Question 1

The image of an external object formed on the retina of the human eye by the eye lens is

Accepted Answer

Introduction / Context: The human eye functions as a natural optical instrument, using a convex lens system to form images of external objects on the retina. Understanding the nature of this image is a standard concept in basic optics and biology. This question asks whether the image on the retina is real or virtual and whether it is upright or inverted. Given Data / Assumptions: The eye contains a convex lens system consisting of the cornea and the crystalline lens. The retina is a light sensitive screen at the back of the eye. Objects are located at normal viewing distances greater than the focal length of the eye lens system. We consider normal healthy vision without defects. Concept / Approach: A convex lens produces a real, inverted image of an object when the object is placed beyond the focal length of the lens. The human eye adjusts its focal length through accommodation to bring light from different distances to focus exactly on the retina. The retina receives this focused light pattern just like a screen, so the image must be real and inverted. Although we perceive objects as upright, this is due to processing by the brain, which interprets and corrects the visual information, not because the retinal image itself is upright. Step-by-Step Solution: Step 1: Model the eye as a convex lens (the eye lens) forming an image on a screen (the retina). Step 2: Recall that for a convex lens, when the object distance is greater than the focal length, it forms a real image on the opposite side of the lens. Step 3: Such a real image is inverted with respect to the object. Step 4: In the human eye, the retina is physically located where this real image forms, so the retinal image must be real and inverted. Step 5: Therefore the correct description of the image on the retina is that it is real and inverted, even though our perception after brain processing is upright. Verification / Alternative check: In laboratory experiments, if you dissect a large animal eye and place a bright object in front of it, you can sometimes see an inverted miniature image on the retina when light passes through the lens system. This demonstrates directly that the retinal image is real and inverted. Eye diagrams in textbooks also show inverted images on the retina along with explanations of how the brain interprets them as upright. These experimental and textbook observations confirm the real and inverted nature of the retinal image. Why Other Options Are Wrong: Virtual and upright would require the retina to be on the same side of the lens as the object or the lens to behave like a simple magnifier, which is not how normal vision works for distant objects. Virtual and inverted is not a standard image type for a single convex lens and does not match the eye structure. Real and upright would require an unusual lens arrangement or object location not typical of the human eye in normal use. Magnified and virtual for all objects describes a simple magnifying glass used very close to the eye, not the natural imaging situation in the eye for everyday distances. Common Pitfalls: Students sometimes confuse the retinal image with perception. Because we always see objects upright, they conclude that the image on the retina must also be upright. Another confusion arises from using magnifying glasses, which do form a virtual, upright image, but that is a separate optical setup. To avoid mistakes, keep in mind the lens formula and remember that the retina plays the role of a screen, so the eye forms a real, inverted image which the brain later interprets as upright. Final Answer: The image of an object formed on the retina of the human eye is real and inverted.

Question 2

In the electromagnetic spectrum and space physics, which of the following radiations has the highest frequency or energy among the options given?

Accepted Answer

Introduction / Context: This general knowledge physics question tests understanding of the broad ordering of different types of radiation by frequency and energy. From radio communication to medical X-rays and high energy cosmic particles arriving from outer space, different radiations span a huge range of frequencies. Competitive exams often ask candidates to identify which region lies at the highest end of this range. Among the options given here, cosmic rays represent extremely high energy radiation or particles that reach the Earth from space and are associated with the highest frequencies or energies, well above X-rays, microwaves and radio waves. Given Data / Assumptions: The options are cosmic rays, X-rays, radio waves and microwaves. Radio waves and microwaves lie in the low frequency, long wavelength part of the electromagnetic spectrum. X-rays lie in the high frequency, short wavelength part of the electromagnetic spectrum. Cosmic rays are extremely energetic radiation or particles arriving from outer space with very high associated frequencies or energies. Concept / Approach: In basic physics teaching, the electromagnetic spectrum is ordered from low frequency, long wavelength to high frequency, short wavelength. The usual order starts from radio waves, then microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays. Above even these in many general knowledge discussions come cosmic rays, which are very energetic particles and radiation from outer space. The higher the frequency, the greater the photon energy. Thus, when we compare the given options, we must choose the radiation associated with the greatest energy and highest frequency, which is cosmic rays in standard GK style questions. Step-by-Step Solution: Step 1: Recall that radio waves have the lowest frequencies and longest wavelengths and are used for radio and television communication. Step 2: Remember that microwaves have higher frequency than radio waves but are still in the lower frequency region of the spectrum and are used in microwave ovens and radar. Step 3: Note that X-rays have much higher frequency and energy than visible light, and they are used in medical imaging and material testing. Step 4: Recognise that cosmic rays, which include very high energy radiation and particles from space, are associated with extremely high energies and effectively the highest frequencies among the options listed. Step 5: Therefore, among the four given options, cosmic rays must be chosen as the radiation with the highest frequency or energy. Verification / Alternative check: A simple check is to arrange the four options in the usual exam order from low frequency to high frequency. Radio waves clearly come at the lowest end, followed by microwaves, then X-rays in the high frequency region. Cosmic rays are always discussed in chapters on high energy astrophysics and radiation hazards, which indicates that they are above X-rays in typical energy comparisons. This ordering confirms that cosmic rays occupy the highest position among these four types of radiation and therefore match the expected answer in general knowledge questions. Why Other Options Are Wrong: X-rays are very high frequency electromagnetic waves, but they do not reach the extreme energies associated with cosmic rays in general knowledge comparisons. Radio waves have the lowest frequencies and longest wavelengths used in everyday communication systems, so they clearly cannot be the highest frequency radiation. Microwaves lie above radio waves but still far below X-rays and cosmic rays in frequency and energy. Common Pitfalls: Students sometimes quickly pick X-rays because they know these waves are more energetic than visible or ultraviolet light. Another mistake is to focus only on the word wave and forget that many textbooks also treat cosmic rays as extremely energetic radiation from space. For competitive exams it is important to remember the extended ordering used in general knowledge tables, where cosmic rays are described at the top end of the energy scale. Whenever cosmic rays appear in an options list with X-rays, microwaves and radio waves, candidates should confidently select cosmic rays as the highest frequency or energy radiation among them. Final Answer: Among the options given, the radiation with the highest frequency or energy is Cosmic rays.

Question 3

At what temperature do the Fahrenheit and Celsius temperature scales show the same numerical reading (that is, the same number on both scales)?

Accepted Answer

Introduction / Context: This question tests basic conceptual understanding of temperature scales used in physics and everyday life. Celsius (also called centigrade) and Fahrenheit are two commonly used scales. Usually, the same physical temperature has different numerical values on these two scales. However, there is one particular temperature at which both scales show the same number. Knowing this special point helps students better understand the linear relationship between the two scales and is often directly asked in competitive exams as a simple but important fact. Given Data / Assumptions: Celsius scale has fixed points: 0 degrees for freezing point of water and 100 degrees for boiling point of water at standard pressure. Fahrenheit scale has 32 degrees for freezing point and 212 degrees for boiling point of water. We are asked to find the temperature at which the numerical value on both scales is the same. The options include 212, 32, -40 and 0 degrees. Concept / Approach: The relationship between temperatures on the Celsius scale (C) and Fahrenheit scale (F) is linear and is given by the formula F = (9 / 5) * C + 32. To find the temperature at which the numbers are equal, we set F equal to C and solve the resulting simple equation. This is a straightforward substitution that shows at which point the two scales intersect. This method is often explained in basic physics and mathematics classes and is an excellent example of how linear relationships work. Step-by-Step Solution: Step 1: Use the conversion formula between Fahrenheit and Celsius: F = (9 / 5) * C + 32. Step 2: At the temperature where the readings are the same, set F = C. Step 3: Substitute F by C in the formula to get C = (9 / 5) * C + 32. Step 4: Rearrange the equation: C - (9 / 5) * C = 32. Step 5: Factor C: C * (1 - 9 / 5) = 32 gives C * ((5 - 9) / 5) = 32, so C * (-4 / 5) = 32. Step 6: Solve for C: C = 32 * (-5 / 4) = -40. Step 7: Therefore Celsius and Fahrenheit scales show the same numerical reading at -40 degrees. Verification / Alternative check: As a quick check, convert -40 degrees Celsius into Fahrenheit using the formula. Substituting C = -40 gives F = (9 / 5) * (-40) + 32 = -72 + 32 = -40. Because the calculated Fahrenheit temperature is also -40, the solution is verified. This double check confirms that -40 is indeed the temperature where both scales coincide numerically. Remembering this special value is very helpful in exams and also in practical scientific work where quick mental conversions are needed. Why Other Options Are Wrong: 212 degrees corresponds to the boiling point of water on the Fahrenheit scale and is equal to 100 degrees on the Celsius scale, so the readings are different. 32 degrees is the freezing point of water in Fahrenheit, which equals 0 degrees Celsius, again giving different numbers. 0 degrees is the freezing point of water in Celsius, equal to 32 degrees Fahrenheit, so the two scales do not match numerically at this point. Common Pitfalls: Many learners try to guess from memory of freezing and boiling points and may choose 0 or 32 degrees because they recognise these as special values. Others may forget the exact conversion formula and become confused between Fahrenheit and Celsius readings. The safest approach in exams is to remember the simple derived fact that the two scales coincide at -40 degrees and, if time permits, quickly verify using the linear formula. This avoids guesswork and builds a deeper understanding of how different temperature scales relate to each other. Final Answer: The Fahrenheit and Celsius scales show the same numerical reading at -40 degrees.

