Question 1

A copper wire of length L and cross-sectional area A carries a current I and dissipates power P as heat. An aluminium wire of the same length L and the same cross-sectional area A is now connected to the same current I. Compared to P, how much power…

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Introduction / Context: This question checks your understanding of electric power dissipation in resistive conductors and how it depends on the material of the wire. Copper and aluminium are common conducting materials, but they have different electrical resistivities. When current is fixed, the power lost as heat in a wire depends directly on its resistance. Recognising which material has higher resistivity helps you decide whether the power dissipation increases or decreases when switching from copper to aluminium while keeping dimensions and current the same. Given Data / Assumptions: • Copper wire: length L, area A, current I, power dissipated P. • Aluminium wire: same length L and same cross-sectional area A. • The same current I is passed through the aluminium wire. • Resistivity of copper is lower than the resistivity of aluminium (rho_cu < rho_al). • Wires are ohmic and obey P = I^2 * R. Concept / Approach: The resistance of a wire is given by R = rho * L / A, where rho is resistivity. For a given length and cross-sectional area, resistance is directly proportional to resistivity. The power dissipated in a resistor carrying a constant current is P = I^2 * R. If the same current I passes through two wires of identical dimensions but different resistivities, then the wire with higher resistivity has higher resistance and therefore dissipates more power for the same current. Aluminium has higher resistivity than copper, so the aluminium wire will have a larger resistance and hence more power dissipation than the copper wire under the same current. Step-by-Step Solution: Step 1: Write the resistance of the copper wire: R_cu = rho_cu * L / A. Step 2: Write the resistance of the aluminium wire: R_al = rho_al * L / A. Step 3: Since L and A are the same, R_al / R_cu = rho_al / rho_cu. Step 4: From material properties, rho_al is greater than rho_cu, so R_al is greater than R_cu. Step 5: Power in copper wire: P = I^2 * R_cu. Step 6: Power in aluminium wire: P_al = I^2 * R_al. Step 7: Because R_al > R_cu and I is the same, P_al > P. Step 8: Therefore, the aluminium wire dissipates more power than the copper wire. Verification / Alternative check: Take simple numbers for illustration. Suppose that for the given dimensions, the copper wire has resistance 1 ohm and aluminium wire has resistance 1.6 ohm (higher because of higher resistivity). With the same current I = 2 A, the copper wire dissipates P_cu = I^2 * R_cu = 4 * 1 = 4 W. The aluminium wire dissipates P_al = 4 * 1.6 = 6.4 W, which is clearly greater than 4 W. This numerical example confirms the general conclusion that power dissipation increases when replacing copper with aluminium under the same current and dimensions. Why Other Options Are Wrong: Option a (Equal to P): Power would be equal only if resistivities were the same or current was adjusted to compensate, which is not the case here. Option b (Less than P): This would require aluminium to have lower resistivity than copper, which is incorrect. Option d (Exactly 2P): There is no information suggesting that the resistance of aluminium is exactly twice that of copper; the actual ratio depends on precise resistivity values. Option e (Zero): Power is zero only when current is zero or resistance is zero, neither of which applies in this situation. Common Pitfalls: A frequent mistake is to focus on the fact that aluminium is also a good conductor and assume it behaves the same as copper. While aluminium is indeed a conductor, its resistivity is higher, leading to higher resistance for identical geometry. Another error is to think in terms of P = V^2 / R and forget that here voltage is not specified; the current I is fixed. In constant current scenarios, power scales directly with resistance, while in constant voltage situations, power scales inversely with resistance. Always pay close attention to whether current or voltage is being held constant. Final Answer: With the same current and same dimensions, the aluminium wire dissipates greater than P (more power) compared to the copper wire.

Question 2

For a static fluid, the pressure at a depth h is given by P = P0 + rho * g * h, where rho is the fluid density and P0 is the pressure at the free surface. This expression is associated with which basic law or principle in fluid mechanics?

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Introduction / Context: The expression P = P0 + rho * g * h describes how pressure increases with depth in a static fluid under gravity. This relationship is fundamental in hydrostatics and is consistent with Pascal's law about transmission of pressure in an enclosed fluid. Understanding this formula is essential for analysing dams, submarines, deep sea diving, and many engineering applications involving liquids at rest. Given Data / Assumptions: • A fluid is at rest under the influence of gravity. • P0 is the pressure at the free surface of the fluid (often atmospheric pressure). • rho is the density of the fluid. • g is the acceleration due to gravity. • h is the depth below the free surface. Concept / Approach: In a static fluid, pressure at a point depends only on the depth and the density of the fluid, not on the shape of the container. The basic hydrostatic equation dP/dh = rho * g integrates to P = P0 + rho * g * h. This result reflects Pascal's law, which states that pressure applied to an enclosed fluid is transmitted undiminished to all parts of the fluid and to the walls of its container. Pascal's law underlies the uniform distribution of pressure in a static fluid and is the conceptual foundation for this pressure depth relationship. Step-by-Step Solution: Step 1: Consider a small fluid element at depth h in a fluid at rest. Step 2: The vertical forces on the element include pressure forces on the top and bottom surfaces, and the weight of the fluid element. Step 3: For equilibrium (no acceleration), the sum of forces must be zero, leading to the differential relation dP/dh = rho * g. Step 4: Integrate this equation from the free surface (h = 0, P = P0) to depth h, giving P = P0 + rho * g * h. Step 5: This result shows that pressure increases linearly with depth in a uniform, incompressible, static fluid. Step 6: This hydrostatic pressure distribution is one of the primary consequences of Pascal's law. Verification / Alternative check: We can test the formula with a simple example. In water (rho ≈ 1000 kg/m^3), at a depth of 10 m, the gauge pressure is approximately rho * g * h ≈ 1000 * 9.8 * 10 ≈ 98 000 Pa, which is close to one extra atmosphere of pressure. This matches what is taught in basic physics and engineering courses and confirms that the relationship is correct. The fact that the same depth in different shaped containers gives the same pressure further illustrates Pascal's law and the independence from container shape. Why Other Options Are Wrong: Option b (Newton's law): Newton's laws of motion describe the relationship between forces and acceleration, not directly the static pressure distribution in a fluid. Option c (Bernoulli's principle): Bernoulli's principle relates pressure, kinetic energy per unit volume, and potential energy per unit volume in a moving fluid, not a static one. Option d (Archimedes' principle): Archimedes' principle deals with buoyant force equal to the weight of displaced fluid, not the pressure depth equation. Option e (Hooke's law): Hooke's law applies to elastic solids, relating stress to strain, and does not govern hydrostatic pressure in fluids. Common Pitfalls: Students sometimes mistakenly associate any fluid formula with Bernoulli's principle, ignoring whether the fluid is moving or at rest. Another confusion is between Pascal's law and Archimedes' principle, since both involve fluids. A quick way to remember is that Pascal's law is about pressure transmission and hydrostatics, while Archimedes deals with buoyancy. Always check whether the fluid is static or dynamic when deciding which principle applies. Final Answer: The pressure depth relation P = P0 + rho * g * h is associated with Pascal's law in fluid mechanics.

Question 3

A group of individual solar cells connected together in a definite pattern to increase the overall output is called what?

