Question 1

In astronomy and classical mechanics, the empirical laws that accurately describe and govern the motion of planets around the Sun are collectively known by which name?

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Introduction / Context: Planetary motion is one of the earliest and most important topics in astronomy and physics. Before modern gravitational theory, astronomers observed how planets move across the sky and summarised these observations in the form of simple laws. These empirical relationships describe how planets orbit the Sun and paved the way for the development of universal gravitation. This question asks you to identify the correct set of laws associated specifically with planetary motion. Given Data / Assumptions: We are dealing with the motion of planets around the Sun. The laws in question are empirical, derived from careful observations. The options include famous laws from mechanics, chemistry, and logic. Concept / Approach: Johannes Kepler, using the observations of Tycho Brahe, formulated three famous laws of planetary motion. These laws describe that planets move in elliptical orbits with the Sun at one focus, sweep out equal areas in equal times, and relate the square of the orbital period to the cube of the semi-major axis. Newton later showed that these laws are a consequence of his law of universal gravitation and his three laws of motion. However, the specific name for the set of planetary motion laws is associated with Kepler. Step-by-Step Solution: Step 1: Recognise that the question is about the motion of planets in their orbits around the Sun. Step 2: Recall that Kepler gave three laws describing orbits, areas swept, and period–radius relationships for planets. Step 3: Match this description with the options given and identify the option that explicitly refers to planetary motion. Step 4: Select “Kepler laws of planetary motion” as the correct answer. Verification / Alternative check: Newton laws of motion are general laws applying to all bodies, not just planets. Avogadro laws relate to gases and the relationship between volume and amount of substance. De Morgan laws are associated with logic and set theory. Only Kepler name is directly and traditionally linked to the laws that describe planetary motion around the Sun. This cross-check ensures the correct identification. Why Other Options Are Wrong: Newton laws of motion: These are fundamental mechanical laws about forces and motion, but they are not the specific empirical laws that originally described planetary orbits. Avogadro gas laws: These refer to relationships in gas behaviour, such as equal volumes of gases containing equal numbers of molecules at the same temperature and pressure, and are unrelated to planetary motion. De Morgan logical laws: These are rules in Boolean algebra and logic concerning negation and logical operators, and have nothing to do with astrophysics. Common Pitfalls: Students sometimes confuse the names Newton and Kepler because both are associated with celestial mechanics. Newton explained why planets follow Kepler laws, but the observational laws themselves carry Kepler name. Another source of confusion is thinking that any famous scientist listed must be the correct choice. Instead, focus on which name is historically tied to planetary motion. Final Answer: The laws that govern planetary motion around the Sun are called Kepler laws of planetary motion.

Question 2

Using the relationship between electric charge, time, and current, find the electric current in amperes when a total charge of 500 coulombs flows through a conductor in 25 seconds.

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Introduction / Context: Electric current is a basic concept in electricity and electronics. It describes how much electric charge flows past a point in a circuit per unit time. This question tests your ability to apply the simple formula relating charge, time, and current. Being comfortable with this relationship is essential for solving many introductory problems in physics and electrical engineering. Given Data / Assumptions: Total electric charge, Q = 500 coulombs. Time taken for this charge to flow, t = 25 seconds. The circuit is assumed to be steady so that current can be treated as constant over this time interval. We are asked to find current I in amperes. Concept / Approach: The definition of electric current is the rate at which charge flows. Mathematically, this is written as I = Q / t, where I is the current in amperes, Q is the charge in coulombs, and t is the time in seconds. One ampere is defined as one coulomb of charge passing through a cross section of a conductor in one second. By substituting the given values of charge and time into this formula, we can directly calculate the current. Step-by-Step Solution: Step 1: Write the formula for current: I = Q / t. Step 2: Substitute Q = 500 coulombs and t = 25 seconds into the formula. Step 3: Compute I = 500 / 25. Step 4: Perform the division: 500 / 25 = 20. Step 5: Therefore, the current I = 20 amperes. Verification / Alternative check: We can quickly check the reasonableness of the result. If 25 seconds correspond to 500 coulombs, then in 1 second the charge flow is 500 / 25 = 20 coulombs per second. Since 1 ampere is 1 coulomb per second, 20 coulombs per second corresponds to 20 amperes. The units match and the arithmetic is straightforward, confirming that the value is correct. Why Other Options Are Wrong: 12500 A: This would correspond to an extremely large current and would arise from incorrectly multiplying rather than dividing the values. 40 A: This suggests a miscalculation, possibly using 500 / 12.5 or another incorrect combination of numbers. 6250 A: This again is unrealistically high and would result from a major arithmetic or formula error, such as using I = Q * t instead of I = Q / t. Common Pitfalls: Students sometimes confuse the formula and use I = t / Q or mistakenly multiply charge and time. Another common mistake is mishandling units, such as mixing coulombs with milli-coulombs or seconds with minutes without converting. Always remember that current in amperes is charge in coulombs divided by time in seconds. Keeping track of units is a powerful way to catch errors before finalising the answer. Final Answer: The electric current flowing is 20 A.

Question 3

In which type of motion of a body does its velocity change with time instead of remaining constant as it moves?

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Introduction / Context: Describing motion accurately is a central part of mechanics. Two important categories of motion are uniform and non-uniform motion. These categories describe how the speed or velocity of an object behaves over time. This question focuses on recognising which term is used when the velocity of a moving body changes as time passes. Given Data / Assumptions: The body is in motion along some path. We are told that its velocity varies with time. We want the standard term used in physics for such motion. Concept / Approach: Uniform motion is defined as motion in which an object covers equal distances in equal intervals of time, implying constant speed and, in a straight line, constant velocity. Non-uniform motion, on the other hand, is motion in which either the speed or direction (or both) of the object change with time. If the velocity is varying, the motion cannot be uniform and must be classified as non-uniform. Terms like linear motion and equilibrium motion describe other aspects and do not directly refer to whether velocity is constant or changing. Step-by-Step Solution: Step 1: Note that the body velocity is changing with time. Step 2: Recall that in uniform motion, velocity remains constant over time, so this description does not fit. Step 3: By definition, non-uniform motion is motion with changing speed or direction, meaning velocity is not constant. Step 4: Conclude that the correct classification is non-uniform motion and select that option. Verification / Alternative check: Consider a car whose speedometer reading increases from 30 km/h to 50 km/h over a few seconds. Its velocity is clearly changing, so its motion is non-uniform. In contrast, a train moving steadily at 60 km/h on a straight track would be in uniform motion. This comparison reinforces that the key requirement for uniform motion is constant velocity, which is not satisfied in the question. Why Other Options Are Wrong: Uniform motion: This refers to motion with constant velocity, which is the opposite of the situation described. Linear motion: This simply indicates motion along a straight line and does not specify whether velocity is constant or changing. Equilibrium motion: This is not a standard category for motion in basic mechanics and does not correctly describe changing velocity. Common Pitfalls: Students sometimes think any motion along a straight line must be uniform, which is not true. A car speeding up or slowing down along a straight road is still undergoing non-uniform motion because its velocity is changing. Confusing path shape (linear or curved) with time variation of velocity is a frequent source of error. Always focus on whether the velocity magnitude and direction remain constant over equal time intervals. Final Answer: The correct description is non-uniform motion.

