Question 1

In thermodynamics, which of the following combinations represent state functions? 1. q + w 2. q (heat exchanged) 3. w (work done) 4. H − T S. Select the correct answer using the code given below.

Accepted Answer

Introduction / Context: Chemical thermodynamics distinguishes between state functions and path functions. State functions depend only on the initial and final states of a system, while path functions depend on the specific way the process is carried out. Examinations often test whether you can correctly classify common thermodynamic quantities such as heat, work, internal energy and Gibbs free energy. This question asks which of the given combinations are state functions. Given Data / Assumptions: q denotes heat exchanged between system and surroundings. w denotes work done by or on the system. q + w is related to change in internal energy. H is enthalpy, T is temperature and S is entropy, so H − T S represents Gibbs free energy G. Concept / Approach: According to the first law of thermodynamics, the change in internal energy ΔU of a system is given by ΔU = q + w (for a particular sign convention). Internal energy U is a state function because it depends only on the state of the system. Heat q and work w individually are not state functions; they depend on the path taken between initial and final states. Gibbs free energy G is defined as G = H − T S, where enthalpy H, temperature T and entropy S are all state functions. Any combination formed from state functions by algebraic addition, subtraction or multiplication by constants is also a state function. Step-by-Step Solution: Step 1: Recognise that q + w equals ΔU (change in internal energy), which is a state function. Therefore, expression 1 represents a state function. Step 2: Heat q by itself is a path function, because the amount of heat exchanged depends on how the process is carried out. Step 3: Work w is also a path function; different processes between the same states can involve different amounts of work. Step 4: H − T S equals Gibbs free energy G, which is a state function because H, T and S are all state functions. Verification / Alternative check: Think of a gas expanding from volume V1 to volume V2. If it expands reversibly, the work and heat have certain values. If it expands irreversibly (for example, free expansion), the values of q and w change, but the initial and final states are the same, so ΔU and ΔG depend only on state and remain unchanged. This confirms that q and w are path dependent, whereas q + w (internal energy change) and H − T S (Gibbs free energy) are state functions. Why Other Options Are Wrong: Option B: 1, 2 and 4 incorrectly includes heat q as a state function. Option C: 2, 3 and 4 incorrectly includes both q and w as state functions. Option D: 1 only ignores the fact that H − T S is also a state function representing Gibbs free energy. Common Pitfalls: A frequent misconception is to assume that any commonly used thermodynamic quantity is a state function. Students often forget that heat and work are process quantities, not properties of state. Remember that any quantity expressible solely in terms of state variables (such as P, V, T, U, H, S) is a state function, while quantities that require knowledge of the path are not. Final Answer: The state functions among the given options are 1 and 4 only.

Question 2

For a certain chemical reaction, the standard Gibbs free energy change ΔG° is −45 kJ mol^-1 and the standard enthalpy change ΔH° is −90 kJ mol^-1 at 0 °C (273 K). Assuming that ΔH° and ΔS° are independent of temperature, at approximately what temper…

Accepted Answer

Introduction / Context: Gibbs free energy is the key criterion for spontaneity of chemical reactions at constant temperature and pressure. A reaction is spontaneous when the change in Gibbs free energy ΔG is negative, non spontaneous when ΔG is positive, and at equilibrium when ΔG is zero. This question gives you ΔG° and ΔH° at 0 °C and asks for the temperature at which ΔG° becomes zero, assuming both ΔH° and ΔS° do not change with temperature. This temperature marks the threshold between spontaneous and non spontaneous behaviour. Given Data / Assumptions: ΔG° at T1 = 273 K is −45 kJ mol^-1. ΔH° is −90 kJ mol^-1 and taken as constant with temperature. ΔS° is assumed constant and can be found from ΔG° = ΔH° − T * ΔS°. We seek the temperature T at which ΔG° = 0. Concept / Approach: The basic relationship is ΔG° = ΔH° − T * ΔS°. At the given temperature T1 = 273 K, ΔG1° and ΔH° are known, so we can calculate ΔS° from the equation ΔS° = (ΔH° − ΔG°) / T1. Once ΔS° is known, we set ΔG° = 0 to find the threshold temperature T_eq at which ΔH° − T_eq * ΔS° = 0. Solving this gives T_eq = ΔH° / ΔS°. Because ΔH° is negative and ΔS° turns out to be negative here, the reaction is spontaneous (ΔG° < 0) at lower temperatures and becomes non spontaneous at sufficiently high temperatures. Step-by-Step Solution: Step 1: Use ΔG° = ΔH° − T * ΔS° at T = 273 K. Step 2: Substitute values: −45 = −90 − 273 * ΔS° (all in kJ mol^-1 and kelvin). Step 3: Rearrange to get −45 + 90 = 273 * ΔS°, therefore 45 = 273 * ΔS°. Step 4: Calculate ΔS° = 45 / 273 kJ mol^-1 K^-1 ≈ 0.165 kJ mol^-1 K^-1 but of opposite sign because the original equation is −45 = −90 − 273 * ΔS°, giving ΔS° ≈ −0.165 kJ mol^-1 K^-1. Step 5: Now set ΔG° = 0 to find the threshold temperature: 0 = ΔH° − T_eq * ΔS°. Step 6: Solve for T_eq: T_eq = ΔH° / ΔS° = (−90) / (−0.165) ≈ 545.45 K. Step 7: On rounding to the nearest whole number in the options, T_eq ≈ 546 K. Verification / Alternative check: At T = 273 K, ΔG° is already negative (−45 kJ mol^-1), so the reaction is spontaneous at that temperature. Since both ΔH° and ΔS° are negative, the reaction is expected to be spontaneous only at sufficiently low temperatures. As T increases, the term −T * ΔS° becomes positive enough to cancel the negative ΔH°, making ΔG° zero and then positive. The calculated threshold temperature around 546 K fits this qualitative behaviour and matches the given option. Why Other Options Are Wrong: Option A: 273 K is the temperature at which ΔG° is given as −45 kJ mol^-1, not zero; the reaction is clearly spontaneous here, not at the threshold. Option B: 298 K is close to room temperature and would not produce ΔG° = 0 according to the calculation using ΔG° = ΔH° − T * ΔS°. Option D: 596 K is higher than the calculated threshold and would make ΔG° positive (non spontaneous), not zero. Common Pitfalls: Students often forget to convert Celsius to kelvin or mishandle signs in the Gibbs free energy equation. Another frequent error is to misinterpret the temperature at which ΔG° changes sign as a minimum temperature for spontaneity, when with negative ΔH° and negative ΔS° the reaction is actually spontaneous below a certain maximum temperature. Always calculate ΔS° carefully and then use ΔG° = 0 to find the threshold temperature. Final Answer: The reaction has ΔG° = 0 at approximately 546 K.

