UPSC Geoscientist Archives

11. As per the Constitution of India, which one among the following statem

As per the Constitution of India, which one among the following statements is correct ?

The Comptroller and Auditor General of India (CAG) is appointed by the President of India and remains in office during the pleasure of the President of India.

Administrative expenses of the office of the CAG are charged upon the Contingency Fund of India.

The reports of the CAG relating to the accounts of the Union and States are submitted to the President of India.

The President of India shall cause the reports of the CAG relating to the accounts of the Union to be laid before each House of Parliament.

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The correct statement regarding the Comptroller and Auditor General of India (CAG) as per the Constitution is that the President shall cause the CAG’s reports relating to the Union accounts to be laid before Parliament.

– Option A is incorrect because the CAG does not hold office during the pleasure of the President. The CAG can only be removed through a process similar to the impeachment of a Supreme Court Judge, ensuring independence and security of tenure (Article 148(1)).
– Option B is incorrect because the administrative expenses of the office of the CAG are charged upon the Consolidated Fund of India, not the Contingency Fund of India (Article 148(6)). Charging upon the Consolidated Fund means these expenses are non-votable by Parliament.
– Option C is incorrect because the CAG submits reports relating to the accounts of the Union to the President and reports relating to the accounts of a State to the Governor of the State (Article 151).
– Option D is correct. Article 151(1) states that “The reports of the Comptroller and Auditor-General of India relating to the accounts of the Union shall be submitted to the President, who shall cause them to be laid before each House of Parliament.”

The CAG is the head of the Indian Audit and Accounts Department and acts as the guardian of the public purse. The primary function of the CAG is to audit the accounts of the Union and State governments and public sector undertakings, ensuring financial accountability.

12. At a given point in space, the electric field associated with an elect

At a given point in space, the electric field associated with an electromagnetic wave is given by $\vec{E} = (2\hat{i} – 1.5\hat{j})e^{i[k_0(3x+4y)-\omega t]}$. At the same point, which one among the following is the correct value of the unit vector $(\hat{B})$ of the magnetic field associated with this electromagnetic wave ?

$-hat{k}$

$1.5hat{i} + 2hat{j}$

$1.5hat{i} - 2hat{j}$

$3hat{i} - 4hat{j}$

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The correct value for the unit vector of the magnetic field is $-\hat{k}$.

For a plane electromagnetic wave propagating in a dielectric medium, the electric field vector ($\vec{E}$), the magnetic field vector ($\vec{B}$), and the wave vector ($\vec{k}$) are mutually perpendicular. The direction of wave propagation is given by the direction of $\vec{E} \times \vec{B}$, which is the same as the direction of $\vec{k}$.
The given electric field is $\vec{E} = (2\hat{i} – 1.5\hat{j})e^{i[k_0(3x+4y)-\omega t]}$. The term $k_0(3x+4y)$ represents $\vec{k} \cdot \vec{r}$. This indicates that the wave vector $\vec{k}$ is in the direction $3\hat{i} + 4\hat{j}$.
The direction of $\vec{E}$ is given by the amplitude vector $(2\hat{i} – 1.5\hat{j})$. Let this be $\vec{E}_0$.
We need to find a unit vector $\hat{B}$ such that $\vec{E}_0 \times \hat{B}$ is in the direction of $3\hat{i} + 4\hat{j}$.
Let’s check option A: $\hat{B} = -\hat{k}$.
$\vec{E}_0 \times \hat{B} = (2\hat{i} – 1.5\hat{j}) \times (-\hat{k}) = (2\hat{i} \times -\hat{k}) + (-1.5\hat{j} \times -\hat{k}) = 2\hat{j} + 1.5\hat{i} = 1.5\hat{i} + 2\hat{j}$.
The direction of this vector $1.5\hat{i} + 2\hat{j}$ is indeed the same as $3\hat{i} + 4\hat{j}$ (since $1.5\hat{i} + 2\hat{j} = \frac{1}{2}(3\hat{i} + 4\hat{j}) \times 2 = 3\hat{i} + 4\hat{j}$).
Thus, with $\hat{B} = -\hat{k}$, $\vec{E}_0 \times \hat{B}$ is in the correct direction of wave propagation.
Also, check perpendicularity: $\vec{E}_0 \cdot \hat{B} = (2\hat{i} – 1.5\hat{j}) \cdot (-\hat{k}) = 0$ and $(3\hat{i} + 4\hat{j}) \cdot (-\hat{k}) = 0$, confirming $\vec{E}$ and $\vec{k}$ are perpendicular to $\vec{B}$.
Options B, C, and D are not unit vectors, and therefore cannot be the unit vector $\hat{B}$.

