Question 1

A man stands on the floor of a lift that accelerates upwards; how does the normal reaction on his feet compare with his weight?

Accepted Answer

Introduction / Context: Situations involving people standing in accelerating lifts are standard examples used in Newtonian mechanics to illustrate how apparent weight changes with acceleration. The normal reaction force from the lift floor is what a weighing machine would read, and it represents how heavy the person feels. This question asks what happens to that reaction force when the lift accelerates upwards. Given Data / Assumptions: A man of mass m stands on the floor of a lift. The lift accelerates upwards with uniform acceleration a. The acceleration due to gravity is g, directed downward. We neglect the mass of the lift and any friction in the cables. Concept / Approach: The key idea is to apply Newton second law to the man in the non accelerating frame of the ground. The forces on the man are his weight mg acting downward and the normal reaction N from the floor acting upward. When the lift accelerates upward with acceleration a, the man also accelerates upward with acceleration a. Newton second law then gives N - mg = m * a. Solving for N shows that N = m(g + a), which is clearly greater than mg as long as a is positive. Thus, the reaction force is greater than the man true weight and he feels heavier. Step-by-Step Solution: Step 1: Draw a free body diagram for the man: weight mg downward, normal reaction N upward. Step 2: The lift and the man accelerate upward with acceleration a, so the man acceleration is upward. Step 3: Apply Newton second law in the vertical direction: net force upward = m * a. Step 4: The net upward force is N - mg, so N - mg = m * a. Step 5: Rearrange to obtain N = mg + m * a = m(g + a). Step 6: Since a is positive for upward acceleration, g + a is greater than g, so N is greater than mg. Verification / Alternative check: If the acceleration a is zero, the lift moves at constant speed or is at rest, and N becomes mg, which matches the usual weight. If the lift accelerates upwards with a small a, the reaction becomes slightly larger than mg, matching the feeling of being pressed harder against the floor. In the extreme case where the acceleration is equal to g upwards, N would be 2mg, so the man would feel twice as heavy. These limits are consistent with the formula N = m(g + a) and confirm that N is greater than mg when a is positive. Why Other Options Are Wrong: Less than the weight of the man: This would occur if the lift accelerates downward, not upward. Equal to the weight of the man: This only holds when the lift has zero acceleration, that is, when it is at rest or moving with constant velocity. Zero, as if the man were weightless: This would correspond to free fall, where the lift accelerates downward with acceleration equal to g, not to an upward accelerating lift. Common Pitfalls: Learners sometimes confuse apparent weight with true weight and forget to consider the direction of acceleration. Another frequent mistake is to write N = mg - m * a regardless of direction, without carefully setting up the sign convention. The safe method is always to draw a free body diagram, define upward as positive and then apply Newton second law systematically. Doing this consistently makes it clear that upward acceleration increases the normal reaction above mg. Final Answer: When the lift accelerates upwards, the reaction force on the floor is greater than the weight of the man.

Question 2

In SI units, what is the derived unit used for measuring the coefficient of viscosity of a fluid?

Accepted Answer

Introduction / Context: Viscosity is a fundamental property of fluids that measures their resistance to flow. Thick fluids such as honey have high viscosity, while thin fluids such as water and alcohol have low viscosity. In physics and engineering, we often work with the coefficient of dynamic viscosity and need to express it in standard SI units. This question tests whether you can correctly identify the SI unit associated with viscosity. Given Data / Assumptions: The quantity of interest is the coefficient of dynamic viscosity. We are using the SI (International System of Units). Candidate units include coulomb, newton second per square metre, watt per metre per degree Celsius and joule per kilogram per kelvin. We assume standard definitions of force, energy and thermal quantities. Concept / Approach: Dynamic viscosity is defined through the relationship between shear stress and velocity gradient in a fluid. In simple form, shear stress tau is proportional to viscosity eta times velocity gradient, so tau = eta * (du/dy). Shear stress has units of force per area (newton per square metre), while velocity gradient has units of per second. Therefore viscosity has units of shear stress divided by velocity gradient, which gives (newton per square metre) / (per second) = newton second per square metre. This is equivalent to the pascal second, because 1 pascal is 1 newton per square metre. Step-by-Step Solution: Step 1: Recall the definition tau = eta * (du/dy), where tau is shear stress and eta is viscosity. Step 2: Identify the units of tau as newton per square metre (N·m^-2), also called the pascal (Pa). Step 3: Identify the units of velocity gradient du/dy as (metre per second) / metre = per second (s^-1). Step 4: Solve for viscosity units: eta has units of tau divided by velocity gradient. Step 5: Divide units: (N·m^-2) / (s^-1) = N·s·m^-2, which is newton second per square metre. Step 6: Conclude that the SI unit of viscosity is newton second per square metre, also written as pascal second. Verification / Alternative check: Standard tables of physical constants list the viscosity of water at room temperature as around 1.0 * 10^-3 Pa·s. Since 1 Pa is 1 N·m^-2, this is the same as 1.0 * 10^-3 N·s·m^-2. No other units in the options match this standard description. Coulomb is the SI unit of electric charge, watt per metre per degree Celsius is a unit of thermal conductivity and joule per kilogram per kelvin is a unit of specific heat capacity, not viscosity. Why Other Options Are Wrong: Coulomb: This is the unit of electric charge and is unrelated to fluid flow and shear stress. Watt per metre per degree Celsius: This unit measures thermal conductivity, describing heat flow rather than resistance to fluid motion. Joule per kilogram per kelvin: This unit is used for specific heat capacity, describing how much energy is needed to raise the temperature of a unit mass. Common Pitfalls: A common confusion arises because many material properties, such as thermal conductivity and specific heat, also have compound SI units. Students may remember the form of one property and incorrectly apply it to another. To avoid this, always start from the defining equation. For viscosity, use tau = eta * (du/dy) and work through the unit analysis from shear stress and velocity gradient. This approach clearly leads to newton second per square metre as the correct unit. Final Answer: In SI units, the coefficient of viscosity is measured in newton second per square metre (N·s·m^-2).

Question 3

In astronomy, which of the following is commonly used as a unit for measuring large astronomical distances between stars and galaxies?

Accepted Answer

Introduction / Context: Distances in astronomy are so enormous that ordinary units like kilometres quickly become inconvenient. Astronomers therefore use special distance units that are better suited to interstellar and intergalactic scales. One of the most widely known of these is the light year. This question asks you to identify which option represents a unit of large astronomical distance, not a unit of wavelength, magnetic flux or illumination. Given Data / Assumptions: We are interested in large distances in space, such as between stars or galaxies. Options include light year, angstrom, weber and lux. Each of these is a real physical unit, but they measure different types of quantities. We assume standard scientific definitions of these units. Concept / Approach: A light year is defined as the distance that light travels in vacuum in one year. Since light travels at about 3 * 10^8 metres per second, the distance covered in a year is roughly 9.46 * 10^15 metres. This huge distance makes the light year a convenient unit for describing how far away stars and galaxies are. In contrast, an angstrom is a very small unit of length used mainly for atomic scale distances, the weber is a unit of magnetic flux and the lux is a unit of illumination or illuminance. Step-by-Step Solution: Step 1: Recall that a light year is the distance travelled by light in one year in vacuum. Step 2: Estimate that this distance is around 9.46 * 10^15 metres, suitable for interstellar scales. Step 3: Identify that astronomers commonly quote distances to nearby stars in light years. Step 4: Recognise that an angstrom is 10^-10 metre, far too small for galaxy scale distances. Step 5: Note that a weber measures magnetic flux and lux measures light falling on a surface, not distance. Step 6: Conclude that among the given options, only the light year is used as a unit of large astronomical distance. Verification / Alternative check: If you look up the distance to well known stars such as Proxima Centauri, you will find values like 4.2 light years. Distances to galaxies are often given in millions of light years. This consistent use of light years in astronomical references confirms that it is a distance unit. By contrast, angstrom units are used when discussing spectral lines and atomic sizes, not galaxy separations. Textbooks on electromagnetism use webers to measure magnetic flux, while lighting engineers use lux to describe how bright a surface is. Why Other Options Are Wrong: Angstrom: This is a unit of length equal to 10^-10 metre, used for atomic and molecular scales, not for astronomical distances. Weber: This is the SI unit of magnetic flux and is unrelated to distance measurement. Lux: This is the SI unit of illuminance, describing how much light falls on a unit area, not a measure of distance. Common Pitfalls: Sometimes students see the word light in lux and assume it must be related to light travel like a light year. In reality, lux quantifies how bright a surface appears, not how far away something is. Another confusion is equating any obscure sounding unit with astronomy. To avoid this, remember the simple definition: a light year is literally the distance light travels in one year, which directly ties it to distance in space. Final Answer: The commonly used unit of large astronomical distance among the options is the light year.

