Question 1

As pure water is cooled from 4 degrees Celsius down to 0 degrees Celsius, how does its density change in this temperature range?

Accepted Answer

Introduction / Context: Water exhibits an unusual behaviour known as anomalous expansion near 4 degrees Celsius. Unlike most substances, whose density increases steadily as they are cooled, water reaches a maximum density at around 4 degrees Celsius and then becomes less dense as it is cooled further toward the freezing point. This question tests whether you know how the density of water changes when it is cooled from 4 degrees Celsius to 0 degrees Celsius. Given Data / Assumptions: We are dealing with pure water at normal atmospheric pressure. Water has its maximum density at approximately 4 degrees Celsius. The temperature range considered is from 4 degrees Celsius down to 0 degrees Celsius. We are asked about the trend in density, not exact numerical values. Concept / Approach: As water is cooled from higher temperatures, its density increases until it reaches a maximum at around 4 degrees Celsius. Below this temperature, the structure of water begins to form more open hydrogen bonded clusters that resemble the structure of ice. These clusters take up more space, causing water to expand slightly as it cools from 4 degrees Celsius to 0 degrees Celsius. Because density is mass divided by volume, an increase in volume for the same mass means a decrease in density. This anomalous behaviour explains why ice floats on water and why lakes freeze from the top downward, which is important for aquatic life in cold climates. Step-by-Step Solution: Step 1: Recall that density is given by rho = mass / volume.Step 2: Note that water reaches its maximum density at about 4 degrees Celsius.Step 3: As water is cooled below 4 degrees Celsius toward 0 degrees Celsius, the formation of more open structures causes it to expand slightly.Step 4: This expansion means that for the same mass of water, the volume increases.Step 5: With mass constant and volume increasing, density must decrease according to rho = mass / volume.Step 6: Therefore, between 4 degrees Celsius and 0 degrees Celsius, the density of water decreases from its maximum value. Verification / Alternative check: Observationally, this behaviour explains why ice floats and why cold water tends to sink until it reaches about 4 degrees Celsius. In a lake, as the surface water cools, the densest water at around 4 degrees Celsius sinks to the bottom, and once the whole body reaches this temperature, further cooling of the surface produces less dense water that remains on top and eventually freezes. This pattern would not happen if density continued to increase all the way down to 0 degrees Celsius. Data tables of water density versus temperature confirm that the maximum occurs near 4 degrees Celsius and that density decreases as temperature approaches 0 degrees Celsius. Why Other Options Are Wrong: The statement that density increases continuously from 4 degrees Celsius to 0 degrees Celsius contradicts the known anomaly of water. Saying that density remains the same ignores measured variations and the physical expansion due to structural changes. The option that none of the above is correct is also wrong because one of the options accurately describes the behaviour. Only the statement that density decreases below its maximum at 4 degrees Celsius matches the observed properties of water. Common Pitfalls: Many learners assume that all substances behave like ideal solids and liquids where cooling always increases density. Others may remember that water is special but forget the specific temperature at which density is maximum. To avoid confusion, remember the key fact: water has its maximum density at about 4 degrees Celsius, and as you cool it further to 0 degrees Celsius, it expands slightly and its density decreases. Final Answer: As water is cooled from 4 degrees Celsius to 0 degrees Celsius, its density decreases below the maximum value it has at 4 degrees Celsius.

Question 2

In electrical engineering, a dynamometer type wattmeter is suitable for measuring power in which type of circuits?

Accepted Answer

Introduction / Context: A dynamometer type wattmeter is a commonly used instrument in electrical engineering laboratories and power systems to measure electric power. This question checks whether you know the important practical fact that such wattmeters are versatile instruments that can work in both direct current (DC) and alternating current (AC) circuits. Knowing where a particular measuring instrument can be used is essential for safe and correct power measurements in real installations. Given Data / Assumptions: The instrument mentioned is a dynamometer type wattmeter. We are concerned with the type of circuit in which it can measure power. No special frequency limitation or power factor condition is given. We assume normal laboratory or power distribution usage conditions. Concept / Approach: A dynamometer wattmeter works on the principle of the interaction between the magnetic fields produced by two coils: a current coil and a pressure (voltage) coil. The deflecting torque is proportional to the product of the instantaneous current and voltage and hence proportional to the power. This principle is valid for both DC and AC, as the instrument essentially responds to the average value of power over time, not just to instantaneous values. Therefore, the same instrument can be used for DC as well as single phase AC power measurement. Step-by-Step Solution: Step 1: Recall that in DC circuits, voltage and current are constant with time, so the power is P = V * I and a dynamometer wattmeter can easily measure this product. Step 2: For AC circuits, both voltage and current vary sinusoidally and may have a phase difference. The wattmeter deflection is proportional to the average value of v(t) * i(t) over a cycle, which gives the real or active power. Step 3: The magnetic interaction between the current coil and pressure coil exists for both DC and AC because the torque depends on instantaneous current and flux, whose average over time gives a steady deflection. Step 4: Therefore, the instrument is suitable for measuring power in both DC and AC circuits without any change in basic construction. Verification / Alternative check: In most standard electrical engineering textbooks, dynamometer type wattmeters are listed as the standard laboratory wattmeters precisely because they can be used for both DC and AC. Permanent magnet moving coil instruments are limited mostly to DC, whereas dynamometer instruments, with air cored coils, avoid problems of hysteresis and eddy currents and therefore are suitable for AC as well. This confirms our conclusion. Why Other Options Are Wrong: Direct current (DC) circuits only: This is incorrect because it ignores the ability of the wattmeter to handle alternating current systems. Alternating current (AC) circuits only: This is also incorrect; nothing in the working principle prevents DC usage. Neither DC nor AC circuits, only for calibration work: This is incorrect because the instrument is specifically built for power measurement in live circuits, not only for calibration. Common Pitfalls: Students sometimes confuse the dynamometer wattmeter with other instruments such as permanent magnet moving coil meters that are restricted mainly to DC. Another confusion is between devices that measure only AC because of their construction and those that can work with both types of currents. Remember that it is the nature of the magnetic circuit and coils that decides this capability. Final Answer: The dynamometer type wattmeter can measure power in both direct current and alternating current circuits. Therefore, the correct option is Both direct current (DC) and alternating current (AC) circuits.

Question 3

In the International System of Units (SI), how many base or fundamental units are defined in total?

