Question 1

In which of the following media does light travel fastest, if we ignore vacuum and consider typical conditions?

Accepted Answer

Introduction / Context: The speed of light is not the same in all materials. In a vacuum, light has its maximum possible speed, and in any material medium such as air, water or glass, light slows down. The amount by which light slows depends on the optical properties of the medium, expressed through its refractive index. This question checks whether you know in which of the listed everyday media light travels fastest. Given Data / Assumptions: - We compare four media: air, water, glass and diamond. - We ignore vacuum in the options, although light is fastest there. - We assume common types of glass and typical atmospheric air at standard conditions. - Refractive index is used to compare the relative speeds of light. Concept / Approach: The speed of light in a medium is given by the relation v = c / n, where c is the speed of light in vacuum and n is the refractive index of the medium. A larger refractive index means that light travels more slowly. Air has a refractive index very close to 1, water has a refractive index around 1.33, common glass is around 1.5, and diamond is around 2.4. Since air has the smallest refractive index of the four, light travels fastest in air compared with water, glass and diamond. Step-by-Step Solution: Step 1: Recall the formula v = c / n, where v is the speed in the medium and n is the refractive index. Step 2: Compare typical refractive indices: air is about 1.00, water about 1.33, glass about 1.5 and diamond about 2.4. Step 3: For air, v_air is approximately c / 1.00, which is almost equal to c. Step 4: For water, v_water is c / 1.33, which is smaller than v_air, so light is slower in water than in air. Step 5: For glass, v_glass is c / 1.5, which is even smaller, so light is slower in glass. Step 6: For diamond, v_diamond is c / 2.4, the smallest of all, so light is slowest in diamond among the given media. Verification / Alternative check: Experiments with refraction support this ordering. Light bends more when entering a medium with a higher refractive index because it slows down more. Glass bends light more than water, and diamond bends light very strongly, which matches the idea that light is slower in these materials. Air causes only a very small bending at the air to vacuum interface, consistent with its refractive index being very close to 1 and therefore its speed being close to that in vacuum. Why Other Options Are Wrong: Water: Light is slower in water than in air because water has a higher refractive index than air. Glass: Common glass has a higher refractive index than water, so light is slower in glass than in water and air. Diamond: Diamond has a very high refractive index, so light travels slowest in diamond among the listed media. Common Pitfalls: Students sometimes assume that denser materials always slow light more, without distinguishing between mass density and optical density. While related, it is the optical property measured by the refractive index that directly determines the speed of light in a medium. Another mistake is forgetting that vacuum is not listed in the options, even though light is fastest there. When only material media like air, water, glass and diamond are provided, always choose the one with the lowest refractive index. Final Answer: Among the given media, light travels fastest in air.

Question 2

For prolonged daily exposure, what sound level is generally considered acceptable for the human ear to avoid a significant risk of hearing damage?

Accepted Answer

Introduction / Context: Human hearing is sensitive but also vulnerable to damage from prolonged exposure to loud sounds. Health and safety agencies provide guidelines on acceptable noise levels to protect workers and the general public from hearing loss. This question asks you to recall the approximate threshold value used in many standards and to understand that safe exposure levels are defined as below that threshold for long periods. Given Data / Assumptions: - Sound intensity is measured in decibels, abbreviated as dB, on a logarithmic scale. - Many occupational safety standards use around 85 dB as a key reference level. - The question refers to prolonged daily exposure, not to very short bursts of sound. - We consider typical adult hearing and do not address special medical conditions. Concept / Approach: Guidelines from health and safety organisations often state that continuous exposure to sound levels at or above about 85 dB over an eight hour workday can increase the risk of noise induced hearing loss. This means that for long durations, sound levels should ideally be kept below this threshold. Occasional peaks above 85 dB may be acceptable if the total daily dose of noise is limited, but as a simple general rule, less than 85 dB is considered an acceptable continuous level for most people. Therefore, the best answer describes sound levels lower than about 85 dB. Step-by-Step Solution: Step 1: Recall that noise level standards define a threshold of concern around 85 dB for long term exposure. Step 2: Understand that levels above this value, especially for many hours per day, can damage the delicate hair cells in the inner ear. Step 3: Recognise that exactly 85 dB is generally treated as the limit where protective measures start to become important, not as a guaranteed safe level for any duration. Step 4: Acceptable levels for long periods are therefore slightly lower than this limit. Step 5: Option C, which describes sound levels lower than about 85 dB, best matches this safety principle. Step 6: Conclude that option C is the correct choice for an acceptable noise level for prolonged exposure. Verification / Alternative check: If you look at typical examples of sound levels, normal conversation is around 60 dB, busy street traffic may be around 70 dB, and a loud motorbike or rock concert can exceed 100 dB. Everyday environments where people spend many hours, such as offices or classrooms, generally try to maintain noise levels well below 85 dB. Ear protection is recommended in factories, construction sites or concerts where levels can easily exceed this threshold, reinforcing the idea that levels below 85 dB are preferred for long term exposure. Why Other Options Are Wrong: Sound levels greater than 85 dB: These levels can be harmful if exposure is prolonged and usually require hearing protection. Exactly 85 dB for any duration: The risk depends on both level and time. Even 85 dB can be harmful if exposure is long enough. Any sound level, there is no guideline: This is incorrect because many national and international bodies provide clear numerical guidelines on safe noise levels. Common Pitfalls: A major misunderstanding is to think that if a person does not feel pain in their ears, the sound level must be safe. In reality, damage can occur gradually at levels that do not feel painful. Another common mistake is to ignore the duration of exposure and focus only on how loud a sound seems at one moment. Always remember that safe exposure guidelines combine sound level and time, and that staying below about 85 dB for extended periods is a sensible rule. Final Answer: For prolonged daily exposure, a noise level lower than about 85 dB is generally considered acceptable for the human ear.

Question 3

In basic thermodynamics, the relationship Q = m c ΔT is the formula for which physical quantity or process?