Question 4

What is the approximate minimum velocity, known as escape velocity, that a rocket must have to overcome Earth gravity and travel into space?

Accepted Answer

Introduction / Context: Escape velocity is an important concept in gravitational physics and space science. It represents the minimum speed that an object must have at the surface of a massive body, such as the Earth, so that it can move infinitely far away without falling back under gravity, assuming no additional propulsion and neglecting air resistance. This question asks for the approximate numerical value of Earth escape velocity expressed in kilometres per second. Given Data / Assumptions: We are discussing escape velocity from the surface of the Earth. Air resistance and atmospheric drag are neglected. The rocket is assumed to move away without further thrust after reaching escape speed. We use the standard formula for escape velocity from a spherical mass. Concept / Approach: Escape velocity v_esc from the surface of a spherical body of mass M and radius R is given by v_esc = sqrt(2 * G * M / R), where G is the universal gravitational constant. For Earth, substituting standard values of G, M and R gives a value of approximately 11.2 kilometres per second. This means any object given this speed at Earth surface and directed outward can theoretically escape Earth gravitational field without additional propulsion, if we ignore the atmosphere. Step-by-Step Solution: Step 1: Write the formula for escape velocity: v_esc = sqrt(2 * G * M / R). Step 2: Insert typical values for Earth: G ≈ 6.67 * 10^-11 N·m^2·kg^-2, M ≈ 5.97 * 10^24 kg, R ≈ 6.37 * 10^6 m. Step 3: Compute the quantity inside the square root: 2 * G * M / R gives a value on the order of 1.25 * 10^8 m^2·s^-2. Step 4: Take the square root to get v_esc ≈ 1.12 * 10^4 m/s. Step 5: Convert metres per second to kilometres per second by dividing by 1000, yielding about 11.2 km/s. Step 6: Compare with the options and identify 11.2 km/s as the correct approximate escape velocity. Verification / Alternative check: Space science references and standard physics textbooks consistently quote Earth escape velocity as about 11.2 km/s. Launch velocities for space missions are designed with this order of magnitude in mind, although real rockets use continuous propulsion and must overcome atmospheric drag, so their actual trajectory planning is more complex. The fact that all reliable sources cluster around 11.2 km/s for Earth escape velocity confirms that this is the correct option. Why Other Options Are Wrong: 18 km/s: This value is significantly higher than the theoretical escape velocity from Earth and would represent excess speed that is not required to just escape gravity. 21 km/s: This is almost twice the correct value and does not match standard calculations or references. 35 km/s: This is far above Earth escape velocity; values like this are associated with orbital speeds around the Sun for inner planets, not escape from Earth surface. Common Pitfalls: Students sometimes confuse orbital speed with escape speed or mix up units, quoting values in metres per second when the question expects kilometres per second. Another mistake is to think escape velocity depends on mass of the rocket itself, whereas it actually depends only on the mass and radius of the planet and the gravitational constant. Remember that for Earth, a good number to keep in mind is approximately 11.2 km/s. Final Answer: The minimum velocity required for a rocket to escape Earth gravity is approximately 11.2 km/s.

Question 5

In basic mechanics, in which direction does the frictional force on a moving object act relative to the object direction of motion?

Accepted Answer

Introduction / Context: Friction is a common contact force that appears whenever two surfaces slide or attempt to slide past one another. It plays an important role in walking, driving and many mechanical systems. A basic point that every physics student must understand is the direction in which friction acts relative to the motion or attempted motion of an object. This question directly tests that idea. Given Data / Assumptions: We consider kinetic friction acting on an object already in motion. The object surface is in contact with another surface, such as a floor. No special cases like rolling resistance or fluid drag are being considered in detail. We focus on the direction of the frictional force, not its exact magnitude. Concept / Approach: Friction is a resistive force that opposes relative motion or the tendency for motion between two contacting surfaces. When an object slides to the right, the frictional force exerted by the surface on the object acts to the left, trying to slow it down. If a force tries to push an object to the left but the object does not yet move, static friction acts to the right to oppose that tendency. In all such cases, friction acts along the surface of contact and opposite to the direction of motion or impending motion. Step-by-Step Solution: Step 1: Identify the direction of motion of the object relative to the surface, for example to the right. Step 2: Recall the definition of friction as a force that opposes relative motion between surfaces. Step 3: Conclude that to oppose motion to the right, the frictional force must act to the left. Step 4: Generalise this reasoning: if the object moves to the left, friction acts to the right, and so on. Step 5: Note that friction acts parallel to the surface of contact, not perpendicular to it. Step 6: Therefore, the correct description is that friction acts in the opposite direction to the motion. Verification / Alternative check: Practical observations confirm this. When you slide a block across a table, it slows down and eventually stops, meaning a force opposite to its motion must be acting. That force is friction. When walking, your foot pushes backward on the ground, and friction pushes you forward; in this case friction opposes the tendency of the foot to slip backward relative to the ground. In every example, friction works to oppose the actual or attempted relative motion along the surface. Why Other Options Are Wrong: In the same direction as the motion: A force in the same direction as the motion would accelerate the object rather than resist its motion, so this does not describe friction. Equally in all directions: Forces that act equally in all directions would not selectively oppose motion; this is not how friction behaves. Always vertically upward: Vertical forces such as normal reaction or weight act perpendicular to the surface, whereas friction acts parallel to the surface and depends on motion direction. Common Pitfalls: Some learners confuse friction with the driving force of an engine or with thrust, which can act in the direction of motion. It is also common to forget that friction can act in the forward direction on a vehicle tyre, because the tyre tries to rotate backward against the road. The consistent rule is that friction always opposes relative sliding, not necessarily the overall motion of the vehicle or system as a whole. Focusing on the local contact motion helps keep this straight. Final Answer: For an object sliding on a surface, the frictional force acts in the opposite direction to the motion.

Question 6

Solids have definite shapes and volumes. Which explanation best describes why, in terms of intermolecular forces inside a solid?

