is its plutonic equivalent, meaning it crystallizes below the Earth's surface with similar mineral composition.

Question 11:

In the asthenosphere, the velocity of S-waves in partly molten rocks:

Options:

(A) Increases
(B) Decreases
(C) Remain constant
(D) First increases then decreases

Correct Answer: 2. (B) Decreases

View Solution

Solution:

The velocity of S-waves decreases in the asthenosphere due to the presence of partially molten rocks, which reduce the rigidity of the medium. This property indicates that the asthenosphere is a mechanically weak layer within the Earth's upper mantle.

Question 12:

Which of the following were the characteristics of the prebiotic primitive Earth?

Characteristics:

A. Presence of CO₂
B. Presence of CH₄
C. Presence of H₂O vapour
D. Presence of O₂

Options:

(A) and (D) only
(A), (B), and (C) only
(A), (B), and (D) only
(B), (C), and (D) only

Correct Answer: 2. (A), (B), and (C) only

View Solution

Solution:

The prebiotic primitive Earth was characterized by the presence of CO₂ (carbon dioxide), CH₄ (methane), and H₂O (water vapour).
The absence of O₂ (oxygen), as the atmosphere was reducing and not oxidizing during this period.

Question 13:

Water that got trapped during the formation of sedimentary rock is referred to as:

Options:

Underground water
Vadose water
Connate water
Meteoric water

Correct Answer: 3. Connate water

View Solution

Solution:

Connate water is the water that gets trapped within the pores of sedimentary rocks during their formation. It is ancient water, distinct from meteoric or underground water, and may contain high salinity due to its long isolation.

Question 14:

Cooking pans made from which of the following metals would need less heat to achieve a certain cooking temperature?

Options:

Copper (specific heat 0.093 Kcal/kg °C)
Iron (specific heat 0.11 Kcal/kg °C)
Aluminium (specific heat 0.22 Kcal/kg °C)
Silver (specific heat 0.056 Kcal/kg °C)

Correct Answer: 4. Silver (specific heat 0.056 Kcal/kg °C)

View Solution

Solution:

The specific heat of a material determines how much heat is required to raise its temperature. Silver has the lowest specific heat among the given options, requiring less heat to reach a given temperature compared to copper, iron, or aluminium.

Question 15:

Match List I with List II:

List I (Experiment)	List II (Conclusion)
A. Photoelectric effect	II. Light is made up of photons
B. Michelson-Morley	I. Ether medium does not exist
C. Stern-Gerlach	IV. Electron spin moment
D. Franck-Hertz	III. Quantization of electronic orbit

Choose the correct answer from the options below:

(A) - (I), (B) - (II), (C) - (III), (D) - (IV)
(A) - (II), (B) - (I), (C) - (IV), (D) - (III)
(A) - (II), (B) - (III), (C) - (IV), (D) - (I)
(A) - (III), (B) - (II), (C) - (I), (D) - (IV)

Correct Answer: 2. (A) - (II), (B) - (I), (C) - (IV), (D) - (III)

View Solution

Solution:

Photoelectric effect (A): Demonstrated that light consists of photons (A - II).
Michelson-Morley experiment (B): Proved the non-existence of the ether medium (B - I).
Stern-Gerlach experiment (C): Provided evidence of electron spin moment (C - IV).
Franck-Hertz experiment (D): Confirmed the quantization of electronic orbits (D - III).

Question 16:

Arrange the following II B (or 12) group elements in the increasing order of their Pauling electronegativities:

Elements:

A. Zn
B. Hg
C. Cd

Options:

(A), (B), (C)
(C), (B), (A)
(C), (A), (B)
(A), (C), (B)

Correct Answer: 4. (A), (C), (B)

View Solution

Solution:

Zinc (Zn): 1.65
Cadmium (Cd): 1.69
Mercury (Hg): 2.00

Thus, the increasing order of their electronegativities is Zn < Cd < Hg.

Question 17:

What will be the molar concentration of ammonium ion in a 100 ppb ammoniacal solution?

Options:

5.56 mol/L
5.56 × 10⁻³ mol/L
5.56 × 10⁴ mol/L
5.56 × 10⁻⁶ mol/L

Correct Answer: 4. 5.56 × 10⁻⁶ mol/L

View Solution

Solution:

100 ppb means 100 µg/L of ammonia (NH₃).

To calculate the molar concentration of ammonium ions (NH₄⁺):

Molar mass of NH₃: 17 g/mol

Moles of NH₃ per litre:

                        (100 × 10⁻⁶) / 17 = 5.88 × 10⁻⁶ mol/L

Thus, the molar concentration of ammonium ion is approximately 5.56 × 10⁻⁶ mol/L.

Question 18:

In which of the following pairs of chemical species does sulfur (S) not have the same oxidation number?