Accepted Answer

Introduction / Context: The formula Q = m c ΔT appears very frequently in physics and engineering problems involving heating or cooling of substances. It expresses a relationship between the heat supplied or removed, the mass of the substance, its specific heat capacity and the resulting change in temperature. This question checks whether you can correctly interpret what physical quantity or process this formula represents, and distinguish it from related concepts such as enthalpy and latent heat. Given Data / Assumptions: - Q represents the amount of heat energy transferred to or from a substance. - m is the mass of the substance, usually in kilograms or grams. - c is the specific heat capacity of the substance, a material property. - ΔT is the change in temperature, final temperature minus initial temperature. - The formula applies where the substance does not change phase over the temperature range considered. Concept / Approach: Specific heat capacity c is defined as the amount of heat required to raise the temperature of unit mass of a substance by 1 degree. When this definition is applied to a mass m undergoing a temperature change ΔT, the total heat required is Q = m c ΔT. This formula therefore gives the heat transfer associated with a temperature change of a given mass, assuming no phase change and constant c over the range. It is not, by itself, the definition of c, which is obtained by rearranging the equation, and it does not represent latent heat, which involves phase changes at constant temperature. Step-by-Step Solution: Step 1: Recall the definition of specific heat capacity: c is heat needed per unit mass per unit temperature rise. Step 2: From this, write Q for a finite mass and temperature change as Q = m c ΔT. Step 3: Recognise that Q is the heat transferred to change the temperature of the mass m by ΔT without a phase change. Step 4: Understand that enthalpy change is a broader concept which can include phase changes and other effects; Q = m c ΔT is a special case. Step 5: Note that latent heat of vaporization uses a different form, Q = m L, with no temperature change while the phase changes. Step 6: Conclude that the formula most directly represents the heat transferred due to a temperature change in a substance of mass m using specific heat capacity. Verification / Alternative check: As a check, consider heating 2 kg of water with c approximately 4200 J kg^-1 K^-1 through 10 degrees. Using Q = m c ΔT gives Q = 2 * 4200 * 10 = 84000 J. This is clearly a calculation of heat transfer. If the same water were boiling at constant temperature, the appropriate formula would involve latent heat, not a temperature difference. This confirms that Q = m c ΔT is linked to sensible heating or cooling, not to latent heat or general enthalpy changes in all processes. Why Other Options Are Wrong: General enthalpy change for any process: Enthalpy can change due to many factors, including pressure and phase change. Q = m c ΔT is only one special case and does not cover all situations. The numerical value of specific heat capacity itself: Specific heat capacity is defined per unit mass and per degree; the formula must be rearranged to c = Q / (m ΔT) to compute c. Latent heat of vaporization during a phase change: Latent heat is associated with phase change at constant temperature and uses Q = m L, not Q = m c ΔT. Common Pitfalls: Learners sometimes apply Q = m c ΔT even when a phase change is taking place, which is not correct. Another common confusion is to label Q = m c ΔT as the specific heat formula, without specifying that it calculates the heat transferred for a given temperature change. Keeping the roles of Q, m, c and ΔT clear, and remembering the separate form for latent heat, helps avoid misapplication of this important equation. Final Answer: The relationship Q = m c ΔT is the formula for heat transferred when the temperature of a mass changes, using specific heat capacity.

Question 4

Sound waves cannot travel through which of the following environments?

Accepted Answer

Introduction / Context: Sound is a mechanical wave that we experience as noise or music. Unlike light, which is an electromagnetic wave, sound requires a material medium to propagate. This basic fact is often tested in physics questions to ensure that students understand the difference between mechanical and electromagnetic waves. The question asks you to identify the environment in which sound waves cannot travel at all. Given Data / Assumptions: - Sound waves are mechanical vibrations of particles in a medium. - The listed environments are a vacuum, water, glass and hydrogen gas. - We assume idealised conditions where the vacuum is perfect, with no particles at all. - We consider audible sound frequencies, not other types of waves. Concept / Approach: Mechanically, sound propagates when particles of the medium vibrate and transfer energy to neighbouring particles. If there are no particles, there can be no vibration and therefore no sound. A perfect vacuum contains no matter and is the absence of particles. In contrast, liquids like water, solids like glass and gases like hydrogen all contain particles that can vibrate. Therefore, sound can travel in these media, although at different speeds, but it cannot travel in a perfect vacuum. Step-by-Step Solution: Step 1: Recall that sound is a mechanical wave and needs a medium such as a gas, liquid or solid. Step 2: Recognise that a vacuum, by definition, contains no particles to vibrate. Step 3: Since there are no particles in a perfect vacuum, sound waves have nothing to compress or rarefy. Step 4: Therefore, sound cannot propagate through a perfect vacuum. Step 5: Consider water, glass and hydrogen gas. All of these are materials made of particles, so sound can travel through them. Step 6: Conclude that the environment in which sound cannot travel is the perfect vacuum. Verification / Alternative check: Simple classroom experiments support this concept. If a ringing bell is placed inside a bell jar and the air is gradually pumped out, the sound becomes fainter and eventually disappears, even though the bell continues to move. This shows that as the medium is removed, sound cannot reach the listener. In contrast, sound travels in water, which is why divers can hear underwater, and in solids such as rails, along which train vibrations can be detected. Why Other Options Are Wrong: Liquid water: Sound travels quite well in water and often moves faster than in air because particles are more closely packed. Solid glass: Sound travels in solids, often faster than in liquids and gases, due to strong interactions between particles. Hydrogen gas: Even though hydrogen gas is very light, it is still a gas with particles, so sound can propagate through it. Common Pitfalls: A common mistake is to confuse sound with light and to think that because light can travel through vacuum, sound can as well. Another pitfall is underestimating how well sound travels through solids like metal or glass. Always remember that sound requires matter, so any gas, liquid or solid can carry sound, while a perfect vacuum cannot. Final Answer: Sound waves cannot travel through a perfect vacuum.

Question 5

Compared to other regions of the electromagnetic spectrum, visible light has a higher frequency than which of the following?

Accepted Answer

Introduction / Context: The electromagnetic spectrum covers a wide range of wave frequencies, from very low frequency radio waves to very high frequency gamma rays. Visible light occupies a narrow band in the middle of this spectrum. Understanding where visible light fits in, and which regions have higher or lower frequencies, is a basic part of physics and general science. This question focuses specifically on comparing the frequency of visible light with radio waves, X rays and ultraviolet radiation. Given Data / Assumptions: - Visible light lies between infrared and ultraviolet in the electromagnetic spectrum. - Radio waves have the lowest frequencies among the options. - X rays and ultraviolet rays are on the high frequency side compared to visible light. - Higher frequency corresponds to higher photon energy. Concept / Approach: From lowest to highest frequency, the spectrum can be ordered approximately as radio waves, microwaves, infrared, visible light, ultraviolet, X rays and gamma rays. Visible light has higher frequency than all types of radio waves and microwaves, but lower frequency than ultraviolet and X rays. Therefore, among the given options, visible light has a higher frequency than radio waves but a lower frequency than ultraviolet rays and X rays. The only correct comparison in which visible light is higher is with radio waves. Step-by-Step Solution: Step 1: Recall the general order of the electromagnetic spectrum by frequency. Step 2: Place radio waves at the low frequency end, below microwaves and infrared. Step 3: Place visible light above infrared but below ultraviolet. Step 4: Place ultraviolet above visible light, and X rays above ultraviolet. Step 5: Compare visible light with each option. Visible light is higher in frequency than radio waves but lower than ultraviolet and X rays. Step 6: Conclude that the only correct statement is that visible light has a higher frequency than radio waves. Verification / Alternative check: Visible light frequencies are roughly in the range of 4 * 10^14 to 7.5 * 10^14 hertz. Radio waves can have frequencies as low as a few kilohertz or megahertz, which are far lower than visible light. Ultraviolet radiation extends from about 7.5 * 10^14 hertz upward into the 10^15 hertz range, and X rays are still higher. These numerical ranges confirm that visible light sits between infrared and ultraviolet, above radio waves but below ultraviolet and X rays. Why Other Options Are Wrong: X rays: X rays have much higher frequencies and energies than visible light, so visible light does not exceed them in frequency. Ultraviolet rays: Ultraviolet radiation is immediately above visible light in the spectrum, so it has higher frequency than visible light. All of the above: This would claim that visible light has a higher frequency than radio, ultraviolet and X rays, which is not correct for the latter two. Common Pitfalls: Students sometimes remember only parts of the spectrum, such as that X rays are dangerous and ultraviolet causes sunburn, but forget the detailed ordering. Another confusion arises because visible light is the most familiar, leading some to think it must be in the middle of everything or higher than most other bands. Memorising the standard order from radio to gamma and practising comparisons helps avoid such mistakes. Final Answer: Visible light has a higher frequency than radio waves.

Question 6

In meteorology, what is the main physical reason that a sea breeze blows from the sea toward the land during the daytime?