Question 4

The force of friction between two solid surfaces in contact will increase under which of the following conditions?

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Introduction / Context: Friction is a contact force that resists the relative motion or tendency of motion between two surfaces in contact. Understanding what increases or decreases friction is essential for engineering, everyday life, and exam questions. This question asks you to identify the factor that will increase the force of friction between two surfaces. Given Data / Assumptions: Two solid surfaces are in contact with each other. The nature of the surfaces and the normal force can be altered. We are asked to determine which specific change increases friction. Concept / Approach: For dry friction between solid surfaces, the maximum static friction and kinetic friction are approximately given by F = mu * N, where mu is the coefficient of friction and N is the normal reaction between the surfaces. If the coefficient of friction remains unchanged, increasing the normal reaction N will increase the frictional force. Lubrication generally reduces friction by introducing a low-shear layer. Removing irregularities can also reduce friction by making surfaces smoother. Creating an air gap removes contact and therefore reduces friction to almost zero. Step-by-Step Solution: Step 1: Recall the relation F = mu * N, where F is friction, mu is the coefficient of friction, and N is the normal force. Step 2: Recognise that increasing N while keeping mu constant will increase F. Step 3: Pressing two surfaces harder together increases the normal reaction between them. Step 4: Therefore, the frictional force increases when the surfaces are pressed harder, making option about increased pressure the correct choice. Verification / Alternative check: Think about trying to push a heavy box on the floor. If someone sits on the box, increasing its weight, the normal force between the box and the floor increases. As a result, it becomes harder to push, indicating that the frictional force has increased. This everyday example supports the theoretical formula and confirms the correct option. Why Other Options Are Wrong: A layer of lubricant is placed between the two surfaces: Lubricants reduce friction by allowing surfaces to slide more easily over a thin fluid film. An air gap is created between the two surfaces: If the surfaces are separated, direct contact and thus solid friction disappear, greatly reducing friction. Irregularities on both surfaces are completely removed: Making surfaces smoother generally reduces friction because there are fewer microscopic interlocks. Common Pitfalls: Some students think that rougher surfaces always mean more friction and that pressing harder is less important. However, the frictional force depends on both the normal force and the coefficient of friction. For a given pair of surfaces, increasing the normal force is a direct and simple way to increase friction. Another mistake is assuming lubrication can increase friction, when its main purpose is in fact to reduce wear and friction. Final Answer: The frictional force will increase when the two surfaces are pressed harder against each other.

Question 5

In materials science, the property of a metal that allows it to be stretched and drawn into long, thin wires without breaking is called what?

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Introduction / Context: Metals have several characteristic mechanical properties that make them suitable for different applications. Two important ones are malleability and ductility. Malleability is related to forming sheets, while ductility is related to drawing wires. This question tests whether you can correctly identify the name of the property that allows metals to be drawn into wires without breaking. Given Data / Assumptions: The material in question is a metal. The process described involves drawing the metal into long, thin wires. The metal must withstand this stretching without fracture. Concept / Approach: Metals possess plasticity, which allows permanent deformation under stress. Ductility specifically refers to the ability of a material to be stretched into a wire, often measured by percentage elongation before fracture. Malleability, in contrast, is the ability to be hammered or rolled into thin sheets. Viscosity is a property of fluids, and tensile strength is a measure of maximum stress before breaking but is not itself the name of the wire-drawing property. Therefore, the correct term for wire drawing capability is ductility. Step-by-Step Solution: Step 1: Identify that the process described is drawing the material into wires. Step 2: Recall that ductility is defined as the ability of a material to be drawn into wires. Step 3: Recall that malleability refers instead to forming thin sheets by hammering or rolling. Step 4: Compare with other options and select ductility as the correct property. Verification / Alternative check: Common ductile metals include copper, aluminium, and gold, all of which are widely used to make electrical wires or jewellery wires. When you see thin metal wires used in power cables or communication lines, the property that allows such shaping is ductility. This real-world observation strongly supports the theoretical definition. Why Other Options Are Wrong: Malleability: This property allows metals to be hammered or rolled into thin sheets, such as aluminium foil, not drawn into wires. Viscosity: Viscosity describes the resistance of a fluid to flow and is primarily a property of liquids and gases, not solid metals. Tensile strength: This is a quantitative measure of the maximum stress a material can withstand while being stretched, but it is not the name of the ability to form wires. Common Pitfalls: Students often confuse malleability and ductility because both involve plastic deformation. A useful memory aid is to associate “d” in ductility with “drawing” into wires and “m” in malleability with “making” sheets. Another mistake is to think any property related to stretching, such as high tensile strength, must be the correct answer. However, terminology in materials science is precise, and ductility is the accepted term for wire-forming capability. Final Answer: The property is called ductility.

Question 6

In atmospheric and environmental physics, the instrument specifically designed to measure the degree of blueness of the sky or the ocean is known by which name?

