UPSC CDS-1

31. Which one of the following is the correct sequence of passage of light

Which one of the following is the correct sequence of passage of light in a compound microscope?

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Condenser—Objective lens—Eyepiece—Body tube
Objective lens—Condenser—Body tube—Eyepiece
Condenser—Objective lens—Body tube—Eyepiece
Eyepiece—Objective lens—Body tube—Mirror
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The correct answer is C. In a compound microscope, light typically passes through the following sequence of components: The light source (often directed by a mirror or built-in lamp) illuminates the specimen, usually passing through a condenser which focuses the light onto the specimen. The light then passes through the specimen, enters the objective lens (which forms a magnified real image), travels up the body tube, and finally passes through the eyepiece (which magnifies the real image to form a virtual image viewed by the observer’s eye).
The correct sequence for light passage in a compound microscope is typically Condenser -> Specimen -> Objective lens -> Body tube -> Eyepiece.
The condenser is located below the stage and its function is to focus the light onto the specimen to provide optimal illumination. The objective lens is located above the specimen and is the primary magnification component. The body tube connects the objective lens to the eyepiece, and the eyepiece is where the user looks to view the final magnified image.

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32. Which one of the following statements regarding haemoglobin is

Which one of the following statements regarding haemoglobin is correct?

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Haemoglobin present in RBC can carry only oxygen but not carbon dioxide.
Haemoglobin of RBC can carry both oxygen and carbon dioxide.
Haemoglobin of RBC can carry only carbon dioxide.
Haemoglobin is only used for blood clotting and not for carrying gases.
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The correct answer is B. Haemoglobin is the protein molecule in red blood cells responsible for transporting oxygen from the lungs to the body’s tissues. While its primary function is oxygen transport, haemoglobin also binds to and transports a small percentage (about 20-25%) of carbon dioxide from the tissues back to the lungs. The majority of carbon dioxide is transported in the blood plasma as bicarbonate ions.
Haemoglobin in RBCs can bind and transport both oxygen and a portion of carbon dioxide.
Oxygen binds to the heme group (iron-containing part) of haemoglobin, forming oxyhaemoglobin. Carbon dioxide binds to the amino groups of the globin protein part of haemoglobin, forming carbaminohemoglobin. The binding of oxygen and carbon dioxide are influenced by factors such as pH, temperature, and the concentration of other molecules like 2,3-bisphosphoglycerate (2,3-BPG).

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33. Which one of the following statements regarding histone proteins is

Which one of the following statements regarding histone proteins is correct?

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Histones are proteins that are present in mitochondrial membrane.
Histones are proteins that are present in nucleus in association with DNA.
Histones are proteins associated with lipids in the cytosol.
Histones are proteins associated with carbohydrates in the cytosol.
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The correct answer is B. Histones are a family of small, positively charged proteins found in the nucleus of eukaryotic cells. They play a crucial role in packaging DNA into structural units called nucleosomes, which are the basic building blocks of chromatin. DNA (which is negatively charged) wraps around histone octamers (eight histone proteins) to form these structures.
Histone proteins are primarily located in the nucleus where they bind to DNA to help package it into chromatin.
There are five main families of histones: H1, H2A, H2B, H3, and H4. H2A, H2B, H3, and H4 form the core particle of the nucleosome, while H1 acts as a linker histone, helping to further compact the chromatin structure. Histone modifications play a significant role in regulating gene expression.

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34. The Hooke’s law is valid for

The Hooke’s law is valid for

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only proportional region of the stress-strain curve
entire stress-strain curve
entire elastic region of the stress-strain curve
elastic as well as plastic region of the stress-strain curve
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The correct answer is A. Hooke’s law states that the stress is directly proportional to the strain within the elastic limit of a material. More specifically, this linear relationship holds true only in the initial part of the elastic region, which is known as the proportional region. Beyond the proportional limit, the material may still behave elastically (return to its original shape upon unloading), but the stress-strain relationship becomes non-linear.
Hooke’s law (Stress ∝ Strain) is strictly valid only in the proportional region of the stress-strain curve.
The stress-strain curve for a ductile material typically shows several regions: proportional limit, elastic limit, yield point, ultimate tensile strength, and fracture point. The elastic limit is the maximum stress a material can withstand without permanent deformation. The proportional limit is the point up to which stress is directly proportional to strain. For many materials, these two points are very close, but Hooke’s law is defined by the proportionality.

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35. What is a constellation?

What is a constellation?

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A particular pattern of equidistant stars from the earth in the sky
A particular pattern of stars that may not be equidistant from the earth in the sky
A particular pattern of planets of our solar system in the sky
A particular pattern of stars, planets and satellites in the sky due to their position in the space
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The correct answer is B. A constellation is defined as a group of stars that appear to form a pattern in the night sky as seen from Earth. The stars within a constellation are typically at vastly different distances from Earth and are not physically related to each other; their apparent pattern is merely a line-of-sight effect from our perspective.
A constellation is an apparent pattern of stars in the sky, where the constituent stars are generally not equidistant from Earth.
The International Astronomical Union (IAU) recognizes 88 official constellations covering the entire celestial sphere. Historically, constellations were used for navigation, timekeeping, and mythology. The stars in a constellation are connected by imaginary lines to form shapes that are often named after animals, mythological figures, or objects.