Accepted Answer

Introduction / Context: Along coastlines, people often experience a cool breeze blowing from the sea toward the land during warm sunny days. This regular daytime wind is called a sea breeze. Understanding why sea breezes occur requires knowledge of how land and water heat at different rates and how this leads to pressure differences that drive air motion. This question asks you to identify the main physical cause behind the formation of a sea breeze. Given Data / Assumptions: - The situation is a sunny daytime along a coast with land on one side and sea on the other. - Land and water have different specific heat capacities and therefore heat at different rates. - Air pressure depends on air temperature and density above land and sea. - Planetary effects like the Coriolis force influence wind direction on large scales but are not the primary local cause here. Concept / Approach: Land surfaces heat up quickly during the day because they have lower specific heat capacity and often lower thermal inertia than water. Water heats more slowly. As a result, air over the land becomes warmer and less dense than air over the sea. Warm air over the land rises, creating a region of lower surface pressure. Cooler, higher pressure air over the sea then flows toward the land to replace the rising warm air. This flow of air from sea to land is the sea breeze. The root cause is therefore a pressure difference created by unequal heating of land and sea, which itself is due to their different specific heats. Step-by-Step Solution: Step 1: During daytime, the sun heats both land and sea, but land warms faster because of lower specific heat capacity. Step 2: The air above the land warms, becomes less dense and rises, producing a local region of lower air pressure near the surface. Step 3: The air above the sea remains relatively cooler and denser, so surface pressure over the water stays relatively higher. Step 4: Air flows from regions of higher pressure to regions of lower pressure, so air moves from the sea toward the land. Step 5: This horizontal movement of cooler air from the sea that replaces rising warm air over land is the sea breeze. Step 6: Therefore, the immediate cause is the pressure difference created by unequal heating of land and sea, controlled by their different specific heats. Verification / Alternative check: Observations show that sea breezes are strongest on clear, hot days when the temperature contrast between land and sea is large. They weaken or disappear on cloudy days or at night when the temperature difference is smaller. This behaviour supports the explanation based on differential heating and pressure differences. Numerical weather models and meteorological diagrams also show pressure gradients from sea to land during typical sea breeze conditions. Why Other Options Are Wrong: Only the Coriolis force: The Coriolis effect turns winds on large scales but does not by itself create the initial pressure difference that drives a sea breeze. Random changes in wind direction: Sea breezes follow a regular pattern based on heating differences and are not random. Only the specific heat of water, without affecting air pressure: Specific heat is important but only because it leads to temperature and pressure differences; ignoring pressure misses the key driving force for the wind. Common Pitfalls: Learners sometimes think of wind as purely random or blame sea breezes entirely on coastal geography. Others focus only on temperature and forget that pressure differences are the actual driving force for air motion. Remember that land and sea heat at different rates due to their specific heats, and this creates pressure gradients that produce well organised local winds like sea breezes. Final Answer: A sea breeze is mainly caused by a pressure difference created by unequal heating of land and sea due to their different specific heats.

Question 7

Which type of mirror is commonly used as a rear view mirror in vehicles to provide a wider field of view?

Accepted Answer

Introduction / Context: Rear view mirrors in cars, buses and other vehicles are essential for safe driving because they allow the driver to see traffic behind and to the sides of the vehicle. The choice of mirror type affects how much of the scene the driver can see and how large or small objects appear. This question asks which type of mirror is most suitable and therefore commonly used as a rear view mirror, given the need for a wide field of view. Given Data / Assumptions: - There are three common mirror types in basic optics: plane, concave and convex. - Rear view mirrors must show a wide area behind and to the side of the vehicle. - Drivers need an image that is upright and reasonably clear, even if the size is reduced. - Safety and practicality are more important than producing life sized images. Concept / Approach: A plane mirror provides an image of the same size as the object and preserves distances, but its field of view is limited by its size. Concave mirrors can magnify images when objects are within focal length, but they can also invert images and give a very limited field of view, which is not suitable for rear view mirrors in vehicles. Convex mirrors, however, produce virtual, upright, diminished images and allow a much wider field of view because they curve outward. This means drivers can see more of the road behind and to the side, even though the images appear smaller. For this reason, convex mirrors are the standard choice for rear view mirrors on the passenger side and often for external mirrors in general. Step-by-Step Solution: Step 1: Describe a plane mirror: it produces an upright virtual image of the same size as the object and has a limited field of view. Step 2: Describe a concave mirror: depending on the object position, it can produce magnified or inverted images and tends to have a narrow field of view. Step 3: Describe a convex mirror: it always produces a virtual, upright, reduced size image and provides a wide field of view. Step 4: Rear view mirrors need to cover as much of the area behind and beside the vehicle as possible for safety. Step 5: A convex mirror achieves this by spreading out the reflected rays and compressing a wide area into a smaller image. Step 6: Conclude that the convex mirror is the preferred and commonly used type for vehicle rear view mirrors. Verification / Alternative check: In many vehicles, you can read a warning printed on the side mirror that says objects in the mirror are closer than they appear. This is a direct consequence of the convex mirror producing reduced size images. Observing the curved outward shape of the mirror also confirms that it is convex. Plane and concave mirrors do not show this characteristic behaviour in standard car mirrors. Why Other Options Are Wrong: Concave mirror: Its field of view is limited, and it can form inverted images of distant objects, which would be confusing and unsafe for drivers. Plane mirror: While it gives correctly sized images, the field of view is too narrow compared with what a convex mirror can provide in the same physical space. Any type of mirror can be used equally well: This is incorrect because different mirror types produce different image properties, and only the convex mirror offers the needed combination of wide field and upright image. Common Pitfalls: A common confusion is between the inside and outside curvature of mirrors. Learners sometimes mix up concave and convex by their shape. Remember that a convex mirror bulges outward like the outside of a sphere and spreads light rays apart, giving a wide view. Associating convex mirrors with rear view mirrors and shop security mirrors helps reinforce this concept. Final Answer: The rear view mirrors in vehicles commonly use a convex mirror to provide a wide field of view.

Question 8

At night along a coastline, what is the main physical reason that a land breeze blows from the land toward the sea?