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Introduction / Context: Many specialised instruments exist in physics, meteorology, and environmental science to measure different properties of the atmosphere and oceans. While common devices like thermometers and barometers are widely known, more specialised instruments are sometimes asked in general knowledge and competitive exams. This question focuses on the name of the instrument used to measure the blueness of the sky or the sea, which is related to how light scatters in the atmosphere or water. Given Data / Assumptions: The instrument is used to measure the degree of blueness. The blueness can refer to the sky or the surface appearance of the ocean. The options list several different scientific instruments. Concept / Approach: The word root “cyano” comes from a Greek term meaning blue. A cyanometer is an instrument with a graded scale of blue colour samples that allows observers to compare and quantify how blue the sky appears. It was historically used to study atmospheric conditions, dust content, and humidity that influence sky colour. Other instruments in the options measure different physical properties, not colour or blueness, so recognising the meaning of the prefixes and common uses of these devices helps to identify the correct answer. Step-by-Step Solution: Step 1: Look for clues in the names of the instruments, such as prefixes that suggest their function. Step 2: Note that “cyano” is associated with the colour blue, suggesting a link to blueness. Step 3: Recall that a cyanometer is historically used to measure and compare sky blue intensity. Step 4: Confirm that other instruments listed are used for different purposes, and select cyanometer as the correct option. Verification / Alternative check: A quick context check for the other instruments helps verify the choice. A barometer measures atmospheric pressure, which is important for weather prediction but not directly for colour measurement. A bathymeter is associated with measuring depth, particularly in oceans. A ceraunograph is related to recording thunder or lightning. Only cyanometer fits the description of measuring blueness, confirming our selection. Why Other Options Are Wrong: Bathymeter: This instrument is used to measure water depth, especially in oceans, and has no direct connection to measuring colour. Ceraunograph: This device is used to record lightning or thunder activity and is associated with storms, not sky colour. Barometer: This measures atmospheric pressure and is widely used in meteorology but does not quantify blueness of the sky or ocean. Common Pitfalls: Many learners are unfamiliar with rare instruments and may guess based on what they have heard most often, such as barometer. Another confusion arises from not recognising common prefixes like “cyano”. Developing the habit of breaking scientific terms into their roots can help deduce meanings even for unfamiliar words. For example, remembering that “baro” relates to pressure, “bathys” to depth, and “cyano” to blue gives a powerful tool for educated guessing. Final Answer: The instrument for measuring the blueness of the sky or ocean is called a cyanometer.

Question 7

In microeconomics, the price of an article decreases from Rs 240 to Rs 220 and, as a result, the quantity demanded increases from 200 units to 210 units. Using the point elasticity of demand formula based on averages, what is the price elasticity of…

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Introduction / Context: Price elasticity of demand measures how responsive quantity demanded is to changes in price. It is a key concept in microeconomics used to understand consumer behaviour, pricing strategies, and tax policies. This question asks you to calculate the point elasticity of demand between two nearby price–quantity combinations using the average or midpoint values. The sign of elasticity is important because it indicates the direction of the relationship between price and quantity demanded. Given Data / Assumptions: Initial price P1 = Rs 240. New price P2 = Rs 220. Initial quantity Q1 = 200 units. New quantity Q2 = 210 units. We use the point elasticity formula based on averages for a small change. Concept / Approach: For small changes, a commonly used approximation of point elasticity of demand is: Elasticity (E) = (ΔQ / ΔP) * (P_avg / Q_avg) where ΔQ = Q2 - Q1, ΔP = P2 - P1, P_avg = (P1 + P2) / 2, and Q_avg = (Q1 + Q2) / 2. Since demand usually falls when price rises and rises when price falls, the elasticity for a normal good is typically negative. This negative sign reflects the inverse relationship between price and quantity demanded. Step-by-Step Solution: Step 1: Compute the change in quantity: ΔQ = Q2 - Q1 = 210 - 200 = 10 units. Step 2: Compute the change in price: ΔP = P2 - P1 = 220 - 240 = -20 rupees. Step 3: Compute the average price: P_avg = (P1 + P2) / 2 = (240 + 220) / 2 = 460 / 2 = 230 rupees. Step 4: Compute the average quantity: Q_avg = (Q1 + Q2) / 2 = (200 + 210) / 2 = 410 / 2 = 205 units. Step 5: Compute the ratio ΔQ / ΔP = 10 / (-20) = -0.5. Step 6: Now calculate elasticity: E = (-0.5) * (P_avg / Q_avg) = (-0.5) * (230 / 205). Step 7: Evaluate 230 / 205 which is approximately 1.12, giving E ≈ -0.5 * 1.12 ≈ -0.56, which rounds to about -0.6. Step 8: Therefore, the price elasticity of demand is approximately -0.6. Verification / Alternative check: We can check the magnitude of elasticity. Since the percentage change in quantity is small compared to the percentage change in price, we expect the elasticity to have magnitude less than 1, indicating inelastic demand. The negative sign is expected because price and quantity demanded move in opposite directions. The computed value of about -0.6 matches these expectations and is close to the option -0.6, confirming the result. Why Other Options Are Wrong: 0.6: This ignores the negative sign and incorrectly suggests a direct relationship between price and quantity demanded, which contradicts the usual law of demand. -1.8: This magnitude is much larger than our calculation and would imply a much more responsive demand than observed. It does not match the correct computation. 1.8: This is not only the wrong magnitude but also positive, which again suggests an incorrect relationship between price and quantity. Common Pitfalls: Students often forget the negative sign in elasticity of demand or mix up the formula by using Q_avg / P_avg instead of P_avg / Q_avg. Another mistake is failing to use averages and instead using only initial values, which gives a less accurate approximation for discrete changes. Careful substitution and attention to signs are crucial to avoid these errors. Final Answer: The point price elasticity of demand is approximately -0.6.

Question 8

A convex lens forms an image at a distance of 16 cm from the lens when an object is placed on the opposite side of the lens at a distance of 20 cm from its optical centre. Using the lens formula and the definition of power, find the power of the len…