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36. For an ideal gas, which one of the following statements does not hol

For an ideal gas, which one of the following statements does not hold true?

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The speed of all gas molecules is same.
The kinetic energies of all gas molecules are not same.
The potential energy of the gas molecules is zero.
There is no interactive force between the molecules.
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An ideal gas is a theoretical model with specific assumptions. Let’s evaluate each statement:
A) The speed of all gas molecules is same. This is false for an ideal gas. At any given temperature, the speeds of the molecules in an ideal gas follow a distribution (like the Maxwell-Boltzmann distribution), meaning molecules have a wide range of speeds.
B) The kinetic energies of all gas molecules are not same. This is true. Since the speeds are not the same (from point A), and kinetic energy is proportional to speed squared ($KE = 1/2 mv^2$), the kinetic energies of individual molecules are also not the same. The average kinetic energy, however, is directly proportional to the absolute temperature.
C) The potential energy of the gas molecules is zero. This is true for an ideal gas. A key assumption of the ideal gas model is that there are no intermolecular forces between the molecules. Potential energy due to interparticle interactions is therefore considered zero.
D) There is no interactive force between the molecules. This is true. This is another fundamental assumption of the ideal gas model, simplifying calculations by ignoring attractions and repulsions.
The question asks which statement does *not* hold true for an ideal gas. Statement A is the one that is false for an ideal gas.
– Ideal gas molecules have a distribution of speeds and kinetic energies.
– Ideal gas molecules have no intermolecular forces.
– Potential energy due to intermolecular forces is zero in an ideal gas.
Real gases deviate from ideal gas behavior, especially at high pressures and low temperatures, where intermolecular forces and the volume of the molecules themselves become significant. The ideal gas law (PV=nRT) is derived based on these ideal gas assumptions.

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37. In spherical polar coordinates (γ, θ, α), θ denotes the polar angle ar

In spherical polar coordinates (γ, θ, α), θ denotes the polar angle around z-axis and α denotes the azimuthal angle raised from x-axis. Then the y-component of P⃗ is given by

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Psinθsinα
Psinθcosα
Pcosθsinα
Pcosθcosα
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In spherical polar coordinates (γ, θ, α), where γ is the magnitude of the vector P⃗ (let’s denote it as P), θ is the polar angle from the positive z-axis, and α is the azimuthal angle from the positive x-axis in the xy-plane, the Cartesian components (Px, Py, Pz) of the vector P⃗ are given by:
$P_x = P \sin\theta \cos\alpha$
$P_y = P \sin\theta \sin\alpha$
$P_z = P \cos\theta$
The question asks for the y-component of P⃗. According to the standard conversion from spherical to Cartesian coordinates, the y-component is $P\sin\theta\sin\alpha$.
– Spherical coordinates typically use (r, θ, φ) or (ρ, θ, φ). The question uses (γ, θ, α) with meanings specified.
– γ (or P) is the magnitude.
– θ is the angle from the z-axis (polar angle).
– α is the angle from the x-axis in the xy-plane (azimuthal angle).
– The projection onto the xy-plane has length $P\sin\theta$.
– This projection is resolved into x and y components using the azimuthal angle α.
The formulas for converting spherical coordinates (P, θ, α) to Cartesian coordinates (Px, Py, Pz) are derived from trigonometry. The projection of the vector onto the z-axis is $P\cos\theta$, giving the z-component. The projection onto the xy-plane has length $P\sin\theta$. This projection forms a right triangle in the xy-plane with the x and y axes, where the hypotenuse is $P\sin\theta$ and the angle with the x-axis is α. The x-component is $(P\sin\theta)\cos\alpha$ and the y-component is $(P\sin\theta)\sin\alpha$.

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38. The PCl$_{5}$ molecule has trigonal bipyramidal structure. Therefore,

The PCl$_{5}$ molecule has trigonal bipyramidal structure. Therefore, the hybridization of p orbitals should be

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sp$^{2}$
sp$^{3}$
dsp$^{2}$
dsp$^{3}$
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The PCl$_{5}$ molecule has a trigonal bipyramidal structure with the Phosphorus atom as the central atom. In PCl$_{5}$, the central Phosphorus atom forms five single bonds with the five Chlorine atoms and has no lone pairs. According to VSEPR theory, a central atom with five electron domains (five bonding pairs) has a trigonal bipyramidal electron geometry and molecular geometry. This geometry is associated with sp$^3$d hybridization of the central atom. The sp$^3$d hybridization involves mixing one s atomic orbital, three p atomic orbitals, and one d atomic orbital to form five hybrid orbitals. The option dsp$^3$ refers to the same type of hybridization, where one d, one s, and three p orbitals are mixed, resulting in five hybrid orbitals oriented in a trigonal bipyramidal arrangement. Among the given options, dsp$^3$ is the only hybridization scheme that corresponds to the trigonal bipyramidal structure of PCl$_{5}$.
– PCl$_{5}$ has a trigonal bipyramidal structure.
– This structure arises from the hybridization of the central atom’s valence orbitals.
– A trigonal bipyramidal geometry corresponds to sp$^3$d (or dsp$^3$) hybridization.
– The central P atom has 5 bonding pairs and 0 lone pairs.
The common notation for main group elements with 5 coordination is sp$^3$d hybridization, involving one s, three p, and one d orbital. The dsp$^3$ notation, while also representing a mix of one d, one s, and three p orbitals, is sometimes used depending on whether the d-orbital is an inner or outer orbital, or simply as an alternative way to list the contributing orbitals. In the context of the given options, dsp$^3$ is the correct choice corresponding to the PCl$_{5}$ structure.

