UPSC Geoscientist Archives

11. Which one of the following ecological adaptations is not ‘dormancy’?

Which one of the following ecological adaptations is not ‘dormancy’?

Hibernation

Aestivation

Diapause

Cyclomorphosis

This question was previously asked in

UPSC Geoscientist – 2023

Download PDF Attempt Online

Cyclomorphosis is not a form of dormancy.

Dormancy is a state of reduced metabolic activity and physiological inactivity in response to unfavorable environmental conditions. Hibernation (winter dormancy), Aestivation (summer dormancy), and Diapause (dormancy in insects, often cued by photoperiod) are all examples of dormancy. Cyclomorphosis refers to seasonal or cyclical changes in the morphology of organisms, particularly plankton, in response to environmental factors like predation pressure or water turbulence, rather than a state of inactivity.

Cyclomorphosis allows organisms to adapt to changing conditions over time, often improving their survival chances against predators or adapting to different water conditions throughout the year. It is an active adaptation rather than a state of metabolic suppression.

12. Which one of the following zones of atmosphere is the farthest from th

Which one of the following zones of atmosphere is the farthest from the earth surface?

Stratosphere

Mesosphere

Ionosphere

Troposphere

This question was previously asked in

UPSC Geoscientist – 2023

Download PDF Attempt Online

The Ionosphere is the farthest zone from the Earth’s surface among the options provided.

The layers of the atmosphere from Earth’s surface upwards are Troposphere, Stratosphere, Mesosphere, Thermosphere, and Exosphere. The Ionosphere is not a distinct layer but a region within the Thermosphere (and parts of the Mesosphere and Exosphere) characterized by ionization due to solar radiation. Among the given options (Stratosphere, Mesosphere, Ionosphere, Troposphere), the Ionosphere is situated highest above the Earth’s surface, being part of the Thermosphere which is above the Mesosphere.

The exosphere is the outermost layer, extending into space. The Thermosphere, where the Ionosphere is located, is known for its high temperatures (though not felt as heat due to low density) and is where phenomena like the aurora occur.

13. Consider the following combination of resistors : [Diagram of resistor

Consider the following combination of resistors :
[Diagram of resistors]
The equivalent resistance of the combination of resistors between P and Q is

2 Ω

3 Ω

3.5 Ω

2.5 Ω

This question was previously asked in

UPSC Geoscientist – 2023

Download PDF Attempt Online

Assuming the diagram represents two parallel branches connected between points P and Q, with one branch consisting of two 3 Ω resistors in series and the other branch being a single 6 Ω resistor:
Let R₁ and R₂ be the resistances in the first branch. Since they are in series, their equivalent resistance R_series1 = R₁ + R₂ = 3 Ω + 3 Ω = 6 Ω.
Let R₃ be the resistance in the second branch. R₃ = 6 Ω.
These two equivalent resistances (R_series1 and R₃) are connected in parallel between P and Q. The equivalent resistance of two resistors in parallel is given by R_eq = (R_a * R_b) / (R_a + R_b).
R_eq = (R_series1 * R₃) / (R_series1 + R₃) = (6 Ω * 6 Ω) / (6 Ω + 6 Ω) = 36 Ω² / 12 Ω = 3 Ω.
The equivalent resistance of the combination of resistors between P and Q is 3 Ω.

Resistors in series add directly (R_total = ΣR_i). Resistors in parallel combine as the reciprocal of the sum of reciprocals (1/R_total = Σ1/R_i).

This problem requires identifying series and parallel combinations within a circuit diagram. Analyzing the connections from one terminal to the other helps in breaking down the circuit into simpler parts. Series connections occur when components are connected end-to-end along a single path. Parallel connections occur when components are connected across the same two points.

14. Which one of the following statements for the emission spectrum of hyd

Which one of the following statements for the emission spectrum of hydrogen is true?

The Lyman series lies in the visible region and the Paschen series lies in the infrared region.

The Lyman series lies in the ultraviolet region and the Paschen series lies in the infrared region.

Both the Lyman and the Paschen series lie in the visible region.

The Lyman series lies in the ultraviolet region and the Paschen series lies in the visible region.

This question was previously asked in

UPSC Geoscientist – 2023

Download PDF Attempt Online

The emission spectrum of hydrogen arises from the de-excitation of electrons from higher energy levels (n_initial) to lower energy levels (n_final). Different series are named based on the final energy level (n_final).
– Lyman series: n_final = 1. Transitions from n_initial = 2, 3, 4,… to n=1. These transitions involve the largest energy drops and thus result in the emission of high-energy photons, which fall in the ultraviolet (UV) region of the electromagnetic spectrum.
– Balmer series: n_final = 2. Transitions from n_initial = 3, 4, 5,… to n=2. These transitions correspond to visible light.
– Paschen series: n_final = 3. Transitions from n_initial = 4, 5, 6,… to n=3. These transitions involve smaller energy drops than Balmer series and fall in the infrared (IR) region of the electromagnetic spectrum.
– Brackett series: n_final = 4. Transitions from n_initial = 5, 6, 7,… to n=4. These are in the far-infrared region.
– Pfund series: n_final = 5. Transitions from n_initial = 6, 7, 8,… to n=5. These are also in the far-infrared region.
The statement that is true is that the Lyman series lies in the ultraviolet region and the Paschen series lies in the infrared region.