Question 3

In standard spherical polar coordinates (r, θ, φ), where θ is the polar angle measured from the positive z axis and φ is the azimuthal angle measured from the positive x axis in the x y plane, the y component of a vector P of magnitude P is given by…

Accepted Answer

Introduction / Context: In physics, especially in electromagnetism and quantum mechanics, spherical polar coordinates are frequently used to exploit symmetry in problems involving spheres or central potentials. It is essential to know how to resolve a vector into its Cartesian components using spherical coordinates. This question asks specifically for the y component of a vector P expressed in terms of its magnitude P and the spherical angles θ and φ. Given Data / Assumptions: Spherical coordinates (r, θ, φ) are used. θ is the polar angle measured from the positive z axis. φ is the azimuthal angle measured in the x y plane from the positive x axis. The vector P has magnitude P and direction specified by (θ, φ). Concept / Approach: In spherical polar coordinates, a point in space is represented by (r, θ, φ). The relationships between Cartesian coordinates (x, y, z) and spherical coordinates are standard and widely used. For a point at distance r from the origin and angles θ and φ defined as above, the Cartesian coordinates are x = r sinθ cosφ, y = r sinθ sinφ and z = r cosθ. For a vector of magnitude P pointing in the same direction, the components scale in the same way, giving P_x = P sinθ cosφ, P_y = P sinθ sinφ and P_z = P cosθ. Therefore, the y component must involve both sinθ and sinφ. Step-by-Step Solution: Step 1: Recall the conversion formulas between spherical and Cartesian coordinates: x = r sinθ cosφ, y = r sinθ sinφ and z = r cosθ. Step 2: Recognise that a vector of magnitude P pointing in the direction (θ, φ) will have components proportional to these direction cosines. Step 3: Write down P_x = P sinθ cosφ, P_y = P sinθ sinφ and P_z = P cosθ. Step 4: Identify from these expressions that the y component of P is P_y = P sinθ sinφ. Verification / Alternative check: If φ = 0, the vector lies in the x z plane, so its y component should be zero. Substituting φ = 0 into P sinθ sinφ gives P sinθ * 0 = 0, which is correct. If φ = 90 degree (or π / 2 radians), the vector lies in the y z plane and its x component should be zero, while y and z remain. Substituting φ = 90 degree into P sinθ sinφ gives P sinθ * 1 = P sinθ, which is non zero in general, consistent with a non zero y component in that plane. Why Other Options Are Wrong: Option B: P sinθ cosφ corresponds to the x component, not the y component. Option C: P cosθ sinφ does not match any standard Cartesian component for the given angular definitions and would give incorrect behaviour, for example non zero y at θ = 0 where the vector should point along z. Option D: P cosθ cosφ similarly fails to vanish when expected and does not correspond to a correct Cartesian component. Common Pitfalls: Many students confuse the roles of sinθ and cosθ or interchange the positions of sinφ and cosφ. A helpful way to remember is that the projection in the x y plane has factor sinθ, while the distribution between x and y involves cosφ for x and sinφ for y. Visualising the geometry in two steps projection down from the z axis and then projection into the x y plane helps prevent errors. Final Answer: The y component of P is P sinθ sinφ.

Question 4

For an ideal gas, which one of the following statements does not hold true according to the kinetic theory of gases?

Accepted Answer

Introduction / Context: The kinetic theory of gases provides a microscopic model for the behaviour of ideal gases. It makes several simplifying assumptions about molecular motion, energy and interactions. Examinations often ask which statements are consistent with this model and which are not. This question targets your understanding of the distribution of molecular speeds and the basic assumptions regarding energy and forces in an ideal gas. Given Data / Assumptions: An ideal gas is being considered. The kinetic theory assumptions apply. We must identify which statement does not correctly describe an ideal gas. Concept / Approach: In an ideal gas, molecules are assumed to be point particles that undergo perfectly elastic collisions and do not exert long range forces on each other. The potential energy of interaction is taken as zero. The molecules have a distribution of speeds described by the Maxwell Boltzmann distribution, even at a fixed temperature. Thus, while the average kinetic energy depends only on temperature, individual molecules have different speeds and kinetic energies at any instant. Therefore, any statement claiming that all molecules have the same speed is inconsistent with kinetic theory. Step-by-Step Solution: Step 1: Recall that, at a given temperature, molecular speeds in an ideal gas are distributed according to the Maxwell Boltzmann distribution. Step 2: This implies that some molecules move slower and some faster than the most probable speed; they do not all share one common speed. Step 3: Therefore, the statement that the speed of all gas molecules is the same is incorrect. Step 4: The other statements are consistent with the ideal gas model: kinetic energies vary, potential energy of interaction is neglected and no long range interactive forces are assumed. Verification / Alternative check: Graphs of the Maxwell Boltzmann distribution show a continuous spread of molecular speeds rather than a single value. Even at constant temperature, there are molecules much faster or slower than the average. Furthermore, in the kinetic theory derivation of pressure, the assumption is not equal speeds but a large number of molecules with a distribution of velocities whose mean square speed relates to temperature. Why Other Options Are Wrong: Option B: The kinetic energies of all molecules are not the same is correct because energy varies with speed, and speeds are distributed. Option C: The potential energy of the gas molecules is zero is an idealisation that the molecules do not have attractive or repulsive potential energy except during collisions, which is part of the ideal gas model. Option D: There is no interactive force between the molecules beyond brief collisions is again an ideal gas assumption, ignoring intermolecular forces. Common Pitfalls: Some students misinterpret the statement that average kinetic energy is proportional to temperature and wrongly assume that every molecule must have exactly that kinetic energy at a given temperature. In reality, temperature relates to the average over many molecules. The key point is that ideal gas theory uses statistical distributions, not identical speeds for all molecules. Final Answer: The statement that does not hold true is The speed of all gas molecules is the same at a given temperature.

Question 5

Hooke's law states that, within certain limits, stress is directly proportional to strain. For a typical ductile material under uniaxial loading, Hooke's law is valid over which portion of the stress strain curve?

Accepted Answer

Introduction / Context: In strength of materials and solid mechanics, Hooke's law is fundamental for describing elastic deformation. It states that, up to a certain limit, the extension or strain is proportional to the applied load or stress. This question asks you to identify the part of the stress strain curve for which this straight line proportionality strictly holds for a typical ductile material, such as mild steel. Given Data / Assumptions: The material is ductile and follows a typical stress strain curve with a proportional limit, an elastic limit, yield point and plastic region. Hooke's law is written as stress proportional to strain or stress = E * strain, where E is Young modulus. We are interested in the regime where stress strain relationship is exactly linear. Concept / Approach: The stress strain curve of a ductile material initially rises as a straight line from the origin. In this region, stress is directly proportional to strain and Young modulus is constant. The upper limit of this perfectly linear portion is called the proportional limit. Beyond this point, the curve may continue to be elastic for a short range, meaning that the material will still return to its original shape when unloaded, but the relationship between stress and strain is no longer strictly linear. Therefore, Hooke's law in its original form applies only to the proportional region, not to the full elastic region and certainly not to the plastic region. Step-by-Step Solution: Step 1: Identify the proportional limit on the stress strain curve as the highest stress value up to which stress and strain are exactly proportional. Step 2: Recognise that Hooke's law describes this linear behaviour, so its domain is the proportional region. Step 3: Beyond the proportional limit but before yielding, the curve may deviate slightly from linearity while still being elastic. In this range, strict proportionality does not hold, so Hooke's law in the simple linear form is not exact. Step 4: In the plastic region, the material undergoes permanent deformation and the stress strain relationship is strongly non linear, so Hooke's law clearly does not apply there. Verification / Alternative check: Experimental stress strain diagrams for mild steel show a straight line from the origin up to a certain stress, followed by a curved region that is still recoverable on unloading, and then a yield plateau and strain hardening region. Textbooks define the proportional limit as the end of the region where Hooke's law is valid. This directly confirms that the law applies only in the initial proportional region, not over the entire elastic span or plastic portion. Why Other Options Are Wrong: Option B: The entire stress strain curve includes large plastic deformations and fracture, where the relationship is highly non linear and irreversible, so Hooke's law does not apply. Option C: The entire elastic region may include slight non linearity beyond the proportional limit, so strict proportionality is not valid throughout. Option D: The plastic region by definition involves permanent deformation and clearly does not satisfy Hooke's law. Common Pitfalls: Students sometimes equate elastic region with proportional region and assume Hooke's law applies up to the elastic limit. In reality, the proportional limit is usually slightly below the elastic limit. For accurate design, we use the strictly linear portion associated with Hooke's law, not the entire elastic span. Final Answer: Hooke's law is valid for only the initial proportional region of the stress strain curve.