Accepted Answer

Introduction / Context: Just as a sea breeze often blows from sea to land during the day, a land breeze commonly blows from land to sea at night in many coastal areas. This daily reversal of wind direction is closely linked to how quickly land and water surfaces heat up and cool down. The question asks for the main physical reason for the formation of a land breeze during the night. Given Data / Assumptions: - We consider night time conditions along a coast with land on one side and sea on the other. - Land and sea have different thermal properties, especially specific heat capacity. - Air pressure near the surface depends on the temperature and density of the air above land and sea. - The situation is essentially the reverse of the daytime sea breeze pattern. Concept / Approach: Land has a lower specific heat capacity than water and generally cools down faster than the ocean after sunset. As the land cools, the air above it becomes cooler and denser, leading to a higher surface air pressure over land. The sea retains heat longer and cools more slowly, so the air above the water stays relatively warmer and less dense, giving a lower surface pressure over the sea. Air naturally flows from high pressure to low pressure, so cool, dense air moves from the land toward the sea. This flow of air from land to sea during the night is the land breeze. Step-by-Step Solution: Step 1: After sunset, the land surface loses heat quickly, while the sea retains heat due to higher specific heat and mixing. Step 2: The air above the land cools, becomes denser and sinks, creating a region of relatively high surface pressure over land. Step 3: The air above the water remains comparatively warmer and less dense, so surface pressure over the sea is relatively lower. Step 4: Air flows from the high pressure region over land toward the low pressure region over the sea. Step 5: This movement of air from land to sea is observed as the land breeze at night. Step 6: Therefore, the key cause is a pressure difference created because land cools faster than sea at night. Verification / Alternative check: Meteorological observations show that land breezes are most pronounced on clear, calm nights, when radiative cooling of land surfaces is strong and temperature contrasts with the sea are clear. Overcast or stormy conditions can disrupt this pattern. This behaviour supports the explanation based on differential cooling and associated pressure gradients. Weather charts for coastal regions often show nighttime wind arrows pointing from land toward sea, consistent with land breeze dynamics. Why Other Options Are Wrong: Heating of the land more than the sea during the night: At night, land usually cools faster than the sea, so this statement reverses the actual physical situation. A uniform temperature with no pressure gradient: Without temperature and pressure differences, there would be no organised flow of air, so a land breeze would not develop. No temperature or pressure difference involved: This ignores the fundamental rule that wind is driven by pressure differences, which in this case arise from differential cooling. Common Pitfalls: Some learners memorise only the idea of a sea breeze and forget that land and sea breeze are a pair of related phenomena that reverse between day and night. Others may remember that sea breezes are linked to heating differences but do not apply the same concept to cooling differences at night. Keeping a clear picture that land heats and cools faster than water, and that air flows from high pressure to low pressure, helps interpret both sea and land breezes correctly. Final Answer: At night, a land breeze occurs because a pressure difference is created when land cools faster than sea, causing air to flow from the land toward the sea.

Question 9

Magnetic domains, which can align to form strong permanent magnets, normally occur in which of the following materials?

Accepted Answer

Introduction / Context: Magnetic materials are often described in terms of magnetic domains, which are small regions where atomic magnetic moments are aligned in the same direction. In some materials, these domains can be aligned by an external magnetic field to produce strong permanent magnets. This question asks you to recall which of the listed metals naturally form such magnetic domains and can therefore exhibit strong ferromagnetism. Given Data / Assumptions: - Copper and silver are common metals used in electrical wiring and jewellery. - Iron is a well known ferromagnetic material used in magnets and transformers. - Magnetic domains are characteristic of ferromagnetic and some ferrimagnetic materials. - We are interested in materials that can show strong permanent magnetisation under ordinary conditions. Concept / Approach: Ferromagnetic materials, such as iron, cobalt and nickel, have atomic structures that allow groups of atomic magnetic moments to align strongly in regions called domains. In an unmagnetised piece of such a material, the domains are randomly oriented, so the net magnetisation is small. When a strong external magnetic field is applied, many domains rotate or grow so that more of them align in one direction, producing a strong overall magnet. Copper and silver, on the other hand, are diamagnetic or weakly paramagnetic; they do not form ferromagnetic domains and do not become strong permanent magnets under normal conditions. Step-by-Step Solution: Step 1: Recall that iron is the classic example of a ferromagnetic material used in school experiments with magnets. Step 2: Recognise that ferromagnetic materials have magnetic domains that can be aligned to create a strong net magnetisation. Step 3: Identify that copper and silver are excellent electrical conductors but are not ferromagnetic; they are not used as permanent magnets. Step 4: Understand that magnetic domains in the sense used in ferromagnetism are not present in copper and silver. Step 5: Therefore, the material among the options that normally exhibits magnetic domains is iron. Step 6: Conclude that the correct answer is iron as a typical ferromagnetic metal. Verification / Alternative check: Practical experience supports this reasoning. Iron nails are strongly attracted by magnets, and iron cores are used inside electromagnets and transformers to enhance magnetic fields. Copper and silver do not cling to magnets in the same way. They are chosen for their electrical conductivity, not for magnetic properties. Physics texts also list iron as a standard domain forming ferromagnet, while copper and silver are described as non magnetic or very weakly magnetic. Why Other Options Are Wrong: Copper: This is a diamagnetic metal and does not form ferromagnetic domains or useful permanent magnets. Silver: This is also diamagnetic and is not used as a ferromagnetic material. All of the above metals: This would incorrectly include copper and silver as domain forming materials, which they are not. Common Pitfalls: Because copper and silver are metals and conduct electricity, some students assume they share all properties with iron, including magnetism. This is not the case. It is important to distinguish electrical conductivity from magnetic behaviour. Remember that only certain metals, mainly iron, cobalt, nickel and some alloys, are ferromagnetic and form strong magnetic domains. Final Answer: Magnetic domains that can align to produce strong permanent magnets normally occur in iron, a typical ferromagnetic metal.

Question 10

Which pair of elements are both malleable metals and good conductors of electricity?

Accepted Answer

Introduction / Context: In basic chemistry and physics, metals are often described as malleable and good conductors of heat and electricity. Non metals and noble gases, by contrast, usually lack these properties. This question tests whether you can correctly identify which pair of elements from the options are metals that show the typical metallic properties of malleability and good electrical conductivity. Given Data / Assumptions: - Tin and silver are both well known metallic elements. - Iodine is a halogen non metal and xenon is a noble gas. - Malleability means the ability to be hammered or rolled into thin sheets without breaking. - Good electrical conductivity is a characteristic property of metals. Concept / Approach: Metals have a structure in which outer electrons are relatively free to move, forming an electron sea that allows electrical conduction and some plastic deformation under stress. This gives rise to properties like malleability and good electrical conductivity. Tin is a soft, malleable metal used in alloys and coatings. Silver is a metal with very high electrical conductivity, better than copper in pure form. Iodine, on the other hand, is a non metal that forms crystals and does not show metallic malleability or high electrical conductivity. Xenon is a noble gas, existing as isolated atoms under normal conditions, and is neither malleable nor a conductor in the solid metallic sense. Step-by-Step Solution: Step 1: Identify tin as a metal in the periodic table, with metallic bonding and typical metallic properties. Step 2: Identify silver as a metal, widely used for jewellery and electrical contacts due to its excellent conductivity. Step 3: Recognise that iodine is a halogen non metal with molecular crystals at room temperature, not malleable and not a good conductor. Step 4: Recognise that xenon is a noble gas and exists as a gas of isolated atoms, not as a solid metal. Step 5: Conclude that the only pair where both elements are malleable metals and good conductors is tin and silver. Step 6: Therefore, option A is the correct answer. Verification / Alternative check: You can check typical uses of each element. Tin is used in solder and tinplate, which require malleability and decent conduction. Silver is used in electrical contacts, high quality connectors and jewellery, all of which rely on its metallic properties. Iodine finds use in medicine and as a disinfectant, not as a structural metal. Xenon is used in lamps and special gas mixtures, again not as a solid metal. These applications confirm which elements are metallic and which are not. Why Other Options Are Wrong: Iodine and silver: Only silver is a malleable, highly conductive metal; iodine is a non metal. Tin and xenon: Tin is a metal, but xenon is a noble gas and not a malleable conductor. Iodine and xenon: Both are non metallic in their usual forms, so neither is a malleable conductor of electricity like a metal. Common Pitfalls: Learners sometimes focus only on one element in a pair, such as silver being a well known conductor, and ignore the nature of the second element. It is crucial to check both elements in the pair. Another source of confusion is assuming that any element associated with lighting or technology, such as xenon in lamps, must be metallic, which is not true. Always check the position of the element in the periodic table and its bonding type. Final Answer: The pair where both elements are malleable metals and good conductors of electricity is tin and silver.

