Question 1

A bullet of mass 10 g is fired horizontally with a velocity of 40 m/s from a gun of mass 8 kg. Using conservation of momentum, what is the recoil velocity (in m/s) of the gun?

Accepted Answer

Introduction / Context: This question is a classic example of the law of conservation of linear momentum applied to a gun firing a bullet. When a bullet is fired forward, the gun recoils backward so that the total momentum of the system gun plus bullet remains constant, provided no external horizontal force acts. Such problems are common in basic mechanics and help students understand how momentum is shared between interacting bodies. Given Data / Assumptions: Mass of bullet, m_bullet = 10 g = 0.01 kg. Velocity of bullet after firing, v_bullet = 40 m/s (forward direction). Mass of gun, m_gun = 8 kg. Initial velocity of both gun and bullet is zero; system is at rest before firing. No external horizontal force acts on the gun bullet system during the short firing interval. Concept / Approach: The law of conservation of linear momentum states that if no external force acts on a system, the total momentum of the system remains constant. Initially, the gun and bullet are at rest, so initial momentum is zero. After firing, the bullet moves forward and the gun recoils backward. The forward momentum of the bullet must be exactly balanced by the backward momentum of the gun so that the vector sum remains zero. We choose the forward direction as positive. If v_gun is the recoil velocity of the gun, then m_bullet * v_bullet + m_gun * v_gun = 0. Solving this relation gives the recoil velocity. The sign will indicate direction; the magnitude is asked in the question. Step-by-Step Solution: Step 1: Write the conservation of momentum equation: initial momentum = final momentum.Step 2: Initially, both gun and bullet are at rest, so initial momentum = 0.Step 3: After firing, bullet momentum = m_bullet * v_bullet = 0.01 * 40 = 0.4 kg m/s (forward).Step 4: Let the recoil velocity of the gun be v_gun (negative for backward motion). Then gun momentum = m_gun * v_gun = 8 * v_gun.Step 5: Apply conservation: 0 = total final momentum = 0.4 + 8 * v_gun.Step 6: Rearrange to get 8 * v_gun = -0.4, so v_gun = -0.4 / 8 = -0.05 m/s.Step 7: The negative sign shows backward direction; the magnitude of recoil velocity is 0.05 m/s. Verification / Alternative check: We can check the reasonableness of the result by comparing masses. The gun is much heavier than the bullet, so its recoil speed should be much smaller than the bullet speed. The bullet moves at 40 m/s, and the gun recoils at only 0.05 m/s, which is 800 times smaller. This ratio is consistent with the mass ratio 8 kg versus 0.01 kg, because momentum m * v must be balanced on both sides. Also, if we used another option, such as 2 m/s or 4 m/s, the recoil momentum of the gun would be huge compared to the bullet momentum, violating conservation. Therefore 0.05 m/s is the only value that keeps total momentum equal to zero. Why Other Options Are Wrong: For a recoil velocity of 2 m/s, gun momentum would be 8 * 2 = 16 kg m/s, far larger than the bullet momentum of 0.4 kg m/s, so total momentum would not be zero. For 0.1 m/s, gun momentum would be 8 * 0.1 = 0.8 kg m/s, still larger than the bullet momentum in magnitude. For 4 m/s, gun momentum becomes 32 kg m/s, which is clearly impossible in this context. Only 0.05 m/s gives gun momentum of 0.4 kg m/s backward, exactly balancing the bullet momentum forward. Common Pitfalls: Students sometimes forget to convert grams into kilograms and obtain wrong numerical values. Others may mix up the sign and think the recoil is in the same direction as the bullet, which would not conserve momentum. Another frequent mistake is to equate the velocities directly instead of momenta, ignoring the different masses. To avoid these errors, always convert units to SI, write the conservation equation with masses and velocities, and solve carefully. Final Answer: The recoil velocity of the gun has magnitude 0.05 m/s in the backward direction.

Question 2

The motion of a freely falling body near the surface of the Earth, when air resistance is neglected, is an example of ________________________ motion.

Accepted Answer

Introduction / Context: Free fall is one of the most important idealised motions studied in basic physics. When an object is dropped from a certain height and air resistance is neglected, it accelerates towards the Earth due to gravity. Understanding the nature of this acceleration and how velocity changes with time forms the basis for many kinematics problems. This question checks whether you recognise that free fall is a case of uniformly accelerated motion. Given Data / Assumptions: The object is falling freely under gravity near the Earth surface. Air resistance and other forces apart from gravity are neglected. Acceleration due to gravity, g, is taken as approximately constant near the surface (about 9.8 m/s^2). The motion is along a straight vertical line. Concept / Approach: Uniformly accelerated motion means that the acceleration remains constant in magnitude and direction throughout the motion. In such motion, velocity changes by equal amounts in equal intervals of time. For a freely falling body near the Earth, the only significant force acting is gravity, which provides a constant downward acceleration g. Because g is effectively constant over the small height range of most school level problems, the body experiences uniform acceleration. Therefore, the motion of a freely falling body without air resistance is a classic example of uniformly accelerated motion. Step-by-Step Solution: Step 1: Recall that free fall is defined as motion under the influence of gravity alone.Step 2: Note that the acceleration due to gravity g near the Earth surface is taken as constant, about 9.8 m/s^2 downward.Step 3: Recognise that when acceleration is constant, the motion is called uniformly accelerated motion.Step 4: Understand that in uniformly accelerated motion, velocity increases or decreases by equal amounts in equal time intervals.Step 5: Apply this to the falling body: at each second, its speed increases by approximately 9.8 m/s as it falls.Step 6: Conclude that the motion of a freely falling body is uniformly accelerated, not constant velocity or non uniform acceleration. Verification / Alternative check: The standard kinematic equations for motion with constant acceleration, such as v = u + a * t and s = u * t + 0.5 * a * t^2, are derived under the assumption of uniform acceleration. These equations are routinely used to solve free fall problems, for example when calculating the time taken to reach the ground or the final velocity just before impact. The success of these equations in predicting experimental results for small height ranges supports the idea that free fall is uniformly accelerated motion when air resistance is negligible. Why Other Options Are Wrong: The option non uniformly accelerated would apply if the acceleration changed with time, as happens when air resistance becomes significant at high speed, but that case is not considered here. Constant velocity means zero acceleration, which would occur if the object were moving at a uniform speed in a straight line, not falling under gravity. Constant speed is also incorrect because the speed of a freely falling body increases with time under gravity. Common Pitfalls: Some learners confuse the term uniform motion with uniformly accelerated motion. Uniform motion means constant velocity and zero acceleration, which does not describe free fall. Others may think acceleration changes as the object moves, especially if they imagine large distances, but near the Earth surface and for moderate heights, the variation in g is extremely small and can be ignored. To avoid these misunderstandings, remember that free fall in introductory physics is treated as motion with constant acceleration g, and is therefore uniformly accelerated. Final Answer: The motion of a freely falling body near the Earth surface is an example of uniformly accelerated motion.

Question 3

For a convex lens, the image is formed at a distance of 25 cm from the lens when the object is placed on the other side of the lens at 12 cm from the optical centre. What is the power of this convex lens (in diopters)?

