Question 1

SOAPS & DETERGENTS — Process understanding.  Question: During soap manufacture (saponification), how does adding common salt (NaCl) help in separating soap from the aqueous mixture?

Accepted Answer

Introduction / Context: In the saponification process, triglycerides react with alkali to form soap (sodium or potassium salts of fatty acids) and glycerol. After the reaction, soap must be separated from the mother liquor. Traditional processes add common salt to “salt out” the soap. Understanding the physical chemistry behind this separation is key to mastering industrial soapmaking. Given Data / Assumptions: Saponification yields a mixture of soap, glycerol, unreacted alkali, and water. Common salt (NaCl) is added to this mixture. We want the mechanism by which separation occurs. Concept / Approach: Adding electrolytes like NaCl reduces the solubility of soap in water (a “salting out” effect). The increased ionic strength favors the precipitation (or separation as curd) of the less soluble soap phase while glycerol and salt remain in the aqueous layer. This is not primarily a density manipulation; rather it is a solubility and activity effect driven by electrolyte interactions. Step-by-Step Solution: Recognize that soap is an amphiphilic salt; its solubility depends on ionic strength.Add NaCl → ionic strength increases.Higher ionic strength reduces soap solubility → soap separates as a distinct phase/precipitate.Glycerol remains in the brine, enabling decantation and recovery. Verification / Alternative check: Classical batch soapmaking demonstrates visible “curd” formation after salting, confirming the salting-out phenomenon common to many colloids and proteins as well. Why Other Options Are Wrong: A, C) Density change is not the operative separation mechanism.D) Increasing solubility would hinder, not aid, separation.E) Evaporation is not the purpose of adding salt. Common Pitfalls: Confusing salting out with flotation or with merely thickening the mixture. The key is reduced solubility due to electrolyte effects. Final Answer: By decreasing the solubility of the soap (salting out)

Question 2

HOUSEHOLD CHEMISTRY — Composition of cleaning agents.  Question: Modern detergents used for cleaning clothes and utensils commonly contain which class of active surfactants?

Accepted Answer

Introduction / Context: Detergents clean by lowering surface tension and forming micelles that trap oily dirt. The key component is the surfactant molecule, with a hydrophobic tail and a hydrophilic headgroup. Recognizing the chemical family used in most household detergents helps explain their effectiveness and compatibility with hard water. Given Data / Assumptions: We focus on the principal active surfactant in typical laundry and dishwashing detergents. Common examples include linear alkylbenzene sulphonates (LAS). Other listed ions (bicarbonate, nitrate, silicate) may appear as builders or additives but are not the core surfactant class. Concept / Approach: Anionic sulphonate surfactants (–SO₃⁻ group) such as sodium dodecylbenzene sulphonate dominate due to good detergency and foaming. They perform well in a range of water conditions and are relatively economical. Bicarbonates and silicates may assist as builders or pH buffers, but they do not provide the primary micelle-forming action. Heavy-metal anions like bismuthates are unrelated to household detergency. Step-by-Step Solution: Identify the functional role needed: surface-active cleaning → surfactant.Match the family widely used in consumer detergents → sulphonate-based anionic surfactants.Eliminate ions used for other roles (buffers/builders) or not used at all in such products. Verification / Alternative check: Product ingredient lists and safety data sheets typically list “linear alkylbenzene sulfonate” or “sodium laureth sulfate,” both sulphonate/sulfate anionic surfactants. Why Other Options Are Wrong: A) Bicarbonates can act as mild alkalis/builders but are not the primary detergents.B) Bismuthates are not used in consumer detergents.D) Nitrates are oxidizing salts/fertilizer components, not detergent actives.E) Silicates can be builders/anti-corrosion agents; not the main surfactants. Common Pitfalls: Confusing builders and fillers with the actual surfactant. The cleaning action fundamentally depends on the sulphonate/sulfate headgroup amphiphiles. Final Answer: sulphonates (anionic surfactants)

Question 3

CEMENT TECHNOLOGY — Role of gypsum.  Question: In Portland cement manufacture, gypsum is interground with clinker primarily to:

Accepted Answer

Introduction / Context: Portland cement sets through hydration reactions of clinker phases (notably tricalcium silicate and tricalcium aluminate). Controlling the early setting time is essential for handling, placing, and finishing concrete. Gypsum plays a specific chemical role in moderating the very fast reaction of aluminate phases. Given Data / Assumptions: Gypsum (calcium sulfate dihydrate) is added in a few percent by mass. Objective: regulate initial set, not long-term strength directly. We consider early hydration chemistry, especially of C₃A. Concept / Approach: Without gypsum, the highly reactive C₃A phase hydrates rapidly, causing “flash set.” Sulfate ions from gypsum combine with C₃A to form ettringite initially, which slows down aluminate hydration, providing workable time. Later, silicate hydration dominates strength development; gypsum’s main effect is on setting regulation rather than tensile strength per se. Step-by-Step Solution: Recognize the problem: C₃A reacts too quickly → flash set.Add gypsum → provides sulfate → forms ettringite with C₃A.Ettringite formation moderates the reaction rate → workable setting time.Therefore, gypsum acts primarily as a set retarder. Verification / Alternative check: Standards limit SO₃ content to balance workability and avoid excessive sulfate, confirming gypsum’s regulatory role. Why Other Options Are Wrong: A) Tensile strength results from hydration products over time, not from gypsum addition directly.C) “Colloidal gel” is not the targeted mechanism here.D) Gypsum does not act as a binder of calcium silicate particles.E) Color effects are incidental; the function is chemical control of set. Common Pitfalls: Confusing setting control with strength gain. Gypsum prevents flash set; it is not a plasticizer or strengthener in itself. Final Answer: decrease the rate of setting of cement (act as a retarder)

Question 4

INDUSTRIAL MATERIALS — Source identification.  Question: Wood is the main raw material for the manufacture of which of the following products at industrial scale?

Accepted Answer

Introduction / Context: Many everyday materials trace back to natural resources. Understanding these value chains helps in environmental awareness and industry basics. Wood, specifically from sustainably managed forests, is a critical feedstock for one major product used globally: paper. Given Data / Assumptions: Wood comprises cellulose, hemicellulose, and lignin. Pulping processes free cellulose fibers for papermaking. Other listed products use different primary feedstocks. Concept / Approach: Pulp and paper mills convert logs or wood chips into pulp via mechanical, chemical (kraft/sulfite), or semi-chemical processes. The cellulose fibers are then processed into a vast range of paper and board grades. Paints rely primarily on resins, pigments, and solvents; inks on pigments/dyes and vehicles; traditional gunpowder uses charcoal (which originates from wood) plus sulfur and potassium nitrate, but the principal bulk raw material for papermaking is unambiguously wood. Step-by-Step Solution: Identify the industry that consumes vast quantities of wood → pulp & paper.Recall that paper is formed from cellulose fibers derived from wood pulp.Eliminate alternatives that depend on other primary inputs (resins, minerals, petrochemicals). Verification / Alternative check: Industry statistics show wood pulp as the dominant raw material in paper manufacture, with increasing recycled fiber content augmenting but not replacing virgin wood in many grades. Why Other Options Are Wrong: A) Paint uses binders and pigments; wood is not its main raw material.C) Ink compositions center on vehicles and pigments, not wood.D) Gunpowder includes charcoal (from wood) but also sulfur and nitrate; the question asks for the main industrial product with wood as the core bulk feedstock, which is paper.E) Glass is made from silica, soda ash, and limestone. Common Pitfalls: Equating “contains some wood derivative” with “main raw material.” Only paper manufacturing directly depends on large-scale wood fiber. Final Answer: paper

Question 5

Which type of electromagnetic waves are used in night vision devices and night vision apparatus?

