Question 1

Which of the following pairs of physical quantities do not have the same dimensions in the dimensional formula sense?

Accepted Answer

Introduction / Context:Dimensional analysis is a powerful tool in physics for checking the consistency of equations and understanding relationships between quantities. Different physical quantities may or may not share the same dimensions even if they appear related in everyday language. This question asks you to identify which pair of quantities has different dimensions, helping you strengthen your grasp of basic dimensional formulas. Given Data / Assumptions:The pairs given are: force and pressure; work and energy; impulse and momentum; weight and force.We use standard dimensional symbols: mass M, length L and time T.We assume Newtonian mechanics definitions for these quantities.We need to find the pair whose members have different dimensional formulas. Concept / Approach:Force is defined as mass times acceleration, so its dimension is [M L T^-2]. Pressure is defined as force per unit area, so its dimension is [M L^-1 T^-2]. Work and energy are both defined as force × distance, giving [M L^2 T^-2]. Impulse is force × time, or change in momentum, and momentum is mass × velocity; both have dimension [M L T^-1]. Weight is the gravitational force acting on a body, equal to m * g, so its dimension is the same as force, [M L T^-2]. Thus, among these pairs, only force and pressure have different dimensions because pressure includes an extra factor of 1/area. Step-by-Step Solution:Step 1: Write the dimensional formula for force: F = m * a, so [F] = [M][L T^-2] = [M L T^-2].Step 2: Write the dimensional formula for pressure: P = F / A, area A has dimension [L^2], so [P] = [M L T^-2] / [L^2] = [M L^-1 T^-2].Step 3: For work and energy: W = F * s, so [W] = [M L T^-2][L] = [M L^2 T^-2]; energy shares the same dimensions.Step 4: For impulse and momentum: impulse J = F * t, [J] = [M L T^-2][T] = [M L T^-1]; momentum p = m * v, [p] = [M][L T^-1] = [M L T^-1].Step 5: For weight and force: weight is a gravitational force (m * g), so it has the same dimension as force, [M L T^-2].Step 6: Conclude that only force and pressure differ in dimensions, [M L T^-2] versus [M L^-1 T^-2]. Verification / Alternative check:Dimensional formulas are often summarised in tables. Consulting such a table shows that work and energy both use joule as a unit, confirming identical dimensions. Impulse and momentum both have units like N·s and kg·m/s, which are dimensionally equivalent. Weight is measured in newtons, just like force. By contrast, pressure uses N/m^2 (pascal), clearly including an extra per square metre factor, confirming that it differs from force. Why Other Options Are Wrong:Work and energy represent different concepts but are measured in the same unit (joule) and share identical dimensions, so they do not form a mismatched pair. Impulse and momentum are closely related; impulse equals change in momentum, so their dimensions must match. Weight is simply a particular force due to gravity, so dimensional formulas for weight and force are the same. Therefore, the pairs in options B, C and D all have matching dimensions. Common Pitfalls:One common mistake is to think that different names automatically imply different dimensions. Another is to confuse pressure with force because both can be involved in contact interactions. To avoid such errors, always fall back on the defining equations: pressure is force per area, which adds an extra L^-2 factor, distinguishing it dimensionally from force. Final Answer:Force and pressure do not have the same dimensions; force is [M L T^-2], while pressure is [M L^-1 T^-2].

Question 2

In basic mechanics, ball bearings are used in cycles, scooters and other wheeled vehicles to achieve what effect between the wheel and the axle?

Accepted Answer

Introduction / Context: Ball bearings are extremely common mechanical components that appear in bicycles, scooters, fans and many rotating machines. They are placed between moving parts in order to improve motion and reduce energy loss. This question tests your understanding of the basic purpose of ball bearings in everyday machines, especially the way they affect friction between a rotating wheel and its supporting axle. Given Data / Assumptions: Ball bearings are placed between the wheel and the axle in cycles, scooters and similar vehicles. The question asks what effect they produce between the wheel and the axle. We consider normal operation, with properly lubricated bearings and no mechanical damage. Concept / Approach: Friction is the force that opposes relative motion between two surfaces in contact. In rotating machines, high friction at the axle wastes energy as heat and makes pedalling or driving harder. Ball bearings replace sliding friction by rolling friction. Small hardened steel balls roll between inner and outer races, separating the wheel hub from the axle. Rolling friction is much smaller than sliding friction, so the main purpose of a ball bearing is to reduce friction and allow smooth rotation with less effort and less wear. Step-by-Step Solution: Step 1: Identify the location of ball bearings in a cycle or scooter. They are placed between the rotating wheel hub and the fixed axle. Step 2: Recall that if metal surfaces slide directly on each other at the axle, friction and wear become large. Step 3: Understand that ball bearings introduce rolling elements between the surfaces, replacing sliding friction with rolling friction. Step 4: Rolling friction is much smaller than sliding friction, so the torque required to turn the wheel reduces significantly. Step 5: Conclude that the main effect is a reduction of friction between the wheel and the axle, leading to smoother motion and higher efficiency. Verification / Alternative check: Practical experience supports this idea. When bearings are worn out or badly lubricated, a bicycle wheel feels tight and hard to rotate, which means friction has increased. After servicing or replacing the bearings, the wheel spins freely with only a small push. Engineers design bearing systems explicitly to minimise friction and wear, not to increase them. Therefore it is clear that the correct role of ball bearings is to reduce friction between wheel and axle. Why Other Options Are Wrong: Increase the friction between the wheel and the axle is wrong because the aim is smooth rotation, not resistance. Reduce the friction between the wheel and the ground is incorrect, since bearings are not placed at the wheel road contact region. Increase both friction and wear at the axle is opposite to the purpose of using precision bearings. Eliminate the need for lubrication at the axle is wrong because most bearings still require lubrication to work properly and to last long. Common Pitfalls: Students sometimes think that more friction is always good because it prevents slipping, but at the axle friction wastes energy. Another mistake is confusing tyre road friction, which should be adequate for grip, with axle friction, which should be as low as possible. Remember that ball bearings are used to reduce friction at the rotating joint, making motion smoother and more efficient. Final Answer: Ball bearings in cycles and scooters are used mainly to reduce the friction between the wheel and the axle.

Question 3

In an experiment with an electric bell placed inside an enclosed evacuated glass bell jar, why is the ringing of the bell not heard outside?