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Introduction / Context: Convex lenses are converging lenses that can form real or virtual images depending on the position of the object. The power of a lens is a measure of how strongly it converges or diverges light and is widely used in optics, particularly in eyeglasses and optical instruments. This question asks you to find the power of a convex lens given the object distance and image distance using basic lens formulas. Given Data / Assumptions: Object distance magnitude = 20 cm. Image distance magnitude = 16 cm. The lens is convex and forms a real image on the opposite side. We use the standard sign convention for thin lenses and the lens formula. Concept / Approach: The thin lens formula is: 1 / f = 1 / v - 1 / u where f is focal length, v is image distance, and u is object distance. Using the usual sign convention for a convex lens, we take u as negative for an object placed on the incoming side and v as positive for a real image on the outgoing side. Once we find f in meters, the power P of the lens is given by: P = 1 / f (in meters) and is expressed in diopters. Step-by-Step Solution: Step 1: Assign signs. For a convex lens, take object distance u = -20 cm and image distance v = +16 cm. Step 2: Use the lens formula: 1 / f = 1 / v - 1 / u. Step 3: Substitute values: 1 / f = 1 / 16 - 1 / (-20) = 1 / 16 + 1 / 20. Step 4: Calculate 1 / 16 + 1 / 20. Use a common denominator 80: 1 / 16 = 5 / 80 and 1 / 20 = 4 / 80. So 1 / f = (5 + 4) / 80 = 9 / 80. Step 5: Therefore, f = 80 / 9 cm, which is approximately 8.89 cm. Step 6: Convert focal length to meters: f ≈ 8.89 cm = 0.0889 m. Step 7: Compute power P = 1 / f ≈ 1 / 0.0889 ≈ 11.25 diopters. Step 8: Since the lens is convex and converging, the power is positive, so the correct answer is 11.25 diopters. Verification / Alternative check: We can quickly check the result by estimating. A lens with focal length around 10 cm has power around 10 diopters. Our calculated focal length is slightly less than 9 cm, so a power just above 11 diopters is reasonable. Also, convex lenses always have positive power values in the usual convention, so negative options can be discarded for a converging lens in this context. Why Other Options Are Wrong: -3.75 diopters: The magnitude is much smaller than calculated and the negative sign would correspond to a diverging lens, not a convex converging lens. -11.25 diopters: The magnitude matches our calculation but the sign is wrong. A convex lens has positive power, so a negative value is not correct here. 3.75 diopters: This is too small in magnitude compared to the value obtained from the correct focal length. Common Pitfalls: Common mistakes include forgetting to use the correct sign convention for u and v, mixing up centimeters and meters when computing power, and inverting the power formula. Some students directly use focal length in centimeters in P = 1 / f, which gives the wrong numerical value because power must be calculated from f in meters. Carefully converting units and applying the formula step by step helps avoid these errors. Final Answer: The power of the convex lens is 11.25 diopters.

Question 9

In ray optics, an image formed by a plane mirror that cannot be received or captured on a physical screen is called what type of image?

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Introduction / Context: When light reflects from mirrors or passes through lenses, different kinds of images can be formed. These images are classified as real or virtual, and also as erect or inverted. Understanding the difference between real and virtual images is fundamental in ray optics. This question asks about the specific nature of the image formed by a plane mirror that cannot be projected onto a screen. Given Data / Assumptions: The reflecting surface is a plane mirror. An image is seen in the mirror. The image cannot be obtained on a physical screen placed in front of the mirror. Concept / Approach: A real image is formed when light rays actually converge to a point, and such an image can be projected onto a screen. A virtual image is formed when light rays appear to diverge from a point behind the mirror or lens; the rays do not come from that point physically, so the image cannot be formed on a screen. Plane mirrors always form images that are virtual, erect, laterally inverted, and of the same size as the object. Therefore, the term that best matches an image that cannot be obtained on a screen is virtual image. Step-by-Step Solution: Step 1: Recall that a plane mirror produces images that appear to be behind the mirror. Step 2: Recognise that since the light rays do not actually meet behind the mirror, the image is virtual. Step 3: Note that virtual images cannot be formed on a screen because there is no real point of convergence of light rays. Step 4: Identify the option that describes an image not obtainable on a screen, which is a virtual image. Verification / Alternative check: You can test this by trying to place a sheet of paper in front of a plane mirror where you see your face. No matter how you position the paper, your face will not appear on the paper as a projected image. This shows that the image in the plane mirror is virtual. In contrast, with a converging lens and a suitably placed object, you can form a real image on a screen, demonstrating the difference. Why Other Options Are Wrong: Real image: A real image can be projected onto a screen, so this does not match the behaviour of a plane mirror. Inverted image: The plane mirror forms erect, not inverted, images. Inversion refers to orientation, not whether the image can be captured on a screen. Erect image: While a plane mirror does form an erect image, erect does not indicate whether it is real or virtual. The key feature in the question is the inability to form the image on a screen, which defines virtual. Common Pitfalls: Students sometimes confuse erect with virtual and inverted with real, but these pairs are not always linked. For example, some virtual images can be inverted in certain optical systems. The correct distinction is whether light actually converges to form the image (real) or appears to originate from a point (virtual). Always focus on ray convergence and the possibility of projection onto a screen when deciding between real and virtual. Final Answer: An image formed by a plane mirror that cannot be obtained on a screen is a virtual image.

Question 10

The attractive or repulsive force exerted by a magnet on another magnetic material is an example of which type of physical force?

Accepted Answer

Introduction / Context: Forces in physics are often classified based on how they act: some require direct contact between objects, while others act at a distance. Magnetism is a familiar phenomenon where magnets can attract or repel each other or attract magnetic materials without touching. This question asks you to identify the correct category for the force exerted by a magnet. Given Data / Assumptions: A magnet exerts a force on another magnet or magnetic object. The objects can interact even when they are not in direct physical contact. The question focuses on the type of force, not its source details. Concept / Approach: Forces that require direct physical contact, such as friction, tension, or muscular pushes and pulls, are called contact forces. Forces like gravity, electrostatic forces, and magnetic forces can act over a distance without contact and are called non-contact or action-at-a-distance forces. The magnetic force between magnets or between a magnet and iron, for example, can be felt even when there is air between them. Therefore, it is correctly classified as a non-contact force. Step-by-Step Solution: Step 1: Observe that magnets can attract or repel each other without touching. Step 2: Recognise that such forces, which act through a field over a distance, are non-contact forces. Step 3: Compare the options and identify which category describes forces that act without direct contact. Step 4: Select non-contact force as the correct classification. Verification / Alternative check: Think of other non-contact forces: gravity attracts objects towards the Earth without any visible connection, and electrostatic charges can attract or repel pieces of paper without touching them. Magnetism behaves in a similar way, which places it in the non-contact category. This grouping is well established in school physics and textbooks. Why Other Options Are Wrong: Muscular force: This is exerted by muscles and always involves physical contact, such as pushing or pulling an object. Contact force: This is the opposite category and requires touching, so it does not correctly describe magnetic interactions at a distance. Electrostatic force: Although electrostatic force is also a non-contact force, the question specifically mentions a magnet, not electric charges, so this option does not match the described situation. Common Pitfalls: Some learners confuse the names of different non-contact forces and think any force that acts at a distance can be called electrostatic or gravitational. It is important to link each type of force to its cause: mass for gravity, charge for electrostatics, and magnetic poles or moving charges for magnetism. Another pitfall is to think forces only occur when there is visible contact. Observing everyday situations with magnets helps correct this misconception. Final Answer: The force exerted by a magnet is an example of a non-contact force.

