Question 1

In the International System of Units (SI), what is the standard unit used to measure heat energy transferred to or from a body?

Accepted Answer

Introduction / Context: This question tests your understanding of SI units in thermodynamics. Heat is a form of energy transfer between systems due to a temperature difference. In modern physics and engineering, all forms of energy, including heat, are measured using a common standard unit. Historically, calories were used for heat, but the SI system has replaced this with a more universal energy unit. Knowing the correct SI unit helps you apply formulas correctly and interpret scientific data. Given Data / Assumptions: Heat energy is considered as energy in transit because of temperature difference. The options include joule, newton, calorie and kelvin. We follow the International System of Units (SI). No numerical calculation is required; this is a definition based question. Concept / Approach: In the SI system, the unit of energy is the joule (symbol J). This unit is used for all types of energy, including mechanical work, electrical energy and heat. One joule is defined as the work done when a force of one newton moves an object through a distance of one metre. Heat energy, although historically measured in calories, is now expressed in joules, and the relation is 1 calorie approximately equals 4.18 joules. Newton is the unit of force, and kelvin is the SI base unit of temperature, not of heat energy. Therefore, the correct SI unit for heat energy is joule. Step-by-Step Solution: Step 1: Recall that energy can exist in several forms such as kinetic, potential, electrical and thermal (heat). Step 2: Recognise that the SI unit for any form of energy is the joule, regardless of how the energy is stored or transferred. Step 3: Note that heat is simply energy in transit due to temperature difference, so it must be measured in the same energy unit. Step 4: Remember that newton is used to measure force, not energy. It is related through the definition of joule but is not itself an energy unit. Step 5: Understand that calorie is a non SI unit traditionally used in older thermodynamics and in food energy labels, but in scientific contexts it is converted to joules. Step 6: Recognise that kelvin measures temperature, which is a state variable, not the amount of heat transferred. Step 7: Conclude that joule is the only correct SI unit for heat energy among the given options. Verification / Alternative check: Look at standard thermodynamic equations such as Q = m * c * ΔT, where Q is the heat energy, m is mass, c is specific heat capacity and ΔT is the temperature change. In SI, mass is in kilogram, c is in joule per kilogram per kelvin, and ΔT is in kelvin. Multiplying these gives Q in joules. This direct appearance of joule in the formula confirms that heat energy is measured in joules when SI units are used. Why Other Options Are Wrong: Newton is the SI unit of force and appears in equations like F = m * a, not in energy balance equations as a unit of heat. Calorie is a traditional unit of heat energy, but it is not part of the SI system; it is now defined in terms of joules. Kelvin is the SI base unit of temperature, which measures the thermal state of a system, not the heat energy transferred. Common Pitfalls: Many learners confuse calories with joules because food packets often display energy in kilocalories, while physics textbooks use joules. Another common error is to mix up temperature (kelvin) with heat, which are related but distinct concepts. To avoid these mistakes, remember that in SI all energies, including heat, are expressed in joules, while kelvin is reserved only for temperature and newton for force. Final Answer: In the SI system, the standard unit of heat energy is the Joule (J).

Question 2

According to the International System of Units (SI), what is the derived unit used to measure power, that is, the rate of doing work or transferring energy per unit time?

Accepted Answer

Introduction / Context: This question focuses on the SI unit of power. Power tells us how quickly work is done or energy is transferred. For example, two machines may do the same amount of work, but the one that does it in less time has higher power. Correctly identifying the SI unit of power is essential in mechanics, electricity and many engineering applications. Given Data / Assumptions: Power is defined as work done or energy transferred per unit time. The options include Boyle, watt, newton and pascal. We are working within the SI system of units. No calculation is required; only recognition of the correct unit is needed. Concept / Approach: In SI, power P is defined as P = W / t, where W is work or energy in joules and t is time in seconds. The derived unit of power is the watt (symbol W), named after James Watt. One watt equals one joule per second. Newton is the unit of force, pascal is the unit of pressure and Boyle is not a unit at all but the name of a scientist associated with gas laws. Therefore, watt is the correct SI unit of power. Step-by-Step Solution: Step 1: Recall that work done W is measured in joules and time t in seconds. Step 2: Power P is defined as P = W / t, which has units of joule per second. Step 3: Recognise that the SI derived unit corresponding to joule per second is called watt (W). Step 4: Note that newton (N) is the SI unit of force and appears in F = m * a, not in the definition of power. Step 5: Understand that pascal (Pa) is the SI unit of pressure, defined as newton per square metre. Step 6: Identify that Boyle refers to Robert Boyle, a scientist known for Boyle law, and is not the name of a unit. Step 7: Conclude that the only correct SI unit of power among the options is watt. Verification / Alternative check: To verify, think about the label on electric appliances such as bulbs, fans or heaters. They are typically rated in watts or kilowatts, for example 60 W, 100 W or 1 kW. These ratings tell you how much electrical power the devices consume. In power plant ratings, you often see megawatt or gigawatt. None of these use newton or pascal as power units, confirming that watt is the standard SI unit for power. Why Other Options Are Wrong: Boyle is the name of a scientist and has no official role as a unit in the SI system. Newton is the SI unit of force and is used in mechanics to measure how strongly one body pushes or pulls another. Pascal is the SI unit of pressure, measuring force per unit area, not the rate of doing work. Common Pitfalls: Students sometimes confuse related physical quantities because the same scientist names appear in multiple contexts, such as Boyle law and Newton laws. Another mistake is to think that any familiar scientific name could be a unit. To avoid this, memorise the core SI units and their meanings: watt for power, newton for force and pascal for pressure. This simple mapping will help you answer unit based questions quickly and accurately. Final Answer: The SI derived unit used to measure power is the Watt (W), equal to one joule per second.

Question 3

When a ray of light passes obliquely from one transparent medium into another and its direction changes at the interface, this phenomenon of bending of light is called what?

Accepted Answer

Introduction / Context: This question is about a basic phenomenon in optics. Light rays often pass from one medium to another, for example from air into glass or water. When this happens obliquely, the direction of the light ray changes. Understanding the correct name of this phenomenon is essential for studying lenses, prisms and many optical instruments that rely on the bending of light at boundaries. Given Data / Assumptions: A ray of light travels from one transparent medium into another (for example, air to water). The ray strikes the boundary at an angle, not along the normal. A change in direction of the ray is observed at the interface. We are asked to name this phenomenon of change in direction. Concept / Approach: The change in direction of light when it passes from one medium to another of different optical density is called refraction. Refraction occurs because the speed of light is different in different media. When light enters a denser medium, it usually slows down and bends towards the normal; when it enters a rarer medium, it speeds up and bends away from the normal. Diffraction, in contrast, is the bending of waves around obstacles or through small openings. Propagation is a general term for the travel of waves and does not necessarily imply bending. Therefore, the correct term for this boundary phenomenon is refraction. Step-by-Step Solution: Step 1: Recall that refraction is defined as the change in direction of a light ray when it passes obliquely from one medium to another. Step 2: Understand that this change in direction is caused by the change in speed of light between the two media. Step 3: Recognise that diffraction describes the spreading of waves around obstacles or through slits, which is a different situation. Step 4: Note that propagation simply means movement or transmission and does not specifically refer to bending. Step 5: Since the question clearly states that the direction of light changes at the boundary between media, identify this as refraction. Verification / Alternative check: A familiar example is the apparent bending of a stick when it is partially immersed in water. The part of the stick under water appears shifted because light from that part is refracted at the water air boundary. Similarly, lenses in spectacles or cameras bend light rays to converge or diverge them by refraction. These everyday observations match the definition and confirm that the described phenomenon is refraction, not diffraction or simple propagation. Why Other Options Are Wrong: Diffraction involves bending and spreading of waves around edges or through narrow openings, not the change in direction at a flat interface between two media. Propagation is a general term for wave travel and does not specifically mean a change in direction due to entering a new medium. No option is correct is wrong because refraction is a standard textbook term that exactly matches the description in the question. Common Pitfalls: Students sometimes confuse refraction and diffraction because both involve some form of bending of waves. A useful way to remember the difference is that refraction happens at a boundary between different media, while diffraction happens around obstacles or through slits in a single medium. Keeping this distinction clear will help you select the correct term in exam questions. Final Answer: The change in direction of light when it passes from one medium to another is called Refraction.