Accepted Answer

Introduction / Context: Lenses are widely used in spectacles, cameras, microscopes and many optical instruments. Two important quantities that describe a lens are its focal length and its power. The power of a lens is the reciprocal of its focal length in metres and is measured in diopters. This numerical problem uses the lens formula to find the focal length of a convex lens from given object and image distances, and then converts that focal length into power. Given Data / Assumptions: The lens is a convex (converging) lens. Image distance from the lens, v = +25 cm (real image on the side opposite to the object). Object distance from the optical centre, u = -12 cm (object placed in front of the lens by the usual sign convention). Lens formula: 1 / f = 1 / v - 1 / u, with distances measured from the optical centre. Power P of a lens: P = 1 / f in metres, measured in diopters (D). Concept / Approach: We first apply the lens formula, taking care of the signs for object and image distances. For a convex lens, conventional sign convention takes distances measured opposite to the direction of incident light as negative. Thus, the object in front of the lens gives a negative u, and a real image on the other side gives positive v. Once we calculate focal length f in centimetres, we convert it to metres and then compute the power P as 1 / f. A positive power indicates a converging or convex lens, which matches the given lens type. Step-by-Step Solution: Step 1: Write the lens formula: 1 / f = 1 / v - 1 / u.Step 2: Substitute the values using sign convention: v = +25 cm, u = -12 cm.Step 3: Compute 1 / f = 1 / 25 - 1 / (-12) = 1 / 25 + 1 / 12.Step 4: Find a common denominator: 1 / 25 = 12 / 300 and 1 / 12 = 25 / 300, so 1 / f = (12 + 25) / 300 = 37 / 300.Step 5: Therefore f = 300 / 37 cm, which is approximately 8.108 cm.Step 6: Convert focal length to metres: f = 8.108 cm = 0.08108 m approximately.Step 7: Calculate power P = 1 / f in diopters: P ≈ 1 / 0.08108 ≈ 12.33 diopters, positive for a convex lens. Verification / Alternative check: The focal length of about 8 cm is reasonable for a fairly strong convex lens. A lens with f around 10 cm would have a power of roughly +10 diopters, so a slightly shorter focal length near 8 cm gives a slightly higher power near +12 diopters. Checking the options, only 12.33 diopters matches the positive value we expect for a converging lens. Both negative power values would correspond to concave lenses, which contradicts the given information. The smaller positive power 4.33 diopters would imply a much longer focal length of about 23 cm, inconsistent with the object and image distances calculated through the lens formula. Why Other Options Are Wrong: -4.33 diopters and -12.33 diopters both imply a diverging lens with negative focal length, which conflicts with the problem statement that the lens is convex. A convex lens must have positive power in the usual sign convention. The value 4.33 diopters is too small; it would correspond to f around 0.23 m, which does not satisfy the given object and image distances when substituted into the lens formula. Therefore, these options do not satisfy the correct optics relationships. Common Pitfalls: Students often mix up the sign convention, forgetting to take u as negative for an object in front of the lens, which can lead to incorrect focal lengths. Another mistake is converting centimetres to metres incorrectly or forgetting this step when computing power in diopters. Some learners also confuse the idea that higher power means shorter focal length; remembering P = 1 / f in metres helps avoid this confusion. Always pay attention to units, signs and the given type of lens when solving such problems. Final Answer: The power of the convex lens is approximately +12.33 diopters.

Question 4

If an object is thrown vertically upwards, what will be its velocity when it reaches its maximum height, assuming motion under gravity and neglecting air resistance?

Accepted Answer

Introduction / Context: Throwing an object vertically upwards is a standard motion studied in kinematics. As the object rises, gravity acts downward, slowing it until it momentarily comes to rest at the highest point before falling back down. Understanding what happens to the velocity at the maximum height is essential for solving many problems involving upward and downward motion under gravity. Given Data / Assumptions: The object is projected vertically upward with some initial velocity. The only significant force acting on the object during its flight is the gravitational force. Air resistance is neglected, so acceleration due to gravity is taken as constant and downward. We are asked for the velocity exactly at the maximum height of the trajectory. Concept / Approach: As the object moves upward, gravity produces a constant downward acceleration, which continuously reduces the upward velocity. At the maximum height, the object changes its direction of motion from upward to downward. At the exact turning point, the instantaneous velocity must be zero, otherwise the object would still be moving upwards or downwards. In equations of motion, this is seen when using v^2 = u^2 + 2 * a * s; at the highest point, we can set v = 0 to find the maximum height. Therefore, the magnitude of velocity at the maximum height is zero. Step-by-Step Solution: Step 1: Consider the motion of the object as it travels upward against gravity.Step 2: Gravity provides a constant downward acceleration g, which reduces the upward velocity over time.Step 3: As long as the object is moving upward, its velocity is positive (upward) but decreasing in magnitude.Step 4: At the maximum height, the object stops rising and is about to start falling back down.Step 5: At this turning point, the instantaneous velocity must be zero; otherwise, motion would continue upward or downward instead of changing direction.Step 6: Therefore, the velocity at the maximum height is 0 m/s. Verification / Alternative check: Using the kinematic equation v = u + a * t, with upward taken as positive and acceleration due to gravity as a = -g, we can describe the upward journey. At some time t_max, the object reaches maximum height, where v = 0. Setting v = 0 gives 0 = u - g * t_max, so t_max = u / g. The equation shows that the final velocity at the highest point is exactly zero, independent of the specific values of u or g. Similarly, using v^2 = u^2 + 2 * a * s and setting v = 0 allows calculation of maximum height, again confirming that v is zero at that point. Why Other Options Are Wrong: The values 4.9 m/s, 14.7 m/s and 20 m/s all represent non zero speeds. Choosing any of these would imply that the object is still moving when it is supposedly at the maximum height, which is not correct for the idealised case of vertical motion under gravity. These numbers may remind students of values like g * t, but they do not apply at the turning point where the direction reverses. Common Pitfalls: Some students confuse the concept of speed with acceleration and think that the object still has some upward speed at the top because gravity is still acting. Gravity indeed acts continuously, but at the highest point, the velocity is momentarily zero while acceleration is still downward. Others may misinterpret diagrams and think the object stops only when it comes back to the starting point. To avoid these mistakes, always remember that at the turnaround point in vertical motion, the instantaneous velocity is zero. Final Answer: At its maximum height, the velocity of the object is 0 m/s.

Question 5

In household electrical safety, fuse wire in a fuse is commonly made of which alloy so that it melts quickly when excessive current flows?

Accepted Answer

Introduction / Context: Fuse wires are vital safety devices in electrical circuits. They are designed to melt and break the circuit when current exceeds a safe limit, protecting appliances and wiring from overheating and possible fire. The material used for fuse wire is not ordinary copper but a special alloy chosen for its electrical and thermal properties. This question checks whether you know which alloy is typically used. Given Data / Assumptions: Fuse wire must melt at a relatively low temperature compared with normal copper conductors. The material should have reasonably high electrical resistance so that it heats up quickly at high current. The alloy must be easy to draw into thin wires and be affordable for widespread use. We are dealing with traditional household fuse wire, not modern circuit breakers. Concept / Approach: A good fuse material has low melting point and high resistivity so that even a small overload current raises its temperature enough to melt it before other parts of the circuit are damaged. An alloy of tin and lead meets these conditions better than pure copper or silver. This alloy has a lower melting point and higher resistivity, so it melts quickly when current is too high. Copper and silver have high melting points and very low resistivity, making them excellent conductors but poor fuse materials. Therefore, traditional fuse wire is usually made from an alloy of tin and lead. Step-by-Step Solution: Step 1: Recall that fuse wire is meant to be the weakest link in the circuit, melting first when current is excessive.Step 2: Identify the properties needed: relatively low melting temperature and higher resistivity than copper.Step 3: Consider the alloy of tin and lead, which has a low melting point and is used in soldering and fuses.Step 4: Compare with tin and copper or lead and copper, which would have higher melting points and be closer to normal conductors.Step 5: Copper and silver together would form a very good conductor with high melting point, which is opposite of what is needed for a fuse.Step 6: Conclude that the common fuse material is an alloy of tin and lead. Verification / Alternative check: In many school laboratories and older household fuse boxes, replacement fuse wire is sold as a soft alloy, often explicitly labelled as tin lead fuse wire. Electrical handbooks describe this alloy as having suitable melting characteristics. Modern cartridge fuses may use refined materials, but the basic principle remains the same: a low melting point alloy such as tin lead combination is used as the fusing element. This supports the selection of tin and lead as the correct answer. Why Other Options Are Wrong: A tin and copper alloy would have higher conductivity and a higher melting point than the tin lead alloy, making it less sensitive as a fuse. A lead and copper alloy would again be too conductive and have too high a melting point to function as a reliable fuse element. Copper and silver are both highly conductive and have relatively high melting temperatures, which is why they are used for permanent wiring and contacts, not for sacrificial fuse links. Common Pitfalls: Some students assume that fuse wires are made of the same material as normal house wiring, usually copper, but simply thinner. This is not correct because the melting characteristics would not be well controlled. Others may think that a very good conductor like silver must be used in all electrical components, forgetting that a fuse is designed to melt, not to survive. To avoid mistakes, remember that fuse wire is intentionally made from a more resistive, lower melting point alloy such as tin and lead. Final Answer: Fuse wire is commonly made of an alloy of tin and lead.

