Question 1

On a boat, where should a magnetic compass ideally be mounted so that it gives the most accurate and reliable readings?

Accepted Answer

Introduction / Context: This question tests practical understanding of how a magnetic compass works and what can disturb its readings. A compass needle aligns itself with the Earth magnetic field. On a boat, local magnetic fields produced by magnets, large iron masses or electric currents can distort the Earth field near the compass, causing errors in the indicated direction. Mariners must therefore choose the mounting location carefully to minimise such interference, which is a common topic in navigation related general knowledge. Given Data / Assumptions: The instrument is a magnetic compass used for navigation. The environment is a boat, which may contain magnets, iron structures and electrical wiring. We need to choose the best place to mount the compass for accurate readings. Options include being near a magnet, near iron, away from wiring or all of these. Concept / Approach: A magnetic compass works by aligning with the resultant magnetic field at its location. Ideally, this should be almost entirely the Earth natural magnetic field. Any strong local fields from nearby magnets, current carrying wires or large ferromagnetic objects such as iron parts will superimpose on the Earth field and deflect the compass needle away from the true direction. Therefore, the compass should be placed as far as practical from these sources of interference. Electrical wiring carrying current produces a magnetic field around the wire, magnets produce static fields and masses of iron can distort field lines. The correct installation principle is to keep the compass away from such influences. Step-by-Step Solution: Step 1: Recall that the compass needle is a small magnet that aligns with the local magnetic field. Step 2: Recognise that the local field near the compass is the vector sum of the Earth magnetic field and any artificial magnetic fields from nearby sources. Step 3: Understand that magnets and large iron objects can greatly distort the field, leading to incorrect readings. Step 4: Note that electrical wiring carrying current generates circular magnetic fields around the wires, which can also disturb the compass. Step 5: Therefore, the best location for mounting the compass is away from electrical wiring and away from magnets and large iron masses, as summarised by the option mentioning distance from electrical wiring and magnetic materials. Verification / Alternative check: In practical navigation, boat builders often mount compasses on dedicated pedestals or stands that are positioned away from engines, electrical panels, loudspeakers and steel structures. If a compass is placed near a metal bulkhead or a powerful magnet such as a loudspeaker magnet, the indicated north direction deviates noticeably from true magnetic north. Professional practice thus confirms that the compass must be isolated as much as possible from nearby magnetic disturbances, reinforcing the choice of the option that emphasizes being away from electrical wiring and magnetic materials. Why Other Options Are Wrong: Mounting the compass near a magnet is the worst choice, since the nearby magnet strongly influences the needle and can almost completely override the Earth field. Mounting the compass near large iron objects allows those objects to distort and concentrate magnetic field lines, again leading to inaccurate direction readings. Saying that all positions are equally suitable ignores the well known fact that local magnetic fields from wiring, magnets and iron seriously affect compass accuracy. Common Pitfalls: Some learners may misinterpret the question and assume that because a compass relies on magnetism, placing it near a magnet will somehow strengthen its function, when in reality this destroys its ability to sense the Earth field correctly. Others may overlook the effect of electric currents on magnetic fields and fail to recognise wiring as a source of disturbance. To avoid such misunderstandings, always remember that a magnetic compass needs a clean magnetic environment and should be kept away from magnets, large iron masses and current carrying conductors. Final Answer: A boat compass should ideally be mounted away from electrical wiring and magnetic materials so that the needle senses mainly the Earth magnetic field and gives accurate readings.

Question 2

In atmospheric electricity, the sound that is heard following a bright flash of lightning is called what?

Accepted Answer

Introduction / Context: When a lightning flash occurs during a thunderstorm, we usually see the bright flash first and then hear a loud sound after a short delay. This common natural event provides a classic example used in physics to explain the behaviour of light and sound. The question checks whether you know the correct term for the sound that follows lightning and understand the basic idea of how it is produced in the atmosphere. Given Data / Assumptions: A flash of lightning occurs in the sky during a thunderstorm. After the flash, a loud sound is heard from the clouds. We assume standard atmospheric conditions, where sound travels much more slowly than light. The question asks for the correct name of this sound phenomenon. Concept / Approach: Lightning is a sudden and powerful electric discharge between clouds or between a cloud and the ground. This discharge rapidly heats the surrounding air to an extremely high temperature. The hot air expands explosively and then contracts, producing pressure waves that travel through the atmosphere. These pressure waves are sound waves, and the term used in physics and meteorology for this sound associated with lightning is “thunder.” The idea is to connect the visual flash with the correct acoustic term. Step-by-Step Solution: Step 1: Recognise that the bright flash in a thunderstorm is called lightning, which is an electrical discharge. Step 2: Understand that the sound that follows lightning is due to the rapid expansion and contraction of air, which generates sound waves. Step 3: Recall that the standard term used in daily life and science for this loud sound is “thunder.” Step 4: Compare with the options: “Stoning,” “Bolting,” and “Cloud clash” are not standard scientific or commonly used terms for this sound. Step 5: Conclude that “Thunder” is the only correct and accepted term for the sound following a flash of lightning. Verification / Alternative check: Textbooks, dictionaries and weather reports consistently pair the words “lightning and thunder” when describing storms. There is no recognised physical phenomenon called “stoning,” “bolting” or “cloud clash” in this context. Also, the separation in time between lightning and thunder is explained by the fact that light travels almost instantaneously to our eyes, whereas sound travels at roughly 340 m/s and reaches us later. This agreement between common language and scientific explanation confirms the correctness of “thunder.” Why Other Options Are Wrong: Stoning: This word does not refer to any atmospheric sound related to lightning in physics or meteorology. Bolting: Sometimes people say “bolt of lightning,” but “bolting” is not the name of the sound itself. Cloud clash: This phrase is informal and not used as a technical or standard term for the sound after lightning. Common Pitfalls: A few learners may confuse the idea of the lightning “bolt” with the sound and think that the sound is also called a bolt. Others might get distracted by invented phrases like “cloud clash.” However, in both everyday language and physics, the correct term is “thunder.” Remember the common pair “lightning and thunder” whenever you encounter similar questions. Final Answer: The sound that follows a flash of lightning is called Thunder.

Question 3

A piece of ice is floating in water in a beaker. When all of the ice has completely melted, what will happen to the level of the water in the beaker?