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Introduction / Context: Solar energy systems commonly use many small photovoltaic cells combined to produce useful levels of electrical power. A single solar cell produces only a small voltage and current. To power homes, street lights, pumps, or other loads, many cells are connected together in an arranged structure. Understanding the correct name for this combined arrangement is important for basic general knowledge about renewable energy and its practical implementation. Given Data / Assumptions: • Many individual solar cells are connected together. • The cells are arranged in a definite pattern, usually series and parallel combinations. • The goal is to increase the total power output and voltage or current level. • The context is basic solar energy terminology. Concept / Approach: A single photovoltaic cell converts sunlight into a small amount of electrical power. By connecting many such cells in series and parallel, we create a larger unit called a solar panel or solar cell panel. When several panels are further connected together in the field, they may form a solar array. A battery is a device that stores electrical energy chemically, not a direct generator from sunlight. Solar heaters and solar cookers are thermal devices that mainly use the heat from sunlight rather than directly converting it into electricity. Therefore, the correct term for a group of solar cells joined together in a definite pattern is solar cell panel. Step-by-Step Solution: Step 1: Recall that a solar cell is the basic photovoltaic unit that produces electricity from light. Step 2: Recognise that a practical power source for homes or devices requires many solar cells wired together. Step 3: The standard name for the resulting assembly of cells is a solar panel or solar cell panel. Step 4: Compare with other devices: batteries store energy, while solar heaters and solar cookers mainly use sunlight for heating. Step 5: Conclude that the correct answer is solar cell panel. Verification / Alternative check: Products sold for rooftop installations are usually described as solar panels of a certain watt rating, such as 250 W solar panels. Technical descriptions mention that each panel contains many individual photovoltaic cells connected in series and parallel to achieve a desired voltage and current. When multiple panels are installed, the structure becomes a solar array, but the basic group of cells is called a panel. This confirms that solar cell panel is the accepted term. Why Other Options Are Wrong: Option a (Battery): A battery stores electrical energy in chemical form and is charged by a panel, but it is not itself a group of solar cells. Option b (Solar heater): A solar heater uses sunlight to heat water or air and does not generate electrical energy directly. Option c (Solar cooker): A solar cooker concentrates sunlight to cook food by heat; again it is a thermal device, not a photovoltaic assembly. Option e (Inverter): An inverter converts direct current (DC) from solar panels or batteries into alternating current (AC), but it is an electronic device, not a collection of solar cells. Common Pitfalls: Sometimes learners mix up the terms solar panel, solar array, and solar battery. A quick memory aid is: solar cell is the smallest unit, solar panel is many cells in one module, solar array is many panels installed together, and battery is for storage. Distinguishing these helps in understanding basic solar power system diagrams and exam questions. Final Answer: A group of solar cells joined together in a definite pattern is called a Solar cell panel.

Question 4

It is difficult to drive a nail into a freely suspended wooden frame with a hammer. Which law of mechanics best explains this observation?

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Introduction / Context: Hammering a nail into wood is a familiar everyday activity that illustrates basic principles of mechanics. When the wooden support is rigid and fixed, the nail enters easily. When the wooden frame is freely suspended and can move, it becomes difficult to drive the nail in. This observation is often linked to Newton's third law of motion, which describes the action and reaction forces between interacting bodies such as the hammer, nail, and frame. Given Data / Assumptions: • A wooden frame is freely suspended and not rigidly fixed to a massive support. • A hammer is used to strike a nail placed against the frame. • The hammer exerts a force on the nail, and the nail interacts with the wooden frame. • We compare this case with the situation where the frame is rigidly fixed to a wall or heavy support. Concept / Approach: Newton's third law states that for every action, there is an equal and opposite reaction. When the hammer strikes the nail, the nail exerts an equal and opposite force on the hammer. Similarly, the nail exerts a force on the wood, and the wood exerts an equal and opposite reaction on the nail. If the wood is rigidly supported, this reaction force has a solid backing and helps the nail penetrate into the wood. However, when the frame is freely suspended, the reaction causes the frame to move away instead of providing a strong opposing force on the nail. In this way, action reaction pairs and the resulting motion explain why the nail does not go in easily. Step-by-Step Solution: Step 1: When the hammer hits the nail, the hammer exerts a force on the nail (action). Step 2: According to Newton's third law, the nail exerts an equal and opposite force on the hammer (reaction). Step 3: The nail also pushes on the wooden frame, trying to penetrate. Step 4: The wooden frame exerts an equal and opposite reaction on the nail. If the frame is rigid and supported, this reaction helps redirect the hammer's impulse into driving the nail inward. Step 5: When the frame is freely suspended, the reaction force causes the entire frame to move backward instead of effectively resisting and helping push the nail into the wood. Step 6: As a result, much of the hammer's energy goes into moving the frame rather than forcing the nail inside, making it difficult to fix the nail. Verification / Alternative check: You can easily verify this by practical experience. Trying to hammer a nail into a loose wooden plank that you are holding in the air is much harder than hammering the same nail when the plank is firmly placed on the ground or against a wall. In the firmly supported case, the support provides an additional reaction force that helps keep the wood from moving away, making the nail enter more effectively. This emphasises the role of action reaction pairs as stated in Newton's third law and how they influence the motion of interacting bodies. Why Other Options Are Wrong: Option a (Law of inertia): Inertia (Newton's first law) deals with the tendency to maintain state of rest or uniform motion but does not directly explain the equal and opposite forces during the hammer nail interaction. Option b (Newton's second law): Newton's second law relates net force and acceleration (F = m * a) but the key idea here is the pair of forces between bodies, which is better captured by the third law. Option d (Pascal's law): This applies to pressure transmission in fluids and is not relevant to hammering a nail in solid wood. Option e (Archimedes' principle): This deals with buoyant force in fluids and has no connection to the situation described. Common Pitfalls: Students sometimes confuse which Newton's law to apply in everyday examples. If the focus is on pairs of forces between two objects (action and reaction), think Newton's third law. If the focus is on how force produces acceleration, think Newton's second law. If the focus is on a body remaining at rest or uniform motion unless disturbed, think Newton's first law. In this nail example, the essential point is the equal and opposite reaction of the frame and support, making Newton's third law the best answer. Final Answer: The difficulty in fixing a nail into a freely suspended frame is best explained by Newton's third law of motion.

Question 5

Which one of the following is NOT a correct property of electromagnetic waves?

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Introduction / Context: Electromagnetic waves, such as light, radio waves, X rays, and microwaves, share a set of common properties described by Maxwell's theory and classical wave optics. Understanding which statements correctly describe these waves is fundamental for topics ranging from communication technology to quantum physics. This question asks you to identify which listed statement is not a true property of electromagnetic waves, so you must recall their key wave characteristics. Given Data / Assumptions: • We are dealing with electromagnetic waves like visible light, radio waves, and X rays. • Statements mention interference, diffraction, field directions, nature of the waves, and propagation without a medium. • We assume standard physics definitions of interference and diffraction. Concept / Approach: Electromagnetic waves consist of oscillating electric and magnetic fields, mutually perpendicular and also perpendicular to the direction of propagation, making them transverse waves. They can travel through vacuum, which is why we receive sunlight from the Sun across empty space. As waves, they show all typical wave phenomena, including reflection, refraction, interference, and diffraction. Interference and diffraction are strong experimental proofs of the wave nature of light. Therefore, any statement claiming that electromagnetic waves do not show interference and diffraction is false. Step-by-Step Solution: Step 1: Recall that in Young's double slit experiment, visible light shows clear interference fringes, proving that electromagnetic waves can interfere. Step 2: Note that light also bends around obstacles and through slits of size comparable to its wavelength, demonstrating diffraction. Step 3: Recognise that electromagnetic waves are transverse; the electric and magnetic fields oscillate perpendicular to each other and to the direction of propagation. Step 4: Understand that electromagnetic waves can travel through vacuum without any material medium and in vacuum they travel at the speed of light, c. Step 5: From these facts, conclude that the incorrect statement is the one claiming electromagnetic waves do not show interference and diffraction. Verification / Alternative check: Numerous experiments confirm interference and diffraction of light. Optical instruments are designed with diffraction limits in mind, and radio antennas are built considering interference and diffraction patterns. These behaviours are not limited to visible light but occur across the full electromagnetic spectrum. Therefore, any claim that electromagnetic waves do not show interference and diffraction contradicts well established experimental results and must be the false property in the list. Why Other Options Are Wrong (i.e., they are actually correct properties): Option b: Electromagnetic waves have electric and magnetic fields oscillating perpendicular to each other, a direct result of Maxwell's equations. Option c: They are transverse waves because field oscillations are perpendicular to the direction of propagation. Option d: They do not require a medium; they can travel through vacuum, which is why sunlight reaches Earth. Option e: In vacuum, all electromagnetic waves travel at the same speed, the speed of light, denoted by c. Common Pitfalls: Some students mistakenly think only water waves or sound waves show interference and diffraction, forgetting that these are general wave phenomena. Another confusion is between sound (which is a mechanical wave and needs a medium) and electromagnetic waves (which do not). Always remember that interference and diffraction are strong evidence for wave nature and apply to electromagnetic waves as well. Final Answer: The statement that is NOT a correct property of electromagnetic waves is: Electromagnetic waves do not show interference and diffraction.