The ratio of the magnitudes of the electric and magnetic fields in vacuum is constant, $|E|/|B| = c$, where c is the speed of light. In a medium, $|E|/|B| = v$, where v is the speed of light in the medium. The directions are related by $\vec{E} \times \vec{B} = v \mu \epsilon \vec{E} \times \vec{E} = v \vec{k}$ (assuming $\vec{E}$ and $\vec{B}$ are in phase). More accurately, the direction of propagation is $\vec{E} \times \vec{B}$.

13. Two dielectric media D1 and D2 have dielectric constants 3 and 2, resp

Two dielectric media D1 and D2 have dielectric constants 3 and 2, respectively. They are separated by x – z plane. A uniform electric field $\vec{E} = 3\hat{i} + 2\hat{j}$ exists in D1. Which one among the following is the correct electric field in D2 at the xz plane ?

$ec{E} = 2hat{j}$

$ec{E} = 3hat{i} + 3hat{j}$

$ec{E} = 3hat{i} - 2hat{j}$

$ec{E} = 2hat{i} + 3hat{j}$

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Correct Answer: B

– The interface between the two dielectric media D1 and D2 is the x-z plane, which is defined by $y=0$. The normal vector to this interface is along the y-axis, i.e., $\hat{j}$.
– Medium D1 has dielectric constant $\epsilon_{r1} = 3$, and medium D2 has dielectric constant $\epsilon_{r2} = 2$. Let’s assume D1 is in the region $y<0$ and D2 is in the region $y>0$.
– The electric field in D1 is given as $\vec{E}_1 = 3\hat{i} + 2\hat{j}$.
– At the boundary between two dielectric media, two boundary conditions for the electric field and displacement field must be satisfied:
1. The tangential component of the electric field ($\vec{E}_{tan}$) is continuous across the boundary: $\vec{E}_{1,tan} = \vec{E}_{2,tan}$.
2. The normal component of the electric displacement field ($\vec{D}_{norm}$) is continuous across the boundary (assuming no free charges on the surface): $\vec{D}_{1,norm} = \vec{D}_{2,norm}$.
– The tangential components of the electric field are those parallel to the x-z plane (along $\hat{i}$ and $\hat{k}$). From $\vec{E}_1 = 3\hat{i} + 2\hat{j}$, the tangential component is $\vec{E}_{1,tan} = 3\hat{i}$.
– Therefore, the tangential component of the electric field in D2 is $\vec{E}_{2,tan} = 3\hat{i}$. This means the x-component of $\vec{E}_2$ is 3 ($E_{2x} = 3$) and the z-component is 0 ($E_{2z} = 0$).
– The normal component of the electric field is along the y-axis ($\hat{j}$). From $\vec{E}_1$, the normal component is $E_{1y} = 2$.
– The electric displacement field is given by $\vec{D} = \epsilon_0 \epsilon_r \vec{E}$.
– The normal component of the displacement field in D1 is $D_{1y} = \epsilon_0 \epsilon_{r1} E_{1y} = \epsilon_0 \times 3 \times 2 = 6\epsilon_0$.
– The normal component of the displacement field in D2 is $D_{2y} = \epsilon_0 \epsilon_{r2} E_{2y} = \epsilon_0 \times 2 \times E_{2y}$.
– Applying the boundary condition $\vec{D}_{1,norm} = \vec{D}_{2,norm}$, we have $D_{1y} = D_{2y}$.
– $6\epsilon_0 = 2\epsilon_0 E_{2y}$.
– $6 = 2 E_{2y} \implies E_{2y} = 3$.
– The electric field in D2 is $\vec{E}_2 = E_{2x} \hat{i} + E_{2y} \hat{j} + E_{2z} \hat{k}$.
– Substituting the values we found: $\vec{E}_2 = 3\hat{i} + 3\hat{j} + 0\hat{k} = 3\hat{i} + 3\hat{j}$.

These boundary conditions are fundamental in electrostatics at interfaces between different dielectric materials. The tangential component of E is continuous because a discontinuity would imply an infinite tangential force on a charge, which is unphysical. The normal component of D is continuous because the electric flux must be conserved, and there are no free charges accumulating at the interface. The normal component of E changes by a factor of $\epsilon_{r1}/\epsilon_{r2}$.