Question 4

In modern communication systems, an optical fibre cable guides light signals over long distances. On which physical principle does an optical fibre mainly work?

Accepted Answer

Introduction / Context: Optical fibre technology is at the heart of modern high speed communication networks, from internet backbones to cable television and medical endoscopes. This question tests whether the learner knows the basic physical principle that allows light to be guided through long, flexible glass or plastic fibres, even when the fibre is bent. Understanding this concept requires familiarity with how light behaves when it tries to pass from one medium to another and under what conditions it is reflected back rather than refracted out. Given Data / Assumptions: Light travels through an optical fibre core made of glass or plastic with a higher refractive index. The core is surrounded by cladding with a lower refractive index. The question asks for the main principle responsible for guiding light along the fibre. Options include scattering, interference, total internal reflection and simple refraction. Concept / Approach: When a ray of light travels from a denser medium (higher refractive index) to a rarer medium (lower refractive index), and if the angle of incidence at the boundary is greater than a certain critical angle, all the light is reflected back into the denser medium. This phenomenon is called total internal reflection. Optical fibres are specially designed so that light entering the core at suitable angles always hits the core cladding boundary at angles greater than the critical angle. As a result, the light is repeatedly totally internally reflected and remains trapped inside the core, even if the fibre is bent gently. This repeated total internal reflection is the working principle of optical fibre communication. Step-by-Step Solution: Step 1: Identify the core and cladding structure of the optical fibre, where the core has higher refractive index than the surrounding cladding. Step 2: Recall the condition for total internal reflection: light must travel from a denser to a rarer medium and must strike the boundary at an angle of incidence greater than the critical angle. Step 3: Understand that in an optical fibre the geometry ensures that rays entering within a certain acceptance angle always meet the core cladding interface at angles above the critical angle. Step 4: Because this condition is satisfied, light rays are not refracted out into the cladding but instead are reflected back into the core again and again. Step 5: Conclude that the basic principle enabling long distance light guidance in optical fibres is total internal reflection, and therefore that option must be chosen. Verification / Alternative check: If one visualises an optical fibre, it behaves like a long, thin light pipe. For such a pipe to guide light, the light must be reflected internally rather than escaping. Simple scattering or interference phenomena cannot explain the tight guidance of rays along the length, especially through curves. Refraction at a single surface would bend the path, but that alone would allow light eventually to leak out. Only total internal reflection, occurring repeatedly at the core cladding boundary, naturally explains why light remains trapped inside the fibre until it exits at the far end, which matches the observed behaviour of optical fibre systems. Why Other Options Are Wrong: Scattering of light involves light being deflected in many directions due to particles or irregularities and would cause signal loss rather than controlled guidance. Interference of light is the phenomenon of overlapping waves creating bright and dark patterns; it is important in many optical instruments but does not explain the guiding mechanism in fibres. Refraction at a single surface describes bending of light when it passes from one medium to another, but does not by itself keep light confined inside a long fibre. Common Pitfalls: Students sometimes confuse simple refraction with total internal reflection and may choose refraction because they remember that light changes direction at interfaces. Others might think of scattering because of loss mechanisms in real fibres. The key for exam success is to remember the precise name of the phenomenon that traps light inside the fibre: total internal reflection. Whenever a question mentions optical fibres, light pipes, periscopes or mirage like phenomena, one should immediately think of total internal reflection as the central concept. Final Answer: An optical fibre works mainly on the principle of total internal reflection of light at the core cladding interface.

Question 5

When a person enters a dark room after being in bright sunlight, why is he or she unable to see objects clearly for a short time?

Accepted Answer

Introduction / Context: This question tests basic understanding of human eye physiology and how the eye adjusts to changes in light intensity. In everyday life, people often experience temporary difficulty seeing when moving from a brightly lit environment into a dark room, such as entering a cinema hall from bright sunlight. This adjustment process is called dark adaptation and is controlled by the iris and pupil as well as by chemical changes in the photoreceptor cells of the retina. Knowing the immediate mechanical reason for the temporary loss of clarity is important in general science and biology questions. Given Data / Assumptions: A person is initially in bright light, for example sunlight. The person then moves suddenly into a dark or dim room. For a short while, he or she cannot see objects clearly. The options mention movements of the retina, iris controlled pupil changes and power of accommodation. Concept / Approach: The iris is the coloured part of the eye that controls the size of the pupil, the opening through which light enters. In bright light, the iris constricts the pupil to a small size to reduce the amount of light entering. In darkness, the iris dilates the pupil to a larger size to allow more light to reach the retina. This adjustment takes some time. When a person suddenly moves from bright light to darkness, the pupil is still small and cannot immediately dilate to the necessary size. As a result, very little light reaches the retina at first, and the person cannot see clearly until the iris has gradually dilated the pupil and the retina has adapted. Step-by-Step Solution: Step 1: In bright light, the iris keeps the pupil constricted so that only a limited amount of light enters the eye. Step 2: When the person suddenly enters a dark room, the light level falls sharply, but the pupil is still small for a brief period. Step 3: Because the pupil is small, very little light reaches the retina, and the visual image is too weak to be perceived clearly. Step 4: The iris then responds by slowly dilating the pupil to let more light enter, a process that takes some time. Step 5: During this adjustment interval, vision is poor, and once the pupil becomes wider and the retina adapts, the person begins to see clearly. Verification / Alternative check: The retina does not physically move forward or backward in ordinary light changes. Instead, the main rapid mechanical change is the variation of pupil size controlled by the iris. The power of accommodation refers to the eye changing the curvature of the lens to focus on near or distant objects and is not directly related to adapting to darkness. Therefore, the only option that correctly describes an immediate cause of temporary poor vision is the limited ability of the iris to dilate the pupil instantly, which matches option about the iris and pupil adjustment. Why Other Options Are Wrong: The retina of the eye moving forward is not a normal physiological response during light changes and does not explain the delay in vision. The retina moving backward is also incorrect; the retina remains fixed while other parts such as iris and lens adjust. The eye does have power of accommodation in dark conditions; accommodation is related to focusing on objects at different distances, not to the quantity of light entering. Common Pitfalls: Students sometimes confuse accommodation with light adaptation and may think that inability to change focus is responsible. Others may not know the exact roles of the iris and retina and may fall for options that sound technical involving retina movements. To avoid such confusion, it is important to remember that the iris and pupil control the amount of light entering the eye, and that dark adaptation is mainly due to delayed dilation of the pupil and chemical adjustments in photoreceptors, not movement of the retina or lack of accommodation. Final Answer: When we move from bright light to a dark room, we cannot see clearly for a while because the iris cannot dilate the pupil immediately to allow more light in.