Accepted Answer

Introduction / Context: This question is a classic conceptual problem in physics that tests understanding of Archimedes principle, buoyancy and the behaviour of floating bodies. It asks what happens to the water level when a piece of ice floating in water melts completely. Such questions are very popular in competitive exams because they appear simple but require clear reasoning about forces and volumes rather than guesswork. Given Data / Assumptions: A beaker contains pure water. A piece of ice is floating freely in this water. The ice melts completely into water. We ignore effects like evaporation and assume the system is closed. We assume normal conditions, with the density of ice less than the density of liquid water. Concept / Approach: Archimedes principle states that a floating body displaces a volume of liquid whose weight is equal to the weight of the body. While the ice is still solid and floating, it displaces a certain volume of water equal in weight to the ice. When the ice melts, it turns into the same mass of water. Because the density of water is higher than that of ice, the melted water occupies exactly the same volume as the water that was displaced by the floating ice. Therefore, the total water level in the beaker does not change when the ice melts. Step-by-Step Solution: Step 1: While the ice is floating, it displaces water of weight equal to its own weight (Archimedes principle). Step 2: Let the mass of the ice be m and the density of water be rho. Then the volume of water displaced is m / rho. Step 3: When the ice melts, its mass remains m but it becomes liquid water with density rho, so the volume of the melted water is also m / rho. Step 4: Therefore, the volume of water formed from the melted ice is equal to the volume of water that had been displaced earlier while the ice was floating. Step 5: Since the volume before and after melting is the same, the water level in the beaker remains unchanged. Verification / Alternative check: We can imagine marking the initial water level on the beaker. At this stage, the floating ice displaces a volume of water, which contributes to the total level. After the ice melts, no solid remains above the surface, but the melted water exactly fills the volume that had been effectively present as displaced water earlier. Many physics textbooks illustrate this by considering numerical values and confirming that the volume calculations match. This cross-check confirms that no net change in water level occurs. Why Other Options Are Wrong: First rise and then fall: There is no physical mechanism for a temporary rise and subsequent fall in the water level in this simple closed system. Fall: The level does not fall because the melted water continues to occupy the same total volume that was previously displaced. Rise: The level does not rise because there is no additional mass of water being added; the ice simply changes its state from solid to liquid. Common Pitfalls: A common mistake is to think that the ice, being larger in volume, must cause the water level to drop when it melts, or that more water appears and raises the level. These confusions occur when students ignore the displacement idea and focus only on visual size. Always remember that for a floating object, the weight of displaced liquid equals the weight of the object. When the object melts into the same liquid, the melted volume exactly replaces the displaced volume, so the level remains the same. Final Answer: When all of the floating ice melts, the water level in the beaker will remain the same.

Question 4

The inability of a body to change its state of rest or of uniform motion in a straight line by itself is called which physical property?

Accepted Answer

Introduction / Context: This question is based on Newtonian mechanics and tests your understanding of a very fundamental concept: why objects tend to remain at rest or in uniform straight-line motion unless acted upon by an external force. The property that describes this tendency is central to Newton's first law of motion and appears frequently in school physics and competitive exams. Given Data / Assumptions: The body may either be at rest or moving with uniform velocity in a straight line. The phrase “inability to change its state” means the body cannot change this state on its own. No external unbalanced force is acting on the body. We must choose the correct name for this property from the options. Concept / Approach: Newton's first law states that a body continues to be in its state of rest or of uniform motion in a straight line unless compelled to change that state by an external unbalanced force. The tendency of a body to resist any change in its state of motion is called inertia. Momentum, velocity and acceleration are all related quantities in mechanics, but they do not describe this “resistance to change” property itself. The correct approach is to recall that inertia is defined as the natural tendency of objects to oppose changes in their state of motion. Step-by-Step Solution: Step 1: Identify the key phrase in the question: “inability of a body to change its state of rest or uniform motion.” Step 2: Recall Newton's first law of motion, which explains that this behaviour is due to inertia. Step 3: Compare the definitions of the options: momentum (mass * velocity), velocity (rate of change of displacement), acceleration (rate of change of velocity). Step 4: Notice that none of these quantities expresses resistance to change of motion; they are measures of motion, not of reluctance to change. Step 5: Conclude that the correct term for this property is inertia. Verification / Alternative check: Textbooks usually define inertia as “the property of a body by virtue of which it cannot change its state of rest or of uniform motion in a straight line by itself.” This wording closely matches the language in the question. Furthermore, examples such as passengers lurching forward when a moving bus stops or backward when it starts are all explained using inertia. Momentum, velocity and acceleration appear in separate formulas and laws (like p = m * v or a = F / m) but do not stand for the property that resists changes in motion. Why Other Options Are Wrong: Momentum: This is the product of mass and velocity and measures the quantity of motion, not resistance to change of motion. Velocity: This is the rate of change of displacement, describing how fast and in which direction a body moves, not the inability to change state. Acceleration: This is the rate of change of velocity and is itself a measure of change, not resistance to change. Common Pitfalls: Some students confuse inertia with momentum because both involve mass, or believe that a high velocity means a body “resists change” more. Although a body with higher mass will indeed have more inertia, the correct term for the property itself remains inertia, while momentum is a separate quantity. Carefully distinguishing these definitions is essential for solving conceptual mechanics questions accurately. Final Answer: The inability of a body to change its state of rest or uniform motion in a straight line by itself is called Inertia.

Question 5

In classical astronomy, Kepler's laws are mainly related to which of the following physical phenomena?

Accepted Answer

Introduction / Context: Johannes Kepler was a German astronomer who formulated three famous empirical laws that describe how planets move around the Sun. These laws provided a precise mathematical description of planetary orbits long before Newton explained the underlying gravitational cause. This question checks whether you can correctly connect Kepler's laws with the physical phenomenon they describe. Given Data / Assumptions: The phrase “Law of Kepler” in the question refers to Kepler's three laws. We must decide whether these laws describe planetary motion, conservation of energy, gravitation in general or something else. We assume a heliocentric solar system model, with planets moving around the Sun. Concept / Approach: Kepler's three laws can be summarised as: (1) planets move in elliptical orbits with the Sun at one focus, (2) the line joining a planet and the Sun sweeps out equal areas in equal times and (3) the square of a planet's orbital period is proportional to the cube of the semi-major axis of its orbit. All three laws are clearly about the motion of planets and other bodies in orbit around the Sun. While Newton later showed that these laws follow from the universal law of gravitation, Kepler's own contribution is specifically about the kinematics of planetary motion, not about general laws of energy or gravity directly. Step-by-Step Solution: Step 1: Recall Kepler's first law: planetary orbits are ellipses with the Sun at one focus. Step 2: Recall the second law: a planet sweeps out equal areas in equal intervals of time, describing how its speed varies along the orbit. Step 3: Recall the third law: T^2 is proportional to a^3, where T is the orbital period and a is the semi-major axis, relating distance from the Sun and orbital period. Step 4: Note that each law describes how planets move rather than why they move that way. Hence they are primarily laws of planetary motion. Step 5: Therefore, the correct option is “Planetary motion.” Verification / Alternative check: Standard astronomy and physics textbooks explicitly label these three relations as “Kepler's laws of planetary motion.” The law of gravitation and the law of conservation of energy are credited mainly to Newton and later physicists, not to Kepler, although they can be used to derive or interpret Kepler's results. Exam questions often contrast Kepler's descriptive laws of orbits with Newton's deeper explanation using forces and energy. This makes it clear that the direct association is with planetary motion. Why Other Options Are Wrong: Law of conservation of energy: This is a fundamental principle of physics, but it was not formulated by Kepler and is much broader than planetary motion alone. Law of gravitation: Newton's universal law of gravitation explains why planets follow Kepler's laws, but Kepler's work itself preceded this and focused on describing orbits rather than stating a universal force law. None: This is incorrect because Kepler's laws clearly have a specific and well-known connection, namely planetary motion. Common Pitfalls: Some students see the word “gravitation” and assume that Kepler's laws must be directly about the gravitational force. In reality, Kepler used observational data (especially from Tycho Brahe) to find patterns in how planets move. Newton later showed that these patterns follow from gravity. Remember that Kepler's role was to describe planetary motion precisely, while Newton provided the theoretical explanation using universal gravitation and energy concepts. Final Answer: Kepler's laws are mainly related to Planetary motion.

Question 6

In acoustics, the decibel (dB) is a commonly used unit for measuring which of the following physical quantities?