Accepted Answer

Introduction / Context: The International System of Units, commonly called the SI system, is the globally accepted standard for scientific and engineering measurements. It is built on a small set of fundamental or base units from which all other derived units are constructed. This question checks your memory of how many such base units exist in the modern SI system, which is important whenever you study dimensional analysis, unit conversion, or physical formulas. Given Data / Assumptions: We are talking about the official SI system, not older metric systems. We need the count of base or fundamental units, not derived units. The latest definition of the SI base units is assumed. Concept / Approach: The SI system defines a fixed set of base quantities, each with a corresponding base unit. These include length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity. Their units are metre, kilogram, second, ampere, kelvin, mole, and candela. All other physical quantities such as force, energy, or pressure are expressed in terms of these base units through derived units like newton, joule, or pascal. Step-by-Step Solution: Step 1: List the seven base quantities: length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity. Step 2: Match each quantity with its SI base unit: metre (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol), and candela (cd). Step 3: Count the number of base units in this list. There are exactly seven distinct base units. Step 4: Conclude that the number of base or fundamental SI units is 7. Verification / Alternative check: You can verify this information from any standard physics or engineering reference, where the SI units table is printed in the introductory chapters. It always lists the same seven base units. Although their definitions have been refined using physical constants like the speed of light or Planck constant, the number of base units itself has remained seven in the modern SI system. Why Other Options Are Wrong: 4: This is too small and would not be enough to describe all fundamental physical quantities. 6: This misses one of the standard base quantities and therefore is incorrect. 5: This also underestimates the total number of SI base units. Common Pitfalls: Learners sometimes confuse base units with derived units and may think quantities like force or energy are fundamental. Another common mistake is to forget either mole or candela when mentally counting the list, leading to an incorrect number such as 5 or 6. To avoid this, it is helpful to remember a simple phrase or mnemonic for all seven basic quantities and their symbols. Final Answer: The International System of Units defines seven base or fundamental units. Therefore, the correct answer is 7.

Question 4

In nuclear and radiation physics, the curie (Ci) is a unit used to measure which physical quantity?

Accepted Answer

Introduction / Context: In nuclear physics, medical imaging, and radiation safety, different units are used to describe radioactive decay and exposure. The curie, symbol Ci, is one of the older but still widely known units. This question tests your knowledge of which physical quantity is measured in curies, which is important for understanding discussions of radioactive sources and their strength. Given Data / Assumptions: The symbol Ci refers to the curie. We are working within the context of radioactivity and nuclear phenomena. We must identify the physical quantity associated with this unit. Concept / Approach: Radioactivity is the process in which unstable atomic nuclei spontaneously emit particles or electromagnetic radiation. The activity of a radioactive source tells us how many disintegrations occur per unit time. Historically, the curie was defined so that 1 curie corresponds to 3.7 * 10^10 disintegrations per second, approximately equal to the activity of one gram of radium. So, the curie is a unit of activity or radioactivity, not of temperature, heat, or energy directly. Step-by-Step Solution: Step 1: Recall that temperature is measured in degrees Celsius, kelvin, or Fahrenheit, not in curies. Step 2: Recall that heat and energy are measured in joules or calories, while electrical resistance is measured in ohms. Step 3: Identify the physical quantity related to the rate of nuclear disintegrations per second. This is called activity or radioactivity of a source. Step 4: Note that the curie was defined as 3.7 * 10^10 disintegrations per second, making it clearly a unit of radioactivity. Verification / Alternative check: Modern SI practice replaces the curie with the becquerel (Bq), where 1 Bq is 1 disintegration per second. The relationship is 1 curie = 3.7 * 10^10 becquerel. Since the becquerel is defined as a unit of activity, the curie must also represent activity. This cross check strongly confirms that the curie measures radioactivity and not temperature, heat, or resistance. Why Other Options Are Wrong: Temperature: Units like kelvin and degree Celsius are used for this; curie is unrelated to thermal state. Heat energy: This uses joule, calorie, or similar units, not the curie. Electrical resistance: The standard unit is the ohm, not the curie. Common Pitfalls: Learners sometimes mix up units when first encountering nuclear physics. It is easy to confuse curie with other nuclear related terms like joule or gray. A helpful strategy is to remember that activity describes how many atoms decay per second and is measured by curie in older literature and becquerel in SI, while absorbed dose uses gray and Sievert for biological effect. Final Answer: The curie is a unit of the activity of a radioactive source, that is, of radioactivity.

Question 5

Which one of the following pairs of physical quantities has the same dimensions in the SI system?

Accepted Answer

Introduction / Context: Dimensional analysis is a powerful tool in physics for checking the correctness of formulas and for comparing different physical quantities. Two quantities that share the same dimensions can often be added or compared meaningfully, while quantities with different dimensions cannot. This question asks which pair of quantities has identical dimensions in the SI system, which is a classic concept tested in exams. Given Data / Assumptions: We have four pairs of physical quantities. We must determine which pair has the same dimensions. Standard SI base dimensions are used: mass (M), length (L), time (T). Concept / Approach: To answer this, we express each quantity in terms of the base dimensions M, L, and T. If two quantities reduce to the same dimensional formula, then they are dimensionally equivalent. Momentum is defined as mass times velocity, and impulse is defined as force times time, or change in momentum. Therefore, their dimensions are closely related. For other pairs like power and Young modulus or energy and angular momentum, we must check their formulas and compare dimensions carefully. Step-by-Step Solution: Step 1: Write dimensions of momentum: p = m * v, so [p] = M * L T^-1 = M L T^-1. Step 2: Write dimensions of impulse: J = F * t, force F has dimensions M L T^-2, so [J] = M L T^-2 * T = M L T^-1. Step 3: Since momentum and impulse both have [M L T^-1], they share the same dimensions. Step 4: Check one incorrect pair as example. Power P = work per unit time, so [P] = (M L^2 T^-2) / T = M L^2 T^-3. Young modulus Y = stress / strain, so [Y] = (force per area) = (M L T^-2) / L^2 = M L^-1 T^-2. These are clearly different. Verification / Alternative check: For energy and angular momentum: energy E has dimensions M L^2 T^-2, while angular momentum L has dimensions M L^2 T^-1. They differ by one power of time. For force constant of a spring k (force per unit extension) dimensions are M T^-2, whereas moment of inertia I has dimensions M L^2, again clearly different. Only momentum and impulse match, confirming our choice. Why Other Options Are Wrong: Power and Young modulus: Power is related to rate of doing work, Young modulus to stiffness; their dimensional formulas are different. Energy and angular momentum: Both involve M and L^2 but differ in the power of T, so they are not dimensionally equal. Force constant and moment of inertia: Force constant has no factor of L, while moment of inertia has L^2, so they differ. Common Pitfalls: Many students assume that if two quantities are related in mechanics, they must have different dimensions, or they misremember the formulas. Another common mistake is to forget that impulse is defined as change in momentum, which strongly suggests that their dimensions are identical. Always build the dimensional formula from the basic definition, rather than relying only on memory. Final Answer: The only pair with the same dimensions is momentum and impulse.