Question 4

In basic mechanics, for an object in a state of rest, the state of motion is considered to correspond to what value of speed?

Accepted Answer

Introduction / Context: This question checks your understanding of the fundamental concept of rest and motion in mechanics. In physics, rest and motion are defined relative to a chosen reference frame. An object may be at rest with respect to one observer and moving with respect to another. However, within a given frame, the definition of rest is precise and linked to the object's speed. Knowing this helps you interpret kinematics problems correctly. Given Data / Assumptions: An object is said to be in a state of rest. We are asked about the value of speed in this state. Options mention increasing speed, decreasing speed, inverse speed and zero speed. We assume a fixed reference frame, such as the ground or a laboratory frame. Concept / Approach: Speed is the rate at which an object covers distance, defined as distance travelled per unit time. If an object is at rest relative to a chosen frame, its position does not change with time in that frame. This implies that the distance travelled is zero over any time interval, so its speed must be zero. Rest means absence of motion, and motion requires non zero speed. Terms like increasing or decreasing speed describe changes in motion, not a state of rest. Therefore, zero speed correctly represents the rest state. Step-by-Step Solution: Step 1: Recall that speed v is defined as v = distance / time. Step 2: If an object is at rest with respect to an observer, its distance from that observer does not change with time. Step 3: Because the distance travelled in any time interval is zero, the calculated speed v must also be zero. Step 4: Understand that increasing speed would imply acceleration, while decreasing speed implies deceleration, both of which describe moving objects. Step 5: Recognise that terms like inverse speed have no standard meaning in basic definitions of rest and motion. Step 6: Conclude that zero speed is the only option that correctly represents a state of rest. Verification / Alternative check: Consider a book lying on a table in a room. For a person standing in the room, the book is at rest because its position relative to the room does not change. If you measure the distance the book travels over 10 seconds, it will be zero, and its speed is zero. However, for an observer on the Sun, both the book and the room are moving due to the Earth rotation and revolution. This shows that rest is relative, but within the chosen frame of the room, the state of rest clearly corresponds to zero speed. Why Other Options Are Wrong: Increasing speed refers to a moving object whose speed is getting larger with time; such an object is not at rest. Decreasing speed describes an object that is slowing down but still moving until its speed reaches zero. Inverse speed has no clear definition in elementary mechanics and does not match the physical meaning of rest. Common Pitfalls: Some learners confuse the words rest and constant speed. An object can move with constant non zero speed, which is still motion, not rest. The key is to check whether the position changes with time. Only when the position is unchanged within a frame do we say the object is at rest, and in that case its speed is zero. Always link the concept of rest with zero speed and zero displacement in the chosen reference frame. Final Answer: For an object in a state of rest, the corresponding speed is Zero speed.

Question 5

In experimental physics, an instrument specially designed for accurately measuring very low temperatures is called what?

Accepted Answer

Introduction / Context: This question comes from the topic of measuring instruments in physics. Different physical quantities such as length, temperature, pressure and electric current require specialised devices for accurate measurement. Ordinary thermometers are suitable for everyday temperatures, but extremely low temperatures, for example in cryogenics experiments, demand special instruments. Knowing the name of the device used for low temperature measurement is useful in general knowledge as well as in basic physics. Given Data / Assumptions: The physical quantity of interest is temperature, specifically very low temperatures. The options list several unusual instrument names. We assume low temperature means values far below the freezing point of water, such as those near liquid nitrogen or liquid helium temperatures. We focus only on the standard terminology used in textbooks. Concept / Approach: The correct term for an instrument used to measure very low temperatures is cryometer. The prefix cryo comes from a Greek word meaning cold or frost and is used in words like cryogenics, cryosurgery and cryostat. A cryometer is designed to remain accurate and responsive at low temperatures where ordinary mercury or alcohol thermometers may not perform well. The other names in the options either do not correspond to standard instruments or are used in different contexts. Therefore, cryometer is the correct choice. Step-by-Step Solution: Step 1: Recognise that instruments for temperature measurement generally include thermometer, pyrometer and cryometer, each for different ranges. Step 2: Recall that pyrometers are used for very high temperatures, while cryometers are designed for very low temperatures. Step 3: Note that the prefix cryo indicates cold, which fits the context of low temperature measurements. Step 4: Examine the options and see that cryometer matches this description directly. Step 5: Verify that the other given names do not represent standard thermometric devices for low temperature measurement. Step 6: Conclude that cryometer is the correct name for an instrument measuring very low temperatures. Verification / Alternative check: In discussions of temperature ranges, textbooks often mention ordinary thermometers for everyday ranges, clinical thermometers for body temperature, pyrometers for furnace temperatures and cryometers for cryogenic temperatures. For example, temperatures near the boiling point of liquid nitrogen (around 77 K) or liquid helium (around 4 K) are measured using cryogenic techniques and instruments. The consistent use of the term cryometer or related cryogenic sensors in this context supports the choice of cryometer as the correct option. Why Other Options Are Wrong: Diagometer is not a standard term associated with low temperature measurement in physics. Chromatoptometer suggests an instrument related to colour or optics, not to temperature, and is not commonly used in basic physics. Cymometer, if encountered, would more likely relate to measuring wave frequencies (cycles), not temperature. Common Pitfalls: Because several of the options sound unfamiliar, students may try to guess based purely on spelling. A helpful strategy is to pay attention to word roots: cryo for cold, pyro for heat and thermo for temperature in general. Recognising these roots allows you to decode many scientific terms and choose the correct option even if you have not memorised all instrument names directly. Final Answer: An instrument designed for accurately measuring very low temperatures is called a Cryometer.

Question 6

In particle physics, what is the name of the antiparticle that corresponds to the electron and has the same mass as an electron but a positive charge?