Accepted Answer

Introduction / Context: The decibel, written as dB, is widely seen on audio equipment, mobile phones and noise-level meters. It is a logarithmic unit that expresses relative levels of power or intensity, especially for sound. This question checks whether you can correctly associate the decibel with the physical quantity it usually measures in general science and everyday applications. Given Data / Assumptions: The unit under discussion is the decibel (dB). Options link it to the speed of light, intensity of sound, intensity of heat or none of these. We consider standard school physics usage where decibels are encountered most often in connection with sound. Concept / Approach: Sound intensity is a measure of the power of sound per unit area. Because the human ear perceives sound across a very wide range of intensities, it is convenient to use a logarithmic scale. The decibel scale expresses sound intensity level relative to a reference intensity. Although decibels can be used in other contexts (for example, in electronics to compare power levels), their primary and most familiar use in basic physics and general knowledge is to quantify sound levels. Therefore, we need to match decibel with the intensity of sound. Step-by-Step Solution: Step 1: Recall that the loudness of sound is often reported as a number in decibels, such as 30 dB, 60 dB or 100 dB. Step 2: Understand that this number represents the sound intensity level compared to a reference intensity, using a logarithmic formula. Step 3: Compare the options: speed of light is measured in metres per second, and heat intensity is not commonly described in decibels in basic physics. Step 4: Recognise that everyday examples, such as noise pollution limits or safe listening levels with headphones, always use decibels for sound. Step 5: Conclude that decibel is a unit connected with the intensity of sound. Verification / Alternative check: Physics textbooks define sound intensity level L in decibels using a formula of the form L = 10 * log10(I / I0), where I is the sound intensity and I0 is a reference intensity. All standard tables of physical quantities list decibel under sound intensity level or audio power levels. The speed of light uses units such as metres per second, while thermal energy and heat flow are measured in joules or watts rather than decibels. This confirms that the most appropriate match in the context of this question is “intensity of sound.” Why Other Options Are Wrong: Speed of light: This is a velocity, measured in units such as m/s, not in decibels. Intensity of heat: Heat or thermal energy is described in joules or watts, and although power ratios can be expressed in decibels in advanced topics, this is not the standard school-level association. None of these: This is incorrect because decibel is definitely used in connection with sound intensity. Common Pitfalls: Some learners confuse decibel with other common units like hertz (Hz), which measures frequency, or they mistakenly think decibel measures loudness directly instead of intensity level. Remember that decibel is a relative logarithmic unit, and in basic physics and GK, it is almost always linked to sound intensity levels. Keeping examples like “noise in a library is about 40 dB” in mind helps anchor this association. Final Answer: The decibel (dB) is commonly used as a unit of intensity of sound.

Question 7

Which one of the following is not a transverse wave under ordinary conditions?

Accepted Answer

Introduction / Context: Waves can be broadly classified as transverse or longitudinal, depending on the direction in which particles of the medium vibrate relative to the direction of wave propagation. This question tests whether you can correctly identify which wave, from a list of common examples, is not a transverse wave under normal circumstances. Understanding this classification is fundamental in wave physics and helps in interpreting many natural phenomena. Given Data / Assumptions: We are comparing light waves, radio waves and sound waves. We assume sound waves in air under ordinary conditions. We are asked to find which is not a transverse wave. Transverse waves have oscillations perpendicular to the direction of propagation, while longitudinal waves have oscillations parallel to that direction. Concept / Approach: Electromagnetic waves, which include light and radio waves, are inherently transverse. In these waves, electric and magnetic field vectors oscillate perpendicular to the direction in which the wave travels. On the other hand, ordinary sound waves in air are mechanical longitudinal waves: air particles oscillate back and forth along the direction of propagation, creating compressions and rarefactions. To answer the question, we must therefore select sound waves as the example that is not transverse. Step-by-Step Solution: Step 1: Recall that light waves are electromagnetic waves. Their electric and magnetic fields oscillate perpendicular to the direction of propagation, so they are transverse. Step 2: Recall that radio waves are also electromagnetic and share the same transverse nature as visible light. Step 3: Consider sound waves in air. These require a material medium and consist of compressions and rarefactions along the line of travel. Step 4: In sound waves in air, particle displacement is parallel to the direction of wave propagation, which is the definition of a longitudinal wave. Step 5: Therefore, sound waves, not light or radio waves, are not transverse under ordinary conditions. Verification / Alternative check: Physics texts clearly state that electromagnetic radiation (light, radio, microwaves, X-rays and so on) is transverse, whereas standard sound waves in gases and liquids are longitudinal. In solids, sound can have both longitudinal and transverse components, but the typical school-level example of sound in air is purely longitudinal. The option “All of the above” would only be correct if each listed wave were not transverse, which contradicts well established theory. Thus, the only correct choice is sound waves. Why Other Options Are Wrong: Light waves: These are transverse electromagnetic waves and do not qualify as “not transverse.” Radio waves: Also electromagnetic and transverse, used for communication over long distances. All of the above: Incorrect because only sound waves in air are longitudinal; light and radio waves remain transverse. Common Pitfalls: Some students assume that all waves are the same and forget the distinction between electromagnetic and mechanical waves. Others may have heard about transverse vibrations in guitar strings and think sound in air behaves similarly. Remember that in electromagnetic waves, no material medium is needed and the fields oscillate perpendicular to the direction of travel, while in ordinary sound in air, compression and rarefaction occur along the direction of propagation, making them longitudinal. Final Answer: The wave that is not transverse under ordinary conditions is Sound waves.

Question 8

Two metallic plates of equal thickness and equal cross-sectional area, with thermal conductivities K1 and K2, are placed together face to face to form a composite slab. What is the equivalent thermal conductivity K of the combined slab?

Accepted Answer

Introduction / Context: Heat conduction through composite slabs is a standard topic in thermal physics and engineering. When two or more layers of different materials are placed together, the effective or equivalent thermal conductivity depends on how they are arranged. This question involves two metallic plates of equal thickness and equal area, placed face to face, and asks for the equivalent thermal conductivity of the resulting composite slab. Given Data / Assumptions: Two plates with thermal conductivities K1 and K2 are used. The plates have equal thickness and equal cross-sectional area. They are placed face to face, so heat flows perpendicular to the layers, passing through one then the other. The temperature difference is applied across the total thickness of the composite slab. Steady state one-dimensional heat conduction is assumed, with no heat loss to surroundings. Concept / Approach: When layers of different materials are arranged in series with respect to heat flow, the total thermal resistance is the sum of the individual resistances. Thermal resistance R for a layer is given by R = L / (K * A), where L is thickness, K is thermal conductivity and A is cross-sectional area. For equal thickness L and equal area A, the equivalent thermal conductivity K_equiv for two layers in series can be obtained by writing the total resistance as the sum of the two and then expressing it in the same form as a single slab of thickness 2L and area A. Step-by-Step Solution: Step 1: Let each plate have thickness L and area A. The conductivities are K1 and K2. Step 2: Thermal resistance of plate 1 is R1 = L / (K1 * A). Thermal resistance of plate 2 is R2 = L / (K2 * A). Step 3: Since they are in series, total resistance R_total = R1 + R2 = L / (K1 * A) + L / (K2 * A). Step 4: Factor out L / A to get R_total = (L / A) * (1 / K1 + 1 / K2). Step 5: For an equivalent single slab of thickness 2L, area A and conductivity K, the resistance would be R_equiv = 2L / (K * A). Step 6: Set R_equiv equal to R_total: 2L / (K * A) = (L / A) * (1 / K1 + 1 / K2). Step 7: Cancel L / A on both sides to obtain 2 / K = 1 / K1 + 1 / K2. Step 8: Rearranging gives 2 / K = (K1 + K2) / (K1 * K2), so K = 2*K1*K2 / (K1 + K2). Verification / Alternative check: The result has the form of a harmonic mean scaled by a factor of 2, which is typical for two equal layers in series. As a quick check, if K1 = K2, say both equal to K0, then K = 2*K0*K0 / (2*K0) = K0, which makes physical sense because the composite of two identical materials should have the same conductivity as each individually. If one conductivity is much smaller than the other, the equivalent conductivity is closer to the smaller value, which also matches physical intuition that the poorer conductor dominates the resistance. These logical checks confirm the correctness of the formula. Why Other Options Are Wrong: K1*K2 / (K1 + K2): This would arise if the total thickness were L instead of 2L, so it does not match the given geometry of equal thickness plates forming a slab of total thickness 2L. (K1 + K2) / (K1*K2): This expression has the wrong dimensional form for conductivity and actually represents the reciprocal of the harmonic mean, not the conductivity itself. None: Incorrect because a valid and standard expression for the equivalent conductivity definitely exists. Common Pitfalls: A common error is to treat the two plates as if they were in parallel for heat flow, which leads to an incorrect arithmetic mean rather than the series combination formula. Another mistake is forgetting that the total thickness is 2L, which changes the equivalent resistance. Students should always draw a simple diagram, mark the thickness of each layer and write thermal resistance expressions correctly before combining them. This habit reduces algebraic mistakes and helps in more complex composite wall problems as well. Final Answer: The equivalent thermal conductivity of the composite slab is 2*K1*K2 / (K1 + K2).