Question 6

Which one of the following groups contains quantities that do not share the same dimensions?

Accepted Answer

Introduction / Context: Dimensional analysis helps us understand whether two physical quantities are of the same kind. Many pairs like pressure and stress or work and energy are closely related and share dimensional formulas, while others differ. This question asks you to identify which group contains quantities that do not have the same dimensions, a common test of understanding in basic physics. Given Data / Assumptions: Four pairs of quantities are given. We must find the pair that is dimensionally different. SI base dimensions mass (M), length (L), and time (T) are used. Concept / Approach: We need to write the dimensional formula for each member of the pair and compare. If they match, the pair has the same dimensions. If they do not match, that pair is the correct answer. Pressure and stress are both force per unit area. Velocity and speed are both displacement per unit time. Work and energy are both force times distance. However, force and impulse, although related, are not dimensionally identical because impulse includes an extra factor of time. Step-by-Step Solution: Step 1: Pressure and stress: both equal force per area, so [P] = [stress] = (M L T^-2) / L^2 = M L^-1 T^-2. Step 2: Velocity and speed: both are distance per time, so [v] = L T^-1 and speed has the same dimensions. Step 3: Work and energy: work W = force * distance, so [W] = M L T^-2 * L = M L^2 T^-2. Energy has the same dimensional formula. Step 4: Force and impulse: force F has [F] = M L T^-2, impulse J = F * time, so [J] = M L T^-2 * T = M L T^-1. These are different because of the exponent of T. Verification / Alternative check: Impulse is also defined as change in momentum, while force is rate of change of momentum. Since rate involves division by time, force will have one extra factor of T in the denominator compared to momentum or impulse. That confirms that force and impulse cannot share the same dimensions, even though they are closely related via Newton laws. Why Other Options Are Wrong: Pressure and stress: Identical definitions in mechanics mean they share the same dimensions. Velocity and speed: Speed is simply the magnitude of velocity; their dimensions are the same. Work and energy: In physics, work done on a system is stored as energy, so they naturally have the same dimensional form. Common Pitfalls: A common error is to focus only on how quantities are used in formulas and not on the definitions. Students may think force and impulse must have the same dimensions because both appear with similar symbols in mechanics. The key is to remember that impulse involves time multiplied by force, while force involves time in the denominator when defined from momentum. Always break down quantities to base units before deciding. Final Answer: The pair that does not have the same dimensions is force and impulse.

Question 7

Why do many trained Indian classical vocalists prefer not to use a harmonium as an accompanying instrument during performance?

Accepted Answer

Introduction / Context: Indian classical music relies heavily on subtle pitch variations, microtones, and flexible intonation. Many serious vocalists are careful about the instruments that accompany them. The harmonium, although popular and convenient, is sometimes disliked by purists. This question explores the main musical reason behind that preference, relating it to scales and tuning systems rather than loudness or shrillness. Given Data / Assumptions: The context is Indian classical vocal music. The instrument in question is the harmonium. The concern is about musical accompaniment and tuning, not simply loudness. We assume knowledge of tempered and just intonation scales. Concept / Approach: Harmoniums used in India are generally tuned to the Western equal tempered scale. In this system, the octave is divided into twelve equal semitones. Indian classical music, however, traditionally uses a system of shrutis and just intonation where intervals between notes are based on simple frequency ratios. Equal temperament slightly compromises these pure ratios to make all keys usable. This mismatch can disturb the exact intonation that an expert vocalist wants to maintain, especially in slow, elaborated ragas. Step-by-Step Solution: Step 1: Recall that in equal tempered tuning, all semitone gaps are identical in terms of frequency ratio, which is convenient for Western harmony and key changes. Step 2: Remember that Indian classical ragas are built on specific scales where certain intervals are meant to be slightly higher or lower than the equal tempered versions, to match natural harmonic ratios and desired emotional colour. Step 3: A harmonium with fixed reeds cannot continuously adjust its pitch to match these subtle intonation demands. It will always play the equal tempered notes. Step 4: As a result, a vocalist who wants to use pure intonation may feel constrained or slightly pulled off key by the fixed tempered tuning of the harmonium. Verification / Alternative check: In many musicological discussions, scholars and senior artists mention that the harmonium is convenient for practice and teaching but is not ideal for capturing the microtonal nuances of certain ragas. String instruments like tanpura, sarangi, or even the human voice can continuously slide and adjust pitch, but harmonium reeds cannot. This supports the view that temperament, not shrillness or loudness, is the core issue. Why Other Options Are Wrong: Because a diatonic scale is used in the harmonium: While diatonic scales do exist, the actual problem is equal tempered tuning rather than the general idea of a diatonic scale. Because the notes of the harmonium are too shrill: Shrillness can be managed by playing softly or adjusting the instrument; it is not the main theoretical objection. Because the intensity of the harmonium notes is too large: Volume can be controlled by the player and is not a fundamental reason for rejection by classical purists. Common Pitfalls: Listeners sometimes think artists dislike the harmonium simply because it sounds loud or metallic. The deeper reason is intonation: the inability of fixed reeds to deliver microtonal shades with precision. Confusing tonal quality (timbre) and tuning systems is another common misunderstanding. When preparing for exams, focus on conceptual reasons like equal tempered scale rather than superficial issues like loudness. Final Answer: The main reason is that the harmonium uses the equal tempered scale, which does not match the subtle pitch requirements of Indian classical music. Therefore, the correct option is because a tempered scale is used in the harmonium.

Question 8

In fundamental physics, which of the following is considered the strongest of the four fundamental forces in the universe?