Question 6

In communication systems, the overall quality of signal transmission between a transmitter and a receiver depends mainly on which of the following factors?

Accepted Answer

Introduction / Context: This question belongs to basic communication and electronics concepts that are often tested in general knowledge and entrance exams. Whenever information is sent from one place to another, the form of the signal (for example, analog or digital, strength and bandwidth) and the physical medium through which it is sent (for example, copper wire, optical fibre, air or space) determine how faithfully the original information reaches the receiver. Understanding that transmission quality is influenced by both signal characteristics and medium characteristics is essential for appreciating real world communication systems. Given Data / Assumptions: The communication process involves a signal that carries information. The signal travels through some medium such as wire, optical fibre, or free space. The question asks what quality of transmission depends on. Options isolate the signal, the medium, both together, or neither. Concept / Approach: Quality of transmission is usually judged by how little the signal is distorted, attenuated or corrupted by noise when it arrives at the receiver. The nature of the signal, such as its frequency range, power level, modulation scheme and whether it is analog or digital, affects how robust it is against noise and interference. At the same time, the nature of the medium, such as its resistance, capacitance, dispersion, attenuation and susceptibility to external noise, also plays a major role. Because both aspects are crucial, engineers design communication systems by matching signal properties to the chosen medium and by selecting or improving the medium to suit the signal. Step-by-Step Solution: Step 1: Consider the signal itself. A weak signal with very large bandwidth may be more vulnerable to noise and distortion than a properly modulated and appropriately powerful signal. Step 2: Think about the medium. A poor quality copper cable or an air path with obstacles introduces attenuation, dispersion and interference that degrade the signal. Step 3: Observe that clearer communication can be achieved by improving the signal properties (for example, using better modulation and coding) even if the medium remains the same. Step 4: Also observe that better performance can be obtained by changing the medium (for example, moving from copper wire to optical fibre) even when the signal format is unchanged. Step 5: Conclude that both the signal and the medium together determine transmission quality, so the correct choice is the option stating that quality depends on both. Verification / Alternative check: A practical way to verify this is to consider a digital signal transmitted over two different media, such as an old noisy cable versus a high quality optical fibre. Even if the signal format is ideal, the bad cable will yield poor performance compared to the fibre, showing the effect of medium. Conversely, if the same fibre is used but with an extremely poorly designed signal (for example, too little power or inappropriate modulation), the received quality will still be bad. These thought experiments show that neither factor can be ignored, confirming that quality depends on both the nature of the signal and the medium. Why Other Options Are Wrong: Nature of the signal only is insufficient because even an excellent signal will suffer if the medium is extremely lossy or noisy. Nature of the medium only is also insufficient because a poor signal design can fail even on a high quality medium. Neither the signal nor the medium characteristics cannot be correct, because these two aspects are precisely what engineers adjust to improve communication quality. Common Pitfalls: Students sometimes focus only on the transmission medium, especially when they have learned about differences between copper wires, radio channels and optical fibres. Others may focus only on the signal side, such as modulation techniques or power levels. In reality, communication system design always considers both together. For exams, remember that whenever a question mentions quality of transmission in a general way, the safest complete answer is that it depends on both the nature of the signal and the nature of the medium. Final Answer: The quality of signal transmission depends on both the nature of the signal and the nature of the transmission medium.

Question 7

In practical lighting applications, carbon arc lamps are most commonly used in which of the following devices or fields?

Accepted Answer

Introduction / Context: This question belongs to basic applied physics and electrical engineering knowledge. Different light sources are suitable for different applications based on their brightness, colour quality, stability and cost. Carbon arc lamps were among the earliest very bright electric light sources and have historically been used for specialised purposes. Understanding where such intense lamps are typically employed, compared to more common sources like incandescent bulbs or modern LEDs, helps learners associate physical principles of electric arcs with real world technology in cinema projection and stage lighting. Given Data / Assumptions: Carbon arc lamps produce light using an electric arc between carbon electrodes. They are capable of producing very bright and intense white light. The options mention cinema projectors, photography, vehicles and domestic lighting. We assume traditional and classic usage patterns of carbon arc lamps. Concept / Approach: Carbon arc lamps operate by passing a strong electric current through a gap between two carbon rods, creating an intense arc that emits bright light. This light is useful where very high brightness and good beam focus are required, even if energy efficiency is not ideal. Historically, such lamps were used in cinema projectors and large searchlights because they could project a powerful beam over long distances or onto large screens. For ordinary car headlights and domestic lighting, carbon arc lamps are not practical due to their complexity, maintenance requirements and cost. Therefore, the most appropriate association is with cinema projectors. Step-by-Step Solution: Step 1: Recall the construction of a carbon arc lamp, which uses two carbon electrodes separated by a small gap. Step 2: When a large current flows, an electric arc forms, generating extremely bright light and significant heat. Step 3: Recognise that this intense light is ideal for projecting images on a cinema screen through lenses and film. Step 4: Note that for still photography, flash units usually use gas discharge tubes or modern electronic flashes rather than heavy carbon arc systems. Step 5: Understand that vehicle headlamps and domestic lighting prefer compact and efficient sources such as halogen, LED or fluorescent lamps, not bulky arc lamps. Step 6: Conclude that cinema projectors are the typical practical application of carbon arc lamps among the options given. Verification / Alternative check: A quick check is to think about where extremely bright, concentrated beams were traditionally required. Cinema projectors need sustained intense light for long periods to illuminate large screens. Stage and searchlights historically used similar arc lamps. In contrast, domestic and vehicle uses require more convenience, safety and efficiency, and have moved towards filament and LED technologies. This mental comparison supports the choice of cinema projectors as the standard answer in general knowledge questions about carbon arc lamp applications. Why Other Options Are Wrong: Still photography generally uses short duration flash tubes or modern electronic flash units rather than noisy and cumbersome carbon arc lamps. Vehicle headlamps need compact, shock resistant and quickly controllable light sources; traditional filament bulbs, halogen lamps and LEDs are far more suitable than carbon arcs. Domestic household lighting prioritises safety, low noise and efficiency, making incandescent bulbs, CFLs and LEDs preferable, not carbon arc lamps. Common Pitfalls: Some learners may assume that any bright light source is used in photography or even vehicles without considering practical design constraints. Others might not differentiate between historical and modern applications. For examination purposes, it is helpful to remember the simple pair: carbon arc lamp and cinema projector. Whenever a question asks about the common use of carbon arc lamps, cinema projectors should immediately come to mind as the correct association. Final Answer: Carbon arc lamps are most commonly used in cinema projectors to provide the very bright light needed to project images onto a large screen.

Question 8

In physics, what is the SI unit used for measuring a solid angle?