Accepted Answer

Introduction / Context: This question belongs to basic physics and applications of the electromagnetic spectrum. Night vision devices are widely used in military operations, surveillance, wildlife observation, and rescue work. They allow us to see in darkness by detecting radiation that is not visible to the human eye. Understanding which region of the spectrum is used in such devices is a common concept tested in school level physics and competitive exams. Given Data / Assumptions: The question asks which type of waves are used in night vision apparatus. Options list radio waves, microwaves, infra red waves, ultraviolet waves, and None of the above. We assume the question refers to standard passive night vision and thermal imaging devices that detect heat signatures. Concept / Approach: All objects with a temperature above absolute zero emit electromagnetic radiation, a significant part of which for everyday temperatures falls in the infra red region. Night vision devices either amplify low levels of visible light or detect infra red radiation emitted by warm objects. Thermal imagers in particular are based on infra red detection, converting these waves into electronic signals and then into a visible image. Radio waves and microwaves are used for communication and radar, while ultraviolet waves are higher energy waves that do not form the basis of common night vision goggles. Step-by-Step Solution: Step 1: Recall that the human eye responds to visible light only, roughly from 400 nm to 700 nm in wavelength. Step 2: Understand that in darkness, there is very little visible light, but warm bodies (humans, animals, vehicles) still emit infra red radiation. Step 3: Night vision and thermal imaging devices are designed with sensors that detect infra red waves instead of or in addition to visible light. Step 4: Radio waves and microwaves are used for wireless communication, broadcasting, and radar, not directly for forming images of warm bodies in standard night vision goggles. Step 5: Ultraviolet radiation is higher energy and is typically blocked by the atmosphere and most glasses, so it is not used in conventional night vision apparatus. Therefore, infra red waves are the correct answer. Verification / Alternative check: A quick way to verify is to think about the term thermal imaging. Thermal implies heat, and heat radiation in physics is strongly associated with infra red emission. Many devices are specifically named infra red cameras or IR goggles. In contrast, there is no common phrase like radio wave night vision or microwave night goggles in basic physics textbooks. This terminology and the connection to body heat confirm that infra red waves are the correct type. Why Other Options Are Wrong: Radio waves: Used for broadcasting and communication, they have wavelengths much longer than needed for thermal imaging. Microwaves: Used in microwave ovens, radar, and communication, not in standard night vision goggles. Ultraviolet waves: High energy radiation used in sterilisation and special lamps, not for ordinary night vision. None of the above: Incorrect because infra red waves, listed in the options, are indeed used in night vision apparatus. Common Pitfalls: Students may confuse radar based detection, which uses microwaves or radio waves, with optical night vision, leading to a wrong choice. Another mistake is guessing ultraviolet because it is also invisible, without considering that the devices are detecting heat, not high energy radiation. To avoid such errors, always link night vision and thermal imaging with infra red, the region most strongly associated with heat radiation in physics. Final Answer: Night vision devices commonly use infra red waves to detect objects in darkness.

Question 6

The blue colour of a clear daytime sky is mainly due to which phenomenon of light?

Accepted Answer

Introduction / Context: The colour of the sky is a classic question in basic physics and general science. Understanding why the clear sky appears blue involves the interaction of sunlight with the Earth's atmosphere. This concept links the nature of white light, atmospheric particles, and the scattering of different wavelengths. Competitive exams frequently ask which phenomenon is responsible for the blue colour of the sky, making this a must know topic for students. Given Data / Assumptions: The question asks why the clear sky appears blue. Options include scattering, diffraction, dispersion, refraction, and None of the above. We assume normal daytime conditions and a clear atmosphere, not sunrise or sunset where colours appear different. Concept / Approach: Sunlight is white light composed of different colours with different wavelengths. As this light passes through the Earth's atmosphere, tiny gas molecules and small particles scatter the shorter wavelength components (blue and violet) more strongly than longer wavelengths (red, orange). This wavelength dependent scattering is called Rayleigh scattering. Because our eyes are more sensitive to blue than violet, and some violet is absorbed, the sky appears predominantly blue to us. Thus, scattering of light is the main phenomenon involved. Step-by-Step Solution: Step 1: Recognise that sunlight is made of multiple colours, each with its own wavelength. Step 2: Understand that when this light enters the atmosphere, it interacts with gas molecules much smaller than the wavelength of visible light. Step 3: In Rayleigh scattering, shorter wavelengths (blue and violet) are scattered much more strongly than longer wavelengths (red and yellow). Step 4: Due to this preferential scattering, light coming from all directions in the sky is rich in blue components, making the sky look blue. Step 5: Therefore, the correct phenomenon is scattering of light, not diffraction, dispersion, or simple refraction. Verification / Alternative check: To verify, think of related phenomena: the reddish colour of the sun at sunrise and sunset is also explained by scattering, where blue light has been scattered away leaving more red light in the direct line of sight. Dispersion explains how white light splits into a spectrum through a prism. Diffraction describes bending of waves around obstacles, and refraction deals with change of direction when light passes between media. Only scattering consistently explains both the blue sky and red sunsets, reinforcing that scattering is the correct choice. Why Other Options Are Wrong: Diffraction of light: Involves bending and spreading of light around edges but does not directly cause the sky's blue colour. Dispersion of light: Occurs in prisms and raindrops to form spectra and rainbows, not the everyday blue sky. Refraction of light: Responsible for bending of light in lenses and mirages, not for the overall sky colour. None of the above: Incorrect because scattering of light, listed as an option, accurately describes the phenomenon. Common Pitfalls: Students sometimes confuse dispersion and scattering because both involve different colours of light. Another common error is to pick refraction simply because it is a frequently mentioned phenomenon in optics. To avoid such mistakes, remember that dispersion is about splitting light through a medium like a prism, while scattering is about light being redirected in many directions by small particles. The blue colour of the sky is a textbook example of scattering, specifically Rayleigh scattering. Final Answer: The blue colour of the clear daytime sky is mainly due to scattering of light in the Earth's atmosphere.