Accepted Answer

Introduction / Context: One of the basic differences between solids, liquids and gases is that solids have definite shapes and volumes, whereas liquids take the shape of their container and gases fill all available space. From a microscopic point of view, this behaviour is explained in terms of intermolecular forces. This question asks you to identify which type of force and what relative magnitude of that force is responsible for the definite shape of solids. Given Data / Assumptions: We compare cohesion (attraction between like molecules) and adhesion (attraction between unlike molecules). We consider the internal structure of a typical crystalline or amorphous solid. Temperature is assumed low enough that the material remains in the solid state. We ignore external forces such as gravity except as they affect the macroscopic shape. Concept / Approach: In solids, molecules or atoms are packed closely together in a fixed arrangement. Strong cohesive forces between neighbouring particles hold them in place, allowing only small vibrations about fixed positions. Because of these strong cohesion forces, the particles do not move freely past one another, and the solid maintains a definite shape and volume. In liquids, cohesion is weaker, so molecules can slide around while remaining close, and in gases, cohesion is negligible, so particles move freely. Adhesion describes how solids and liquids stick to other materials, but it does not primarily determine the shape of the solid itself. Step-by-Step Solution: Step 1: Recall that cohesion refers to attractive forces between molecules of the same substance. Step 2: In a solid, particles are closely packed in a regular or semi regular arrangement. Step 3: Strong cohesive forces keep each particle near its equilibrium position, allowing little relative motion. Step 4: Because the particles cannot easily move past one another, the solid retains a definite shape and volume. Step 5: Adhesion involves attraction between different materials, such as glass and water, and affects wetting and sticking but not the basic solid shape. Step 6: Therefore, the best explanation is that the forces of cohesion are very large in solids. Verification / Alternative check: Everyday experience supports this reasoning. Ice cubes, metal blocks and wooden pieces all retain their shapes even when moved from one container to another, showing that internal structure holds them together. When a solid is heated strongly enough to melt, its internal cohesive forces are partially overcome, and it becomes a liquid that flows and takes the shape of its container. This change confirms that the key difference lies in how strongly the particles are bound together by cohesive forces. Why Other Options Are Wrong: Because the forces of cohesion are very small: If cohesion were small, particles would move freely and the material would behave like a liquid or gas, not a solid. Because the forces of adhesion are very large: Adhesion describes interactions with other materials and is more relevant to phenomena like wetting; it does not explain the internal rigidity of a solid. Because the forces of adhesion are very small: This says little about the internal structure of the solid and does not account for its definite shape. Common Pitfalls: A common mistake is to confuse cohesion and adhesion. Students may recall that water droplets form due to cohesion and that water sticks to glass due to adhesion, but they do not always connect cohesion to the rigidity of solids. Another error is to think that solids are hard simply because particles are closer together, without considering the strength of the attractive forces holding them in place. Remember that closeness plus strong cohesive forces together give solids their definite shape. Final Answer: Solids have definite shapes mainly because the forces of cohesion between their molecules are very large.

Question 7

A narrow capillary tube is dipped vertically into water and water rises in the tube due to capillarity. How can the height to which water rises be increased?

Accepted Answer

Introduction / Context: Capillarity is the phenomenon in which a liquid rises or falls in a very thin tube relative to the surrounding liquid level. This effect is important in many natural and technological processes, such as the movement of water in soil and plants, and ink in capillary pens. In the case of water in a clean glass capillary tube, the liquid rises above the outside water level. This question asks how we can increase the height of that capillary rise by changing the geometry of the tube or the vessel. Given Data / Assumptions: A clean, narrow glass capillary tube is dipped vertically into pure water. The water wets the glass, so the contact angle is acute. Capillary rise occurs due to surface tension and adhesion between water and glass. Other conditions such as temperature and liquid properties are kept constant. Concept / Approach: For a wetting liquid like water in a glass capillary tube, the height h of capillary rise is given by the formula h = (2 * T * cos(theta)) / (rho * g * r), where T is surface tension, theta is the contact angle, rho is density of the liquid, g is acceleration due to gravity and r is the radius of the capillary tube. All other quantities being constant, the formula shows that h is inversely proportional to r. That means if the radius of the tube is made smaller, the height of the liquid column increases. Step-by-Step Solution: Step 1: Write the capillary rise formula for a wetting liquid: h = (2 * T * cos(theta)) / (rho * g * r). Step 2: Assume T, theta, rho and g remain unchanged for the same water and glass at the same temperature. Step 3: Observe that h is inversely proportional to r, the radius of the capillary tube. Step 4: Conclude that if r decreases, h must increase, and if r increases, h must decrease. Step 5: Changing the depth of water in the vessel does not appear in the formula and therefore does not directly affect capillary rise height. Step 6: Therefore, the height of water rise can be increased by decreasing the radius of the capillary tube. Verification / Alternative check: This prediction matches everyday observations. Very fine tubes or narrow gaps between plates show a noticeable rise of water, whereas wider tubes show little capillary effect. In plant xylem vessels, which are extremely narrow, significant capillary rise combines with transpiration pull to lift water to great heights. Demonstrations with tubes of different internal diameters dipped into the same water clearly show higher columns in the thinner tubes, confirming the inverse relationship between h and r. Why Other Options Are Wrong: By increasing the radius of the capillary tube: Increasing r reduces h according to the formula, so the rise becomes smaller, not larger. By increasing the depth of water in the vessel only: The formula does not involve the depth of water in the reservoir; the external level only serves as a reference for measuring the rise. By no practical change in radius or vessel height: If no relevant parameter is changed, the height of capillary rise will remain the same under identical conditions. Common Pitfalls: Some learners think that more water in the vessel will somehow push more water up the tube, forgetting that capillary rise is driven by surface tension and adhesion, not by hydrostatic pressure from depth for the small heights considered. Others may misremember the formula and think h is proportional to r instead of inversely proportional. Always look at the correct formula and focus on which variables are in the numerator and which are in the denominator. Final Answer: The height of capillary rise of water can be increased by decreasing the radius of the capillary tube.

Question 8

Two large ships travel on close, parallel courses in the same direction and tend to move towards each other. Which physical principle best explains this effect?

Accepted Answer

Introduction / Context: When two large ships move in the same direction on close, parallel paths, observers sometimes notice that the ships seem to be drawn toward each other. This can be dangerous if they get too close. Understanding the physics behind this phenomenon helps in safe navigation and illustrates an important principle of fluid dynamics. This question asks which physical principle gives the correct qualitative explanation. Given Data / Assumptions: Two large ships are moving in the same direction along nearly parallel paths. The distance between their hulls is relatively small compared to open water around them. Water flows between and around the ships as they move. We neglect small effects such as magnetism and treat the main fluid as water. Concept / Approach: Bernoulli principle states that in an ideal flowing fluid, an increase in the speed of the fluid accompanies a decrease in the fluid pressure, provided elevation changes are small. Between two closely spaced moving ships, water must rush through a narrower gap. Because the same volume of water has to pass through a smaller cross section, its speed in the gap increases. According to Bernoulli principle, this higher speed corresponds to lower pressure between the ships compared to the outer sides. The resulting pressure difference produces a net force that pushes the ships toward each other. Step-by-Step Solution: Step 1: Picture two large hulls moving side by side with a narrow channel of water between them. Step 2: As the ships move forward, water is forced to flow through this narrow gap. Step 3: The restriction in cross sectional area between the hulls causes water speed there to increase. Step 4: Apply Bernoulli principle: higher fluid speed implies lower fluid pressure in the region of faster flow. Step 5: On the outer sides of the ships, water speed is lower and pressure is relatively higher. Step 6: The higher external pressure and lower internal pressure create a net sideways force that pushes the ships toward each other. Verification / Alternative check: This effect is similar to the attraction felt when two sheets of paper hang loosely and air is blown between them. The faster air between the sheets lowers the pressure there, and the sheets move closer together. Many fluid dynamics demonstrations show the same principle with suspended balls or streamlines around obstacles. Navigation guidelines warn ship captains about the risk of bank suction and ship interaction in narrow channels, which are explained using Bernoulli principle and related fluid dynamic concepts. Why Other Options Are Wrong: Magnetic properties of the ship material: Most ships are made of steel, which can be magnetic, but the distances and weak fields involved cannot produce significant mutual attraction compared with hydrodynamic forces. Newton third law of motion directly: While all forces obey Newton third law, this law explains action reaction pairs, not the specific origin of this sideways attraction; it does not directly account for the pressure differences in the water. Earth gravitational attraction: Although there is a tiny gravitational attraction between any two masses, the effect is negligible compared with hydrodynamic forces from the moving water. Common Pitfalls: A common misunderstanding is to attribute any attraction between masses solely to gravity or magnetism, ignoring fluid effects. Others may think of Newton third law as a catch all explanation. To avoid this, always ask what medium is involved and how it behaves. Here, the key is the water flow pattern and the pressure changes it causes. Bernoulli principle gives the simplest and most direct explanation of the ships moving toward each other. Final Answer: The tendency of the two ships to move towards each other is best explained by Bernoulli principle on fluid pressure and speed.