Accepted Answer

Introduction / Context: This question tests a basic but important fact from units and measurements in physics. Just as ordinary plane angles are measured in degrees or radians, three dimensional angular spreads are measured using solid angles. Solid angle helps describe how large an object appears from a given point in three dimensional space and is used in areas like radiometry, astronomy and nuclear physics. The SI system introduces a special unit for solid angle that differs from the units used for ordinary plane angles. Given Data / Assumptions: We are dealing with solid angle, not simple plane angle. The options are degree, radian, steradian and radian second. Degree and radian are familiar from plane geometry. We need the dedicated SI unit for solid angle. Concept / Approach: A plane angle measures the opening between two lines in a plane and is measured in radians or degrees. In contrast, a solid angle measures the portion of space subtended by a surface at a point, such as how large the Sun appears in the sky. The solid angle is defined as the area cut out on a sphere divided by the square of the radius of that sphere. In SI, the unit for plane angle is the radian, and the corresponding unit for solid angle is the steradian. One steradian is the solid angle that cuts off an area on the sphere equal to the square of the radius. This is a standard definition in textbooks on physics. Step-by-Step Solution: Step 1: Recall that plane angle is measured in degrees or radians, while solid angle is a three dimensional analogue of angle. Step 2: Remember that the SI unit of plane angle is the radian and the SI unit of solid angle is the steradian. Step 3: Compare the options. Degree and radian correspond to plane angle, not solid angle. Step 4: Radian second is not a standard unit for angle; seconds are used for time, not for measuring angles in SI naming. Step 5: Therefore, select steradian as the correct SI unit of solid angle. Verification / Alternative check: The full sphere around a point in three dimensional space subtends a solid angle of 4 * pi steradian. This formula is often given in physics problems involving radiation and flux. Remembering this relationship reinforces the fact that steradian is the appropriate unit. Additionally, one can recall that steradian is abbreviated as sr in SI tables, whereas radian is abbreviated as rad. This distinction between radian and steradian is standard across physics and engineering references. Why Other Options Are Wrong: Degree is an older unit for plane angle and is not the SI unit for solid angle. Radian is the SI unit for plane angle, not for solid angle, although it is sometimes loosely mentioned in three dimensional contexts. Radian-second is not recognised as a unit of angle or solid angle in the SI system and mixes time units incorrectly. Common Pitfalls: Students sometimes confuse radian and steradian because of their similar sounding names and because both relate to angular measures. It is important to remember that radian is for two dimensional angles in a plane, while steradian is for three dimensional solid angles. A simple memory aid is that the extra syllable in steradian hints at the extra dimension considered. In exams, any question that clearly mentions solid angle should point directly to steradian as the correct unit. Final Answer: The SI unit used for measuring a solid angle is the steradian.

Question 9

Among the following physical quantities, which one is a scalar quantity (that is, it has magnitude only and no direction)?

Accepted Answer

Introduction / Context: This question tests a fundamental concept in physics: the difference between scalar and vector quantities. Scalars are described completely by magnitude alone, while vectors need both magnitude and direction for full description. In mechanics and electromagnetism, many commonly used quantities fall clearly into one of these two categories. Being able to classify typical quantities correctly is essential for solving problems and understanding how physical laws are expressed in vector form. Given Data / Assumptions: The options are electric current, electric field, acceleration and linear momentum. Vector quantities require direction as well as magnitude. Scalar quantities do not have an inherent direction in space. We must identify which quantity is treated as scalar in standard physics. Concept / Approach: Acceleration and linear momentum are clearly vector quantities in mechanics. Acceleration describes rate of change of velocity and points in the direction of the change. Linear momentum equals mass times velocity and therefore has the same direction as velocity. Electric field at a point in space also has both magnitude and direction, indicating the force per unit positive charge at that point, and is thus a vector field. Electric current, although it involves charges moving in a certain direction, is defined in terms of the rate of flow of charge through a cross section and is treated as a scalar quantity in basic circuit theory. Current has magnitude and a sign (for conventional direction), but it is not represented as a vector in three dimensional space in elementary physics. Step-by-Step Solution: Step 1: Consider acceleration. It changes with direction when an object turns, even at constant speed, so it must be a vector quantity. Step 2: Consider linear momentum, defined as p = m * v, where v is velocity. Because velocity is a vector, momentum is also a vector. Step 3: Examine electric field, which is defined as force per unit charge. The direction of the field indicates the direction of the force on a positive test charge, so it is clearly a vector. Step 4: Examine electric current, which in elementary circuit theory is described by a magnitude (for example, 2 ampere) and a sign that indicates conventional direction but is not usually handled as a vector in space. Step 5: Conclude that among the listed options, electric current is treated as a scalar quantity. Verification / Alternative check: A practical check is to think about how these quantities are written in formulas and problem statements. Acceleration, momentum and electric field are often shown with arrows over them in textbooks and used with vector addition rules. Electric current, on the other hand, is typically written without vector notation and combined algebraically as a scalar when applying Kirchhoff laws in circuit analysis. This difference in notation and use supports the classification of electric current as scalar in standard introductory physics. Why Other Options Are Wrong: Electric field is a vector quantity because it has a specific direction at each point and is represented as a vector field. Acceleration is a vector because it indicates not just how fast velocity changes but also in which direction the change occurs. Linear momentum is a vector since it depends on velocity, which itself is directional. Common Pitfalls: Some students get confused because electric current flows in a particular direction in a wire and think it must therefore be a vector. However, in basic physics and engineering, current is treated as a scalar magnitude with a chosen reference direction, not as a full vector in three dimensional space. Understanding this distinction helps avoid mistakes in vector related questions and clarifies why certain quantities appear in vector equations while others do not. Final Answer: Among the given options, the quantity that is treated as a scalar is Electric current.

Question 10

For long distance data transmission, which of the following communication channels has the highest propagation speed for signals?

Accepted Answer

Introduction / Context: This question examines understanding of how fast signals travel in different communication channels. Modern communication uses radio waves for broadcasting, microwaves for satellite and line of sight links and light pulses in optical fibres for high speed internet backbones. Many learners assume that one technology must be dramatically faster than another. In reality, all of these signals are electromagnetic in nature and propagate at speeds close to the speed of light. Hence their propagation speeds are nearly the same, and the main advantages of one channel over another come from bandwidth, noise performance and capacity rather than basic propagation speed. Given Data / Assumptions: The channels mentioned are microwave, optical fibre and radio wave links. All three rely on electromagnetic waves to carry information. Electromagnetic waves in vacuum travel at the speed of light, approximately 3 * 10^8 m/s. In materials such as glass or air, speed is slightly reduced but still of the same order. Concept / Approach: Microwaves and radio waves are simply electromagnetic waves of different frequencies propagating mostly through air or space. Optical fibres use light, which is also electromagnetic radiation at higher frequencies, traveling through glass or plastic. The propagation speed of electromagnetic waves in vacuum is a constant, and in media it is reduced by the refractive index but still remains very high. Differences between these methods are more about how much data can be packed into the signal (bandwidth), how much noise is present and how well the medium shields against interference, not about basic propagation speed. Therefore, when asked which channel is fastest in terms of propagation, the correct conceptual answer is that they are all nearly equally fast. Step-by-Step Solution: Step 1: Recognise that microwave links use microwave frequency electromagnetic waves traveling through air or space. Step 2: Note that radio communication uses radio frequency electromagnetic waves, also traveling through air or space. Step 3: Understand that optical fibre uses light pulses, which are also electromagnetic waves, traveling through glass with a speed somewhat less than that in vacuum. Step 4: Recall that electromagnetic waves of any frequency have the same speed in vacuum and similar very high speeds in materials, with variations due to refractive index but not by orders of magnitude. Step 5: Conclude that there is no major difference in propagation speed among these channels, so the best description is that all operate at nearly the same propagation speed. Verification / Alternative check: One way to verify is to consider that a satellite link using microwaves and a terrestrial fibre optic link can both transmit signals across continents in fractions of a second. Any delays noticeable to users are dominated by switching, routing and processing times rather than by the basic wave propagation speed. For example, crossing an ocean of thousands of kilometres at close to 3 * 10^8 m/s takes only a few tens of milliseconds. This observation confirms that in practical communication systems, the propagation speed of microwave, radio and optical fibre channels is very similar. Why Other Options Are Wrong: Microwave links alone cannot be called fastest because their waves travel at essentially the same speed as other electromagnetic waves. Optical fibre has advantages in bandwidth and noise immunity, but its signal speed in glass is slightly less than the speed of light in vacuum and not vastly different from other channels. Radio wave links also use electromagnetic waves, but they are not significantly faster than microwaves or light in fibre; they all propagate at similar high speeds. Common Pitfalls: Students often mix up propagation speed with data rate or bandwidth. Optical fibre can carry more data per second than many radio links because of higher frequency and larger usable bandwidth, but the speed at which individual wavefronts propagate is still close to the speed of light, similar to radio and microwaves. To avoid confusion, always separate the idea of propagation speed from the amount of information that can be encoded per second on that signal. Final Answer: All three channels use electromagnetic waves and hence all of them operate at nearly the same propagation speed.

Question 11

In India, on which date is National Science Day celebrated every year to commemorate the discovery of the Raman Effect?