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39. For a certain reaction, $\Delta G^0 = -45$ kJ/mol and $\Delta H^0 = -9

For a certain reaction, $\Delta G^0 = -45$ kJ/mol and $\Delta H^0 = -90$ kJ/mol at 0 °C. What is the minimum temperature at which the reaction will become spontaneous, assuming that $\Delta H^0$ and $\Delta S^0$ are independent of temperature?

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273 K
298 K
546 K
596 K
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The spontaneity of a reaction is determined by the change in Gibbs Free Energy, $\Delta G$. A reaction is spontaneous if $\Delta G < 0$. The relationship between $\Delta G$, $\Delta H$, and $\Delta S$ is given by $\Delta G = \Delta H - T\Delta S$. We are given $\Delta G^0 = -45$ kJ/mol and $\Delta H^0 = -90$ kJ/mol at 0 °C (273 K). Assuming $\Delta H^0$ and $\Delta S^0$ are independent of temperature, we can calculate $\Delta S^0$ at 273 K: $\Delta G^0_{273} = \Delta H^0 - (273 \text{ K})\Delta S^0$ $-45 \text{ kJ/mol} = -90 \text{ kJ/mol} - 273\Delta S^0$ $273\Delta S^0 = -90 + 45 = -45 \text{ kJ/mol}$ $\Delta S^0 = \frac{-45}{273} \text{ kJ/(mol·K)}$ The reaction is spontaneous when $\Delta G < 0$: $\Delta H^0 - T\Delta S^0 < 0$ $-90 \text{ kJ/mol} - T \left(\frac{-45}{273} \text{ kJ/(mol·K)}\right) < 0$ $-90 + T \left(\frac{45}{273}\right) < 0$ $T \left(\frac{45}{273}\right) < 90$ $T < 90 \times \frac{273}{45}$ $T < 2 \times 273$ $T < 546 \text{ K}$ So, the reaction is spontaneous at temperatures below 546 K. The temperature at which $\Delta G$ becomes zero (equilibrium) is $T = 546$ K. The question asks for the "minimum temperature at which the reaction will become spontaneous". While the phrasing is awkward for a reaction that is spontaneous below a certain temperature, it likely refers to the boundary temperature (546 K) where spontaneity starts or ends depending on the temperature direction, or potentially the lowest temperature among options where it is spontaneous. However, the calculation directly yields 546 K as the key temperature threshold.
– Spontaneity requires $\Delta G < 0$. - $\Delta G = \Delta H - T\Delta S$. - Calculate $\Delta S$ from the given data at 273 K. - Find the temperature range where $\Delta G < 0$. - The transition temperature where $\Delta G = 0$ is $T_{eq} = \Delta H / \Delta S$.
For reactions with $\Delta H < 0$ and $\Delta S < 0$, the reaction is spontaneous at low temperatures ($T < T_{eq}$) and non-spontaneous at high temperatures ($T > T_{eq}$). The boundary temperature is $T_{eq}$.

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40. Which of the following is/are state function/functions? 1. q+w 2.

Which of the following is/are state function/functions?

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  • 1. q+w
  • 2. q
  • 3. w
  • 4. H-TS

Select the correct answer using the code given below.

1 and 4 only
1, 2 and 4
2, 3 and 4
1 only
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A state function is a property whose value depends only on the state of the system, not on the path taken to reach that state.
1. q+w: According to the first law of thermodynamics, $\Delta U = q+w$, where U is internal energy. Internal energy (U) is a state function, so the change in internal energy ($\Delta U$) is also a state function. Thus, q+w represents $\Delta U$ and is a state function.
2. q: Heat (q) is a path-dependent quantity; the amount of heat transferred depends on the process followed.
3. w: Work (w) is a path-dependent quantity; the amount of work done depends on the process followed.
4. H-TS: This expression is the definition of Gibbs Free Energy (G). Gibbs Free Energy (G) is a state function, as it is defined in terms of state functions (Enthalpy H, Temperature T, and Entropy S).
Therefore, q+w and H-TS are state functions.
– State functions are independent of the path.
– Internal energy ($\Delta U = q+w$) is a state function.
– Gibbs Free Energy ($G = H-TS$) is a state function.
– Heat (q) and Work (w) are path functions.
Examples of other state functions include pressure (P), volume (V), temperature (T), enthalpy (H), entropy (S), and internal energy (U). Examples of path functions include heat (q) and work (w).

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