Question 9

In a solar cooker, why is the transparent cover usually made of glass rather than left open or covered with metal?

Accepted Answer

Introduction / Context: Solar cookers use sunlight as their only energy source to heat and cook food. Their design must therefore maximise the amount of solar energy absorbed and minimise heat losses to the surroundings. One common design feature is a transparent glass cover over the dark interior cooking area. This question asks why glass is used and what physical effect it exploits. Given Data / Assumptions: The solar cooker has a dark interior surface that absorbs sunlight. A transparent glass sheet covers the top of the cooker. Sunlight includes visible and short wavelength radiation from the Sun. The hot interior surfaces emit mainly longer wavelength infrared radiation. Concept / Approach: Glass is transparent to most of the visible light and short wavelength solar radiation, so these rays pass through the glass and are absorbed by the dark interior surfaces, heating them. The heated surfaces and the air inside then emit infrared radiation at longer wavelengths. Glass is much less transparent to this longer wavelength infrared radiation and tends to reflect or absorb it, reducing the rate at which heat escapes. In addition, the glass lid reduces convective heat losses by trapping warm air inside. Together, these effects create a greenhouse effect: sunlight enters easily, but heat is retained, so the temperature inside rises. Step-by-Step Solution: Step 1: Sunlight passes through the glass cover and reaches the dark interior of the cooker. Step 2: The dark surfaces absorb the solar energy and convert it into heat, raising their temperature. Step 3: The hot interior and the food emit infrared radiation at longer wavelengths. Step 4: Glass is relatively opaque to these longer wavelengths, so much of the infrared radiation is reflected back or absorbed and re emitted inside. Step 5: The glass lid also physically traps warm air, reducing heat loss by convection to the outside. Step 6: As a result, the interior temperature increases more than it would without the glass, allowing effective cooking. Verification / Alternative check: The same principle explains why a car interior can become very hot when parked in sunlight with the windows closed. Visible solar radiation enters, heats the interior surfaces and the air inside, but much of the resulting infrared radiation is trapped, so heat builds up. Greenhouses for plants operate similarly with glass or transparent plastic roofs. Experiments with a solar box cooker show significantly lower internal temperatures if the glass cover is removed, confirming the importance of the greenhouse effect. Why Other Options Are Wrong: Because it absorbs all of the Sun rays: If the glass absorbed all sunlight, very little energy would reach the interior, and cooking would be inefficient. Because it reflects most of the Sun rays: Reflecting sunlight away would reduce, not increase, the energy available for cooking. Because it radiates heat better than metals: Metals usually have different radiative properties, but the key role of the glass cover is trapping heat, not radiating it away. Common Pitfalls: Some learners think the glass cover simply prevents dust or insects from entering, overlooking its thermal role. Others confuse absorption with transmission and believe that absorbing sunlight in the glass is desirable. The most useful way to remember the function of the glass cover is to associate it with the greenhouse effect: let in short wavelength sunlight and restrict the escape of long wavelength heat radiation and warm air. Final Answer: The glass cover is used because it traps heat inside by a greenhouse effect, allowing sunlight in but reducing heat loss.

Question 10

Tap water is placed in an open pan and allowed to evaporate naturally. After some time, how does the temperature of the remaining water typically change?

Accepted Answer

Introduction / Context: Evaporation is a common everyday process in which a liquid slowly changes into vapour at the surface, even at temperatures below its boiling point. People often notice that water or sweat on the skin feels cool as it evaporates. This question asks what happens to the temperature of tap water left in an open pan when evaporation is allowed to occur over some time under ordinary conditions. Given Data / Assumptions: Tap water is placed in a shallow, open pan exposed to air. The water is allowed to evaporate at room temperature without external heating. The process is slow and occurs over a moderate time interval. Ambient air conditions are reasonably constant. Concept / Approach: Evaporation is a cooling process. The most energetic molecules near the surface of the liquid escape into the vapour phase, taking away more kinetic energy on average than the molecules that remain. As a result, the average kinetic energy of the remaining molecules decreases, which means the temperature of the liquid falls. However, heat is also exchanged with the surrounding air and container, so the temperature does not keep falling indefinitely. Under normal room conditions, the water temperature drops slightly below ambient, producing mild cooling, but it does not decrease rapidly to very low values. Step-by-Step Solution: Step 1: Recognise that during evaporation, high energy molecules leave the liquid surface and become vapour. Step 2: Understand that the remaining liquid molecules have lower average kinetic energy after these energetic molecules depart. Step 3: Lower average kinetic energy corresponds to a lower temperature of the remaining liquid. Step 4: At the same time, the liquid can absorb heat from the surrounding air and the pan, partially offsetting the cooling. Step 5: In typical room conditions, the overall effect is a modest drop in water temperature below the starting value. Step 6: Therefore, the best description is that the temperature decreases slightly, not that it increases or drops extremely rapidly. Verification / Alternative check: Practical experience supports this explanation. If you wet your hand and hold it in moving air, it feels cool but does not freeze. Similarly, water in an evaporating dish used in laboratories feels slightly cooler than the surrounding air while evaporation is active. If the water is left long enough, it eventually comes back close to room temperature as the rate of evaporation decreases, but during active evaporation the net effect is mild cooling. Why Other Options Are Wrong: It increases to a very high value: There is no external heat source to raise the temperature significantly; evaporation removes energy rather than adding it. It increases slightly: This contradicts the fundamental idea that evaporation is a cooling process where higher energy molecules leave the liquid. It decreases extremely rapidly to freezing: Under normal room conditions, evaporation is not fast enough, and heat exchange with the surroundings prevents such a rapid temperature drop. Common Pitfalls: Some learners confuse boiling with evaporation and think any phase change to vapour must involve heating, leading them to expect a temperature increase. Others know evaporation causes cooling but overestimate the effect, imagining that water could quickly become very cold or freeze purely due to evaporation in ordinary conditions. Remember that evaporation cools the liquid by a modest amount, and the extent of cooling depends on humidity, air flow and heat supply from the environment. Final Answer: When tap water evaporates in an open pan, its temperature usually decreases slightly below the starting value.

Question 11

Long distance radio waves sent out by a broadcasting station are mainly reflected back to Earth by which layer of the atmosphere?