Accepted Answer

Introduction / Context: Nature is governed by four fundamental interactions or forces: gravitational, electromagnetic, strong nuclear, and weak nuclear. Each has a characteristic range and relative strength. This multiple choice question tests your knowledge of which of these forces is the strongest when particles are at very small separations, a key fact in particle physics and nuclear physics. Given Data / Assumptions: The four fundamental forces listed are gravity, electromagnetism, strong nuclear force, and weak nuclear force. We are comparing their relative strengths at typical subatomic distance scales. We ignore macroscopic effects like the long range nature of gravity. Concept / Approach: The strong nuclear force is responsible for binding protons and neutrons within atomic nuclei. It overcomes the strong repulsive electromagnetic force between positively charged protons at very short distances. The electromagnetic force is much stronger than gravity at atomic scales, and the weak force controls certain radioactive decays. Gravity is extremely weak at the level of elementary particles. Therefore, the strong nuclear force is considered the strongest among the four, though it acts only over very short ranges inside the nucleus. Step-by-Step Solution: Step 1: Recall that gravitational force is attractive between masses but extremely weak for subatomic particles. Step 2: Recognise that electromagnetic force acts between charges and is much stronger than gravity for protons and electrons, but still weaker than the strong nuclear force at nuclear distances. Step 3: Understand that the strong nuclear force binds nucleons together; it must be very strong to overcome proton proton repulsion. Step 4: Note that the weak nuclear force is responsible for beta decay and similar processes, and is weaker than the strong and electromagnetic forces. Step 5: Conclude that the strong nuclear force has the greatest relative strength at very small distances. Verification / Alternative check: Standard physics texts often quote relative strengths at particle scales, with the strong force taken as 1, electromagnetic force roughly 1/137, weak force much smaller, and gravity drastically smaller still. This numerical comparison reinforces the qualitative fact that the strong nuclear force dominates at subatomic ranges, even though its range is very short, about the size of a nucleus. Why Other Options Are Wrong: Gravitational force: Extremely weak between individual particles, significant only when huge masses like planets or stars are involved. Electromagnetic force: Stronger than gravity but weaker than the strong nuclear force at nuclear distances. Weak nuclear force: As the name suggests, it is weaker than the strong and electromagnetic forces and is important mainly for particle decay processes. Common Pitfalls: The name weak nuclear force and strong nuclear force can be confusing if one does not remember the context. Another mistake is to think gravity is strongest because it operates on large scales, but that is due to mass accumulation and infinite range rather than intrinsic strength at particle levels. Always distinguish between range and intrinsic coupling strength when comparing these forces. Final Answer: The strongest of the four fundamental forces is the strong nuclear force.

Question 9

Which one of the following liquids typically shows capillary fall rather than capillary rise in a clean glass tube?

Accepted Answer

Introduction / Context: Capillary action is a familiar phenomenon in daily life, observed when liquids rise or fall in thin tubes. Whether a liquid rises or falls depends on the interplay between cohesive forces within the liquid and adhesive forces between the liquid and the tube material. This question asks which liquid shows capillary fall, not rise, in a clean glass tube, which is a standard conceptual point in fluid mechanics. Given Data / Assumptions: The tube is made of clean glass. The liquids considered are common ones such as water, alcohol, mercury, and kerosene. We are focusing on the direction of capillary movement relative to the general liquid surface. Concept / Approach: If adhesive forces between the liquid and the tube are stronger than the cohesive forces within the liquid, the liquid wets the tube, forming a concave meniscus and rising in the capillary. If cohesive forces dominate, the liquid does not wet the tube, forms a convex meniscus, and experiences capillary depression or fall. Water and most organic liquids wet clean glass and rise. Mercury, however, does not wet glass and therefore shows capillary fall. Step-by-Step Solution: Step 1: Recall that water rises in a thin glass tube, forming a concave meniscus due to strong adhesive forces with glass. Step 2: Alcohols and kerosene also generally wet glass and show a rise rather than a fall in narrow tubes. Step 3: Mercury has strong cohesive forces among its own atoms and poor adhesion to glass, so it forms a convex meniscus and the level inside the capillary is lower than the outside level. Step 4: This lowering of the mercury column is called capillary depression or capillary fall. Verification / Alternative check: In standard physics experiments, students observe that a thin glass capillary dipped in water shows a column of water higher than the outer water level, whereas a capillary dipped in mercury shows the mercury level inside the tube lower than that in the reservoir. This direct observation supports the idea that mercury exhibits capillary fall in glass tubes. Why Other Options Are Wrong: Water: It wets glass strongly, leading to capillary rise and a concave meniscus. Alcohol: It also generally wets glass and tends to rise in capillary tubes. Kerosene: Like many organic liquids, it shows capillary rise in glass rather than fall. Common Pitfalls: A student may guess water because it is the liquid used in many capillary examples, but the question specifically asks about capillary fall, not rise. Another mistake is to assume that all liquids behave the same in glass tubes. Remember the key rule: if the liquid wets the surface, you get rise and a concave meniscus; if it does not wet the surface, you get fall and a convex meniscus. Final Answer: The liquid that shows capillary fall in a clean glass tube is mercury.

Question 10

A car parked in bright sunlight with all its windows closed becomes extremely hot inside. Which phenomenon mainly explains this effect?

Accepted Answer

Introduction / Context: Many people have experienced how hot the interior of a parked car can become on a sunny day, especially when all windows are closed. This is not mainly due to the engine but due to a physical process similar to the greenhouse effect. Understanding this everyday phenomenon helps link basic physics concepts about radiation and heat transfer to real life safety issues. Given Data / Assumptions: The car is parked in sunlight with its engine off. All windows are closed, and sunlight enters through the glass. We are concerned with the reason the interior air temperature rises significantly. Concept / Approach: Sunlight, especially visible and short wavelength solar radiation, passes easily through glass windows and is absorbed by the car interior surfaces such as seats and dashboard. These surfaces heat up and re radiate energy mainly as longer wavelength infrared radiation. Glass is relatively less transparent to this infrared radiation, so the heat tends to be trapped inside. This trapping of heat due to selective transmission and absorption is known as the greenhouse effect, analogous to what happens in a greenhouse or in the Earth atmosphere with certain gases. Step-by-Step Solution: Step 1: Sunlight enters the car through transparent windows and is absorbed by interior surfaces. Step 2: The absorbed energy raises the temperature of the seats, dashboard, and other interior elements. Step 3: These hot surfaces emit infrared radiation, which is partly blocked or reflected by the glass windows. Step 4: Because the heat gets trapped and cannot easily escape, the interior temperature climbs much higher than the outside temperature. Verification / Alternative check: You can verify this by comparing a car with windows slightly open to one with windows fully closed under the same conditions. The car with slight ventilation will usually be cooler, because hot air can escape. The strong effect of glass in trapping heat can be observed similarly in greenhouses where plants are grown in cold environments. The same physical principles apply, reinforcing that this is an example of the greenhouse effect. Why Other Options Are Wrong: Heat produced by the engine: In a parked car with the engine off, engine heat is not the main source; the major heating comes from solar radiation. Good conducting power of the metal body: Metal conduction distributes heat but does not by itself create the strong temperature rise inside. The double layered windscreen: This can influence insulation, but the key mechanism is selective transmission and trapping of radiation, which is the greenhouse effect. Common Pitfalls: People often think heat simply accumulates because the car is a closed space, ignoring the role of radiation and glass properties. Others may confuse conduction with radiation. Remember that the greenhouse effect involves short wavelength solar radiation entering easily and long wavelength infrared radiation being partially blocked, causing heat build up. This concept appears in many exam questions about climate change and everyday physics. Final Answer: The main reason a closed car becomes so hot in sunlight is due to the greenhouse effect.