Question 6

Which of the following is a characteristic feature of an exothermic reaction?

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Introduction / Context: Chemical reactions are often classified as exothermic or endothermic based on how they exchange energy with their surroundings. Exothermic reactions release energy, usually in the form of heat, while endothermic reactions absorb energy. Recognising whether heat is released or absorbed is a fundamental part of understanding reaction energetics, fuel combustion, and many everyday processes like burning, neutralisation, and respiration. Given Data / Assumptions: • The question specifically asks about a characteristic of exothermic reactions. • We consider simple chemical and physical examples, such as combustion and neutralisation. • The primary focus is on heat exchange, not detailed thermodynamic equations. Concept / Approach: An exothermic reaction is one in which the energy of the products is lower than the energy of the reactants, with the difference released to the surroundings, typically as heat. In energy diagrams, the energy level drops from reactants to products, and the environment receives the released energy. This often leads to a temperature rise in the surroundings or the reaction mixture. Common examples include burning of fuels, many oxidation reactions, and neutralisation of acids and bases. Therefore, the central characteristic is the release of heat. Step-by-Step Solution: Step 1: Recall the definition: exothermic reactions release energy to the surroundings. Step 2: Identify that this energy is usually observed as heat, sometimes accompanied by light (as in flames). Step 3: Compare with endothermic reactions, which absorb heat and often cause cooling of the surroundings. Step 4: Look at the options and find the statement that best matches the idea of releasing heat energy. Step 5: Conclude that "Release of heat" is the correct characteristic feature of exothermic reactions. Verification / Alternative check: Consider the combustion of methane: CH4 + 2O2 → CO2 + 2H2O + energy. This reaction is strongly exothermic and is used in domestic gas burners. The flame and hot cooking vessel are direct evidence that heat is being released. Similarly, mixing a strong acid with a strong base causes the solution to warm up, showing that neutralisation is exothermic. These everyday examples confirm that exothermic reactions are associated with heat release. Why Other Options Are Wrong: Option b (Absorption of heat): This describes endothermic reactions, such as melting of ice or photosynthesis, not exothermic ones. Option c (No change in temperature at all): In real systems, exothermic reactions usually cause an observable temperature rise unless heat is immediately removed; no change in temperature is not their defining feature. Option d (None of the options is correct): This is incorrect because option a accurately describes exothermic reactions. Option e (Formation of only cold products): Products of exothermic reactions can be hot or cold depending on conditions; the key point is energy release, not that products are always "cold." Common Pitfalls: Students sometimes mix up exothermic and endothermic terms because both involve heat change. A simple memory aid is: "Exo" means "out," so energy goes out to the surroundings in exothermic reactions. "Endo" means "in," so energy goes into the system in endothermic reactions. Also remember that exothermic reactions often feel hot to the touch because they warm the surroundings. Final Answer: A characteristic of an exothermic reaction is the Release of heat.

Question 7

A water tank filled with water appears shallower than its actual depth when viewed from above. Which phenomenon of light is mainly responsible for this apparent shallowing?

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Introduction / Context: When you look into a pool, pond, or water tank from above, the bottom often appears closer to the surface than it really is. This is a classic demonstration of refraction of light at a boundary between two media, water and air. Understanding this effect is important for explaining apparent depth, design of optical instruments, and everyday visual illusions involving transparent materials. Given Data / Assumptions: • The tank is filled with water and observed from the air above. • Light rays travel from water (denser medium) into air (rarer medium) and then to the observer's eye. • The observer views the bottom from a position nearly above the tank, not from the side. • The water surface is relatively flat and calm. Concept / Approach: Refraction is the bending of light as it passes from one medium to another with different optical density. When light travels from a denser medium (water) to a rarer medium (air), it bends away from the normal to the interface. The human eye traces back the refracted rays in a straight line, assuming they have travelled in a straight path. Because the rays have bent away from the normal, the eye perceives the origin of the rays at a point higher than the actual bottom, causing the bottom to appear shallower. This phenomenon is known as apparent depth due to refraction. Step-by-Step Solution: Step 1: Light rays from a point on the bottom of the tank travel upward through water towards the surface. Step 2: At the water air boundary, the rays pass from denser water to rarer air. Step 3: According to refraction rules, when light passes from denser to rarer medium, it bends away from the normal. Step 4: The observer in air receives these refracted rays and traces them backward in straight lines. Step 5: The backward extensions meet at a point that is closer to the surface than the true bottom, giving an apparent depth less than the actual depth. Step 6: Therefore, the bottom appears raised and the tank looks shallower due to refraction. Verification / Alternative check: If you insert a stick into water at an angle, it seems bent or broken at the surface. This is also due to refraction. Similarly, a coin at the bottom of a cup may become visible when water is added, again illustrating refraction effects. In all these cases, light rays bend at the interface, and the brain misinterprets their origin. No significant reflection, diffraction, or scattering is needed to explain the apparent shallowing of the tank. Why Other Options Are Wrong: Option a (Reflection): Reflection involves light bouncing back from a surface. While some reflection occurs at the water surface, it does not cause the bottom to appear shallower. Option c (Diffraction): Diffraction is bending of waves around obstacles or through small openings and is not responsible for apparent depth changes. Option d (Total internal reflection): Total internal reflection occurs when light is completely reflected inside a denser medium beyond a critical angle, as in optical fibers. It does not primarily explain the shallow appearance in normal viewing conditions. Option e (Scattering): Scattering can make water appear coloured or hazy but does not by itself create the apparent shift in depth. Common Pitfalls: Students sometimes confuse refraction with reflection because both occur at surfaces. A helpful clue is that apparent depth and bending of submerged objects are classic refraction examples. Another misconception is to think diffraction applies whenever light bends, but diffraction specifically involves obstacles and slits of size comparable to wavelength. For transparent media boundaries, always think of refraction first. Final Answer: A water tank appears shallower from above mainly due to Refraction of light at the water air interface.

Question 8

In the International System of Units (SI), what is the base unit of temperature?