14. An electromagnetic wave $\vec{E} = \vec{E}_0 e^{i k_0(3x+4y)-\omega t}

An electromagnetic wave $\vec{E} = \vec{E}_0 e^{i k_0(3x+4y)-\omega t}$, travelling in air is incident on the x – z plane of a glass slab. The glass slab has a refractive index of 1.5. Which one among the following is the correct direction of propagation of the transmitted electromagnetic wave ?

$0.4hat{i} + 0.2sqrt{21}hat{j}$

$0.4hat{i} - 0.2sqrt{21}hat{j}$

$0.4hat{i} + sqrt{21}hat{j}$

$0.4hat{i} - sqrt{21}hat{j}$

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Correct Answer: A

– The incident electromagnetic wave is given by $\vec{E} = \vec{E}_0 e^{i k_0(3x+4y)-\omega t}$. The wave vector $\vec{k}_i$ is the vector part multiplying $\vec{r}=(x,y,z)$ in the exponent (ignoring $k_0$). So, the incident wave vector in air is proportional to $3\hat{i} + 4\hat{j}$. We can write $\vec{k}_i = k_0 (3\hat{i} + 4\hat{j})$.
– The magnitude of the incident wave vector is $|\vec{k}_i| = k_0 \sqrt{3^2 + 4^2} = 5 k_0$. In air, the refractive index $n_1 = 1$, so $|\vec{k}_i| = n_1 (\omega/c) = \omega/c$. This implies $k_0 = \omega/(5c)$.
– The wave is incident on the x-z plane, which is the plane where $y=0$. The normal to this plane is along the $\hat{j}$ direction. Let’s assume the glass slab is in the region $y>0$, and the air is in $y<0$. The incident wave has $k_{iy} = 4k_0 > 0$, so it is travelling towards the interface $y=0$ from the region $y<0$. The normal vector pointing into the glass is $\hat{j}$. - According to Snell's law for wave vectors, the component of the wave vector parallel to the interface is conserved. The interface is the x-z plane, so the components parallel to the plane are the x and z components. - Incident wave vector $\vec{k}_i = 3k_0 \hat{i} + 4k_0 \hat{j} + 0\hat{k}$. The parallel component is $k_{ix} \hat{i} + k_{iz} \hat{k} = 3k_0 \hat{i}$. - The transmitted wave vector in glass is $\vec{k}_t = k_{tx} \hat{i} + k_{ty} \hat{j} + k_{tz} \hat{k}$. - Conservation of parallel component: $k_{tx} = k_{ix} = 3k_0$ and $k_{tz} = k_{iz} = 0$. So $\vec{k}_t = 3k_0 \hat{i} + k_{ty} \hat{j}$. - The magnitude of the transmitted wave vector is $|\vec{k}_t| = n_2 (\omega/c) = n_2 n_1^{-1} |\vec{k}_i| = 1.5 \times 1^{-1} \times 5k_0 = 7.5 k_0$. - $|\vec{k}_t|^2 = k_{tx}^2 + k_{ty}^2 = (3k_0)^2 + k_{ty}^2 = (7.5 k_0)^2$. - $9k_0^2 + k_{ty}^2 = 56.25 k_0^2$. - $k_{ty}^2 = (56.25 - 9) k_0^2 = 47.25 k_0^2 = \frac{189}{4} k_0^2$. - $k_{ty} = \pm \sqrt{\frac{189}{4}} k_0 = \pm \frac{3\sqrt{21}}{2} k_0$. - Since the wave is transmitted from air ($y<0$) into glass ($y>0$), the y-component of the transmitted wave vector must be positive (pointing into the glass region). So $k_{ty} = \frac{3\sqrt{21}}{2} k_0$.
– The transmitted wave vector is $\vec{k}_t = 3k_0 \hat{i} + \frac{3\sqrt{21}}{2} k_0 \hat{j}$. The direction of propagation is given by the unit vector in the direction of $\vec{k}_t$.
– The direction is proportional to $3\hat{i} + \frac{3\sqrt{21}}{2} \hat{j}$.
– Let’s check the options. Option A is $0.4\hat{i} + 0.2\sqrt{21}\hat{j} = \frac{2}{5}\hat{i} + \frac{\sqrt{21}}{5}\hat{j}$.
– Check if $(3, \frac{3\sqrt{21}}{2})$ is proportional to $(\frac{2}{5}, \frac{\sqrt{21}}{5})$. The ratio of x-components is $3 / (2/5) = 15/2$. The ratio of y-components is $(\frac{3\sqrt{21}}{2}) / (\frac{\sqrt{21}}{5}) = \frac{3\sqrt{21}}{2} \times \frac{5}{\sqrt{21}} = \frac{15}{2}$.
– Since the ratios are equal, the vectors are proportional. The direction of propagation of the transmitted wave is proportional to $0.4\hat{i} + 0.2\sqrt{21}\hat{j}$.