Accepted Answer

Introduction / Context: Radio communication over very long distances, such as between countries or across oceans, often uses sky waves that travel up to the upper atmosphere and are reflected back toward Earth. This reflection allows radio signals to reach far beyond the horizon. The property of reflecting certain radio frequencies depends on a specific ionised layer of the atmosphere. This question asks which named layer is responsible for that reflection. Given Data / Assumptions: We are talking about radio waves used in long distance radio communication. The waves leave the broadcasting station, travel upward and are reflected back toward Earth. The atmosphere contains several layers: troposphere, stratosphere, mesosphere and ionosphere. We assume typical conditions during day and night that allow ionisation to exist. Concept / Approach: The ionosphere is a region of the upper atmosphere, roughly from about 60 kilometres to several hundred kilometres altitude, where solar ultraviolet radiation ionises atoms and molecules, creating free electrons and ions. This layer of charged particles can reflect or refract certain radio frequencies, especially in the high frequency bands used for long distance communication. Lower layers such as the troposphere and stratosphere mostly contain neutral air and do not reflect radio waves in the same way. The mesosphere lies below the ionosphere but does not have sufficient ionisation to act as the main reflecting layer. Step-by-Step Solution: Step 1: Identify that sky wave radio communication relies on reflection of radio signals by charged layers in the upper atmosphere. Step 2: Recall that the ionosphere is the part of the atmosphere that contains significant numbers of ions and free electrons. Step 3: Recognise that these charged particles interact with radio waves, causing them to bend and, for suitable frequencies, return toward the Earth. Step 4: Troposphere and stratosphere are mostly neutral and are more important for weather and ozone, not for radio reflection. Step 5: The mesosphere is a transitional region with limited ionisation; it is not the principal reflecting layer for radio. Step 6: Therefore, the correct answer is that radio waves are reflected by the ionosphere. Verification / Alternative check: Descriptions of shortwave and high frequency radio communication mention the ionosphere explicitly as the reflecting layer. Variations in ionospheric conditions due to solar activity, day night cycles and seasons directly affect long distance radio reception, causing fading or improved propagation. No such direct link exists between tropospheric weather patterns and the reflection of these specific radio frequencies, confirming that it is the ionosphere that plays the central role. Why Other Options Are Wrong: Troposphere: This is the lowest layer of the atmosphere, containing most of the weather, clouds and water vapour, but it does not reflect radio waves in the way needed for long distance sky wave communication. Stratosphere: This layer contains the ozone layer and is important for absorbing ultraviolet radiation but is not the main reflecting region for radio waves. Mesosphere: Although it lies above the stratosphere, it does not have the high electron densities that characterise the ionosphere. Common Pitfalls: Some students assume that because weather and storms affect radio reception near the ground, the troposphere must be responsible for reflections. Others may think the ozone rich stratosphere is involved because it interacts with ultraviolet light. The key is to remember that reflection of radio waves in this context depends on ionisation, and the word ion in ionosphere directly hints that this is the layer filled with charged particles. Final Answer: Long distance radio waves are mainly reflected back to Earth by the ionosphere.

Question 12

A rapid series of still pictures can be made to appear as a smoothly moving picture in cinema and animation because the human eye and brain have which property?

Accepted Answer

Introduction / Context: Movies, television and many forms of animation rely on the fact that a rapid sequence of still images can create the illusion of continuous motion. This basic optical illusion depends on the way the human visual system processes information over time. The underlying phenomenon is often described as persistence of vision. This question asks which description best captures that effect. Given Data / Assumptions: We are considering a series of still pictures shown in quick succession. The time interval between successive frames is small, typically less than a tenth of a second. The human eye and brain work together to form a visual impression. We neglect details of modern display technologies and focus on human perception. Concept / Approach: Persistence of vision refers to the tendency of the eye and brain to retain a visual image for a short interval after the object is removed. If a second image appears within this retention period, the two images overlap in perception. When many such images are shown quickly one after another, the brain blends them into a smooth, continuous sequence. The critical interval is around one tenth to one fifteenth of a second for most people, which is why traditional film and video formats use frame rates of about 24 frames per second or more. Step-by-Step Solution: Step 1: Recognise that each frame of a movie is actually a separate still photograph. Step 2: These frames are projected in rapid succession, for example 24 frames per second. Step 3: The eye and brain do not instantly erase one image when the next appears; they retain each image briefly. Step 4: This retention time, on the order of one tenth of a second, means that successive images overlap in perception. Step 5: As a result, the sequence of slightly different still images is interpreted as continuous motion rather than separate pictures. Step 6: Therefore, the property responsible is the short persistence of visual impressions in the eye and brain. Verification / Alternative check: Simple experiments support this idea. If you spin a sparkler in a circle, you see a continuous ring of light rather than a single moving point, because the light impression persists briefly as the sparkler moves. Toys such as thaumatrope discs, where two images on opposite sides of a disc appear merged when the disc is spun quickly, also rely on persistence of vision. In all of these cases, the visual system blends closely spaced images into one continuous perception. Why Other Options Are Wrong: They can refocus extremely quickly: While the eye can refocus on objects at different distances, this is not the main reason still images appear as motion. The eye is physically faster than the brain: The brain and eye work together, but the crucial point is not speed difference; it is the persistence of the signal in the visual system. They can shut out some of the light: The eye does not deliberately shut out images to create motion; rather, it retains them briefly, causing overlap in perception. Common Pitfalls: Students sometimes think movies work by literally moving pictures, not by flashing separate frames, so they overlook the role of human perception. Others may recall the term persistence of vision but confuse it with the speed of eye movement or focusing. Remember that what matters is the short time during which an image remains in the visual system after it disappears, allowing rapidly changing images to blend into a smooth sequence. Final Answer: Fast series of still pictures appear as motion because the eye and brain retain an image for about one tenth of a second, so two images close in time merge into one.

Question 13

Why are integrated circuits (ICs) generally considered more advantageous than conventional circuits built from many discrete components?

Accepted Answer

Introduction / Context: Integrated circuits revolutionised electronics by allowing thousands or even millions of components to be fabricated on a single small semiconductor chip. Compared with older circuits made from separate resistors, capacitors and transistors wired together, ICs offer many practical benefits. This question asks you to identify the best overall statement describing why integrated circuits are more advantageous. Given Data / Assumptions: Conventional circuits are built from individual, discrete components mounted and wired on boards. Integrated circuits contain many components fabricated together on a single chip. We compare reliability, ease of servicing and replacement in both approaches. We assume modern manufacturing techniques for ICs and typical consumer electronics. Concept / Approach: Because integrated circuits are manufactured in a highly controlled environment, the probability of individual component failure is reduced, and the overall reliability of the device increases. When faults do occur, replacing one small IC is often easier and cheaper than diagnosing and replacing individual discrete components spread across a complex board. In conventional circuits, identifying which resistor, transistor or connection has failed can be time consuming. Integrated designs also reduce size, power consumption and cost. The options in this question list several of these advantages, and the best choice is the one that combines them all. Step-by-Step Solution: Step 1: Consider reliability: ICs contain many components built together with uniform processes, reducing the chance of loose connections and wiring errors. Step 2: Recognise that this often makes equipment using ICs more reliable than those built from many discrete parts. Step 3: In conventional circuits, a single faulty resistor or transistor may be physically small and hard to locate, especially on dense boards. Step 4: With IC based designs, technicians can replace a single chip that contains the suspect function instead of chasing individual components. Step 5: Manufacturers design boards so that if a particular IC fails, that entire package can be swapped, simplifying servicing. Step 6: Therefore, all the listed statements contribute to the overall advantage of integrated circuits, making the combined option correct. Verification / Alternative check: Real world experience with consumer electronics supports these points. Modern devices such as smartphones, televisions and computers rely heavily on ICs and can pack huge computing power into small volumes. Repairs often involve replacing a faulty module or chip rather than checking each resistor. Additionally, failure rates have decreased over time as integrated manufacturing has improved. Textbooks on electronics highlight benefits such as miniaturisation, lower power consumption, higher reliability and easier mass production for integrated circuits. Why Other Options Are Wrong: They make equipment more reliable and easier to service: This is true but does not include the other specific advantages listed. A faulty part of a conventional circuit may be difficult to locate: This is also true but explains only one aspect of why ICs are preferred. If part of an IC fails, the entire circuit is replaced: This is correct in practice but again is only one part of the full picture. Common Pitfalls: Some learners think integrated circuits are advantageous only because they are smaller or cheaper, overlooking reliability and maintenance benefits. Others may focus only on the ease of replacement without appreciating that conventional circuits often require more manual assembly and testing. In exams, when several individually correct statements are combined into one comprehensive option, it is usually a hint that the combined option is the best answer. Final Answer: Integrated circuits are more advantageous mainly because all of the listed reasons apply: they improve reliability, simplify servicing and allow easy replacement of faulty sections.