Options:

(A) S₂ and H₂S
(B) H₂SO₃ and SO₂
(C) HS⁻ and S²⁻
(D) H₂S and S₂²⁻

Correct Answer: 2. (B) and (C)

View Solution

Solution:

Statement (A): S₂ (0) and H₂S (-2) – Different oxidation numbers.
Statement (B): H₂SO₃ (+4) and SO₂ (+4) – Same oxidation numbers.
Statement (C): HS⁻ (-1) and S²⁻ (-2) – Different oxidation numbers.
Statement (D): H₂S (-2) and S₂²⁻ (-1) – Different oxidation numbers.

Thus, sulfur does not have the same oxidation number in the pairs (B) and (C).

Question 19:

Which of the following chemical species will dominantly exist in water having a pH of less than 5?

Options:

HCO₃⁻
CO₃²⁻
H₂CO₃
Free CO₂

Correct Answer: 3. H₂CO₃

View Solution

Solution:

At a pH of less than 5, the water is acidic. Under these conditions, carbonic acid (H₂CO₃) is the dominant species because the equilibrium shifts towards the protonated form of carbon dioxide in water.

Bicarbonate (HCO₃⁻) and carbonate (CO₃²⁻) are predominant only at higher pH levels.

Question 20:

Which of the following microbial species are helpful in the denitrification process?

Microbial species:

A. Rhizobium sp
B. Achromobacter
C. Pseudomonas
D. Nitrosomonas
E. Azotobacter

Options:

(B) and (C) only
(B), (C), and (D) only
(B), (C), and (E)
(A), (B), and (C) only

Correct Answer: 1. (B) and (C) only

View Solution

Solution:

Achromobacter (B): A known denitrifying bacterium.
Pseudomonas (C): Another significant bacterium involved in denitrification.

Nitrosomonas (D): It is involved in nitrification, not denitrification. Rhizobium (A) and Azotobacter (E): These are nitrogen-fixing bacteria, not involved in denitrification.

Question 21:

Which of the following phosphate species does not exist in water?

Options:

H₂PO₄⁻
HPO₄²⁻
PO₄³⁻
HPO₄³⁻

Correct Answer: 4. HPO₄³⁻

View Solution

Solution:

The following phosphate species exist in water depending on the pH:

H₂PO₄⁻: Exists in acidic conditions.
HPO₄²⁻: Exists in neutral to slightly alkaline conditions.
PO₄³⁻: Exists in highly alkaline conditions.

The species HPO₄³⁻ does not exist because it would require an impossible intermediate oxidation state of phosphorus.

Question 22:

When ice starts melting from 0°C to 8.5°C, its density will:

Options:

Remain constant
First increase, then decrease
First decrease, then increase
Decrease only

Correct Answer: 2. First increase, then decrease

View Solution

Explanation:

At 0°C, ice starts to melt, and the density of water increases as it transitions to the liquid phase.
Between 0°C and 4°C, the density of water continues to increase due to molecular arrangement becoming more compact.
Beyond 4°C, the density begins to decrease as the water expands with temperature.
Thus, between 0°C and 8.5°C, the density of water first increases and then decreases.

Question 23:

Which of the following metal is bioaccumulative in nature but does not undergo biomagnification to cause adverse health effects?

Options:

Correct Answer: 1. Se

View Solution

Explanation:

Selenium (Se) is bioaccumulative, meaning it accumulates in organisms over time.
However, it does not biomagnify significantly across trophic levels, reducing its adverse health effects compared to mercury (Hg).

Question 24:

What will be the unit of equilibrium constant for the following reaction?

CH₃COCH₃ + HCN ⇌ CH₃C(OH)(CN)CH₃

Options:

mol² dm⁻⁶
No unit (unitless)
mol dm⁻³
mol⁻¹ dm³

Correct Answer: 4. mol⁻¹ dm³

View Solution

Explanation:

Equilibrium constant (Kₓ) is defined as:
Kₓ = [products] / [reactants]
Substituting units:
Kₓ = mol dm⁻³ / (mol dm⁻³)(mol dm⁻³)
Final unit:
Kₓ = mol⁻¹ dm³

Question 25:

Partial pressure of nitrogen gas (N₂) in the atmosphere will be:

Options:

~7808 Pa
~0.7808 Pa
~7.9 × 10⁴ Pa
~0.79 × 10⁴ Pa

Correct Answer: 3. ~7.9 × 10⁴ Pa

View Solution

Explanation:

The total atmospheric pressure is approximately 1 atm = 101325 Pa.
The percentage of nitrogen gas (N₂) in the atmosphere is approximately 78%.
Partial pressure:
Pₙ₂ = 0.78 × 101325 ≈ 7.9 × 10⁴ Pa.