Accepted Answer

Introduction / Context: The classic school demonstration with an electric bell placed inside a glass bell jar helps show that sound needs a material medium for its transmission. When air is gradually removed from the jar by a vacuum pump, the sound of the bell becomes fainter and finally cannot be heard. This question checks if you understand the basic requirement of a medium for sound waves and why a ringing bell in a vacuum is not heard outside. Given Data / Assumptions: An electric bell is ringing inside a glass bell jar. The bell jar has been evacuated so that there is almost no air inside. Observers stand outside the jar and try to hear the sound. The electric supply to the bell continues, so the hammer still strikes the bell. Concept / Approach: Sound is a mechanical wave. It consists of compressions and rarefactions of particles of a medium such as air, water or a solid. If there is no medium, compressions and rarefactions cannot travel. Vacuum is a region with no material particles, so sound cannot propagate through it. In the experiment, the bell continues to vibrate mechanically, but because the air has been removed from the jar, there are no air particles inside to carry the sound to the glass walls and then to the air outside. Step-by-Step Solution: Step 1: Recognise that the bell mechanism continues to work, because it is connected to an electric supply. Step 2: Recall that sound requires a material medium to travel, usually air in such experiments. Step 3: Understand that when the bell jar is evacuated, air is removed and the inside space becomes almost a vacuum. Step 4: In vacuum, there are no particles to vibrate and pass on sound energy from the bell to the walls of the jar. Step 5: Therefore, even though the bell vibrates, sound waves do not reach the outside, so the ringing is not heard. Verification / Alternative check: When the jar is first filled with air, the bell is clearly audible, which shows that the jar glass itself does not completely block sound. As air is pumped out, the sound gradually decreases, matching the idea that fewer air particles are available to transmit the wave. If air is let back into the jar, the sound becomes loud again. This change happens even though the bell volume and glass remain the same, proving that the absence or presence of air is the key factor. Why Other Options Are Wrong: The ringing is not loud enough is incorrect because the bell is loud when air is present. The glass of the bell jar absorbs the sound waves is wrong, since sound is heard through the same glass before evacuation. The bell jar is too small is irrelevant, because size does not suddenly change as air is removed. The electric circuit stops working in vacuum is incorrect; the bell continues to vibrate mechanically. Common Pitfalls: Students sometimes think that glass blocks sound completely because it looks solid and transparent. In reality, some sound can pass through glass. Another common confusion is between sound waves and electromagnetic waves such as light. Light can travel through vacuum, but sound cannot. Keeping this difference clear helps in many physics questions about waves and media. Final Answer: The ringing is not heard because there is no air or any other gas in the bell jar to transmit sound.

Question 4

When a lift moves upward with an acceleration, how does the apparent weight of a person or body inside the lift compare with its true weight?

Accepted Answer

Introduction / Context: Questions involving a person standing in an accelerating lift are standard applications of Newton laws of motion. They help students understand the idea of apparent weight and how it changes when the lift accelerates upward or downward. This question asks what happens to the apparent weight of a body when the lift is going up with an acceleration, a case that occurs in many real life situations such as high speed elevators. Given Data / Assumptions: A body of mass m is standing on a weighing machine inside a lift. The lift is moving upward with acceleration a, in addition to the acceleration due to gravity g. The apparent weight is the normal reaction shown by the weighing machine. We neglect friction and assume vertical motion only. Concept / Approach: The true weight of a body is given by W = m * g, where g is acceleration due to gravity. In an accelerating frame like a lift, the normal reaction between the body and the floor changes. When the lift accelerates upward, the floor must push harder on the body to produce the net upward acceleration. The weighing machine measures this normal reaction. Therefore, in an upward accelerating lift, the apparent weight R becomes m * (g + a), which is greater than m * g. Step-by-Step Solution: Step 1: Draw a free body diagram for the body in the lift, showing weight m * g downward and normal reaction R upward. Step 2: Since the lift accelerates upward with acceleration a, the body also accelerates upward with acceleration a. Step 3: Apply Newton second law in the upward direction: R minus m * g equals m * a. Step 4: Rearrange to get R = m * g + m * a = m * (g + a). Step 5: Because a is positive when the lift accelerates upward, g + a is greater than g, so R is greater than the true weight m * g. Verification / Alternative check: You can check limiting cases. If the acceleration a is zero, the lift moves at constant velocity and R becomes m * g, equal to the true weight. If the lift suddenly accelerates upward strongly, you feel heavier and the weighing machine would show a larger reading, which matches the formula R = m * (g + a). These physical sensations confirm that apparent weight increases when the lift accelerates upward. Why Other Options Are Wrong: It may be more or less than the true weight is too vague and does not describe the specific case of upward acceleration. It is less than the true weight is the situation for downward acceleration, not upward acceleration. It is equal to the true weight is correct only when the lift has no acceleration. It becomes zero for any acceleration is incorrect, as zero apparent weight occurs only in free fall when acceleration equals g downward. Common Pitfalls: A common mistake is to think that velocity and acceleration have the same effect. Students sometimes confuse a lift moving upward at constant speed with a lift accelerating upward. Another pitfall is forgetting the direction when applying Newton second law, which can lead to sign errors. Always set up the free body diagram, choose a consistent direction as positive and write the equation of motion carefully. Final Answer: When the lift goes up with acceleration, the apparent weight of the body is more than the true weight.

Question 5

When a man jumps out of a moving train and lands on the ground, in which direction is he likely to be thrown due to inertia of motion?

Accepted Answer

Introduction / Context: This question is a simple but important application of Newton first law of motion, also called the law of inertia. It deals with what happens when a person jumps from a moving train. Many accident safety instructions in rail travel are based on this idea. The question asks in which direction the man is likely to be thrown when he jumps out of a moving train and lands on the ground. Given Data / Assumptions: The train is moving in a straight line with some horizontal velocity. The man jumps out and lands on the ground beside the track. We ignore air resistance and assume a short jump close to the ground. The main consideration is the horizontal motion due to inertia. Concept / Approach: According to the law of inertia, a body in motion continues to move with the same velocity in a straight line unless acted upon by an external unbalanced force. When the man is inside the moving train, he shares the horizontal velocity of the train. At the instant he jumps, his body still has this horizontal velocity. When he lands on the ground, his feet suddenly experience friction with the stationary ground, but his upper body still tends to move forward. This mismatch makes him topple or be thrown forward in the direction of the train motion. Step-by-Step Solution: Step 1: When the man is standing in the moving train, he has the same horizontal velocity as the train. Step 2: At the moment of jumping, his body does not instantly lose this horizontal velocity; inertia keeps him moving forward. Step 3: As he lands, his feet come into sudden contact with the ground, which is at rest, so friction acts on his feet to reduce their forward motion. Step 4: The upper part of his body is still moving forward due to inertia, so his body tends to rotate and he is thrown forward relative to the ground. Step 5: Therefore, the most likely direction in which he is thrown is forward, in the direction of train motion. Verification / Alternative check: You can compare this with similar experiences, such as suddenly stopping while running. When you try to stop quickly, your body tends to lean or fall forward, not backward, because your centre of mass continues its forward motion. The same principle applies when getting down from a bus or train; you are advised to face forward and run a few steps to reduce relative speed, which shows that the natural tendency is to move forward due to inertia. Why Other Options Are Wrong: Backward, opposite to the motion of the train is wrong, because nothing gives the man a sudden strong backward push. Sideways, perpendicular to the track is incorrect, since there is no reason for a sideways force in this idealised situation. Randomly in any direction ignores the predictable effect of inertia. He will remain exactly at the point of contact is impossible, because his body already has forward velocity before contact. Common Pitfalls: Some students think that jumping out cancels the forward velocity completely, as if the act of jumping suddenly removes inertia. Others visualise the man leaving the train and immediately standing still on the ground, which is unrealistic. Remember that, horizontally, nothing stops the man instantly when he leaves the train, so he carries forward motion with him and will tend to fall or be thrown forward. Final Answer: A man jumping out of a moving train is likely to be thrown forward, in the direction of motion of the train due to inertia.

Question 6

The direction of heat flow between two objects in thermal contact depends mainly on which property of the objects?