Question 14

A bimetal strip is made of copper and iron welded side by side. It is straight at room temperature and held vertically with the iron strip on the left and the copper strip on the right. When the strip is heated, in which direction will it bend?

Accepted Answer

Introduction / Context: Bimetal strips are widely used in thermostats and temperature sensing devices because two different metals bonded together expand by different amounts when heated. This difference in expansion causes the strip to bend in a predictable direction as temperature changes. The question describes a copper iron bimetal strip and asks you to determine which way it bends when heated. Given Data / Assumptions: The strip is made of two metals, copper and iron, welded along their length. At room temperature the strip is straight. The strip is held vertically, with iron on the left side and copper on the right side. The strip is heated uniformly along its length. Concept / Approach: Different metals have different coefficients of linear expansion. Copper has a higher coefficient of expansion than iron, which means that for the same rise in temperature and the same original length, copper expands more than iron. In a bimetal strip, the two metals are rigidly bonded, so they must stay attached along their common surface. When heated, the metal that tries to expand more becomes the outer, longer side of the bend, while the metal that expands less becomes the inner, shorter side. Therefore, when a copper iron strip is heated, the copper side, which expands more, will be on the convex (outer) side, and the strip will bend towards the iron side. Step-by-Step Solution: Step 1: Recall that the coefficient of linear expansion of copper is greater than that of iron. Step 2: When the strip is heated, copper attempts to increase its length more than iron does. Step 3: Because the two metals are welded together, they cannot separate and must bend to accommodate the different expansions. Step 4: The metal that expands more will form the outer side of the curve, while the one that expands less will form the inner side. Step 5: In this case the copper strip is on the right and expands more, so it becomes the outer, longer side of the bend. Step 6: Therefore, the strip bends towards the left, where the iron strip is located and acts as the inner side of the curve. Verification / Alternative check: Thermostats in some appliances use bimetal strips specifically because they bend predictably when heated. If you look up tables of coefficients of thermal expansion, you will see that copper has a value around 16 to 17 * 10^-6 per kelvin, while iron is around 11 to 12 * 10^-6 per kelvin. This confirms that copper length increases more for each degree of temperature rise. Experimental demonstrations with bimetal strips taped or clamped at one end show that the strip curves toward the metal with the smaller expansion coefficient when heated. Why Other Options Are Wrong: It will remain straight: If the strip remained straight, it would imply that both metals expanded by exactly the same amount, which is not true for copper and iron. It will bend towards the right (copper side): This would require iron to expand more than copper, opposite to the actual material properties. Its bending direction is completely random: The direction is determined by the relative expansion coefficients and is entirely predictable. Common Pitfalls: Students sometimes remember that one side expands more but then incorrectly assume that the strip bends towards that side. It is important to visualise the geometry: the longer side must lie on the outside of the curve, so the strip bends toward the shorter side. Another mistake is to confuse which material has the higher expansion coefficient. Consulting standard material property tables helps keep these facts clear. Final Answer: When heated, the copper iron bimetal strip will bend towards the left, that is, towards the iron side.

Question 15

In basic physics, consider the statement: "The ability to do work is called energy." How should this statement be classified?

Accepted Answer

Introduction / Context: Many textbook definitions introduce the concept of energy in simple language to make it accessible to beginners. One such popular definition is that energy is the ability or capacity to do work. This question presents that statement and asks you to decide whether it correctly describes the physical quantity called energy, or whether it should instead be associated with some other quantity such as power, force or momentum. Given Data / Assumptions: The statement under consideration is: "The ability to do work is called energy." We use standard physics meanings of the words work, energy, power, force and momentum. We are looking for a conceptual definition, not a detailed mathematical formula. The context is introductory physics and general science education. Concept / Approach: Work in physics is defined as the product of force and displacement in the direction of the force. Energy, in simple terms, is the capacity of a system to perform work. Different forms of energy such as kinetic, potential, thermal and chemical can all be transformed into work under suitable conditions. Power, on the other hand, is the rate of doing work, force is a push or pull that can cause acceleration and momentum is mass times velocity. None of these alternate quantities match the simple idea of ability to do work as closely as energy does. Step-by-Step Solution: Step 1: Recall that energy is often introduced as the capacity or ability of a body or system to do work. Step 2: Note that when a system has energy, such as a raised weight or a charged battery, it can perform work on other objects. Step 3: Compare this with power, which is defined as work done per unit time and measures how fast work is done, not simply the ability to do it. Step 4: Force is a vector quantity that can cause a change in motion but does not, by itself, guarantee that work is done unless there is displacement. Step 5: Momentum measures the quantity of motion (mass times velocity) and is conserved in collisions; it is not usually described as the ability to do work. Step 6: Therefore, the statement that the ability to do work is called energy is a correct introductory description of energy. Verification / Alternative check: Consider a stretched spring, a raised mass and a charged battery. All of these can do work when allowed to act: the spring can move a mass, the raised object can fall and drive a machine, and the battery can power a circuit. Each situation illustrates stored energy. When the stored energy is used, work is done. If a body has no energy at all, it cannot do work on other bodies. These examples support the idea that energy is closely linked to the ability to do work. Why Other Options Are Wrong: Incorrect, because it defines power: Power is defined as work done per unit time, not as the capacity to do work. Incorrect, because it refers to force: Force is a push or pull; without displacement, a force does not necessarily result in work, so it is not simply the ability to do work. Incorrect, because it should define momentum instead: Momentum is mass times velocity and describes motion, not the capacity to perform work. Common Pitfalls: Students sometimes mix up energy and power, using the words interchangeably in everyday language. For example, they may say that a device uses a lot of energy when they really mean it uses power quickly. In physics, however, energy, work and power are distinct but related concepts. Remember that energy and work share the same unit, the joule, while power has the unit watt. Keeping these distinctions clear helps you recognise that the ability to do work is a good conceptual definition of energy, not of power or other quantities. Final Answer: The statement is a correct description of the physical quantity energy.

Question 16

In the flame of a typical Bunsen burner or candle, which region is the hottest part of the flame?