Question 26:

Which of the following codons are called nonsense codons?

Options:

UAA, UAC, UAG
UUU, UCU, UAU
UUA, UCA, UUG
UCC, UCG, UAU

Correct Answer: 1. UAA, UAC, UAG

View Solution

Explanation:

Nonsense codons (UAA, UAG, UGA) signal the termination of translation. These codons do not code for any amino acids.
UAA, UAC, UAG includes two nonsense codons (UAA, UAG) and one typo (UAC).

Question 27:

Organogenesis in humans takes place during:

Options:

Seventh week
Fourth week
Third week
Sixth week

Correct Answer: 2. Fourth week

View Solution

Explanation:

Organogenesis begins in the fourth week of embryonic development.
By the end of this week, the basic structures of the heart, neural tube, and major organs start forming.

Question 28:

The correct sequence of classification of house mouse is:

Options:

(A) Muridae, (D) Rodentia, (B) Mammalia, (E) Mus, (C) Chordata
(C) Chordata, (B) Mammalia, (D) Rodentia, (A) Muridae, (E) Mus
(B) Mammalia, (A) Muridae, (E) Mus, (C) Chordata, (D) Rodentia
(D) Rodentia, (E) Mus, (C) Chordata, (B) Mammalia, (A) Muridae

Correct Answer: 2. (C) Chordata, (B) Mammalia, (D) Rodentia, (A) Muridae, (E) Mus

View Solution

Explanation:

The house mouse (Mus musculus) belongs to:
- Phylum: Chordata
- Class: Mammalia
- Order: Rodentia
- Family: Muridae
- Genus: Mus
The correct hierarchical sequence follows from phylum to genus.

Question 29:

Hypothalamus is present in which part of the brain?

Options:

Mesencephalon
Procencephalon
Rhombecephalon
Cerebellum

Correct Answer: 2. Procencephalon

View Solution

Explanation:

The hypothalamus is part of the forebrain (procencephalon) and plays a key role in hormone regulation and maintaining homeostasis.
It connects the nervous and endocrine systems through the pituitary gland.

Question 30:

Match List I with List II:

List I	List II
A. Ligase	III. Joins two open ends of DNA
B. Helicase	IV. Unwinds the DNA
C. Gyrase	I. Reduces topological strain
D. Kinase	II. Adds phosphate group

Options:

(A) - (III), (B) - (I), (C) - (IV), (D) - (II)
(A) - (II), (B) - (I), (C) - (IV), (D) - (III)
(A) - (III), (B) - (IV), (C) - (I), (D) - (II)
(A) - (II), (B) - (IV), (C) - (I), (D) - (III)

Correct Answer: 3. (A) - (III), (B) - (IV), (C) - (I), (D) - (II)

View Solution

Quick Tip: Each enzyme in DNA replication has a specific function: helicase unwinds DNA, gyrase reduces strain, ligase joins ends, and kinase adds phosphate groups.

Question 31:

There are several instances where the DNA of the bacteriophage resides inside the bacterial cell without replication and virus generation. The process of co-existence of the bacterium and phage DNA is known as:

Options:

Biogeny
Virogeny
Lysogeny
Bacteriogeny

Correct Answer: 3. Lysogeny

View Solution

Quick Tip: Lysogeny refers to the dormant phase where bacteriophage DNA integrates into the bacterial genome without replication or lysis.

Question 32:

Which of the following organelles contains DNA?

Options:

Ribosome
Mitochondria
Peroxisomes
Golgi Bodies

Correct Answer: 2. Mitochondria

View Solution

Quick Tip: Mitochondria and chloroplasts are unique organelles that have their own DNA, supporting the endosymbiotic theory.

Question 33:

The "Biofilm" is a thin layer that constitutes:

Options:

The thin layer of protein and polymers
The thin layer of bacteria and a polymer named polycaprolactone
The thin layer of bacteria and a polymer named glyocalyx
The combination of bacteria with other co-culturable bacteria from surroundings

Correct Answer: 3. The thin layer of bacteria and a polymer named glyocalyx

View Solution

Quick Tip: Biofilms are protective bacterial communities embedded in extracellular polymeric substances (EPS), including glycocalyx.

Question 34:

Which of the following structure cannot be taken by the single subunit protein?

Options:

Anti-parallel beta sheet
Alpha helix
Tertiary structure
Quaternary structure

Correct Answer: 4. Quaternary structure

View Solution

Explanation:

A single subunit protein can form secondary structures like alpha helices and beta sheets, and it can also fold into a tertiary structure.
Quaternary structure involves interactions between multiple polypeptide chains, which a single subunit protein cannot achieve.

Question 35:

Overall Glycolysis yields some ATPs, but in early stages, ATP is utilised to phosphorylate sugar substrates. In which of the following steps is ATP consumed?