Question 7

As you go down into a deep well from the Earth's surface, how does your weight change?

Accepted Answer

Introduction / Context: This question tests understanding of how gravitational acceleration varies with depth inside the Earth. Many students know that weight is the force with which the Earth attracts a body, but they often assume it is constant near the surface. In reality, gravitational acceleration changes slightly with height above the surface and with depth below it. The question asks specifically what happens when you go down into a well, which is a conceptual application of the formula for gravity inside the Earth. Given Data / Assumptions: We are considering going down into a well, i.e., moving slightly below the Earth's surface. The depth of the well is very small compared to the Earth's radius. Weight is defined as W = m * g, where g is the local acceleration due to gravity. The Earth is assumed approximately spherical and of uniform density for the basic concept. Concept / Approach: Outside or on the surface of the Earth, g is approximately constant. Inside a uniformly dense sphere, gravitational acceleration decreases linearly with distance from the centre. As you move from the surface into the interior, the distance from the centre decreases, and only the mass enclosed within that radius contributes to gravity. Thus, g becomes smaller inside the Earth. Since weight is proportional to g, going down into a well results in a slight decrease in weight, not an increase. Step-by-Step Solution: Step 1: Recall that weight W = m * g, where m is constant and g depends on position. Step 2: At the Earth's surface, g has its familiar value of about 9.8 m/s^2. Step 3: Inside the Earth, for a radius r less than the Earth's radius R, the effective mass producing gravity decreases, so g(r) is less than g at the surface. Step 4: When you go down a small depth h in a well, your distance from the centre becomes R - h, and the local g decreases slightly. Step 5: Because W = m * g and g is now slightly smaller, your weight decreases slightly as you go down the well. Verification / Alternative check: We can compare this with what happens when you go up a mountain. At higher altitudes, your distance from the Earth's centre increases, and g also decreases slightly, so your weight becomes a bit less. Inside the Earth, a similar reduction in g occurs, but for a different physical reason (less enclosed mass). Both situations lead to a reduction in weight, though the change is very small for moderate heights or depths. This consistency of reasoning supports the conclusion that your weight decreases slightly when you go down into a well. Why Other Options Are Wrong: It increases slightly: Incorrect, because moving towards the centre inside the Earth reduces the effective gravitational pull. It remains exactly the same: Not true; while the change is small, it is not zero according to the theory of gravity inside a sphere. None of the above: Incorrect because one of the given options correctly describes the behaviour. It first increases then decreases: This pattern is not valid for small depths in a well; g inside a uniform Earth does not increase below the surface. Common Pitfalls: Many students think that moving closer to the Earth's centre must always increase gravity, forgetting that inside a spherical shell the outer layers do not contribute to the gravitational pull on an interior point. Another common misconception is that g is completely constant near the surface, leading to the answer that weight remains exactly the same. Remembering that only the mass enclosed within a given radius contributes to gravity inside the Earth helps avoid such mistakes. Final Answer: As you go down into a deep well, your weight decreases slightly because the local gravitational acceleration becomes a little smaller inside the Earth.

Question 8

At which location will the weight of an object be minimum?

Accepted Answer

Introduction / Context: This question explores how gravitational force varies with position relative to the Earth. It is common in basic physics and general science to ask where on or inside the Earth a body would weigh the most or the least. The options include points on the surface such as the poles and the equator, as well as the Earth's centre. Knowing the correct location where weight is minimum requires understanding both gravity inside a sphere and the effects of rotation and shape. Given Data / Assumptions: We consider weight W = m * g, where m is constant and g varies with position. Locations given: North Pole, South Pole, Equator, centre of the Earth, and a high mountain peak. The Earth is approximated as a sphere with uniform or smoothly varying density. Rotational effects act mainly at the surface, especially at the equator. Concept / Approach: On the Earth's surface, g is not exactly the same everywhere. It is slightly higher at the poles and slightly lower at the equator due to the Earth's rotation and its equatorial bulge. As you go above the surface, g decreases with height. Inside a uniform sphere, g decreases linearly with decreasing radius from the centre. At the very centre of the Earth, the gravitational pulls from all directions cancel, making net g equal to zero. Since weight is m * g, the minimum possible weight (actually zero) occurs at the centre of the Earth. Step-by-Step Solution: Step 1: Recognise that weight depends directly on the local value of g. Step 2: At the poles, g is slightly larger than at the equator, so weight is slightly higher, not lower. Step 3: At the equator, rotation reduces effective g a little, so weight is less than at the poles but still not zero. Step 4: On high mountain peaks, the distance from the Earth's centre is slightly larger, so g decreases a bit, but again not to zero. Step 5: At the centre of the Earth, gravitational forces from all directions balance and g becomes zero, making weight W = m * 0 = 0, which is the minimum possible value. Verification / Alternative check: We can compare the magnitude of g at various locations. Standard values show g ≈ 9.83 m/s^2 at the poles and ≈ 9.78 m/s^2 at the equator. On high mountains it is slightly less than at sea level but still close to 9.7–9.8 m/s^2. The theory of gravity inside a sphere states that g is proportional to r (distance from the centre), and at r = 0, g = 0. This mathematical relationship confirms that the centre of the Earth is the unique location where weight becomes zero and hence is minimum. Why Other Options Are Wrong: The North Pole: g is actually slightly higher here than at the equator, so weight is not minimum. The South Pole: Same reasoning as the North Pole; weight is slightly higher than at the equator. The Equator: Weight is lower than at the poles but still not the minimum possible because g is not zero. On a high mountain peak: g decreases with height but does not reach zero on any ordinary mountain, so weight is reduced but not minimum. Common Pitfalls: The most common mistake is to choose the equator, because students remember that g is smallest there on the surface due to centrifugal effects. However, the question is about the absolute minimum weight anywhere, not just among surface points. Another pitfall is to ignore the possibility of going inside the Earth altogether. Remembering that g becomes zero at the centre, giving zero weight, is crucial to arriving at the correct answer. Final Answer: The weight of an object is minimum (actually zero) at the center of the Earth.

Question 9

In car safety airbag systems, which chemical compound is commonly used inside the airbag to rapidly produce gas on impact?