Question 11

For a body moving in a straight line with uniform (constant) acceleration, its final velocity can be expressed in terms of its average velocity and initial velocity by which of the following relations?

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Introduction / Context: In uniformly accelerated motion along a straight line, there are several useful kinematic equations that relate displacement, time, initial velocity, final velocity, and acceleration. One of these relations connects the average velocity to the initial and final velocities. From that, we can also express the final velocity in terms of the average and initial velocities. This question tests your understanding of these basic kinematic relationships. Given Data / Assumptions: Motion is along a straight line. Acceleration is uniform (constant) during the time interval considered. We have an initial velocity u, a final velocity v, and an average velocity v_avg. We want an expression for v in terms of v_avg and u. Concept / Approach: For uniformly accelerated motion in a straight line, the average velocity over a time interval is simply the arithmetic mean of the initial and final velocities: v_avg = (u + v) / 2 From this equation, we can solve for v in terms of v_avg and u by rearranging. This algebraic manipulation gives a direct formula for final velocity in the desired form, which we can then compare with the options. Step-by-Step Solution: Step 1: Start with the relationship v_avg = (u + v) / 2. Step 2: Multiply both sides by 2 to remove the denominator: 2 * v_avg = u + v. Step 3: Rearrange to solve for v: v = 2 * v_avg - u. Step 4: Express this in words: final velocity equals two times the average velocity minus the initial velocity. Step 5: Compare this expression with the options and choose the one that matches. Verification / Alternative check: As a quick check, imagine a case where the initial velocity u = 0 and the final velocity v is some positive value. In uniformly accelerated motion, the average velocity would then be v_avg = (0 + v) / 2 = v / 2. Applying our relation v = 2 * v_avg - u gives v = 2 * (v / 2) - 0 = v, which is correct. This confirms that the algebra and the relationship are consistent with a simple example. Why Other Options Are Wrong: Final velocity = average velocity minus initial velocity: This does not follow from the equation v_avg = (u + v) / 2 and gives wrong values when tested with numerical examples. Final velocity = 2 × average velocity plus initial velocity: This would give v = 2 * v_avg + u, which is inconsistent with the derivation and leads to an incorrect relationship. Final velocity = average velocity plus initial velocity: This relation has no basis in standard kinematic equations and is dimensionally misleading in this context. Common Pitfalls: A common error is to misremember the formula for average velocity, sometimes writing v_avg = (v - u) / 2 instead of (u + v) / 2. Another frequent mistake is incorrect algebra when rearranging equations, such as forgetting to multiply both sides by 2 or incorrectly isolating v. Careful step-by-step manipulation and checking with simple test values help avoid these issues. Final Answer: For uniformly accelerated motion, the final velocity is final velocity = 2 × average velocity minus initial velocity.

Question 12

A body moving in a straight line speeds up from an initial velocity of 25 m/s to a final velocity of 30 m/s in 10 seconds. What is the magnitude of its acceleration in m/s^2?

Accepted Answer

Introduction / Context: Acceleration is the rate of change of velocity with respect to time. It is a core concept in kinematics and dynamics. This question provides initial and final velocities and the time taken for the change, then asks you to calculate the acceleration. Such straightforward problems reinforce how to use the basic definition of acceleration in numerical form. Given Data / Assumptions: Initial velocity u = 25 m/s. Final velocity v = 30 m/s. Time interval t = 10 seconds. Motion is along a straight line and acceleration is assumed constant during this interval. Concept / Approach: For motion with constant acceleration, the acceleration a is defined as the change in velocity divided by the time taken: a = (v - u) / t Here, v is the final velocity, u is the initial velocity, and t is the time interval. Positive values of a indicate speeding up in the direction of motion, while negative values indicate slowing down. Step-by-Step Solution: Step 1: Write the formula for acceleration: a = (v - u) / t. Step 2: Substitute the given values: u = 25 m/s, v = 30 m/s, and t = 10 s. Step 3: Compute the change in velocity: v - u = 30 - 25 = 5 m/s. Step 4: Divide by the time: a = 5 / 10 = 0.5 m/s^2. Step 5: Therefore, the acceleration magnitude is 0.5 m/s^2. Verification / Alternative check: If the body increases its speed by 5 m/s over 10 seconds, then each second the speed increases by 0.5 m/s on average. This intuitive interpretation matches the calculated acceleration of 0.5 m/s^2. The units also work out correctly: (m/s) divided by s gives m/s^2, which is the correct unit for acceleration. Why Other Options Are Wrong: 5.5 m/s^2: This is much larger than the correct value and cannot be obtained by correctly applying the formula. 1 m/s^2: This would correspond to a 10 m/s increase in velocity over 10 seconds, which does not match the given data. 11 m/s^2: This would require a very large change in velocity over a short time and is incompatible with the numbers given in the question. Common Pitfalls: Errors often arise from reversing u and v, leading to incorrect signs, or from dividing the wrong way round, such as t / (v - u). Some students also forget to include the correct units in the final answer. Always remember that acceleration is change in velocity per unit time and that the direction and sign can be important in more complex problems. Final Answer: The acceleration of the body is 0.5 m/s^2.

Question 13

During uniform motion of an object along a straight line, which physical quantity associated with its motion remains constant with time?