Accepted Answer

Introduction / Context: Heat transfer is a fundamental concept in physics and everyday life. When two objects at different temperatures are placed in contact, energy in the form of heat flows between them until equilibrium is reached. The question asks which property determines the direction of heat flow between two objects when they touch, an idea that is essential for understanding thermodynamics and daily applications such as heating and cooling. Given Data / Assumptions: Two objects are in thermal contact so that heat can flow between them. They may have different masses, materials and heat capacities. We want to know what decides the direction in which heat energy moves. We assume no external work is done and no phase change for simplicity. Concept / Approach: The second law of thermodynamics states that heat naturally flows from a body at higher temperature to a body at lower temperature, and not spontaneously in the opposite direction. Temperature is the measure of the average kinetic energy of particles, while heat content depends on mass, specific heat capacity and temperature. Even if a large body at slightly lower temperature has more total heat content, heat still flows from the hotter body to the colder one. Therefore the direction of heat flow depends only on the temperature difference, not on mass or state alone. Step-by-Step Solution: Step 1: Identify the temperatures T1 and T2 of the two objects in contact. Step 2: If T1 is greater than T2, heat flows from object 1 to object 2. Step 3: If T2 is greater than T1, heat flows from object 2 to object 1. Step 4: When T1 equals T2, the objects are in thermal equilibrium and there is no net heat flow between them. Step 5: Conclude that temperature difference alone decides the direction of heat flow, independent of their masses or physical state. Verification / Alternative check: Consider placing a small hot metal spoon into a large bucket of cold water. Even though the water contains more total heat energy, heat flows from the hotter spoon to the colder water, causing the spoon to cool and the water to warm slightly. Similarly, when you touch a hot object, heat flows into your hand because the object temperature is higher, regardless of which has greater mass. These examples confirm that temperature difference governs the direction of heat transfer. Why Other Options Are Wrong: Their masses only is wrong because a heavier object at lower temperature does not automatically send heat to a lighter hotter object. Their total heat contents only is incorrect, since a body with more total heat can still receive heat if its temperature is lower. Whether they are solid, liquid or gas affects rate of heat transfer but not the direction. Their shapes and surface areas influence how fast heat is exchanged but not the fundamental direction of flow. Common Pitfalls: A common misunderstanding is to think that heat flows from a body with more thermal energy to one with less. In reality, it is the temperature difference that matters. Students also confuse heat and temperature as the same quantity. Remember that temperature is an intensive property and decides the direction of heat flow, while heat is an extensive form of energy transferred due to that difference. Final Answer: The direction of heat flow between two objects depends on their temperatures.

Question 7

A potentiometer wire is 100 cm long with a constant potential difference maintained across it. Two cells are connected in series, first supporting each other and then opposing each other. The balance points are found at 50 cm and 10 cm from the posi…

Accepted Answer

Introduction / Context: Potentiometers are precision instruments used to compare emfs of cells and to measure internal resistance without drawing current from the source. This question describes a standard experiment where two cells are connected in series, first in the same direction so that their emfs support each other, and then in opposite directions so that they oppose each other. The balance lengths on the potentiometer wire are given for both cases, and you need to find the ratio of the emfs of the two cells. Given Data / Assumptions: Total potentiometer wire length is 100 cm with uniform potential gradient. When two cells are in series aiding, balance point is at 50 cm from the positive end. When the same two cells are in series opposing, balance point is at 10 cm from the positive end. Let the emfs of the two cells be E1 and E2, with E1 greater than E2 for convenience. Potential gradient along the wire is constant and equal in both experiments. Concept / Approach: In a potentiometer, emf is proportional to the balancing length, because E equals k times L, where k is the potential gradient and L is the balance length. When cells are connected in series supporting each other, the effective emf is E1 plus E2. When they are connected in series opposing, the effective emf is E1 minus E2, assuming E1 is greater than E2. Using the two balance lengths, we can form two proportionality equations involving the same constant k and then divide them to eliminate k and find the ratio E1 to E2. Step-by-Step Solution: Step 1: For the series aiding case, write (E1 + E2) = k * 50. Step 2: For the series opposing case, write (E1 - E2) = k * 10. Step 3: Divide the first equation by the second: (E1 + E2) / (E1 - E2) = (k * 50) / (k * 10) = 50 / 10 = 5. Step 4: So (E1 + E2) / (E1 - E2) = 5, which implies E1 + E2 = 5 * (E1 - E2). Step 5: Expand and rearrange: E1 + E2 = 5E1 - 5E2, so 6E2 = 4E1, giving E1 / E2 = 6 / 4 = 3 / 2. Verification / Alternative check: To check the result, assume E2 equals 2 units. Then E1 equals 3 units from the ratio 3 : 2. In the aiding case, total emf is 5 units, proportional to 50 cm, so 1 unit corresponds to 10 cm of length. In the opposing case, effective emf is 1 unit, which should correspond to 10 cm, exactly matching the given data. This consistency confirms that the ratio E1 : E2 equals 3 : 2 and validates the algebraic steps. Why Other Options Are Wrong: 3 : 4 and 4 : 3 do not satisfy the relation (E1 + E2) / (E1 - E2) = 5 when substituted. 5 : 4 and 5 : 1 also fail to produce the correct ratio of balance lengths and do not agree with the derived equations. Only 3 : 2 remains consistent with the experimental conditions and the proportionality of emf and balance length. Common Pitfalls: A frequent mistake is to mix up which combination of cells (aiding or opposing) corresponds to which length, or to forget that the effective emf in the opposing connection is the difference E1 minus E2, not the sum. Another error is to attempt direct substitution without using the proportionality factor k, leading to confused algebra. Always write expressions for the effective emfs in both cases, set them equal to k times the measured lengths, and then divide to eliminate k before solving for the ratio of emfs. Final Answer: The ratio of the emfs of the two cells is 3 : 2.

Question 8

If a musical band is played on the surface of the Moon, how will the sound be heard by an observer standing some distance away?

Accepted Answer

Introduction / Context: The Moon provides an excellent natural example to explain how sound travels. Unlike Earth, the Moon has no significant atmosphere. This question asks what happens if a band plays music on the Moon and an observer stands some distance away. The answer illustrates the basic requirement of a medium for sound transmission and helps distinguish sound waves from electromagnetic waves such as light and radio. Given Data / Assumptions: A musical band is playing on the surface of the Moon. An observer is located some distance away on the lunar surface. The Moon has essentially no atmosphere and no air between the band and the observer. No special communication equipment is mentioned, only natural hearing by ears. Concept / Approach: Sound is a mechanical wave that needs a material medium, such as air, water or solid rock, for its propagation. The oscillating source creates vibrations in the medium, which travel as compressions and rarefactions. In a vacuum there are no particles to vibrate, so sound cannot travel. The Moon surface is in a near vacuum; there is no air between the band and the listener. Therefore the sound vibrations produced by the instruments will not reach the observer through the empty space, and the music cannot be heard. Step-by-Step Solution: Step 1: Recall that human ears detect sound only when pressure waves in a material medium reach the eardrum. Step 2: Note that the Moon has almost no atmosphere, so there is effectively vacuum between distant points on the surface. Step 3: When the band plays, the instruments vibrate and can set any nearby solid surface into vibration, but there is no air to carry the sound over open space. Step 4: Without a continuous material medium between source and listener, mechanical sound waves cannot propagate. Step 5: Therefore, regardless of distance, an observer cannot hear the band through the vacuum and the sound will not be heard at all. Verification / Alternative check: Astronauts on the Moon communicated using radio systems built into their suits. Although they were often only a few metres apart, they could not talk to each other directly through the lunar vacuum. This real historical fact supports the principle that sound cannot travel through vacuum. The only way to hear would be through direct vibrations conducted by a solid, such as touching helmets, which is not mentioned in the question. Thus, the correct conclusion is that no sound is heard at any distance in open lunar vacuum. Why Other Options Are Wrong: It will reverberate suggests multiple reflections in air, but there is no air to produce such effects. It will be heard faintly or clearly assumes that sound can travel through vacuum, which is false. Mention of a specific distance like 10 kilometres does not change the basic rule that sound needs a medium. It will be converted into radio waves automatically is wrong because sound does not convert itself into electromagnetic waves without special equipment. Common Pitfalls: Students sometimes confuse sound with light. They know that sunlight reaches Earth through space and wrongly assume that sound behaves the same way. Others think that the Moon thin exosphere can carry sound, but it is far too rarefied for meaningful propagation. To avoid such confusion, remember that sound is mechanical and needs matter, while light and radio waves are electromagnetic and can move through vacuum. Final Answer: If a band is played on the Moon, the sound will not be heard at any distance at all in the surrounding vacuum.