Question 9

A convex mirror is used to form an image of a real object. Which of the following statements about the image are correct?

Accepted Answer

Introduction / Context: Convex mirrors are commonly used in vehicles as rear view mirrors and at road junctions because of their special image forming properties. This question tests your understanding of the characteristics of the image formed by a convex mirror when the object is real and placed in front of the mirror. Knowing these properties is important in basic geometrical optics. Given Data / Assumptions: The reflecting surface is convex, bulging outward toward the object. The object is real and placed in front of the convex mirror. Standard sign conventions for spherical mirrors apply. We have several statements about the image and must decide which are correct. Concept / Approach: For a convex spherical mirror, irrespective of the position of a real object along the principal axis, the image has certain fixed qualitative features. The image formed is always virtual (behind the mirror), upright or erect, and diminished in size compared to the object. Moreover, its position is always between the pole (the central point on the mirror surface) and the focus of the mirror. By recalling the ray diagrams and mirror formula, we can check which statements in the options match these known properties. Step-by-Step Solution: Step 1: Recall that convex mirrors always form virtual images of real objects, and these images appear behind the mirror. Step 2: Because the image is virtual, it is also erect, meaning it has the same orientation as the object and is not inverted. Step 3: For any object position in front of a convex mirror, the image size is smaller than the object, so it is diminished. Step 4: Ray diagrams show that the image position is located between the pole of the mirror and its focal point on the reflected side. Step 5: Therefore, all three given statements (diminished size, erect image and location between pole and focus) are correct, so the correct composite option is “All of the above.” Verification / Alternative check: Using the mirror formula 1 / f = 1 / v + 1 / u with a positive focal length for a convex mirror (when using the sign convention where distances measured opposite to incident light are negative) always leads to an image distance behind the mirror, that is, v is positive for convex mirrors. This means the image is virtual and erect. Calculations for different object distances u show that the magnification m = v / u is less than 1 in magnitude, indicating a reduced image. Graphical ray constructions also confirm that the image lies between the pole and the focus. All these checks agree with the qualitative statements above. Why Other Options Are Wrong: It is diminished in size: This is true but incomplete as an answer because the question asks which description is fully correct; other true statements also exist. It is erect: Also true, but again only part of the full description. It always lies between the pole and the focus: This is correct, yet not the only true property. The best answer must include all true properties together, which is option D. Common Pitfalls: Some students mix up the properties of convex and concave mirrors. For concave mirrors, image properties change drastically depending on object distance, but for convex mirrors the behaviour is more uniform. Another confusion is thinking that “virtual” automatically means “same size,” which is not true here because the convex mirror produces a smaller image. Always link the type of mirror to a standard set of properties learned from ray diagrams. Final Answer: The image formed by a convex mirror of a real object is diminished, erect and located between the pole and the focus, so the correct choice is All of the above.

Question 10

In information theory, the capacity of a communication channel is commonly measured in terms of which of the following?

Accepted Answer

Introduction / Context: In communication systems, the term “channel capacity” refers to the maximum rate at which information can be transmitted over a communication channel with an arbitrarily low probability of error. At a basic level, many general knowledge questions link this idea with the concept of bandwidth, which plays a key role in determining how much data can be carried. This question asks you to identify the quantity with which channel capacity is most closely associated in common usage. Given Data / Assumptions: We are dealing with a communication channel, such as a telephone line or radio link. The question asks about the measure related to the “capacity” of this channel. Options include band rate, bandwidth, data flow and bit capacity. We assume basic understanding from digital communications and general science. Concept / Approach: In technical terms, Shannon's formula for channel capacity states that capacity C (in bits per second) depends on bandwidth B (in hertz) and signal-to-noise ratio. Although capacity and bandwidth are not identical, bandwidth is the primary physical resource that limits how much information can be transmitted. In many general knowledge contexts, channel capacity is loosely said to be measured or constrained by the bandwidth of the channel. Other terms in the options are either informal or less standard. Therefore, we associate channel capacity with bandwidth. Step-by-Step Solution: Step 1: Recall that bandwidth represents the range of frequencies that a channel can carry effectively, often measured in hertz. Step 2: Recognise that a larger bandwidth allows more independent signal components to be transmitted per unit time, leading to a higher possible data rate. Step 3: In Shannon's theory, channel capacity is proportional to bandwidth for a given signal-to-noise ratio, emphasizing their close relationship. Step 4: Compare the options: “band rate,” “data flow” and “bit capacity” are either vague or are consequences of channel capacity rather than primary measures. Step 5: Conclude that bandwidth is the correct choice when asked what the capacity of a communication channel is measured in or strongly related to. Verification / Alternative check: Elementary communication engineering texts show graphs and formulas where channel capacity curves are plotted as functions of bandwidth and signal-to-noise ratio. Even everyday technology marketing refers to “broadband” to signal high data capacity, directly highlighting available bandwidth. The other terms in the options do not appear as standard units or primary parameters for channel capacity. This confirms that bandwidth is the best match for the question asked. Why Other Options Are Wrong: Band rate: This term is not standard; there is “baud rate” in digital communications, but it refers to symbol rate, not channel capacity itself. Data flow: This is a descriptive phrase, not a specific measurable quantity with a standard unit. Bit capacity: This is vague and not a conventional term in communication theory; capacity is measured in bits per second, but the controlling physical parameter is bandwidth. Common Pitfalls: Students sometimes confuse “baud rate” with bandwidth or assume that any phrase containing “bit” automatically refers to capacity. Another source of confusion is mixing up the maximum theoretical capacity with the actual data flow in a specific system. While actual data flow depends on many factors, the underlying limit placed by the channel is strongly tied to bandwidth. Remembering Shannon's formula and the everyday use of “broadband” as a sign of higher capacity helps fix this concept. Final Answer: The capacity of a communication channel is most directly associated with its Bandwidth.