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Introduction / Context: The International System of Units (SI) defines base units for fundamental physical quantities such as length, mass, time, and temperature. Knowing the correct SI unit for temperature is important in physics, chemistry, engineering, and many scientific calculations involving thermodynamic equations and gas laws. Everyday scales like Celsius and Fahrenheit are useful in daily life, but the SI base unit is specific and must be used in scientific work. Given Data / Assumptions: • We are asked specifically for the SI base unit of temperature. • Options include kelvin, joule, Celsius, Fahrenheit, and calorie. • We assume standard SI definitions adopted in science and engineering. Concept / Approach: In SI, the base unit of thermodynamic temperature is the kelvin (symbol K). The Celsius scale is related to the kelvin by a simple offset: temperature in kelvin = temperature in degrees Celsius + 273.15. Fahrenheit is an older scale mainly used in some countries for weather and cooking. Joule is the SI unit of energy, and calorie is a non SI unit of energy often used in nutrition. None of these except kelvin is a base unit of temperature in SI. Step-by-Step Solution: Step 1: Recall that SI uses seven base units, one of which is for thermodynamic temperature. Step 2: Identify that this base unit is kelvin, with symbol K. Step 3: Recognise that degrees Celsius and degrees Fahrenheit are temperature scales, but not SI base units. Step 4: Note that joule and calorie measure energy, not temperature directly. Step 5: Therefore, the correct SI unit of temperature is kelvin. Verification / Alternative check: Gas law equations such as PV = n * R * T use absolute temperature in kelvin. The Stefan Boltzmann law for radiation and many thermodynamic relations also require temperature in kelvin. Scientific references, data tables, and fundamental constants list temperatures in kelvin, reinforcing that kelvin is the standard SI unit. Celsius is often used for weather reports and laboratory readings, but those values are converted to kelvin for theoretical calculations. Why Other Options Are Wrong: Option b (Joule): This is the SI unit of energy or work, not temperature. Option c (Celsius): Degree Celsius is a commonly used temperature scale but not the SI base unit; it is derived from the kelvin scale. Option d (Fahrenheit): Degree Fahrenheit is an older temperature scale used mainly in a few countries, not part of SI. Option e (Calorie): This is an energy unit often used in nutrition and heat calculations, not a temperature unit. Common Pitfalls: Students sometimes answer Celsius because it is familiar from daily life and laboratory thermometers. To avoid confusion, remember that SI base units are defined for fundamental quantities and appear in most theoretical formulas. Whenever you see T in gas law equations, it is assumed to be in kelvin, not in degrees Celsius or Fahrenheit. Final Answer: The SI base unit of temperature is the Kelvin.

Question 9

When a piece of pure silicon is doped with aluminium (a trivalent impurity), what type of semiconductor is formed?

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Introduction / Context: Doping is the process of adding small amounts of impurity atoms to pure semiconductors like silicon to control their electrical properties. Depending on whether the dopant has three or five valence electrons, the resulting material becomes p-type or n-type. Understanding this classification is crucial for learning how diodes, transistors, and integrated circuits work in modern electronics. Given Data / Assumptions: • The base material is pure silicon, which has four valence electrons. • Aluminium is used as the dopant. • Aluminium has three valence electrons (it is trivalent). • The amount of dopant is small compared to the number of silicon atoms. Concept / Approach: In a silicon crystal, each silicon atom forms four covalent bonds with neighbouring silicon atoms. When a trivalent impurity such as aluminium, boron, or gallium is introduced, it has only three valence electrons to form bonds. As a result, one covalent bond remains incomplete, creating a "hole" (an effective positive charge carrier). These holes can move through the lattice when an electric field is applied, and they act as majority carriers. A semiconductor in which holes are the majority carriers is called a p-type semiconductor. Step-by-Step Solution: Step 1: Note that silicon is tetravalent and forms four covalent bonds in its crystal structure. Step 2: Aluminium is trivalent, having only three valence electrons available for bonding. Step 3: When an aluminium atom replaces a silicon atom in the lattice, it forms only three complete bonds, leaving one bond incomplete and creating a hole. Step 4: These holes act as positive charge carriers that can move when an external electric field is applied. Step 5: Since holes become the majority carriers, the doped semiconductor is classified as p-type. Verification / Alternative check: Standard semiconductor physics categorises trivalent dopants (boron, aluminium, gallium, indium) as acceptor impurities because they can accept an electron into the incomplete bond, leaving behind a mobile hole. In contrast, pentavalent dopants (phosphorus, arsenic, antimony) donate extra electrons and produce n-type materials. As aluminium is trivalent, silicon doped with aluminium must behave like a p-type semiconductor, confirming our conclusion. Why Other Options Are Wrong: Option a: Doping significantly changes conductivity by increasing the number of charge carriers; it does not leave conductivity unchanged. Option b: n-type semiconductors are formed by pentavalent dopants that donate extra electrons, not by trivalent aluminium. Option d: Doping with aluminium introduces more charge carriers (holes) and typically reduces resistivity, increasing conductivity. Option e: The doped silicon does not become an insulator; it actually becomes more conductive than pure silicon. Common Pitfalls: A common confusion is mixing up donor and acceptor impurities. To avoid this, remember that trivalent impurities (three valence electrons) cannot fully bond with four neighbours, so they create holes and lead to p-type behaviour. Pentavalent impurities (five valence electrons) have one extra electron and therefore produce n-type behaviour. A quick memory trick is "p-type → positive (holes), n-type → negative (electrons)." Aluminium, with three valence electrons, clearly belongs on the p-type side. Final Answer: When pure silicon is doped with aluminium, it becomes a p-type semiconductor.

Question 10

A voltage V is applied across a single resistor of resistance R, and the power dissipated is P. The same voltage V is now applied across a parallel combination of three equal resistors, each of resistance R. What is the total power dissipated in the…

Accepted Answer

Introduction / Context: This problem examines how electric power dissipation changes when we replace a single resistor with several equal resistors in parallel, while keeping the applied voltage the same. Understanding series and parallel combinations of resistors and how they affect equivalent resistance and power is fundamental in DC circuit analysis and practical electrical design. Given Data / Assumptions: • Case 1: A single resistor of resistance R has voltage V across it and power dissipated P. • Case 2: Three equal resistors, each of resistance R, are connected in parallel across the same voltage V. • Power dissipation in both cases is resistive heating, obeying P = V^2 / R_eq. • All resistors are ohmic and have the same resistance value R. Concept / Approach: When voltage is fixed, the power dissipated in a resistor or network is P = V^2 / R_eq, where R_eq is the equivalent resistance seen by the source. For a single resistor, R_eq = R. For three equal resistors in parallel, the reciprocal of the equivalent resistance is the sum of reciprocals: 1 / R_eq = 1 / R + 1 / R + 1 / R = 3 / R, which gives R_eq = R / 3. Because the equivalent resistance decreases in parallel, the total current drawn increases and therefore total power increases. By comparing P in the two situations, we can find the new power in terms of P. Step-by-Step Solution: Step 1: For the single resistor case, power is P = V^2 / R. Step 2: For three resistors of resistance R in parallel, compute equivalent resistance. Step 3: Use 1 / R_eq = 1 / R + 1 / R + 1 / R = 3 / R. Step 4: Solve for R_eq: R_eq = R / 3. Step 5: In the parallel case, the power dissipated is P_parallel = V^2 / R_eq. Step 6: Substitute R_eq = R / 3 to get P_parallel = V^2 / (R / 3) = 3 * V^2 / R. Step 7: Since P = V^2 / R, we have P_parallel = 3P. Verification / Alternative check: We can check with simple numeric values. Let R = 3 ohm and V = 3 V. In the single resistor case, P = V^2 / R = 9 / 3 = 3 W. For three 3 ohm resistors in parallel, each branch gets 3 V. Power in each resistor is P_branch = 9 / 3 = 3 W. There are three such branches, so total power is 3 W + 3 W + 3 W = 9 W. This is exactly 3 times the original 3 W, confirming that the parallel combination dissipates 3P. Why Other Options Are Wrong: Option a (P): This would mean the parallel network draws the same power as a single resistor, which is not true because equivalent resistance is smaller. Option c (P/3): This suggests power decreases when adding parallel resistors, which contradicts P = V^2 / R_eq and the reduced resistance. Option d (2P/3): This is an arbitrary fraction and does not follow from the series parallel rules or the calculations. Option e (9P): This would correspond to nine times the original power, which requires either nine resistors or different values; three parallel equal resistors give three times, not nine times, the power. Common Pitfalls: A common mistake is to mix up the power formulas P = I^2 * R and P = V^2 / R without paying attention to which quantity (voltage or current) is held constant. In this question, voltage is the same in both cases, so using P = V^2 / R_eq is appropriate. Another error is doing the parallel resistance calculation incorrectly, such as adding resistances instead of their reciprocals. Always remember: series resistances add directly, while parallel resistances use reciprocal addition. Final Answer: With the same applied voltage, the parallel combination of three equal resistors dissipates 3P total power.