Question 9

An empty upright cylinder is half filled with a liquid. How does adding the liquid affect the position of the combined centre of gravity of the cylinder–liquid system?

Accepted Answer

Introduction / Context: Centre of gravity is an important concept in mechanics and stability. It is the point at which the entire weight of a body or system can be considered to act. This question considers an empty cylinder that is then half filled with a liquid. You need to predict how the position of the combined centre of gravity of the cylinder and the liquid changes compared to the original empty cylinder. Given Data / Assumptions: An upright hollow cylinder is initially empty and standing on its base. The cylinder is uniform, so its own centre of gravity lies at mid height of its axis. The cylinder is then half filled with a liquid poured into it. The liquid has density greater than that of air and occupies the lower half of the cylinder. We consider the centre of gravity of the combined system of cylinder plus liquid. Concept / Approach: The combined centre of gravity of a system of bodies lies at the weighted average of their individual centres of gravity, with weights equal to their masses. Initially only the cylinder contributes, with its centre at mid height. After liquid is added, there is additional mass located in the lower part of the cylinder. The centre of gravity of the liquid in a half filled cylinder lies at the centre of the liquid column, which is at one quarter of the cylinder height from the base. The combined centre of gravity must lie between the cylinder centre and the liquid centre. Since the new mass is below the old centre, the overall centre of gravity shifts downward. Step-by-Step Solution: Step 1: Let the cylinder height be H. The centre of gravity of the empty cylinder is at H / 2 from the base. Step 2: When the cylinder is half filled, the liquid occupies height H / 2, so its centre of gravity is at H / 4 from the base. Step 3: Denote the mass of the cylinder as mc and the mass of the added liquid as ml. Step 4: The combined centre of gravity position y measured from the base is y = (mc * (H / 2) + ml * (H / 4)) / (mc + ml). Step 5: Since H / 4 is less than H / 2 and ml is positive, the weighted average y must be less than H / 2, meaning the centre of gravity moves downward from its original position. Verification / Alternative check: Consider extreme cases. If the mass of the liquid is very small compared to the cylinder, the position of the combined centre will be only slightly lower than H / 2, but still lower. If the mass of the liquid is very large compared to the cylinder, the combined centre will come close to H / 4, the position of the liquid centre. In every possible case with some liquid present, the combined centre must lie between H / 4 and H / 2 and therefore below H / 2. This confirms that the centre of gravity always shifts downward when a liquid is added to the lower part of the cylinder. Why Other Options Are Wrong: Shifted upwards is impossible, because all added mass lies below the original centre of gravity. Shifted neither upward nor downward can occur only if no new mass is added or it is added exactly at the original centre height, which is not the case. Shifted upward or downward depending on density is incorrect because any finite liquid mass at a lower level pulls the centre down, regardless of exact density. Moves to the very bottom is wrong; the centre of gravity will approach the lower region but cannot lie at the extreme bottom while some mass remains above. Common Pitfalls: Students sometimes think that if the cylinder itself is heavy, the centre of gravity cannot move, which is not correct. Adding even a small mass in a lower position will still pull the centre slightly downward. Others may incorrectly assume the location depends on density direction in a more complicated way. The key idea is that any additional mass below the original centre will lower the combined centre of gravity, improving stability. Final Answer: When the cylinder is half filled with liquid, the combined centre of gravity is shifted downwards compared to the empty cylinder.

Question 10

A pure capacitor has a capacitance of 1 farad. In a direct current (DC) circuit under steady state conditions, what is its effective resistance?

Accepted Answer

Introduction / Context: Capacitors are widely used components in electrical and electronic circuits. Their behaviour is different in alternating current circuits and direct current circuits. This question focuses on a pure capacitor connected in a DC circuit and asks for its effective resistance in steady state. Understanding this helps clarify why capacitors block DC but pass AC signals under suitable conditions. Given Data / Assumptions: We have an ideal or pure capacitor of capacitance 1 farad. The capacitor is connected in a DC circuit. We are interested in the long term or steady state behaviour, not the initial transient. We neglect leakage and internal resistance, treating the capacitor as ideal. Concept / Approach: In a DC circuit, when a capacitor is first connected to a voltage source, there is a transient current as the capacitor charges. The relationship between charge Q, capacitance C and voltage V is Q = C * V. As the capacitor charges, voltage across it increases and current decreases. After sufficient time, the capacitor becomes fully charged such that the voltage across it equals the supply voltage. At that moment, no further current flows through the circuit. An element that does not allow any steady current to pass behaves like an open circuit, which corresponds to infinite resistance in DC steady state. Step-by-Step Solution: Step 1: Recognise that in DC, current i in a capacitor is related to rate of change of voltage by i = C * dV / dt. Step 2: In steady state DC, the voltage across the capacitor is constant with time, so dV / dt equals zero. Step 3: Substituting into the relation gives i = C * 0 which is zero current. Step 4: A circuit element through which zero current flows under a finite applied DC voltage acts like an open circuit. Step 5: In terms of resistance, an open circuit is represented as having infinite resistance, so the effective resistance of the pure capacitor in DC is infinite. Verification / Alternative check: In practical DC circuits, if you place an ideal capacitor in series with a lamp and battery, the lamp may flash briefly when the capacitor starts charging, and then it goes dark permanently. The initial flash corresponds to transient current, and the final darkness corresponds to zero steady current. An ohmmeter connected across a large DC blocking capacitor also typically reads a very high resistance after charging, matching the idea that in DC steady state the capacitor path is effectively open. These observations support the conclusion of infinite effective resistance for DC. Why Other Options Are Wrong: Zero is incorrect because a path with zero resistance would allow maximum current, opposite to what actually happens. 1 ohm or 1 divided by 2 ohm are arbitrary finite resistances and cannot represent the complete blocking of DC current. Equal to its reactance at 50 hertz is irrelevant, because reactance applies to AC, and in DC the frequency is zero. Common Pitfalls: Students sometimes confuse AC and DC behaviour. They know that capacitors can pass AC and may incorrectly assume they also allow DC current indefinitely. Others forget that once a capacitor is charged, it acts like a break in the circuit for DC. The key is to focus on steady state conditions: after transients die out, the current in a pure capacitor under DC goes to zero, which corresponds to infinite effective resistance. Final Answer: In a DC circuit under steady state, a pure capacitor behaves as an open circuit and has infinite effective resistance.