Question 11

During a thunderstorm you are travelling in a car. What is the safest action to protect yourself from lightning?

Accepted Answer

Introduction / Context: Thunderstorms and lightning can be dangerous, and basic safety knowledge is important for everyone. Many people travel by car during such weather conditions. This question tests whether you understand why the interior of a closed car is usually a safe place during lightning, thanks to the physics of electric fields and conductors. Given Data / Assumptions: There is a thunderstorm with the possibility of lightning strikes. You are already inside a car when the storm occurs. The car has a metal body and closed windows, which is typical of most vehicles. Concept / Approach: A car with a conducting metal body acts as a kind of Faraday cage. When lightning strikes the car, the electric charge flows over the outer surface of the metal body and then to the ground, rather than passing through the interior passenger space. As long as the occupants remain inside, not touching metal parts connecting to the exterior, they are usually safe. The dangerous situations are standing in open ground, near tall objects like trees or poles, or touching conductors that can carry the lightning current through the body. Step-by-Step Solution: Step 1: Recall that lightning tends to strike tall or isolated objects and follow conductive paths to ground. Step 2: A car body made of metal provides a continuous conducting shell around the occupants. Step 3: If lightning strikes the car, current flows around the outer surface of the metal and to the ground, keeping the interior relatively safe. Step 4: Therefore, the safest action is to remain inside the car with windows closed and avoid touching exposed metal parts. Verification / Alternative check: Safety guidelines from meteorological and disaster management agencies often recommend that during lightning storms, one should stay inside fully enclosed vehicles rather than standing under trees or in open fields. The idea of the vehicle acting as a protective cage is well established and used as a standard precaution in many countries. Why Other Options Are Wrong: Get out of the car and lie flat on the ground: Lying flat can increase the risk of ground current passing through the body and is not recommended when a safer shelter is already available. Abandon the car and take shelter under a nearby tree: Trees are often struck by lightning, and standing under one is dangerous due to side flashes and ground currents. Touch the nearest electric pole: Electric poles are tall conductors and can channel lightning, making them very unsafe contact points. Common Pitfalls: Many people wrongly believe that staying inside a car is dangerous simply because it has metal. The key point is that the metal shell protects the interior by directing current around it. Another mistake is to assume that lying directly on wet ground is safer than staying in a car, which is not true in this situation. Always seek a proper enclosed shelter during thunderstorms. Final Answer: The safest action during a thunderstorm when you are already in a car is to remain inside the closed car.

Question 12

An atom bomb (fission bomb) works mainly on which nuclear principle?

Accepted Answer

Introduction / Context: Modern nuclear weapons and reactors rely on nuclear reactions to release very large amounts of energy. There is an important distinction between fission bombs, commonly called atom bombs, and fusion bombs, often called hydrogen bombs. This question asks about the basic principle behind an atom bomb, which is a key concept in nuclear physics and general knowledge. Given Data / Assumptions: The device mentioned is an atom bomb, understood as a fission based weapon. Typical fissile materials are uranium 235 or plutonium 239. We consider the basic underlying nuclear principle, not the engineering details. Concept / Approach: Nuclear fission is the process in which a heavy nucleus splits into smaller nuclei, neutrons, and energy. In a chain reaction, neutrons from one fission event trigger further fissions. If this chain reaction is uncontrolled and grows very rapidly, it releases an enormous amount of energy in a very short time, which is the working principle of an atom bomb. In contrast, a nuclear reactor uses a controlled chain reaction where the rate is carefully limited to produce steady power. Fusion weapons are based on nuclear fusion, which is different again. Step-by-Step Solution: Step 1: Recognise that atom bombs historically used uranium 235 or plutonium 239 undergoing fission. Step 2: In fission, a heavy nucleus splits into lighter nuclei and more neutrons, releasing large amounts of energy according to E = m c^2. Step 3: If the emitted neutrons are not absorbed or controlled, they cause further fissions, leading to a rapidly multiplying chain reaction. Step 4: In an atom bomb, this chain reaction is deliberately made uncontrolled, so the energy is released almost instantaneously, causing an explosion. Verification / Alternative check: Descriptions of the first nuclear weapons, such as those used in the Second World War, clearly state that they relied on a supercritical mass of fissile material and a rapidly growing fission chain reaction. Nuclear power plants, on the other hand, use control rods and moderators to keep the reaction rate stable, showing the contrast between controlled and uncontrolled fission. Why Other Options Are Wrong: Nuclear fusion of light nuclei: This is the principle of hydrogen bombs and stellar energy, not typical atom bombs. Controlled chain reaction in a nuclear reactor: This describes nuclear power plants, not weapons. Purely chemical explosive reaction: Chemical explosives release much less energy per unit mass than nuclear reactions and are not the main mechanism of an atom bomb. Common Pitfalls: A frequent confusion is to mix up fission and fusion. An easy way to remember is that fission means splitting heavy nuclei, while fusion means joining light nuclei like hydrogen isotopes. Another mistake is to think that all nuclear devices use the same principle. For exams, always connect atom bomb with uncontrolled fission chain reaction and power reactor with controlled fission. Final Answer: An atom bomb works mainly on nuclear fission with an uncontrolled chain reaction.

Question 13

Lightning can heat the surrounding air to approximately what maximum temperature?

Accepted Answer

Introduction / Context: Lightning is a powerful natural electrical discharge that produces intense brightness and very high temperatures. Knowing the approximate temperature of air around a lightning channel helps us appreciate how energetic these events are and why they can cause fires and severe damage. This question asks you to select the most realistic value among the given options. Given Data / Assumptions: The question refers to the temperature of air heated by a lightning bolt. We are asked for an approximate maximum temperature in degree Celsius. The values range from hundreds to hundreds of thousands of degrees. Concept / Approach: Measurements and estimates of lightning channels show that the air can be heated to a temperature many times higher than the surface of the Sun. Commonly cited values are around 30000 degree Celsius or more. This extreme heating causes rapid expansion of air, which we hear as thunder. Lower values like 300 or 3000 degree Celsius underestimate the intensity, while 300000 degree Celsius is much larger than typical estimates used in standard references and school level texts. Step-by-Step Solution: Step 1: Recall that the surface temperature of the Sun is roughly 5500 degree Celsius. Step 2: Lightning is often described as being about five to six times hotter than the Sun surface, suggesting a range near 30000 degree Celsius. Step 3: Compare the options: 300 and 3000 degree Celsius are too low to match descriptions of lightning temperature. Step 4: A value like 300000 degree Celsius would be far higher than typical textbook values and is not generally quoted in basic physics or geography references. Step 5: Therefore, 30000 degree Celsius is the most appropriate estimate. Verification / Alternative check: Popular science materials and meteorology sources usually state that lightning can heat the surrounding air to about 30000 Kelvin or degree Celsius, which is roughly five times hotter than the Sun surface. This widely accepted figure confirms that 30000 degree Celsius is the standard approximate value used in educational contexts. Why Other Options Are Wrong: 300 degree Celsius: This is comparable to very hot oven temperatures and cannot explain the explosive nature of thunder. 3000 degree Celsius: Still far below commonly quoted lightning temperatures and not consistent with comparisons to the Sun. 300000 degree Celsius: This is unrealistically high for school level approximations and not usually reported in standard references. Common Pitfalls: Students may be tempted to pick extremely large numbers, assuming that bigger means more impressive. However, exam questions often use values that match widely known approximate figures. Another pitfall is to confuse Kelvin and degree Celsius units; at these very high temperatures, numerically they are almost the same, because the difference of 273 is relatively small compared to 30000. Final Answer: Lightning can heat the surrounding air to about 30000 degree Celsius.