Accepted Answer

Introduction / Context: This question tests awareness of an important national observance related to science in India. National Science Day marks the discovery of the Raman Effect by the Indian physicist C. V. Raman, a discovery that later earned him the Nobel Prize in Physics. The day is used to promote scientific temper, encourage young students to take interest in science and celebrate scientific achievements. Competitive exams frequently ask for the exact date of this observance, closely linking current affairs with basic science history. Given Data / Assumptions: National Science Day is connected with the Raman Effect. It is celebrated annually in India. The options provide four dates in February: 23, 25, 26 and 28. We must recall the standard officially recognised date. Concept / Approach: The Raman Effect, which describes the inelastic scattering of light by molecules, was discovered by C. V. Raman on 28 February 1928. To honour this discovery and promote scientific awareness, the Government of India designated 28 February as National Science Day. Every year, programmes, exhibitions, lectures and competitions are organised on this date in schools, colleges and research institutions. Therefore, when asked about National Science Day, one should directly associate it with 28 February. Step-by-Step Solution: Step 1: Remember that National Science Day in India is linked to the discovery of the Raman Effect. Step 2: Recall that C. V. Raman announced his discovery on 28 February, which later became the chosen date for the observance. Step 3: Look at the options: February 23, 25, 26 and 28. Step 4: Identify February 28 as the recognised date of both the original discovery announcement and the official National Science Day celebration. Step 5: Select February 28 as the correct answer. Verification / Alternative check: A simple way to verify this date is to recall that many schools in India hold science exhibitions or special assemblies near the end of February, specifically on 28 February, and news reports mention this date every year. Examination guides and general knowledge books also list 28 February as National Science Day, often alongside other important observance days such as National Technology Day and World Environment Day. This repeated exposure reinforces the correct date in memory. Why Other Options Are Wrong: February 23 is not associated with the Raman Effect and is not recognised as National Science Day in India. February 25 does not correspond to any major national science observance linked to C. V. Raman in standard references. February 26 is also not officially designated as National Science Day and appears only as a distractor in this multiple choice question. Common Pitfalls: Because all the given dates fall close together in late February, some candidates rely on guesswork and confuse 25, 26 and 28. Others mix up National Science Day with other international days related to science or technology. To avoid such errors, learners should firmly memorise the association: Raman Effect, C. V. Raman and 28 February as National Science Day in India. Creating a quick mental link between Raman and the end of February can help ensure accurate recall in exam situations. Final Answer: National Science Day is celebrated in India every year on February 28.

Question 12

In plant physiology, photosynthesis in green plants generally takes place at the fastest rate in which colour of light among the options given?

Accepted Answer

Introduction / Context: This question comes from basic biology and plant physiology, a topic that often appears even in general knowledge sections. Photosynthesis is the process by which green plants use light energy to convert carbon dioxide and water into sugars and oxygen. The rate of photosynthesis depends on several factors, including the intensity and wavelength (colour) of light. Chlorophyll pigments absorb some colours more efficiently than others. Understanding which colour leads to the fastest photosynthesis helps students connect the absorption spectrum of chlorophyll with practical plant growth and exam questions. Given Data / Assumptions: The process discussed is photosynthesis in green plants. The question compares different light conditions: darkness, red, white and yellow light. Chlorophyll pigments strongly absorb red and blue light and reflect green. Darkness obviously provides no light energy for photosynthesis. Concept / Approach: The absorption spectrum of chlorophyll shows strongest absorption in the red and blue regions of visible light, with a peak in the red region. Because photosynthesis is driven by absorbed light, the rate is highest under light whose wavelengths are efficiently absorbed. In many standard biology experiments and textbooks, red light is described as producing the highest rate of photosynthesis, followed closely by blue light. White light contains all colours, but if intensity is kept the same, the effective absorption per unit energy may be less strongly concentrated in the most efficient bands. Therefore, among the given options, red light is the most appropriate answer according to classic exam oriented biology teaching. Step-by-Step Solution: Step 1: Recognise that photosynthesis requires light; in darkness, there can be no light dependent reactions, so the rate becomes almost zero. Step 2: Recall that chlorophyll appears green because it reflects green light and absorbs mainly red and blue wavelengths. Step 3: Understand that strong absorption of red light means more energy is available to drive the light dependent reactions when plants are exposed to red light. Step 4: Compare red, white and yellow light. Yellow light falls in a region with less chlorophyll absorption than red, and white light spreads energy across many colours, including some that are not used efficiently. Step 5: Conclude that red light provides the most efficient use of light energy for photosynthesis among the given options, leading to the fastest rate. Verification / Alternative check: In laboratory demonstrations, students often perform experiments using aquatic plants such as Hydrilla under different coloured filters to measure oxygen bubble release as an indicator of photosynthetic rate. Results typically show highest activity in red light, significant activity in blue, moderate in white and very low in green or yellow. General biology textbooks also state that the maximum rate of photosynthesis occurs in red light. These experimental and textbook observations confirm that red light should be chosen as the correct option for such questions. Why Other Options Are Wrong: Darkness provides no light energy, so the light dependent stage of photosynthesis cannot proceed at all, making this clearly incorrect. White light contains all colours; while it can support photosynthesis, the question specifically asks in which colour it is fastest, and standard teaching places red above white in effectiveness per unit energy. Yellow light is not as strongly absorbed by chlorophyll as red or blue light, so the rate of photosynthesis in pure yellow light is lower than in red light. Common Pitfalls: Students may be tempted to choose white light thinking that since it contains all colours, it must be the best. This can be misleading if one forgets that the intensity and absorption at specific effective wavelengths matter more than simply having many colours present. Others might incorrectly associate green light with plant colour and assume it is best for growth, when in fact green is mostly reflected and not used. Remembering that chlorophyll absorbs red and blue most strongly and that red is usually cited as giving the maximum photosynthetic rate will help avoid such mistakes. Final Answer: Photosynthesis generally takes place at the fastest rate in red light.

Question 13

In household electrical wiring, a fuse is used in the main supply as a safety device. Which of the following statements correctly describes a good fuse wire?

Accepted Answer

Introduction / Context: This question tests basic knowledge of safety devices in electrical circuits. A fuse is a simple but vital protection component in domestic and industrial installations. Its purpose is to break the circuit automatically when current exceeds a safe limit, preventing overheating and possible fire or equipment damage. To play this sacrificial protective role, a fuse wire must be chosen with specific material properties. Understanding these properties is important for safely designing or using electrical systems. Given Data / Assumptions: A fuse is connected in the mains supply of a circuit as a safety device. The options describe how it is connected, what it is made of, its melting point and its resistance. We must identify the correct essential property of a fuse wire. We assume a standard domestic alternating current supply and conventional fuse design. Concept / Approach: A fuse wire is designed to melt and break the circuit when the current exceeds a predetermined safe value. This means that the fuse should heat up quickly when current is too high. For this to happen reliably, the material should have a low melting point and a suitable resistance so that excess current causes sufficient heating. It is always connected in series with the load, not in parallel. While expensive metals like silver may be used in special fuses, common domestic fuses use inexpensive alloys like tin and lead. Very high resistance would cause significant energy loss even at normal currents, which is undesirable. Therefore, the key correct statement is that a fuse wire must have a low melting point. Step-by-Step Solution: Step 1: Recall that the function of a fuse is to melt and open the circuit when current becomes abnormally high. Step 2: For the fuse to melt quickly during overload, the material should have a low melting point and appropriate cross sectional area so that heating is concentrated in the fuse. Step 3: The fuse is always placed in series with the live supply line so that all current flows through it; if it were in parallel, current could bypass it and it would not provide protection. Step 4: The material used for domestic fuses is usually an inexpensive alloy with a definite and relatively low melting point, not costly metals like silver in ordinary installations. Step 5: The resistance of the fuse should be low under normal current so that it does not waste power significantly; extremely high resistance is not desired. Step 6: Conclude that the statement about having a low melting point is the correct one among the options. Verification / Alternative check: One can verify this by considering what happens during an overload due to a short circuit or too many appliances. The current suddenly increases, causing rapid heating in the fuse wire. Because its melting point is low compared to other conductors, it melts first and breaks the circuit before the main wiring overheats. If the fuse wire had a high melting point, other parts of the circuit might get damaged before the fuse reacts. This practical insight confirms that low melting point is an essential property of fuse wires. Why Other Options Are Wrong: If the fuse were connected in parallel with the main switch, current could bypass it and the fuse would not effectively protect the circuit; correct installation is in series. Common domestic fuse wires are not made mainly from silver alloys; they use cheaper alloys like tin lead combinations tuned to specific melting points. A very high resistance fuse would cause unacceptable voltage drop and power loss even at normal currents, which is not desirable in household circuits. Common Pitfalls: Students sometimes confuse properties needed for efficient conductors with those needed for fuse materials and may think that high melting point or very low resistance is always best. It is important to remember that a fuse is intentionally designed to fail safely and quickly under overload, so a controlled low melting point is crucial. Another common error is misunderstanding series and parallel connections. For exam preparation, always associate fuses with series connection and low melting point materials. Final Answer: The correct statement about a good fuse wire is that it must have a low melting point so that it melts and breaks the circuit when the current becomes too high.