Question 11

In astronomy, one astronomical unit (1 AU) is defined as the average distance between which two celestial bodies?

Accepted Answer

Introduction / Context: Astronomical distances are very large, so it is convenient to use special units rather than everyday units like kilometres. One common unit is the astronomical unit, abbreviated as AU. This unit is defined based on the average distance between two familiar bodies in the Solar System. The question checks whether you know which pair of bodies is used to define 1 AU. Given Data / Assumptions: We want the definition of one astronomical unit. It is based on the average separation of a planet and the Sun. The unit is widely used in describing distances inside the Solar System. We neglect details of orbital eccentricity and focus on the definition only. Concept / Approach: By convention, astronomers define one astronomical unit as the average distance between Earth and the Sun over one complete orbit. This distance is roughly equal to 1.5 times 10 to the power 8 kilometres and serves as a convenient yardstick. Distances of other planets from the Sun or between planets are often expressed in multiples of AU. For example, Jupiter orbits at about 5.2 AU from the Sun. Therefore, among the options, the correct pair is Earth and the Sun. Step-by-Step Solution: Step 1: Recall that AU is a unit chosen to relate directly to Earth orbit, because Earth is our reference planet. Step 2: Remember that 1 AU represents the average distance between Earth and the Sun, not between Earth and its Moon. Step 3: Understand that other planet Sun distances are measured relative to this standard unit. Step 4: Check the options and locate Earth and the Sun as option A. Step 5: Select Earth and the Sun as the correct answer for the definition of one astronomical unit. Verification / Alternative check: Common astronomy references and textbooks state that 1 AU is about 149.6 million kilometres, the mean distance from Earth to the Sun. The distance between Earth and the Moon is about 384 thousand kilometres, far smaller and not used as the AU definition. Distances for Jupiter, Mars or Pluto are several AU and not used as the base unit. This confirms that Earth Sun distance is the correct defining pair for 1 AU. Why Other Options Are Wrong: Earth and the Moon correspond to a much smaller distance and are not used as the basis of the AU. Jupiter and the Sun are separated by several AU and that distance varies; it is expressed using AU rather than defining AU. Pluto and the Sun likewise are many AU apart and do not define the unit. Mars and the Sun form a pair whose separation is about 1.5 AU, not the definition of 1 AU. Common Pitfalls: Some learners confuse 1 AU with any planet Sun distance and think it may vary for each planet. Others mix up Earth Moon distance with Earth Sun distance, especially if they do not remember the approximate numerical values. To avoid errors, remember that AU is anchored to the Earth Sun system, making it a natural standard for measuring relative positions of planets in the Solar System. Final Answer: One astronomical unit is defined as the average distance between Earth and the Sun.

Question 12

The quality or timbre of a musical sound produced by an instrument mainly depends on which characteristic of the sound wave?

Accepted Answer

Introduction / Context: When you hear the same note played on a flute, a violin and a piano, you can easily tell which instrument is producing the sound even if the note and loudness are the same. This distinctive character of sound is called quality or timbre. The question asks which feature of the sound wave determines this quality for a musical instrument. Given Data / Assumptions: Different musical instruments can produce the same fundamental frequency. The loudness or intensity can also be controlled to be similar. Yet each instrument has a different quality that allows us to identify it. We consider physical properties of sound such as frequency, intensity and overtones. Concept / Approach: Any musical sound is not a single pure sine wave. It consists of a fundamental frequency plus additional higher frequency components called overtones or harmonics. The combination, relative strengths and phases of these overtones determine the waveform shape. This unique waveform gives each instrument its characteristic timbre. Frequency decides pitch, intensity decides loudness, but timbre or quality depends on the pattern of overtones. Therefore the number and arrangement of overtones is the main factor controlling sound quality. Step-by-Step Solution: Step 1: Recognise that when two instruments play the same musical note, they have the same fundamental frequency. Step 2: Note that if they are adjusted to the same loudness, the intensity is also similar. Step 3: Despite this, their sounds are easily distinguishable, so another property must be responsible. Step 4: Understand that each instrument has a different set of overtones with different amplitudes, which changes the overall waveform. Step 5: Conclude that the quality or timbre depends on the number and relative strengths of these overtones. Verification / Alternative check: If you analyse the sound from a violin and a tuning fork on an oscilloscope, the tuning fork waveform looks almost like a pure sine wave with very few overtones, while the violin waveform is more complex. When equalised for loudness and pitch, the only clear difference lies in the pattern of harmonics. Audio engineers use this fact when synthesising instrument sounds, adjusting overtone content to mimic real instruments. These practical observations support the idea that overtones govern sound quality. Why Other Options Are Wrong: Frequency alone determines pitch, telling us whether a note is high or low, not which instrument plays it. Intensity alone decides loudness or softness, not the characteristic tone colour. Speed of sound in air depends mainly on temperature and medium and is almost the same for different instruments in the same room. Direction of propagation does not change the inherent quality of the sound produced by an instrument. Common Pitfalls: Students often confuse pitch and quality, thinking that any change in frequency changes quality. While extreme changes may alter our perception, timbre refers specifically to differences at the same pitch and loudness. Another mistake is to use the word overtones without connecting it to the actual physical concept of harmonics. Remember that quality or timbre is directly linked to the harmonic content of the sound wave. Final Answer: The quality or timbre of sound from an instrument mainly depends on the number and arrangement of overtones in the sound wave.

Question 13

In the CGS system of units, 1 joule of energy is equal to how many ergs?

Accepted Answer

Introduction / Context: Different unit systems are used in physics to measure the same physical quantities. In mechanics, the SI system uses joule as the unit of energy, while the older CGS system uses erg as the unit of energy. Converting between these units is a standard general knowledge point and is often tested in basic physics questions. This question asks how many ergs are contained in one joule. Given Data / Assumptions: We know that SI base units for energy involve metre, kilogram and second. CGS base units for energy involve centimetre, gram and second. We want the numerical factor that converts 1 joule into ergs. We assume the standard conversion of metre to centimetre and kilogram to gram. Concept / Approach: In SI, one joule is defined as one newton metre. One newton is one kilogram metre per second squared. In CGS, one erg is defined as one dyne centimetre. One dyne is one gram centimetre per second squared. To convert between joules and ergs, we use the relations 1 kilogram equals 1000 grams and 1 metre equals 100 centimetres. Applying these conversions shows that one joule equals 10 raised to power 7 ergs. Step-by-Step Solution: Step 1: Write 1 joule as 1 newton metre, and 1 newton as 1 kilogram metre per second squared. Step 2: So 1 joule equals 1 kilogram metre squared per second squared. Step 3: Express kilogram in grams and metre in centimetres: 1 kilogram equals 10^3 grams and 1 metre equals 10^2 centimetres. Step 4: Therefore, 1 joule equals (10^3 grams) times (10^2 centimetres)^2 per second squared, which is 10^3 times 10^4 grams centimetre squared per second squared, giving 10^7 grams centimetre squared per second squared. Step 5: Since 1 erg is defined as 1 gram centimetre squared per second squared, it follows that 1 joule equals 10^7 ergs. Verification / Alternative check: You can remember this conversion using standard physics tables, which consistently list 1 joule equal to 10^7 ergs. The reverse conversion then becomes 1 erg equal to 10^-7 joules. This relation is widely used when comparing older CGS results with modern SI results in textbooks and exam problems. Checking multiple sources will always confirm the factor 10^7 as correct. Why Other Options Are Wrong: 10^3 ergs, 10^4 ergs, 10^5 ergs and 10^6 ergs are too small by several orders of magnitude compared to the correct factor of 10^7. If any of these smaller values were correct, the reverse conversion from ergs to joules would be inconsistent with known reference data and dimensional analysis. Common Pitfalls: Students often get confused by powers of ten, especially when dealing with multiple unit conversions at once. A frequent error is to write 1 joule equals 10^6 ergs or 10^5 ergs due to skipping one step in the squared metre to squared centimetre conversion. Always convert metre to centimetre first, then square the factor, and finally multiply by the kilogram to gram factor to avoid missing a power of ten. Final Answer: In the CGS system, 1 joule is equal to 10^7 ergs.