Accepted Answer

Introduction / Context: Uniform motion is one of the simplest types of motion studied in physics. It is important to clearly understand what remains constant and what can change in such motion. This question focuses on uniform motion along a straight line and asks you to identify which physical quantity stays constant with time. Given Data / Assumptions: The motion is along a straight line. The motion is described as uniform. We compare quantities such as acceleration, deceleration, force, and velocity. Concept / Approach: Uniform motion in a straight line means the object covers equal distances in equal intervals of time in a fixed direction. Therefore, both the speed and direction of motion remain constant, which means that the velocity remains constant. If velocity is constant, then acceleration, which is the rate of change of velocity, must be zero. There is no deceleration, and the net force along the line of motion is zero according to Newton second law. Step-by-Step Solution: Step 1: Recall that uniform motion implies constant speed and direction. Step 2: Therefore, the velocity, which includes both speed and direction, remains constant. Step 3: Since velocity does not change, acceleration and deceleration are both zero and do not remain non-zero constants. Step 4: According to Newton second law, if acceleration is zero, the net force must also be zero, so force is not a non-zero constant in this context. Step 5: Conclude that the correct constant quantity is velocity. Verification / Alternative check: Imagine a car on a straight highway moving at a steady 60 km/h without changing direction or speed. Its velocity remains constant. There is no change in speed, so acceleration is zero. There is no continuous non-zero force pushing it to increase speed; instead, forces are balanced. This simple example confirms that in uniform straight-line motion, velocity is the quantity that stays constant over time. Why Other Options Are Wrong: Acceleration: In uniform motion, acceleration is zero because velocity does not change with time. It does not remain a non-zero constant. Deceleration: Deceleration refers to negative acceleration and is absent in uniform motion, where speed is not decreasing. Force: The net force is zero in uniform motion, but the question is about the motion quantity that remains constant. Velocity describes the motion directly, not the force. Common Pitfalls: Some students confuse the idea of uniform motion with uniformly accelerated motion. In uniformly accelerated motion, acceleration is constant but velocity changes. In uniform motion, it is velocity that is constant and acceleration that is zero. Keeping these two cases conceptually separate is important for solving kinematics problems correctly. Final Answer: During uniform straight-line motion, the velocity of the object remains constant with time.

Question 14

A constant force acts on two different objects that are initially at rest and have different masses, but the force is applied for the same duration of time on each. Which physical quantity will be the same for both objects after the force has acted?

Accepted Answer

Introduction / Context: This question combines basic ideas from Newton laws of motion with the concept of momentum. It asks you to reason about what happens when the same force is applied for the same time to objects of different masses that start from rest. Understanding the relationship between force, time, and change in momentum is crucial for many problems in mechanics and collisions. Given Data / Assumptions: Two objects have different masses. Both objects are initially at rest, so their initial velocities are zero. The same constant force F acts on each object. The force is applied for the same time interval t for both objects. We are asked which physical quantity is the same for both after the interaction. Concept / Approach: Newton second law in impulse form states that the change in momentum (Δp) of an object is equal to the impulse applied to it: Δp = F * t If the same force F acts for the same time t on both objects, then F * t is the same for both, so the change in momentum is the same. Since both start from rest, their initial momentum is zero, which means their final momentum magnitudes are equal. However, because their masses differ, their accelerations and final velocities will be different, and thus their kinetic energies will also differ. Step-by-Step Solution: Step 1: Use the impulse–momentum relation: Δp = F * t. Step 2: For each object, the same force F acts for the same time t, so the impulse F * t is identical. Step 3: Since both objects start from rest, their initial momentum is zero. Step 4: Therefore, final momentum p_final = Δp = F * t is the same for both objects. Step 5: Conclude that momentum is the physical quantity that will be equal for both objects after the force acts. Verification / Alternative check: Let object 1 have mass m1 and object 2 have mass m2. After time t under the same force F, accelerations are a1 = F / m1 and a2 = F / m2, which are different if the masses differ. The velocities after time t are v1 = a1 * t and v2 = a2 * t. However, the momentum p1 = m1 * v1 = m1 * (F / m1) * t = F * t, and p2 = m2 * v2 = m2 * (F / m2) * t = F * t. Thus, p1 and p2 are equal, confirming that the momentum is the same. Why Other Options Are Wrong: Acceleration: Since acceleration a = F / m, different masses experience different accelerations under the same force, so acceleration is not equal. Kinetic energy: Kinetic energy depends on mass and velocity (KE = 0.5 * m * v^2). Because masses and velocities differ, kinetic energies will generally be different. Velocity: Objects with different masses under the same force will have different accelerations and therefore different velocities after the same time interval. Common Pitfalls: Students sometimes assume that because the same force and time are used, velocities must be the same, overlooking how acceleration depends on mass. Others think kinetic energy, rather than momentum, is equal because the same work might be assumed to be done, which is not true in this setup. Remember that impulse directly relates to change in momentum, not to velocity or kinetic energy alone. Final Answer: The physical quantity that will be the same for both objects is momentum.

Question 15

In basic kinematics, which of the following correctly gives the formula for distance travelled by an object moving at constant speed for a given time?

Accepted Answer

Introduction / Context: One of the first relationships students learn in physics is the simple formula connecting distance, speed, and time. It is widely used in everyday problems involving travel, such as finding how far a vehicle goes in a certain time at a given speed. This question checks your understanding of the correct algebraic relationship among these quantities when speed is constant. Given Data / Assumptions: The object moves with constant speed. We consider a time interval t for which speed does not change. We want the formula that expresses distance travelled in terms of speed and time. Concept / Approach: Speed is defined as the distance travelled per unit time. In symbols, this is: speed = distance / time Rearranging this equation to make distance the subject gives: distance = speed × time This relation holds when speed is constant over the time interval considered, which is the case in this simple kinematics situation. Step-by-Step Solution: Step 1: Start from the definition of speed: speed = distance / time. Step 2: Multiply both sides by time to solve for distance. Step 3: This gives distance = speed × time. Step 4: Compare this final expression with the options and select the matching one. Verification / Alternative check: Consider an example: if a car travels at a constant speed of 60 km/h for 2 hours, the distance travelled is 60 × 2 = 120 km. This matches everyday reasoning and confirms that multiplying speed and time gives distance. Using any of the other suggested formulas would produce incorrect results that do not match reality. Why Other Options Are Wrong: Distance = time ÷ speed: This is a rearrangement that would make units inconsistent, giving time squared per length, which is not appropriate for distance. Distance = speed × acceleration: Multiplying speed and acceleration gives units of m^2/s^3 in SI units, which does not correspond to distance and has no standard physical meaning in this context. Distance = velocity ÷ speed: Velocity and speed have the same units, so their ratio is dimensionless and cannot represent distance, which must have units of length. Common Pitfalls: Students sometimes mix up the roles of variables and may incorrectly believe that distance is speed divided by time, which reverses the correct relationship. Dimensional analysis is a good tool to catch such mistakes: distance must have units of length, while speed has length per time and time has units of time, so multiplying speed by time gives length. Remembering and checking units will often prevent formula errors. Final Answer: The correct formula is distance = speed × time.

Question 16

From the point of view of atomic structure, which of the following is the main reason why metallic solids are good conductors of electricity?