Question 14

No electric current will flow between two charged bodies if both bodies have the same value of which of the following physical quantities?

Accepted Answer

Introduction / Context: This question addresses a core concept in basic electricity: the driving cause of electric current between two points. Current flows when there is a potential difference, similar to how water flows when there is a difference in water level. Understanding that equal potentials imply no net driving force for charge flow is essential for students learning about circuits, capacitors and electrostatics. Many general knowledge and entrance exam questions test whether learners correctly identify electric potential, not just charge, as the deciding factor. Given Data / Assumptions: Two bodies are charged and may have different or equal charges. Electric current will flow only if there is some driving difference between them. The options include resistance, charge, potential and charge to potential ratio. We assume that the bodies can be connected by a conductor that allows charge to move. Concept / Approach: Electric potential (or electric potential energy per unit charge) at a point determines how much work is done in bringing a small test charge from a reference point to that point. When two conductors are at different potentials and connected by a path for charge, charges will flow from higher potential to lower potential until the potentials become equal. The actual amount of charge each body carries is not decisive; what matters is the difference in potential. Electric resistance determines how much current flows for a given potential difference but does not by itself cause current when the potential difference is zero. Therefore, if two bodies have the same potential, no current will flow between them when they are connected. Step-by-Step Solution: Step 1: Recall Ohm law in basic form: I = V / R, where I is current, V is potential difference and R is resistance. Step 2: Notice that if V, the potential difference, is zero, then I becomes zero regardless of the value of R, meaning no current flows. Step 3: Understand that two bodies can have different amounts of charge yet be at the same potential, especially when their shapes or sizes differ. Step 4: Recognise that equal potential implies equal electric potential energy per unit charge, so there is no net driving force pushing charges from one body to the other. Step 5: Conclude that no current flows when the electric potentials of the two bodies are equal, making potential the correct quantity in the question. Verification / Alternative check: A helpful check is to think about charging two identical metal spheres. If both are brought to the same potential using a voltage source and then connected with a wire, no spark is observed and no current flows because there is no potential difference. On the other hand, if they have different potentials and are then connected, a transient current flows until their potentials equalise. This practical behaviour confirms that equal potential, not equal charge or equal resistance, is the condition for no current flow between the bodies. Why Other Options Are Wrong: Equal electrical resistance does not prevent current; current depends on both resistance and potential difference. Two resistors with equal resistance can still have current flowing through them if connected to a voltage source. Equal charge on both bodies does not guarantee zero potential difference, because potential also depends on size, shape and environment. Equal charge to potential ratio is not a standard criterion used in electrostatics, and in any case the flow is determined directly by potential difference, not by this ratio. Common Pitfalls: A common mistake is to think that if two bodies carry the same amount of charge, there is no tendency for charge to move between them. However, if their capacities differ, they may still be at different potentials, and current can flow. Another confusion arises when students focus on resistance and overlook the crucial role of potential difference. To avoid such errors, always remember that electric current flows due to potential difference, and if potentials are equal, current cannot flow between the bodies through a conductor. Final Answer: No current will flow between two charged bodies if they have the same electric potential.

Question 15

At the Curie temperature, a ferromagnetic material undergoes a transition and becomes which type of magnetic material?

Accepted Answer

Introduction / Context: This question comes from the topic of magnetism and material science. Ferromagnetic materials such as iron, cobalt and nickel exhibit strong magnetisation and can form permanent magnets at ordinary temperatures. However, their magnetic behaviour changes with temperature. There is a characteristic temperature, called the Curie temperature, above which ferromagnetic order is lost. Understanding what happens at this temperature is important in physics, electrical engineering and materials technology, and it is a popular concept in multiple choice questions. Given Data / Assumptions: We start with a ferromagnetic material, meaning it shows strong spontaneous magnetisation below a certain temperature. At the Curie temperature, a change in magnetic ordering occurs. Options include diamagnetic, non-magnetic, paramagnetic and none of the others. We assume the material structure remains intact; only magnetic ordering changes. Concept / Approach: Ferromagnetism arises because atomic magnetic moments in certain materials tend to align parallel to each other in regions called domains, leading to a net magnetisation. As temperature increases, thermal agitation tends to randomise these moments. At a specific temperature, known as the Curie temperature (or Curie point), thermal energy becomes strong enough to break the long range alignment of domains. Above this temperature, the material no longer shows strong spontaneous magnetisation and behaves like a paramagnetic substance, where magnetic moments are more or less randomly oriented but can align weakly with an external field. The material does not become diamagnetic or entirely non-magnetic; rather, it becomes paramagnetic. Step-by-Step Solution: Step 1: Identify that below the Curie temperature, ferromagnetic materials have aligned domains that produce strong magnetisation. Step 2: Recognise that increasing temperature increases random thermal motion of atoms and magnetic moments. Step 3: At the Curie temperature, thermal energy disrupts the cooperative alignment responsible for ferromagnetism. Step 4: Above this temperature, the material still has atomic magnetic moments, but they no longer align spontaneously to form strong domains. Step 5: This new state matches the behaviour of paramagnetic materials, which show weak magnetisation only in the presence of an external magnetic field. Step 6: Therefore, we conclude that the ferromagnetic material becomes paramagnetic at the Curie temperature. Verification / Alternative check: One way to verify is to recall that magnetic susceptibility of ferromagnets is very high below the Curie temperature but drops sharply and follows a paramagnetic like Curie law above that temperature. Experimental plots of magnetisation versus temperature show a steep fall at the Curie point, beyond which the behaviour is similar to paramagnets. These standard graphs in physics textbooks confirm that ferromagnets do not become diamagnetic or fully non-magnetic; instead, they become paramagnetic when heated beyond the Curie temperature. Why Other Options Are Wrong: Diamagnetic materials are those in which induced magnetisation is opposite to the applied field and very weak; ferromagnets do not convert to purely diamagnetic state at the Curie temperature. Non-magnetic material is an imprecise term; ferromagnets heated above the Curie temperature still respond weakly to magnetic fields, so describing them as non-magnetic is inaccurate. None of the others cannot be correct because paramagnetic material is explicitly listed and is the correct description of their behaviour above the Curie temperature. Common Pitfalls: Students sometimes misinterpret the phrase non-magnetic to mean that the material loses all magnetic properties at high temperature, which is not true. Others confuse diamagnetism, which is present to some extent in all materials, with the change from ferromagnetic to paramagnetic ordering. To avoid mistakes, remember the clear sequence: ferromagnetic below the Curie point, paramagnetic above it, with the Curie temperature marking a magnetic phase transition between these two states. Final Answer: At the Curie temperature, a ferromagnetic material loses its strong ordered magnetisation and becomes paramagnetic.

Question 16

A solid disk and a solid sphere of the same radius but possibly different masses roll without slipping down identical inclined planes of the same height and length. Which object will reach the bottom of the plane first?