Accepted Answer

Introduction / Context: When studying combustion and flames, it is useful to know that temperature is not the same everywhere in a flame. Different coloured regions correspond to different temperatures and combustion conditions. In laboratory burners and candles, we often distinguish between a dark inner zone, a luminous bright zone and a non luminous blue outer zone. This question asks which of these regions is the hottest. Given Data / Assumptions: We consider a steady, well adjusted Bunsen burner or candle flame. The flame has an inner dark zone, a luminous yellow region and an outer blue, non luminous region. Complete combustion occurs where fuel mixes well with air. We neglect special flames with unusual fuel mixtures. Concept / Approach: The hottest part of a flame is usually where the fuel has the best supply of oxygen and combustion is most complete. In a Bunsen burner with adequate air supply, the outer blue region is where gas and oxygen mix thoroughly and burn completely, reaching high temperatures. The luminous yellow region contains glowing soot particles and often indicates incomplete combustion, with lower temperature than the hottest outer zone. The dark inner zone near the burner mouth or wick is composed mostly of unburnt gas or vapour and is relatively cool. Step-by-Step Solution: Step 1: Identify the main regions of the flame: inner dark zone, middle luminous zone and outer non luminous blue zone. Step 2: Recognise that the inner dark zone contains unburnt fuel and little oxygen, so combustion is minimal and temperature is relatively low. Step 3: In the luminous zone, incomplete combustion produces glowing carbon particles, giving yellow light but not the highest temperature. Step 4: The outer blue non luminous zone is where fuel mixes efficiently with surrounding air and combustion is nearly complete. Step 5: Complete combustion releases more energy per unit mass of fuel and leads to higher temperatures. Step 6: Therefore, the hottest part of the flame is the blue outer non luminous zone. Verification / Alternative check: Practical tests support this explanation. When heating objects in a laboratory, instructions often recommend placing the object in the outer blue cone of a Bunsen burner flame for the greatest heating effect. If you hold a piece of wire in different parts of the flame, it becomes red hot quickest in the outer blue region. In a candle flame, experiments with thermocouples or heat sensitive paints similarly show higher temperatures near the outer edges compared with the luminous inner part. Why Other Options Are Wrong: Bright yellow luminous zone in the middle: Although bright and easy to see, this region contains incomplete combustion and hot soot particles; it is not the hottest region. Inner dark zone: This zone has unburnt gas and relatively little combustion, so it is the coolest part of the flame. A special non luminous zone that is cooler: This option contradicts the observation that the non luminous blue outer zone is hottest; there is no cooler non luminous region that is hotter than the blue outer zone. Common Pitfalls: Students often assume that the brightest or most luminous part of a flame must be the hottest. In reality, brightness depends on glowing particles and emission, not purely on temperature. Another confusion arises when comparing different flames; for example, a smoky yellow candle flame and a clean blue gas flame. It is important to focus on the internal structure of a single flame and remember that complete combustion in the blue outer region produces the highest temperatures. Final Answer: The hottest region of a typical laboratory or candle flame is the blue outer non luminous zone of the flame.

Question 17

Compressions and rarefactions are characteristic features of which type of mechanical wave?

Accepted Answer

Introduction / Context: Mechanical waves transfer energy through a medium by the oscillatory motion of particles. There are two basic types of mechanical waves: longitudinal and transverse. Understanding how the medium particles move in each case helps you identify which features belong to which wave type. This question asks about compressions and rarefactions, which are key concepts in sound and other longitudinal waves. Given Data / Assumptions: We consider mechanical waves propagating through a medium such as air, a spring or a rope. In longitudinal waves, particle displacement is parallel to the direction of wave travel. In transverse waves, particle displacement is perpendicular to the direction of wave travel. The terms compressions and rarefactions refer to regions of higher and lower density or pressure. Concept / Approach: In a longitudinal wave, particles of the medium move back and forth along the direction of propagation. This motion creates regions where particles are closer together (compressions) and regions where they are farther apart (rarefactions). Sound waves in air are the standard example: compressions correspond to high pressure regions and rarefactions to low pressure regions. Transverse waves, such as waves on a rope, have crests and troughs instead, where the displacement is up and down relative to the direction of travel. They do not create alternating compressions and rarefactions along the direction of propagation. Step-by-Step Solution: Step 1: Define a longitudinal wave as one where particle motion is parallel to the direction of wave propagation. Step 2: Recognise that as particles move back and forth, they bunch together in some regions and spread out in others. Step 3: Identify regions of high particle density and pressure as compressions. Step 4: Identify regions of low particle density and pressure as rarefactions. Step 5: Contrast this with a transverse wave, where particles move perpendicular to the direction of travel, creating crests and troughs instead. Step 6: Conclude that compressions and rarefactions are characteristic of longitudinal waves, not transverse waves. Verification / Alternative check: When a toy spring or slinky is compressed and released, you can see regions where coils are close together (compressions) moving along the spring, followed by regions where coils are farther apart (rarefactions). This clearly illustrates a longitudinal wave. By contrast, when you flick a rope up and down, you see peaks and valleys moving along the rope, with the rope segments moving perpendicular to the wave direction, which indicates a transverse wave. Sound wave diagrams in textbooks likewise show alternating high and low pressure bands, consistent with longitudinal compressions and rarefactions. Why Other Options Are Wrong: Transverse waves: These waves are described by crests and troughs, not compressions and rarefactions along the direction of travel. Both longitudinal and transverse waves equally: This is incorrect because the specific terminology of compressions and rarefactions is reserved for longitudinal waves. Neither longitudinal nor transverse waves: This would deny a fundamental description of sound waves and other longitudinal wave phenomena. Common Pitfalls: Students sometimes memorise the terms crest and trough and then incorrectly attach them to all types of waves, or they may use compressions and rarefactions when talking about water waves, which are actually surface waves with transverse and longitudinal components. To avoid confusion, remember that compressions and rarefactions describe variations in density and pressure along the direction of propagation and are therefore specific to longitudinal waves like sound in fluids. Final Answer: Compressions and rarefactions are characteristic of longitudinal waves, such as sound waves in air.

Question 18

Which of the following molecular compounds is a nonelectrolyte and therefore does not conduct electricity in aqueous solution?

Accepted Answer

Introduction / Context: Electrolytes are substances that produce ions when dissolved in water and thus conduct electricity in aqueous solution. Nonelectrolytes dissolve without forming ions and do not conduct electricity. Recognising which compounds behave as electrolytes or nonelectrolytes is important in chemistry, biology and many practical applications. This question asks you to identify a nonelectrolyte among a set of common substances. Given Data / Assumptions: The compounds listed are C2H6O (ethanol), HCl, Na2CO3 and NaCl. We consider their behaviour when dissolved in water. An electrolyte forms ions in solution and allows electric current to pass. A nonelectrolyte remains as neutral molecules and does not conduct electricity significantly. Concept / Approach: Strong electrolytes include soluble ionic compounds such as sodium chloride and sodium carbonate, and strong acids such as hydrochloric acid. These dissociate almost completely into ions in water. Ethanol (C2H6O), on the other hand, is a covalent organic molecule that does not ionise significantly in water. It dissolves by forming hydrogen bonds with water but remains as neutral ethanol molecules. Because there are very few ions present, aqueous ethanol is essentially a nonelectrolyte and conducts electricity very poorly compared with ionic solutions. Step-by-Step Solution: Step 1: Classify each substance as ionic or molecular and as acid, base or neutral compound. Step 2: HCl is a strong acid that ionises in water to give H+ and Cl- ions, making it a strong electrolyte. Step 3: NaCl is an ionic salt that dissociates into Na+ and Cl- ions, also a strong electrolyte. Step 4: Na2CO3 is another ionic salt that dissociates into Na+ and CO3^2- ions in solution, again an electrolyte. Step 5: C2H6O (ethanol) is a covalent organic compound that dissolves in water as neutral molecules with negligible ionisation. Step 6: Because ethanol does not form ions in solution, it is a nonelectrolyte and does not conduct electricity significantly. Verification / Alternative check: Conductivity experiments with a simple circuit and a light bulb show that solutions of NaCl, Na2CO3 or HCl cause the bulb to glow brightly, indicating good conduction. In contrast, a solution of pure ethanol in water produces little or no glow, showing very low conductivity. Textbooks on chemistry list ethanol and most other simple alcohols as nonelectrolytes, while strong acids and soluble ionic compounds are consistently listed as strong electrolytes. Why Other Options Are Wrong: HCl: This is a strong acid that ionises completely in water, making it a strong electrolyte, not a nonelectrolyte. Na2CO3: This is a salt composed of Na+ and CO3^2- ions, which dissociate in solution and conduct electricity well. NaCl: This is a common table salt that dissociates into Na+ and Cl- ions and is a classic example of an electrolyte. Common Pitfalls: Some learners mistakenly assume that any substance that dissolves in water is an electrolyte. In reality, many organic molecules like sugars and alcohols dissolve readily but do not ionise, so they do not conduct electricity. Others may be misled by the fact that ethanol can burn and has chemical energy, but this has nothing to do with ionic conduction. Always check whether the compound forms ions in solution when deciding if it is an electrolyte. Final Answer: Among the given compounds, C2H6O (ethanol) is a nonelectrolyte and does not conduct electricity significantly in aqueous solution.