Accepted Answer

Introduction / Context: This question tests your knowledge of fundamental particles and antiparticles in modern physics. According to quantum theory and relativistic physics, every particle has a corresponding antiparticle with the same mass and opposite charge, if it is charged. For the electron, which is a negatively charged lepton, the corresponding antiparticle plays an important role in processes such as pair production and beta plus decay. Recognising its correct name is part of basic particle physics general knowledge. Given Data / Assumptions: The particle of interest is the electron, a light, negatively charged particle. The antiparticle must have the same mass as the electron but opposite (positive) charge. Options include positron, proton, alpha particles and beta particles. We assume standard definitions from particle physics. Concept / Approach: The antiparticle of the electron is called the positron. It has the same rest mass as the electron but carries a positive electric charge of the same magnitude. When an electron and a positron meet, they can annihilate, converting their mass into energy, usually in the form of gamma ray photons. Protons and alpha particles are much more massive than electrons and are not antiparticles of electrons. The term beta particles can refer to either electrons or positrons emitted in nuclear beta decay, but it does not uniquely name the antiparticle of the electron. Therefore, the correct and specific name is positron. Step-by-Step Solution: Step 1: Recall that an antiparticle has the same mass and spin as its corresponding particle but opposite charge and some other quantum numbers. Step 2: The electron has negative charge and small mass, so its antiparticle must have positive charge and the same small mass. Step 3: The particle with these properties is known as the positron. Step 4: Check the other options: the proton is positively charged but about 1836 times heavier than the electron, so it cannot be the electron antiparticle. Step 5: Alpha particles are helium nuclei composed of two protons and two neutrons, with much larger mass and charge +2e. Step 6: Beta particles is a generic term that can refer to fast electrons or positrons emitted in beta decay, not a specific name for the antiparticle. Step 7: Conclude that positron is the only correct, specific antiparticle of the electron. Verification / Alternative check: In many nuclear physics examples, beta plus decay is written as p → n + e^+ + ν, where e^+ denotes the positron. This notation makes it clear that e^+ is the positive counterpart of the electron e^-. Experiments involving pair production also show a gamma photon converting into an electron (e^-) and a positron (e^+), again highlighting the particle antiparticle pair. Such standard reactions confirm that the antiparticle of the electron is indeed the positron. Why Other Options Are Wrong: Proton is a much heavier positively charged particle that forms part of the nucleus; it is not the antiparticle of the electron. Alpha particles are helium nuclei with charge +2e and significant mass, completely different from a single electron or its antiparticle. Beta particles is a general term for fast electrons or positrons from radioactive decay, but it is not a unique name for the electron antiparticle. Common Pitfalls: Students sometimes confuse proton with positron because both are positively charged and their names sound similar. The key difference is that a proton is a baryon in the nucleus, while a positron is a lepton with the same mass as the electron. Remembering that the prefix posi in positron indicates positive electron helps distinguish it from proton. Paying attention to mass as well as charge is important when identifying antiparticles. Final Answer: The antiparticle corresponding to the electron is called the Positron.

Question 7

An object has a mass of 60 kg on the Earth. Approximately what will be its weight on the Moon, where the acceleration due to gravity is about one sixth of that on Earth (answer in newton, N)?

Accepted Answer

Introduction / Context: This question examines your understanding of the difference between mass and weight and how weight depends on the local acceleration due to gravity. Mass is a measure of the amount of matter in an object and remains the same everywhere. Weight, however, is the gravitational force acting on that mass and changes when the gravitational field changes, such as moving from Earth to the Moon. Many competitive questions ask you to estimate weight on different celestial bodies using simple ratios. Given Data / Assumptions: Mass of the object, m = 60 kg. Acceleration due to gravity on Earth, g_E, is about 9.8 m/s^2, commonly approximated as 10 m/s^2 in simple problems. Acceleration due to gravity on the Moon, g_M, is about one sixth of that on Earth. We are asked for the weight on the Moon in newton (N), where weight W = m * g. Concept / Approach: Weight is defined as the force with which a planet or celestial body attracts a mass, given by W = m * g. On Earth, a 60 kg mass has a weight approximately W_E = 60 * 10 = 600 N. On the Moon, where gravity is about one sixth of Earth gravity, the weight will be roughly one sixth of 600 N. That is 600 / 6 = 100 N. The mass of the object does not change, only the local gravitational acceleration does. Therefore, the approximate weight on the Moon is 100 N. Step-by-Step Solution: Step 1: Write the expression for weight: W = m * g. Step 2: For Earth, take g_E ≈ 10 m/s^2. Then W_E = 60 kg * 10 m/s^2 = 600 N. Step 3: The Moon gravity is g_M = g_E / 6, which is about 10 / 6 ≈ 1.67 m/s^2. Step 4: Compute the weight on the Moon: W_M = m * g_M = 60 kg * (10 / 6) m/s^2 = 600 / 6 N. Step 5: Evaluate 600 / 6 = 100 N. Step 6: Compare with options and select 100N as the best approximate value. Verification / Alternative check: A quick ratio method can also be used. Since weight is proportional to g and the Moon gravity is one sixth of Earth gravity, the weight on the Moon must be one sixth of the weight on Earth. If the weight on Earth is about 600 N, dividing by 6 gives 100 N. This simple proportion directly yields the same answer without detailed arithmetic, confirming the result. Why Other Options Are Wrong: 60N is too small because it corresponds to one tenth of the Earth weight, whereas the Moon gravity is one sixth, not one tenth, of Earth gravity. 600N is the approximate weight on Earth, not on the Moon. Since Moon gravity is weaker, the weight must be less than this. 10N is far too small and would correspond to a mass of only about 1 kg under Earth gravity, not 60 kg under Moon gravity. Common Pitfalls: Some students mistakenly think that mass changes when moving from Earth to the Moon and try to adjust the mass instead of adjusting gravity. Always remember that mass remains constant, while weight changes with local gravitational acceleration. Another common error is to forget the one sixth factor and to apply the full Earth gravity value. Using the ratio method and keeping the distinction between mass and weight clear helps avoid these mistakes. Final Answer: A mass of 60 kg will have an approximate weight of 100N on the Moon.

Question 8

Reflection of light from a perfectly smooth and polished surface, such as a plane mirror, in which parallel incident rays remain parallel after reflection, is specifically called what type of reflection?

Accepted Answer

Introduction / Context: This optics question asks you to recall the correct term for reflection from a smooth surface like a mirror. Light can reflect in different ways depending on the nature of the surface. For highly polished surfaces, the reflected rays preserve their relative orientation, producing clear images. Recognising the standard terminology for this type of reflection is important for understanding image formation, mirrors and optical devices. Given Data / Assumptions: The reflecting surface is smooth and polished, like a plane mirror. Parallel rays of light fall on the surface. After reflection, the rays remain parallel and follow the law of reflection. We are asked to name this kind of reflection. Concept / Approach: When light falls on a smooth, polished surface, all the microscopic points on the surface have nearly the same orientation. As a result, parallel incident rays are reflected in a single, well defined direction, and the law of reflection (angle of incidence equals angle of reflection) holds at every point. This is called regular reflection or specular reflection. On rough surfaces, irregularities cause the reflected rays to scatter in many directions; this is called diffuse or irregular reflection. Therefore, reflection from a mirror like surface is termed regular reflection. Step-by-Step Solution: Step 1: Recall that a plane mirror provides a clear image because reflected rays remain well organised and predictable. Step 2: Understand that this behaviour occurs when the surface irregularities are much smaller than the wavelength of light, making the surface effectively smooth. Step 3: Recognise that this type of reflection is called regular or specular reflection in textbooks. Step 4: Compare with rough surfaces such as walls or paper, where incoming parallel rays are scattered in many directions, giving no clear image; this is diffuse or irregular reflection. Step 5: Since the question refers to a smooth mirror like surface, identify the appropriate term as regular reflection. Verification / Alternative check: Think about what happens when you look at yourself in a mirror versus in a sheet of paper. The mirror shows a sharp image due to regular reflection, while the paper does not show an image because it reflects light diffusely. Diagrams of mirror reflection in physics textbooks also label reflection from plane mirrors as regular reflection. This consistent usage confirms that regular is the correct option for a smooth surface. Why Other Options Are Wrong: Irregular reflection refers to reflection from rough surfaces where light is scattered in many directions and no clear image is formed. Diffused reflection is another term for irregular reflection and does not describe the mirror like reflection that forms sharp images. Fused reflection is not a standard term in basic optics and does not correspond to any commonly described type of reflection. Common Pitfalls: Students sometimes confuse the terms regular and diffuse reflection because both obey the law of reflection at each microscopic point. The key difference is the smoothness of the surface and the overall pattern of reflected rays. Remember that mirrors and shiny metal surfaces produce regular reflection, while walls, cloth and paper produce diffuse reflection. Associating the word regular with mirror like reflection helps you choose the correct option. Final Answer: Reflection from a smooth, polished mirror like surface is called regular reflection.