Question 11

A parallel-plate capacitor with air between its plates has capacitance C. If the space between the plates is completely filled with a dielectric material of dielectric constant 7, what will be the new capacitance?

Accepted Answer

Introduction / Context: Capacitors store electric charge and energy, and their capacitance depends on geometry and the material between the plates. Replacing air with a dielectric material increases capacitance by a factor equal to the dielectric constant of that material. This concept is crucial in designing capacitors for electronic circuits, tuning circuits, and energy storage applications. Given Data / Assumptions: • Original capacitor: parallel plates with air as dielectric, capacitance C. • New capacitor: the same plates and separation, but the space is filled with a dielectric. • Dielectric constant (relative permittivity) of the material k = 7. • Plate area and separation remain unchanged. Concept / Approach: For a parallel-plate capacitor, capacitance with a dielectric fully filling the space between plates is given by C = k * epsilon0 * A / d, where k is the dielectric constant, epsilon0 is permittivity of free space, A is plate area, and d is plate separation. When air (k ≈ 1) is replaced by a dielectric with constant k, the capacitance increases by the factor k, assuming air is approximated as vacuum. Thus, if the original capacitance was C with air, the new capacitance with dielectric becomes k * C. Step-by-Step Solution: Step 1: Write the capacitance with air (or vacuum): C_air = epsilon0 * A / d. Step 2: The given capacitance C corresponds to C_air. Step 3: When a dielectric with constant k = 7 fully fills the space, capacitance becomes C_dielectric = k * epsilon0 * A / d. Step 4: Substitute C_air into this expression: C_dielectric = k * C_air = 7 * C. Step 5: Therefore, the new capacitance is 7C. Verification / Alternative check: If C = 2 microfarad with air, filling the space with a dielectric of k = 7 would give C_dielectric = 7 * 2 microfarad = 14 microfarad. This simple scaling behaviour is consistent with theory and with experimental use of dielectric materials to build high capacitance devices in smaller volumes. Why Other Options Are Wrong: Option a (C): This would mean the dielectric has no effect, which contradicts the basic definition of dielectric constant. Option b (C/7): This implies capacitance decreases, which is opposite to what happens when a dielectric is inserted. Option d (14C): This would require an effective dielectric constant of 14, not 7. Option e (C/14): This is also a decrease and has no basis in the formula for capacitance with a dielectric. Common Pitfalls: Students sometimes confuse relative permittivity (dielectric constant) with absolute permittivity epsilon0 or misremember whether the capacitance increases or decreases. A helpful reminder is that inserting a dielectric material increases the ability of the capacitor to store charge at the same voltage, so capacitance increases by the factor k. Be careful not to confuse this with effects on electric field strength or energy density, which involve related but different formulas. Final Answer: The capacitance of the capacitor becomes 7C when the dielectric with k = 7 completely fills the space between the plates.

Question 12

For which of the following objects does the centre of mass lie outside the material of the body itself?

Accepted Answer

Introduction / Context: The centre of mass of a body is a point at which the entire mass of the body may be considered to be concentrated for translational motion analysis. For many simple solids, the centre of mass lies inside the material. However, for some hollow or ring-shaped bodies, the centre of mass can lie in empty space. Recognising such cases is helpful for understanding rotational motion, stability, and balancing of objects. Given Data / Assumptions: • We compare several everyday objects: a pen, a cricket ball, a ring, a book, and a solid cube. • For each object, we consider its standard, uniform mass distribution. • The question asks where the centre of mass lies with respect to the material of the object. Concept / Approach: For a uniform solid body, such as a cricket ball, book, or solid cube, the centre of mass lies at the geometric centre, which is inside the material. For a ring or thin circular hoop, the mass is distributed along a circle and there is essentially empty space in the middle. The geometric centre of this circle is a point in space that has no material but is still the location where the mass distribution balances in all directions. This point is the centre of mass and lies outside the material of the ring itself. Step-by-Step Solution: Step 1: Consider a cricket ball, book, fountain pen, and solid cube. Each is mostly solid and filled, so their centre of mass is somewhere inside the material. Step 2: For uniform objects, the centre of mass usually coincides with the geometric centre; for the ball, cube, and book, the geometric centre is clearly within the object. Step 3: A fountain pen has an irregular shape but its centre of mass will still lie somewhere within the physical body, provided mass distribution is typical. Step 4: For a ring or thin hoop, the mass lies around a circular path, leaving an empty region at the centre. Step 5: The centre of mass of a uniform ring lies at the geometric centre of the circle, i.e., in the empty space inside the ring. Step 6: Therefore, among the given options, only the ring has its centre of mass outside the material of the body. Verification / Alternative check: If you support a uniform ring at a single point on its circumference and let it hang freely, it will orient so that the lowest point on the ring lies directly below its centre of mass. Observations show that the centre of each circle (a point with no material) is vertically above the point of support, indicating that the centre of mass is at this geometric centre in empty space. No such behaviour is seen for solid objects like a cricket ball or cube, where the centre of mass is inside the material. Why Other Options Are Wrong: Option a (A fountain pen): Although its mass distribution may be uneven, the centre of mass lies somewhere inside the pen body. Option b (A cricket ball): Being nearly spherical and solid, its centre of mass lies at its geometric centre inside the ball. Option d (A book): A book is a solid rectangular body whose centre of mass lies inside, at the intersection of its diagonals. Option e (A solid cube): Like the book, the cube's centre of mass is inside the material, at the geometric centre. Common Pitfalls: Beginners sometimes think the centre of mass must always lie within the physical material, but that is not true for hollow objects like rings and thin shells. Other examples where the centre of mass may lie outside the material include boomerangs and certain bent rods. The key idea is that centre of mass depends on mass distribution, not merely on the shape boundary. Final Answer: Among the given options, the object whose centre of mass lies outside the material is A ring.

Question 13

Two parallel-plate capacitors C1 and C2 are made using plates of gold and aluminium respectively, but all four plates have the same area and the same plate separation with air between them. If rho_g and rho_a are the electrical resistivities of gold…