Question 6

In the history of modern physics, who is credited with formulating the special and general theories of relativity?

Accepted Answer

Introduction / Context: The theory of relativity is one of the most revolutionary developments in twentieth century physics. It fundamentally changed our understanding of space, time, gravity and energy. Knowing which scientist proposed these ideas is a standard part of general knowledge as well as science education. This question asks you to identify the physicist associated with the special and general theories of relativity. Given Data / Assumptions: The question refers to the theory of relativity in the context of modern physics. This includes both special relativity, dealing with high speed motion, and general relativity, dealing with gravity and curved spacetime. We consider well established historical facts about major physicists. The options list several famous scientists from different periods and fields. Concept / Approach: Albert Einstein is the scientist who introduced the special theory of relativity in 1905, explaining how measurements of space and time change for observers moving at constant high velocities relative to each other. He later extended these ideas to include gravity in the general theory of relativity, published in 1915, where gravity is described as the curvature of spacetime caused by mass and energy. Isaac Newton formulated classical mechanics and the law of universal gravitation centuries earlier, but his framework is different from relativity. Niels Bohr contributed mainly to atomic structure and quantum theory, and Michael Faraday made fundamental discoveries in electromagnetism. Thus, the correct answer is Albert Einstein. Step-by-Step Solution: Step 1: Recall that the theory of relativity is a twentieth century development, not from the era of classical mechanics.Step 2: Identify Albert Einstein as the physicist who proposed special relativity in 1905 as part of his famous papers.Step 3: Remember that Einstein later published the general theory of relativity in 1915, which redefined gravity.Step 4: Note that Isaac Newton developed classical mechanics and the inverse square law of gravitation, not relativity.Step 5: Recognise that Niels Bohr is known for the Bohr model of the atom and quantum ideas, not relativity.Step 6: Recognise that Michael Faraday is famous for work on electromagnetic induction and field concepts, not for relativity.Step 7: Conclude that Albert Einstein is the scientist who discovered and developed the theory of relativity. Verification / Alternative check: Standard physics textbooks and historical accounts all associate relativity with Einstein. Special relativity is built on two postulates about the constancy of the speed of light and the equivalence of physical laws in inertial frames. General relativity describes gravity as geometry of spacetime and successfully predicts phenomena such as the bending of light near massive objects and the precession of Mercury orbit. These predictions were confirmed by observations and cemented Einstein reputation. No other scientist among the options is credited with this theory, although many have extended and tested it. Why Other Options Are Wrong: Isaac Newton discovered the law of gravitation and developed calculus and laws of motion, but his view treats space and time as absolute, which relativity modifies. Niels Bohr is closely associated with early quantum theory, including quantised orbits in hydrogen, not relativity. Michael Faraday laid foundations of electromagnetic theory through experiments, but the mathematical theory of electromagnetism was largely formulated by James Clerk Maxwell, who is not in the options. Therefore, none of these scientists discovered relativity. Common Pitfalls: Some learners mix up major scientific names and may respond with the first familiar physicist they recall without linking the name to the correct theory. Others may confuse relativity with classical gravity and choose Newton. To avoid such confusion, remember that relativity is the modern replacement and extension of Newton gravity and is strongly associated with Einstein name, while Newton work belongs to classical physics. Final Answer: Albert Einstein is credited with formulating the special and general theories of relativity.

Question 7

Which device is used for the detection and measurement of all common types of nuclear radiation such as alpha, beta and gamma radiations?

Accepted Answer

Introduction / Context: Radioactive substances emit different kinds of nuclear radiation, mainly alpha particles, beta particles and gamma rays. Measuring this radiation is important in nuclear physics, medical imaging, radiation safety and environmental monitoring. Several instruments exist for measuring various physical quantities, but only some are designed to detect ionising radiation. This question asks you to identify the device commonly used to detect and count alpha, beta and gamma radiations. Given Data / Assumptions: We want a device that can detect nuclear radiation types such as alpha, beta and gamma. The instrument should provide counts or intensity of radiation. Other listed instruments measure different physical properties, not nuclear radiation intensity. We assume a basic level of knowledge of laboratory instruments. Concept / Approach: The Geiger counter, more fully called the Geiger Muller counter, is a widely used device for detecting ionising radiation. It contains a Geiger tube filled with gas and electrodes. When radiation enters the tube, it ionises the gas, causing a momentary current pulse that can be counted. By counting pulses per second, the Geiger counter measures the intensity of radiation. A polarimeter is used to measure optical rotation of plane polarised light in optically active substances. A calorimeter measures heat changes in physical and chemical processes. A radiometer typically measures radiant energy or light intensity but is not used to count nuclear particles in the way a Geiger counter does. Therefore, the Geiger counter is the correct answer. Step-by-Step Solution: Step 1: Identify that alpha, beta and gamma radiation are all forms of ionising radiation emitted by radioactive sources.Step 2: Recall that the Geiger counter is specifically designed to detect ionising radiation using gas ionisation.Step 3: Note that the Geiger counter can detect many different types of ionising radiation, including alpha (with suitable window), beta and gamma.Step 4: Recognise that a polarimeter is used in chemistry labs to study optically active solutions, not nuclear radiation.Step 5: Recognise that a calorimeter is used to measure heat and energy changes, for example in combustion or dissolution experiments.Step 6: Recognise that a radiometer usually measures light or radiant flux and is not the standard tool for detecting alpha or beta particles.Step 7: Conclude that the instrument used for detection and measurement of alpha, beta and gamma radiation is the Geiger counter. Verification / Alternative check: In many images of nuclear laboratories or safety inspections, personnel carry handheld Geiger counters, which make clicks when they detect radiation. Textbooks on nuclear physics and radiation safety list Geiger Muller counters as basic detection instruments alongside scintillation counters and ionisation chambers. These texts explain that with proper calibration and shielding, a Geiger counter responds to various types of radiation and is used for survey work. The other devices listed are not described in nuclear radiation contexts, confirming that Geiger counter is the correct choice. Why Other Options Are Wrong: A polarimeter measures the angle of rotation of plane polarised light, helping to determine concentration or purity of optically active compounds, not radiation. A calorimeter measures heat energy gained or lost by a system, not nuclear particle counts. A radiometer can measure radiant energy of light or electromagnetic waves in some ranges, but it is not a standard detector for alpha and beta particles and is not used like a Geiger counter for radiation protection. Common Pitfalls: Students sometimes confuse radiometer and Geiger counter because radiometer sounds like it should measure radiation. However, in common laboratory usage, radiometers relate to light and radiant energy, while Geiger counters relate to nuclear radiation. Another mistake is to choose calorimeter because nuclear processes involve energy, but a calorimeter measures heat in chemical or physical changes, not radioactive emissions directly. Always connect alpha, beta and gamma detection with Geiger Muller counters in basic questions. Final Answer: The device used to detect and measure alpha, beta and gamma radiations is the Geiger counter.