Question 11

In physical chemistry and physics, fog is an example of which type of colloidal system?

Accepted Answer

Introduction / Context: Colloids are mixtures where tiny particles of one substance are dispersed in another substance. Different everyday phenomena such as fog, smoke, foam and emulsions are all examples of colloidal systems. This question tests whether you can correctly identify the type of dispersion that fog represents, combining basic ideas from atmospheric science and colloid chemistry. Given Data / Assumptions: We are considering ordinary fog that appears near the ground, especially in cool and humid conditions. Fog consists of water droplets suspended near the earth's surface. The continuous phase is the surrounding air. We must classify fog in terms of which state is dispersed and which state forms the medium. Concept / Approach: Fog is essentially a cloud formed at ground level. It consists of tiny liquid water droplets suspended in air. In colloid terminology, the dispersed phase is liquid and the dispersion medium (continuous phase) is gas, which is mainly air. Such a system, where liquid is dispersed in a gas, is called a liquid aerosol. The question uses more general phrases like “liquid dispersed in gas,” so we match fog to that description. The other options involve solids or gases in liquids or solids and do not fit the actual composition of fog. Step-by-Step Solution: Step 1: Identify the visible component of fog: tiny droplets of liquid water. Step 2: Identify the medium in which these droplets are suspended: the air, which is a gas. Step 3: Therefore, the dispersed phase is liquid, and the dispersion medium is gas. Step 4: In colloid classification, this combination is described as a liquid dispersed in a gas (aerosol). Step 5: Match this classification with the options and choose “Liquid dispersed in gas.” Verification / Alternative check: Standard tables of colloid types list fog and mist under aerosols, specifically as examples of liquid in gas systems. Smoke, by contrast, is listed as solid particles dispersed in gas. Milk is a liquid in liquid emulsion, and foam is gas in liquid. Since fog is visually similar to clouds and arises from condensation of water vapour into droplets, it clearly belongs with the liquid in gas category. This confirms that the correct classification is liquid dispersed in gas. Why Other Options Are Wrong: Solid dispersed in liquid: This describes suspensions such as muddy water or colloidal sols, not fog. Liquid dispersed in solid: This is typical of gels or certain solid emulsions, for example, jelly, not atmospheric fog. Gas dispersed in liquid: This corresponds to foams, such as whipped cream or shaving foam, again very different from fog. Common Pitfalls: Because fog is often associated with humidity and water, some students may focus only on the liquid part and forget that air is the continuous medium. Others may confuse fog with smoke, which is solid particles in gas. A clear mental picture helps: fog consists of tiny liquid water droplets floating in air, so think “liquid in gas.” Remembering related examples like mist and spray, which are also liquid aerosols, makes this classification easier to recall. Final Answer: Fog is an example of a colloidal system where liquid is dispersed in gas.

Question 12

If the reading of a barometer is observed to be steadily going down, what kind of weather condition does this generally indicate?

Accepted Answer

Introduction / Context: A barometer is an instrument used to measure atmospheric pressure. Changes in barometric pressure provide important clues about upcoming weather. This question asks what it generally means when the barometer reading is going down. Understanding this link between pressure and weather patterns is a key part of basic meteorology and general knowledge. Given Data / Assumptions: The barometer reading is decreasing over time rather than remaining steady. We are considering typical mid-latitude weather conditions. We must link a falling pressure to likely weather outcomes. Options include storm, rainfall, snow and intense heat. Concept / Approach: In general, high atmospheric pressure is associated with clear, calm weather, while low pressure is associated with unsettled, cloudy and rainy conditions. A falling barometer indicates that a low pressure system is approaching or intensifying. Low pressure encourages rising air masses that cool, condense and form clouds, often leading to rain. While storms can also be associated with low pressure, many general knowledge questions focus on the idea that a falling barometer is a sign of rainfall or wet weather in general. Therefore, the most direct match among the options is rainfall. Step-by-Step Solution: Step 1: Recognise that a barometer measures atmospheric pressure. Step 2: Recall that a falling barometer reading means atmospheric pressure is decreasing. Step 3: Understand that decreasing pressure usually indicates that a low pressure system is moving in. Step 4: Low pressure encourages air to rise, expand and cool, which leads to condensation of water vapour and cloud formation. Step 5: Cloud formation under such conditions often results in rainfall, so a falling barometer is commonly taken as an indication of likely rain. Verification / Alternative check: Weather guides and textbooks often list “falling pressure” as a sign of worsening or wetter weather. Maritime traditions also teach sailors to watch the barometer because a sudden drop may signal rain and strong winds. While storms are indeed associated with very low pressure, not every fall in barometer reading is accompanied by a full storm, whereas prolonged or steady falls are strongly correlated with rain-bearing systems. Because exam questions usually expect the simpler and more general association, rainfall is considered the standard answer. Why Other Options Are Wrong: Storm: Severe storms are linked with very low pressure, but the question does not specify a rapid or extreme fall, and in many GK sources, rainfall is presented as the main outcome of a falling barometer. Snow: Snow can occur under certain low pressure conditions, but it depends heavily on temperature. The question does not refer to cold climates specifically. Intense heat: Hot, dry weather is more often associated with stable high pressure, not falling pressure. Common Pitfalls: Students sometimes equate any low pressure with “storm” and may choose that option immediately. However, examination keys for introductory level often associate falling barometer readings with rain or wet weather in a broad sense. Another pitfall is to ignore regional climate and think of snow as the default precipitation, but this is not correct for most parts of the world. Remember the simple rule: falling pressure is a warning of unsettled, likely rainy weather, while rising pressure suggests clearing conditions. Final Answer: A steadily falling barometer reading generally indicates the approach of Rainfall or wet weather.

Question 13

For an electric bulb operated from a given supply, the brightness of the bulb effectively depends on which of the following electrical quantities?

Accepted Answer

Introduction / Context: The brightness of an electric bulb is related to the electrical power it consumes. Power, in turn, depends on both the voltage applied across the bulb and the current flowing through it. This question checks whether you can connect brightness with these basic electrical quantities rather than thinking of voltage and current in isolation. Given Data / Assumptions: We consider an ordinary filament or incandescent type bulb or any resistive lamp. The bulb is connected to an electrical supply with some voltage V and current I. Brightness is assumed to be proportional to the electrical power converted into light and heat. We must choose whether brightness depends on current, voltage, both or none. Concept / Approach: The electrical power P consumed by a device is given by formulas such as P = V * I, P = I^2 * R or P = V^2 / R, where V is voltage, I is current and R is resistance. Brightness of a bulb increases with the power dissipated in its filament. Because power can be expressed using both voltage and current together, it is clear that brightness depends on both quantities rather than on only one of them in a simple way. In practice, if voltage increases across a fixed resistance, current also increases and thus power goes up, making the bulb brighter. The key idea is that voltage and current are linked and together determine power and brightness. Step-by-Step Solution: Step 1: Write the basic relation for electrical power: P = V * I. Step 2: Observe that this formula directly involves both voltage V and current I, and power is what determines brightness. Step 3: If we hold the resistance constant, increasing voltage increases current, and both changes together cause an increase in power. Step 4: Similarly, if for some reason current is increased at the same voltage, the power and brightness increase. Step 5: Therefore, brightness cannot be described as depending solely on current or solely on voltage; it depends on both voltage and current. Verification / Alternative check: Consider a bulb rated at 60 W, 230 V. Using P = V * I, the rated current is I = P / V. If the applied voltage is reduced significantly, the current will fall and the bulb will glow dimly because power is lower. If voltage is increased beyond the rating, current will rise and the bulb glows brighter until it risks burning out. These examples show that both voltage and current together set the power level. Therefore, treating brightness as a function of only one of these quantities is incomplete for general questions like this. Why Other Options Are Wrong: Current: Focusing only on current ignores the fact that current itself depends on voltage and resistance, and power is the product V * I. Voltage: Voltage alone does not determine brightness because different bulbs with different resistances can draw different currents at the same voltage. None: Brightness clearly depends on electrical quantities, so this is not correct. Common Pitfalls: A frequent mistake is to memorize that “brightness depends on current” without considering the role of voltage. Another is to think that standard mains voltage is fixed, so only current matters. In truth, the correct physical relationship for power includes both V and I, and exam questions at this level expect the answer that reflects the complete dependency. Keeping the formula P = V * I in mind provides a quick and reliable way to answer similar questions. Final Answer: The brightness of a bulb effectively depends on Both A & B, that is, on both current and voltage.