Question 9

In atmospheric and fluid convection, which statement best describes how air typically moves during a convection current formed by heating the air near the ground?

Accepted Answer

Introduction / Context: This question deals with the basic idea of convection, an important mode of heat transfer in fluids such as air and water. Convection is responsible for many natural phenomena, including sea breezes, thunderstorms and the working of chimneys. Understanding how warm and cool air move relative to each other in a convection current is essential for interpreting weather patterns and basic physics experiments involving heated fluids. Given Data / Assumptions: Air near a heat source (such as the ground warmed by the Sun) is heated. Cooler air exists in the surrounding region. We are asked to choose the best description of how air moves during convection. We consider typical convection in the atmosphere or in a heated room, ignoring complex turbulence. Concept / Approach: When air is heated, its temperature increases, causing its density to decrease. As a result, warm air becomes lighter and tends to rise. Cooler air surrounding the warm region is denser and moves in to take the place of the rising warm air. This continuous cycle of rising warm air and sinking cool air forms a convection current. The most accurate verbal description among the options is that cool air is replaced by warm air, capturing both the upward movement of warm air and the downward movement of cool air that completes the cycle. Step-by-Step Solution: Step 1: When air near the ground is heated, its molecules move faster and spread apart, reducing its density. Step 2: Due to lower density, the warm air experiences a net upward buoyant force and rises. Step 3: As warm air rises, it leaves behind a region of lower pressure near the ground. Step 4: Cooler, denser air from nearby regions moves in horizontally to fill this low pressure area. Step 5: This cooler air is then heated in turn, rises, and the process repeats, creating a convection current. Step 6: The overall effect is that cool air is continually being replaced by warm, rising air above the heat source. Verification / Alternative check: A common example is a candle placed below a small paper spiral or a smoke stream. The warm air from the candle flame rises, carrying smoke upward and drawing cooler air in from below and around the flame. Similarly, in sea breezes, land heats faster than sea during the day, warm air rises over land and cool sea air flows in to replace it. These observations match the description that cool air moves in while warm air rises, which is summarised as cool air is replaced by warm air in simple multiple choice questions. Why Other Options Are Wrong: Cool air falls down is incomplete; while cool air does sink, convection also involves warm air rising and replacing it. Warm air falls down is incorrect because warm air is less dense and tends to rise, not fall. Air always falls down is wrong because air can also rise when it is heated; convection currents involve both upward and downward motion. Common Pitfalls: Students sometimes focus only on one part of the convection cycle, such as warm air rising, and forget that the cycle must be completed by cooler air moving in. In multiple choice questions, partial descriptions can be tempting but are not fully correct. To avoid confusion, remember that convection is a continuous cycle: warm air rises, cool air moves in to replace it and then itself becomes warm and rises. The option that captures both replacement and temperature difference is usually the best choice. Final Answer: During convection, cool air is replaced by warm air as warm, less dense air rises and cooler, denser air moves in beneath it.

Question 10

Among the following physical quantities, which one is a vector quantity, that is, it has both magnitude and direction and obeys vector addition?

Accepted Answer

Introduction / Context: This question tests your ability to distinguish between scalar and vector quantities in physics. A vector quantity has both magnitude and direction and follows the rules of vector addition, while a scalar has only magnitude. Many physical quantities may sound similar but differ in whether direction is an essential part of their definition. Correctly identifying vectors is important in mechanics and field theory. Given Data / Assumptions: The quantities listed are pressure, impulse, gravitational potential and coefficient of friction. We are asked to pick the only vector quantity among them. No numerical data is provided; this is a concept based classification question. We assume standard definitions from classical mechanics. Concept / Approach: Impulse is defined as the change in momentum of a body when a force acts on it for a short time. Since momentum is a vector quantity (mass times velocity) and force is also a vector, impulse must be a vector quantity with both magnitude and direction. Pressure is force per unit area and is treated as a scalar in basic physics (in more advanced treatments it is part of a tensor but not a simple vector here). Gravitational potential is scalar energy per unit mass. Coefficient of friction is a dimensionless scalar constant. Therefore, impulse is the only vector quantity in the list. Step-by-Step Solution: Step 1: Recall that a vector quantity must have a direction that is physically meaningful and must obey vector addition rules. Step 2: Define impulse J as J = F * Δt, where F is the force vector and Δt is the time interval. Step 3: Recognise that impulse has the same direction as the force and represents the change in momentum, which is also a vector. Step 4: Consider pressure P, defined as normal force per unit area. In elementary physics, P is treated as a scalar because only its magnitude matters in basic problems. Step 5: Gravitational potential is energy per unit mass at a point in a field and is a scalar; it does not have a direction by itself. Step 6: The coefficient of friction is a dimensionless scalar describing the ratio of frictional force to normal reaction, not a vector. Step 7: Conclude that among the given options, impulse is the only vector quantity. Verification / Alternative check: An additional way to verify is to think about how these quantities combine. If you apply two impulses in different directions, the net effect on momentum is obtained by vector addition of the two impulses. On the other hand, pressures at different points are usually combined by arithmetic or weighted averages, not by vector addition. Similarly, gravitational potentials and coefficients of friction are added as scalars. This behaviour confirms that impulse behaves as a vector, while the others behave as scalars in typical problems. Why Other Options Are Wrong: Pressure is treated as a scalar quantity in basic mechanics, describing magnitude of normal force per unit area, without a direction that obeys vector addition rules at this level. Gravitational potential is a scalar field quantity; it provides a value of potential energy per unit mass at each point, not a direction. Coefficient of friction is a scalar constant between two surfaces and has no direction. Common Pitfalls: Students may confuse pressure with force because both involve the idea of push and area. While force is clearly a vector, pressure is defined as a scalar in basic courses. Another pitfall is to think of gravitational potential as a vector because gravity has a direction. However, gravitational field or acceleration is the vector quantity; gravitational potential is scalar. Focusing on definitions and how quantities add helps avoid these misunderstandings. Final Answer: The only vector quantity among the options is impulse.

Question 11

A domestic fluorescent tube light produces white light. Which option best explains the physical reason for the white fluorescent light emitted by such a tube?