Question 8

In elasticity, what is the SI unit of the physical quantity known as Youngs modulus, which measures the ratio of stress to strain in a material?

Accepted Answer

Introduction / Context: Youngs modulus is a fundamental mechanical property that measures the stiffness of a material. It is defined as the ratio of tensile or compressive stress to the corresponding strain within the elastic limit. Because stress is force per unit area and strain is dimensionless, the unit of Youngs modulus must be the same as the unit of stress. Understanding this helps connect concepts in mechanics with units used in engineering. Given Data / Assumptions: Youngs modulus E is defined as E = stress / strain. Stress is force per unit area, so stress has units of newton per square metre. Strain is the ratio of change in length to original length and is a pure number with no unit. The SI unit for pressure and stress is the pascal (Pa). Concept / Approach: Since Youngs modulus is the ratio of stress to strain and strain has no unit, the unit of Youngs modulus is the same as the unit of stress. In SI units, stress is measured in newton per square metre (N/m^2). This unit is given a special name, pascal (Pa). Therefore, the correct unit for Youngs modulus is the pascal. Newton alone is a unit of force, not of stress. Joule is a unit of energy. Erg is an old CGS unit of energy, not used as an SI unit in this context. Hence, pascal is the correct SI unit. Step-by-Step Solution: Step 1: Recall the definition of Youngs modulus E = stress / strain.Step 2: Recognise that stress = force / area, with SI units newton per square metre.Step 3: Note that strain is dimensionless because it is a ratio of two lengths.Step 4: Therefore, the unit of E equals the unit of stress, which is N/m^2.Step 5: Remember that in SI, one pascal is defined as one newton per square metre.Step 6: Conclude that Youngs modulus is measured in pascals. Verification / Alternative check: Engineering tables list Youngs modulus for steel as about 2 * 10^11 N/m^2 or 200 gigapascals (200 GPa). The fact that both forms are used interchangeably shows that pascal and N/m^2 are the same unit. No standard reference expresses Youngs modulus in joules or ergs, because it is not energy but stress. This reinforces that pascal is the correct SI unit. Why Other Options Are Wrong: Newton is a unit of force, suitable for measuring weight or other forces, but Youngs modulus relates to stress, not pure force. Joule is a unit of work or energy, related to force times distance, not stress. Erg is a CGS unit of energy and is again not used to measure stress or Youngs modulus. None of these match the definition or dimensional form of Youngs modulus. Common Pitfalls: Students sometimes confuse units of stress and energy, especially because both involve force and distance in different combinations. Another mistake is assuming that any mechanical quantity may be expressed in newtons alone. To avoid these errors, always analyse the definition of the quantity: because Youngs modulus is stress divided by a dimensionless number, its unit must be the same as stress, which is pascal in SI. Final Answer: The SI unit of Youngs modulus is the pascal (Pa).

Question 9

In elasticity, the statement that the strain produced in a body is directly proportional to the stress applied on it, within the elastic limit, is known as ________.

Accepted Answer

Introduction / Context: When forces act on a solid body, it deforms. For small deformations, there is a simple linear relationship between the stress applied and the resulting strain, as long as the material remains within its elastic limit. This relationship is a cornerstone of the theory of elasticity and is commonly stated as stress is directly proportional to strain. The law that describes this behaviour has a specific name that students must remember. Given Data / Assumptions: We are dealing with elastic deformation of materials where they return to original shape once the load is removed. Within the elastic limit, the material follows a linear stress strain relationship. The statement refers to strain being directly proportional to stress. Several laws with different names are given in the options, some from other areas of physics. Concept / Approach: The law describing linear proportionality between stress and strain is known as Hookes law, named after the scientist Robert Hooke. In mathematical form it is often written as stress proportional to strain or stress = E * strain, where E is Youngs modulus. This relationship holds only within the elastic limit; beyond that, the material may deform plastically and not obey Hookes law. Dollar law and Miller law are not standard names in elasticity, and Kepler laws belong to planetary motion, not material deformation. Step-by-Step Solution: Step 1: Recall the standard statement in elasticity: within the elastic limit, stress is directly proportional to strain.Step 2: Remember that this fundamental relationship is known as Hookes law.Step 3: Note that in formula form, stress = E * strain, where E is Youngs modulus.Step 4: Recognise that Dollar law and Miller law are not standard physics laws associated with elasticity.Step 5: Recognise that Kepler laws describe planetary motion and orbits around the sun.Step 6: Conclude that the correct name for the law relating stress and strain is Hookes law. Verification / Alternative check: Physics and engineering textbooks on strength of materials always introduce Hookes law early in the discussion of elastic behaviour. Stress strain graphs show a straight line region from the origin up to the proportional limit. This linear section represents Hookes law region. The term is widely used in mechanics, civil engineering, mechanical engineering and materials science to describe how materials behave under small loads. No other law name in the options is used in this context, which confirms Hookes law is the correct choice. Why Other Options Are Wrong: Dollar law and Miller law do not correspond to any major principle in basic physics related to elasticity; they are distractor names. Kepler laws describe the motion of planets and satellites, such as that planets sweep out equal areas in equal times, and have nothing to do with stress or strain. Therefore, these options do not match the definition in the question. Common Pitfalls: Some learners may confuse Hookes law with Youngs modulus or think that Hookes law is simply F = k * x for springs. In fact, F = k * x is a specific application of Hookes law to springs, where k is the spring constant. The more general form relates stress and strain in materials. Another mistake is to think Hookes law applies beyond the elastic limit, which is not true. Remember that the direct proportionality holds only as long as the material behaves elastically. Final Answer: The statement that strain is directly proportional to stress within the elastic limit is known as Hookes law.

Question 10

In density and buoyancy calculations, what is the unit of relative density (also called specific gravity)?

Accepted Answer

Introduction / Context: Relative density, also known as specific gravity, is a useful concept in fluid mechanics and material science. It compares the density of a substance to the density of a reference substance, usually water for liquids and solids, or air for gases. Because it is a ratio of two densities, its unit behaviour is different from absolute density. This question asks you to identify whether relative density has an SI unit. Given Data / Assumptions: Relative density is defined as density of a substance divided by density of a reference substance at a specified temperature. Density itself has units of mass per unit volume, for example kg/m3. The reference density uses the same units as the substance density. We are working in SI units and standard definitions. Concept / Approach: Relative density is the ratio of two densities with the same unit. When one quantity with units is divided by another quantity with the same units, the units cancel. The result is a pure number that tells how many times denser a substance is compared with the reference. Therefore, relative density has no units. For example, if a liquid has density 1200 kg/m3 and water has density 1000 kg/m3, the relative density is 1200 / 1000 = 1.2, and this 1.2 is a dimensionless number. Step-by-Step Solution: Step 1: Write the definition: relative density = density of substance / density of reference.Step 2: Use symbols: RD = rho_substance / rho_reference.Step 3: Note that both numerators and denominators have the same SI unit, usually kg/m3.Step 4: When dividing kg/m3 by kg/m3, the units cancel completely.Step 5: The result is a pure number with no units attached.Step 6: Therefore, relative density has no unit. Verification / Alternative check: Many data tables for specific gravity list values without any units, for example specific gravity of mercury approximately 13.6, specific gravity of gasoline around 0.7, etc. Laboratories measure specific gravity using hydrometers and report dimensionless values. If relative density truly had units such as kg/m3, these would be included in tables and would not cancel. This practical usage confirms that relative density is dimensionless and has no unit. Why Other Options Are Wrong: kg/m is the unit of linear mass density, which is mass per unit length, not relevant for volumetric density ratios. kg/m2 is unit of surface mass density, again unrelated. kg/m3 is the unit of absolute density, not of relative density. While density itself is measured in kg/m3, relative density is a ratio of two densities and therefore does not carry this unit. Common Pitfalls: A frequent mistake is to assume that because relative density involves density, it must inherit the same units. Another error is to confuse absolute density with relative density and use them interchangeably. To avoid these mistakes, remember that any time you see a ratio of two quantities with the same units, the units cancel, making the ratio dimensionless. Relative density is always a pure number and never carries units. Final Answer: Relative density has no unit; it is a dimensionless quantity.