Question 11

In mechanics, what primarily distinguishes a second class lever from a third class lever?

Accepted Answer

Introduction / Context: Levers are simple machines that allow a small effort to move a larger load by using a pivot point called the fulcrum. There are three standard classes of levers, and they are classified according to the relative positions of the fulcrum, the load and the effort. Understanding these classes helps in analysing everyday tools such as crowbars, wheelbarrows and tongs. This question asks what primarily distinguishes a second class lever from a third class lever. Given Data / Assumptions: - A lever has three key points: the fulcrum, the load and the point where effort is applied. - First class, second class and third class levers differ in the order of these three points along the lever arm. - We focus on the distinction between second class and third class levers. - Examples include wheelbarrows, nutcrackers and human forearms. Concept / Approach: In a second class lever, the load lies between the fulcrum and the effort. A wheelbarrow is a classic example: the wheel is the fulcrum, the load is in the tray and the effort is applied at the handles. In a third class lever, the effort is applied between the fulcrum and the load. The human forearm lifting a weight is a good example: the elbow is the fulcrum, the biceps apply effort partway along the arm and the load is in the hand. Therefore, the primary distinction is the relative positions of load and effort relative to the fulcrum, not the material or the mere presence of multiple loads. Step-by-Step Solution: Step 1: Recall that a second class lever has the sequence fulcrum, load, effort along the lever. Step 2: Recall that a third class lever has the sequence fulcrum, effort, load along the lever. Step 3: Observe that in a second class lever, the load is between fulcrum and effort, giving a mechanical advantage. Step 4: Observe that in a third class lever, the effort is between fulcrum and load, often providing speed or range of motion rather than force advantage. Step 5: Recognise that this classification is based only on the relative positions of these three points. Step 6: Conclude that the main distinguishing factor between second and third class levers is the relative position of load and effort with respect to the fulcrum. Verification / Alternative check: Consider some concrete examples. A wheelbarrow (second class) has the load in the middle and effort at the end. A nutcracker is another second class lever. In contrast, the human arm lifting a weight (third class) has the muscle force applied between the elbow and the hand. The fishing rod or a pair of tweezers also act as third class levers. These standard textbook examples confirm that the lever class depends on where the load and effort are placed relative to the fulcrum. Why Other Options Are Wrong: Only the location of the fulcrum: In all lever classes the fulcrum is at one end or between the other forces; classification depends on all three points, not just the fulcrum position alone. The presence of multiple loads: A lever can act on more than one load, but this does not define the lever class; the key factor is the order of fulcrum, load and effort. The material of the fulcrum support: The class of a lever does not depend on what it is made of but on the geometry of forces and positions. Common Pitfalls: Students sometimes focus on how the lever is used in practice, such as whether it is for lifting or cutting, instead of its geometric configuration. Others think the names of the classes refer to the strength or efficiency of the lever. To avoid confusion, always sketch the positions of fulcrum, load and effort in a simple diagram and classify the lever by the sequence of these points. Final Answer: Second and third class levers are primarily distinguished by the relative position of the load and the effort with respect to the fulcrum.

Question 12

In human vision, what does long sightedness (hypermetropia) mean?

Accepted Answer

Introduction / Context: Common vision defects include short sightedness and long sightedness. Understanding these conditions is important not only for physics of lenses but also for general health knowledge. Long sightedness is also known as hypermetropia. This question asks you to recall what a long sighted person experiences when looking at near and distant objects. Given Data / Assumptions: - Long sightedness is referred to as hypermetropia in medical and physics contexts. - The question concerns clarity of near versus distant vision. - We consider the usual case without any corrective lenses in place. - Short sightedness (myopia) is the opposite condition, where far objects are not clear. Concept / Approach: In a normal eye, the lens changes shape to focus images of objects at different distances on the retina. In long sightedness, the eyeball is too short or the eye lens is not strong enough, so light from near objects is not converged sufficiently and the image tends to form behind the retina. As a result, near objects appear blurry, while distant objects may still be focused correctly on the retina and appear reasonably clear. Corrective convex lenses are used in glasses or contact lenses to converge light more before it enters the eye, allowing near objects to be focused properly. Step-by-Step Solution: Step 1: Define hypermetropia as a condition where the eye has difficulty focusing on nearby objects. Step 2: Recall that in hypermetropia, the image of a near object would form behind the retina if no correction is used. Step 3: Recognise that distant rays are nearly parallel and require less convergence, so the eye may still focus distant objects well on the retina. Step 4: Conclude that near objects appear blurred, while distant objects can appear clear. Step 5: Compare this with the options and see that option A matches this behaviour. Step 6: Therefore, long sightedness means near images are blurry, while far images remain clearer. Verification / Alternative check: People with hypermetropia commonly report difficulty reading small print or doing close work, such as sewing, especially without glasses, while they may see road signs or distant scenery fairly well. Eye doctors prescribe converging lenses for such patients. Physics diagrams of hypermetropic eyes in textbooks show that the image of a near object lies behind the retina and is brought forward by a convex lens so that it falls on the retina. Why Other Options Are Wrong: Far images are blurry while near images remain clear: This describes short sightedness or myopia, not long sightedness. Both near and far images are equally blurry: This could be due to other conditions, but it is not the standard definition of hypermetropia. There is no blurring at any distance: That would correspond to near normal vision for a given age and is not a defect. Common Pitfalls: Learners sometimes reverse the definitions of long sightedness and short sightedness because the terms can be confusing. A simple way to remember is that long sighted people see long distance objects better than near ones. In contrast, short sighted people see near objects clearly and distant objects poorly. Linking these terms to the need for reading glasses or distance glasses can help fix the concepts. Final Answer: Long sightedness (hypermetropia) means near images are blurry while distant images remain clear.

Question 13

In weather systems, middle latitude cyclones (mid latitude depressions) are mainly fueled by which energy source?