Question 14

In the CGS system, the unit erg second (erg multiplied by second) is the unit of which physical quantity?

Accepted Answer

Introduction / Context: Units in physics often reveal the nature of the physical quantity they measure. In the CGS system, energy is measured in ergs and time in seconds. When these units are combined as erg second, they represent a quantity with dimensions of energy multiplied by time. This question asks which physical quantity uses erg second as its unit. Given Data / Assumptions: Energy in CGS is measured in ergs. Time is measured in seconds. The combined unit erg second has dimensions of energy multiplied by time. We compare this with known physical quantities such as momentum and Planck constant. Concept / Approach: Angular momentum and linear momentum involve mass, length and velocity, and their units are of the form mass times length squared per time or mass times length per time. Energy has units of mass times length squared per time squared. When we multiply energy by time, we obtain mass times length squared per time. This is the dimension of action, a quantity that appears in quantum mechanics, and of Planck constant h. In the CGS system, Planck constant is often expressed in erg second. Therefore, the unit erg second corresponds to Planck constant or more generally to action. Step-by-Step Solution: Step 1: Write the CGS unit of energy as 1 erg, which has dimensions of mass times length squared per time squared. Step 2: Multiply energy by time to get erg second, which has dimensions mass times length squared per time. Step 3: Recall that Planck constant h connects energy and frequency through the relation E = h * f, so h has units of energy divided by frequency. Step 4: Frequency has units of 1 divided by time, so h as energy divided by frequency has units of energy times time. Step 5: Therefore, Planck constant in CGS units is measured in erg second, making erg second the unit of action or Planck constant. Verification / Alternative check: Look at standard values: in CGS, Planck constant h is often written as approximately 6.63 times 10 to the power minus 27 erg second. In SI, the same constant is expressed in joule second, showing that in both systems the unit is energy multiplied by time. Linear momentum uses gram centimetre per second, and angular momentum uses gram centimetre squared per second, which are clearly different from erg second. This confirms that erg second belongs to Planck constant and action, not to momentum or pure energy. Why Other Options Are Wrong: Angular momentum has units mass times length squared per time, which in CGS is gram centimetre squared per second, not erg second. Linear momentum has units mass times length per time, that is gram centimetre per second. Energy alone uses ergs, not erg second, and power uses energy per time, not energy times time. Common Pitfalls: Students sometimes see the word erg and immediately think of energy, ignoring the extra factor of second. Others group angular momentum and action together because they have similar dimensions in SI, but in CGS the presence of explicit ergs makes it clear that energy is involved. Always check the dimensions carefully and remember that Planck constant links energy and time or energy and frequency, so its unit is energy multiplied by time. Final Answer: The unit erg second in the CGS system is used for Planck constant or action.

Question 15

A simple periscope used in submarines and observation instruments works mainly on which optical principle?

Accepted Answer

Introduction / Context: Periscopes are devices that allow an observer to see over obstacles or out of a submerged submarine without exposing themselves. They use mirrors or prisms to redirect light from the object to the eye. Although simple periscopes can use plane mirrors, many optical periscopes and related instruments rely on total internal reflection inside glass prisms. This question asks which optical principle is mainly used in a standard periscope design taught in basic physics. Given Data / Assumptions: We consider a typical prism based periscope as described in school physics. Light from an object enters a glass prism and is turned through right angles. The prism is designed so that light undergoes total internal reflection at its faces. The device allows the observer to see around an obstruction by redirecting the ray path. Concept / Approach: Total internal reflection occurs when light travels inside a denser medium towards a boundary with a rarer medium at an angle greater than the critical angle. In this case, all the light is reflected back into the denser medium with no loss, acting like a perfect mirror surface. Prisms in periscopes are cut so that light striking the internal faces at suitable angles experiences total internal reflection. This produces a brighter image than ordinary mirrors, which may absorb some light. Therefore the main working principle behind a prism periscope is total internal reflection. Step-by-Step Solution: Step 1: Recall that a basic periscope uses two right angled prisms placed at the top and bottom. Step 2: Light from the distant object enters the first prism and strikes the hypotenuse face at an angle greater than the critical angle. Step 3: At this face, the light undergoes total internal reflection and is turned through about 90 degrees. Step 4: The light then travels down the periscope tube and enters the second prism, where it again experiences total internal reflection and is turned toward the observer eye. Step 5: Because both reflections are total internal reflections, the image remains bright and undistorted, showing that total internal reflection is the key principle. Verification / Alternative check: If ordinary plane mirrors were used instead of prisms, the image would be formed by simple reflection at the mirror surface, but some light would be absorbed and the mirrors could tarnish over time. In high quality optical instruments such as binoculars and periscopes, prisms are preferred for their durability and brightness. The internal faces of these prisms are not silvered; they work purely by total internal reflection. This confirms that periscopes in standard physics illustrations are examples of total internal reflection in action. Why Other Options Are Wrong: Refraction only refers to bending of light at the interface, but in a periscope the main turning of the ray is due to reflection inside the prism. Diffraction of light occurs when waves bend around obstacles or pass through narrow apertures and is not the primary effect here. Simple combination of reflection and refraction at a single surface does not describe the ideal prism behaviour, which relies on complete internal reflection. Interference of light concerns the superposition of waves and does not explain the functioning of a periscope. Common Pitfalls: Students sometimes remember that periscopes use mirrors and mistakenly think of ordinary reflection rather than total internal reflection. Others may confuse refraction through lenses with the reflection process in prisms. To keep concepts clear, associate prisms in optical instruments with total internal reflection, which gives a bright, undimmed image by reflecting almost all the incident light inside the denser medium. Final Answer: A standard prism based periscope works mainly on the principle of total internal reflection.

Question 16

In a cinema hall, if the distance between the projector and the screen is increased by 2 percent, how does the intensity of illumination on the screen change approximately?