Question 6

In a standard compound microscope, considering only the main optical components and ignoring the mirror and the observer's eye, which one of the following gives the correct sequence in which light passes through the instrument from the specimen towa…

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Introduction / Context: A compound microscope uses several optical components to magnify a small specimen. Understanding the order in which light passes through these components is important for both basic physics and practical laboratory use. This question asks you to identify the correct sequence of main optical elements (excluding the mirror and the observer's eye) through which light travels from the specimen towards the observer. Given Data / Assumptions: We consider a typical compound microscope setup. Light from a source is reflected by a mirror, passes through the condenser and specimen, then through the objective lens and eyepiece. The body tube simply holds the optical alignment between objective and eyepiece. The options list the order of components after the condenser. Concept / Approach: In a compound microscope, the condenser collects and focuses light onto the specimen placed on the stage. The objective lens, located just above the specimen, forms a real, inverted and magnified image of the specimen inside the body tube. This intermediate image lies near the focal plane of the eyepiece. The eyepiece (eye lens) then acts as a magnifier, producing a further magnified virtual image that the observer sees. Thus, after the condenser, light must pass first through the objective lens, then travel through the body tube where the intermediate image is formed, and finally through the eyepiece. Step-by-Step Solution: Step 1: Starting from the specimen side, light is concentrated by the condenser, so the first element in the sequence is the condenser. Step 2: Immediately above the specimen and condenser is the objective lens, which gathers light and forms the first magnified image. Step 3: This intermediate image lies in the body tube, which simply provides mechanical support and the correct distance between objective and eyepiece. Step 4: Finally, the eyepiece receives the intermediate image and produces a virtual, magnified image for the observer. Verification / Alternative check: Textbook diagrams of the compound microscope clearly show the order: mirror (or light source), condenser, specimen, objective lens, body tube and eyepiece. The options that place the condenser first, followed by the objective and then body tube and eyepiece, align with this standard layout. Including the eyepiece before the body tube or placing the condenser after the objective would contradict the usual design and ray diagrams. Why Other Options Are Wrong: Option A: Condenser → Objective lens → Eye piece → Body tube incorrectly places the eyepiece before the body tube, reversing the natural order of internal components. Option B: Objective lens → Condenser → Body tube → Eye piece reverses the positions of objective and condenser, which is not how light is focused onto the specimen. Option D: Eye piece → Objective lens → Body tube → Mirror reverses the entire logical order and even introduces the mirror after the eyepiece, which is physically impossible. Common Pitfalls: Students sometimes confuse the roles of objective lens and condenser. The condenser is always nearer to the light source and the specimen, focusing illumination, while the objective is the primary image forming lens closest to the specimen. Remembering this distinction helps in correctly ordering the components in any microscope question. Final Answer: The correct sequence is Condenser → Objective lens → Body tube → Eye piece.

Question 7

In meteorology and experimental physics, humidity (the amount of water vapour present in air) is measured using which one of the following instruments?

Accepted Answer

Introduction / Context: Humidity is an important parameter in weather forecasting, comfort studies, and many industrial processes. It measures the amount of water vapour present in the air. Several instruments are used in meteorology and environmental monitoring, each with a specific purpose. This question asks you to identify the instrument whose primary function is to measure humidity. Given Data / Assumptions: Humidity refers to the water vapour content of air, often expressed as relative humidity. We consider common instruments: hydrometer, hygrometer, psychrometer and anemometer. Only one of these is the general name for instruments that measure humidity directly. Concept / Approach: The term hygrometer is used for instruments designed to measure humidity. There are several types of hygrometers, including hair hygrometers, electronic capacitance hygrometers and psychrometers. A psychrometer is a specific type of hygrometer that uses the difference between wet bulb and dry bulb thermometer readings to determine humidity. Hydrometers, on the other hand, measure the relative density or specific gravity of liquids, not humidity. Anemometers measure wind speed. Thus, the general answer to which device measures humidity is hygrometer. Step-by-Step Solution: Step 1: Recall that hydrometer is used for measuring relative density of liquids such as milk, battery acid or alcohol solutions. Step 2: Recognise that anemometer is used in meteorology to measure wind speed and is usually a cup or vane type device. Step 3: Hygrometer is the broad name for instruments used to measure humidity or moisture content in air. Step 4: Psychrometer is a specific type of hygrometer, but in general usage, the term hygrometer is the standard answer for a humidity measuring device. Verification / Alternative check: Weather stations often report humidity measured by electronic hygrometers or psychrometers. The instrument manuals and meteorology textbooks refer to these devices as hygrometers. Hydrometers are found in laboratories and industries handling liquids, not air humidity. Anemometers are mounted on masts for wind measurement, not for moisture content, confirming the identification. Why Other Options Are Wrong: Option A: Hydrometer is for liquid density, not for atmospheric humidity. Option C: Psychrometer is indeed used to determine humidity but is specifically a type of hygrometer using two thermometers. In many exam questions, the standard and more general term sought is hygrometer. Option D: Anemometer measures wind speed, not humidity. Common Pitfalls: Students may confuse hydrometer and hygrometer because their names sound similar. Remember that hydro relates to liquid water and density, while hygro relates to moisture or humidity. Also, some texts emphasise psychrometers, which can cause confusion. Keeping clear associations hydrometer with liquids, hygrometer with humidity, and anemometer with wind helps avoid these mistakes. Final Answer: The instrument used to measure humidity is the Hygrometer.

Question 8

Which one of the following physical quantities is not a vector quantity and therefore has magnitude only but no direction associated with it?