Accepted Answer

Introduction / Context: Middle latitude cyclones, also called mid latitude depressions or extratropical cyclones, are large low pressure systems that bring much of the day to day weather in temperate regions. They differ from tropical cyclones in how they obtain their energy. While tropical cyclones draw energy mainly from warm ocean surfaces and latent heat release, middle latitude cyclones are driven by contrasts between warm and cold air masses. This question asks you to identify the primary energy source that fuels these cyclones. Given Data / Assumptions: - The systems in question are middle latitude cyclones, not tropical storms. - They form along fronts where warm and cold air masses meet, such as the polar front. - Temperature differences create density and pressure gradients in the atmosphere. - Gravity and rotation convert these gradients into motion. Concept / Approach: Middle latitude cyclones are examples of baroclinic instability, a process in which potential energy stored in horizontal temperature gradients is converted into kinetic energy of winds and the organised circulation of the cyclone. Strong contrasts between warm and cold air masses along fronts mean that warm, light air lies next to cold, dense air. The atmosphere is not fully in its minimum energy state, and under appropriate conditions, waves grow along the front, leading to the development of cyclones. As the system evolves, warm air rises over cold air and cold air sinks under warm air, converting potential energy into motion. Step-by-Step Solution: Step 1: Identify that middle latitude cyclones form along fronts with strong temperature contrasts between warm and cold air masses. Step 2: Recognise that these temperature contrasts imply differences in air density and pressure at different levels of the atmosphere. Step 3: Understand that the atmosphere tends toward lower potential energy states by moving air masses so that warm air rises and cold air sinks. Step 4: During cyclone development, warm air from lower latitudes is lifted over colder polar air, and cold air spreads equatorward near the surface. Step 5: This rearrangement converts gravitational potential energy associated with the temperature gradients into kinetic energy of winds and the rotating storm system. Step 6: Therefore, the main energy source is the conversion of potential energy from temperature contrasts between air masses into the kinetic energy of the cyclone. Verification / Alternative check: Meteorological models show that when horizontal temperature gradients are strong, disturbances along the front can grow rapidly into large cyclonic systems. If the temperature contrast weakens, the available potential energy decreases, and the systems become weaker. This behaviour confirms that strong temperature gradients are the primary fuel. While latent heat release from condensation adds energy and can intensify storms, the fundamental driver distinguishing middle latitude cyclones is baroclinic conversion of potential energy due to temperature contrasts. Why Other Options Are Wrong: Only the motion of air high in the atmosphere: Upper level winds shape and steer cyclones but do not provide the fundamental energy source without underlying temperature contrasts. Only the energy from falling precipitation: The energy associated with raindrops falling is relatively small compared with the energy in large scale temperature gradients. Random changes in wind direction: Cyclones are highly organised systems driven by systematic gradients, not by random fluctuations without temperature structure. Common Pitfalls: Some learners confuse the energy sources of middle latitude cyclones with those of tropical cyclones and focus only on warm ocean water or latent heat. While latent heat plays a role, the key distinguishing idea for middle latitude systems is baroclinic conversion of potential energy due to temperature contrasts. Remember that middle latitude cyclones form along fronts and owe their existence to boundaries between warm and cold air, whereas tropical cyclones grow over warm oceans with relatively uniform warm air. Final Answer: Middle latitude cyclones are mainly fueled by conversion of potential energy from strong temperature contrasts between warm and cold air masses into kinetic energy of winds.

Question 14

How do magnetic field lines behave around a long, straight current carrying wire?

Accepted Answer

Introduction / Context: A current carrying conductor produces a magnetic field in the space around it. For a long, straight wire carrying a steady current, the pattern of the magnetic field lines is simple and is a standard result in electromagnetism. This question checks whether you know the shape of these field lines and how they are arranged around the wire. Given Data / Assumptions: - The conductor is a long, straight wire carrying a steady current in one direction. - The environment is otherwise empty space or air, with no nearby magnetic materials. - We consider the magnetic field produced by the current, not by external magnets. - Right hand rule conventions are used to determine the direction of the field. Concept / Approach: According to Ampere's law and the Biot Savart law, the magnetic field around a long, straight current carrying wire forms concentric circles around the wire. The direction of the field lines can be found using the right hand grip rule: if you grasp the wire with your right hand so that your thumb points in the direction of the current, your fingers curl in the direction of the magnetic field lines. The field strength decreases with distance from the wire, but the lines form closed loops circling the wire, not radial spokes. Step-by-Step Solution: Step 1: Recall that a steady current in a straight wire produces a magnetic field in the surrounding space. Step 2: Use the right hand grip rule: thumb in the direction of the current, fingers curling around the wire. Step 3: Recognise that the curling of the fingers represents magnetic field lines forming circles centered on the wire. Step 4: Understand that each field line is a closed loop that circles the wire and does not start or end at the wire. Step 5: Note that there is no radial component pointing directly away from or toward the wire in this idealised case. Step 6: Conclude that the correct description is that the magnetic field lines circle the wire in closed loops concentric with the wire. Verification / Alternative check: Experiments using iron filings and a compass around a current carrying wire show that compass needles align tangentially to circles around the wire. If you sprinkle iron filings on a card pierced by a vertical wire carrying current and gently tap the card, the filings arrange themselves in circular patterns centered on the wire. This experimental evidence matches the theoretical prediction of circular field lines. Why Other Options Are Wrong: They extend radially outward from the wire: Radial field lines would point straight away from or toward the wire, which is not observed for a current carrying conductor. They both circle and extend radially outward: In the idealised case of a long straight wire in free space, the field is purely circular, not partially radial. They do not exist around the wire at all: This contradicts basic electromagnetic theory and many experiments showing that currents create magnetic fields. Common Pitfalls: Some learners imagine electric field lines, which can be radial around a point charge, and incorrectly transfer that picture to magnetic fields around a wire. Another confusion is to think that current produces only electric effects and to forget the associated magnetic field. Keeping separate mental pictures for electric fields of charges and magnetic fields of currents, and using the right hand grip rule consistently, helps avoid these errors. Final Answer: Magnetic field lines around a long, straight current carrying wire circle the wire in closed loops that are concentric with the wire.

Question 15

Hydroelectric power stations ultimately obtain their energy from the movement of which natural resource?

Accepted Answer

Introduction / Context: Hydroelectric energy is one of the major forms of renewable energy used around the world. In hydroelectric power stations, mechanical energy is converted into electrical energy using turbines and generators. To understand the physics and geography of power generation, it is important to know what immediate natural resource provides the energy that drives the turbines in a hydroelectric plant. This question asks you to identify that resource. Given Data / Assumptions: - The focus is on hydroelectric power, not on wind, solar or other forms. - Hydroelectric power stations typically use dams, reservoirs and rivers. - Turbines are turned by a moving fluid that has kinetic and potential energy. - We consider the immediate mechanical source of energy at the plant. Concept / Approach: In a hydroelectric power station, water is collected at a higher elevation in a dam or reservoir, giving it gravitational potential energy. When the water is allowed to flow down through pipes or channels, this potential energy is converted into kinetic energy of moving water. The moving water strikes the blades of a turbine, causing it to rotate. The rotating turbine shaft drives a generator that converts mechanical energy into electrical energy. Although the water cycle is ultimately driven by solar energy, the direct resource harnessed at the plant is the flowing water itself. Step-by-Step Solution: Step 1: Recognise that hydroelectric power depends on water stored at a height in reservoirs behind dams. Step 2: When water is released, gravity pulls it downward, increasing its speed and converting potential energy into kinetic energy. Step 3: The fast moving water hits turbine blades, transferring kinetic energy and making the turbine rotate. Step 4: The rotating turbine drives an electric generator, which converts mechanical energy into electrical energy. Step 5: The immediate natural resource that provides the mechanical energy at the site is the flowing water in rivers and reservoirs. Step 6: Therefore, the correct answer is the movement of water, not air or direct solar radiation. Verification / Alternative check: If the flow of water stops, for example during a severe drought when reservoirs are empty, hydroelectric power generation drops or ceases. This direct dependence on water flow confirms that water is the immediate resource. Wind turbines, in contrast, depend on moving air, and solar power plants depend on sunlight on panels. Visits to hydroelectric plants or diagrams in textbooks always show water channels, turbines and generators linked directly in the power generation process. Why Other Options Are Wrong: Moving air in the atmosphere: This is the resource for wind power, not hydroelectric power. Direct solar radiation on the panels: This describes solar photovoltaic power plants, not hydroelectric stations. All natural resources equally: Hydroelectric plants specifically rely on flowing water; other forms of power generation use different resources. Common Pitfalls: Students sometimes mix up the ultimate energy source with the immediate one. It is true that the water cycle that fills reservoirs is driven by solar energy causing evaporation, but the direct mechanical energy harnessed at the power station is from flowing water. The question is concerned with the immediate resource used at the plant, not with the very long chain of processes that originally supplied that resource. Final Answer: Hydroelectric energy comes from the movement of flowing water in rivers and reservoirs.