Question 11

In the formula for uniformly accelerated motion, average velocity = (u + v) / 2, what does the symbol u represent?

Accepted Answer

Introduction / Context: In one dimensional motion with constant acceleration, several useful kinematic equations relate displacement, velocity, acceleration and time. One of these is the formula for average velocity when acceleration is uniform: average velocity equals half the sum of initial and final velocities. In this formula, specific symbols are used for initial and final velocities. Understanding what u and v stand for is basic but very important. Given Data / Assumptions: We are dealing with motion along a straight line with constant acceleration. The formula given is average velocity = (u + v) / 2. Standard notation in kinematics uses u for initial velocity and v for final velocity. Displacements are represented by s or x, not by u or v. Concept / Approach: In kinematics, u is traditionally used to denote the initial velocity of an object at the start of the time interval considered. The symbol v represents the velocity at the end of that time interval, called the final velocity. When acceleration is constant, the average velocity over the time interval is simply the arithmetic mean of these two velocities, given by (u + v) / 2. Displacement is a separate quantity with different symbols. Therefore, in the given formula, u clearly represents initial velocity. Step-by-Step Solution: Step 1: Recall the standard symbols used in equations of motion: u for initial velocity, v for final velocity, a for acceleration, t for time.Step 2: Recognise the formula average velocity = (u + v) / 2 as valid for uniform acceleration.Step 3: Understand that this formula computes the mean of the starting and ending velocities.Step 4: Since v is the final velocity, u must correspond to the starting velocity.Step 5: Note that displacement is not represented by u in any standard equations; displacement uses s or x.Step 6: Conclude that in this formula, u represents the initial velocity. Verification / Alternative check: Consider a motion starting with velocity u and ending with velocity v after time t under constant acceleration a. Other equations of motion are v = u + a * t and s = u * t + 0.5 * a * t^2. These equations all treat u as the velocity at time t = 0. When plotting velocity against time, the graph is a straight line from u to v, and the average of the two endpoints is (u + v) / 2. This reinforces that u is initial velocity and fits consistently with all standard kinematic relations. Why Other Options Are Wrong: Final velocity is represented by v, not u, in standard notation. Initial displacement and final displacement are not represented by u and are usually written as x0 and x or s. Average velocity is calculated using displacements and time, but the symbol u itself refers to velocity, not displacement. Therefore, interpreting u as any type of displacement is incorrect. Common Pitfalls: Students occasionally forget the standard symbols and may mix up u and v, especially if they are new to kinematics. Others might think that u is a generic symbol and assign it different meanings in different questions. To avoid confusion, always remember the standard convention: u is initial velocity, v is final velocity, and these definitions remain consistent across all basic equations of motion. Final Answer: In the formula average velocity = (u + v) / 2, u represents the initial velocity.

Question 12

If an object covers unequal distances in equal intervals of time along a path, its motion is said to be in ____________ motion.

Accepted Answer

Introduction / Context: Describing the type of motion of an object requires understanding how its position changes with time. One basic classification is between uniform and non uniform motion. This question tests whether you can recognise non uniform motion by reading a simple description focusing on distances covered in equal time intervals. Given Data / Assumptions: The object moves along a path, possibly straight or curved. We look at equal intervals of time, such as every second or every minute. In those equal time intervals, the distances covered are unequal. The options include uniform, linear, non uniform and equilibrium as possible descriptions. Concept / Approach: Uniform motion is defined as motion in which an object covers equal distances in equal intervals of time, no matter how small those intervals are. In contrast, non uniform motion occurs when the object covers unequal distances in equal time intervals, meaning its speed is changing. The term linear motion describes motion along a straight line and does not by itself tell us whether speed is constant. Equilibrium refers to balanced forces, not directly to the pattern of distance versus time. Therefore, when an object covers unequal distances in equal times, its motion is non uniform. Step-by-Step Solution: Step 1: Recall the definition of uniform motion: equal distances in equal intervals of time.Step 2: Compare this with the description in the question, which specifically says unequal distances in equal intervals of time.Step 3: Recognise that unequal distances in equal times imply the speed is changing.Step 4: A changing speed means the motion is not uniform; instead, it is non uniform motion.Step 5: Linear motion only specifies a straight line path, not constancy of speed, so it does not match the description given.Step 6: Equilibrium relates to forces balancing to produce zero acceleration, not directly to the pattern of distances described.Step 7: Conclude that the correct term is non uniform motion. Verification / Alternative check: Imagine a car that travels 10 metres in the first second, 20 metres in the second second and 5 metres in the third second. In each one second interval, the distances are different, so the car is speeding up and slowing down. This is clearly non uniform motion. If it had been uniform motion, the car would cover, for example, 10 metres every second consistently. This practical example matches the verbal definition and confirms that unequal distances in equal times correspond to non uniform motion. Why Other Options Are Wrong: Uniform motion would require equal distances in equal time intervals, which is the opposite of the description given. Linear motion could be uniform or non uniform, depending on whether the speed is constant; the question specifically mentions unequal distances, so linear alone is not adequate. Equilibrium concerns the net force on an object being zero, leading to constant velocity, and is not the direct term used to describe the pattern of distances in time. Common Pitfalls: Students may sometimes focus only on the word motion and choose uniform by habit, without carefully noting the unequal distances in the description. Others may think linear automatically implies uniform, which is not true. To avoid mistakes, always match the verbal description exactly with the definitions: equal distances in equal times for uniform motion, and unequal distances in equal times for non uniform motion. Final Answer: If an object covers unequal distances in equal time intervals, it is said to be in non uniform motion.

Question 13

The impression of an image on the retina of the human eye persists for about what fraction of a second, allowing us to perceive smooth motion in cinema and animation?