Accepted Answer

Introduction / Context: Capacitance of a parallel-plate capacitor depends primarily on the geometry of the plates and the dielectric material between them. It does not depend on which conducting material the plates themselves are made of, as long as the plates are good conductors. This question checks whether you can separate the role of conductor properties (like resistivity) from the factors that determine capacitance. Given Data / Assumptions: • Capacitor C1: made with two gold plates. • Capacitor C2: made with two aluminium plates. • All four plates have the same area A. • Plate separation d is the same for both capacitors. • The space between plates is air (same dielectric). • rho_g and rho_a are electrical resistivities of gold and aluminium. Concept / Approach: For a parallel-plate capacitor with air (or vacuum) as dielectric, capacitance is given by C = epsilon0 * A / d, where epsilon0 is permittivity of free space, A is area of each plate, and d is the plate separation. This formula does not involve the resistivity or conductivity of the plate material. As long as the plates are conductors, the electric field between them and the capacitance are determined solely by geometry and dielectric properties. Because both capacitors have the same plate area, same separation, and same dielectric, their capacitances must be equal, regardless of whether the plates are gold or aluminium. Step-by-Step Solution: Step 1: Write the capacitance formula for a parallel-plate capacitor: C = epsilon0 * A / d for air as dielectric. Step 2: For capacitor C1 (gold plates), C1 = epsilon0 * A / d. Step 3: For capacitor C2 (aluminium plates), C2 = epsilon0 * A / d. Step 4: Because A and d are identical and dielectric is air in both cases, C1 and C2 must be equal. Step 5: Therefore, C1 = C2, independent of rho_g and rho_a. Verification / Alternative check: Imagine gradually replacing gold with aluminium while keeping the same shape and size of plates and the same gap. The electric field in the dielectric region depends on how charges arrange on the surfaces, but both gold and aluminium are excellent conductors and will distribute charges on their surfaces to maintain equipotential conditions. As long as they behave as ideal conductors, the capacitance remains unchanged. Practical capacitors often use aluminium foils instead of more expensive metals because capacitance is determined by geometry and dielectric, not by plate resistivity. Why Other Options Are Wrong: Option a (C1 > C2) and option b (C2 > C1): These suggest that plate material changes capacitance, which is not true when geometry and dielectric are identical. Option c (C1 * rho_a = C2 * rho_g): This incorrectly mixes resistivity with capacitance; there is no such direct relation in basic capacitor theory. Option e (C1 / C2 = rho_g / rho_a): Again, this implies capacitance depends on resistivity ratio, which is not the case. Common Pitfalls: Students often confuse properties of conductors relevant to resistance (which involves resistivity and length) with properties relevant to capacitance. While resistivity affects how much a conductor heats up or how much voltage drop occurs along its length, capacitance in a parallel-plate configuration depends only on plate area, separation, and dielectric constant. Remember: geometry and dielectric determine capacitance; material resistivity affects resistance, not capacitance directly. Final Answer: Because both capacitors have the same geometry and dielectric, their capacitances are equal, so C1 = C2.

Question 14

A photon of X-ray has energy 1 keV, while a photon of visible light has energy 3 eV. In this context, which one of the following statements is NOT correct?

Accepted Answer

Introduction / Context: Different regions of the electromagnetic spectrum, such as X rays and visible light, differ in energy, frequency, and wavelength but share some common properties in vacuum. This question compares an X-ray photon and a visible photon, asking which statement about their properties is not correct. Understanding the relationships between energy, frequency, wavelength, and speed for photons is fundamental in modern physics and electromagnetic theory. Given Data / Assumptions: • X-ray photon energy: 1 keV (1000 eV). • Visible photon energy: 3 eV. • Both are electromagnetic waves and photons. • We consider their behaviour in vacuum. Concept / Approach: For a photon, energy E is related to frequency f by E = h * f, where h is Planck constant. Because 1 keV is much greater than 3 eV, the X-ray photon has much higher frequency. Wave speed v in a given medium is related to frequency and wavelength by v = f * lambda. In vacuum, all electromagnetic waves travel at the same speed, the speed of light c. Therefore, if energy and frequency increase, wavelength must decrease to keep v constant. This means X rays have shorter wavelengths than visible light, and they are generally more penetrating. The only statement that contradicts these principles is any claim that their speeds in vacuum are different. Step-by-Step Solution: Step 1: Compare energies: E_X-ray = 1 keV = 1000 eV, E_visible = 3 eV. So the X-ray photon has much higher energy. Step 2: Use E = h * f. Higher energy implies higher frequency. So f_X-ray > f_visible. Step 3: Use v = f * lambda. In vacuum, v = c for both photons. Step 4: With the same speed c and higher frequency, the X-ray photon must have shorter wavelength than the visible photon: lambda_X-ray < lambda_visible. Step 5: Because both are electromagnetic waves in vacuum, their speeds are identical: v_X-ray = v_visible = c. Step 6: Therefore, any statement claiming different speeds in vacuum is not correct. Verification / Alternative check: Electromagnetic theory and many experiments show that all electromagnetic waves travel at the same speed c in vacuum, approximately 3 * 10^8 m/s, regardless of frequency or energy. This is true for radio waves, microwaves, infrared, visible, ultraviolet, X rays, and gamma rays. Differences arise in frequency, wavelength, and interaction with matter, not in their speed in vacuum. So the idea that their speeds in vacuum are different is definitely incorrect. Why Other Options Are Correct (and therefore not the answer): Option a: With higher energy and frequency, X rays have shorter wavelength than visible light; this is a standard fact of the spectrum. Option b: By definition, 1 keV and 3 eV are very different energies, so the photons do have different energies. Option d: Higher energy implies higher frequency via E = h * f, so the X-ray photon frequency is higher than that of visible light. Option e: X-ray photons interact differently with matter and are generally far more penetrating than visible photons, which are easily absorbed or scattered. Common Pitfalls: Students sometimes incorrectly believe that higher energy photons must travel faster, but for electromagnetic waves in vacuum, speed is fixed at c. The extra energy appears as higher frequency (and shorter wavelength), not increased speed. Another confusion is mixing up behaviour in matter versus vacuum; speeds can differ in materials, but in vacuum all electromagnetic waves share the same speed. Final Answer: The incorrect statement is: The speeds of the two photons in vacuum are different.

Question 15

Jet streams are upper-level tropospheric winds. Which one of the following statements about jet streams is NOT correct?

Accepted Answer

Introduction / Context: Jet streams are fast flowing, narrow air currents found in the upper levels of the atmosphere, particularly near the tropopause. They play a crucial role in weather patterns, storm tracks, and aviation. Understanding their typical characteristics, such as location, continuity, and seasonal variation, is important in both meteorology and general geography. This question asks you to identify which statement about jet streams is not correct. Given Data / Assumptions: • Jet streams are upper-level winds, especially in the troposphere near the tropopause. • They are associated with strong horizontal temperature gradients, often near frontal zones. • We focus on the polar front jet stream as a typical example. • We consider seasonal variations between winter and summer. Concept / Approach: Jet streams are relatively narrow bands of high-speed winds that flow from west to east along wave-like paths. They are typically located near the tropopause at altitudes roughly between 8 km and 15 km. These currents can extend over thousands of kilometres, making them continuous over long distances, though their position and strength vary. The polar front jet stream is stronger in winter when the temperature contrast between polar and mid-latitude air masses is greatest. In summer, this temperature contrast weakens, and the polar front jet generally becomes weaker and shifts poleward. Therefore, any statement claiming that the polar front jet reaches maximum strength in summer is not correct. Step-by-Step Solution: Step 1: Confirm that jet streams are narrow, high-velocity wind bands; this is a standard definition. Step 2: Recognise that they occur near the tropopause, typically between 8 km and 15 km altitude, and follow wave-like (meandering) paths. Step 3: Note that jet streams are often continuous over long distances, sometimes encircling the globe in segments. Step 4: Remember that the polar front jet is strongest in winter due to the large temperature gradient between polar and mid-latitude air. Step 5: In summer, this temperature gradient is weaker, so the polar front jet stream usually weakens instead of strengthening. Step 6: Thus, the statement that the polar front jet achieves its maximum force (strength) in summer is not correct. Verification / Alternative check: Meteorological data and textbooks describe how jet streams shift and change strength with the seasons. The polar front jet intensifies in winter in both hemispheres, while subtropical jets behave differently. Airline flight planning also uses stronger winter jet streams for faster eastbound flights. The seasonal pattern is therefore opposite of what the incorrect statement suggests, confirming that summer is not when the polar front jet is at its maximum strength. Why Other Options Are Correct (and therefore not the answer): Option a: Jet streams are indeed narrow bands of very high-speed winds compared to surrounding airflows. Option b: They run near the tropopause and follow meandering, wave-like paths influenced by large-scale atmospheric waves. Option c: Jet streams typically extend over thousands of kilometres and can be considered continuous over long stretches. Option e: They are associated with strong horizontal temperature gradients, especially around fronts and the polar jet region. Common Pitfalls: A frequent confusion is mixing up which season has stronger jet streams. Because winter has larger temperature differences between air masses, the polar front jet is stronger in winter, not summer. Another mistake is thinking of jet streams as short or local winds, when they actually extend over large distances. Keeping the association between strong temperature gradients and stronger jets helps avoid these errors. Final Answer: The incorrect statement is: In summer, the polar front jet achieves its maximum strength.