Accepted Answer

Introduction / Context: In mechanics, physical quantities are classified as scalars or vectors. Scalars have magnitude only, while vectors have both magnitude and direction and obey the rules of vector addition. This classification is fundamental for correctly applying Newton laws and kinematics. The question asks you to select the quantity that is not a vector from the given list. Given Data / Assumptions: Momentum is considered as mass times velocity. Displacement is the change in position from initial to final point. Torque is associated with rotational effect of a force about a point or axis. Speed is the rate of change of distance with time. Concept / Approach: A vector quantity has both magnitude and direction and transforms according to vector laws. Displacement has a specific direction from the initial to the final position, so it is a vector. Momentum is mass times velocity; since velocity is a vector, momentum is also a vector. Torque is defined as the cross product of position vector and force, and hence is a vector (more precisely a pseudovector). Speed, however, is defined as the magnitude of velocity without any directional information; it is a scalar quantity. Therefore, among the options, speed is the only non vector quantity. Step-by-Step Solution: Step 1: Identify displacement as a directional change in position, which clearly qualifies as a vector. Step 2: Recognise that momentum p = m * v, where v is velocity, so p has both magnitude and direction. Step 3: Recall torque τ = r × F, where r and F are vectors. The cross product yields a vector quantity pointing along the axis of rotation. Step 4: Note that speed is defined simply as distance travelled per unit time and does not specify any direction, so it is scalar. Verification / Alternative check: Consider algebraic operations. Vectors must be added using head to tail or component wise rules. You cannot meaningfully add two speeds with directions to get a resultant speed vector; instead you would add velocities if direction matters. In contrast, displacement, momentum and torque can be broken into components along axes and added as vectors, which further confirms their vector nature. Why Other Options Are Wrong: Option A: Momentum has direction opposite to applied force during deceleration and along the direction of motion when moving, demonstrating vector character. Option B: Displacement, unlike distance, depends on direction from start to end point and can cancel when a path returns to the origin. Option C: Torque produces rotation about an axis and changes when the force direction changes, exhibiting vector behaviour. Common Pitfalls: Students occasionally confuse speed with velocity. Velocity is a vector because it includes direction; speed is the magnitude of velocity only. Remember that every vector has a corresponding scalar magnitude, and confusing the two can lead to errors in conceptual questions. Carefully read whether the question refers to speed or velocity. Final Answer: The quantity that is not a vector is Speed.

Question 9

On the Fahrenheit temperature scale, at what temperature does pure water freeze under standard atmospheric pressure?

Accepted Answer

Introduction / Context: Different temperature scales are used in science and everyday life, such as Celsius, Fahrenheit and Kelvin. Knowing the key fixed points on these scales, like the freezing and boiling points of water, is important for conversions and basic understanding of thermometry. This question asks for the freezing point of pure water on the Fahrenheit scale under standard atmospheric conditions. Given Data / Assumptions: We consider pure water in equilibrium between solid and liquid phases. The pressure is one atmosphere (standard atmospheric pressure). The Fahrenheit scale is used, not Celsius or Kelvin. Concept / Approach: On the Celsius scale, pure water freezes at 0 °C and boils at 100 °C under one atmosphere. On the Fahrenheit scale, these two fixed points are 32 °F for freezing and 212 °F for boiling. The interval between 32 °F and 212 °F is divided into 180 equal parts, making the Fahrenheit degree smaller than the Celsius degree. Therefore, the freezing point of water corresponds to 32 °F on this scale. Step-by-Step Solution: Step 1: Recall the two fixed points used in defining the Fahrenheit scale: freezing of water and boiling of water at one atmosphere. Step 2: Remember that on this scale, water freezes at 32 °F and boils at 212 °F. Step 3: The question asks for the freezing point, so we select 32 °F. Step 4: Therefore, the correct numerical value on the Fahrenheit scale is 32. Verification / Alternative check: Temperature conversion formulas between Celsius and Fahrenheit are given by F = (9 / 5) * C + 32. Substituting C = 0 °C (freezing point on Celsius) gives F = (9 / 5) * 0 + 32 = 32 °F. This confirms that water freezes at 32 °F. Substituting C = 100 °C gives F = (9 / 5) * 100 + 32 = 212 °F, matching the boiling point on Fahrenheit scale. Why Other Options Are Wrong: Option B: 0 °F is a colder temperature on this scale and does not correspond to the freezing point of water; it was historically defined using a brine mixture. Option C: 48 °F is above the freezing point; at this temperature, water is in liquid form. Option D: 37 °F is slightly above freezing; water would be cold but not necessarily frozen under normal conditions. Common Pitfalls: Students sometimes confuse the zero points of Celsius and Fahrenheit scales. On Celsius, 0 °C corresponds to freezing, while on Fahrenheit, freezing occurs at 32 °F. Mixing these two values or assuming that 0 means freezing on all scales leads to incorrect answers. It helps to remember at least the pair (0 °C, 32 °F) for freezing and (100 °C, 212 °F) for boiling as anchor points. Final Answer: Pure water freezes at 32 °F on the Fahrenheit scale.

Question 10

In semiconductor physics, the process of deliberately adding small amounts of suitable impurity atoms to a pure (intrinsic) semiconductor crystal in order to modify its electrical conductivity is called what?

Accepted Answer

Introduction / Context: Semiconductors such as silicon and germanium are the foundation of modern electronics. Their electrical properties can be precisely controlled by introducing very small amounts of impurity atoms. This process allows the creation of p type and n type materials used in diodes, transistors and integrated circuits. The question asks for the technical term used to describe this intentional addition of impurities to a pure semiconductor. Given Data / Assumptions: We start with a pure or intrinsic semiconductor crystal. Small quantities of impurity atoms (donors or acceptors) are introduced. The goal is to change carrier concentration and hence conductivity. The process is controlled and deliberate, not accidental contamination. Concept / Approach: The intentional addition of impurity atoms to a semiconductor is called doping. Donor impurities (such as phosphorus in silicon) create extra free electrons, producing n type material, while acceptor impurities (such as boron in silicon) create additional holes, producing p type material. Doping levels can range from very light to very heavy, allowing fine tuning of device characteristics. The other words in the options, like mixing, forming or diluting, are generic and do not capture the specific semiconductor engineering meaning. Step-by-Step Solution: Step 1: Recall that intrinsic semiconductors have relatively low conductivity due to limited numbers of charge carriers. Step 2: When specific impurity atoms are added in controlled amounts, the number of electrons or holes increases dramatically. Step 3: The technical term for this process of adding impurities is doping, and the impurities themselves are called dopants. Step 4: Therefore, among the options, doping is the correct term. Verification / Alternative check: Electronics textbooks consistently describe n type and p type semiconductors as doped semiconductors. Processes such as diffusion doping and ion implantation are standard in semiconductor fabrication. None of these processes are referred to as simply mixing or diluting, which further confirms that doping is the accepted term. Why Other Options Are Wrong: Option B: Mixing is a very general word and does not convey the precise meaning of controlled impurity addition in crystal lattices. Option C: Forming may refer to shaping or manufacturing devices but is not the specific term for adding impurity atoms. Option D: Diluting would suggest reducing concentration by adding more of a solvent or host, which is the opposite of what is done in doping, where small quantities of impurities are added to increase carrier concentration. Common Pitfalls: Some learners confuse doping in semiconductor context with everyday use of the term in sports. In electronics, doping is a neutral technical term with no negative connotation. Another point of confusion is to think any contamination is doping, but in device fabrication, doping is carefully controlled in type and concentration, unlike accidental contamination which is undesirable. Final Answer: The process is called Doping.

Question 11

Sound waves are mechanical longitudinal waves and require a material medium for their propagation. Through which one of the following arrangements can sound waves not travel at all?