Accepted Answer

Introduction / Context: The human eye and brain do not respond to visual changes instantaneously. When an image forms on the retina, its impression persists for a short time even after the object is removed. This phenomenon, sometimes called persistence of vision, allows us to experience a sequence of still images shown rapidly as continuous motion, as in films and animated videos. This question asks you to recall the approximate duration of this persistence. Given Data / Assumptions: The question refers to the persistence time of an image on the retina. It is asking for an approximate fractional value in seconds. Common textbook value used in basic physics is about one sixteenth of a second. The values provided include several simple fractions of a second. Concept / Approach: Persistence of vision refers to the short time that an image continues to appear in the eye after the actual object has been removed. Experiments and practical experience in cinema show that if images are flashed on a screen quickly enough, the brain blends them into smooth motion. For standard school level physics, it is accepted that the impression on the retina persists for about 1/16 second. This is why showing 16 or more frames per second can create the illusion of continuous movement. Therefore, among the options, 1/16th of a second is the correct approximate duration. Step-by-Step Solution: Step 1: Remember that persistence of vision is a short duration, much less than a full second.Step 2: Recall the commonly quoted textbook value, which is about 1/16 second for image persistence on the retina.Step 3: Compare this with the options 1/10th, 1/8th, 1/16th and 1/5th of a second.Step 4: Recognise that 1/16th matches the standard physics teaching value.Step 5: Understand that longer times like 1/5th second would give too much overlap and would not match actual cinematic practice.Step 6: Conclude that the correct answer is 1/16th of a second. Verification / Alternative check: Traditional films use 24 frames per second, and older silent films often used about 16 frames per second or slightly more. At these rates, the change between consecutive frames happens faster than the persistence time, so the brain blends them into apparent continuous motion. If persistence were much longer, the moving images would smear together and appear blurred. If it were much shorter, a higher frame rate would be necessary to achieve smooth motion. The commonly accepted estimate of about 0.06 seconds, roughly 1/16 second, fits this practical experience. Why Other Options Are Wrong: 1/10th and 1/8th of a second are both longer than 1/16th, and while they are of the same order, the standard value taught in introductory physics is 1/16th. The fraction 1/5th of a second is significantly larger and would produce a different visual effect, leading to more noticeable afterimages and motion blur. Since the question is based on the textbook value and common approximation, 1/16th is correct and the other fractions are considered incorrect here. Common Pitfalls: Some students may not remember the exact fraction and may guess a value like 1/10th or 1/8th because they sound familiar. Others may confuse persistence of vision with reaction time of the eye, which is typically longer and describes how quickly we respond to stimuli, not how long a static image stays. To avoid confusion, link persistence of vision specifically with frame rates in movies and the rough value 1/16 second that corresponds with early cinematic technology. Final Answer: The impression of an image on the retina persists for about 1/16th of a second.

Question 14

Which one among the following is not an electromagnetic wave?

Accepted Answer

Introduction / Context: Electromagnetic waves form a broad spectrum that includes radio waves, microwaves, infrared, visible light, ultraviolet, X rays and gamma rays. These waves consist of oscillating electric and magnetic fields and can travel through vacuum. In contrast, some types of radiation are streams of particles with mass and charge. This question asks you to distinguish between true electromagnetic waves and particle rays such as cathode rays. Given Data / Assumptions: X rays and gamma rays are high energy parts of the electromagnetic spectrum. Infrared rays are lower frequency electromagnetic waves associated with heat radiation. Cathode rays are streams of electrons emitted from the cathode in a discharge tube. We are choosing which option is not an electromagnetic wave. Concept / Approach: Electromagnetic waves are massless and electrically neutral disturbances that propagate at the speed of light in vacuum. They do not consist of moving material particles. Cathode rays, on the other hand, are actual electrons moving at high speed in a vacuum tube. Because electrons have mass and negative charge, cathode rays are considered streams of charged particles, not electromagnetic waves. X rays, gamma rays and infrared rays all belong to the electromagnetic spectrum, differing only in frequency and wavelength. Therefore, cathode ray is the correct choice as not being an electromagnetic wave. Step-by-Step Solution: Step 1: Recall that X rays are high energy electromagnetic waves used in medical imaging.Step 2: Recall that gamma rays are even higher energy electromagnetic waves emitted in nuclear transitions.Step 3: Recall that infrared rays are lower energy electromagnetic waves associated with heat and thermal radiation.Step 4: Understand that cathode rays are electrons emitted from a heated or negatively charged cathode in a discharge tube.Step 5: Note that electrons have rest mass and electric charge, so a beam of electrons is not an electromagnetic wave.Step 6: Conclude that cathode ray is the only option that is not an electromagnetic wave. Verification / Alternative check: Historical experiments with cathode ray tubes showed that cathode rays are deflected by electric and magnetic fields in a way consistent with negatively charged particles. J J Thomson measured the charge to mass ratio of cathode ray particles and identified them as electrons. Electromagnetic waves, in contrast, are not deflected in the same way because they are not charged particles. X rays, gamma rays and infrared rays are all produced and detected by methods involving accelerating charges and transitions in atoms, consistent with electromagnetic theory. Why Other Options Are Wrong: X rays are electromagnetic waves with very short wavelength and high energy, widely used in medicine and material testing. Gamma rays are even more energetic electromagnetic waves associated with nuclear processes. Infrared rays are part of the electromagnetic spectrum with longer wavelength than visible light and are felt as heat. All three are genuine electromagnetic waves and therefore cannot be the correct answer for a question asking which is not an electromagnetic wave. Common Pitfalls: Students may be misled by the word ray, assuming that anything called a ray must be electromagnetic. However, cathode rays are an important exception because they are beams of electrons. It is important to distinguish naming conventions from physical nature. Remember that electromagnetic waves are massless and neutral, while cathode rays are made of particles with mass and charge. Final Answer: Cathode ray is not an electromagnetic wave; it is a stream of electrons.

Question 15

If the speed of an object moving along a straight line is constant, its motion is said to be _________.

Accepted Answer

Introduction / Context: Describing motion precisely requires clear definitions. One fundamental idea is uniform motion, in which an object moves at a constant speed along a straight path. This question asks you to recall the correct term for motion along a straight line with constant speed, a basic concept in kinematics. Given Data / Assumptions: The object is moving along a straight line, so its direction is fixed. The speed of the object remains constant with time. No mention is made of oscillation, rotation or changes in direction. We need the correct descriptive term from the given options. Concept / Approach: Uniform motion is defined as motion in which an object covers equal distances in equal intervals of time in a straight line. This implies a constant speed and unchanging direction, so the velocity is constant in magnitude and direction. Periodic motion refers to motion that repeats itself after equal intervals of time, such as oscillations, which may or may not have constant speed. Circular motion involves movement along a circular path, where direction changes continuously even if speed is constant. Non uniform motion means either speed, direction or both change with time. Therefore, the appropriate term here is uniform motion. Step-by-Step Solution: Step 1: Note that the motion is along a straight line, so there is no change in direction.Step 2: The speed is constant, meaning the object covers equal distances in equal time intervals.Step 3: These two conditions together match the definition of uniform motion.Step 4: Periodic motion would require repeating cycles such as oscillations, which are not mentioned.Step 5: Circular motion would involve a curved path; here the path is straight, so circular does not apply.Step 6: Non uniform motion would have variable speed or direction, which contradicts the constant speed stated.Step 7: Conclude that the correct description is uniform motion. Verification / Alternative check: Imagine a car moving on a straight highway at exactly 60 km/h with no speed changes and no turns. Over every one minute interval, it travels the same distance, for example one kilometre each minute. This is uniform motion. In contrast, a car moving around a circular track at constant speed has constant speed but changing direction, so its motion is uniform circular motion, not uniform motion along a straight line. The question clearly specifies straight line, reinforcing that uniform motion is the correct description. Why Other Options Are Wrong: Periodic motion focuses on repetition in time, such as a pendulum swinging back and forth. It does not necessarily imply constant speed along a straight line. Circular motion implies a curved path, not a straight line as given. Non uniform motion means speed or direction changes over time, which contradicts the condition that speed is constant and direction is fixed. Common Pitfalls: Some students may confuse uniform motion with periodic motion because both involve regular patterns. However, uniform motion is about equal distances in equal times, while periodic motion is about repetition of whole cycles in equal times. Others may think that any motion with constant speed is uniform, ignoring direction changes. The definition of uniform motion in simple kinematics includes straight line motion with constant speed. Final Answer: If an object moves along a straight line with constant speed, its motion is called uniform motion.