Question 14

When an object covers equal distances in equal intervals of time in a straight line, the object is said to be in which type of motion?

Accepted Answer

Introduction / Context: Motion can be classified in many ways, and one of the simplest and most important distinctions is between uniform and non-uniform motion. This question uses a standard textbook definition involving equal distances in equal intervals of time to test whether you can correctly name the type of motion described. Given Data / Assumptions: The object moves along a path, which we can consider as a straight line for simplicity. In any equal time intervals (for example, every second or every minute), the object covers exactly the same distance. There is no mention of changing speed or acceleration. We must select the appropriate type of motion from the options given. Concept / Approach: Uniform motion is defined as motion in which an object covers equal distances in equal intervals of time, no matter how small those intervals are, provided they are equal. In such motion, the speed of the object remains constant. Non-uniform motion, by contrast, involves varying speeds and unequal distances in equal times. The options “linear” and “translational” describe direction and type of motion but not whether the speed is constant, and “equilibrium” refers to forces rather than motion. Therefore, the correct term here is uniform motion. Step-by-Step Solution: Step 1: Focus on the key phrase in the question: “equal distances in equal intervals of time.” Step 2: Recall the standard definition from physics: motion is called uniform when equal distances are covered in equal times. Step 3: Recognise that in uniform motion, the speed of the object remains constant. Step 4: Compare with “linear motion,” which only indicates that the path is a straight line, and with “translational motion,” which refers to motion where all parts of a body move in the same direction. Step 5: See that the question is clearly about speed constancy, not about the shape of the path, so “uniform” is the correct answer. Verification / Alternative check: Consider a car traveling along a straight road at a constant speed of 40 km/h. Every hour it moves 40 km, so the distance covered in equal time intervals is equal, fitting the definition of uniform motion. If the car speeds up or slows down, the distances covered in each hour become different, and the motion is no longer uniform. Textbooks explicitly state “equal distances in equal intervals of time” as the hallmark of uniform motion, which directly matches the wording of this question. Why Other Options Are Wrong: Linear: Linear motion simply means motion along a straight line; it can be either uniform or non-uniform, depending on whether speed is constant. Translational: Translational motion refers to overall movement of a body from one location to another, where all parts of the body move parallel to each other; it does not specify constant speed. Equilibrium: This term refers to the balance of forces (net force equal to zero) and does not by itself describe the pattern of distances covered in time. Common Pitfalls: Some learners may fixate on the idea of motion in a straight line and choose “linear” without noticing that the key feature in the statement is about equal distances and time intervals. Others may recall the term “translational motion” from rigid body dynamics and confuse it with uniformity. Always pay attention to the definition given in the question; whenever equal distances in equal time intervals appear, think of uniform motion. Final Answer: An object that covers equal distances in equal intervals of time is in Uniform motion.

Question 15

Water kept in an earthen pot is observed to be less cool on rainy days than in summer. What is the main reason for this difference?

Accepted Answer

Introduction / Context: Earthen pots are widely used in many regions to cool drinking water naturally. They work on the principle of evaporation. However, people often notice that water stored in such pots feels less cool during the rainy season than in the hot, dry days of summer. This question tests whether you understand how humidity and evaporation affect the cooling process. Given Data / Assumptions: An earthen pot is porous, allowing some water to seep through and wet the outer surface. In summer, the surrounding air is usually hot and relatively dry. On rainy days, the air is cooler but contains a high amount of water vapour (high humidity). We compare how effectively the pot cools water in summer versus during the rainy season. Concept / Approach: Evaporation is a surface phenomenon in which high energy molecules escape from a liquid into the air. Evaporation causes cooling because it removes heat from the remaining liquid. The rate of evaporation depends on temperature, surface area, wind speed and, very importantly, the humidity (moisture content) of the air. When humidity is high, the air is already close to saturation, so it cannot accept much more water vapour, and evaporation slows down. In an earthen pot, less evaporation from the outer surface means less cooling of the water inside. Step-by-Step Solution: Step 1: Note that the earthen pot has tiny pores through which water seeps and wets the outer surface. Step 2: In summer, the hot and dry air allows rapid evaporation of this water, which absorbs heat from the pot and the water inside, making the stored water cool. Step 3: During the rainy season, the air has a high percentage of moisture and is much more humid. Step 4: Because the air is nearly saturated with water vapour, the evaporation of water from the pot surface becomes slow. Step 5: With reduced evaporation, less heat is drawn away, so the cooling effect is weaker, and the water in the pot is less cool. Verification / Alternative check: Everyday observations support this explanation. Clothes dry quickly on a hot, dry day because evaporation is fast, but they remain damp for longer in humid or rainy weather. The same principle applies to earthen pots. Scientific discussions of evaporation always list “dry air” as a factor enhancing evaporation and “humid air” as a factor reducing it. Options that involve pores closing or rain water blocking them are not consistent with how earthen pots and porous materials behave in practice. Why Other Options Are Wrong: The pores in the pot are closed: The pores do not normally close simply because it is rainy; the pot remains porous. Rain water stops oozing out from the pores: The oozing of water is from inside to outside due to capillary action, not directly controlled by raindrops. The surface of the pot becomes damp: While the surface may be damp, the key factor is that evaporation from this damp surface is slow due to humidity, which is already captured in option C. Common Pitfalls: Learners sometimes imagine mechanical effects like pores being blocked rather than thinking of the evaporation process. Another confusion arises from feeling that rainy weather is cooler, so they expect water to be cooler too, forgetting that evaporation is more important here than air temperature. Always connect “earthen pot cooling” with “evaporation,” and remember that high humidity slows evaporation and therefore reduces cooling. Final Answer: Water in an earthen pot is less cool on rainy days mainly because the water that oozes out does not evaporate quickly due to the large percentage of moisture in the air.

