Question 1

In physics, one billionth of a second, that is 10^-9 second, is known by which standard unit of time?

Accepted Answer

Introduction / Context: Modern physics and electronics often deal with very small intervals of time when studying high speed events such as switching in digital circuits, light pulses, or nuclear reactions. This question checks whether you can correctly match the size of a time interval, one billionth of a second, with its standard name in the metric system. Given Data / Assumptions: The time interval in question is one billionth of a second. One billionth means 1 divided by 1 000 000 000. This is written in scientific notation as 10^-9 second. Concept / Approach: The metric system uses prefixes for powers of ten. Common time related prefixes are milli for 10^-3, micro for 10^-6, nano for 10^-9, and pico for 10^-12. The key idea is to memorise these prefixes and associate each one with the corresponding power of ten. Then, when a question states the power, you can directly choose the correct prefix and unit name. Step-by-Step Solution: Step 1: Translate the phrase one billionth of a second into scientific notation. One billionth is 1 divided by 10^9, so it equals 10^-9 second. Step 2: Recall the metric prefixes: 10^-3 is milli, 10^-6 is micro, 10^-9 is nano, and 10^-12 is pico. Step 3: Match 10^-9 with the correct prefix, which is nano. Step 4: Attach this prefix to the base unit second to obtain the term nanosecond. Verification / Alternative check: If you work the other way round, a nanosecond is one billionth of a second, a microsecond is one millionth of a second, and a millisecond is one thousandth of a second. Only nanosecond fits the description of one billionth of a second. Why Other Options Are Wrong: Option A: Microsecond corresponds to 10^-6 second, which is one millionth of a second, not one billionth. Option C: Millisecond corresponds to 10^-3 second, which is one thousandth of a second. Option D: Picosecond corresponds to 10^-12 second, which is one trillionth of a second and is even smaller than a nanosecond. Common Pitfalls: Students often mix up micro and nano because both represent very small times. Confusing 10^-6 and 10^-9 is a common mistake. Building a simple memory aid for the order milli, micro, nano, pico helps avoid this error. Always relate the prefix back to its power of ten rather than just relying on feeling that a word sounds small. Final Answer: The correct unit is the nanosecond.

Question 2

Among the common types of radioactive radiation, which ray has the least penetrating power when passing through matter such as paper, metal, or human tissue?

Accepted Answer

Introduction / Context: Radioactive substances emit different types of radiation, mainly alpha, beta, and gamma rays. Each type interacts with matter in a different way and has different penetrating power. This question checks whether you can identify which radiation is least able to pass through materials and is therefore easiest to stop. Given Data / Assumptions: The options include alpha rays, beta rays, gamma rays, and X rays. We compare only their penetrating powers in ordinary matter. We assume typical energies as encountered in basic nuclear physics. Concept / Approach: Alpha particles are relatively heavy and carry a double positive charge. Because of this, they interact very strongly with matter and lose energy quickly. Beta particles are fast moving electrons or positrons and interact less strongly than alpha particles. Gamma rays and X rays are high energy electromagnetic waves that can penetrate deep into materials. Thus, to find the least penetrating radiation, we look for the radiation that is most easily absorbed or stopped by a thin barrier. Step-by-Step Solution: Step 1: Recall that alpha particles can be stopped by a sheet of paper or by a few centimetres of air. Step 2: Beta particles have greater penetrating power and can pass through paper but are stopped by a few millimetres of aluminium. Step 3: Gamma rays and X rays have the highest penetrating power and require thick lead or concrete to significantly reduce their intensity. Step 4: Therefore, among the given options, alpha rays have the least penetrating power. Verification / Alternative check: Many safety charts show typical shielding materials. Alpha radiation is blocked by paper, beta by thin metal, and gamma by thick dense shielding. This order confirms that alpha radiation is the easiest to stop and thus the least penetrating. Why Other Options Are Wrong: Option A: Gamma rays have very high penetrating power and are used in medical imaging and industrial inspection because they pass through many materials. Option B: Beta rays penetrate more than alpha rays and can pass through paper, so they cannot be the least penetrating. Option D: X rays are similar to gamma rays in nature and are also highly penetrating. Common Pitfalls: Students sometimes confuse ionising power with penetrating power. Alpha particles have very high ionising power but low penetrating power. Gamma rays are the opposite, with low ionising power per unit path but high penetration. Keeping these two ideas distinct helps to avoid confusion in exam questions. Final Answer: The least penetrating radiation among the options is alpha rays.

Question 3

Graphite is often described as a good conductor of electricity even though it is a non metallic form of carbon. Decide whether this statement about the electrical conductivity of graphite is correct or incorrect.