Accepted Answer

Introduction / Context: This question tests understanding of rotational motion and energy distribution in rolling bodies. When different objects roll down an inclined plane without slipping, their acceleration depends not only on gravity and the angle of the plane but also on how their mass is distributed, which is captured by the moment of inertia. Comparing a solid disk and a solid sphere of the same radius is a classic conceptual problem in mechanics that illustrates how rotational inertia affects translational motion. Given Data / Assumptions: There are two rigid bodies: a solid disk and a solid sphere. Both have the same radius but may have different masses. They roll without slipping down inclined planes of the same height and length. Friction is sufficient only to prevent slipping; it does no net work. We neglect air resistance and assume uniform gravitational field. Concept / Approach: When a body rolls without slipping, gravitational potential energy is converted into both translational kinetic energy and rotational kinetic energy. The total acceleration of the body down the plane depends on its moment of inertia about the centre. For a solid disk of mass m and radius R, the moment of inertia is I = (1 / 2) * m * R^2. For a solid sphere, the moment of inertia is I = (2 / 5) * m * R^2. The expression for linear acceleration a of a rolling object on an incline is a = g * sin(theta) / (1 + I / (m * R^2)). A smaller ratio I / (m * R^2) gives a larger acceleration. Because the solid sphere has a smaller I / (m * R^2) than the solid disk, it experiences greater acceleration and reaches the bottom first. The result is independent of mass. Step-by-Step Solution: Step 1: Write the expression for the moment of inertia of a solid disk: I_disk = (1 / 2) * m * R^2. Step 2: Write the expression for the moment of inertia of a solid sphere: I_sphere = (2 / 5) * m * R^2. Step 3: Compute the ratio I / (m * R^2) for each shape. For the disk, I_disk / (m * R^2) = 1 / 2. For the sphere, I_sphere / (m * R^2) = 2 / 5. Step 4: Compare these ratios: 1 / 2 is 0.5 and 2 / 5 is 0.4, so the solid sphere has a smaller I / (m * R^2). Step 5: Use the formula for linear acceleration of a rolling object on an incline: a = g * sin(theta) / (1 + I / (m * R^2)). A smaller value of I / (m * R^2) gives a larger acceleration. Step 6: Therefore, the solid sphere has greater acceleration than the solid disk and reaches the bottom of the incline first, regardless of mass differences, as long as rolling without slipping occurs. Verification / Alternative check: An alternative way to see this is through energy distribution. Both objects start with the same potential energy if their centres start at the same height. At the bottom, this potential energy is converted to translational kinetic energy (1 / 2 * m * v^2) plus rotational kinetic energy (1 / 2 * I * omega^2). The object with smaller moment of inertia uses less energy in rotation for the same angular speed and can therefore have more translational kinetic energy, meaning higher linear speed at the bottom. Because the sphere has a smaller moment of inertia factor than the disk, it attains a higher linear speed and reaches the bottom earlier. Why Other Options Are Wrong: Saying it depends on their masses is incorrect because the acceleration for rolling objects on an incline, under the stated assumptions, is independent of mass and depends only on shape through I / (m * R^2). The solid disk does not reach first; its larger relative moment of inertia leads to smaller acceleration compared to the sphere. Both reaching at the same time would be true only if their effective acceleration down the incline were identical, which is not the case for these different shapes. Common Pitfalls: Many students assume that if the bodies have different masses, the heavier one will roll faster, which is not correct for pure rolling without slipping in this idealised scenario. Others may believe that identical heights and radii must lead to identical times. The key point is that distribution of mass (moment of inertia) affects rotational dynamics. Remembering the specific values of I for common shapes and the form a = g * sin(theta) / (1 + I / (m * R^2)) helps avoid such misconceptions. Final Answer: Under the given conditions, the object that reaches the bottom first is the solid sphere.

Question 17

In electrical energy calculations, a kilowatt-hour (kWh) is a practical unit used for measuring which physical quantity?

Accepted Answer

Introduction / Context: This question focuses on the distinction between power and energy in the context of household electricity usage. Electricity bills often show consumption in kilowatt-hours, so many people encounter this term in daily life. However, learners sometimes confuse kilowatt, which is a unit of power, with kilowatt-hour, which is a unit related to energy. Understanding the difference is essential for correctly interpreting electric bills and solving numerical problems in physics and electrical engineering. Given Data / Assumptions: The unit mentioned is kilowatt-hour, abbreviated as kWh. Power is the rate of doing work or using energy, measured in watts or kilowatts. Energy is the total work done or total electrical energy consumed, measured in joules in SI units. The options include power, electric charge, energy and electric current. Concept / Approach: A kilowatt-hour is defined as the amount of energy consumed when a device with a power rating of one kilowatt (1000 watt) operates for one hour. Power is energy per unit time, so multiplying power by time gives energy. In symbols, energy E = P * t. If P is in kilowatt and t is in hour, then E is in kilowatt-hour. Thus, kilowatt-hour is a unit of energy, not of instantaneous power, charge or current. Electric energy companies use kWh because it is a convenient measure for the relatively large amounts of energy consumed in homes and industries. Step-by-Step Solution: Step 1: Recall that power P is measured in watt (W) or kilowatt (kW) and represents energy used per second. Step 2: Recall that energy E can be calculated from power and time using E = P * t. Step 3: Consider a device rated at 1 kW running for 1 hour. The total energy used is E = 1 kW * 1 hour = 1 kilowatt-hour. Step 4: Observe that kilowatt-hour therefore represents a quantity of energy, equal in SI units to 1000 watt * 3600 second = 3.6 * 10^6 joule. Step 5: Conclude that kilowatt-hour is a unit of energy, and select the option corresponding to energy. Verification / Alternative check: A simple way to confirm is to look at an electricity bill, which shows how many kilowatt-hour have been consumed in a month. Bills charge for energy used, not for power capacity or instantaneous current. Also, physics textbooks explicitly state that 1 kWh is a commercial unit of electric energy. Since both everyday usage and textbook definitions align, we can be confident that kilowatt-hour is indeed a unit of energy. Why Other Options Are Wrong: Power is measured in watt or kilowatt, not in kilowatt-hour. Kilowatt-hour includes time and therefore belongs to energy, not power. Electric charge is measured in coulomb in SI units, not in kilowatt-hour; it represents quantity of charge, not energy. Electric current is measured in ampere and refers to the rate of flow of charge, again not represented by kilowatt-hour. Common Pitfalls: Many learners mistakenly think that because the word kilowatt appears inside kilowatt-hour, both must be units of power. The key to avoid this confusion is to remember that adding hour converts a rate into a total amount, just as kilometres per hour is a speed while kilometre is a distance. Kilowatt describes how fast energy is used, while kilowatt-hour describes how much energy has been used over a period of time. Final Answer: A kilowatt-hour (kWh) is a practical unit of energy, commonly used for measuring electrical energy consumption.

Question 18

In very cold countries, alcohol is often used as a thermometric liquid in thermometers instead of mercury mainly because of which of the following reasons?