Accepted Answer

Introduction / Context: This question relates to the working principle of fluorescent tube lights, which are widely used in homes and offices. Unlike incandescent bulbs that produce light by heating a filament, fluorescent lamps rely on gas discharge and fluorescence. Understanding why the light appears white requires knowing how ultraviolet radiation and phosphor coatings interact inside the tube. Given Data / Assumptions: The device considered is a domestic fluorescent tube light. The tube contains a low pressure gas (often mercury vapour) and has a white coating on the inner surface of the glass. Electric current is passed through the tube using a starter and choke. We are asked why the tube finally emits white fluorescent light. Concept / Approach: Inside a fluorescent tube, the electric discharge through low pressure mercury vapour produces mainly ultraviolet (UV) radiation, not visible light. The inner surface of the glass tube is coated with a phosphor material, which absorbs the ultraviolet photons and re emits energy as visible light of various wavelengths. The mixture of these visible wavelengths appears as white light to our eyes. The key step is the conversion of invisible ultraviolet radiation into visible light by the phosphor coating. Simply having heavy current or a vacuum does not explain the white fluorescence. Step-by-Step Solution: Step 1: When the tube is switched on, a high voltage starts an electric discharge through low pressure mercury vapour inside the tube. Step 2: Collisions between electrons and mercury atoms excite the atoms, which then emit photons primarily in the ultraviolet range. Step 3: Ultraviolet light produced inside the tube cannot be seen directly by the human eye and would be harmful if emitted strongly. Step 4: The inner surface of the tube is coated with a white phosphor material that is sensitive to ultraviolet radiation. Step 5: When ultraviolet photons fall on the phosphor coating, the coating absorbs this energy and re emits it as visible light of many wavelengths. Step 6: The combination of these visible wavelengths forms a white appearing light, which is what we see coming from the fluorescent tube. Verification / Alternative check: If the inner coating is damaged or missing, the tube would not produce proper white light. In such a case, more ultraviolet radiation would escape, and the brightness would decrease or the colour would change. Also, compact fluorescent lamps (CFLs) use the same principle: a gas discharge produces UV, which is converted to visible light by a phosphor coating. These practical observations support the explanation that the key step involves ultraviolet radiation falling on a white phosphor coating, not just collisions or vacuum alone. Why Other Options Are Wrong: Collision between molecules of filled gas under electric current is part of the process that produces ultraviolet radiation, but by itself it does not explain the white fluorescent light observed; the role of the phosphor coating is missing. Heavy current is not the main reason for fluorescence; even moderate currents can produce light if the discharge and coating are correct. Vacuum inside the tube is incorrect; fluorescent tubes contain low pressure gas, not a complete vacuum, and the emission mechanism relies on gas discharge. Common Pitfalls: Students often give overly generic answers such as electric discharge or heavy current without mentioning the critical role of ultraviolet radiation and the phosphor coating. Another confusion is to assume that the gas itself directly emits white light. Remember that in fluorescent tubes, the gas primarily produces UV, and the coating converts it to visible light. Recognising this two step process will help you answer similar questions correctly. Final Answer: A fluorescent tube gives white light mainly because ultra violet rays produced inside the tube fall on the white inner coating (phosphor), which then emits visible white light.

Question 12

A capillary tube is partially dipped vertically into a vessel containing water. Because of capillarity, water rises in the tube. How can the height to which water rises in the capillary tube be increased?

Accepted Answer

Introduction / Context: This question examines the concept of capillarity, which is the rise or fall of a liquid in a narrow tube due to surface tension and adhesive forces. Capillary action is important in many natural and practical situations, such as water rising in plant xylem, ink moving in a thin tube and oil lamps drawing fuel up a wick. Understanding which factors affect the height of capillary rise helps you apply the formula correctly in problems. Given Data / Assumptions: A narrow capillary tube is partially dipped vertically in water. Water wets glass, so the liquid rises in the capillary. We want to increase the height h of the water column in the tube. Options involve changing the radius of the tube or the water level in the vessel. Concept / Approach: The height h of capillary rise for a liquid that wets the tube walls is given by the relation h = 2 * T * cosθ / (ρ * g * r), where T is surface tension, θ is the contact angle, ρ is the liquid density, g is acceleration due to gravity and r is the radius of the capillary. All other factors being constant, h is inversely proportional to r, the radius of the tube. This means that if you decrease the radius, the height of rise increases. Increasing the radius will reduce the height, and changing the water level in the vessel does not significantly affect the capillary rise formula. Step-by-Step Solution: Step 1: Write the formula for capillary rise: h = (2 * T * cosθ) / (ρ * g * r). Step 2: Observe that surface tension T, contact angle θ, liquid density ρ and g are properties of the liquid and environment and are not changed in the options. Step 3: Notice that the only variable mentioned in the options that appears in the formula is the radius r of the capillary tube. Step 4: Since h is inversely proportional to r, decreasing r will increase h, while increasing r will reduce h. Step 5: Realise that increasing the height of water in the vessel does not change r, T, θ, ρ or g, so it does not affect capillary rise. Step 6: Therefore, the way to increase the height of water rise is to use a capillary tube with a smaller radius. Verification / Alternative check: Capillary rise can be observed experimentally using glass tubes of different diameters dipped in the same liquid. You will see that the narrower tube shows a higher water column, while the wider tube shows a lower rise. This is a standard demonstration in school laboratories and matches the mathematical relationship h ∝ 1 / r. It confirms that decreasing the radius increases the height of the capillary rise. Why Other Options Are Wrong: Increasing the radius of the capillary tube increases r in the denominator, which decreases the height h of capillary rise. Increasing the height of water in the vessel changes the external water level but does not alter the capillary action formula, so it does not significantly change the additional rise inside the tube. None of these is incorrect because decreasing the radius is a valid and effective way to increase the height of capillary rise. Common Pitfalls: Students sometimes assume that pouring more water into the vessel will raise the level in the capillary tube, confusing the base water level with capillary rise. Another mistake is to think that wider tubes should allow more water to rise higher, which is opposite to what the formula shows. Remember the key relationship: for a given liquid and conditions, a narrower capillary means higher capillary rise. Writing down the formula and focusing on the dependence on r helps avoid these misconceptions. Final Answer: The height of water rise in the capillary tube can be increased by decreasing the radius of the capillary tube.

Question 13

Hot water is poured simultaneously into four identical metallic tumblers that are painted on the outside with different types of surfaces. After some time, in which tumbler will the water be found to have cooled the most due to maximum heat loss by …

Accepted Answer

Introduction / Context: This question is about heat transfer by radiation and how the nature and colour of a surface affect its ability to emit and absorb thermal radiation. In everyday life, black and rough surfaces are known to get hotter in the Sun and to cool more quickly in the shade. Physics explains this behaviour by the concepts of good and poor radiators. Understanding which type of surface radiates heat most effectively helps answer questions involving cooling and heating of bodies. Given Data / Assumptions: Four identical metallic tumblers contain hot water initially at the same temperature. The tumblers differ only in the nature and colour of their outer surfaces: rough white, rough black, shining white and shining black. The tumblers are kept under identical environmental conditions. We are asked in which tumbler the water cools the most, that is, where heat loss by radiation is maximum. Concept / Approach: According to thermal radiation principles, a good absorber of heat is also a good emitter, and poor absorbers are also poor emitters. Black, dull and rough surfaces are excellent absorbers and emitters, while white, shiny and polished surfaces are poor absorbers and emitters. Therefore, a rough black surface will radiate heat away more effectively than a shining or white surface. As a result, the tumbler with rough black paint will allow the hot water to lose heat faster and cool more in a given time. Step-by-Step Solution: Step 1: Recall that black surfaces absorb and emit thermal radiation more strongly than white surfaces, which mostly reflect radiation. Step 2: Understand that rough or matte surfaces behave as better radiators than smooth, shiny surfaces with the same colour. Step 3: Combine these ideas: a rough black surface is both black (good emitter) and rough (good radiator), making it the best radiator among the given options. Step 4: Recognise that rough white is a good radiator compared to shiny surfaces but still less effective than rough black because white reflects more radiation. Step 5: Shining white and shining black surfaces reflect more radiation and emit less, making them poorer radiators than rough surfaces. Step 6: Therefore, the water in the rough black painted tumbler will lose heat fastest and cool the most. Verification / Alternative check: An experimental way to verify this is to fill identical containers with hot water and paint them with different surface finishes. After some time, measuring the temperature of the water will show that the container with a dull black surface leads to the greatest temperature drop. This experiment is often demonstrated in school laboratories and supports the theoretical prediction that rough black is the best radiator. The same idea explains why solar cookers and radiators are often painted black and rough. Why Other Options Are Wrong: Rough white, although rough, is white and therefore reflects more radiation than black surfaces; it is a poorer radiator than rough black. Shining white is both reflective and smooth, making it one of the poorest radiators among the options. Shining black, while black, is shiny and reflects more radiation than a rough surface, so it is less effective at radiating heat than rough black. Common Pitfalls: Students sometimes give importance only to colour and ignore the effect of surface roughness, or vice versa. The correct understanding is that both colour and surface texture influence radiation: black is better than white, and rough is better than smooth for emission. Another confusion is between absorption and emission; however, good absorbers are also good emitters. Keeping these relationships clear helps you choose rough black as the best radiator for cooling in such questions. Final Answer: The water will cool the most in the tumbler painted rough black, because it is the best radiator of heat.