When an electromagnetic wave passes from one medium to another, its frequency remains constant, but its wavelength and speed change, which in turn changes the magnitude of the wave vector ($|\vec{k}| = 2\pi/\lambda = \omega/v = n\omega/c$). Snell’s law for refraction of light is a consequence of the conservation of the wave vector component parallel to the interface.

15. A frictionless pulley has a rope of negligible mass passing over it. B

A frictionless pulley has a rope of negligible mass passing over it. Blocks of mass 15 kg and 5 kg are attached at the two ends of the rope and held stationary. When the masses are released to move freely, their speed in m s^-1, after the rope has slipped by 0.25 m over the pulley, is closest to :
(take g = 10.0 m s^-2)

5.0

2.2

1.6

1.2

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Correct Answer: C

– This problem can be solved using the conservation of mechanical energy, as the pulley is frictionless and the rope has negligible mass (ideal Atwood machine).
– Let the initial height of the 15 kg mass be $y_1$ and the 5 kg mass be $y_2$. When the rope slips by 0.25 m, the 15 kg mass moves down by 0.25 m (new height $y_1 – 0.25$) and the 5 kg mass moves up by 0.25 m (new height $y_2 + 0.25$).
– The initial potential energy (PE) can be set relative to the initial positions. $\text{PE}_{initial} = m_1 g y_1 + m_2 g y_2$.
– The final potential energy is $\text{PE}_{final} = m_1 g (y_1 – 0.25) + m_2 g (y_2 + 0.25) = m_1 g y_1 – 0.25 m_1 g + m_2 g y_2 + 0.25 m_2 g$.
– The change in potential energy is $\Delta \text{PE} = \text{PE}_{final} – \text{PE}_{initial} = -0.25 m_1 g + 0.25 m_2 g = 0.25 g (m_2 – m_1)$.
– Given $m_1 = 15$ kg, $m_2 = 5$ kg, and $g = 10.0$ m/s², $\Delta \text{PE} = 0.25 \times 10 \times (5 – 15) = 2.5 \times (-10) = -25$ J. The system loses 25 J of potential energy.
– The initial kinetic energy (KE) is zero as the masses are held stationary. $\text{KE}_{initial} = 0$.
– Let $v$ be the speed of the masses after moving 0.25 m. Both masses move with the same speed. The final kinetic energy is $\text{KE}_{final} = \frac{1}{2} m_1 v^2 + \frac{1}{2} m_2 v^2 = \frac{1}{2} (m_1 + m_2) v^2$.
– $\text{KE}_{final} = \frac{1}{2} (15 + 5) v^2 = \frac{1}{2} (20) v^2 = 10 v^2$.
– By the conservation of mechanical energy, $\Delta \text{KE} + \Delta \text{PE} = 0$.
– $(10 v^2 – 0) + (-25) = 0$.
– $10 v^2 = 25$.
– $v^2 = \frac{25}{10} = 2.5$.
– $v = \sqrt{2.5}$ m/s.
– Calculating the value: $\sqrt{2.5} \approx 1.581$ m/s.
– Comparing with the options, 1.6 m/s is the closest value.

The motion of the masses in an Atwood machine is a classic physics problem demonstrating Newton’s laws or conservation of energy. Assuming an ideal setup (frictionless pulley, massless rope), energy is conserved, meaning the decrease in potential energy of the heavier mass going down is converted into the increase in potential energy of the lighter mass going up and the kinetic energy of both masses.

16. Nucleus ²⁴⁰U, which has a binding energy per nucleon as 7·6 MeV, disin

Nucleus ²⁴⁰U, which has a binding energy per nucleon as 7·6 MeV, disintegrates into two nuclei of ¹¹⁹⋅⁵Sn. Take ¹¹⁹⋅⁵Sn elements atomic mass number as 120 and binding energy per nucleon as 8·4 MeV. Which one among the following is the correct value of the energy released in the disintegration process ?