Question 16

Which of the following phenomena most clearly demonstrates the particle nature of light?

Accepted Answer

Introduction / Context: Light exhibits both wave-like and particle-like behaviour, a concept known as wave particle duality. Different experiments highlight different aspects of this duality. Wave phenomena such as interference and diffraction show light's wave nature, while some effects, like the photoelectric effect, strongly support the idea that light is made of discrete energy packets called photons. This question asks you to identify which phenomenon most directly demonstrates the particle nature of light. Given Data / Assumptions: • We consider several optical phenomena: diffraction, interference, photoelectric effect, polarisation, and refraction. • We assume standard textbook interpretations of these experiments. • We focus on which effect requires light to be quantised into photons to be explained correctly. Concept / Approach: The photoelectric effect involves the emission of electrons from a metal surface when light of sufficient frequency falls on it. Classical wave theory could not explain why there is a threshold frequency, why increasing intensity below this threshold does not cause emission, and why emitted electrons appear almost instantaneously. Einstein explained the effect by proposing that light comes in discrete quanta (photons) with energy E = h * f. Only if a single photon has enough energy to overcome the metal work function can it eject an electron. This explanation strongly supports the particle nature of light. Step-by-Step Solution: Step 1: Recall that diffraction and interference are classic wave phenomena, where waves overlap and produce patterns of maxima and minima. Step 2: Polarisation shows that light is a transverse wave with oscillations in specific directions, again supporting wave behaviour. Step 3: Refraction is the bending of light when it passes between media of different optical densities and can be explained using wavefronts or ray models. Step 4: The photoelectric effect, however, involves discrete emission of electrons depending on light frequency, not just intensity. Step 5: Einstein's photon model, where light energy is quantised as photons, successfully explains the existence of a threshold frequency and the immediate emission of electrons. Step 6: Therefore, the phenomenon that clearly demonstrates the particle nature of light is the photoelectric effect. Verification / Alternative check: Experiments show that increasing the intensity of light below a certain frequency does not cause photoemission, which contradicts classical wave expectations that more energy should eventually knock out electrons. However, if light is treated as photons, each with energy proportional to frequency, then light below threshold frequency has photons that individually lack the energy to liberate electrons, no matter how many are present. Only when photon energy exceeds the work function do electrons get emitted. This discrete, packet based behaviour is distinctly particle like. Why Other Options Are Wrong: Option a (Diffraction): Diffraction patterns arise from wave bending around obstacles and support the wave nature of light. Option b (Interference): The formation of interference fringes in double slit experiments demonstrates the superposition of waves. Option d (Polarisation): Polarisation shows that light has transverse oscillations and is a wave property, not a particle specific phenomenon. Option e (Refraction): Refraction can be described using wavefronts and Snell's law and does not by itself demand a particle model. Common Pitfalls: Learners sometimes think that any "special" optical effect must be evidence of particle behaviour, but many of the famous experiments (interference, diffraction, polarisation) actually confirm the wave model. The key is to identify which effect could not be explained by classical wave theory alone. The photoelectric effect is historically and conceptually the most important evidence for quantisation of light into photons and thus for its particle nature. Final Answer: The phenomenon that most clearly shows the particle nature of light is the Photoelectric effect.

Question 17

A hydraulic lift used in garages and workshops works based on which fundamental principle of fluid mechanics?

Accepted Answer

Introduction / Context: Hydraulic lifts are widely used to raise heavy vehicles and loads with relatively small applied forces. They use an incompressible fluid and different piston areas to multiply force. The basic principle underlying their operation is a key law of fluid mechanics that relates pressure in an enclosed fluid. Recognising this principle is important for understanding many hydraulic devices such as brakes, jacks, and presses. Given Data / Assumptions: • The device is a hydraulic lift, using a fluid in a closed system. • A small force is applied on a small piston, generating pressure in the fluid. • This pressure is transmitted to a larger piston supporting a heavy load. • The fluid is assumed incompressible and at rest (no significant flow). Concept / Approach: Pascal's law states that when pressure is applied to a confined fluid, that pressure is transmitted undiminished in all directions throughout the fluid and to the walls of the container. In a hydraulic lift, a small input force on a small-area piston produces a pressure P = F1 / A1. This same pressure acts on a larger-area piston, generating a larger output force F2 = P * A2. Because A2 is much greater than A1, F2 can be much larger than F1, enabling the lift to raise heavy loads with relatively little effort. This is the exact working principle of hydraulic machines. Step-by-Step Solution: Step 1: Use Pascal's law: pressure applied to an enclosed fluid is transmitted equally in all directions. Step 2: For the input piston, pressure P = F1 / A1, where F1 is the applied force and A1 is the piston area. Step 3: The same pressure P is transmitted through the fluid to the output piston. Step 4: On the output piston, the force is F2 = P * A2 = (F1 / A1) * A2. Step 5: If A2 > A1, then F2 > F1, providing mechanical advantage and allowing heavy loads to be lifted. Step 6: This use of pressure transmission and force multiplication is exactly the application of Pascal's law. Verification / Alternative check: Real-world hydraulic systems such as car jacks, brake systems, and industrial presses all rely on the same principle: fluid pressure is transmitted uniformly. If Newton's laws alone were sufficient, the fluid step of equal pressure transmission would be missing. Engineers design hydraulic systems by choosing suitable piston areas so that Pascal's law leads to the required force multiplication. This confirms that Pascal's law is the foundation of hydraulic lifts. Why Other Options Are Wrong: Option a (Newton's law): While Newton's laws describe motion and forces generally, they do not specifically explain the pressure transmission in fluids that hydraulic lifts rely on. Option c (Archimedes' principle): This principle deals with buoyant force on bodies immersed in a fluid and does not explain force multiplication in hydraulic machines. Option d (Joule's law): This relates to heat produced by electric current in a conductor and is unrelated to hydraulics. Option e (Bernoulli's principle): Bernoulli's principle applies to moving fluids and relates pressure to fluid speed and elevation; hydraulic lifts involve mostly static fluids. Common Pitfalls: Students sometimes confuse Archimedes' principle with Pascal's law because both concern fluids. A simple way to distinguish them is: Archimedes' principle explains why objects float or sink (buoyancy), while Pascal's law explains how pressure in a confined fluid is transmitted and used to multiply forces. For hydraulic lifts, always think of Pascal's law and equal pressure transmission. Final Answer: A hydraulic lift works on Pascal's law, which states that pressure applied to a confined fluid is transmitted equally in all directions.

Question 18

What is the process called in which a substance changes directly from the solid state to the gaseous state without passing through the liquid state?