Question 19

On a hot day, a mirage may appear on a road or desert as if water were present. This optical illusion is mainly caused by which condition in the lower atmosphere?

Accepted Answer

Introduction / Context: Mirages are fascinating optical illusions often observed in deserts or on sun heated roads. A common type is the inferior mirage, where the road ahead appears wet or reflective, as if covered by a pool of water, even though it is dry. Understanding mirages requires knowledge of how light bends when it passes through air layers of different temperatures and densities. This question asks what atmospheric condition leads to the formation of such mirages. Given Data / Assumptions: The observation is made on a hot day, over a surface such as sand or asphalt that is strongly heated by the Sun. The air near the ground is heated unevenly, with temperature varying with height. Light travels from distant objects through this layered air to the observer. We consider basic refraction in a medium with varying refractive index. Concept / Approach: The key idea is that the refractive index of air depends on its density, which in turn depends on temperature. Hot air near the ground is less dense and has a slightly lower refractive index than cooler air above it. When there is a strong temperature gradient, light rays travelling from distant objects bend gradually as they pass through layers of air with different refractive indices. In an inferior mirage, some rays curve upward after approaching the hot layer near the ground and reach the observer eye as if they came from the ground. The brain interprets these rays as reflections, creating the illusion of water or a shiny surface. Step-by-Step Solution: Step 1: Recognise that the ground becomes very hot in strong sunlight and heats the air just above it. Step 2: The air close to the surface becomes hotter and less dense than the cooler air higher up. Step 3: Because refractive index of air decreases slightly with decreasing density, the hot near surface air has a lower refractive index than the cooler air above. Step 4: Light rays from distant objects entering this region experience a continuous change in refractive index and bend according to Snell law in a graded medium. Step 5: Some rays bend upward after approaching the hot layer, so they reach the observer from directions that would normally correspond to reflections from the ground. Step 6: The observer brain interprets these rays as if they were coming from a shiny or watery surface, producing the mirage of water. Verification / Alternative check: Mirages are rarely seen on cool, cloudy days when the ground and air temperatures are more uniform, but they are common on hot sunny days when asphalt or sand is strongly heated. This correlation between strong heating and mirage formation supports the explanation based on unequal heating and resulting density gradients. Physics demonstrations using tanks of liquid with carefully layered refractive indices can reproduce similar bending of light and inverted images, showing that gradients in refractive index are sufficient to create mirage like effects. Why Other Options Are Wrong: Magnetic disturbances: Earth magnetic field and ordinary magnetic disturbances do not significantly affect the path of visible light in air and are not responsible for mirages. Equal heating of different parts of the atmosphere: If the air were heated uniformly, there would be no strong refractive index gradients and no mirage of this type. Depletion of the ozone layer: Ozone layer changes affect ultraviolet radiation absorption high in the atmosphere but do not directly cause the local refractive effects that produce road level mirages. Common Pitfalls: Some learners think the apparent water must be due to actual moisture on the road or to reflections from the sky alone. While the sky reflection contributes to the visual impression, the essential mechanism is the bending of light in layers of air with different refractive indices created by unequal heating. Another confusion is to attribute many atmospheric optical phenomena, such as rainbows or mirages, to a single cause; in reality, rainbows are due to refraction and reflection in water droplets, whereas mirages arise from refractive index gradients in air. Final Answer: A mirage on a hot road is mainly caused by unequal heating of different parts of the atmosphere, producing layers of air with different densities and refractive indices.

Question 20

In basic physics and weather science, which instrument is specifically used to measure atmospheric air pressure?

Accepted Answer

Introduction / Context: Measuring air pressure is essential in physics, meteorology and everyday weather forecasting. Atmospheric pressure influences wind patterns, cloud formation and storm systems. Instruments that measure this pressure help meteorologists predict weather changes and allow scientists to study the behaviour of gases. This question checks whether you can correctly identify the standard instrument used to measure atmospheric air pressure from a list of scientific devices. Given Data / Assumptions: We are dealing with atmospheric air pressure, not pressure in a laboratory gas container. The options include several scientific instruments commonly mentioned in physics. Each instrument has a specific and different purpose. We assume normal school level definitions of these devices. Concept / Approach: The instrument that is traditionally and most widely used to measure atmospheric air pressure is the barometer. A barometer may be of mercury type, aneroid type or digital type, but its primary function remains the same, to give a reading of the surrounding air pressure. A manometer is related but is mainly used for measuring gas or liquid pressure in closed systems, such as pressure inside a tank or pipeline. A spectrometer is used to analyse light spectra and identify elements or compounds. A bolometer measures the power of incident electromagnetic radiation, often in infrared astronomy. Therefore the only option that directly matches atmospheric air pressure measurement is the barometer. Step-by-Step Solution: Step 1: Focus on the phrase atmospheric air pressure in the question.Step 2: Recall that weather reports often mention barometric pressure, meaning pressure measured by a barometer.Step 3: Match barometer with air pressure measurement used in meteorological stations and laboratories.Step 4: Recognise that a manometer usually measures pressure differences in gases or liquids confined in tubes or vessels.Step 5: Recall that a spectrometer deals with light spectra and is not a pressure measuring device.Step 6: Note that a bolometer is used to measure the power of incoming radiation, not air pressure.Step 7: Conclude that barometer is the only correct answer. Verification / Alternative check: In weather charts and instruments, barometric pressure is usually recorded in units such as millibars or hectopascals. The barometer reading helps to identify high pressure and low pressure systems that drive weather changes. Textbooks and general knowledge references describe a barometer specifically as an instrument for measuring atmospheric pressure. Manometers are shown as U shaped tubes to compare pressures of gases or liquids, and spectrometers are drawn as optical devices splitting light into bands. These descriptions confirm that barometer is the correct choice for measuring air pressure around us. Why Other Options Are Wrong: Manometer measures pressure in closed systems, not typically open atmospheric pressure as in weather observations. Spectrometer is used to study the spectrum of light and to analyse the composition of substances, not to measure pressure. Bolometer measures the power of electromagnetic radiation, often in research applications, and has no direct role in routine air pressure measurements. Common Pitfalls: Students sometimes confuse barometer and manometer because both relate to pressure. The key difference is that barometers are designed for atmospheric pressure, whereas manometers are used for relative or gauge pressure in controlled systems. Others may choose spectrometer simply because the name sounds technical. To avoid errors, remember barometer for barometric or atmospheric pressure and manometer for gas pressure in pipes and vessels. Final Answer: The instrument that measures atmospheric air pressure is the barometer.

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