Accepted Answer

Introduction / Context: Modern cars are equipped with airbag systems designed to protect passengers during collisions. These airbags must inflate within fractions of a second when an impact is detected. To achieve this, a chemical compound inside the airbag rapidly decomposes to produce a large volume of gas. General science questions often ask which compound is used in airbags because it connects basic chemistry with an everyday safety device. Given Data / Assumptions: The question asks which compound is used in car airbags to generate gas. Options include sodium bicarbonate, sodium azide, sodium nitrite, sodium peroxide, and None of the above. We assume traditional gas generating airbag designs that rely on rapid decomposition of a solid chemical. Concept / Approach: Sodium azide (NaN3) has historically been used in many airbag systems. When triggered by a sensor in a collision, an igniter heats the sodium azide, causing it to decompose quickly into sodium metal and nitrogen gas. The large volume of nitrogen fills the airbag almost instantly. While newer systems may use alternative compounds for safety and environmental reasons, sodium azide is the standard answer in exam oriented general science questions on airbags. Step-by-Step Solution: Step 1: Recognise that the key requirement for an airbag is very rapid gas generation. Step 2: Recall that sodium azide decomposes explosively when heated, releasing nitrogen gas. Step 3: The reaction produces mostly nitrogen, which is inert and safe to use as a filling gas in the bag. Step 4: Compare other compounds: sodium bicarbonate is common in baking powder and fire extinguishers, not in airbags; sodium nitrite and peroxide have different uses and reaction behaviours. Step 5: Conclude that sodium azide is the compound associated with traditional airbag systems. Verification / Alternative check: Standard general science references and safety engineering texts describe early and widely used airbags as sodium azide based. The rapid decomposition of NaN3 to N2 gas is well documented. Although environmental concerns have led to research on azide free systems, exam questions continue to refer to sodium azide as the canonical example. None of the other sodium salts listed are routinely cited as the primary gas generating compound in airbags, supporting this conclusion. Why Other Options Are Wrong: Sodium bicarbonate: Used in baking powders and some fire extinguishers, but not as the main gas generator in airbags. Sodium nitrite: Has roles in food preservation and industry but is not standard in airbag inflation chemistry. Sodium peroxide: A strong oxidising agent, dangerous for rapid oxygen release, not used in nitrogen based airbag systems. None of the above: Incorrect because sodium azide is correctly included and is the expected answer. Common Pitfalls: Some students may confuse sodium azide with more familiar compounds like sodium bicarbonate simply because they have seen the latter in household contexts. Others might overthink and worry about modern replacements for azide, leading them to choose None of the above. For exam purposes, however, sodium azide remains the standard textbook answer for the compound used in car airbags. Final Answer: Car safety airbags traditionally use sodium azide to generate nitrogen gas rapidly on impact.

Question 10

In fireworks, which chemical element or its compounds are mainly responsible for producing a green coloured flame?

Accepted Answer

Introduction / Context: Fireworks display brilliant colours due to the presence of specific metal salts that emit characteristic colours when heated. Understanding which element produces which colour is a common general science and chemistry question. In particular, green flames are strongly associated with compounds of certain metals. This question checks whether you can correctly link the green colour in fireworks to the element barium. Given Data / Assumptions: The question asks which element (or its compounds) produces a green flame in fireworks. Options include sodium, potassium, barium, mercury, and None of the above. We assume standard pyrotechnic formulations used in festival and display fireworks. Concept / Approach: Different metal salts emit different colours in a flame test because their electrons absorb and re emit energy at specific wavelengths. Sodium salts produce a bright yellow flame, potassium salts give a lilac or light violet colour, and barium salts are well known for their green emission in pyrotechnics. Mercury is not used for colour effects in fireworks due to its toxicity and other issues. Therefore, barium is the correct element for green flames. Step-by-Step Solution: Step 1: Recall the basic flame test colours from chemistry: sodium gives yellow, potassium gives lilac, and barium gives green. Step 2: Fireworks use similar principles but with more complex mixtures to intensify these colours. Step 3: Green colours in fireworks are typically produced by barium salts such as barium nitrate or barium chlorate. Step 4: Confirm that sodium based fireworks would appear yellow or golden, not green. Step 5: Conclude that barium is the element responsible for the green flame, making it the correct option. Verification / Alternative check: Chemistry textbooks and reference charts for flame tests consistently list barium compounds as giving a green flame. Pyrotechnic guides similarly mention barium compounds as key ingredients for green fireworks. Sodium and potassium have well documented but different colours, while mercury is avoided due to toxicity and is not a normal colourant in fireworks. These multiple sources align with the choice of barium as the correct answer. Why Other Options Are Wrong: Sodium: Produces a bright yellow flame, which is used for golden or yellow effects but not green. Potassium: Produces a lilac or pale violet flame, not the vivid green seen in some fireworks. Mercury: Toxic and not used as a standard colour producing element in fireworks. None of the above: Incorrect because barium, the correct element, is already present in the options. Common Pitfalls: Some students might guess sodium simply because it is the most familiar metal from salt, or potassium because they have seen its flame test. Others might overthink the question and choose None of the above. To avoid these mistakes, it helps to memorise a short mapping: sodium–yellow, potassium–lilac, barium–green, and strontium–red. This makes colour based questions about fireworks and flame tests easy to answer. Final Answer: In fireworks, the green coloured flame is mainly produced by compounds of barium.

Question 11

Potassium permanganate (KMnO₄) is used for purifying drinking water mainly because it acts as which of the following?

Accepted Answer

Introduction / Context: Potassium permanganate (KMnO₄) is a dark purple compound commonly used in laboratories and in household water treatment in small quantities. It appears in general science questions because of its strong chemical properties and its use in disinfection. Understanding whether it acts as an oxidising or reducing agent and how that relates to water purification is important for both chemistry and everyday awareness of safe water practices. Given Data / Assumptions: The question asks why potassium permanganate is used to purify drinking water. Options describe it as a sterilising agent, impurity dissolver, reducing agent, oxidising agent, or None of the above. We assume low, controlled doses of KMnO₄ added to water to kill germs and oxidise contaminants. Concept / Approach: Chemically, potassium permanganate is a powerful oxidising agent. An oxidising agent is a substance that accepts electrons from other species, thereby oxidising them. In water treatment, KMnO₄ oxidises organic matter, iron, manganese, and some microorganisms, helping to remove odours and kill pathogens. While we may casually say that it sterilises water, the underlying reason is its oxidising power. Therefore, the correct conceptual answer is that it is an oxidising agent, which in turn provides a sterilising effect. Step-by-Step Solution: Step 1: Recall that in redox chemistry, KMnO₄ is almost always cited as a classic oxidising agent. Step 2: In aqueous solution, permanganate ions (MnO₄⁻) accept electrons and get reduced, while other substances are oxidised. Step 3: When added to water, KMnO₄ helps oxidise organic impurities and can damage cell walls of microorganisms, effectively disinfecting the water. Step 4: The phrase dissolves impurities is misleading because the main action is chemical oxidation, not simple dissolution. Step 5: Hence, the best answer is that potassium permanganate is used because it is an oxidising agent. Verification / Alternative check: Textbooks and practical manuals mention KMnO₄ in titrations as a standard example of a strong oxidising agent. Water treatment literature also describes its use in oxidising iron and manganese in groundwater. It is never described as a reducing agent in these contexts. While some sources may informally call it a sterilising agent, that term describes the effect, not the chemical property. Exam oriented questions generally expect the precise chemical term oxidising agent as the answer. Why Other Options Are Wrong: A sterilising agent: Describes the effect but does not capture the specific chemical nature that exams are testing. It dissolves impurities of water: Incorrect because the key process is oxidation, not mere dissolution. A reducing agent: Opposite of its real behaviour; KMnO₄ is well known as an oxidiser, not a reducer. None of the above: Incorrect because oxidising agent is explicitly listed and is the correct concept. Common Pitfalls: Some students choose sterilising agent because they think in everyday terms instead of chemical terminology. Others may be unsure of the difference between oxidising and reducing agents. To avoid such mistakes, remember that in most school level examples KMnO₄, K₂Cr₂O₇, and concentrated nitric acid are strong oxidising agents. Linking potassium permanganate with oxidation, disinfection, and colour changes in titrations helps fix this concept. Final Answer: Potassium permanganate is used to purify drinking water mainly because it acts as an oxidising agent.