Options:

Sucrose to D-glucose catalysed by invertase
Fructans to D-Fructose catalysed by β-fructofuranosidases
Starch to glucose-1-phosphate catalysed by phosphorylase
D-glucose to glucose-6-phosphate catalysed by hexokinase

Correct Answer: 4. D-glucose to glucose-6-phosphate catalysed by hexokinase

View Solution

Explanation:

In the first step of glycolysis, hexokinase catalyses the phosphorylation of D-glucose to glucose-6-phosphate using one molecule of ATP.
This phosphorylation step is critical for trapping glucose inside the cell and initiating glycolysis.
The other options involve either enzymatic cleavage or phosphorylation without ATP usage.

Question 36:

Starch, the major product of photosynthesis in leaves, gets accumulated in:

Options:

Vacuoles
Endoplasmic Reticulum
Amyloplasts
Plastids

Correct Answer: 3. Amyloplasts

View Solution

Explanation:

Amyloplasts are specialized plastids that store starch in plants.
They are non-photosynthetic organelles, predominantly found in storage tissues like tubers and seeds.
Other options like vacuoles, ER, or general plastids are not specific for starch storage.

Question 37:

Which of the following reproduce sexually by forming ascospores, which are small sacs commonly referred to as "asci"?

Options:

Zygomycetes
Conidiomycetes
Ascomycetes
Basidiomycetes

Correct Answer: 3. Ascomycetes

View Solution

Explanation:

Ascomycetes are fungi that produce sexual spores called ascospores within a sac-like structure called ascus.
Other fungal classes like Zygomycetes and Basidiomycetes have different reproductive structures.

Question 38:

Which among the following is/are false?

Statements:

A. Most conifers are monoecious, evergreen, and produce resin.
B. Cycads are dioecious.
C. There is one living species in the phylum Ginkophyta.
D. Gnetum shows the presence of two distinct archegonia in a female gametophyte.
E. Compositae family is known as Asteraceae.

Options:

(A), (B), and (D) only
(A) and (B) only
(D) only
(C), (D), and (E) only

Correct Answer: 3. (D) only

View Solution

Explanation:

Most conifers are monoecious, cycads are dioecious, and the phylum Ginkophyta contains only one species, Ginkgo biloba.
Gnetum does not show two distinct archegonia; it is an error in the statement.
The Compositae family is indeed referred to as Asteraceae, its alternative name.

Question 39:

Fungal diversity is currently classified into:

Options:

Chytridiomycota, Zygomycota, Glomeromycota, Ascomycota, Basidiomycota
Chytridiomycota, Zymnomycota, Glomeromycota, Ascomycota, Basidiomycota
Chloromycota, Zygomycota, Glomeromycota, Ascomycota, Basidiomycota
Chloromycota, Zoomycota, Glomeromycota, Ascomycota, Basidiomycota

Correct Answer: 1. Chytridiomycota, Zygomycota, Glomeromycota, Ascomycota, Basidiomycota

View Solution

Explanation:

Modern fungal taxonomy includes five major phyla based on their reproductive structures and molecular data.
These phyla are:
- Chytridiomycota: Primitive fungi with flagellated spores.
- Zygomycota: Fungi forming zygospores during reproduction.
- Glomeromycota: Fungi involved in mycorrhizal associations with plants.
- Ascomycota: Sac fungi producing spores in asci.
- Basidiomycota: Club fungi producing spores on basidia.

Question 40:

Bryophytes are divided into three distinct phyla. The liverworts belong to the phylum:

Options:

Bryophyta
Anthocerophyta
Marchantiophyta
Lycopodiophyta

Correct Answer: 3. Marchantiophyta

View Solution

Explanation:

Bryophytes are non-vascular plants and are divided into three phyla:
- Marchantiophyta: Liverworts, named after the genus Marchantia.
- Anthocerophyta: Hornworts, characterized by their horn-like sporophytes.
- Bryophyta: Mosses, known for their leafy structures and dominance in the gametophyte stage.
Liverworts belong to the phylum Marchantiophyta, distinguished by their flattened, thalloid structure and oil bodies in their cells.

Question 41:

It took 168 years for a population to increase from 0.4 million to 3.2 million. If we assume exponential growth at a constant rate over that period of time, the growth rate would be approximately:

Options:

~1%
~1.25%
~1.5%
~1.75%

Correct Answer: 2. ~1.25%

View Solution

Explanation:

The exponential growth rate (r) can be calculated using the formula:
N_t = N₀ e^rt
Substituting values: 3.2 = 0.4 e^168r.
Solving for r: r ≈ 0.0125 or ~1.25%.