Question 16

An in car GPS navigation device determines its position by receiving signals via which type of communication link?

Accepted Answer

Introduction / Context: Global Positioning System devices, commonly called GPS, are now built into many cars and smartphones. They can determine position anywhere on the globe with good accuracy, provided they can receive signals from satellites. Understanding the basic communication channel used by GPS helps clarify how these devices work and why they require a clear view of the sky. This question asks you to identify the type of link that an in car GPS uses to obtain its location information. Given Data / Assumptions: - The device is a standard GPS receiver installed in an automobile. - GPS satellites orbit the Earth and transmit signals continuously. - The receiver is passive; it listens to signals but does not need to transmit back for basic positioning. - The communication medium is wireless and operates through the atmosphere. Concept / Approach: GPS satellites broadcast precise timing and orbital information using radio signals in the microwave region of the electromagnetic spectrum. A GPS receiver in the car picks up these signals from multiple satellites. By measuring the travel time of the signals and knowing the positions of the satellites, the receiver computes its own position on the Earth's surface. The channel used is therefore a satellite based radio communication link, not infrared reflection from signs, wired cables or short range local signals. Step-by-Step Solution: Step 1: Recall that GPS stands for Global Positioning System and relies on a constellation of satellites. Step 2: Understand that each GPS satellite broadcasts radio signals containing timing and positional data. Step 3: The GPS receiver in the car receives these signals from several satellites at once. Step 4: By comparing the arrival times of the signals and using known satellite positions, the receiver calculates its distance from each satellite. Step 5: Using these distances, it performs a trilateration calculation to determine its own exact location. Step 6: Thus, the communication channel is radio communication from satellites, specifically in microwave frequency bands. Verification / Alternative check: Practical observations confirm this explanation. GPS reception is poor inside tunnels, inside thick walled buildings or under heavy tree cover because the receiver cannot see enough satellites and cannot receive their radio signals clearly. No connection to underground cables or road signs is needed, and the system works on the ocean or in remote deserts where there are no such infrastructure elements. Technical documentation on GPS systems always mentions radio frequency bands such as the L1 and L2 bands, confirming that radio signals from satellites are used. Why Other Options Are Wrong: Infrared beams reflected from road signs: GPS does not rely on roadside reflectors; it needs signals from orbiting satellites and works even where there are no signs. Electrical signals sent through underground cables: GPS receivers are not physically connected to cables, and the system is designed to work wirelessly anywhere on the globe. Short range Bluetooth links from smartphones only: While some apps can share location via Bluetooth, the original position information still comes from satellite radio signals, not purely from nearby phones. Common Pitfalls: Some learners confuse GPS with mobile data or internet services and think the device needs a cellular network to know its position. In reality, GPS positioning uses direct satellite radio signals, and mobile data is used only to download maps or traffic information. Remember that the core of GPS is satellite based radio communication, and that is the link the question is asking about. Final Answer: An in car GPS navigation device determines its position using radio signals transmitted from navigation satellites orbiting the Earth.

Question 17

In gas laws, the combined gas law relates which set of state variables in a single relationship?

Accepted Answer

Introduction / Context: The behaviour of an ideal gas can be described by several gas laws, such as Boyle law, Charles law and Gay Lussac law. Each of these laws relates two variables while keeping the third constant. The combined gas law merges these individual relationships into one equation that links pressure, volume and temperature for a fixed amount of gas. This question asks which set of variables the combined gas law relates. Given Data / Assumptions: - We are dealing with an ideal gas and a fixed amount of gas (constant number of moles). - Pressure is usually denoted by P, volume by V and temperature by T (in kelvin). - Boyle law relates P and V at constant T, Charles law relates V and T at constant P, and Gay Lussac law relates P and T at constant V. - The combined gas law collects these into a single equation. Concept / Approach: The combined gas law is often written as P1 V1 / T1 = P2 V2 / T2 for a fixed amount of gas changing from one state to another. This shows that the product of pressure and volume divided by temperature is constant for a given quantity of gas. Thus, the combined gas law simultaneously relates pressure, volume and temperature. It does not hold any of these three fixed; instead, it describes how a change in one affects the others when the amount of gas remains constant. Step-by-Step Solution: Step 1: Recall Boyle law: P is inversely proportional to V at constant T. Step 2: Recall Charles law: V is directly proportional to T at constant P. Step 3: Recall Gay Lussac law: P is directly proportional to T at constant V. Step 4: Combine these relationships to obtain P V / T = constant for a fixed amount of gas. Step 5: Express this as P1 V1 / T1 = P2 V2 / T2, showing that P, V and T are all linked. Step 6: Conclude that the combined gas law relates volume, temperature and pressure together. Verification / Alternative check: The ideal gas equation P V = n R T, where n is the number of moles and R is the gas constant, also links pressure, volume and temperature. For a fixed amount of gas, n and R are constant, so P V / T is constant. This is consistent with the combined gas law and again shows that P, V and T are all involved. Solving the ideal gas equation for any two variables requires knowledge of the third, reinforcing the idea that all three are interconnected. Why Other Options Are Wrong: Volume and temperature only: This describes Charles law, not the combined gas law. Temperature and pressure only: This describes Gay Lussac law, not the combined law on its own. Volume and pressure only: This describes Boyle law, which holds temperature constant, not the combined law. Common Pitfalls: A common error is to memorise individual gas laws without seeing how they fit together. Students sometimes think the combined gas law is just one more formula rather than a summary of the three basic laws. Remember that combined gas law is essentially the ideal gas equation rearranged for a fixed amount of gas and that it connects P, V and T in one expression. Final Answer: The combined gas law relates volume, temperature and pressure together for a fixed amount of gas.

Question 18

Which of the following observations best supports the dynamo theory for the origin of the Earth magnetic field?

Accepted Answer

Introduction / Context: The dynamo theory explains the Earth magnetic field as the result of electric currents generated by the motion of conducting fluids within the planet. Understanding what kind of interior structure is needed for this process is an important topic in geophysics. This question asks which observation provides the strongest support for the dynamo theory as the origin of the Earth global magnetic field. Given Data / Assumptions: - The Earth has several layers: crust, mantle, outer core and inner core. - The dynamo theory requires a region of electrically conducting fluid in motion. - Seismic studies provide information about which layers are solid and which are liquid. - The composition of the core is mostly iron and nickel with some lighter elements. Concept / Approach: According to dynamo theory, a global magnetic field is generated when electrically conducting fluid moves in the presence of rotation and existing magnetic fields. In the Earth, the outer core is believed to be a liquid layer of iron-rich material that can conduct electricity and convect due to thermal and compositional gradients. These flows, combined with the rotation of the planet, maintain the geomagnetic field. The most directly supportive evidence is therefore the existence of a liquid, electrically conducting outer core. Other facts, such as the presence of iron or convection, matter as background, but without a liquid conducting layer, the dynamo could not operate. Step-by-Step Solution: Step 1: Identify the basic requirement of a dynamo: moving electrically conducting fluid. Step 2: Consider the internal structure of the Earth: a solid inner core, a liquid outer core, and a solid mantle and crust. Step 3: Seismic wave studies indicate that the outer core is liquid because certain seismic waves do not pass through it. Step 4: The composition of the outer core is mostly iron and nickel, which are good electrical conductors when molten. Step 5: Convection in this liquid metal layer, combined with Earth rotation, can generate and maintain a magnetic field according to dynamo theory. Step 6: Therefore, the observation that most directly supports dynamo theory is the existence of a liquid, electrically conducting outer core. Verification / Alternative check: Measurements of variations in the geomagnetic field over time, known as secular variation, are consistent with flows in the outer core. Computer simulations of geodynamos that use parameters similar to a liquid iron outer core can reproduce many features of the observed field, including polarity reversals. These lines of evidence, combined with seismic data, all point toward a convecting, conducting outer core as the source of the field. Why Other Options Are Wrong: The core of the Earth contains a large amount of solid iron: While iron is important for conductivity, the solid inner core alone cannot generate a dynamo because it lacks large scale fluid motion; the key is the liquid outer core. Granite and basalt in the crust conduct electricity very well: The crust is mostly solid and does not provide the kind of deep, global scale fluid motion needed for a dynamo, and these rocks are not highly conductive compared with molten iron. Convection occurs only in the inner solid core of the Earth: The inner core is believed to be solid; convection in the liquid outer core, not the solid inner core, is central to dynamo theory. Common Pitfalls: Students sometimes focus on the presence of iron alone and overlook the need for fluid motion. Other times they may confuse conduction in solid rocks with the dynamic behaviour of a fluid core. To support a planetary dynamo, both conductivity and motion are required. Remember that a liquid, convecting outer core of conducting material is the key ingredient that makes the Earth magnetic dynamo plausible. Final Answer: The best evidence supporting dynamo theory is that the outer core of the Earth contains liquid material that conducts electricity and can flow.