Question 12

Which part of the Sun is most easily visible only during a total solar eclipse?

Accepted Answer

Introduction / Context: Total solar eclipses provide rare opportunities to observe outer layers of the Sun that are normally hidden by its intense brightness. The question asks which part of the Sun becomes easily visible during such an eclipse. This is a standard general science topic linking basic astronomy with observational phenomena seen from Earth, and it often appears in school exams and competitive tests. Given Data / Assumptions: The event in question is a total solar eclipse, when the Moon completely covers the Sun's bright disc as seen from Earth. Options include the core, photosphere, sunspots, corona, and None of the above. We assume typical viewing with proper safety measures, not specialised instruments. Concept / Approach: Under normal conditions, the bright photosphere of the Sun overwhelms the light from its outer atmosphere. During a total solar eclipse, the Moon covers the photosphere, and the sky darkens enough that the faint outer atmosphere, called the corona, becomes visible as a glowing halo around the eclipsed Sun. The core is never directly visible, sunspots belong to the photosphere and are more easily studied with filters, and the photosphere itself is what gets blocked during totality. Thus, the corona is the part revealed most dramatically during a total solar eclipse. Step-by-Step Solution: Step 1: Recall the structure of the Sun: core, radiative zone, convective zone, photosphere, chromosphere, and corona. Step 2: Recognise that the photosphere is the visible surface that we normally see as the bright solar disc. Step 3: During a total solar eclipse, the Moon moves in front of the Sun and completely covers the photosphere for a short time. Step 4: With the bright disc hidden, the faint outer layers, particularly the corona, become visible as a whitish, extended halo. Step 5: Therefore, the corona is the part most easily visible only during totality, making it the correct answer. Verification / Alternative check: Astronomy photographs and descriptions of total solar eclipses almost always focus on images of the corona, which are difficult to capture under normal conditions. Scientists historically used eclipses to study the corona and discover elements such as helium. The core is buried deep inside the Sun and cannot be directly observed in visible light. Sunspots, though on the photosphere, are actually harder to see during totality because the entire disc is blocked. This pattern of evidence confirms that corona is the right choice. Why Other Options Are Wrong: Core: Located deep inside the Sun; never directly visible from Earth. Photosphere: This is the bright surface that gets covered by the Moon during a total solar eclipse. Sunspots: These are dark spots on the photosphere; when the Sun is completely eclipsed, these are not visible. None of the above: Incorrect because the corona, correctly listed, is the well known feature visible during totality. Common Pitfalls: Some students may misinterpret the question and choose photosphere, forgetting that this is the part that is actually hidden during an eclipse. Others might not remember the term corona and be tempted by None of the above. To avoid confusion, recall standard eclipse descriptions that emphasise the dramatic appearance of the solar corona as a crown or halo around the dark lunar disc. Final Answer: The part of the Sun most easily visible only during a total solar eclipse is the corona.

Question 13

The core material used in ordinary lead pencils is mainly which of the following?

Accepted Answer

Introduction / Context: Although we commonly speak of the lead in a pencil, modern pencils do not actually contain metallic lead. Instead, their writing core is made from a form of carbon that leaves dark marks on paper. This simple everyday misconception is frequently used in general science questions to test whether students know the correct material in pencil cores and to clear up confusion about the term lead pencil. Given Data / Assumptions: The question asks which material is used in pencils. Options include silicon, phosphorus, graphite, charcoal, and None of the above. We assume a standard writing pencil used in schools and offices. Concept / Approach: The dark writing core of a pencil is composed of graphite mixed with clay and other binders, not metallic lead. Graphite is an allotrope of carbon with layers that can slide over each other, allowing the material to leave a mark on paper. The term lead pencil is historical, from a time when similar looking materials were confused with lead ore. Silicon and phosphorus are not used for this purpose, and while charcoal is also a form of carbon, it is not the standard core material of ordinary pencils. Step-by-Step Solution: Step 1: Recall that modern writing pencils use a graphite and clay mixture in their core. Step 2: Understand that graphite is a crystalline form of carbon with a layered structure, making it soft and able to slide off onto paper. Step 3: Recognise that the word lead in lead pencil is a misnomer; metallic lead is toxic and not used for writing cores. Step 4: Compare with charcoal, which is a different form of carbon used in art pencils and drawing sticks, not usually in standard writing pencils. Step 5: Conclude that graphite is the correct core material in ordinary pencils. Verification / Alternative check: Manufacturers and encyclopaedia entries consistently describe pencil leads as mixtures of graphite and clay, with varying proportions to adjust hardness. There is no mention of silicon or phosphorus in the core of everyday pencils. Art supplies may use charcoal sticks, but these are marketed separately as charcoal pencils or sticks, not as regular graphite pencils. This consistent usage confirms that graphite is the correct answer. Why Other Options Are Wrong: Silicon: A semiconductor material used in electronics, not for pencil cores. Phosphorus: Used in matches and certain industrial applications, not in pencil leads. Charcoal: A carbonaceous material used in drawing, but standard pencils are specifically graphite based. None of the above: Incorrect because graphite, the correct material, is explicitly listed. Common Pitfalls: Many people assume that because we say lead pencil, the core must contain lead. This is historically inaccurate and scientifically wrong. Some may also confuse charcoal drawing tools with graphite pencils and choose charcoal. To avoid these errors, remember the key point: ordinary pencils use a graphite and clay core, which is safer and provides controlled hardness for writing. Final Answer: The core of an ordinary pencil is mainly made of graphite, not metallic lead.

Question 14

A married couple with blood groups AB positive and O negative adopt a male child. Later, twin boys are born to them. The three sons have blood groups A positive, B positive, and O positive. What is the blood group of the adopted son?