Accepted Answer

Introduction / Context: Matter can change from one physical state to another when energy is absorbed or released. Common changes include melting, freezing, evaporation, and condensation. Some substances, however, can change directly from solid to gas without becoming liquid in between. Recognising the correct name for this process is important in basic physics and chemistry, especially when studying phase changes and phase diagrams. Given Data / Assumptions: • The change described is from solid directly to gaseous state. • There is no intermediate liquid state in this process. • Examples include substances like dry ice (solid carbon dioxide) and naphthalene under appropriate conditions. Concept / Approach: The direct transition from solid to gas is known as sublimation. During sublimation, molecules at the surface of a solid gain enough energy to break free from the solid structure and enter the gaseous phase, bypassing the liquid state. This usually occurs when the substance has a relatively high vapour pressure at temperatures below its melting point or when pressure conditions favour the gas phase. Evaporation is the change from liquid to gas, condensation is gas to liquid, distillation is a separation technique based on different boiling points, and fusion (melting) is solid to liquid. Step-by-Step Solution: Step 1: Identify the initial and final states: solid to gas. Step 2: Recall that the term for solid to liquid is melting or fusion and for liquid to gas is evaporation or boiling. Step 3: Recognise that the special case where solid directly becomes gas is called sublimation. Step 4: Compare with other terms: condensation is gas to liquid, distillation is a purification method, and evaporation is liquid to gas. Step 5: Therefore, the correct term for direct solid to gas change is sublimation. Verification / Alternative check: Consider dry ice (solid CO2) at atmospheric pressure. It does not melt into liquid CO2 under normal conditions; instead, it directly forms CO2 gas, producing the foggy effect commonly seen in stage performances. This is a textbook example of sublimation. Similarly, iodine crystals can slowly turn into iodine vapour, and naphthalene balls gradually disappear from cupboards as they sublimate. These familiar examples confirm that the process is known as sublimation. Why Other Options Are Wrong: Option a (Evaporation): This is the process where a liquid changes to gas at temperatures below its boiling point; it does not start from the solid state. Option b (Condensation): This is gas changing to liquid, the opposite direction of what is described. Option d (Distillation): This is a separation or purification technique involving boiling and condensation, not a fundamental phase change name. Option e (Fusion): Fusion or melting is the change from solid to liquid, not directly to gas. Common Pitfalls: Learners sometimes confuse sublimation with evaporation because both involve a transition to the gaseous state. The key difference is the starting state: sublimation starts from a solid, while evaporation starts from a liquid. It is also important not to mix up condensation (gas to liquid) with freezing (liquid to solid). Paying attention to the initial and final phases in the question helps keep the terms straight. Final Answer: The direct conversion of a solid into a gas without passing through the liquid state is called Sublimation.

Question 19

In thermodynamics, supercooling refers to the cooling of a liquid below its normal freezing point without it actually solidifying.

Accepted Answer

Introduction / Context: Supercooling is an important concept in thermodynamics and phase change physics. It describes a situation in which a liquid is cooled below its normal freezing point but does not immediately change into a solid. Understanding supercooling helps students connect ideas about freezing, melting, and the role of impurities and disturbances in starting crystallisation. This question checks whether you can distinguish between simple cooling, freezing at the normal freezing point, and the special phenomenon of cooling below that point without solidification. Given Data / Assumptions: A substance is in the liquid state. The liquid is cooled gradually. The focus is on the temperature region around its freezing or melting point. No external disturbance or impurity is assumed that would trigger immediate crystallisation. Concept / Approach: Every pure substance has a characteristic freezing point at a given pressure. At this temperature, the liquid normally begins to change into a solid. However, if the liquid is very pure and undisturbed, it can be cooled below this freezing point and still remain liquid. This unusual state is called supercooled. The key idea is that freezing requires not only low temperature but also nucleation sites where crystals can begin to form. Without such nucleation, the liquid can exist in a metastable state below its usual freezing point. Step-by-Step Solution: Step 1: Recall that the freezing point is the temperature at which a liquid normally starts to solidify at a given pressure. Step 2: Supercooling describes a liquid that has been cooled below this freezing point but has not yet formed a solid phase. Step 3: The phrase that correctly captures this behaviour is “cooling a liquid below its freezing point without it solidifying”. Step 4: Other descriptions that involve melting point or immediate solidification do not match the precise definition of supercooling. Verification / Alternative check: A common classroom demonstration of supercooling is with very pure water. Under ideal conditions, water can be cooled below 0 degree Celsius while still remaining liquid. When the supercooled water is disturbed or a crystal seed is introduced, it suddenly freezes. This confirms that supercooling refers to the liquid state persisting below the freezing point and not simply cooling to or slightly around that temperature. Why Other Options Are Wrong: Cooling a liquid below its melting point and immediately solidifying it: This describes normal freezing, not supercooling, because the liquid does not remain liquid below the freezing point. Cooling a liquid exactly at its melting or freezing point: This only refers to the phase change temperature itself, not to temperatures below it while staying liquid. Heating a solid above its melting point until it becomes a liquid: This describes melting, the opposite phase change, and is unrelated to supercooling. Common Pitfalls: Students often confuse melting point and freezing point, since for a pure substance at standard pressure they have the same numerical value. Another mistake is thinking that as soon as the liquid goes below the freezing point it must instantly become solid. In reality, the formation of a solid phase requires nucleation and growth, and without these, supercooling can occur. It is also important not to mix supercooling with superheating, which refers to heating a liquid above its boiling point without boiling. Final Answer: The correct description is cooling a liquid below its freezing point without it solidifying.

Question 20

In geometrical optics, the angle of incidence is defined as the angle between the incident ray of light and the normal drawn to the surface at the point where the ray strikes.

Accepted Answer

Introduction / Context: In geometrical optics, several angles are defined to describe how light interacts with surfaces such as plane mirrors, lenses, and transparent boundaries. One of the most fundamental is the angle of incidence, which appears in laws such as the law of reflection and Snell law of refraction. This question checks whether you remember the correct geometric definition of the angle of incidence. Given Data / Assumptions: A ray of light (incident ray) strikes a surface, such as a mirror or glass boundary. A normal line is drawn at the point where the ray meets the surface. The normal is a line perpendicular to the surface at that point. Concept / Approach: By definition, the angle of incidence is the angle between the incident ray and the normal to the surface at the point of incidence. It is not measured with respect to the surface itself. Measuring with respect to the normal gives a consistent framework for both reflection and refraction, and is used in formulas such as Snell law (n1 sin i = n2 sin r), where i is the angle of incidence and r is the angle of refraction. Step-by-Step Solution: Step 1: Identify the incident ray, which is the ray of light approaching and striking the surface. Step 2: Draw the normal, which is a line perpendicular to the surface at the point of incidence. Step 3: Define the angle of incidence as the angle between the incident ray and this normal line. Step 4: Compare this definition with the given options and choose the one that mentions “normal to the surface and the incident ray”. Verification / Alternative check: The law of reflection states that the angle of incidence equals the angle of reflection, and both of these angles are measured with respect to the normal. Textbooks and standard diagrams always show the incident ray, reflected ray, and normal forming a plane, and the angles are marked between the rays and the normal. This consistent treatment confirms that the correct definition uses the normal, not the surface or tangent. Why Other Options Are Wrong: Surface of the mirror and the incident ray: This would measure the angle from the surface, not from the normal, and is not the standard definition used in optics. Tangent to the surface and the incident ray: A tangent is parallel to the surface, so this option again effectively measures from the surface, not from the normal. Reflected ray and the incident ray: The angle between these two rays is sometimes of interest, but it is not called the angle of incidence. It is related but distinct. Common Pitfalls: Many learners mistakenly think angles are always measured from the surface because it feels more intuitive visually. However, using the normal simplifies the mathematical treatment of both reflection and refraction. Confusing the angle of incidence with the angle between incident and reflected rays can also cause errors when applying formulas. Always remember to draw the normal and measure angles from that line. Final Answer: The angle of incidence is the angle between the normal to the surface and the incident ray.

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