UPSC NDA-2 Archives

21. Which one of the following is the correct relation between Å and nm?

Which one of the following is the correct relation between Å and nm?

[amp_mcq option1=”1 nm = 10⁻¹ Å” option2=”1 nm = 10 Å” option3=”1 nm = 1 Å” option4=”1 nm = 10⁻² Å” correct=”option2″]

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UPSC NDA-2 – 2018

The Angstrom (Å) is a unit of length equal to 10⁻¹⁰ meters. The nanometer (nm) is a unit of length equal to 10⁻⁹ meters. To find the relation between them, we can write 1 nm = 10⁻⁹ m = 10¹ × 10⁻¹⁰ m = 10 × (10⁻¹⁰ m) = 10 Å. Therefore, 1 nm = 10 Å.

This question tests the knowledge of common units of length used in physics and chemistry, particularly at the atomic and molecular scales, and the ability to convert between them based on their definitions in terms of meters.

The Angstrom unit, though not part of the SI system, is widely used, especially in crystallography and spectroscopy. The nanometer is an SI unit and is commonly used in nanotechnology and measuring wavelengths of visible light.

22. A stainless steel chamber contains Ar gas at a temperature T and press

A stainless steel chamber contains Ar gas at a temperature T and pressure P. The total number of Ar atoms in the chamber is n. Now Ar gas in the chamber is replaced by CO₂ gas and the total number of CO₂ molecules in the chamber is n/2 at the same temperature T. The pressure in the chamber now is P’. Which one of the following relations holds true? (Both the gases behave as ideal gases)

[amp_mcq option1=”P’ = P” option2=”P’ = 2P” option3=”P’ = P/2″ option4=”P’ = P/4″ correct=”option3″]

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The ideal gas law states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. For the first scenario with Ar gas, we have PV = n_Ar RT. Since the number of atoms is n, and assuming R is taken such that n represents the total number of particles, the equation can be written as PV = nRT. For the second scenario with CO₂ gas, the number of molecules is n/2, and the temperature T and volume V (of the chamber) are the same. So, P’V = (n/2)RT. Dividing the second equation by the first gives (P’V)/(PV) = ((n/2)RT)/(nRT), which simplifies to P’/P = (n/2)/n = 1/2. Therefore, P’ = P/2.

The key principle is the ideal gas law (PV = nRT or PV = NkT, where N is the number of particles and k is Boltzmann’s constant). The pressure of an ideal gas depends only on the number of particles, volume, and temperature, not on the type of gas (as long as it behaves ideally). Since the number of particles is halved while temperature and volume are kept constant, the pressure must also be halved.

Avogadro’s hypothesis states that equal volumes of all ideal gases, at the same temperature and pressure, contain the same number of molecules. This question applies the ideal gas law inversely, showing that for a fixed volume and temperature, pressure is directly proportional to the number of moles or molecules.

23. The magnetic field strength of a current-carrying wire at a particular

The magnetic field strength of a current-carrying wire at a particular distance from the axis of the wire

[amp_mcq option1=”depends upon the current in the wire” option2=”depends upon the radius of the wire” option3=”depends upon the temperature of the surroundings” option4=”None of the above” correct=”option1″]

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UPSC NDA-2 – 2018

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The magnetic field strength of a current-carrying wire at a particular distance from the axis of the wire depends upon the current in the wire.

– The magnetic field strength (B) around a long straight current-carrying wire is given by the formula B = (μ₀ * I) / (2π * r).
– In this formula, I represents the current in the wire, and r represents the perpendicular distance from the axis of the wire.
– The magnetic field strength is directly proportional to the current (I).
– The magnetic field strength is inversely proportional to the distance (r).

μ₀ is the permeability of free space, a constant. The formula shows the direct dependence on current and distance, but not explicitly on the wire’s radius or surrounding temperature in the basic model for an external point. Temperature could indirectly affect the magnetic field by influencing the wire’s resistance and thus the current (if voltage is constant), but the fundamental relationship is B ∝ I for a given geometry.