Question 14

When a person enters a dark room after being in bright sunlight, they are not able to see objects clearly for a short time. What is the main reason for this temporary inability to see?

Accepted Answer

Introduction / Context: This question deals with the adaptation of the human eye to changes in light intensity. Our eyes can operate over a wide range of brightness levels, from bright sunlight to dim starlight. However, the adjustment to new light levels is not instantaneous. When moving from a bright environment into a dark one, people often experience a brief period where they cannot see clearly. Understanding the role of the iris and pupil in this process is essential for basic biology and physics of vision. Given Data / Assumptions: The person moves suddenly from bright sunlight into a dark room. Initially, they are unable to see objects clearly in the dark room. The options mention the retina moving, the iris and the power of accommodation. We assume a normal healthy eye without disease. Concept / Approach: The iris is the coloured part of the eye surrounding the pupil. It controls the size of the pupil and hence the amount of light entering the eye. In bright light, the pupil becomes small (constriction) to protect the retina from excess light. In dim light, the pupil must widen (dilate) to allow more light to enter. When you move suddenly from bright to dark, the iris cannot immediately dilate the pupil to its new required size. Until the pupil has widened sufficiently and chemical changes in the photoreceptor cells occur, vision is poor. This delay in dilation and adaptation is the main reason for the temporary loss of clear vision. Step-by-Step Solution: Step 1: In bright sunlight, the pupil is constricted to a small size to reduce the amount of light reaching the retina. Step 2: When the person suddenly enters a dark room, the available light is much less, but the pupil is still small for a brief period. Step 3: Because the pupil is small, very little light reaches the retina, so the image formed is too dim to be perceived clearly. Step 4: The iris muscles gradually respond to the reduced light by dilating the pupil, increasing its size. Step 5: As the pupil widens and photochemical adaptation in rods occurs, more light reaches the retina and sensitivity improves, restoring clear vision. Step 6: Therefore, the temporary inability to see clearly is due to the iris not being able to widen the pupil instantly. Verification / Alternative check: This effect can be felt in daily life. When entering a dark cinema hall from a sunny street, you initially struggle to see the steps and seats, but after a short time your eyes adapt and you can see quite well. Conversely, when you come out into bright light from a dark room, you may be dazzled until the pupils constrict. In both cases, the iris and pupil size adjustment takes finite time. The retina does not move forwards or backwards; its position is fixed on the back of the eye. These observations support the explanation based on iris and pupil dilation. Why Other Options Are Wrong: The retina of the eye moving forward or backward is incorrect; the retina is a fixed light sensitive layer and does not move in normal eye function. The eye having no power of accommodation would affect the ability to focus on near or far objects, not the ability to see in low light. The key process here is the time taken by the iris and associated muscles to dilate the pupil and for photoreceptors to adapt, not a change in retinal position. Common Pitfalls: Students may confuse accommodation, which is the adjustment of the lens to focus on objects at different distances, with adaptation to different light levels, which mainly involves changes in pupil size and retinal sensitivity. It is important to remember that accommodation affects focus (near versus far), while pupil dilation and photochemical changes affect brightness sensitivity. In this question, the issue is brightness, so the iris and pupil are the central elements. Final Answer: The temporary inability to see clearly in a dark room after bright light occurs because the iris cannot immediately dilate the pupil to let in enough light.

Question 15

A boy sitting in a train moving with uniform horizontal velocity drops a coin out of the window. As seen by a man standing on the ground, what trajectory does the coin follow as it falls to the Earth?

Accepted Answer

Introduction / Context: This question tests your understanding of projectile motion and relative observation from different frames of reference. When an object is dropped from a moving vehicle, it has both horizontal and vertical motion components. A passenger inside the vehicle and an observer on the ground may describe the path differently. Recognising the shape of the trajectory as seen from the ground is a standard kinematics concept. Given Data / Assumptions: The train moves horizontally with uniform velocity, meaning constant speed and direction. A boy in the train drops a coin from some height. Air resistance is neglected for simplicity. We consider the path of the coin as seen by a person standing on the ground. Concept / Approach: At the instant the coin is released, it already has the same horizontal velocity as the train, because it was moving with the train. When the coin is dropped, there is no horizontal force acting on it (ignoring air drag), so it continues to move horizontally with the same uniform speed. At the same time, gravity acts downward, giving the coin a vertical acceleration. The combination of uniform horizontal motion and uniformly accelerated vertical motion produces a parabolic path, which is the characteristic trajectory of a projectile in a uniform gravitational field. Step-by-Step Solution: Step 1: At the moment of release, the coin has a horizontal velocity equal to the uniform velocity of the train. Step 2: Immediately after release, the horizontal component of velocity remains constant because no horizontal force acts on the coin. Step 3: Gravity provides a constant downward acceleration g, giving the coin a vertical velocity component that increases with time. Step 4: The horizontal displacement increases linearly with time, while the vertical displacement increases with time squared due to constant acceleration. Step 5: When you plot vertical displacement against horizontal displacement, the resulting curve is a parabola, which is the path seen by the ground observer. Step 6: Therefore, the man standing outside the train will find the trajectory of the coin to be parabolic. Verification / Alternative check: In mathematical terms, horizontal motion can be described by x = u * t, where u is constant horizontal speed. Vertical motion follows y = (1/2) * g * t^2 downward (if initial vertical velocity is zero). Eliminating time t between these two equations leads to y being proportional to x^2, which is the equation of a parabola. This confirms that the path is not a straight line but a curved parabolic trajectory. Numerous experiments, such as dropping a ball from a moving cart, show similar behaviour. Why Other Options Are Wrong: A horizontal straight line would require no vertical acceleration, which is impossible when gravity acts. A vertical straight line would be seen by the boy in the train if he ignores the horizontal motion, but not by the ground observer, who sees both components. A circle does not arise from combining uniform horizontal motion with uniform vertical acceleration; the equations of motion do not match circular motion conditions. Common Pitfalls: Students sometimes think that dropping an object from a moving vehicle makes it fall straight down because they imagine only the vertical motion. This error usually comes from forgetting that the object already has horizontal velocity at the moment of release. Another confusion arises from mixing the viewpoints of the train passenger and the ground observer. Always specify the reference frame: to a person in the train, the coin may appear to fall nearly straight down; to the ground observer, the path is clearly parabolic. Final Answer: To a man standing on the ground, the coin follows a parabola as it falls.

Question 16

For bodies freely falling under gravity near the Earth at a given place (neglecting air resistance), which of the following physical quantities remains constant during the fall?