Accepted Answer

Introduction / Context: This question applies basic genetics of ABO blood groups and Rh factor to a family scenario. By using the possible genotypes of the parents, we can work out which blood groups their biological children can have and then deduce which child must be adopted. Such questions are popular in biology sections of competitive exams because they test reasoning, not just memorisation of blood group facts. Given Data / Assumptions: Father's blood group: AB positive (genotype ABO: IAIB; Rh: Rh⁺, likely ++ or +-). Mother's blood group: O negative (genotype ABO: ii; Rh: Rh⁻, genotype --). Three sons have blood groups: A positive, B positive, and O positive. One son is adopted and two are biological twins; all are male, which does not affect ABO or Rh inheritance. Concept / Approach: For ABO inheritance, the father with AB has alleles IA and IB, and the mother with O has alleles i and i. Therefore, all biological children must inherit IA or IB from the father and i from the mother, giving genotypes IAi (group A) or IBi (group B). It is impossible for them to be group O (genotype ii) because neither parent can provide a second i allele. For the Rh factor, a positive father (+-) and negative mother (--) can have positive or negative children, but this does not change the ABO combinations. Thus, any biological child can only be A or B, never O, which identifies the adopted child. Step-by-Step Solution: Step 1: Determine possible ABO genotypes of biological children from AB (IAIB) × O (ii) parents. Step 2: Possible offspring genotypes are IAi (group A) or IBi (group B), because the child always gets i from the O parent and IA or IB from the AB parent. Step 3: Conclude that the biological children can only have blood groups A or B, not O or AB. Step 4: Examine the sons: A positive, B positive, and O positive. A and B are consistent with biological inheritance; O is not. Step 5: Therefore, the son with O positive blood group must be the adopted child. Verification / Alternative check: To further verify, we can look at the Rh factor. The father is positive (could be ++ or +-), and the mother is negative (--). A cross of +- × -- can produce +- (Rh positive) or -- (Rh negative) children. Thus, biological children may be Rh positive or negative. The question states that all sons are Rh positive, which is entirely possible for both adopted and biological children. However, the Rh data do not contradict the ABO conclusion that only A and B are biologically possible; O must therefore belong to the adopted son. Why Other Options Are Wrong: A positive: This can be a biological child since IAi (group A) is a valid genotype from AB × O parents. B positive: This can also be a biological child since IBi (group B) is possible from the same cross. Cannot be determined on the basis of the given data: Incorrect because the ABO inheritance clearly rules out O as a biological child in this parental combination. None of the above: Incorrect because O positive is a listed option and fits the adopted child logically. Common Pitfalls: Some students focus too much on the Rh factor and get confused, forgetting that ABO inheritance alone is enough to decide this question. Another common mistake is assuming that because the parents include group O, they can have a child with group O, ignoring the fact that the other parent is AB and cannot provide the second i allele needed to form ii. Carefully listing possible genotypes from the cross helps avoid such reasoning errors. Final Answer: The adopted son is the one with blood group O positive.

Question 15

For equal masses of substances, which of the following has the highest specific heat capacity?

Accepted Answer

Introduction / Context: This question deals with the concept of specific heat capacity in physics. Specific heat capacity tells us how much heat a substance must absorb to raise the temperature of 1 kg of the substance by 1 degree Celsius. Water is famous for having a high specific heat compared to most metals and many common liquids. Questions like this connect basic thermodynamics to everyday phenomena, such as why coastal climates are milder or why water heats and cools slowly. Given Data / Assumptions: We compare equal masses of four substances: iron, water, gold, and benzene. The question asks which has the highest heat capacity; in exam context this usually means highest specific heat capacity. We assume standard temperature ranges where tabulated specific heat values apply. Concept / Approach: Specific heat capacity (c) is defined by the relation Q = m * c * ΔT, where Q is heat absorbed, m is mass, and ΔT is temperature change. For equal masses and equal temperature rise, the substance with the largest c requires the most heat. Water has a specific heat capacity of about 4.18 J/g°C, which is much higher than that of metals like iron and gold and generally higher than many organic liquids. Therefore, among the listed substances, water has the greatest specific heat capacity. Step-by-Step Solution: Step 1: Recall approximate specific heat values: water ≈ 4.18 J/g°C; iron ≈ 0.45 J/g°C; gold ≈ 0.13 J/g°C. Step 2: Recognise that metals typically have relatively low specific heats compared to water. Step 3: Organic liquids like benzene have specific heats lower than water, typically around 1–2 J/g°C. Step 4: Compare all four: water's specific heat is clearly the highest among them. Step 5: Conclude that for equal masses and equal temperature changes, water will absorb the most heat, so it has the greatest heat capacity. Verification / Alternative check: We can cross check this with everyday observations. Metals such as iron and gold heat up quickly when exposed to a flame, indicating they need relatively little energy to change temperature. Water, by contrast, takes a long time to boil, showing that it requires a lot of energy to raise its temperature. Coastal areas, where large bodies of water moderate the climate, also illustrate water's high heat capacity. Benzene and similar liquids warm faster than water under the same heating, again suggesting lower specific heat. All this supports water as the correct answer. Why Other Options Are Wrong: An iron piece: Iron has a much lower specific heat capacity than water, so it heats and cools more quickly. A gold piece: Gold has an even lower specific heat capacity than iron, making this option clearly incorrect. Benzene: An organic liquid with lower specific heat than water; it does not store as much heat per unit mass per degree. None of the above: Incorrect because water is included and is known to have the highest specific heat among the given choices. Common Pitfalls: Some students may misinterpret heat capacity as simply relating to how hot something feels to the touch, without considering mass and specific heat. Others might be unsure about benzene's properties and overestimate its specific heat. To avoid such mistakes, remember the general rule: water has a very high specific heat compared to most substances, particularly metals, which is why it is often used as a coolant and temperature stabiliser in nature and technology. Final Answer: For equal masses, water has the highest specific heat capacity among the substances listed.

Question 16

Which radioactive element is most commonly detected naturally inside the human body?