Different series in the hydrogen emission spectrum are defined by the principal quantum number of the final energy level of the electron transition. The energy and thus the region of the spectrum depend on the energy difference between the initial and final levels.

The energy of the emitted photon is given by the difference in energy between the initial and final states, E = R_H * (1/n_final² – 1/n_initial²), where R_H is the Rydberg constant. Larger energy differences (smaller n_final) correspond to shorter wavelengths (higher energy photons).

15. Which one of the following is the correct electronic configuration of

Which one of the following is the correct electronic configuration of copper?

1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹

1s²2s²2p⁶3s²3p⁶3d⁹4s²

1s²2s²2p⁶3s²3p⁶3d¹⁰4s²

1s²2s²2p⁶3s²3p⁶3d⁸4s²

This question was previously asked in

UPSC Geoscientist – 2023

Download PDF Attempt Online

Copper (Cu) has an atomic number of 29. The Aufbau principle and Hund’s rule predict the filling of orbitals in increasing order of energy. The expected electronic configuration based on the strict Aufbau principle would be 1s²2s²2p⁶3s²3p⁶4s²3d⁹.
However, elements like Copper (and Chromium) are exceptions to this rule. There is a slight energy difference between the 4s and 3d orbitals. The configuration with a completely filled or half-filled d subshell is more stable than one that is nearly filled.
In the case of Copper, promoting one electron from the 4s orbital to the 3d orbital results in the configuration 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹. This configuration has a stable, completely filled 3d subshell and a half-filled 4s subshell, which is energetically more favorable than the 3d⁹4s² configuration.
Therefore, the correct electronic configuration of copper is 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹.

Copper is an exception to the standard Aufbau principle; its ground state electronic configuration involves the transfer of an electron from the 4s to the 3d orbital to achieve a more stable completely filled 3d subshell.

Other elements exhibiting similar exceptions include Chromium (Cr, Z=24) with configuration [Ar] 3d⁵ 4s¹ instead of [Ar] 3d⁴ 4s². These exceptions highlight the complex interplay of electron-electron repulsion and orbital energies.

16. The phenomenon of radioactivity was first discovered by

The phenomenon of radioactivity was first discovered by

Marie Curie

Henri Becquerel

Frederick Soddy

Ernest Rutherford

This question was previously asked in

UPSC Geoscientist – 2023

Download PDF Attempt Online

The phenomenon of radioactivity was first discovered by French physicist Henri Becquerel in 1896. He observed that uranium salts spontaneously emitted radiation that could expose photographic plates, even when the salts were wrapped in opaque paper.

Henri Becquerel’s accidental discovery of radiation from uranium marked the beginning of the study of radioactivity.

Marie Curie and her husband Pierre Curie followed up on Becquerel’s work, isolating new radioactive elements like polonium and radium and coining the term “radioactivity”. They, along with Becquerel, shared the Nobel Prize in Physics in 1903 for their work on radioactivity. Ernest Rutherford later characterized the different types of radiation (alpha, beta, gamma) and developed the nuclear model of the atom. Frederick Soddy is known for his work on isotopes and their role in radioactivity.

17. What is the total number of orbitals associated with the principal qua

What is the total number of orbitals associated with the principal quantum number 3?

This question was previously asked in

UPSC Geoscientist – 2023

Download PDF Attempt Online

The principal quantum number (n) defines the energy level or shell. For a given principal quantum number n, the number of possible subshells is equal to n. These subshells are characterized by the azimuthal quantum number (l), which can take integer values from 0 to n-1.
For n = 3, the possible values of l are 0, 1, and 2.
l = 0 corresponds to the s subshell. The number of orbitals in an s subshell is 1 (m_l = 0).
l = 1 corresponds to the p subshell. The number of orbitals in a p subshell is 3 (m_l = -1, 0, +1).
l = 2 corresponds to the d subshell. The number of orbitals in a d subshell is 5 (m_l = -2, -1, 0, +1, +2).
The total number of orbitals associated with a principal quantum number n is the sum of the number of orbitals in each subshell:
Total orbitals for n=3 = (number of orbitals for l=0) + (number of orbitals for l=1) + (number of orbitals for l=2)
Total orbitals for n=3 = 1 (3s) + 3 (3p) + 5 (3d) = 9.
Alternatively, the total number of orbitals in a shell with principal quantum number n is given by n².
For n = 3, total orbitals = 3² = 9.

For a given principal quantum number n, the total number of atomic orbitals in that shell is n².

Each atomic orbital can hold a maximum of 2 electrons with opposite spins (Pauli Exclusion Principle). Thus, the maximum number of electrons in a shell with principal quantum number n is 2n². For n=3, the maximum number of electrons is 2 * 3² = 18.