Accepted Answer

Introduction / Context: Unlike electromagnetic waves, sound waves need a material medium such as gas, liquid or solid to propagate. They travel by setting particles of the medium into vibration. A classic conceptual test in physics exams is to ask which situation would prevent sound from propagating because of the absence of a suitable medium. This question presents different materials placed in different environments and asks where sound waves cannot travel at all. Given Data / Assumptions: Sound waves are mechanical and require matter to propagate. Solids, liquids and gases can act as media for sound. Vacuum does not provide particles for vibration, so sound cannot pass through vacuum. The options involve different combinations of materials and surroundings. Concept / Approach: Sound propagation depends on the presence of material particles that can oscillate and transmit the disturbance. Metals like copper and silver, glass, water and wood are all capable of transmitting sound because they contain atoms and molecules. However, in a perfect vacuum, there are no particles to support the mechanical vibrations required for sound. Even if you place a hollow object in a vacuum, the space inside and outside is void of air or any material medium, so sound cannot travel through that region. Step-by-Step Solution: Step 1: Consider a copper wire in air. Both copper and air are materials, so sound can travel through the wire and the surrounding air. Step 2: A silver slab in air similarly provides a continuous solid medium for sound and is surrounded by air, so sound can propagate. Step 3: A glass prism placed in water gives two different media, glass and water, both of which support sound transmission. Step 4: A wooden hollow pipe placed in vacuum means that both inside and outside the pipe there is vacuum. Without air or other material filling the hollow space, sound cannot propagate through the vacuum region. Verification / Alternative check: The classic school experiment placing a ringing bell inside a bell jar and gradually evacuating the air demonstrates that as air is removed, the sound becomes fainter and ultimately disappears when near vacuum is achieved. This shows clearly that sound cannot travel through vacuum. In contrast, solid bars and liquids are known to transmit sound effectively, often better than gases. Why Other Options Are Wrong: Option A: Copper wire in air provides two media, solid and gas, both capable of carrying sound, so sound can propagate. Option B: Silver slab in air also offers solid and gaseous media and does not prevent sound transmission. Option C: Glass prism in water combines a solid and a liquid, both of which can transmit sound waves. Common Pitfalls: Some students may incorrectly think that sound cannot travel in solids or that only air transmits sound. In fact, sound travels fastest in solids and more slowly in liquids and gases. The only environment in which sound cannot propagate is vacuum, where there are no particles to oscillate. Always look for the presence or absence of a material medium when answering propagation questions. Final Answer: Sound waves cannot travel through a wooden hollow pipe placed in vacuum.

Question 12

In units of centimetre, one nanometre (1 nm) is equal to which of the following numerical values?

Accepted Answer

Introduction / Context: Nanometre is a very small unit of length widely used in atomic physics, semiconductor technology and nanoscience. Being comfortable with converting nanometres into metres and centimetres helps in understanding orders of magnitude and in comparing sizes at microscopic scales. This question asks you to express one nanometre in terms of centimetre using scientific notation. Given Data / Assumptions: 1 nanometre (1 nm) is defined as 10^-9 metre. 1 metre equals 100 centimetre. We need to convert 10^-9 metre into centimetre. Concept / Approach: First, recall the definitions: 1 nm = 10^-9 m and 1 m = 100 cm = 10^2 cm. To convert nm to cm, we multiply 10^-9 m by 10^2 cm per metre. This gives 10^-9 * 10^2 cm = 10^-7 cm. Hence, one nanometre is equal to 10^-7 centimetre. The other options correspond to different powers of ten and represent much larger lengths than a nanometre. Step-by-Step Solution: Step 1: Start from the definition 1 nm = 10^-9 m. Step 2: Use the relation 1 m = 100 cm = 10^2 cm. Step 3: Replace metre in the nanometre definition: 1 nm = 10^-9 * (1 m) = 10^-9 * (10^2 cm). Step 4: Multiply the powers of ten: 10^-9 * 10^2 = 10^(-9 + 2) = 10^-7. Step 5: Therefore, 1 nm = 10^-7 cm. Verification / Alternative check: As an alternative, convert everything through metres. Knowing that 1 cm = 10^-2 m, we can write 1 nm / 1 cm = (10^-9 m) / (10^-2 m) = 10^-9 / 10^-2 = 10^(-9 + 2) = 10^-7. This means that one nanometre is 10^-7 of a centimetre, confirming the earlier calculation. Why Other Options Are Wrong: Option B: 10^-6 cm corresponds to a length ten times larger than a nanometre. Option C: 10^-4 cm is much larger still, equal to 1 micrometre (10^-6 m), not a nanometre. Option D: 10^-3 cm equals 10^-5 m, which is 10 000 times larger than a nanometre. Common Pitfalls: Students sometimes mix up micrometre (μm) and nanometre (nm). Remember that 1 μm = 10^-6 m, whereas 1 nm = 10^-9 m, so 1 μm is 1000 nm. Another common error is mishandling powers of ten when converting between metre and centimetre. Always track exponents carefully adding when multiplying powers of ten and subtracting when dividing. Final Answer: One nanometre is equal to 10^-7 cm.

Question 13

Consider the following statements about mechanics: 1. There is no net moment (torque) on a rigid body which is in equilibrium. 2. The momentum of a single body is always conserved in all situations. 3. The kinetic energy of an object is always conse…

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Introduction / Context: Classical mechanics deals with equilibrium, conservation laws and energy changes. This question tests your understanding of when torques, momentum and kinetic energy are conserved, and in what sense. It presents three statements and asks which are correct, focusing on the conditions for equilibrium and conservation. Given Data / Assumptions: We consider rigid bodies and particles in Newtonian mechanics. Equilibrium implies no change in the state of motion of the body. Conservation of momentum is considered in the presence or absence of external forces. Conservation of kinetic energy is considered in various interactions or collisions. Concept / Approach: For a rigid body to be in complete mechanical equilibrium, two conditions must hold: the net force acting on it must be zero, and the net moment (or net torque) about any point must also be zero. Thus statement 1 is correct. Momentum conservation applies to an isolated system with no external net force; the momentum of an individual body can change due to forces from other bodies, so statement 2 is generally false. Kinetic energy is conserved only in perfectly elastic collisions; in inelastic processes it is not conserved, so statement 3 is also false. Therefore, only statement 1 is correct. Step-by-Step Solution: Step 1: For equilibrium, recall the conditions: ΣF = 0 (no net force) and Στ = 0 (no net torque). Statement 1 says there is no net moment on a body in equilibrium, which is consistent with this definition. Step 2: Consider statement 2. Momentum p of a body changes when a net external force acts on it according to F = dp / dt. Thus, momentum of a single body is not always conserved; only the total momentum of an isolated system is conserved. Step 3: For statement 3, kinetic energy is conserved only in special cases such as ideal elastic collisions. In general, kinetic energy can be transformed into other forms (heat, potential energy, sound) and is not always conserved. Step 4: Therefore, only statement 1 is correct. Verification / Alternative check: Think of a block on a table at rest. The net force is zero (weight balanced by normal reaction) and the net torque about any point is also zero, satisfying statement 1. Now imagine a ball falling under gravity; its momentum clearly changes with time, contradicting statement 2. For statement 3, consider a perfectly inelastic collision where two objects stick together; the kinetic energy decreases and is partly converted into heat and deformation, showing that kinetic energy is not always conserved. Why Other Options Are Wrong: Option A: 1, 2 and 3 claims that momentum and kinetic energy are always conserved individually, which is false. Option B: 2 and 3 only ignores the correct statement 1 and incorrectly accepts both 2 and 3. Option C: 1 and 2 only wrongly accepts the idea that the momentum of a single body is always conserved. Common Pitfalls: Students sometimes misapply conservation laws, forgetting the requirement of isolation and absence of external forces. Momentum conservation always applies to a closed system, not necessarily to each body inside it. Similarly, mechanical energy and kinetic energy are conserved only in certain idealised interactions. Distinguishing between system level conservation and single body changes is essential to answer such questions correctly. Final Answer: The correct choice is 1 only.