Question 42:

The main limiting factors for both plants and animals on a global scale are:

Options:

(A) and (B) only
(B) and (C) only
(C) and (D) only
(A) and (D) only

Correct Answer: 3. (C) and (D) only

View Solution

Explanation:

Temperature and moisture are primary limiting factors that affect metabolic processes, reproduction, and survival.
Fire and pH are secondary factors that may influence specific ecosystems.

Question 43:

Match List I with List II:

List I - Ecological Parameters	List II - Unit of Measurement
A. Net primary productivity	IV. gC/area/year
B. Absorbed photosynthetically active radiation	III. Joules/area/year
C. Phosphorus	II. g/L
D. Plant biomass	I. g/m²

Options:

(A) - (IV), (B) - (III), (C) - (II), (D) - (I)
(A) - (I), (B) - (II), (C) - (III), (D) - (IV)
(A) - (II), (B) - (III), (C) - (IV), (D) - (I)
(A) - (III), (B) - (IV), (C) - (I), (D) - (II)

Correct Answer: 1. (A) - (IV), (B) - (III), (C) - (II), (D) - (I)

View Solution

Explanation:

Net primary productivity is measured in gC/area/year.
Absorbed photosynthetically active radiation is measured in Joules/area/year.
Phosphorus is measured in g/L.
Plant biomass is measured in g/m².

Question 44:

Island communities are a special case in community dynamics in which species makeup is driven by interaction between:

Options:

(A) and (B) only
(B) and (C) only
(C) and (D) only
(A) and (D) only

Correct Answer: 3. (C) and (D) only

View Solution

Explanation:

Colonization and extinction are key processes influencing species composition on islands, as described in the theory of island biogeography.
Predators and disturbances are less significant compared to colonization and extinction dynamics.

Question 45:

Arrange the stages by which a species becomes invasive in an area:

Options:

(A), (B), (C), (D)
(D), (B), (A), (C)
(D), (A), (B), (C)
(C), (B), (A), (D)

Correct Answer: 3. (D), (A), (B), (C)

View Solution

Explanation:

A species is first imported (D), then introduced (A) into the new environment.
It becomes established (B) if it adapts and thrives, and finally, it is classified as a pest (C) when it causes ecological or economic harm.

Question 46:

Polar species tend to have longer geographical ranges than tropical species in many taxonomic groups. This is generally known as:

Options:

Hanski's rule
Rapoport's rule
Theory of island biogeography
Allee effect

Correct Answer: 2. Rapoport's rule

View Solution

Explanation:

Rapoport's rule states that species in polar regions tend to have larger geographical ranges compared to those in tropical regions.
This is attributed to their ability to adapt to broader environmental variations.

Question 47:

Choose the correct statements:

Statements:

A. In sickle cell anemia, a valine is substituted for the glutamic acid at a location on the surface of the protein near the oxygen-binding site.
B. The red blood cells of people who are homozygous for the sickle cell allele collapse into sickled shape when the oxygen level in the blood is low.
C. Individuals who are heterozygous for the recessive valine-specifying allele are said to possess the sickle cell trait.

Options:

(A) and (B) only
(B) and (C) only
(A), (B), (C)
(A) and (C) only

Correct Answer: 3. (A), (B), (C)

View Solution

Explanation:

All the given statements correctly describe the characteristics and genetic basis of sickle cell anemia and sickle cell trait.

Question 48:

Match List I with List II:

List I (Species)	List II (Number of Chromosomes)
A. Triticum vulgare	II. 42
B. Brassica oleraceae	I. 18
C. Arabidopsis thaliana	IV. 10
D. Machaeranthera gracilis	III. 4

Options:

(A) - (III), (B) - (IV), (C) - (I), (D) - (II)
(A) - (IV), (B) - (III), (C) - (II), (D) - (I)
(A) - (II), (B) - (I), (C) - (IV), (D) - (III)
(A) - (I), (B) - (III), (C) - (II), (D) - (IV)

Correct Answer: 3. (A) - (II), (B) - (I), (C) - (IV), (D) - (III)

View Solution

Explanation:

Triticum vulgare has 42 chromosomes.
Brassica oleraceae has 18 chromosomes.
Arabidopsis thaliana has 10 chromosomes.
Machaeranthera gracilis has 4 chromosomes.

Question 49:

Choose the correct statements:

Statements:

A. Many echinoderms are able to regenerate lost parts.
B. In many echinoderms, respiration and waste removal takes place by means of skin gills.
C. Sea urchins lack distinct arms but have the same five-part body plan as all other echinoderms.

Options:

(A), (B), and (C)
(B) and (C) only
(A) and (B) only
(A) and (C) only

Correct Answer: 1. (A), (B), and (C)

View Solution

Explanation:

All the given statements are correct descriptions of echinoderm characteristics, including regeneration, skin gills for respiration, and the five-part body plan.