Question 16

A hydrogen atom is excited from the ground state to a higher energy level with principal quantum number n = 4. How many distinct spectral lines will appear in the emission spectrum when the atom returns to the ground state through all possible trans…

Accepted Answer

Introduction / Context: The line spectrum of the hydrogen atom arises because electrons can occupy only discrete energy levels and emit photons when they jump from a higher level to a lower level. This question explores how many different spectral lines can result when an atom initially excited to a particular energy level makes all possible downward transitions. It tests both your understanding of quantum energy levels and your ability to apply a simple combinational formula. Given Data / Assumptions: We have a hydrogen atom with one electron. The electron is excited from the ground state (n = 1) to a higher energy level with principal quantum number n = 4. The electron can return to the ground state through various intermediate energy levels, emitting photons at each transition. We assume all possible downward transitions occur at some stage. Concept / Approach: When an electron is at the energy level n, it can make transitions to any lower level (n - 1, n - 2 and so on down to 1). Each distinct pair of energy levels (n2, n1) with n2 > n1 corresponds to one possible spectral line, because the energy difference and therefore the wavelength of the emitted photon is unique. For a given upper level n, the total number of spectral lines that can be produced when the electron completes all possible transitions down to the ground level is given by the formula n * (n - 1) / 2. This counts all unique downward transitions between levels 1 to n. Step-by-Step Solution: Step 1: Identify the highest level that the electron reaches: n = 4. Step 2: Recall the formula for the number of possible spectral lines when all transitions from level n down to 1 are allowed: Number of lines = n * (n - 1) / 2. Step 3: Substitute n = 4 into the formula: Number of lines = 4 * (4 - 1) / 2. Step 4: Simplify: 4 * 3 / 2 = 12 / 2 = 6. Step 5: Therefore, a total of 6 distinct spectral lines can appear in the emission spectrum. Verification / Alternative check: We can list all downward transitions explicitly to confirm the count. From n = 4, the electron can drop to n = 3, 2 or 1. From n = 3, it can further drop to n = 2 or 1. From n = 2, it can drop to n = 1. So the possible transitions are: 4 → 3, 4 → 2, 4 → 1, 3 → 2, 3 → 1 and 2 → 1. Counting these gives 6 transitions, each corresponding to one spectral line at a different wavelength. This matches the formula result and confirms that 6 is correct. Why Other Options Are Wrong: 2: This would be the number of transitions if only a couple of specific jumps were allowed, but here all possible transitions are considered. 5: This is one less than the correct count and might arise from forgetting one of the possible transitions, such as 4 → 2 or 3 → 1. 3: This could be guessed by taking n - 1 instead of n * (n - 1) / 2, but it does not count all pairs of levels. Common Pitfalls: Students often confuse the number of possible final levels (which is n - 1) with the number of distinct transitions. Another mistake is counting only transitions that end at the ground state and ignoring intermediate transitions between excited levels. Remember that each drop from any higher level to any lower level generates a spectral line. Using the formula n * (n - 1) / 2 is a reliable shortcut, and listing transitions explicitly is a good way to verify your answer in small cases. Final Answer: The number of spectral lines obtained when the hydrogen atom drops from n = 4 back to the ground state through all possible paths is 6.

Question 17

When an electric bulb accidentally breaks, a small explosive-like sound is heard. This sound is mainly due to which of the following reasons?

Accepted Answer

Introduction / Context: Many electric bulbs are either evacuated or filled with gas at a pressure different from the surrounding air. When such a bulb breaks, people often hear a sharp, explosive-like sound. This question tests your understanding of pressure differences and how they can cause a sudden movement of air that produces noise when the thin glass envelope of a bulb shatters. Given Data / Assumptions: An electric bulb has a thin glass shell. Inside the bulb there is usually a partial vacuum or low pressure gas. The surrounding atmosphere is at normal atmospheric pressure, which is higher than the pressure inside the bulb. The bulb breaks suddenly, creating an opening between inside and outside. Concept / Approach: If there is a pressure difference between two regions separated by a barrier, removing or breaking that barrier causes the fluid (here, air) to rush from the higher pressure region to the lower pressure region. In the case of a typical bulb, the inside is at lower pressure, and the outside is at higher atmospheric pressure. When the glass breaks, external air rapidly moves inward to fill the low pressure region, producing a sharp sound due to the sudden rush and the shattering glass. There is no significant chemical reaction between gases, and the inner gas is not highly compressed in the way described in some of the incorrect options. Step-by-Step Solution: Step 1: Recognise that many bulbs are partially evacuated or contain a small amount of inert gas at a pressure lower than atmospheric pressure. Step 2: Understand that the outside air pressure is higher than the pressure inside the bulb. Step 3: When the glass bulb breaks, the barrier between these two pressure regions disappears almost instantly. Step 4: High pressure air from the outside rushes into the bulb to equalise the pressure, and this sudden inward flow produces a sharp sound and contributes to the shattering effect. Step 5: Therefore, the correct explanation is that the air outside rushes in to fill the vacuum or low pressure region inside the bulb. Verification / Alternative check: If the bulb interior were at higher pressure than the outside, we would expect gases to rush outward, and the sound pattern and glass fragments would be different. However, typical incandescent and some other types of bulbs are designed with low pressure or slight vacuum to reduce oxidation of the filament and improve efficiency. Physics and engineering references confirm that the slight “pop” heard is mainly due to air rushing inward and the mechanical fracture of the glass. There is no rapid chemical reaction that would resemble an explosion in such bulbs under normal conditions. Why Other Options Are Wrong: A chemical reaction between the gases inside the bulb: The gases inside (often inert gases like argon or nitrogen) are chosen specifically for their low reactivity, so explosive chemical reactions are not expected. Compressed gases inside rushing out suddenly: The interior is at lower, not higher, pressure than the outside. Thus, the direction of flow is from outside to inside, not the reverse. None of these: This is incorrect because there is a clear physical explanation involving air rushing in. Common Pitfalls: Students sometimes imagine that the bulb contains highly compressed gas which “explodes” outward, but this is not typical for household bulbs. Another misunderstanding is blaming vague “chemical reactions” whenever a sudden sound occurs. In reality, pressure differences are often enough to cause loud noises when containers rupture. Remember that the greater outside atmospheric pressure forces air into the bulb when it breaks, producing the characteristic popping sound. Final Answer: The explosive-like sound when a bulb breaks is mainly due to air outside rushing in suddenly to fill the vacuum inside the bulb.

Question 18

Energy is continuously generated in the Sun mainly due to which of the following processes occurring in its core?