Question 19

Among the following colours of visible light, which has the highest photon energy and therefore the highest frequency?

Accepted Answer

Introduction / Context: Visible light is made up of a spectrum of colours, often remembered by the sequence red, orange, yellow, green, blue, indigo and violet. Each colour corresponds to light waves of different wavelengths and frequencies. Photon energy is directly proportional to the frequency of light and inversely proportional to its wavelength. This question asks which listed colour among violet, red, blue and white corresponds to the highest photon energy. Given Data / Assumptions: - The visible spectrum runs from red at the long wavelength end to violet at the short wavelength end. - Photon energy E is related to frequency f by E = h * f, where h is Planck constant. - Frequency and wavelength are related by c = f * λ, with c being the speed of light and λ the wavelength. - White light is a mixture of many colours, not a single wavelength. Concept / Approach: Since E = h * f, higher frequency means higher photon energy. Because c = f * λ, for a constant speed of light in vacuum, shorter wavelength corresponds to higher frequency. Within the visible range, red light has the longest wavelength and lowest frequency, while violet light has the shortest wavelength and highest frequency. Blue light lies between green and violet, so its frequency and energy are higher than red but lower than violet. White light is not a single frequency; it is a combination of many wavelengths and thus does not correspond to a single photon energy. Therefore, among the options, violet light has the highest photon energy. Step-by-Step Solution: Step 1: Recall the order of colours in the visible spectrum: red, orange, yellow, green, blue, indigo, violet. Step 2: Recognise that as you move from red to violet, wavelength decreases and frequency increases. Step 3: Use E = h * f to conclude that higher frequency light has higher photon energy. Step 4: Compare red and blue. Blue has a shorter wavelength and higher frequency than red, so its photons have more energy. Step 5: Compare blue and violet. Violet has an even shorter wavelength and higher frequency than blue, so its photons carry more energy. Step 6: Note that white light is a mixture of colours and does not correspond to a single energy; it is not a separate spectral colour. Verification / Alternative check: Approximate wavelength ranges confirm this ordering. Red light is around 620 to 750 nanometres, green around 495 to 570 nanometres, blue around 450 to 495 nanometres and violet around 380 to 450 nanometres. Since violet has the smallest wavelength among these, it has the highest frequency and thus the highest photon energy. Spectral diagrams and physics texts consistently place violet at the high energy end of the visible band. Why Other Options Are Wrong: Red: Red has the longest wavelength and lowest frequency among common visible colours, so its photons have lower energy compared with blue and violet. Blue: Blue photons are more energetic than red photons but less energetic than violet photons because blue wavelengths are longer than violet wavelengths. White: White is not a single spectral colour; it is a combination of many wavelengths, so it does not represent a single photon energy. Common Pitfalls: A common mistake is to think that white is the highest energy colour because it appears brightest. However, brightness is related to intensity and perception, not directly to the energy per photon of a specific wavelength. Another confusion is mixing up wavelength and frequency trends, forgetting that shorter wavelength means higher frequency and higher photon energy. Memorising the red to violet order and remembering that violet is the high energy end of visible light helps avoid these errors. Final Answer: Among the listed colours, violet light has the highest photon energy and the highest frequency.

Question 20

In basic physics, the rate at which energy is transferred or work is done per unit time is called which physical quantity?

Accepted Answer

Introduction / Context: The idea of how quickly energy is transferred or how fast work is done is fundamental in physics, engineering and everyday life. When we compare a powerful machine with a less powerful one, we are really talking about the rate of doing work, not just the total amount of work. This question tests your understanding of the basic definition of power as a physical quantity, and asks you to distinguish it from units or other related terms such as watt, volt and potential. Given Data / Assumptions: Energy transfer and work done are measured in joules in the SI system. Time is measured in seconds in the SI system. Power is defined as work done per unit time or energy transferred per unit time. Watt is a unit of power, volt is a unit of electric potential difference and potential is a related concept, not the rate itself. Concept / Approach: In physics, power P is defined by the formula P = W / t, where W is the work done or energy transferred, and t is the time taken. This definition captures the idea of how fast energy is being converted or used. If the same amount of work is done in less time, power is higher. The watt (symbol W) is the SI unit that measures power numerically, where 1 watt equals 1 joule per second. Volt is the unit of electric potential difference, which relates to energy per unit charge, not directly to the rate of energy transfer. The word potential by itself refers to potential energy or potential difference, not to a rate. Step-by-Step Solution: Step 1: Recall the definition P = W / t for power, where W is work or energy and t is time.Step 2: Recognise that this formula clearly shows power is a rate: energy per unit time.Step 3: Identify that the name of the physical quantity described by this definition is power.Step 4: Understand that watt is the SI unit used to express numerical values of power, not the name of the quantity itself.Step 5: Note that volt relates to potential difference, not directly to rate of energy transfer per unit time.Step 6: Conclude that the rate at which energy is transferred is called power. Verification / Alternative check: Consider two bulbs, 40 W and 100 W, connected to the same voltage supply. The 100 W bulb converts electrical energy to light and heat at a faster rate than the 40 W bulb. Here, the watt values measure the power of the bulbs. Similarly, when lifting weights, a person who lifts the same weight to the same height in half the time develops twice the power. These examples confirm that we use the concept of power to talk about the rate of doing work or transferring energy. Why Other Options Are Wrong: Watt is a unit of power, not the name of the physical quantity itself. Volt is the unit of electric potential difference, describing energy per unit charge, not energy per unit time. The term potential is incomplete and usually refers to potential energy or electric potential, not to a rate. Only power correctly names the quantity that measures how fast energy is transferred or work is done. Common Pitfalls: Students often confuse a quantity with its unit and may say that watt is the rate at which energy is transferred. While watts do measure that rate, the underlying quantity is power. Others mix up volt and watt because both appear in electrical formulas. To avoid these mistakes, clearly separate the concept (power) from the units used (watt, kilowatt and so on). Final Answer: The rate at which energy is transferred or work is done per unit time is called power.

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