Accepted Answer

Introduction / Context: The brightness of an image on a cinema screen depends on the light intensity reaching it from the projector. As the distance between the projector and the screen changes, the intensity changes according to the inverse square law, which is a key idea in optics and wave physics. This question asks what happens to the intensity of illumination when the projector screen distance is increased slightly by 2 percent. Given Data / Assumptions: Initial distance between projector and screen is some value d. The distance is increased by 2 percent, making it 1.02 times d. Light spreads roughly uniformly over the screen area as distance changes. The inverse square law applies: intensity is proportional to 1 divided by distance squared. We seek the approximate percentage change in intensity. Concept / Approach: The inverse square law states that intensity I of light from a point source is inversely proportional to the square of the distance r from the source, so I is proportional to 1 divided by r^2. When distance increases by a small factor, intensity decreases by the square of that factor. For a 2 percent increase in distance, the new distance is 1.02 times the old distance, so the intensity becomes 1 divided by 1.02 squared times the original intensity. We can compute this factor and convert it to a percentage decrease to get an approximate answer. Step-by-Step Solution: Step 1: Let the original distance be d and original intensity be I. Step 2: New distance d new equals 1.02 times d because of the 2 percent increase. Step 3: By the inverse square law, new intensity I new is proportional to 1 divided by (d new)^2 which is 1 divided by (1.02^2) times 1 divided by d^2. Step 4: Therefore I new equals I divided by 1.02^2, since I is proportional to 1 divided by d^2. Step 5: Compute 1.02^2 which is approximately 1.0404, so I new is about I divided by 1.0404, or about 0.96 times I, implying a decrease of roughly 4 percent. Verification / Alternative check: Another way is to use a small change approximation. For small percentage changes, the relative change in 1 divided by r^2 is about minus 2 times the relative change in r. A 2 percent increase in distance means a relative change of plus 0.02. The relative change in 1 divided by r^2 is therefore approximately minus 2 times 0.02, which is minus 0.04 or minus 4 percent. This matches the direct calculation using 1 divided by 1.0404 and gives confidence that the correct answer is a 4 percent decrease in intensity. Why Other Options Are Wrong: Decreases by about 2 percent underestimates the effect because intensity depends on the square of distance, not linearly on distance. Increases by about 2 percent or 4 percent are opposite to the expected inverse relationship between intensity and distance. Remains exactly unchanged would be correct only if distance did not change, but here we specifically increase the distance, so intensity must change. Common Pitfalls: A typical mistake is to treat intensity as inversely proportional to distance rather than to distance squared, leading to a 2 percent estimate instead of 4 percent. Some students also forget to square the factor 1.02 properly or to interpret the result as a decrease rather than simply quoting the factor 1 divided by 1.0404. To avoid such errors, always write the inverse square relation explicitly and carefully square the distance factor when calculating percentage changes. Final Answer: When the projector screen distance is increased by 2 percent, the intensity of illumination on the screen decreases by about 4 percent.

Question 17

Which of the following physical quantities are dimensionless, that is, have no fundamental physical dimensions?

Accepted Answer

Introduction / Context: Physical quantities are often classified by their dimensions and units. Some quantities have dimensions of length, mass or time, while others are ratios of similar quantities and therefore dimensionless. This question asks which of the listed quantities angle, strain and specific gravity are dimensionless in the sense that they do not have fundamental physical dimensions, even though they may have units for convenience. Given Data / Assumptions: Angle can be measured in radians or degrees. Strain is defined as change in length divided by original length. Specific gravity is defined as a ratio of densities. We examine each quantity in terms of base dimensions like length, mass and time. Concept / Approach: A dimensionless quantity is one that can be expressed as a pure number, usually as a ratio of two quantities that share the same dimensions. Strain is change in length divided by original length, so both numerator and denominator have the dimension of length, which cancel out. Specific gravity is the ratio of the density of a substance to the density of a reference substance, usually water, so mass per volume in numerator and denominator cancel. Angle in radians is defined as arc length divided by radius, again a ratio of lengths. As a result, all three quantities angle, strain and specific gravity are dimensionless, even though we sometimes attach words like radian to angle or specify conditions for specific gravity. Step-by-Step Solution: Step 1: Consider strain, which equals change in length divided by original length. The dimension of both lengths is L, so strain has dimension L divided by L, which equals 1, making it dimensionless. Step 2: Consider specific gravity, defined as density of a substance divided by density of water. Each density has dimension mass per length cubed, or M divided by L^3, so the ratio again has dimension 1. Step 3: Consider angle in radians, defined as arc length divided by radius. Both arc length and radius have dimension L, so their ratio has dimension 1. Step 4: Since all three quantities are ratios of similar dimensional quantities, they all are dimensionless. Step 5: Therefore the correct option is that all of these quantities are dimensionless. Verification / Alternative check: Dimensional analysis tables in physics books usually list examples of dimensionless quantities. Strain is always listed with no dimensions, just a number. Specific gravity is given without fundamental dimensions, though it may sometimes be described as a number such as 0.8 or 1.2. Angle in radians is treated as a pure number, because one full circle is 2 times pi radians, and pi is a dimensionless constant. These references confirm that all three quantities are dimensionless as far as physical dimensions are concerned. Why Other Options Are Wrong: Angle only, strain only or specific gravity only each wrongly suggest that the other two quantities have dimensions, which is not correct. None of these is clearly false because standard definitions show all three as ratios of like quantities and therefore dimensionless. Common Pitfalls: Students sometimes think that angle has a dimension because it is measured in radians or degrees, but these are just convenient labels for a pure ratio. Similarly, they may treat specific gravity as having mass or volume units, forgetting that it is a ratio of two densities. To avoid confusion, always check whether a quantity definition involves a ratio of identical dimensions; if so, the resulting quantity is dimensionless. Final Answer: All three listed quantities angle, strain and specific gravity are dimensionless, so the correct choice is that all of these are dimensionless.

Question 18

Fish and other aquatic creatures are able to survive in a deep frozen pond during winter mainly because

Accepted Answer

Introduction / Context: In cold regions, lakes and ponds can freeze at the surface during winter, yet fish and other aquatic animals often survive until spring. This interesting observation involves properties of water and heat transfer. The question asks for the main reason why aquatic creatures can live in a deep frozen pond, which is an important application of basic physics combined with a little biology. Given Data / Assumptions: A pond is exposed to very low temperatures in winter. The surface water can freeze and form an ice layer. Fish and other aquatic animals are present below the surface. We consider the typical behaviour of water as it cools. Concept / Approach: Water has an unusual property: its density is maximum at about 4 degrees Celsius. When the surface water cools below this temperature, it becomes less dense and stays near the surface, eventually freezing to form ice. The deeper water remains at around 4 degrees Celsius and stays liquid. Ice is also a poor conductor of heat and acts as an insulating layer, which slows further cooling of the water below. As a result, there is always a region of liquid water under the ice where fish can survive, so the key reason is the presence of this unfrozen water layer immediately below the ice sheet. Step-by-Step Solution: Step 1: As air temperature drops, the top layer of water in the pond cools down. Step 2: When water cools to around 4 degrees Celsius, it becomes densest and sinks, causing slightly warmer water from below to rise and cool in turn. Step 3: Once the overall pond water reaches near 4 degrees Celsius, further cooling at the surface creates water colder than 4 degrees Celsius, which is less dense and remains at the top. Step 4: This surface layer can freeze into ice while the lower water stays at around 4 degrees Celsius and remains liquid. Step 5: The ice layer insulates the water below, maintaining a region of liquid water under the ice where fish and other animals can survive. Verification / Alternative check: In very cold winters, you may see a thick ice sheet on lakes, but fishermen can still catch live fish by making a hole in the ice. This shows that water beneath the ice is unfrozen and supports life. If the whole pond froze solid from top to bottom, fish could not survive. Experiments and observations confirm that ice floats and forms at the top, while deeper water remains liquid due to the density anomaly of water at 4 degrees Celsius. This supports the idea that the crucial factor is the liquid water layer below the ice. Why Other Options Are Wrong: Fish are cold blooded animals is true, but being cold blooded alone cannot allow survival if all water turns to solid ice. Ice is a poor conductor of heat and does help insulation, but insulation by itself would not matter if there were no liquid water under the ice. Adapting to live completely inside ice is incorrect; most fish cannot live frozen solid for long periods. Water at the bottom becoming hotter than at the top is misleading; the bottom water is slightly warmer than freezing but still cold, around 4 degrees Celsius, and does not become hot. Common Pitfalls: Students sometimes pick the option about fish being cold blooded because it seems like a biological explanation. While this helps fish tolerate lower temperatures, physics of water density and ice formation is the primary reason that any water at all remains in liquid form. Another common confusion is to focus only on the insulating effect of ice without recognising that the structure of the water column keeps a reservoir of liquid water below the ice sheet. Final Answer: Fish and other aquatic creatures survive in frozen ponds mainly because there is always a layer of liquid water just beneath the ice where they can live.