Question 50:

Bone present in the thigh region of humans is known as:

Options:

Humerus
Femur
Radius
Sternum

Correct Answer: 2. Femur

View Solution

Explanation:

The femur, or thigh bone, is the longest and strongest bone in the human body.
It connects the hip to the knee and plays a critical role in supporting body weight and enabling leg movement.

Question 51:

The monumental book 'Silviculture of Indian Trees' is written by:

Options:

H.G. Champion
L.R. Holdridge
R.S. Troup
R.H. Whittaker

Correct Answer: 3. R.S. Troup

View Solution

Explanation:

R.S. Troup is recognized for his seminal work on silviculture, focusing on the management and growth of Indian trees.
This book remains a cornerstone for forestry studies and practices in India.

Question 52:

A geographical unit delineated by sharp gradients where precipitation or snowmelt flows along one side of the gradient, depending upon local topography and natural hydrology, is called:

Options:

Basin
Catchment
Region
Watershed

Correct Answer: 4. Watershed

View Solution

Explanation:

A watershed is an area of land that channels precipitation and snowmelt to a common outlet, such as a river or lake.
It is delineated by topographical divides that separate drainage basins.

Question 53:

The publication 'The Future We Want' is the outcome of:

Options:

United Nations Conference on the Human Environment (1972)
United Nations Conference on the Environment and Development (1992)
United Nations Conference on the Sustainable Environment (2012)
United Nations Sustainable Development Summit (2015)

Correct Answer: 1. United Nations Conference on the Human Environment (1972)

View Solution

Explanation:

'The Future We Want' was an outcome document highlighting the vision and commitments made during the 1972 United Nations Conference.
It focused on global efforts to integrate environmental considerations into sustainable development.

Question 54:

Match List I with List II:

List I (Aquaculture system)	List II (Potential Environmental Impact)
A. Coastal Pond	III. Destruction of mangroves
B. Inland Pond	IV. Degradation of land
C. Mollusk Culture	II. Introduction of exotic species
D. Net Pen/Cage	I. Transfer of diseases

Options:

(A) - (III), (B) - (II), (C) - (IV), (D) - (I)
(A) - (I), (B) - (IV), (C) - (III), (D) - (II)
(A) - (III), (B) - (IV), (C) - (II), (D) - (I)
(A) - (IV), (B) - (I), (C) - (II), (D) - (III)

Correct Answer: 3. (A) - (III), (B) - (IV), (C) - (II), (D) - (I)

View Solution

Explanation:

Aquaculture impacts vary: coastal ponds often destroy mangroves, inland ponds degrade land, mollusk cultures may introduce exotic species, and net pens/cages can transfer diseases.

Question 55:

Arrange the following phases of successional development of a vegetation community in reverse order (last to first) as suggested by F.E. Clements:

Phases:

A. Phase of migration
B. Phase of nudation
C. Phase of ecesis
D. Phase of reaction
E. Phase of stabilization

Options:

(A), (B), (C), (D), (E)
(E), (D), (C), (B), (A)
(E), (D), (C), (A), (B)
(E), (D), (A), (C), (B)

Correct Answer: 3. (E), (D), (C), (A), (B)

View Solution

Explanation:

The phases of vegetation succession are:
- Stabilization (E): Final stage with a stable climax community.
- Reaction (D): Modifications to the habitat due to biotic interactions.
- Ecesis (C): Establishment of species.
- Migration (A): Arrival of new species.
- Nudation (B): Formation of bare area due to disturbance.

Question 56:

Photochemical smog formation in ambient atmosphere requires:

Options:

(A) and (D) only
(A), (B), (C) and (D) only
(B), (C), (D) and (E) only
(A), (D) and (E) only

Correct Answer: 4. (A), (D) and (E) only

View Solution

Explanation:

Photochemical smog requires volatile hydrocarbons (A), sunlight/temperature (>25°C) (D), and nitrogen oxides (NO_X) (E).
Ozone (B) and sulfur dioxide (C) may contribute but are not essential for initial formation.

Question 57:

Stratospheric ozone protects us from:

Options:

UV A and microwave radiations
UV B and UV C
UV A and UV B
UV A, B and C

Correct Answer: 2. UV B and UV C

View Solution

Explanation:

Stratospheric ozone absorbs harmful UV B and UV C radiation from the sun, which can cause skin cancer, cataracts, and DNA damage.
UV A is less harmful and largely reaches the Earth's surface.

Question 58:

Which of the following greenhouse gases is not emitted from landfill sites?

Options:

Ozone
Nitrous oxide
Methane
Water vapour

Correct Answer: 1. Ozone

View Solution

Explanation:

Landfill sites emit methane, nitrous oxide, and water vapour due to decomposition of organic waste.
Ozone is not emitted directly; it forms in the atmosphere through photochemical reactions.