Question 14

The absorption of ink into blotting paper is mainly due to which physical phenomenon?

Accepted Answer

Introduction / Context: Blotting paper is commonly used to soak up excess ink from writing. The way ink quickly spreads into the paper is a visible example of a microscopic physical process. This question asks which phenomenon is primarily responsible for ink absorption into blotting paper, which connects the idea of capillarity with everyday observations. Given Data / Assumptions: Blotting paper is a porous material with many tiny pores or capillaries. Ink is a liquid that can flow into these pores. We are asked to identify the dominant physical phenomenon that draws ink into the paper. Concept / Approach: Capillary action occurs when liquid rises or moves in narrow tubes or pores due to the balance between cohesive forces within the liquid and adhesive forces between the liquid and the solid surface. Blotting paper has a network of microscopic capillaries that draw liquid ink into them. While viscosity, diffusion, and other effects are present, the main cause of the rapid, upward and lateral movement of ink in the paper is capillary action, not siphon action or simple diffusion alone. Step-by-Step Solution: Step 1: Observe that ink does not just slowly spread by random motion; it quickly rises and spreads through the pores, even against gravity to some extent. Step 2: Blotting paper is designed to be highly porous, providing many narrow pathways similar to capillary tubes. Step 3: Capillary action in these tiny pores pulls the ink into the paper due to adhesive attraction between ink and fibre and cohesive forces within the liquid. Step 4: Therefore, the dominant phenomenon is capillary action, which explains both the speed and direction of the ink movement. Verification / Alternative check: Experiments in school often show that when the lower edge of blotting paper is dipped into coloured water, the liquid climbs upward through the paper. This is similar to the capillary rise of liquids in narrow tubes. The same behaviour is visible when a paper towel soaks up water from a surface. These examples are always explained using capillary action, confirming that this is the relevant phenomenon here as well. Why Other Options Are Wrong: Viscosity of ink: Viscosity describes resistance to flow; high viscosity would actually slow down the movement rather than cause it. Diffusion of ink through the paper: Diffusion is random motion of molecules and is a much slower process, not responsible for the rapid rise of ink against gravity. Siphon action: A siphon involves a continuous tube and pressure difference between two reservoirs; this does not match the situation inside blotting paper. Common Pitfalls: Students sometimes confuse any spreading of liquid with diffusion and any movement through a tube with siphon action. The key recognition is that capillary action can pull liquid upward against gravity in very narrow spaces due to surface tension and adhesion. Whenever you see liquid rising in narrow pores or thin tubes, think of capillary action first. Final Answer: Ink is absorbed into blotting paper mainly due to the capillary action phenomenon.

Question 15

A ball is thrown vertically upward from the ground and passes a point 25 m above the ground twice, the two crossings being 4 s apart. What was the initial speed of projection? (Take g ≈ 10 m/s^2.)

Accepted Answer

Introduction / Context: Vertical motion under gravity is a standard topic in kinematics. When a ball is thrown straight up, it passes the same height twice, once while going up and once while coming down. The time gap between these two events can be related to the initial speed. This question uses that idea to find the initial speed of projection from the given height and time interval. Given Data / Assumptions: Initial velocity of the ball is u (upwards). Acceleration due to gravity g is taken as 10 m/s^2 downward. The ball passes height h = 25 m above the ground twice. The time between the two crossings of height 25 m is 4 s. Motion is vertical with no air resistance. Concept / Approach: For vertical motion with constant acceleration, the displacement y at time t is given by y = u t - (1/2) g t^2 when upward is taken as positive. When the ball passes a given height h twice, the corresponding times t1 and t2 are roots of a quadratic equation. For such a quadratic, the sum and difference of roots have simple relations. The time gap t2 - t1 depends only on u, h, and g. By using this relation along with the known interval, we can solve for u without first finding t1 or t2 individually. Step-by-Step Solution: Step 1: Write the equation for height h = 25 m: 25 = u t - (1/2) g t^2. Step 2: Rearrange to standard form: (1/2) g t^2 - u t + 25 = 0. With g = 10 m/s^2, this becomes 5 t^2 - u t + 25 = 0. Step 3: The two roots of this quadratic are t1 and t2, the times when the ball is at 25 m. For a quadratic a t^2 + b t + c = 0, the difference between roots is t2 - t1 = sqrt(b^2 - 4 a c) / a. Step 4: Here, a = 5, b = -u, and c = 25. The time interval between the two crossings is given as 4 s, so 4 = sqrt(b^2 - 4 a c) / a. Step 5: Substitute a and c: 4 = sqrt(u^2 - 4 * 5 * 25) / 5 = sqrt(u^2 - 500) / 5. Step 6: Multiply both sides by 5: 20 = sqrt(u^2 - 500). Square both sides: 400 = u^2 - 500, so u^2 = 900. Step 7: Therefore u = 30 m/s, taking the positive root because the ball is thrown upward. Verification / Alternative check: We can verify quickly by computing the times with u = 30 m/s. Use 25 = 30 t - 5 t^2. Rewrite as 5 t^2 - 30 t + 25 = 0. Solving gives t = [30 ± sqrt(900 - 500)] / 10 = [30 ± sqrt(400)] / 10 = (30 ± 20) / 10, so t1 = 1 s and t2 = 5 s. The time gap is 5 - 1 = 4 s, which matches the given condition. This confirms that u = 30 m/s is correct. Why Other Options Are Wrong: 18 m/s: For this speed, the maximum height itself would be too small to pass 25 m twice, so no real solution exists. 25 m/s: Solving the quadratic with this value leads to a different time gap than 4 s. 36 m/s: This would produce a time interval greater than 4 s between the two crossings of 25 m. Common Pitfalls: One common mistake is to treat 4 s as the total time of flight rather than the interval between two specific heights. Another is to try to find t1 and t2 separately without using the quadratic root relations, which can lead to algebra errors. Remember that when a projectile passes the same height twice, using the sum and difference of roots is an efficient and elegant method. Final Answer: The initial speed of projection was 30 m/s.