Question 14

The operation of an electric safety fuse in a domestic circuit depends mainly on which of the following features of the current flowing through it? 1. Magnetic effect of the current 2. Chemical effect of the current 3. Magnitude of the current 4. He…

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Introduction / Context: Safety fuses are protective devices used in electrical circuits to prevent damage from excessive current. When current exceeds a certain safe value, the fuse element melts and breaks the circuit, thereby guarding appliances and wiring. This question asks which properties or effects of the current are essential for the correct functioning of a fuse. Given Data / Assumptions: A fuse consists of a thin wire made of a material with relatively low melting point. The fuse is connected in series with the circuit it protects. If current exceeds a rated value, the fuse wire heats up and melts. We consider magnetic, chemical, magnitude and heating effects of current. Concept / Approach: The basic principle of a fuse is the heating effect of electric current. According to Joule heating, the heat produced per unit time in a conductor is proportional to I^2 * R, where I is current and R is resistance. If the current magnitude is too high, the heat generated increases sharply, causing the fuse wire to melt when it reaches its melting point. Thus, both the magnitude of the current and the heating effect are crucial. Magnetic and chemical effects are not directly relevant to the simple melting operation of a fuse wire in domestic wiring. Step-by-Step Solution: Step 1: Recall the formula for heat generated in a conductor: H proportional to I^2 * R * t, showing dependence on current magnitude. Step 2: Recognise that safety fuses are designed so that when current exceeds a specific magnitude, the wire heats to its melting point and breaks. Step 3: The magnetic effect of current is important in devices such as relays, solenoids and electromagnets, but a simple fuse wire does not rely on magnetism to operate. Step 4: The chemical effect of current applies to electrolysis and electroplating, which are not part of a fuse action. Therefore, the working of safety fuses depends mainly on magnitude of current and heating effect of current. Verification / Alternative check: Notice that fuses are rated in amperes, such as 5 A or 15 A, indicating the maximum safe current magnitude. If current remains below this value, the heating effect is not sufficient to melt the fuse. When a fault occurs and current rises sharply, the I^2 dependence ensures rapid temperature rise and melting. This clearly shows that magnitude and heating effect of current are the critical factors for fuse operation. Why Other Options Are Wrong: Option A: 1, 2, 3 and 4 incorrectly includes magnetic and chemical effects, which are not basic to fuse operation. Option B: 1, 2 and 3 only omits heating effect, which is the central principle behind melting of the fuse element. Option D: 4 only acknowledges heating effect but ignores the explicit role of current magnitude; fuse ratings are defined in terms of maximum current. Common Pitfalls: Students sometimes focus solely on the heating effect and forget that it depends on the square of current, making the magnitude of current essential. Others may see sparks and think of magnetic effects, but these are secondary phenomena. Always connect the idea of fuse operation to Joule heating and the current rating marked on the fuse. Final Answer: The working of safety fuses depends on 3 and 4 only, that is, magnitude of the current and the heating effect of the current.

Question 15

In atmospheric physics and meteorology, adiabatic temperature changes in rising and sinking air masses are primarily caused by which one of the following processes?

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Introduction / Context: In meteorology, the temperature of air parcels often changes as they move vertically in the atmosphere. These changes can be adiabatic, meaning they occur without heat exchange with the surroundings, or diabatic, involving heating and cooling from external sources. Understanding adiabatic processes is essential for explaining cloud formation, atmospheric stability and lapse rates. This question asks which process is responsible for adiabatic temperature changes in the atmosphere. Given Data / Assumptions: Adiabatic processes involve no net heat exchange (q = 0) with the environment. We consider rising and descending air parcels in the troposphere. Pressure decreases with height in the atmosphere. Air behaves approximately as an ideal gas over the scales considered. Concept / Approach: When an air parcel rises, it moves into a region of lower pressure and expands. This expansion requires the parcel to do work on its surroundings, and since no heat is added in an adiabatic process, the internal energy of the parcel decreases, leading to a drop in temperature. Conversely, when air sinks, it encounters higher pressure, is compressed, work is done on the parcel and its temperature increases. Thus, adiabatic temperature changes are directly tied to expansion and compression of the air, not to processes like radiation absorption or latent heat release, which are non adiabatic and involve energy exchange. Step-by-Step Solution: Step 1: Recall the first law of thermodynamics: ΔU = q − w, where q is heat added and w is work done by the system. Step 2: For an adiabatic process, q = 0, so ΔU = −w. If the air parcel expands and does work on the surroundings, w is positive and internal energy decreases, lowering temperature. Step 3: When air compresses, work is done on the parcel (w is negative in the chosen convention), so ΔU becomes positive and temperature increases. Step 4: Therefore, expansion and compression of air due to changing pressure with altitude are the mechanisms that cause adiabatic temperature changes in the atmosphere. Verification / Alternative check: Adiabatic lapse rates, such as the dry adiabatic lapse rate of about 9.8 °C per kilometre for dry air, are derived assuming no heat exchange but allowing expansion work. Cloud physics and mountain weather patterns rely on this concept. Latent heat release, on the other hand, modifies the lapse rate and leads to moist adiabatic processes which are not strictly adiabatic in the thermodynamic sense because heat is released or absorbed during phase changes. Why Other Options Are Wrong: Option A: Deflection and advection describe horizontal movements and changes in wind direction, not the direct thermodynamic cause of adiabatic temperature changes. Option B: Latent heat of condensation releases heat and is a diabatic process; it does affect temperature but does not fit the strict definition of adiabatic (q = 0). Option D: Partial absorption of solar radiation by the atmosphere is a radiative heating process, again not adiabatic because external heat is added. Common Pitfalls: Students sometimes use the term adiabatic loosely to include processes where latent heat is involved. In strict thermodynamic terms, adiabatic means no heat exchange with the environment, so latent heat effects actually make the process pseudo adiabatic or moist adiabatic. For exam purposes, always associate adiabatic temperature changes in air with expansion and compression due to pressure variations. Final Answer: Adiabatic temperature changes in the atmosphere are caused by expansion and compression of the air.

Question 16

In cloud physics, the collision coalescence process of precipitation (also called the warm rain process) is primarily applicable to which type of clouds?