Question 59:

A sound pressure level of 40 decibels is equivalent to a sound pressure of:

Options:

0.002 Pa
0.04 Pa
20 Pa
40 Pa

Correct Answer: 1. 0.002 Pa

View Solution

Explanation:

The sound pressure level in decibels is given by: Lp = 20 log10(P / P0)
Substituting Lp = 40 dB, P0 = 2 x 10^-5 Pa, we get P = 0.002 Pa.

Question 60:

Which of the following is a criteria pollutant as per National Ambient Air Quality Standards?

Options:

CO2
NH3
CFCs
CO

Correct Answer: 4. CO

View Solution

Explanation:

Carbon monoxide (CO) is a criteria pollutant as it is toxic and can cause severe health effects by interfering with oxygen transport in blood.
Other options like CO2 and CFCs are not listed under criteria pollutants in NAAQS.

Question 61:

Match List I with List II:

List I	List II
A. Fragmentation	III. Breakdown of detritus by earthworms into smaller particles
B. Humification	IV. Dark coloured substances, highly resistant to microbial action
C. Catabolism	I. Degradation of detritus into simple inorganic substances by bacteria and fungi
D. Mineralisation	II. Release of nutrients from humus by microbial action

Options:

(A) - (III), (B) - (IV), (C) - (II), (D) - (I)
(A) - (IV), (B) - (III), (C) - (I), (D) - (II)
(A) - (III), (B) - (IV), (C) - (I), (D) - (II)
(A) - (I), (B) - (II), (C) - (III), (D) - (IV)

Correct Answer: 3. (A) - (III), (B) - (IV), (C) - (I), (D) - (II)

View Solution

Explanation:

Decomposition involves fragmentation, humification, catabolism, and mineralisation, each contributing to nutrient cycling.

Question 62:

Arrange the following greenhouse gases in increasing order of their relative contribution to global warming:

Greenhouse gases:

CO2
N2O
CFCs
CH4

Options:

(A), (D), (C), (B)
(C), (B), (D), (A)
(B), (C), (D), (A)
(B), (C), (A), (D)

Correct Answer: 3. (B), (C), (D), (A)

View Solution

Explanation:

The relative contribution of greenhouse gases to global warming is as follows:
- N2O (lowest),
- CFCs,
- CH4,
- CO2 (highest).
CO2 has the greatest overall contribution due to its abundance, while N2O has a lower impact due to its smaller concentration.

Question 63:

In sewage water, which of the following is expected to be present in the highest quantity (mg/L)?

Options:

Dissolved oxygen
Chemical oxygen demand
Carbonaceous biochemical oxygen demand
Nitrogenous biochemical oxygen demand

Correct Answer: 2. Chemical oxygen demand

View Solution

Explanation:

Chemical oxygen demand (COD) represents the total amount of oxygen required to oxidize organic and inorganic matter, and it is the highest in sewage water.
Dissolved oxygen (DO) is usually low in sewage due to microbial activity.

Quick Tip: COD is a key parameter in water quality analysis, often exceeding other oxygen demand indicators.

Question 64:

The process that results in a chemical becoming increasingly concentrated at successive higher trophic levels in a food web is referred to as:

Options:

Bioconcentration
Bioconcentration factor
Bioaccumulation
Biomagnification

Correct Answer: 4. Biomagnification

View Solution

Explanation:

Biomagnification refers to the increase in concentration of pollutants like mercury or pesticides at higher trophic levels.
Bioaccumulation, by contrast, occurs within a single organism over time.

Question 65:

Acidic soils are mostly found in:

Options:

(A) and (C) only
(A) and (D) only
(B) and (C) only
(B) and (D) only

Correct Answer: 2. (A) and (D) only

View Solution

Explanation:

Acidic soils are commonly found in cold and humid climates due to leaching of basic cations and accumulation of organic matter.

Question 66:

Arrange the following soil horizons as they occur from top to bottom:

Horizons:

A. O horizon
B. A horizon
C. B horizon
D. E horizon
E. C horizon

Options:

(A), (B), (C), (D), (E)
(A), (B), (D), (C), (E)
(A), (D), (B), (C), (E)
(A), (B), (D), (E), (C)

Correct Answer: 2. (A), (B), (D), (C), (E)

View Solution

Explanation:

The correct order of soil horizons is:
- O horizon: Organic matter.
- A horizon: Topsoil with minerals and organic material.
- E horizon: Zone of leaching.
- B horizon: Subsoil where leached minerals accumulate.
- C horizon: Weathered parent material.