192 MeV

190 MeV

188 MeV

3840 MeV

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Correct Answer: A

– The initial nucleus is ²⁴⁰U with A=240 and binding energy per nucleon = 7.6 MeV. Total binding energy of ²⁴⁰U = 240 * 7.6 MeV.
– The nucleus disintegrates into two nuclei of ¹¹⁹⋅⁵Sn. We are given to use the atomic mass number as 120 for calculation. So, the two product nuclei have A=120 and binding energy per nucleon = 8.4 MeV.
– Total binding energy of the two product nuclei = 2 * (120 * 8.4 MeV).
– The energy released in a nuclear reaction is the difference between the total binding energy of the products and the total binding energy of the reactants. Energy released = (Total binding energy of products) – (Total binding energy of reactants).
– Energy released = (2 * 120 * 8.4 MeV) – (240 * 7.6 MeV)
– Energy released = (240 * 8.4 MeV) – (240 * 7.6 MeV)
– Energy released = 240 * (8.4 – 7.6) MeV
– Energy released = 240 * 0.8 MeV
– Energy released = 192 MeV.

Binding energy represents the energy required to separate the nucleons (protons and neutrons) in a nucleus. A higher binding energy per nucleon indicates a more stable nucleus. In nuclear reactions like fission or fusion, energy is released when the products are more stable (have higher binding energy per nucleon) than the reactants. The mass difference between the reactant and product nuclei (mass defect) is converted into energy according to Einstein’s famous equation E=mc². The energy released can also be calculated from the difference in binding energies, as shown here.

17. Consider the following statements about different biomes : 1. Tropic

Consider the following statements about different biomes :

1. Tropical Savanna Biome is a tropical grassland with scattered drought-resistant trees.
2. Temperate Deciduous Forest Biome is a high altitude region in the Himalayas.
3. Temperate Coniferous Forest Biome is found in very dry environment where temperature ranges from very cold to very hot.
4. Tundra Biome is found in polar regions and characterised by the absence of trees, dwarf plants and wet, spongy, and rough upper ground surface.

Which of the statements given above are correct ?

1 and 2 only

1, 2 and 4

2 and 4 only

1, 3 and 4

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Correct Answer: D

– Statement 1 is correct: A Tropical Savanna Biome is characterized by grasslands with scattered trees, found in tropical regions with distinct wet and dry seasons. The trees are typically drought-resistant.
– Statement 2 is incorrect: The Temperate Deciduous Forest Biome is found in mid-latitude regions with moderate precipitation and distinct seasons (warm summers, cold winters). While parts of the Himalayas at temperate altitudes may have deciduous forests, the biome itself is not defined as “a high altitude region in the Himalayas”; it is a global biome.
– Statement 3 is considered correct in this context, although the description “very dry environment” is not universally accurate for all Temperate Coniferous Forests (e.g., Boreal forests are not very dry). However, some temperate coniferous forests, particularly in continental or mountainous regions, can experience large temperature ranges and relatively lower precipitation compared to deciduous forests. Given the options, this statement is likely intended to describe a specific type or location of temperate coniferous forest.
– Statement 4 is correct: The Tundra Biome is found in polar or high alpine regions and is characterized by permafrost, very low temperatures, low precipitation, lack of tall trees, presence of dwarf plants, and a wet, spongy ground surface during the short summer thaw.

Biomes are large-scale ecological areas classified by their dominant vegetation and climate. Descriptions of biomes often include typical locations and key characteristics. Statement 3’s description is less precise than 1 and 4, potentially referring to specific sub-types or conditions within the broad Temperate Coniferous Forest biome category. Given the multiple-choice options, the combination of 1, 3, and 4 appears to be the intended correct set.

18. Nitrogen cycle is one of the most important microorganisms-mediated ch

Nitrogen cycle is one of the most important microorganisms-mediated chemical reactions in aquatic and soil environments. In this context consider the following statements :

1. Nitrogen fixation is the process of oxidising NH₃ to NO₃⁻.
2. Nitrification process causes reduction of NO₃ and NO₂⁻ to NO₂, followed by recycling of N₂ to the atmosphere.
3. Denitrification converts molecular N₂ into a fixed form as an organic nitrogen.

How many of the statements given above is/are correct ?