Accepted Answer

Introduction / Context: The Sun is the primary source of energy for life on Earth, and understanding how it produces this enormous amount of energy is a key concept in astrophysics and general science. This question asks you to identify the main physical process responsible for the Sun's continuous energy generation deep inside its core. Given Data / Assumptions: The Sun is a star composed mainly of hydrogen and helium. The core of the Sun has extremely high temperature and pressure. Energy is produced continuously over billions of years. We must choose between nuclear fusion, nuclear fission, natural radioactivity and artificial radioactivity. Concept / Approach: Nuclear fusion is the process in which light atomic nuclei combine to form heavier nuclei, releasing large amounts of energy in the process. In the Sun's core, hydrogen nuclei (protons) fuse through a multi-step sequence known as the proton–proton chain to form helium. Nuclear fission, in contrast, is the splitting of heavy nuclei such as uranium, which is used in many nuclear power plants on Earth but is not the dominant process in the Sun. Radioactivity describes spontaneous decay of unstable nuclei and is not the main energy source in normal stars. Therefore, nuclear fusion is the correct answer. Step-by-Step Solution: Step 1: Recall that the Sun consists mainly of hydrogen gas, especially in its core. Step 2: Under the extreme pressure and temperature in the core, hydrogen nuclei are forced close enough to overcome electrostatic repulsion. Step 3: These hydrogen nuclei fuse in a series of reactions (proton–proton chain) to form helium, with some mass converted into energy according to E = m * c^2. Step 4: The huge energy released in each fusion reaction sustains the Sun's bright radiation over billions of years. Step 5: Recognise that this is nuclear fusion, not fission or simple radioactivity, and select fusion as the correct process. Verification / Alternative check: Astrophysical models and observations of solar neutrinos strongly support the nuclear fusion model for energy production in the Sun. The proton–proton chain, along with other fusion cycles like the CNO cycle in more massive stars, explains the observed luminosity and lifetime of stars. Nuclear fission, on the other hand, would consume heavy nuclei and is not consistent with the Sun's composition. Natural radioactivity does occur in some cosmic environments but cannot account for the enormous and sustained energy output of the Sun. Why Other Options Are Wrong: Nuclear fission: This process powers many nuclear reactors on Earth but is not responsible for the Sun's main energy output, because the Sun is not primarily composed of heavy fissile elements like uranium. Radioactivity: Although some radioactive processes occur in the universe, they are not dominant in the Sun's core energy generation. Artificial radioactivity: This is induced in laboratories and reactors on Earth and has no connection to natural processes inside the Sun. Common Pitfalls: Students sometimes mix up fission and fusion because both involve nuclear reactions and energy release. A helpful memory trick is: fusion “fuses” light nuclei like hydrogen into helium and is what powers stars, whereas fission “splits” heavy nuclei and is used in many human-made nuclear power plants and weapons. Remember that stars are mostly hydrogen, which naturally suggests fusion as the main energy process. Final Answer: Energy in the Sun is continuously generated mainly due to Nuclear fusion reactions in its core.

Question 19

Rolling friction that acts when a body rolls over a surface is primarily caused by which of the following?

Accepted Answer

Introduction / Context: Friction plays a crucial role in everyday motion. While sliding friction is often discussed, rolling friction, which acts when wheels or balls roll over surfaces, is also very important in engineering and physics. This question asks about the main physical cause of rolling friction, a subtle concept that goes beyond the simple idea of surfaces rubbing against each other. Given Data / Assumptions: A rigid body such as a wheel, cylinder or ball is rolling over a surface. There is some resistance to rolling motion, known as rolling friction or rolling resistance. We must choose which factor best explains this resistance among non-elastic effects, gravity, power dissipation or buoyancy. We ignore sliding friction and assume pure rolling without slipping at the point of contact. Concept / Approach: In an ideal world with perfectly rigid bodies and surfaces, a wheel rolling without slipping would experience no energy loss due to friction at the point of contact, and rolling friction would be zero. In reality, both the wheel and the surface deform slightly due to the load. These deformations are not perfectly elastic, meaning that some energy is lost in each cycle of compression and recovery. This non-elastic behaviour, often called hysteresis loss, is the primary cause of rolling friction. Gravity provides the normal force but does not by itself explain friction, and “power dissipation” is a result rather than a cause. Buoyancy acts in fluids, not in typical rolling situations on solid ground. Step-by-Step Solution: Step 1: Recognise that as a wheel or ball rolls, the part in contact with the surface is slightly flattened due to the load. Step 2: Both the rolling body and the surface undergo elastic and plastic deformation at the contact zone. Step 3: Because the materials are not perfectly elastic, they do not recover all the energy stored during deformation; some energy is lost as heat. Step 4: This non-elastic behaviour leads to a small resistive torque that opposes rolling, which we perceive as rolling friction. Step 5: Hence, non-elastic effects in the materials are the main cause of rolling friction. Verification / Alternative check: Engineers who design tyres, bearings and road surfaces pay close attention to material properties such as elasticity, hysteresis and internal damping because these control rolling resistance. Experiments show that wheels made from very elastic materials and running on smooth, rigid surfaces have significantly lower rolling friction. This observation is consistent with the idea that reduced non-elastic deformation lowers rolling resistance. Gravity is still present in these experiments but does not change, whereas changes in material elasticity clearly change rolling friction. Why Other Options Are Wrong: Gravity: Gravity provides the normal force and influences how strongly two surfaces press together, but it does not directly cause rolling friction. Power dissipation: This is a consequence of friction, not a fundamental cause. Rolling friction leads to power dissipation, not the other way around. Buoyancy: Buoyant forces act on bodies immersed in fluids and are unrelated to the friction encountered in normal rolling motion on solid surfaces. Common Pitfalls: Students sometimes think of friction only as a product of roughness and forget that deformation and material properties also play a role. Another mistake is to pick “gravity” whenever normal forces are involved, even though gravity alone does not explain why energy is lost in rolling. The key idea to remember is that rolling friction arises mainly from non-elastic deformations and internal energy losses in the materials at the rolling contact. Final Answer: Rolling friction is primarily caused by Non-elastic effects in the rolling body and the surface.

Question 20

On the Kelvin temperature scale, the melting point of ice (equivalent to 0 °C) is equal to how many kelvin?

Accepted Answer

Introduction / Context: Temperature can be measured on different scales, such as Celsius, Fahrenheit and Kelvin. The Kelvin scale is an absolute temperature scale widely used in physics and chemistry because it starts from absolute zero. This question asks you to convert the familiar melting point of ice from degrees Celsius to the Kelvin scale, a simple but frequently tested concept. Given Data / Assumptions: The melting point of pure ice at standard atmospheric pressure is 0 °C. The Kelvin scale is related to the Celsius scale by a fixed offset. We must compute the corresponding temperature in kelvin. We assume standard conversion without any experimental variation. Concept / Approach: The Kelvin scale is defined so that absolute zero, the lowest possible temperature where molecular motion is minimal, is 0 K. The relationship between Celsius and Kelvin is given by: K = °C + 273 (more precisely, 273.15, but many basic problems use 273 for simplicity). Therefore, to convert a temperature from Celsius to Kelvin, we simply add 273. Applying this to 0 °C directly gives the answer. Step-by-Step Solution: Step 1: Write the conversion formula between the Celsius and Kelvin scales: K = °C + 273. Step 2: Substitute the melting point of ice, 0 °C, into the formula. Step 3: Calculate K = 0 + 273. Step 4: This gives K = 273. Step 5: Therefore, 0 °C corresponds to 273 K on the Kelvin scale. Verification / Alternative check: We also know that the boiling point of water at standard pressure is 100 °C, which corresponds to approximately 373 K (since 100 + 273 = 373). The difference between the freezing and boiling points of water is 100 °C on the Celsius scale and also 100 K on the Kelvin scale, showing that both scales use the same size degree, just shifted by 273. This consistency confirms that 0 °C must be 273 K. Any other value would disrupt the known relationship between the two fixed points. Why Other Options Are Wrong: 373 K: This corresponds to about 100 °C, the boiling point of water, not the melting point of ice. 100 K: This would imply a Celsius temperature of -173 °C, far below the freezing point of water. 0 K: This is absolute zero, the theoretical temperature at which all classical motion of particles stops. It is not the melting point of ice. Common Pitfalls: Some learners either forget the exact offset between Celsius and Kelvin or confuse the boiling point and freezing point values. A quick memory aid is: freezing point of water is 0 °C or 273 K, boiling point is 100 °C or 373 K. Remember that the scale shift is always 273 (or 273.15 in more precise work), and the size of one degree is the same in both Celsius and Kelvin. Final Answer: On the Kelvin scale, the melting point of ice is 273 K.

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