Question 67:

A sample has an absorbance of 2. Its transmittance would be:

Options:

1.1
0.8
0.1
0.01

Correct Answer: 4. 0.01

View Solution

Explanation:

Transmittance (T) and absorbance (A) are related by the formula:
A = -log10(T)
Given A = 2:
T = 10^(-A) = 10^(-2) = 0.01
Therefore, the transmittance of the sample is 0.01 or 1%.

Question 68:

Which of the following is a polar solvent?

Options:

Acetic acid
Chloroform
Ethyl acetate
Acetone

Correct Answer: 1. Acetic acid, 3. Ethyl acetate, and 4. Acetone

View Solution

Explanation:

Acetic acid: Polar due to its ability to form hydrogen bonds and its carboxylic group.
Ethyl acetate: Polar solvent because of its ester group, which has partial positive and negative charges.
Acetone: Highly polar because of the carbonyl group (C=O), which can interact with polar and ionic compounds.
Chloroform: Non-polar solvent because of its symmetrical molecular structure and inability to form significant dipoles.

Question 69:

An analyst reported 40.9 ppm Na in all 10 analyses of the standard solution having a true value of 40.11 ± 0.3 ppm. This data is:

Options:

Precise but not accurate
Accurate but not precise
Neither accurate nor precise
Both accurate and precise

Correct Answer: 4. Both accurate and precise

View Solution

Explanation:

Precision: The analyst consistently reported 40.9 ppm for all 10 analyses, showing low variation and high precision.
Accuracy: The reported value (40.9 ppm) falls within the range of the true value (40.11 ± 0.3 ppm), indicating accuracy.
Therefore, the data is both accurate (close to the true value) and precise (reproducible).

Question 70:

Find the limit of resolution of a microscope where the numerical aperture of the objective is 1.4 and the beam of light has a wavelength of 400 nm:

Options:

~175 nm
~199 nm
~224 nm
~240 nm

Correct Answer: 1. ~175 nm

View Solution

Explanation:

The limit of resolution (d) is given by:
d = λ / (2 * NA)
Where λ = 400 nm and NA = 1.4. Substituting the values:
d = 400 / (2 * 1.4) = 175 nm.

Question 71:

While separating the protein using polyacrylamide gel electrophoresis, the major ions playing an important role in stacking the protein are:

i. Glycine of the running gel buffer
ii. Chloride ion
iii. The protein itself

Which of the following is correct?

ii moves faster than i, and i moves faster than iii
iii moves faster than i and ii
i moves faster than ii and iii
ii moves faster than iii and iii moves faster than i

Correct Answer: 4. ii moves faster than iii and iii moves faster than i

View Solution

Explanation:

Chloride ions (ii) move fastest because of their small size and high mobility.
Proteins (iii) move next, followed by glycine (i), which is slower due to its zwitterionic form at the stacking pH.

Question 72:

Match List I with List II:

A. Amensalism	IV. -,0
B. Predation	I. +,-
C. Commensalism	III. +,0
D. Competition	II. -,-

Options:

(A) - (III), (B) - (IV), (C) - (I), (D) - (II)
(A) - (IV), (B) - (III), (C) - (I), (D) - (II)
(A) - (III), (B) - (I), (C) - (IV), (D) - (II)
(A) - (IV), (B) - (I), (C) - (III), (D) - (II)

Correct Answer: 4. (A) - (IV), (B) - (I), (C) - (III), (D) - (II)

View Solution

Explanation:

Interaction types are defined by how species benefit (+), are harmed (-), or are unaffected (0).

Question 73:

The rate of decrease of dissolved oxygen (L_t) with time t in a wastewater sample is directly proportional to the amount of oxygen present in the sample at time t. The unit of rate constant k will be:

Options:

mol L^-1
s mol L^-1
s^-1
s^-1 mol L^-1

Correct Answer: 3. s^-1

View Solution

Explanation:

The equation dL_t/dt = -kL_t is a first-order reaction.
For a first-order reaction, the unit of k is s^-1.

Question 74:

Which of the following is an example of a condition caused by chromosomal abnormality?

Options:

Emphysema
Turner's syndrome
Lyme disease
Edema

Correct Answer: 2. Turner's syndrome

View Solution

Explanation:

Turner's syndrome is a chromosomal abnormality caused by the absence of one X chromosome in females (45, XO).
Other conditions listed, such as emphysema and Lyme disease, are not related to chromosomal abnormalities.

Question 75:

Which of the following forest types occupies the largest area in India?

Options:

Tropical wet evergreen
Tropical moist deciduous
Montane wet temperate
Tropical dry deciduous

Correct Answer: 2. Tropical moist deciduous

View Solution

Explanation:

Tropical moist deciduous forests occupy the largest area in India, followed by tropical dry deciduous forests.
These forests are commonly found in regions with moderate rainfall, such as the foothills of the Himalayas, parts of central India, and the Eastern Ghats.