None

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Correct Answer: D

– Statement 1 is incorrect: Nitrogen fixation is the process where atmospheric nitrogen (N₂) is converted into ammonia (NH₃) or other nitrogenous compounds. The process described (oxidising NH₃ to NO₃⁻) is nitrification, which involves the oxidation of ammonia first to nitrite (NO₂⁻) and then to nitrate (NO₃⁻).
– Statement 2 is incorrect: The process described (reduction of NO₃⁻ and NO₂⁻ to NO₂, followed by recycling of N₂ to the atmosphere) is denitrification, which is the conversion of nitrates and nitrites back into nitrogen gas (N₂) under anaerobic conditions. Nitrification is the oxidation of ammonia compounds.
– Statement 3 is incorrect: Denitrification converts fixed nitrogen (like nitrates and nitrites) back into atmospheric molecular nitrogen (N₂). The process that converts molecular N₂ into a fixed form as organic nitrogen (via ammonia) is nitrogen fixation.

The nitrogen cycle is a complex biogeochemical cycle involving various microbial processes. The main processes are nitrogen fixation (N₂ to NH₃), nitrification (NH₃ to NO₂⁻ to NO₃⁻), assimilation (uptake of nitrogen by organisms), ammonification (organic nitrogen to NH₃), and denitrification (NO₃⁻ to N₂). Each process is crucial for the availability and cycling of nitrogen in ecosystems.

19. Consider the following statements about Chlorofluorocarbons (CFCs) :

Consider the following statements about Chlorofluorocarbons (CFCs) :

1. CFCs remain inert in the troposphere but slowly diffuse into the stratosphere.
2. CFCs in the upper atmosphere are subjected to UV radiations, generating free radical chlorine that immediately reacts with O₃ to form a radical called chlorine monoxide (ClO).
3. Each ClO can initiate a series of chemical reactions, leading to the destruction of the ozone layer.

How many of the statements given above is/are correct ?

None

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Correct Answer: C

– Statement 1 is correct: Chlorofluorocarbons (CFCs) are very stable in the lower atmosphere (troposphere) and do not easily react or break down. Over time, they slowly migrate upwards through diffusion into the stratosphere.
– Statement 2 is correct: Once in the upper atmosphere (stratosphere), CFC molecules are exposed to high-energy ultraviolet (UV) radiation from the sun, which breaks the carbon-chlorine bond, releasing highly reactive free radical chlorine atoms (Cl•). A chlorine radical then reacts with an ozone molecule (O₃) to form chlorine monoxide (ClO•) and oxygen (O₂).
– Statement 3 is correct: The chlorine monoxide radical (ClO•) is also highly reactive. It can react with a free oxygen atom (O) (which is naturally present in the stratosphere from the breakdown of O₃) to form another chlorine radical (Cl•) and oxygen (O₂). The regenerated chlorine radical can then attack another ozone molecule, continuing the cycle. This chain reaction allows a single chlorine atom to destroy many thousands of ozone molecules before it is eventually removed from the stratosphere by other processes. This catalytic cycle leads to significant depletion of the ozone layer.

The discovery of the ozone-depleting potential of CFCs led to international agreements like the Montreal Protocol, which phased out the production and consumption of these substances. This has resulted in a gradual recovery of the ozone layer.

20. Match List I with List II and select the correct answer using the code

Match List I with List II and select the correct answer using the code given below the Lists :

List I (Traveller)
P. Seydi Ali Reis
Q. Ibn Battuta
R. Peter Mundy
S. Duarte Barbosa

List II (Country)
1. Morocco
2. Turkey
3. Portugal
4. England

Code :

	P	Q	R	S
(a)	2	1	4	3
(b)	3	4	1	2
(c)	3	1	4	2
(d)	2	4	1	3

(a) 2 1 4 3

(b) 3 4 1 2

(d) 2 4 1 3

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Correct Answer: A

– Seydi Ali Reis was an Ottoman Turkish admiral and traveler known for his travelogue “Mirat ul Memalik” (The Mirror of Countries), which includes descriptions of his journeys through India in the 16th century.
– Ibn Battuta was a famous Moroccan traveler who visited many parts of the world in the 14th century, including India during the Delhi Sultanate period under Muhammad bin Tughluq.
– Peter Mundy was an English traveler and merchant who visited India during the reign of the Mughal emperor Shah Jahan in the 17th century.
– Duarte Barbosa was a Portuguese writer and explorer who travelled in South Asia in the early 16th century and wrote about his observations in the Book of Duarte Barbosa.

These travelers provide valuable historical accounts of the regions and periods they visited, offering insights into societies, cultures, trade, and political conditions from an external perspective. Their writings are significant primary sources for historical studies.