Accepted Answer

Introduction / Context: This question examines basic knowledge of natural radioactivity in living organisms. Even without exposure to man made radiation sources, humans contain small amounts of naturally occurring radioactive isotopes. Understanding which of these isotopes is most common in the human body helps connect nuclear physics to biology and health science, and it commonly appears in general science sections of competitive exams. Given Data / Assumptions: We are asked which radioactive element is most commonly detected in humans under normal conditions. Options include potassium 40, cobalt 60, iodine 131, plutonium 238, and None of the above. We assume ordinary, healthy individuals without unusual contamination or medical treatment. Concept / Approach: Potassium is an essential element for life, present in all cells. A small fraction of natural potassium consists of the radioactive isotope potassium 40 (K 40). Because our bodies contain significant potassium, we naturally contain K 40 as well. This makes potassium 40 the most common radioactive element in humans. Cobalt 60 and iodine 131 are used in medical treatments and diagnostics but are not naturally present in large amounts. Plutonium 238 is an artificial transuranic element not normally found in the body. Step-by-Step Solution: Step 1: Recall that potassium is a major intracellular ion; all humans contain several tens of grams of potassium. Step 2: Recognise that natural potassium includes a small percentage of K 40, which is radioactive. Step 3: Therefore, anyone with normal potassium levels also carries K 40 in their tissues. Step 4: Compare with cobalt 60 and iodine 131, which are manufactured isotopes used in radiotherapy and imaging and do not accumulate naturally in healthy individuals. Step 5: Conclude that potassium 40 is the naturally occurring radioactive element most commonly detected in humans. Verification / Alternative check: Background radiation studies often mention that a significant portion of internal radiation exposure comes from K 40 inside the body. Another contributor is carbon 14, but it is not listed among the options here. Cobalt 60 and iodine 131 appear mainly in the context of cancer therapy and thyroid imaging, respectively, and decay away over time after treatment. Plutonium 238 is related to nuclear power and space power sources, not to normal human biochemistry. These comparisons confirm that K 40 is the correct answer. Why Other Options Are Wrong: Cobalt-60: A man made isotope used for radiotherapy; it is not naturally abundant in the human body. Iodine-131: Used in thyroid diagnosis and treatment; present only temporarily and not in all humans. Plutonium-238: A synthetic element used in specialised applications; not normally found in biological tissues. None of the above: Incorrect because potassium 40 is correctly listed and is the natural internal source of radiation in humans. Common Pitfalls: Students may be distracted by isotopes like iodine 131 and cobalt 60 because they are frequently mentioned in medical contexts. However, these are artificial and episodic in the body, not constant background sources. To answer correctly, remember that potassium and carbon are fundamental elements in biology, and their natural radioactive isotopes (K 40 and C 14) are the main contributors to internal radioactivity under normal conditions. Final Answer: The radioactive element most commonly detected naturally in humans is potassium-40.

Question 17

Which combination of gases is commonly used in different types of welding processes?

Accepted Answer

Introduction / Context: Welding is a key industrial process used to join metals. Different types of welding processes, such as oxy fuel welding and arc welding, use various gases either as fuel, oxidisers, or shielding gases. General science and technical aptitude questions often ask about common gas combinations used in welding, especially the classic oxygen acetylene combination and shielding gases like argon. Given Data / Assumptions: The question asks about gases used in different types of welding. Options provide several combinations involving oxygen, hydrogen, acetylene, argon, and nitrogen. We assume common welding methods: oxy acetylene welding and arc welding processes like MIG and TIG. Concept / Approach: Oxy acetylene welding uses a mixture of oxygen and acetylene to produce a very hot flame for melting and joining metals. In addition, modern arc welding processes such as MIG (Metal Inert Gas) and TIG (Tungsten Inert Gas) use inert gases like argon as shielding gases to protect the molten weld from atmospheric contamination. Thus, across different welding types, oxygen, acetylene and argon are all important. Hydrogen and nitrogen are specialised or less desirable in many common welding procedures, making the combination that includes oxygen, acetylene and argon the best match to the phrase different types of welding. Step-by-Step Solution: Step 1: Recognise that oxygen and acetylene together form the oxy acetylene flame, widely used in gas welding and cutting. Step 2: Recall that argon is a noble gas used as a shielding gas in MIG and TIG welding to prevent oxidation and contamination. Step 3: Understand that the question mentions different types of welding, suggesting we should think beyond just one process. Step 4: Observe that the combination oxygen, acetylene and argon covers both oxy fuel welding and shielded arc welding. Step 5: Conclude that oxygen, acetylene and argon is the most appropriate answer among the given choices. Verification / Alternative check: Standard welding handbooks describe oxy acetylene welding as the most common gas welding process, using a combination of oxygen and acetylene. For arc welding processes, shielding gases such as argon, helium, or CO₂ are essential. Very few common processes rely primarily on nitrogen or hydrogen as core welding gases; nitrogen can even cause porosity and brittleness in some welds. Therefore, a combination that includes oxygen, acetylene and argon correctly reflects gases used across multiple welding methods. Why Other Options Are Wrong: Oxygen and hydrogen: This combination can produce a flame but is not the standard for common welding due to safety and material concerns. Oxygen and acetylene: Correct for oxy acetylene welding alone but does not reflect gases used in other common welding types such as MIG or TIG. Oxygen, hydrogen, acetylene and nitrogen: Includes gases not typically desirable as general welding gases; nitrogen is not a standard shielding gas. None of the above: Incorrect because oxygen, acetylene and argon together accurately represent gases used in multiple welding processes. Common Pitfalls: Many students immediately choose oxygen and acetylene because it is the most famous pair and forget that the question refers to different types of welding, not just oxy fuel. Others might be confused by the presence of nitrogen and hydrogen, thinking more gases must be better. To avoid such mistakes, remember that argon is a key shielding gas in arc welding, and combining oxygen, acetylene and argon covers both major categories: gas welding and arc welding. Final Answer: The gases commonly used across different welding processes include oxygen, acetylene and argon.

Question 18

When a gas is converted into a liquid, what is this process called?

Accepted Answer

Introduction / Context: Phase changes between solid, liquid, and gas are fundamental concepts in physics and chemistry. Each change has a specific name, such as melting, freezing, evaporation, and condensation. This question asks for the term used when a gas turns into a liquid. Recognising these terms is important for understanding weather phenomena, refrigeration, and many industrial processes, and it is frequently tested in school and competitive exams. Given Data / Assumptions: A gas changes state and becomes a liquid. The options are condensation, evaporation, deposition, sublimation, and None of the above. We assume the process occurs due to cooling or compression at suitable conditions. Concept / Approach: When a gas loses energy (for example, when it cools) and its molecules come closer together to form a liquid, the process is called condensation. This is the reverse of evaporation, which is the conversion of liquid to gas. Deposition is the direct change from gas to solid without passing through the liquid state, and sublimation is solid to gas directly. Therefore, among the provided options, condensation is the correct term for gas turning into liquid. Step-by-Step Solution: Step 1: Identify the direction of phase change in the question: gas → liquid. Step 2: Recall that condensation is defined as the process by which a gas becomes a liquid. Step 3: Recognise that evaporation is the opposite process (liquid → gas), so it cannot be correct here. Step 4: Understand that deposition is gas → solid and sublimation is solid → gas, both bypassing the liquid phase. Step 5: Conclude that condensation is the only option that correctly names gas changing into liquid. Verification / Alternative check: Everyday examples confirm this terminology. Water vapour in the air condenses on cold surfaces as droplets; we call this condensation. In the water cycle, formation of clouds from water vapour is also described as condensation. Refrigerators and air conditioners cause moisture in air to condense on cooling coils. In all of these cases, the term condensation is used, and it always refers to gas turning into liquid, matching the question exactly. Why Other Options Are Wrong: Evaporation: Refers to a liquid changing into a gas at the surface, the reverse of the process described. Deposition: Describes a gas directly becoming a solid, as in frost formation, not gas to liquid. Sublimation: Refers to a solid directly becoming a gas, such as dry ice turning to CO₂ gas. None of the above: Incorrect because condensation is correctly listed and matches the definition. Common Pitfalls: A common mistake is confusing condensation with evaporation because both involve liquids and gases. Students sometimes remember one term but not the direction of change. Another error is mixing up deposition and condensation since both involve gas losing energy, but deposition skips the liquid phase. To avoid this, keep a clear mental map of phase changes and their names: melting, freezing, evaporation, condensation, sublimation, and deposition. Final Answer: The process of a gas changing into a liquid is called condensation.