Accepted Answer

Introduction / Context: This question examines your understanding of motion under gravity. When an object falls freely near the Earth surface, its motion is influenced by the gravitational field. Some quantities change during the fall, while others remain constant. Being able to identify the constant acceleration due to gravity g and distinguish it from changing energy values is fundamental in kinematics and energy conservation problems. Given Data / Assumptions: A body is falling freely under gravity near the Earth at a fixed location. Air resistance is neglected. We consider kinetic energy, potential energy and acceleration as the body falls. The local value of gravitational acceleration g is taken as constant. Concept / Approach: For free fall near the Earth surface, the acceleration due to gravity g is approximately constant at that location, independent of the mass of the body and its instantaneous speed. As the body falls, its velocity increases linearly with time, so its kinetic energy, which depends on the square of velocity, increases. At the same time, gravitational potential energy decreases because height decreases. Although the sum of kinetic and potential energies remains constant in the absence of air resistance, the individual values of kinetic and potential energy do not. Therefore, among the given options, acceleration is the quantity that remains constant during the fall. Step-by-Step Solution: Step 1: For free fall, write the equation of motion: v = u + g * t, where u is initial velocity, v is velocity at time t and g is acceleration due to gravity. Step 2: Notice that g appears as a constant in this equation, meaning acceleration remains the same throughout the motion. Step 3: Kinetic energy is given by K = (1/2) * m * v^2. As v increases with time, K increases, so kinetic energy is not constant. Step 4: Potential energy is given by U = m * g * h, where h is height above a reference level. As the body falls and h decreases, U decreases, so it is not constant either. Step 5: The sum of K and U remains constant (if air resistance is neglected), but the question asks specifically about the listed quantities, not their sum. Step 6: Conclude that only acceleration remains constant among kinetic energy, potential energy and acceleration. Verification / Alternative check: Graphical analysis also supports this conclusion. A velocity versus time graph for free fall is a straight line with slope equal to g, indicating constant acceleration. A kinetic energy versus time graph curves upward because kinetic energy depends on v^2. Potential energy versus height is linear, but since height decreases with time in a nonlinear way during fall, potential energy versus time is not constant. These visualisations confirm that acceleration is constant while the energies change. Why Other Options Are Wrong: Kinetic energy does not remain constant because the speed of the falling body increases, causing its kinetic energy to increase. Potential energy decreases as the body loses height, so it is also not constant. All the above is incorrect because both kinetic and potential energies change, leaving only acceleration as constant. Common Pitfalls: Some students mix up the idea of constant total mechanical energy with individual energy components. While the total energy K + U is conserved in ideal free fall, kinetic energy and potential energy separately change with time. Another common confusion is to think acceleration decreases as the object speeds up, which is not true in uniform gravitational fields near the Earth surface. Remember: for free fall at a fixed location, g is constant and independent of velocity or mass. Final Answer: For bodies falling freely under gravity at a given place, the quantity that remains constant is the acceleration due to gravity.

Question 17

A boy is swinging a ball attached to a string in a horizontal circle. If the string suddenly snaps at some instant, how will the ball move just after the string breaks?

Accepted Answer

Introduction / Context: This mechanics question tests your understanding of circular motion and the concept of inertia. When an object moves in a circle, a centripetal force is needed to constantly change its direction. If this centripetal force is removed suddenly, the object does not continue in a circle but follows a new path determined by its velocity at that instant. Recognising that this path is tangential to the circle is key to answering the question correctly. Given Data / Assumptions: A ball is attached to a string and is being rotated in a horizontal circle. The motion is uniform circular motion just before the string breaks. The string suddenly snaps at some point in the motion. Air resistance and other forces besides gravity are neglected immediately after the break. Concept / Approach: In uniform circular motion, the velocity of the ball is always tangential to the circle, while the centripetal force provided by the string acts towards the center. This inward force keeps changing the direction of the velocity vector. If the string breaks, the centripetal force instantly becomes zero. According to Newton first law of motion (the law of inertia), the ball will continue to move in a straight line in the direction of its velocity at that instant. That straight line is the tangent to the circle at the point where the string snapped. Gravity then acts downward, so the ball follows a projectile like path starting along that tangent. Step-by-Step Solution: Step 1: Identify the forces acting during circular motion: tension in the string provides centripetal force towards the center, and gravity acts vertically. Step 2: The instantaneous velocity of the ball is always along a tangent to the circular path at every point. Step 3: When the string snaps, the tension force disappears suddenly, removing the centripetal force. Step 4: By inertia, the ball will continue to move in the direction of its instantaneous velocity, which is along the tangent at the breaking point. Step 5: After that, gravity starts to pull the ball downward, causing it eventually to fall to the ground along a curved trajectory starting from the tangential direction. Step 6: Therefore, the correct description is that the ball continues along the tangent to the circle and then falls down. Verification / Alternative check: You can visualise a similar situation by swinging a stone tied to a string and then releasing it at some point. The stone does not fly toward the center or continue in a circle; it shoots off along a straight line path that is tangential to the circle at the release point. This behaviour is often demonstrated in physics classes and matches the prediction of Newton laws. The same principle is used in sports such as hammer throw, where the athlete swings the hammer in a circle and releases it to send it off tangentially. Why Other Options Are Wrong: The ball will travel towards the center is incorrect because once the string breaks, there is no inward force to pull the ball toward the center. It will fall down at the same place where the string breaks is wrong because the ball has non zero tangential velocity, so it will move horizontally before falling. None of these is incorrect because option describing tangential motion is a correct and precise statement of what happens. Common Pitfalls: Students sometimes think that because centripetal force was pointing towards the center just before the break, the ball will move inward when the string snaps. This ignores the difference between force and velocity direction. The velocity is tangential, not radial, in circular motion. Once the centripetal force vanishes, there is no reason for the ball to move towards the center. Remembering that velocity is always tangent to the circle and that objects move in straight lines when unforced helps avoid this error. Final Answer: When the string snaps, the ball will continue to travel along a tangent to the circle at the point of snapping and then fall down under gravity.

Question 18

In the case of uniform circular motion of a body along a circular path, what is the nature of its acceleration vector throughout the motion?

Accepted Answer

Introduction / Context: This question concerns uniform circular motion, a common topic in mechanics. In uniform circular motion, an object moves around a circle at constant speed. Even though the speed is constant, the velocity vector changes direction continuously, which implies the presence of acceleration. Knowing the direction and magnitude behaviour of this acceleration, called centripetal acceleration, is essential for understanding circular motion, orbits and rotating systems. Given Data / Assumptions: The motion is uniform circular motion, meaning constant speed along a circular path. The path radius is constant. We are asked about the nature of acceleration, specifically its magnitude and direction. We neglect any tangential acceleration, since speed is constant. Concept / Approach: In uniform circular motion, the speed v of the body remains constant, but the direction of velocity changes continuously. Acceleration is defined as the rate of change of velocity. Because only the direction of velocity changes while magnitude is constant, the acceleration must be perpendicular to the velocity at every point. For circular motion, this acceleration points towards the center of the circle and is called centripetal acceleration. Its magnitude is given by a_c = v^2 / r, where r is the radius of the circle. Since v and r are constant in uniform circular motion, a_c has constant magnitude and is always directed radially inwards. Step-by-Step Solution: Step 1: Recognise that in uniform circular motion, speed v is constant but velocity vector changes direction. Step 2: Because velocity direction changes, there must be non zero acceleration; otherwise, velocity would remain unchanged. Step 3: The required force and acceleration to keep the body on a circular path are directed towards the center, known as centripetal force and centripetal acceleration. Step 4: The magnitude of centripetal acceleration is a_c = v^2 / r, where v and r are both constant for uniform circular motion. Step 5: Since v and r are constant, a_c has constant magnitude; its direction is always inward, pointing towards the center of the circle. Step 6: Therefore, acceleration is constant in magnitude and directed radially inwards throughout the motion. Verification / Alternative check: A vector diagram of velocity vectors at two neighbouring points on the circle shows that the change in velocity Δv points towards the center. Repeating this reasoning at different points confirms that acceleration always points to the center. Also, if the acceleration were zero, the object would travel in a straight line, not a circle. If the acceleration were tangential, the speed would change, contradicting uniform speed. Hence, centripetal acceleration must be purely radial and of constant magnitude in uniform circular motion. Why Other Options Are Wrong: Variable in magnitude but constant in direction is incorrect because the direction of acceleration in circular motion changes as the radius direction changes, though it always points towards the center. Variable in magnitude but tangential to the circle is wrong because in uniform circular motion, there is no tangential acceleration; tangential acceleration would change the speed. Zero acceleration is incorrect because a change in direction of velocity requires non zero acceleration; without acceleration, the path would be a straight line. Common Pitfalls: Students sometimes think that constant speed means zero acceleration, forgetting that acceleration depends on changes in velocity, including direction. Another mistake is to assume that acceleration must act along the direction of motion; in circular motion, acceleration is perpendicular to velocity. Remember: in uniform circular motion, acceleration is centripetal, always directed towards the center, and has constant magnitude v^2 / r. Final Answer: In uniform circular motion, the acceleration is constant in magnitude and directed radially inwards toward the center of the circle.