24. The frequency of ultrasound waves is

The frequency of ultrasound waves is

[amp_mcq option1=”less than 20 Hz” option2=”between 20 Hz and 2 kHz” option3=”between 2 kHz and 20 kHz” option4=”greater than 20 kHz” correct=”option4″]

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The frequency of ultrasound waves is greater than 20 kHz.

– The audible range of sound frequencies for humans is typically considered to be from 20 Hz to 20,000 Hz (20 kHz).
– Sound waves with frequencies below 20 Hz are called infrasound.
– Sound waves with frequencies above 20 kHz are called ultrasound.

Ultrasound has various applications in medicine (imaging, therapy), industry (cleaning, welding), and technology (sonar, pest control). The exact upper limit of human hearing varies with age and individual.

25. Which one of the following is called ‘syngas’?

Which one of the following is called ‘syngas’?

[amp_mcq option1=”C(s) + H₂O(g)” option2=”CO(g) + H₂O(g)” option3=”CO(g) + H₂(g)” option4=”NO₂(g) + H₂(g)” correct=”option3″]

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‘Syngas’ is a mixture of carbon monoxide (CO) and hydrogen (H₂).

– Syngas, or synthesis gas, is a fuel gas consisting primarily of carbon monoxide (CO) and hydrogen (H₂).
– It is produced through processes like gasification of coal, biomass, or waste, or steam reforming of natural gas.
– Syngas is a crucial intermediate in the production of synthetic fuels, ammonia, methanol, and other chemicals.

Option A, C(s) + H₂O(g), represents a reaction (steam-carbon reaction or gasification) used to produce syngas. Option B, CO(g) + H₂O(g), represents reactants in the water-gas shift reaction, which can be used to adjust the CO:H₂ ratio in syngas. Option D is a mix of nitrogen dioxide and hydrogen, which is not syngas.

26. What is the formula mass of anhydrous sodium carbonate? (Given that th

What is the formula mass of anhydrous sodium carbonate? (Given that the atomic masses of sodium, carbon and oxygen are 23 u, 12 u and 16 u respectively)

[amp_mcq option1=”286 u” option2=”106 u” option3=”83 u” option4=”53 u” correct=”option2″]

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The formula mass of anhydrous sodium carbonate (Na₂CO₃) is 106 u.

– The chemical formula for anhydrous sodium carbonate is Na₂CO₃.
– The formula mass is the sum of the atomic masses of all atoms in the formula unit.
– Atomic masses given: Sodium (Na) = 23 u, Carbon (C) = 12 u, Oxygen (O) = 16 u.
– Formula mass of Na₂CO₃ = (2 × Atomic mass of Na) + (1 × Atomic mass of C) + (3 × Atomic mass of O)
– Formula mass = (2 × 23 u) + (1 × 12 u) + (3 × 16 u)
– Formula mass = 46 u + 12 u + 48 u = 106 u.

Anhydrous means without water. Sodium carbonate also exists as hydrated forms, like sodium carbonate decahydrate (washing soda, Na₂CO₃·10H₂O), which would have a significantly higher formula mass due to the inclusion of water molecules.

27. Which one of the following is a heterogeneous mixture?

Which one of the following is a heterogeneous mixture?

[amp_mcq option1=”Hydrochloric acid” option2=”Vinegar” option3=”Milk” option4=”Soda water” correct=”option3″]

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UPSC NDA-2 – 2018

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Milk is a heterogeneous mixture (specifically, a colloid/emulsion).

– A heterogeneous mixture has a non-uniform composition where different components can often be seen or separated.
– A homogeneous mixture (solution) has a uniform composition throughout.
– Hydrochloric acid (HCl in water), vinegar (acetic acid in water), and soda water (CO₂ in water) are all solutions, meaning the solute is dissolved at a molecular level, forming homogeneous mixtures.
– Milk is an emulsion of fat globules and proteins dispersed in water. While it may appear homogeneous, these particles are large enough to scatter light (Tyndall effect) and are not dissolved at a molecular level, making it heterogeneous.