Question 16

In vehicles such as cars and motorcycles, the side rearview mirrors are usually made of which type of mirror?

Accepted Answer

Introduction / Context: Rearview mirrors in vehicles are designed to give drivers a wide field of view so they can see traffic behind and to the sides. The type of mirror used affects how large the field of view is and how images appear. Knowing which type of mirror is used for side mirrors is a common application of basic optics in everyday life. Given Data / Assumptions: The question specifically refers to side mirrors of vehicles, often called rearview mirrors on the doors. These mirrors should cover a wide area behind and beside the vehicle. Image needs to be upright so that orientation is natural for the driver. The options include convex, concave, plane and an incorrect descriptive term inverted. Concept / Approach: Convex mirrors bulge outward and have the property of diverging light rays. They form virtual, diminished and upright images of objects located in front of them. This allows a convex mirror to show a wider field of view than a plane mirror of the same size, which is very useful for vehicle mirrors. Although objects appear smaller and therefore seem farther away, more area around the vehicle is visible. Concave mirrors, in contrast, can form magnified or inverted images depending on object distance, which is confusing for driving. Plane mirrors provide a life size view but a limited field of view. Therefore, side mirrors are made convex. Step-by-Step Solution: Step 1: Identify the main requirement for side mirrors: a wide field of view to see adjacent and rear traffic.Step 2: Recall that convex mirrors produce virtual, diminished, upright images and show a wider area.Step 3: Compare with plane mirrors, which produce life size virtual images but show less area.Step 4: Compare with concave mirrors, which can invert images and produce large distortions depending on distance, undesirable in traffic situations.Step 5: Note that inverted is not a type of mirror but a property of some images produced by concave mirrors.Step 6: Conclude that the correct type used in side mirrors is the convex mirror. Verification / Alternative check: If you look carefully at side mirrors on many vehicles, there is often a warning printed such as objects in mirror are closer than they appear. This warning is characteristic of convex mirrors because they shrink images, making them appear more distant. The wide viewing angle is easily observed by comparing the area visible in a convex mirror versus a plane mirror of the same size. Vehicle safety regulations and design manuals also specify convex side mirrors to reduce blind spots. Why Other Options Are Wrong: Concave mirrors can form inverted real images when objects are beyond the focal length, which would be confusing for drivers because the scene behind might appear upside down or distorted. Plane mirrors show a smaller field of view, so more vehicles would remain in blind spots. Inverted is not a mirror type at all, so it cannot be correct. Only convex mirrors combine a wide field of view with upright images, which is ideal for side mirrors. Common Pitfalls: A common mistake is to think that plane mirrors are used everywhere because rearview mirrors inside the car sometimes appear flat. In fact, many interior mirrors may be slightly curved, and side mirrors are clearly convex. Another pitfall is mixing up concave and convex mirrors; remembering that convex mirrors bulge outward and are used when a wide field of view is needed helps avoid confusion. Final Answer: The side mirrors of vehicles are usually made of convex mirrors.

Question 17

During _____________ motion of an object along a straight line, the velocity remains constant with time and equal distances are covered in equal intervals of time.

Accepted Answer

Introduction / Context: Classifying motion involves looking at how an object position and velocity change over time. When an object moves along a straight line with constant velocity, it is an important special case used in many physics problems and definitions. This question asks which term best describes motion in which velocity remains constant with time along a straight line. Given Data / Assumptions: The object moves along a straight line path. Velocity remains constant in magnitude and direction with time. Equal distances are covered in equal intervals of time. We choose from terms linear, translational, uniform and equilibrium. Concept / Approach: Uniform motion is defined as motion in which an object covers equal distances in equal intervals of time along a straight line. This implies constant velocity. Linear motion refers to motion along a straight path but does not specify whether speed is constant or changing. Translational motion means that all points of a body move in parallel paths; this can occur with variable speed, so it is not automatically constant velocity. Equilibrium refers to a state where net force is zero, which implies constant velocity but does not specifically describe motion pattern in the same way. The correct term that directly captures constant velocity in a straight line is uniform motion. Step-by-Step Solution: Step 1: Recall the definition of uniform motion: equal distances covered in equal time intervals in a straight line.Step 2: Relate constant velocity to equal distances per equal times; if velocity is constant, speed and direction do not change.Step 3: Recognise that this matches exactly the description of uniform motion.Step 4: Note that linear motion describes the path shape but not constancy of speed.Step 5: Note that translational motion describes overall body movement and can be non uniform or uniform.Step 6: Equilibrium is a condition on forces, not the standard descriptive term for this distance time pattern.Step 7: Conclude that uniform motion is the correct completion of the sentence. Verification / Alternative check: A car moving at a constant 60 km/h in a straight line on a highway is in uniform motion. In every 10 minutes, it covers the same distance. This situation is best described as uniform motion rather than simply linear or translational, because those words do not highlight the constancy of velocity. While it is true that such a car is also in translational motion and its forces may sum to zero, the special term uniform motion is what textbooks use for equal distances in equal times. Why Other Options Are Wrong: Linear motion could be uniform or non uniform depending on whether speed changes, so it is too general. Translational motion is also general and does not imply constant speed. Equilibrium, while related to balanced forces, is not the phrase normally used in the definition given in the question and can apply even when motion is not along a simple straight line segment. Therefore, uniform is the best and standard answer. Common Pitfalls: Students sometimes choose linear because they focus only on the straight line part of the description and ignore the constant velocity condition. Others may be attracted to translational because it sounds technical. To avoid such errors, always match the entire description, including the equal distances in equal times, with the definition of uniform motion. Final Answer: When velocity remains constant along a straight line, the motion is called uniform motion.

Question 18

In gas laws and physical chemistry, the molecular mass of a gas is ________.

Accepted Answer

Introduction / Context: In physical chemistry, several different properties describe gases, including molecular mass, vapour pressure, density and temperature. It is important not to confuse these properties or assume simple proportional relationships where none exist. Molecular mass is an intrinsic property of the gas molecules, while vapour pressure is a property of a liquid in equilibrium with its vapour at a given temperature. This question checks whether you understand that molecular mass and vapour pressure are not directly proportional in the simple ways suggested. Given Data / Assumptions: Molecular mass (also called molecular weight) is the mass of one mole of gas molecules, measured in kg per mole or gram per mole. Vapour pressure is the pressure exerted by a vapour in equilibrium with its liquid at a given temperature. The options suggest simple proportional relationships between molecular mass and vapour pressure such as twice, equal to or half. We assume ideal or near ideal behaviour but focus on conceptual relationships rather than exact equations. Concept / Approach: Molecular mass depends on the types and numbers of atoms in a molecule and is fixed for a given substance. Vapour pressure depends mainly on the temperature and the nature of intermolecular forces in the liquid form of the substance. For example, a liquid with weak intermolecular forces tends to have a higher vapour pressure at a given temperature. There is no universal rule that vapour pressure is equal to, half of or twice the molecular mass. While lighter molecules may sometimes correspond to higher vapour pressures, many exceptions occur because intermolecular forces and molecular structure are more important. Therefore, molecular mass and vapour pressure are not directly related by any simple multiplicative factor. Step-by-Step Solution: Step 1: Define molecular mass as the mass of a single molecule times Avogadro number, specific to each gas.Step 2: Define vapour pressure as the pressure of gas in equilibrium with its liquid at a given temperature.Step 3: Note that molecular mass is measured in g/mol or kg/mol, while vapour pressure is measured in pascal or related units.Step 4: Recognise that having different dimensions means they cannot be numerically equal in a meaningful unit consistent way.Step 5: Realise that there is no general law stating vapour pressure is half or twice molecular mass; such relationships would be arbitrary.Step 6: Conclude that, in general, the molecular mass of a gas is not directly related to its vapour pressure by a simple fixed factor. Verification / Alternative check: Consider water and ethanol. Water has a molecular mass of about 18 g/mol, while ethanol has about 46 g/mol. At room temperature, ethanol has a significantly higher vapour pressure than water despite having a larger molecular mass. This simple comparison shows that vapour pressure does not increase or decrease uniformly with molecular mass. Instead, hydrogen bonding and other intermolecular interactions control vapour pressure. Many pairs of substances show different trends, confirming that no simple proportionality connects molecular mass and vapour pressure. Why Other Options Are Wrong: The statements that molecular mass is twice its vapour pressure, equal to its vapour pressure or half its vapour pressure are not based on any accepted physical law. They mix quantities with different units and dimensional forms. Even if one tried to assign numerical values, these relationships would fail across different substances and temperatures. Therefore, these options are incorrect general statements about gases. Common Pitfalls: Students sometimes assume that every property of a substance must have a direct simple relationship with every other property. In reality, some properties are independent. Another mistake is to confuse vapour density, which does have a relation with molecular mass under some conditions, with vapour pressure. Vapour density is proportional to molecular mass for gases at the same temperature and pressure, but vapour pressure is a different concept. To avoid confusion, keep clear definitions of each property and avoid assuming proportionality without a known law. Final Answer: The molecular mass of a gas is not directly related to its vapour pressure by any simple fixed factor.