Question 19

In optics, the refractive index of a transparent medium is defined in terms of the speed of light. Which of the following correctly gives this definition?

Accepted Answer

Introduction / Context: This question tests your knowledge of the basic definition of refractive index in geometrical optics. The refractive index describes how much the speed of light is reduced in a medium compared to vacuum and determines how much light bends when entering or leaving that medium. Many formulas in optics, including Snell law, use refractive index as a key parameter, so it is important to know its correct definition in terms of speeds. Given Data / Assumptions: We consider a transparent medium such as glass, water or air. c denotes the speed of light in vacuum, and v denotes the speed of light in the medium. The refractive index is represented by n. We are asked to choose the correct expression relating n, c and v. Concept / Approach: The absolute refractive index n of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in that medium: n = c / v. Since light travels fastest in vacuum, c is the maximum speed. In any material medium, v is less than c, so n is greater than or equal to 1. This definition reflects how much the medium slows down light. Expressions involving the ratio reversed (v / c) or the product c * v are incorrect. Therefore, refractive index is correctly defined as c divided by v, or in words, the ratio of speed of light in vacuum to the speed of light in the medium. Step-by-Step Solution: Step 1: Let c be the speed of light in vacuum, which is approximately 3 * 10^8 m/s. Step 2: Let v be the speed of light in the given medium, for example, in glass or water. Step 3: The absolute refractive index n of the medium is defined as n = c / v. Step 4: Because v is always less than or equal to c, n is always greater than or equal to 1, which matches physical expectations. Step 5: Compare this with the options; the one that states the ratio of speed of light in vacuum to the speed of light in the medium is correct. Step 6: Reject options mentioning the inverse ratio or product, as they do not match the standard definition. Verification / Alternative check: Consider light in water, where the refractive index is approximately n ≈ 1.33. This means that light travels about 1.33 times faster in vacuum than in water, so v = c / 1.33. If we were to define n as v / c, then n would be less than 1, which contradicts the standard values listed in optical tables. Similarly, defining n as a product has no physical meaning in this context. The fact that known refractive indices for common materials are greater than 1 confirms that n must be defined as c / v, not its inverse. Why Other Options Are Wrong: Defining refractive index as the ratio of speed of light in the medium to the speed of light in vacuum (v / c) would produce values less than 1 for typical media, in conflict with standard refractive index data. Defining it as the product of speeds (c * v) has no conventional meaning in optics and is dimensionally inconsistent with a dimensionless refractive index. None of above is incorrect because the option defining refractive index as the ratio c / v is correct and widely accepted. Common Pitfalls: Students sometimes invert the ratio mistakenly and think n = v / c, often due to misremembering the definition. A useful way to avoid this is to recall that refractive index is greater than or equal to 1 for any medium, with n = 1 in vacuum. This can only be true if n = c / v, since v cannot exceed c. Focusing on this inequality helps you remember the correct ratio even if you forget the exact wording. Final Answer: The refractive index of a transparent medium is defined as the ratio of the speed of light in vacuum to the speed of light in the medium, that is n = c / v.

Question 20

A hunter on level ground wants to shoot a bird sitting on a distant tree with his rifle. In order to compensate for the effect of gravity on the bullet, in which direction should he aim the rifle relative to the bird?

Accepted Answer

Introduction / Context: When a projectile such as a bullet is fired from a rifle, it does not travel in a perfectly straight line. Instead, the bullet follows a curved path under the influence of gravity. This question uses the situation of a hunter shooting at a bird on a distant tree to test your understanding of projectile motion and how gravity affects the path of a bullet in flight. Given Data / Assumptions: The hunter is on level ground, and the bird is on a tree at some horizontal distance. The bullet leaves the rifle with high speed and is affected by Earth's gravity. Air resistance is either neglected or assumed small compared with gravitational effects. The hunter wants the bullet to hit the bird's position on the tree. Concept / Approach: Once the bullet leaves the gun barrel, gravity acts downward on it throughout its flight. Even if the bullet is fired horizontally, gravity pulls it down, causing the bullet to drop below the straight line extending from the barrel. To make sure the bullet arrives at the bird's position, the hunter must aim slightly above the target so that the downward drop during flight brings the bullet onto the bird's actual location. This is the basic idea of compensating for bullet drop in projectile motion. Step-by-Step Solution: Step 1: Imagine drawing a straight line from the rifle to the bird. This is the line of sight. Step 2: If the hunter aims exactly along this line and fires, the bullet immediately begins to fall under gravity as it travels forward. Step 3: Because of this downward acceleration, the bullet's path will curve below the straight line of sight, and at the distance of the tree the bullet may pass under the bird. Step 4: To counteract this drop, the hunter must aim the rifle slightly above the bird so that the bullet starts on a higher line and then curves downward into the bird's position. Step 5: Aiming vertically upwards or below the bird would cause even larger deviations from the desired path, reducing the chance of hitting the bird. Verification / Alternative check: The trajectory of the bullet can be treated as a horizontal projectile. The vertical displacement of the bullet is given by y = (1/2)*g*t^2 downward from the initial line of the barrel, where g is acceleration due to gravity and t is time of flight. The longer the distance to the bird, the longer the time in the air and the more the bullet drops. Professional shooters routinely adjust their aim above the target at long range, a practice known as holding "over" the target, which confirms the reasoning behind aiming higher than the bird. Why Other Options Are Wrong: At a slightly lower elevation than the bird: This would make the bullet drop even further below the bird, almost guaranteeing a miss. Directly along the line of sight to the bird's position: This might work only at very short distances, but at larger distances the bullet drop will cause a miss below the target. Vertically upwards, above both the hunter and the bird: This is unrealistic and would cause the bullet to rise and fall in a path that does not intersect the bird's position at all. Common Pitfalls: A common misconception is to assume that a very fast bullet travels in a straight line and that gravity can be ignored. In reality, gravity acts continuously, and even high speed projectiles experience noticeable drop over distance. Another mistake is to think that aiming directly at the target is always correct; in many practical shooting and sports situations, players and hunters must aim above the target to allow for gravitational effects. Final Answer: To compensate for bullet drop, the hunter should aim the rifle at a slightly higher elevation than the bird.

Physics Questions