Question 19

The rotational effect of a force acting on a body about a given axis of rotation is described in physics in terms of which quantity?

Accepted Answer

Introduction / Context: When a force acts on a rigid body, it can cause translation, rotation or both, depending on the line of action and point of application. To describe the tendency of a force to produce rotation about a specific axis, physics uses a special quantity. This question asks for the correct term used to express the rotational effect of a force about an axis of rotation. Given Data / Assumptions: A force is acting on a body that can rotate about a fixed axis. The line of action of the force is at some perpendicular distance from this axis. We want the quantity that measures how effectively this force tends to rotate the body. We assume basic rigid body mechanics and ignore friction for the definition. Concept / Approach: The rotational effect of a force about an axis is measured by the moment of force, also called torque. It is defined as the product of the magnitude of the force and the perpendicular distance from the axis of rotation to the line of action of the force. A larger force or a larger perpendicular distance produces a larger moment and stronger tendency to rotate. Other terms like centripetal or centrifugal force describe different concepts related to circular motion, not the rotation causing ability of a single force about a pivot. Step-by-Step Solution: Step 1: Consider a lever or a door hinged at one side. When you push at the handle, the door rotates about its hinges. Step 2: The force applied at the handle has a certain magnitude and acts at a distance from the hinge line. Step 3: The rotational tendency depends on both how large the force is and how far from the hinge it is applied. Step 4: In physics, this product of force and perpendicular distance is called the moment of force or torque. Step 5: Therefore, the correct term for the rotational effect of a force about an axis is moment of force. Verification / Alternative check: The formula for moment of force is often written as tau equals F times d, where tau is torque or moment, F is the applied force and d is the perpendicular distance from the axis. This concept explains many real devices such as spanners, levers and seesaws. The greater the torque, the easier it is to rotate the object. Physics texts consistently use the term moment of force or torque for this effect, while reserving centripetal and centrifugal forces for circular motion dynamics, confirming the correct option. Why Other Options Are Wrong: Centre of gravity is the point where the weight of the body effectively acts; it is not a measure of rotational effect of an external force. Centripetal force is the force that keeps a body moving in a circular path, directed toward the centre, not a measure of torque. Centrifugal force is a perceived outward force in a rotating frame and does not quantify the turning effect of a single applied force about a pivot. Linear momentum deals with straight line motion (mass times velocity) and is not a measure of rotation about a point. Common Pitfalls: Students sometimes mix up centripetal force and torque because both are associated with circular motion. However, centripetal force is about maintaining circular motion, while moment of force is about starting or influencing rotation about an axis. Another mistake is to think centre of gravity is automatically related to any rotational question, when in fact it only describes where the weight acts. Always remember that the term torque or moment of force is specifically about the rotational effect of a force around a pivot or axis. Final Answer: The rotational effect of a force about an axis is described in terms of the moment of force (torque).

Question 20

Two bodies A and B have the same mass and receive the same amount of heat. If the temperature of A rises more than that of B due to this heating, what can you conclude about their specific heat capacities?

Accepted Answer

Introduction / Context: Specific heat capacity tells us how much heat is needed to raise the temperature of a unit mass of a substance by one degree. When equal amounts of heat are supplied to equal masses of different substances, they generally show different temperature rises depending on their specific heat capacities. This question uses that idea to compare two bodies A and B when A shows a larger temperature rise than B for the same supplied heat and equal mass. Given Data / Assumptions: Bodies A and B have the same mass m. The same amount of heat Q is given to both bodies. The temperature rise of A is larger than the temperature rise of B. We assume no heat loss to surroundings, so all supplied heat goes into raising temperatures. We neglect phase changes or chemical reactions. Concept / Approach: The relation between heat, specific heat capacity and temperature change is Q = m * c * delta T, where m is mass, c is specific heat capacity and delta T is temperature change. For given Q and m, the temperature rise delta T is inversely proportional to c. A substance with a lower specific heat capacity will have a larger temperature rise for the same heat input. Since A shows a greater temperature increase than B while having the same mass and receiving the same heat, it must have a smaller specific heat capacity than B. Step-by-Step Solution: Step 1: Write the heating equation for body A: Q = m * cA * delta TA. Step 2: Write the heating equation for body B: Q = m * cB * delta TB. Step 3: Since Q and m are the same for both bodies, divide the two equations to obtain cA * delta TA equals cB * delta TB. Step 4: Rearrange to get cA divided by cB equals delta TB divided by delta TA. Step 5: Because delta TA is greater than delta TB, the ratio delta TB divided by delta TA is less than one, so cA is less than cB, meaning body A has smaller specific heat capacity than body B. Verification / Alternative check: Consider a simple numerical example. Take equal masses of water and copper. Water has a large specific heat, while copper has a much smaller one. If you supply the same amount of heat to both equal masses, copper temperature rises more than the water temperature. This real life example follows exactly the pattern given in the question and fits the conclusion that the body showing larger temperature increase has lower specific heat capacity. This supports the reasoning that cA is less than cB. Why Other Options Are Wrong: The specific heat capacity of A is greater than that of B would make delta TA smaller than delta TB for the same Q and m, opposite to what is described. Options suggesting both A and B have the same specific heat capacity rely on thermal conductivity, which affects rate of heat flow, not the final temperature rise for a fixed amount of heat absorbed. Specific heat capacities of both must be zero is impossible, since then any finite heat would cause infinite temperature change, which does not match observation. Common Pitfalls: Students sometimes confuse thermal conductivity, which controls how fast heat flows through a material, with specific heat capacity, which controls how much heat is needed to change temperature. The question concerns final temperature rise for given heat input, where conductivity does not matter if each body is heated uniformly. Another mistake is to reason qualitatively without using the simple formula Q = m * c * delta T, which immediately clarifies the inverse relationship between specific heat capacity and temperature rise. Final Answer: Since body A warms more for the same heat and mass, the specific heat capacity of A is less than that of B.

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