Accepted Answer

Introduction / Context: This question relates to the choice of thermometric liquids in thermometers and how physical properties of liquids affect their usefulness. In extremely cold regions, thermometers must measure temperatures well below the freezing point of water and also below the freezing point of mercury. Using a liquid that solidifies at a higher temperature would make the thermometer useless in such conditions. Alcohol based thermometers solve this problem and are common in very cold climates. Understanding the key property that makes alcohol suitable is an important concept in basic physics and everyday science. Given Data / Assumptions: We are considering thermometers used in very cold countries. Alcohol is compared with mercury as a thermometric liquid. Options mention conductivity, colouring, freezing point and sticking to glass. We must identify the most important reason for preferring alcohol in this specific context. Concept / Approach: A thermometric liquid should remain in liquid form over the temperature range to be measured. If it freezes within that range, it cannot expand or contract properly and the thermometer fails. Mercury freezes at a temperature around -39 degrees Celsius. In some very cold regions, air temperatures can drop below this limit, risking solidification of mercury. Alcohol (such as ethanol) has a much lower freezing point, around -114 degrees Celsius, and therefore remains liquid at temperatures where mercury would freeze. While it is true that alcohol can be coloured for visibility and has other convenient properties, the main reason it is favoured in very cold climates is its very low freezing point. Step-by-Step Solution: Step 1: Identify the requirement: a thermometric liquid that remains liquid at very low temperatures common in cold countries. Step 2: Recall that mercury freezes near -39 degrees Celsius, which is insufficient for extremely low temperature measurements. Step 3: Note that alcohol, such as ethanol, has a much lower freezing point (well below -100 degrees Celsius), so it remains liquid in extreme cold. Step 4: Recognise that this very low freezing point is the primary property that allows alcohol thermometers to function where mercury thermometers would fail. Step 5: Therefore, among the options, the correct main reason is that alcohol has a very low freezing point. Verification / Alternative check: To verify, imagine using a mercury thermometer in a polar region where air temperatures might fall below -40 degrees Celsius. Once mercury freezes, it cannot expand or contract, and the thermometer reading becomes stuck. In contrast, alcohol remains fluid at these temperatures and can continue to respond to temperature changes. Other properties such as ease of colouring and wetting of glass are helpful for readability and construction, but they do not address the fundamental requirement of staying liquid. Hence, the low freezing point is the decisive factor in this context. Why Other Options Are Wrong: Alcohol being a poor conductor of heat is not particularly relevant; thermometer response actually benefits from reasonable thermal conduction to reach equilibrium quickly. Alcohol can indeed be coloured to see its level more easily, but this is a secondary convenience, not the main reason for its use in very cold countries. The idea that alcohol does not stick to the glass tube is not the primary consideration; both mercury and alcohol can be made to wet or not wet glass with proper design. Common Pitfalls: Students sometimes latch onto secondary advantages like easy colouring or visibility and choose those options, forgetting the fundamental requirement that the thermometric liquid must not freeze within the measurement range. Others may focus on conductivity without realising that the key limiting factor in extreme cold is freezing point. To avoid this, always consider the operational temperature range first when evaluating why one liquid is chosen over another in thermometer design. Final Answer: Alcohol is used as a thermometric liquid in very cold countries mainly because alcohol has a very low freezing point and remains liquid at temperatures where mercury would freeze.

Question 19

In the visible region of the electromagnetic spectrum, which colour of light has the longest wavelength?

Accepted Answer

Introduction / Context: This question tests basic knowledge of the visible spectrum of light, a topic commonly illustrated by the rainbow or a prism experiment. Visible light consists of a range of colours from violet to red, each associated with a different wavelength and frequency. Knowing which colour corresponds to the longest wavelength is important in optics, communication and general science questions. This fact also connects with the well known ordering of colours in a rainbow. Given Data / Assumptions: We are considering only the visible region of the electromagnetic spectrum. The common rainbow order of colours is violet, indigo, blue, green, yellow, orange and red. The options given are violet, yellow, red and white. We must identify which single colour has the longest wavelength. Concept / Approach: The visible spectrum can be arranged in order of increasing wavelength or increasing frequency. When arranged in order of increasing wavelength, violet has the shortest wavelength and red has the longest wavelength. Frequency is inversely related to wavelength, so violet has the highest frequency and red the lowest frequency among visible colours. White light is not a single wavelength but a mixture of many visible wavelengths. Therefore, among the individual colours listed, red light clearly corresponds to the longest wavelength in the visible range. Step-by-Step Solution: Step 1: Recall the standard order of colours in a rainbow using the mnemonic such as VIBGYOR: Violet, Indigo, Blue, Green, Yellow, Orange, Red. Step 2: Remember that in this order, violet is at the short wavelength, high frequency end, and red is at the long wavelength, low frequency end. Step 3: Compare the options. Violet is at the short wavelength end, so it cannot have the longest wavelength. Step 4: Yellow lies between green and orange and therefore has a wavelength intermediate between those colours, less than red. Step 5: White light is a mixture of all visible wavelengths and is not a single wavelength colour. Step 6: Conclude that red light has the longest wavelength among the visible colours listed. Verification / Alternative check: As a cross check, think about dispersion through a glass prism. When white light passes through a prism, violet deviates the most and red deviates the least. This behaviour occurs because violet light has a shorter wavelength and therefore experiences greater change in speed and bending in the glass compared to red. The smaller deviation of red corresponds to its longer wavelength. Such prism experiments are standard in school physics and confirm the wavelength ordering of visible colours. Why Other Options Are Wrong: Violet light has the shortest visible wavelength and therefore cannot be the longest. Yellow light has a wavelength in the middle of the visible spectrum and is shorter than red. White light is not a single colour but a mixture of many wavelengths, so talking about a single longest wavelength for white is not meaningful. Common Pitfalls: Some learners get confused between wavelength and frequency and mistakenly think that the colour that appears most intense or brightest must have the longest wavelength. Others may not clearly remember the order of colours in the spectrum. To avoid such errors, memorise the simple rule that in visible light violet has the shortest wavelength and red has the longest, and remember the standard VIBGYOR ordering from violet to red. Final Answer: In the visible spectrum, the colour with the longest wavelength is Red.

Question 20

Approximately how much time does light from the Sun take to travel through space and reach the Earth?

Accepted Answer

Introduction / Context: This question relates to basic astronomy and the finite speed of light. Although sunlight appears to reach us instantly, it actually takes a measurable amount of time to travel the average distance from the Sun to the Earth. Knowing the approximate travel time helps students appreciate both the enormous speed of light and the large scale of the solar system. This fact often appears in general knowledge and science quizzes. Given Data / Assumptions: The speed of light in vacuum is approximately 3 * 10^8 metres per second. The average distance between the Sun and the Earth is about 1.5 * 10^11 metres, often called one astronomical unit. We are asked for an approximate time, so a rounded value is acceptable. The options include 5 minutes, 4.5 hours, 8 minutes and 5.5 hours. Concept / Approach: The time taken by light to travel a given distance is given by time = distance / speed. Using the approximate values, time = (1.5 * 10^11 metres) / (3 * 10^8 metres per second). This calculation gives around 500 seconds, which is a little over 8 minutes. Therefore, the standard rounded statement in textbooks is that sunlight takes about 8 minutes to reach the Earth. The options involving hours are clearly far too large compared to this value. Step-by-Step Solution: Step 1: Write the formula for travel time: time = distance / speed. Step 2: Substitute approximate distance from Sun to Earth: distance is 1.5 * 10^11 metres. Step 3: Substitute speed of light in vacuum: speed is 3 * 10^8 metres per second. Step 4: Compute the ratio: time = (1.5 * 10^11) / (3 * 10^8) = (1.5 / 3) * 10^(11 - 8) = 0.5 * 10^3 seconds. Step 5: 0.5 * 10^3 seconds is 500 seconds. Divide by 60 to convert seconds to minutes: 500 / 60 is approximately 8.3 minutes. Step 6: Round to a simple value suitable for general knowledge: about 8 minutes. Verification / Alternative check: A quick mental check is to remember that light can circle the Earth multiple times in one second, so taking several hours to reach us from the Sun would be inconsistent with known scales of the solar system. Many introductory astronomy and physics books directly state that it takes about 8 minutes for sunlight to arrive at Earth, and this is a widely accepted educational fact. This confirms that the correct approximate time is about 8 minutes rather than any of the options involving hours or 5 minutes. Why Other Options Are Wrong: About 5 minutes is too short, since a simple calculation shows the travel time is closer to 8 minutes. About 4.5 hours is far too long and would imply an unrealistically huge distance or a much lower speed of light, contradicting well established physics. About 5.5 hours is also far too long and inconsistent with basic astronomical data. Common Pitfalls: Some learners may confuse the average distance from the Sun to the Earth with larger astronomical distances and overestimate travel time. Others may not be comfortable with powers of ten and therefore fail to perform a quick approximate calculation. To avoid mistakes, it is helpful to memorise the simple fact that light from the Sun takes about 8 minutes to reach Earth and to understand this as a direct consequence of the large distance and finite speed of light. Final Answer: Light from the Sun takes about 8 minutes to reach the Earth.

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