Accepted Answer

Introduction / Context: Metals are widely used in electrical wiring and electronic components because they conduct electricity very well. This good conductivity is closely tied to the internal structure of metals at the atomic and electronic level. The question asks you to identify the primary reason that explains why metals allow electric current to flow so easily. Given Data / Assumptions: We are considering metallic solids such as copper, aluminium, and silver. Electricity conduction refers to the movement of electric charge through the metal. Several material properties are proposed as possible reasons. Concept / Approach: In metallic bonding, atoms in the metal lattice contribute some of their outer electrons to form a sea of delocalised or free electrons that can move throughout the crystal. These mobile electrons act as charge carriers when a potential difference is applied, resulting in electric current. Properties such as melting point, tensile strength, or atomic density may influence other characteristics, but they are not the fundamental cause of electrical conductivity in metals. Step-by-Step Solution: Step 1: Recall that electric current in a metal is due to the motion of electrons. Step 2: In metals, the outer electrons are not tightly bound to individual atoms and can move relatively freely. Step 3: When a voltage is applied across a metal, these free electrons drift in a preferred direction, forming an electric current. Step 4: Thus, the key reason for good conductivity is the presence of many free or delocalised electrons, not properties like melting point or tensile strength. Verification / Alternative check: Non-metals, such as rubber or glass, do not have a comparable sea of free electrons and are typically insulators. Alloys or materials with fewer free electrons have lower conductivity. This contrast supports the idea that mobile electrons are the crucial factor. Experiments on temperature dependence of resistance in metals also show behaviour that fits the free electron model. Why Other Options Are Wrong: Because they have a low melting point: Many metals, like iron and tungsten, have high melting points yet conduct electricity well. Melting point is not the key factor for conduction. Because they have high tensile strength: Tensile strength measures resistance to pulling forces and is mainly mechanical, not directly related to electrical conductivity. Because they have very high atomic density: While metals often have closely packed atoms, high density alone does not guarantee good electrical conduction. Some dense materials are poor conductors. Common Pitfalls: Students sometimes mix up physical properties and assume that any strong or dense material must be a good conductor. Another confusion is thinking that ions, rather than electrons, carry current in metals. In metallic solids, conduction is dominated by electrons, while ionic conduction is more relevant in solutions and molten salts. Always link metallic conductivity to the presence of free electrons. Final Answer: Metals conduct electricity mainly because they contain many free electrons.

Question 17

Within the electromagnetic spectrum, which of the following types of radiation has the lowest frequency and therefore the longest wavelength?

Accepted Answer

Introduction / Context: The electromagnetic spectrum covers a wide range of electromagnetic waves, from very low frequency radio waves to extremely high frequency gamma rays. Each region of the spectrum has different properties and applications. This question tests your knowledge of the relative order of frequency among visible light, gamma rays, X-rays, and ultraviolet rays. Given Data / Assumptions: We are comparing the frequencies of four types of electromagnetic radiation. The options are visible light, gamma rays, X-rays, and ultraviolet rays. The question specifically asks for the lowest frequency among these. Concept / Approach: In the electromagnetic spectrum, frequency increases as we move from radio waves to microwaves, infrared, visible light, ultraviolet, X-rays, and finally gamma rays. Consequently, wavelength decreases in this direction, because the speed of light c is given by c = frequency × wavelength. Among the four given options, visible light lies at a lower frequency region compared to ultraviolet, X-rays, and gamma rays, which are progressively higher in frequency and energy. Step-by-Step Solution: Step 1: Recall the basic order of the electromagnetic spectrum from low to high frequency: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma ray. Step 2: Locate each option within this order. Step 3: Visible light comes before ultraviolet, X-rays, and gamma rays, so it has a lower frequency than all of them. Step 4: Therefore, among the provided options, visible light has the lowest frequency. Verification / Alternative check: You can think of energy per photon, which is proportional to frequency (E = h * f). Gamma rays are known to be extremely energetic, used in nuclear and high energy processes. X-rays, used in medical imaging, are also highly energetic. Ultraviolet rays have enough energy to cause chemical changes and sunburn. Visible light is less energetic than ultraviolet and therefore has a lower frequency. This energy-based reasoning matches the spectral ordering. Why Other Options Are Wrong: Gamma rays: These have very high frequencies and the shortest wavelengths in the electromagnetic spectrum among the given options, not the lowest frequency. X-rays: These have frequencies higher than ultraviolet and visible light and are used for penetrating materials, indicating high energy. Ultra violet rays: These have frequencies higher than visible light but lower than X-rays and gamma rays, so they do not have the lowest frequency among the four. Common Pitfalls: Students often remember that visible light sits in the middle of the entire spectrum, but forget its position relative to ultraviolet and X-rays. Confusing wavelength and frequency is also common: lower frequency means longer wavelength, and higher frequency means shorter wavelength. Keeping the full order of the spectrum in mind and linking it to common examples (sunlight, medical X-rays, nuclear gamma rays) helps prevent these mistakes. Final Answer: Among the given options, visible light has the lowest frequency.

Question 18

In electrical science, what is the SI unit used to measure the electric resistance of a conductor?

Accepted Answer

Introduction / Context: Electric resistance is a fundamental quantity in circuit theory, describing how strongly a component opposes the flow of electric current. It plays a key role in Ohm law and in calculating current, voltage, and power in electrical networks. This question asks you to recall the correct SI unit used to measure resistance. Given Data / Assumptions: We are dealing with electric resistance of a conductor. The unit requested is the standard SI unit. The options list several different physical units. Concept / Approach: According to Ohm law, resistance R is defined by the relation: R = V / I where V is potential difference in volts and I is current in amperes. The SI unit of resistance is derived from this relation and is called the ohm, symbolised by the Greek letter omega in notation, but written as “ohm” in text. Other listed units, such as dyne, pascal, and joule, belong to different physical quantities. Step-by-Step Solution: Step 1: Recall Ohm law, which relates voltage, current, and resistance. Step 2: From R = V / I, note that units are volt per ampere. Step 3: Recognise that the SI unit corresponding to volt per ampere is named ohm. Step 4: Compare with the options and select ohm as the correct unit. Verification / Alternative check: In circuit diagrams and component specifications, resistances are given in ohms, kilo-ohms, or mega-ohms (for example, 100 ohm resistor, 1 kilo-ohm resistor). This consistent usage in electronics and physics confirms that ohm is the correct SI unit for resistance. None of the other units listed are ever used to describe resistance in standard practice. Why Other Options Are Wrong: Dyne: This is a unit of force in the cgs system, not an SI unit and not related to resistance. Pascal: This is the SI unit of pressure, equal to one newton per square meter, and is unrelated to electrical resistance. Joule: This is the SI unit of energy or work, not of resistance. Common Pitfalls: Some students may confuse ohm with other electrical units such as volt or ampere, especially when learning multiple concepts at once. Remember that volts measure potential difference, amperes measure current, and ohms measure resistance. A helpful memory aid is the simple phrase from Ohm law: V (volt) = I (ampere) × R (ohm), which links all three units together. Final Answer: The SI unit of electric resistance is the ohm.