Question 19

Two vectors are said to be equal if which of the following conditions is satisfied?

Accepted Answer

Introduction / Context: In vector physics, quantities such as displacement, velocity and force are represented by vectors, which have both magnitude and direction. Defining when two vectors are equal is essential for adding vectors, comparing physical quantities and solving problems in mechanics and fields. This question checks whether you remember the precise condition for vector equality. Given Data / Assumptions: We are working with geometric vectors in a plane or in space. Each vector has magnitude (length) and direction. The question asks for the condition under which two vectors are considered equal. Options distinguish between matching magnitudes, directions or both. Concept / Approach: By definition, two vectors are equal if they have the same magnitude and the same direction, regardless of their initial points or positions in space. Vectors can be moved parallel to themselves without changing their identity. If only magnitudes match but directions differ, they are not equal, even if one is opposite; they might be negatives of each other. If only directions match but lengths differ, they represent different physical quantities. Therefore, equality requires both equal length and equal direction. Step-by-Step Solution: Step 1: Recall that a vector is fully specified by magnitude and direction.Step 2: Understand that equality of vectors means complete coincidence of these two characteristics.Step 3: If only magnitudes are equal, vectors with different directions can represent quite different physical effects, so they are not equal.Step 4: If only directions are equal but magnitudes differ, the vectors are parallel but not equal.Step 5: If magnitudes are equal but directions are opposite, the vectors are negatives of each other, not equal.Step 6: Therefore, two vectors are equal only when both their magnitudes and directions are the same. Verification / Alternative check: Consider two displacement vectors representing walking 5 metres north from different starting points. Both have magnitude 5 metres and direction north, so they are equal as vectors even if they start at different locations. If one displacement were 5 metres north and the other 5 metres south, they would have equal magnitudes but opposite directions, representing different physical moves, so they would not be equal. This example illustrates that both magnitude and direction must match for vectors to be equal. Why Other Options Are Wrong: Option a, only magnitudes are same, ignores direction and would treat opposite vectors as equal, which is incorrect. Option b, only directions are same, would treat vectors of different lengths as equal, which is also wrong. Option d, magnitudes same but directions opposite, describes vectors that are equal in magnitude but negative of each other, not equal vectors. Common Pitfalls: Students sometimes confuse equal vectors with parallel vectors while doing vector diagrams. Parallel vectors have the same or opposite direction but can have different lengths. Equal vectors are a stricter concept requiring both same length and same direction. To avoid confusion, always check both properties when comparing vectors. Final Answer: Two vectors are said to be equal when both their magnitudes and directions are the same.

Question 20

In nuclear physics, the working principle of an atomic bomb is based on which type of nuclear process?

Accepted Answer

Introduction / Context: An atomic bomb is an extremely powerful explosive device that derives its destructive energy from nuclear reactions. In basic nuclear physics, it is very important to distinguish between fission, which is the splitting of heavy nuclei, and fusion, which is the joining of light nuclei. The question tests whether you remember which specific nuclear process, and which mode of control, is used in an atomic bomb as opposed to a peaceful nuclear reactor. Understanding this difference helps in separating concepts related to nuclear weapons and nuclear power plants. Given Data / Assumptions: We are dealing with an atomic bomb, not a hydrogen bomb. Atomic bombs use heavy nuclei such as uranium 235 or plutonium 239. The focus is on the principle or basic working process of the device. Controlled and uncontrolled reactions are different ways of managing a chain reaction. Concept / Approach: An atomic bomb is based on a nuclear fission chain reaction. In nuclear fission, a heavy nucleus absorbs a neutron, becomes unstable, and splits into smaller nuclei, releasing energy and more neutrons. If these additional neutrons go on to split more nuclei without any control mechanism, the process rapidly accelerates. This fast, self sustaining chain reaction is called uncontrolled nuclear fission and results in a massive, explosive release of energy. In contrast, nuclear power reactors operate on controlled nuclear fission, where moderators and control rods regulate the rate of the chain reaction. Step-by-Step Solution: Step 1: Recall that an atomic bomb uses uranium 235 or plutonium 239 as fuel, both of which are fissionable materials.Step 2: When one such nucleus undergoes fission, it splits into smaller fragments and releases a large amount of energy and several neutrons.Step 3: These emitted neutrons strike other nearby nuclei, causing further fission events and releasing more neutrons.Step 4: In an atomic bomb, this chain reaction is not slowed down or regulated; it grows at an extremely rapid rate.Step 5: Because the reaction is uncontrolled, the energy release occurs in a tiny fraction of a second, producing an enormous explosion. This behaviour matches the description of uncontrolled nuclear fission. Verification / Alternative check: You can verify the answer by comparing atomic bombs with nuclear power reactors. Power reactors safely generate energy over long periods using controlled nuclear fission, whereas atomic bombs release energy almost instantaneously. Hydrogen bombs, on the other hand, are based mainly on nuclear fusion of light elements such as isotopes of hydrogen. Since the question clearly refers to an atomic bomb, not a hydrogen bomb, and to its primary operating principle, the only correct process is uncontrolled nuclear fission. This cross comparison confirms the answer. Why Other Options Are Wrong: Uncontrolled nuclear fusion is associated with thermonuclear or hydrogen bombs, not traditional atomic bombs, so option A is incorrect. Controlled nuclear fission is used in nuclear reactors to generate electricity safely, not in bombs, so option C does not match. Controlled nuclear fusion has not yet been achieved on a continuous, commercial scale and is not the principle behind atomic bombs, so option D is also incorrect. Only uncontrolled nuclear fission correctly describes the working process of an atomic bomb. Common Pitfalls: Students often confuse atomic bombs with hydrogen bombs and mix up fission and fusion. Another common mistake is to ignore the key words controlled and uncontrolled. Remember that whenever you see the term atomic bomb, you should think of heavy nuclei undergoing fission in an uncontrolled chain reaction. When you see hydrogen bomb or thermonuclear bomb, you should think of fusion. Carefully reading those adjectives in the options helps avoid such confusion in multiple choice questions. Final Answer: Uncontrolled nuclear fission

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