Question 16

Why does a typical female human voice sound shriller and higher pitched than a typical male voice?

Accepted Answer

Introduction / Context: Everyday experience tells us that most adult female voices sound higher pitched or more shrill compared to most adult male voices. In physics, pitch is directly related to the frequency of sound waves. This question connects basic wave concepts to human speech and vocal physiology, asking you to identify the correct physical reason for the difference in perceived pitch. Given Data / Assumptions: We are comparing typical adult male and adult female speaking voices. The word shrill implies a higher perceived pitch. We focus on the main physical factor rather than cultural or training effects. Concept / Approach: Pitch of a sound is primarily determined by its frequency. Higher frequency waves are heard as higher pitch, while lower frequency waves are heard as deeper or more bass like. The human voice is produced by vibration of the vocal folds in the larynx. Female vocal folds are generally shorter and thinner than male vocal folds, which allows them to vibrate at higher frequencies. As a result, the sound waves produced have higher frequency, and the voice is perceived as sharper or more shrill. Step-by-Step Solution: Step 1: Recall that pitch and frequency are closely related; greater frequency implies higher pitch. Step 2: Consider that physical properties of vocal folds such as length, mass, and tension influence their vibration frequency. Step 3: In general, adult female vocal folds are smaller and tend to vibrate faster for the same tension, producing higher frequency sound. Step 4: Therefore, a typical female voice has higher fundamental frequency than a typical male voice and is perceived as more shrill. Verification / Alternative check: Measured average speaking frequencies support this explanation. Male voices often have fundamental frequencies around 100 to 150 hertz, while female voices are commonly around 180 to 250 hertz or higher. These values confirm that female voices have higher frequency sound waves, matching the perception of higher pitch. Amplitude, by contrast, relates to loudness rather than pitch. Why Other Options Are Wrong: Because it has lower frequency: Lower frequency would produce a deeper voice, not a shrill one. Because it has greater amplitude only: Amplitude affects loudness, not pitch; a loud voice can be high or low in pitch. Because of vocal training alone: Training can modify quality and control, but the basic pitch difference comes from anatomy and frequency, not training by itself. Common Pitfalls: A common confusion is between loudness and pitch. Many students think a louder sound is always shriller, which is not true. Another misunderstanding is attributing vocal differences mainly to practice or style. While training influences timbre and range, the fundamental pitch characteristics are rooted in the physical properties of the vocal folds and hence in frequency differences. Final Answer: A typical female voice sounds shriller because it has higher frequency than a typical male voice.

Question 17

If a spacecraft could travel through space at the speed of light in vacuum, how long would it take to cover a distance of 4 light years?

Accepted Answer

Introduction / Context: The term light year is widely used in astronomy to describe large distances between stars and galaxies. Many learners confuse it with a unit of time because of the word year. In fact, it is a unit of distance defined in terms of how far light travels in one year. This question tests whether you can correctly relate distance measured in light years to travel time when moving at the speed of light. Given Data / Assumptions: One light year is the distance that light travels in vacuum in one year. The spacecraft is assumed to travel at the speed of light in vacuum. The distance to be covered is 4 light years. Relativistic complications are ignored; we use the simple definition. Concept / Approach: By definition, if an object moves at the speed of light, it covers one light year in one year of time. Therefore, to cover 4 light years at the same speed, it would take four times as long, that is, four years. The important idea is that a light year is a measure of distance, not time, and that distance divided by speed gives the travel time. Step-by-Step Solution: Step 1: Recall the definition: distance of 1 light year = distance travelled by light in 1 year at speed c in vacuum. Step 2: If a spacecraft also travels with speed equal to c, it will cover 1 light year in 1 year of travel time. Step 3: The required distance is 4 light years, which is four times the distance for 1 light year. Step 4: Time taken = distance / speed. When distance is quadrupled and speed is unchanged, time is also quadrupled, so time = 4 years. Verification / Alternative check: You can think of this like everyday motion with simpler units. If a car covers 1 kilometre every minute, then to cover 4 kilometres at the same speed, it needs 4 minutes. The same proportional reasoning applies for light years and years, just with much larger scales. The math is simple scaling with distance proportional to time for constant speed. Why Other Options Are Wrong: 4 seconds: This would mean that light travels 1 light year in 1 second, which contradicts the definition. 4 minutes: Similar reasoning; the unit year in light year would lose its meaning. 40 years: This would correspond to travelling at one tenth of light speed, not at the full speed of light, so it does not match the given condition. Common Pitfalls: Students often treat light year as a time interval rather than a distance. Another confusion arises when they ignore the definition and try to overcomplicate the problem. The key is to remember that if you move at the same speed as the reference (light) and cover a multiple of the light year distance, the time in years scales in the same way. Final Answer: At the speed of light, it would take 4 years to travel 4 light years.

Question 18

From the point of view of chemistry, how is air best described?

Accepted Answer

Introduction / Context: Air is all around us and is essential for life, but its chemical nature is often misunderstood. In chemistry and environmental science, it is important to classify substances correctly as elements, compounds, or mixtures. This question checks your understanding of how air is categorised based on its composition. Given Data / Assumptions: Air refers to the Earth atmosphere near the surface. It contains gases like nitrogen, oxygen, argon, carbon dioxide, and small amounts of others. We are describing air at normal temperature and pressure. Concept / Approach: An element is a substance made of only one type of atom. A compound has a fixed chemical formula and definite proportions of elements chemically bonded together. A mixture contains two or more substances combined physically, not chemically, with variable composition. Air clearly contains nitrogen, oxygen, argon, carbon dioxide, water vapour, and traces of many other gases, and their proportions can vary from place to place and time to time. Thus air is a mixture of gases rather than a pure element or compound. Step-by-Step Solution: Step 1: List major components of dry air: about 78 percent nitrogen, 21 percent oxygen, around 1 percent argon, and small fractions of carbon dioxide and other gases. Step 2: Note that water vapour content in air varies widely, from very dry desert air to very humid tropical air. Step 3: Because the components are not chemically bonded in a fixed ratio and can be separated by physical methods such as fractional distillation of liquefied air, air is not a compound. Step 4: Since air clearly contains several different chemical species, it cannot be a single element. Step 5: Therefore, air fits the definition of a mixture, specifically a mixture of several gases. Verification / Alternative check: Industrial processes for separating air into its components rely on the fact that air is a mixture. For example, oxygen plants cool and liquefy air, then separate nitrogen, oxygen, and argon by fractional distillation. This would not be possible if air were a single pure compound with fixed composition and chemical bonds between nitrogen and oxygen molecules. Why Other Options Are Wrong: A pure compound: A compound has a fixed chemical formula such as H2O or CO2, which air does not have. A single element: Air has multiple elements present (nitrogen, oxygen, argon, etc.), so it cannot be a single element. A liquid under normal conditions: At standard temperature and pressure, air is a gas, not a liquid. Common Pitfalls: One mistake is to associate the word air with oxygen alone, because we breathe it, and then to call it an element. Another is to think that because air has a fairly constant average composition, it must be a compound. Remember that mixtures can also have relatively stable average compositions while still varying locally and being separable by physical means. Final Answer: Chemically, air is best described as a mixture of several gases.