Colloids like milk are considered heterogeneous mixtures because the dispersed phase (e.g., fat particles) is not uniformly distributed throughout the dispersion medium at the molecular level and can be observed or separated under certain conditions (e.g., centrifuging milk to separate cream).

28. A sample of oxygen contains two isotopes of oxygen with masses 16 u an

A sample of oxygen contains two isotopes of oxygen with masses 16 u and 18 u respectively. The proportion of these isotopes in the sample is 3 : 1. What will be the average atomic mass of oxygen in this sample?

[amp_mcq option1=”17.5 u” option2=”17 u” option3=”16 u” option4=”16.5 u” correct=”option4″]

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The average atomic mass of oxygen in this sample is 16.5 u.

– Average atomic mass is calculated as the weighted average of the masses of its isotopes based on their relative abundance.
– The sample contains isotopes with masses 16 u and 18 u in a proportion of 3:1.
– This means out of 4 parts, 3 parts have mass 16 u and 1 part has mass 18 u.
– Relative abundance of 16u isotope = 3 / (3+1) = 3/4 = 0.75
– Relative abundance of 18u isotope = 1 / (3+1) = 1/4 = 0.25
– Average atomic mass = (Mass₁ × Abundance₁) + (Mass₂ × Abundance₂)
– Average atomic mass = (16 u × 0.75) + (18 u × 0.25)
– Average atomic mass = 12 u + 4.5 u = 16.5 u.

Naturally occurring oxygen is predominantly isotope ¹⁶O (around 99.76%), with smaller amounts of ¹⁷O (around 0.038%) and ¹⁸O (around 0.205%). The standard average atomic mass of oxygen listed on the periodic table is approximately 15.999 u, reflecting the natural isotopic abundance. This question presents a hypothetical sample with a specific, non-natural isotopic ratio.

29. The atomic number of an element is 8. How many electrons will it gain

The atomic number of an element is 8. How many electrons will it gain to form a compound with sodium?

[amp_mcq option1=”One” option2=”Two” option3=”Three” option4=”Four” correct=”option2″]

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The element with atomic number 8 (Oxygen) will gain two electrons to form a compound with sodium.

– The element with atomic number 8 is Oxygen (O). Its electronic configuration is 2, 6.
– To achieve a stable octet configuration in its outermost shell, Oxygen needs to gain 8 – 6 = 2 electrons, forming the O²⁻ ion.
– Sodium (Na, atomic number 11) has electronic configuration 2, 8, 1 and tends to lose 1 electron to form the Na⁺ ion.
– In the formation of sodium oxide (Na₂O), two sodium atoms each lose one electron, and one oxygen atom gains two electrons.

Elements form chemical bonds (ionic or covalent) to achieve a stable electron configuration, usually a full valence shell (like noble gases). Oxygen is a non-metal in Group 16 and typically forms -2 ions by gaining two electrons. Sodium is an alkali metal in Group 1 and typically forms +1 ions by losing one electron.

30. Which one of the following is a waterborne disease?

Which one of the following is a waterborne disease?

[amp_mcq option1=”Jaundice” option2=”Tuberculosis” option3=”Rabies” option4=”Arthritis” correct=”option1″]

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Jaundice, when caused by Hepatitis A or Hepatitis E, is a symptom related to waterborne transmission.

– Waterborne diseases are caused by pathogens transmitted through contaminated water.
– Hepatitis A and Hepatitis E viruses, which cause liver inflammation leading to jaundice (yellowing of skin/eyes), are typically spread through the faecal-oral route, often via contaminated water or food.
– Tuberculosis is airborne.
– Rabies is spread through animal bites.
– Arthritis is joint inflammation with various causes, not typically infectious or waterborne.

While jaundice is a symptom and not a disease itself, among the given options, it is the only condition directly linked to waterborne pathogens (specifically, the viruses causing Hepatitis A and E). Therefore, in the context of potential UPSC-style options, Jaundice (as a manifestation of waterborne Hepatitis) is the intended answer.