Question 19

Which scientific device or instrument is commonly used to measure the depth of the ocean or sea bed?

Accepted Answer

Introduction / Context: This general knowledge physics question asks about the specialised instrument used to measure the depth of the ocean or sea bed. Measuring sea depth is important for navigation, submarine operations, laying underwater cables, and understanding ocean topography. Different scientific instruments are used to measure length, density, or very small sizes, so it is important to connect each instrument name with its correct application. Given Data / Assumptions: We are interested in measuring the depth of the ocean or sea bed below the water surface. The options list four instruments: Lexometer, Nanometer, Fathometer, and Hydrometer. Each instrument is associated with a different type of physical measurement. Standard textbook definitions of these devices are assumed. Concept / Approach: A fathometer is an instrument that measures depth using the echo sounding principle. It sends ultrasonic or sound pulses downward into the water. These pulses reflect from the sea bed and return to the instrument. By measuring the time taken for the echo to return and knowing the speed of sound in water, the depth of the ocean is calculated. Other instruments in the list are used in very different contexts and are not suitable for depth sounding in oceans. Step-by-Step Solution: Step 1: Recall that ocean depth is measured using echo sounding, a technique based on sound waves in water. Step 2: Identify the instrument associated with echo sounding and depth measurement, namely the fathometer. Step 3: Recognise that a nanometer is a unit or instrument related to very small distances on the order of 10^-9 metre, not kilometres of ocean depth. Step 4: Note that a hydrometer is used to measure the relative density or specific gravity of liquids, not depth. Step 5: Conclude that the correct choice for measuring ocean depth is the fathometer. Verification / Alternative check: Ship navigation charts and descriptions of marine exploration often mention echo sounders and fathometers as standard equipment. Historical depth units included the fathom, which gives the instrument its name. Modern sonar systems are based on the same principle, confirming that fathometers are indeed depth measuring devices. The other instruments have well established uses that do not match ocean depth measurement, which supports the conclusion. Why Other Options Are Wrong: Lexometer: This name is rarely used in standard physics contexts for ocean depth and is not the recognised echo sounding instrument. Nanometer: Linked with measuring or indicating nanometre scale lengths, used in optics or materials science, not in marine depth measurement. Hydrometer: Designed to measure the relative density of liquids such as acids, milk, or battery electrolyte, not vertical distance below water surface. Common Pitfalls: A common confusion is to pick hydrometer because it is clearly related to liquids. However, hydrometer measures density, not depth. Another error is to be distracted by the word nano and think about precision, but nanometer refers to extremely small lengths, which is the opposite scale from ocean depths. Remember that ocean depth is measured by echo sounding, and fathometer is the classic name for such an instrument on ships. Final Answer: The device used to measure the depth of the ocean is the fathometer.

Question 20

Which of the following statements about X rays is not true according to basic physics and medical imaging?

Accepted Answer

Introduction / Context: This conceptual physics question tests knowledge of the basic properties of X rays. X rays are a form of high energy electromagnetic radiation used widely in medical imaging, security scanners, and crystallography. Knowing which statements accurately describe X rays helps avoid misconceptions about their behaviour, risks, and applications. Given Data / Assumptions: Four statements describe properties of X rays. We must identify the statement that is not true. Standard behaviour of X rays in vacuum and in matter is assumed. Basic facts from school level physics about electromagnetic waves are used. Concept / Approach: X rays are electromagnetic waves with short wavelength and high frequency. Like visible light, they travel at the speed of light in vacuum. They have strong penetrating power and can pass through many materials that visible light cannot, which is why they are used to see inside the human body or luggage. X rays can cause chemical changes and can expose photographic plates or films. Under special arrangements, such as crystals or multilayer mirrors, X rays can also show reflection, refraction and diffraction effects. Therefore, any statement claiming that X rays have low penetrating power is not correct. Step-by-Step Solution: Step 1: Recall that X rays are electromagnetic waves, so in vacuum they travel at the speed of light, about 3 x 10^8 m/s. Thus, the statement about speed of light is true. Step 2: Recognise that X rays can expose and darken photographic plates, which is why early X ray images were captured on film. So this statement is also true. Step 3: Understand that X rays can show reflection or refraction at grazing angles or in crystal lattices, so with suitable setups, reflection and refraction can be observed. This statement is acceptable. Step 4: Compare this with the first statement that X rays have low penetrating power. In reality X rays have high penetrating power compared with visible light, so this statement is false. Step 5: Therefore, the statement about low penetrating power is the one that is not true. Verification / Alternative check: Medical X ray machines can produce images of bones and internal organs because X rays pass through soft tissues more easily than through bone, demonstrating significant penetration. Security scanners at airports also rely on the ability of X rays to pass through bags. If X rays truly had low penetrating power, they would be stopped at the surface and could not be used for these purposes. This practical evidence confirms that the statement claiming low penetrating power is incorrect. Why Other Options Are Wrong: X rays travel with the speed of light in vacuum: This is correct for all electromagnetic waves, including X rays. X rays can be reflected, refracted or diffracted: Under special conditions, particularly with crystals and grazing incidence optics, X rays show these wave behaviours, so this is true. X rays can affect photographic plates: X rays cause chemical changes similar to visible light, so they can expose photographic film, which is a true statement. Common Pitfalls: Students sometimes think that because X rays are dangerous, they might be easily blocked, and therefore assume low penetration. In fact, the danger partly arises from their high penetration and ionising ability. Another confusion is to think only visible light gets reflected or refracted, forgetting that all electromagnetic waves can show these effects under suitable conditions. Keeping in mind that X rays are simply higher energy electromagnetic waves helps correct these misunderstandings. Final Answer: The statement that is not true is: X rays have low penetrating power in matter.

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