Accepted Answer

Introduction / Context: Precipitation in the atmosphere can form through different microphysical processes. Two major mechanisms are the collision coalescence process (warm rain process) and the ice crystal or Bergeron process (cold cloud process). Understanding which process dominates in which type of cloud is important in meteorology and climatology. This question focuses on the collision coalescence process. Given Data / Assumptions: Collision coalescence involves growth of liquid water droplets by collision and merging. The freezing level is the altitude where temperature drops to 0 °C. Clouds may be entirely above 0 °C (warm clouds) or extend into subfreezing regions (cold or mixed phase clouds). Concept / Approach: The collision coalescence process occurs in warm clouds composed entirely of liquid water droplets, where temperatures are above 0 °C throughout the cloud depth. In such clouds, tiny droplets are carried by turbulent air, repeatedly colliding and merging to form larger drops that eventually grow heavy enough to fall as rain. In contrast, clouds that extend beyond the freezing level contain ice crystals and supercooled droplets, and precipitation formation is more strongly influenced by the Bergeron ice crystal process. Thus, collision coalescence is particularly applicable to clouds that do not extend beyond the freezing level. Step-by-Step Solution: Step 1: Identify that collision coalescence is a mechanism where larger droplets fall faster, collide with smaller droplets and coalesce, increasing in size. Step 2: Recognise that this process requires a population of liquid water droplets and is most efficient when ice crystals are absent or negligible. Step 3: Warm clouds, which do not extend into subzero temperatures, consist entirely of liquid droplets and therefore favour collision coalescence. Step 4: Cold or mixed phase clouds that extend above the freezing level will involve significant ice processes and the Bergeron mechanism, not pure collision coalescence. Verification / Alternative check: Textbooks on atmospheric science describe maritime and tropical warm clouds, which often do not reach high altitudes, as primary sites for the warm rain process via collision coalescence. Radar and aircraft observations show that in such clouds, raindrop size distributions can be explained by droplet collision and growth. High altitude clouds with ice crystals show different precipitation signatures, indicating dominance of ice phase processes. Why Other Options Are Wrong: Option B: Clouds extending beyond the freezing level typically involve ice crystals and supercooled droplets, where the Bergeron process or mixed processes dominate precipitation formation. Option C: Saying all types of clouds ignores the clear distinction between warm and cold cloud microphysics, so this is too broad and incorrect. Option D: Cirrocumulus clouds are high altitude, cold clouds generally composed of ice crystals, not warm, collision coalescence dominated clouds. Common Pitfalls: Students may confuse the two main precipitation mechanisms and assume they operate equally in all clouds. A clear mental separation helps: warm rain process (collision coalescence) acts in clouds above freezing temperature throughout their depth, while the Bergeron process requires coexistence of ice crystals and supercooled water droplets in cold clouds. Remembering this distinction makes such questions straightforward. Final Answer: The collision coalescence process is applicable primarily to clouds which do not extend beyond the freezing level (warm clouds).

Question 17

When light passes by the edge of an obstacle or through a narrow slit and bends into the region of geometrical shadow, the phenomenon responsible for this bending of light is called what?

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Introduction / Context: Wave optics explains several phenomena that cannot be understood by treating light purely as rays. One such phenomenon is the bending and spreading of light when it encounters obstacles or apertures comparable in size to its wavelength. This question asks for the correct term used to describe the bending of light around a corner or through a narrow slit into the shadow region. Given Data / Assumptions: Light behaves as a wave with a definite wavelength. Light encounters an obstacle edge or passes through a narrow opening. The bending happens into regions that would be dark according to simple ray optics. Concept / Approach: Diffraction is the phenomenon in which waves bend around obstacles or spread after passing through narrow apertures. It is significant when the size of the obstacle or aperture is of the same order as the wavelength of the wave. Reflection involves bouncing of light from a surface, refraction involves change in direction at an interface due to change in refractive index, and total internal reflection is a special case of refraction at large incident angles in denser media. None of these describe bending into geometrical shadow caused by obstacles or slits; that is specifically diffraction. Step-by-Step Solution: Step 1: Recognise that the question mentions light bending around a corner or through a slit into what ray optics would predict as shadow. Step 2: Recall that such spreading and bending effects are characteristic of diffraction, a wave phenomenon. Step 3: Note that reflection implies light bouncing from surfaces without entering shadow regions behind obstacles. Step 4: Refraction involves bending at an interface between two media, not around obstacles, and total internal reflection occurs within a denser medium, not in free space around a slit. Verification / Alternative check: Classic experiments with single and double slits demonstrate diffraction patterns, where light spreads out and interferes, producing bright and dark fringes on a screen. These effects cannot be explained by straight line ray diagrams. Huygens principle and wavefront analysis show that each point on a wavefront acts as a secondary source, leading to bending around edges, which is the essence of diffraction. Why Other Options Are Wrong: Option A: Reflection describes light returning from a surface with the angle of incidence equal to the angle of reflection; it does not explain bending behind obstacles. Option B: Refraction is bending at a boundary between different media due to a change in speed of light, not bending into shadow regions behind an obstacle. Option D: Total internal reflection is a special refraction effect where light is completely reflected back into a denser medium at angles above the critical angle. Common Pitfalls: Students sometimes mix diffraction and refraction, especially because both involve changes in direction. A useful distinction is that refraction involves interfaces between media, whereas diffraction involves obstacles and apertures causing spreading of wavefronts. Remember that diffraction explains why you can hear sound around corners and why light forms fringes in slit experiments. Final Answer: The bending of light around a corner or slit is due to diffraction.

Question 18

On a hot day in a desert, distant objects such as trees or the sky sometimes appear as if reflected in water on the ground, forming a mirage. This optical illusion is mainly produced because of which combination of phenomena of light?

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Introduction / Context: Mirages are common optical phenomena observed in deserts and on hot roads. They create the illusion of water or inverted images of distant objects on the ground. Understanding mirages helps illustrate how temperature gradients in air affect the path of light. This question asks which optical phenomena of light are mainly responsible for forming a mirage in a desert environment. Given Data / Assumptions: We consider a hot desert surface under strong sunlight. Air near the surface is much hotter and therefore less dense than air higher up. Refractive index of air decreases with increasing temperature. Light from distant objects travels through layers of air with varying refractive index. Concept / Approach: In a desert, strong heating of the ground warms the air just above it, creating a vertical gradient in temperature and hence refractive index. Light rays from distant objects entering this region undergo continuous refraction as they pass through layers of air with slightly different refractive indices. When the angle of incidence on a layer of lower refractive index becomes large enough, the condition for total internal reflection is satisfied at some level, causing light to be reflected upward towards the observer. The observer interprets these rays as if they come from the ground, producing an apparent inverted image resembling a water reflection. Thus, mirage formation involves both refraction in a gradient medium and total internal reflection at a critical layer. Step-by-Step Solution: Step 1: Recognise that temperature gradients create layers of air with different refractive indices, with hotter, less dense air near the surface having lower refractive index. Step 2: As a light ray from a distant object travels downward towards the hot surface, it continuously bends away from the normal due to refraction in layers of decreasing refractive index. Step 3: At some depth, the incident angle becomes greater than the critical angle for the interface between denser and rarer air layers, and total internal reflection occurs, sending the light back upward. Step 4: The observer's eye traces these rays back in straight lines and perceives them as originating from the ground, creating the illusion of a reflected image or water surface. Verification / Alternative check: Ray diagrams of mirage formation in textbooks show curved light paths bending near the ground and then reflecting upward due to total internal reflection. The presence of both continuous refraction (due to gradual index change) and a critical angle leading to total internal reflection is emphasised. Pure reflection from a mirror like surface is not involved because the ground is not highly reflecting like a calm water surface; instead, the illusion arises from atmospheric optics. Why Other Options Are Wrong: Option A: Refraction alone cannot explain the abrupt change in direction that produces a clear inverted image; total internal reflection at some level is also needed. Option B: Reflection from the ground surface is not the primary cause; otherwise, mirages would appear even when the air is uniform in temperature. Option C: Total internal reflection of light only ignores the initial bending of rays caused by refraction in the temperature gradient, which is crucial to bring rays to the critical angle. Common Pitfalls: Students often over simplify mirages as just refraction or just reflection. The correct picture is a combination: continuous refraction in a graded index medium plus total internal reflection at a certain depth. Another confusion is to think there is actual water present; in reality, the mirage is purely an optical illusion created by atmospheric conditions. Final Answer: The formation of a desert mirage is mainly due to both refraction and total internal reflection of light.