18. The oxidation state and covalency of Al in [AlCl (H₂O)₅]²⁺ are

The oxidation state and covalency of Al in [AlCl (H₂O)₅]²⁺ are

+ 3 and 3

+ 3 and 6

+ 2 and 6

+ 2 and 1

This question was previously asked in

UPSC Geoscientist – 2023

Download PDF Attempt Online

The complex is [AlCl (H₂O)₅]²⁺.
Let the oxidation state of Aluminum (Al) be x.
The charge of the chloride ligand (Cl⁻) is -1.
The charge of the water ligand (H₂O) is 0 (neutral molecule).
The overall charge of the complex ion is +2.
The sum of the oxidation states of the central metal ion and the charges of the ligands equals the overall charge of the complex:
x + (charge of Cl) + 5 * (charge of H₂O) = +2
x + (-1) + 5 * (0) = +2
x – 1 = +2
x = +3
So, the oxidation state of Al in [AlCl (H₂O)₅]²⁺ is +3.
The covalency of the central metal atom/ion in a coordination complex is equal to its coordination number, which is the total number of sigma bonds formed between the metal and the ligands. This is typically the number of ligands directly attached to the central metal.
In [AlCl (H₂O)₅]²⁺, the Al³⁺ ion is bonded to one Cl⁻ ligand and five H₂O ligands.
Total number of ligands attached = 1 + 5 = 6.
The coordination number of Al is 6.
Therefore, the covalency of Al in this complex is 6.
The oxidation state is +3 and the covalency is 6.

Oxidation state is the charge the central atom would have if all ligands were removed along with the electron pairs they donate. Covalency in coordination complexes is the number of coordinate bonds formed, equal to the coordination number.

Aluminum typically exhibits an oxidation state of +3. It is a common central metal in coordination chemistry, often forming complexes with a coordination number of 6, resulting in an octahedral geometry.

19. Which one of the following is the correct order of increase of the ato

Which one of the following is the correct order of increase of the atomic radius of the elements?

[amp_mcq option1=”C < B < Si < Al" option2="C < B < Al < Si" option3="C < Si < B < Al" option4="Si < Al < C < B" correct="option1"]

This question was previously asked in

UPSC Geoscientist – 2023

Download PDF Attempt Online

The elements are C (Carbon), B (Boron), Si (Silicon), and Al (Aluminum).
C and B are in Period 2, Group 14 and 13 respectively.
Si and Al are in Period 3, Group 14 and 13 respectively.
Atomic radius generally decreases across a period from left to right due to increasing nuclear charge pulling the electron cloud closer. So, B > C.
Atomic radius generally increases down a group due to the addition of electron shells. So, Si > C and Al > B.
Within Period 3, Al is to the left of Si, so Al > Si.
Comparing elements across periods and groups: Period 3 elements (Si, Al) are generally larger than Period 2 elements (C, B).
Combining these trends:
Within Period 2: C < B Within Period 3: Si < Al Comparing across periods: C and B are smaller than Si and Al. Approximate covalent radii (in picometers, pm): C: 70 pm B: 85 pm Si: 110 pm Al: 125 pm Ordering them in increasing radius: C (70) < B (85) < Si (110) < Al (125). The correct order is C < B < Si < Al.

Atomic radius decreases across a period and increases down a group. The increase in radius down a group due to adding a new electron shell is usually more significant than the decrease across a period due to increasing nuclear charge.

Exceptions and nuances exist in atomic radius trends, especially for transition metals and noble gases. The type of radius (covalent, ionic, van der Waals) also matters depending on the bonding situation.

20. Which one of the following quantities is dimensionless?

Which one of the following quantities is dimensionless?

Specific weight

Specific heat

Specific density

Specific gravity

This question was previously asked in

UPSC Geoscientist – 2023

Download PDF Attempt Online

A dimensionless quantity is a quantity without any physical units or dimensions.
Specific weight is weight per unit volume (Force/Volume), with units like N/m³ or lb/ft³. It has dimensions [M L⁻² T⁻²].
Specific heat (or specific heat capacity) is the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree. It has units like J/(kg·K) or cal/(g·°C), and dimensions [L² T⁻² θ⁻¹].
Specific density is often used interchangeably with relative density or specific gravity, which is the ratio of the density of a substance to the density of a reference substance (like water). In this context, it is dimensionless. However, sometimes ‘specific density’ might be used ambiguously.
Specific gravity is precisely defined as the ratio of the density of a substance to the density of a reference substance. As it is a ratio of two quantities with the same dimensions (density), it is dimensionless.
Given the options, Specific Gravity (D) is unambiguously a dimensionless quantity. Specific density (C) would also be dimensionless if it means relative density, but specific gravity is the more commonly used term for this dimensionless ratio. If specific density were interpreted as density per unit mass (1/volume), it would be dimensional. Therefore, Specific Gravity is the correct answer.

Specific gravity is defined as the ratio of the density of a substance to the density of a standard substance, making it a dimensionless quantity.

Dimensionless quantities are useful as they can be used in ratios and provide relative measures without depending on the system of units. Examples include Reynolds number, Mach number, friction coefficient, and angles (in radians).