Question 19

Which is the largest natural satellite (moon) in our solar system?

Accepted Answer

Introduction / Context: The solar system contains many natural satellites, commonly called moons, orbiting different planets. Knowing which moon is the largest helps in understanding comparative planetology and is a favourite topic in astronomy based general knowledge questions. This question asks you to identify the largest natural satellite among well known examples such as Titan, Ganymede, and Earth's Moon. Given Data / Assumptions: We are asked for the largest natural satellite in the solar system by physical size (diameter). Options include Titan, Ganymede, Callisto, Earth's Moon, and None of the above. We consider recognised major moons of the planets, not dwarf planet satellites. Concept / Approach: Several moons in the outer solar system are larger than Earth's Moon. Ganymede, a moon of Jupiter, has the largest diameter, slightly bigger than that of Titan (Saturn's moon) and also larger than Mercury. Titan is the second largest moon, followed by Callisto and Io. Earth's Moon is smaller than both Ganymede and Titan. Therefore, among the given options, Ganymede is the correct answer as the largest natural satellite in the solar system. Step-by-Step Solution: Step 1: Recall approximate sizes: Ganymede has a diameter of about 5,268 km, making it the largest moon. Step 2: Titan's diameter is about 5,150 km, placing it very close to but still smaller than Ganymede. Step 3: Callisto has a diameter of around 4,820 km, smaller than both Ganymede and Titan. Step 4: Earth's Moon has a diameter of about 3,474 km, which is clearly smaller than all the above. Step 5: Conclude that Ganymede is the largest natural satellite in the solar system and select that option. Verification / Alternative check: Astronomy references and data tables list Ganymede as larger in diameter than the planet Mercury, which is a useful memory aid. Titan, although notable for its thick atmosphere, is slightly smaller in diameter. Earth's Moon, while familiar to us, is not the largest moon overall. None of the recent discoveries of outer system moons have surpassed Ganymede in size. This consistent ranking across scientific sources verifies that Ganymede is indeed the largest natural satellite in our solar system. Why Other Options Are Wrong: Titan (moon of Saturn): Second largest moon, slightly smaller in diameter than Ganymede. Callisto (moon of Jupiter): Large but smaller than both Ganymede and Titan. Earth's Moon: Smaller than the three large outer moons listed, especially Ganymede and Titan. None of the above: Incorrect because Ganymede, the correct moon, is explicitly given as an option. Common Pitfalls: Some students may think Titan is the largest because it is often emphasised for its atmosphere and surface lakes. Others may assume Earth's Moon is the largest simply because it is the most familiar. To avoid confusion, remember this hierarchy: Ganymede is largest, Titan is second, followed by Callisto and then Io, with Earth's Moon further down the list. This ranking makes the question straightforward once memorised. Final Answer: The largest natural satellite in our solar system is Ganymede (moon of Jupiter).

Question 20

Which of the following plants is not capable of manufacturing its own food by photosynthesis?

Accepted Answer

Introduction / Context: This question tests your understanding of autotrophic and heterotrophic nutrition in biology. Green plants and some plant like organisms are capable of photosynthesis, using chlorophyll, sunlight, carbon dioxide, and water to make their own food. Other organisms lack chlorophyll and must obtain food from external sources. The question asks which listed organism cannot manufacture its own food, highlighting the difference between true plants and fungi. Given Data / Assumptions: Options are algae, mushroom, carrot, and cabbage, with an additional None of the above choice. We assume normal, healthy examples of each organism. We use the standard definition of photosynthesis: conversion of light energy into chemical energy using chlorophyll. Concept / Approach: Algae, carrot plants, and cabbage are all photosynthetic organisms that contain chlorophyll in their green parts. Algae, though diverse, are generally considered autotrophic and capable of photosynthesis. Carrot and cabbage are typical green plants; their leaves produce food through photosynthesis, although the carrot root stores food. Mushrooms, however, belong to the kingdom Fungi. Fungi do not have chlorophyll and cannot photosynthesise; they obtain nutrients by decomposing organic matter or forming parasitic or symbiotic relationships. Thus, mushrooms cannot manufacture their own food and are heterotrophic. Step-by-Step Solution: Step 1: Evaluate algae: Most algae contain chlorophyll and live in water, performing photosynthesis like plants. Step 2: Evaluate carrot: The carrot plant has green leaves that photosynthesise; the orange root is a storage organ for food made by the leaves. Step 3: Evaluate cabbage: Cabbage leaves are green and rich in chlorophyll, so the plant manufactures its own food. Step 4: Evaluate mushroom: Mushrooms are fungi, lack chlorophyll, and cannot use sunlight to make food. Step 5: Conclude that mushroom is the only organism in the list that is not capable of manufacturing its own food. Verification / Alternative check: Biology textbooks classify organisms by their mode of nutrition. Autotrophs, such as green plants and algae, synthesise food using photosynthesis. Heterotrophs, including animals and fungi, rely on external organic sources. Mushrooms are regularly cited as classic examples of saprophytic fungi that decompose dead organic matter. Carrot and cabbage appear as typical garden vegetables and are always included among plants that produce their own food. This classification strongly supports mushroom as the correct answer. Why Other Options Are Wrong: Algae: Most algae are photosynthetic autotrophs and can manufacture their own food. Carrot: The carrot plant is a green plant; though the edible root is underground, the plant as a whole performs photosynthesis. Cabbage: A leafy vegetable with chlorophyll containing leaves that photosynthesise and produce food for the plant. None of the above: Incorrect because one listed organism, the mushroom, clearly cannot photosynthesise. Common Pitfalls: Some students may be misled by the fact that mushrooms grow from the ground like plants and are eaten as vegetables, causing them to overlook that mushrooms are fungi, not plants. Others might mistakenly think carrots cannot photosynthesise because the root is underground, forgetting that photosynthesis occurs in the leaves. To answer correctly, remember that green colour from chlorophyll is a strong indicator of photosynthesis, and fungi like mushrooms lack this pigment. Final Answer: The plant among the options that is not capable of manufacturing its own food is the mushroom, because it is a fungus and heterotrophic.

General Science Questions