Question 19

A body of mass m moves in a vertical circular loop of radius R. To just complete the loop, what should be the minimum value of v^2 at the lowest point in terms of g and R?

Accepted Answer

Introduction / Context: The motion of a body in a vertical circular loop is a classic problem in mechanics. It combines concepts of energy conservation and circular motion. To just complete the loop without losing contact at the top, the body must have a certain minimum speed. This question asks for the minimum value of v^2 at the lowest point expressed in terms of g and R, where R is the radius of the loop. Given Data / Assumptions: Mass of the body is m. The body moves in a vertical circle of radius R. g is the acceleration due to gravity. We are interested in the minimum speed at the lowest point so that the body just maintains contact at the top of the loop. Air resistance is neglected and the track is rigid and frictionless. Concept / Approach: Two conditions are used. First, at the top of the loop, the normal reaction must not become negative, so the centripetal force requirement is m v_top^2 / R ≥ m g, giving a minimum v_top^2 = g R. Second, mechanical energy is conserved between the bottom and the top of the loop. The loss in kinetic energy equals the gain in potential energy when the body rises by 2 R from bottom to top. Using these two ideas together, we relate v_bottom^2 to v_top^2 and then find the required expression for v_bottom^2, which is the v^2 asked in the question. Step-by-Step Solution: Step 1: Apply the condition at the top of the loop. At minimum speed, centripetal force is provided entirely by weight, so m v_top^2 / R = m g. Step 2: Simplify this to obtain v_top^2 = g R. Step 3: Use conservation of mechanical energy between bottom and top. At bottom: kinetic energy is (1/2) m v_bottom^2 and potential energy can be taken as zero reference. At top: kinetic energy is (1/2) m v_top^2 and potential energy is m g (2 R). Step 4: Write energy conservation: (1/2) m v_bottom^2 = (1/2) m v_top^2 + m g (2 R). Step 5: Cancel m from all terms and multiply the entire equation by 2: v_bottom^2 = v_top^2 + 4 g R. Step 6: Substitute v_top^2 = g R from Step 2. Then v_bottom^2 = g R + 4 g R = 5 g R. Step 7: Therefore, to just complete the loop, the minimum value of v^2 at the lowest point must be 5 g R. Verification / Alternative check: Many standard textbooks present this result as a key formula: minimum speed at lowest point v_bottom = sqrt(5 g R). Squaring both sides gives v_bottom^2 = 5 g R, which matches what we obtained by detailed energy and force analysis. This consistency with known results is a strong verification of the calculation. Why Other Options Are Wrong: 0: A zero speed at the bottom obviously would not move the body at all, so the loop would not be completed. gR: This corresponds to v_top^2, the minimum speed squared at the top, not at the bottom. 2gR: This value is too low; using it would not satisfy both energy conservation and the contact condition at the top simultaneously. Common Pitfalls: A typical mistake is to forget that the body rises by a height of 2 R from bottom to top, using R instead of 2 R in potential energy. Another error is either ignoring the contact condition at the top or using it incorrectly. Always use both the centripetal force condition at the top and energy conservation between bottom and top to derive the correct expression for v_bottom^2. Final Answer: The required minimum value of v^2 at the lowest point is 5gR.

Question 20

Which of the following phase changes is an example of an exothermic process where a substance releases heat to its surroundings?

Accepted Answer

Introduction / Context: Phase changes involve energy exchange between a substance and its surroundings. Some processes absorb heat (endothermic), while others release heat (exothermic). Understanding which changes are exothermic is important in chemistry, meteorology, and everyday situations like freezing water or condensation on a cold glass. This question asks you to identify which listed phase changes are exothermic. Given Data / Assumptions: We consider phase changes at constant pressure. The listed changes include liquid to solid, gas to solid, and gas to liquid. We are asked which involve release of heat to the surroundings. Concept / Approach: When a substance moves from a higher energy phase to a lower energy phase, it releases latent heat and the process is exothermic. Gas has more energy than liquid, and liquid has more energy than solid for the same substance at a given temperature. Therefore, gas to liquid and liquid to solid are exothermic. Gas to solid is also exothermic because it skips the intermediate liquid stage but still moves from high energy gas to low energy solid. All three changes listed move to a lower energy phase and thus release heat. Step-by-Step Solution: Step 1: Liquid to solid (freezing): When water freezes into ice, it releases latent heat to the surroundings, so freezing is exothermic. Step 2: Gas to liquid (condensation): When water vapour condenses into liquid water, it releases latent heat, warming the surrounding air; thus condensation is exothermic. Step 3: Gas to solid (deposition): Examples include frost formation when water vapour converts directly to ice on cold surfaces. This also releases heat to the surroundings. Step 4: Since all individual processes listed are exothermic, the correct combined choice is that all of them are examples of exothermic changes. Verification / Alternative check: You can remember that melting, evaporation, and sublimation (solid to gas) are endothermic, as they require adding heat to move to a higher energy phase. The reverse processes, such as freezing, condensation, and deposition, are exothermic because the substance gives up energy. Weather forecasts often mention latent heat release during condensation and cloud formation, confirming that these phase changes warm the surroundings. Why Other Options Are Wrong: Liquid to solid (freezing) only: While this is exothermic, it is not the only exothermic process listed. Gas to solid (deposition) only: This is also exothermic but does not include the other correct options. Gas to liquid (condensation) only: Condensation is exothermic, but again, the other listed processes share this property. Common Pitfalls: A common confusion is to assume that gas to solid must be endothermic because it sounds like two steps combined. In reality, deposition is just the reverse of sublimation and is exothermic. Another pitfall is to memorise only one or two examples without understanding the general rule that moving to a more ordered, lower energy phase releases heat. Final Answer: All the listed phase changes involve release of heat to the surroundings, so the correct choice is All of the above.

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