Question 19

Two metallic copper wires have the same length, but the radius of wire 1 is half of the radius of wire 2. The resistance of wire 1 is R. If both wires are connected in series, what is the total resistance of the combination?

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Introduction / Context: This problem tests your understanding of how resistance depends on the dimensions of a conductor and how resistances combine in series. For metallic wires of the same material and same length, resistance changes with the cross sectional area, which in turn depends on the radius of the wire. Once the individual resistances are known, the total resistance for a series connection is simply the sum. Such questions are common in basic electricity and help build intuition for designing circuits and choosing conductor sizes. Given Data / Assumptions: • Both wires are made of copper (same resistivity). • Both wires have the same length L. • Radius of wire 1 is r1 and radius of wire 2 is r2, with r1 = r2 / 2. • Resistance of wire 1 is given as R. • The two wires are connected in series, so the same current flows through both. Concept / Approach: The resistance of a uniform wire is given by R = rho * L / A, where rho is resistivity, L is length, and A is cross sectional area. For a circular wire, A = pi * r^2, so resistance is inversely proportional to r^2 when material and length are fixed. If one wire has half the radius of another but the same length and material, its area is one quarter, so its resistance is four times larger. After finding the relationship between the resistance of wire 1 and wire 2, we simply add the two resistances in series to find the total resistance. Step-by-Step Solution: Step 1: Write the resistance formula for a wire: R = rho * L / A. Step 2: For circular cross section, area A = pi * r^2. Step 3: Let the radius of wire 2 be r2. Then the radius of wire 1 is r1 = r2 / 2. Step 4: Area of wire 1: A1 = pi * (r2 / 2)^2 = pi * r2^2 / 4. Step 5: Area of wire 2: A2 = pi * r2^2. Step 6: Since resistivity rho and length L are the same for both, R1 / R2 = A2 / A1 = (pi * r2^2) / (pi * r2^2 / 4) = 4. Step 7: Therefore, R1 = 4 * R2. Given R1 = R, we get 4 * R2 = R so R2 = R / 4. Step 8: In series, total resistance R_total = R1 + R2 = R + R / 4 = 5R / 4. Verification / Alternative check: We can quickly check with simple numbers. Assume rho * L / (pi * r2^2) = 1 ohm so that wire 2 has resistance 1 ohm. Then wire 1, with one quarter the area, would have R1 = 4 ohm. The question states R1 = R, so R = 4 ohm and R2 = 1 ohm = R / 4. The total series resistance is 4 + 1 = 5 ohm = 5R / 4. This numerical example matches the algebraic result and confirms the reasoning. Why Other Options Are Wrong: Option a (2R): This would correspond to R2 = R, which ignores the dependence on the radius squared. Option b (R/2): This suggests the total resistance is less than that of wire 1 alone, which is impossible for series connection with a positive resistance wire. Option d (3R/4): This is even smaller than R and contradicts the idea that adding a resistance in series increases the total resistance. Option e (R/4): This equals the resistance of the thinner wire divided by 4 and does not follow from the geometry. Common Pitfalls: Students often forget that resistance is inversely proportional to the square of the radius, not directly proportional to radius. Another common mistake is to assume that having the same length automatically means equal resistance, ignoring the change in cross sectional area. In series combinations, some learners also incorrectly apply formulas for parallel resistors. To avoid errors, always relate area to radius carefully and then use the correct series rule, R_total = R1 + R2. Final Answer: The total resistance of the series combination of the two copper wires is 5R/4.

Question 20

When the Sun is near the horizon during sunrise or sunset, it appears reddish in colour. Which phenomenon of light is mainly responsible for this observation?

Accepted Answer

Introduction / Context: The reddish appearance of the Sun near the horizon at sunrise and sunset is a classic question in wave optics and atmospheric physics. It helps students understand how Earth atmosphere affects the light coming from the Sun. The key idea is that different colours of light are affected differently when they pass through air and its suspended particles. This question focuses on identifying the main optical phenomenon responsible for the reddish colour seen near the horizon. Given Data / Assumptions: • The Sun is observed when it is close to the horizon, either in the early morning or late evening. • Sunlight has to travel a longer path through the atmosphere at these times than at noon. • The atmosphere contains molecules, dust, and other small particles that interact with sunlight. • We assume normal clear sky conditions without extreme pollution or clouds blocking the Sun. Concept / Approach: White sunlight is a mixture of many colours, each corresponding to a different wavelength. Shorter wavelengths (blue and violet) are scattered much more strongly by air molecules than longer wavelengths (red and orange). This wavelength dependent scattering by molecules and tiny particles is called Rayleigh scattering. When the Sun is near the horizon, sunlight passes through a much thicker layer of atmosphere, so most of the blue and green components are scattered out of the direct path. The light that finally reaches our eyes directly from the Sun is therefore richer in red and orange components, making the Sun appear reddish. The dominant phenomenon here is scattering of light, not reflection, refraction, or dispersion in a prism. Step-by-Step Solution: Step 1: Recognise that at sunrise and sunset, sunlight travels a longer path through the atmosphere compared to midday. Step 2: Recall that atmospheric particles scatter shorter wavelength light (blue and violet) more strongly than longer wavelength light (red). Step 3: Understand that because of this scattering, much of the blue and green light is removed from the direct beam before it reaches the observer. Step 4: The remaining direct sunlight is dominated by red and orange wavelengths, causing the Sun to appear reddish. Step 5: Conclude that the main phenomenon involved is scattering of light in the atmosphere. Verification / Alternative check: The same scattering mechanism explains why the sky appears blue during the day. When the Sun is high, we receive scattered blue light from all directions while looking away from the Sun. Near the horizon, the light has travelled more through the atmosphere, so so much of the blue light has been scattered away that the Sun itself appears red. No significant refraction or dispersion processes are required to explain the colour, although small refraction effects do shift the apparent position of the Sun slightly. This reinforces that scattering is the correct answer. Why Other Options Are Wrong: Option a (Reflection of light): Reflection occurs at surfaces like mirrors or water, and is not the dominant effect in the open sky between the Sun and observer. Option b (Refraction of light): Refraction bends the path of light when it passes between media of different densities; it slightly changes the apparent position of the Sun but does not primarily cause the red colour. Option c (Dispersion of light): Dispersion is splitting of white light into different colours by a prism or raindrops. It explains rainbows, not the overall reddish tint of the Sun at the horizon. Option e (Total internal reflection of light): This is a phenomenon where light is completely reflected inside a denser medium at large incidence angles and is not relevant to sunlight travelling through air. Common Pitfalls: Many students confuse dispersion and scattering, thinking that because we see colours, dispersion must be involved. However, dispersion requires a medium like a prism where different wavelengths refract by different amounts. In the case of the reddish Sun, the main process is selective scattering by atmospheric particles. Another confusion is to attribute almost every optical effect in air to refraction, but here refraction is minor compared to scattering. Remember: blue sky and red sunsets are classic signatures of Rayleigh scattering. Final Answer: The reddish appearance of the Sun near the horizon is mainly due to scattering of light.

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