Accepted Answer

Introduction / Context: Carbon exists in different forms called allotropes, such as diamond and graphite. These forms have very different physical properties. This question focuses on the electrical property of graphite and asks you to decide whether the statement that graphite is a good conductor of electricity is correct or not. Given Data / Assumptions: Graphite is a crystalline form of carbon. The statement claims that graphite is a good conductor of electricity. No special conditions such as very high temperature or solution state are mentioned. Concept / Approach: In graphite, each carbon atom is covalently bonded to three other carbon atoms in a planar hexagonal arrangement, forming layers. Each carbon atom has one electron that is delocalised and free to move within the layer. These mobile electrons enable electrical conduction along the layers. In contrast, diamond has each carbon bonded to four neighbours in a rigid three dimensional network with no free electrons, so diamond acts as an insulator. By comparing these structures, we can decide whether the given statement about graphite is correct. Step-by-Step Solution: Step 1: Recall the bonding in graphite where each carbon atom forms three sigma bonds, leaving one electron delocalised in a pi system. Step 2: Recognise that these delocalised electrons can move freely along the layers, providing a path for electric current. Step 3: Note that graphite is widely used as an electrode material in dry cells and other electrochemical cells, which confirms its good conductivity. Step 4: Conclude that the statement that graphite is a good conductor of electricity is correct under normal conditions. Verification / Alternative check: Everyday examples include the use of graphite rods as electrodes in batteries and in electrolysis experiments in school laboratories. A material selected for electrodes must conduct electricity. This practical use confirms that graphite conducts electric current well enough for many applications. Why Other Options Are Wrong: Option B: The statement is incorrect contradicts the known fact that graphite is conductive, so it is wrong. Option C: Graphite conducts only when it is red hot is incorrect because graphite electrodes conduct at room temperature. Option D: Graphite conducts only when dissolved in water is also wrong because graphite is not used as an ionic solution but as a solid electrode. Common Pitfalls: Students sometimes generalise that all non metals are poor conductors and therefore assume graphite must also be an insulator. Graphite is an important exception. Another confusion arises from mixing properties of diamond and graphite. Remember that structure determines properties, and the layered structure of graphite with delocalised electrons explains its conductivity. Final Answer: The correct evaluation is that The statement is correct.

Question 4

A siphon is used to transfer liquid from one vessel to another placed at a lower level. Under which condition will the siphon fail to work and the liquid will no longer flow through the tube?

Accepted Answer

Introduction / Context: A siphon is a simple device used to transfer liquid from a higher container to a lower container using a bent tube. The working of a siphon often appears almost magical to beginners, but it is based on pressure differences and gravity. This question asks under what condition the siphon action will stop and fail to work. Given Data / Assumptions: A siphon connects two vessels containing the same liquid. The outlet of the siphon is initially placed below the level of liquid in the first vessel. We consider steady conditions without air leaks into the tube. Concept / Approach: A siphon works because the outlet end of the tube is lower than the liquid surface in the source vessel. This creates a difference in hydrostatic pressure and allows gravity to pull the liquid through the tube. The continuous column of liquid in the tube transmits pressure. When the levels become equal, the pressure difference that drives the flow disappears, so the flow stops. Temperature or density equality alone does not destroy the pressure difference created by height, so these factors do not by themselves cause failure of the siphon. Step-by-Step Solution: Step 1: Recall that pressure at a depth h in a liquid is given by p = rho * g * h, where rho is density and g is acceleration due to gravity. Step 2: In a working siphon, the outlet liquid surface is lower, so the pressure at the outlet side of the tube is lower than the pressure at the inlet side at the same depth. Step 3: This pressure difference, together with gravity, pulls the liquid from the higher vessel to the lower vessel. Step 4: If the liquid levels in both vessels reach the same height, the pressure difference between the two sides vanishes, so the driving force for flow disappears and the siphon stops working. Verification / Alternative check: You can test this with a simple home experiment using two containers and a tube. Once the levels become equal, you will see the flow cease even though the tube is still filled with liquid. Changing temperature but keeping height difference will not instantly stop the siphon, which confirms that height difference is the controlling factor. Why Other Options Are Wrong: Option B: When the temperatures of the liquids are the same does not affect the pressure difference directly, so the siphon can still operate. Option C: When the densities of the liquids are equal is usually the case anyway, as the siphon typically carries the same liquid; equal density does not stop the flow. Option D: Unequal limb lengths do not by themselves prevent siphon action, as long as the outlet end is lower than the inlet liquid level and the tube remains filled without air gaps. Common Pitfalls: A common misconception is that the shape or length of the tube controls the siphon. In reality, the key factors are the relative heights of the liquid surfaces and the continuity of the liquid column. Students should focus on the pressure difference created by height rather than on geometric details of the tube. Final Answer: The siphon will fail when the liquid levels in the two vessels are at the same height.

Question 5

In the context of waves, amplitude is a measure of the maximum displacement of the medium from its rest position. A wave with high amplitude will have which characteristic?

Accepted Answer

Introduction / Context: Waves appear in many forms, such as water waves, sound waves, and light waves. Important properties of waves include amplitude, frequency, and wavelength. This question checks whether you know how amplitude is related to the energy carried by a wave and whether it has any direct link to pitch. Given Data / Assumptions: Amplitude measures the maximum displacement of the oscillating medium from its equilibrium position. We consider simple sinusoidal waves. Pitch is related to frequency in the case of sound waves. Concept / Approach: For mechanical waves, such as sound and water waves, the energy transported by the wave is proportional to the square of the amplitude. A larger amplitude means that particles of the medium move farther from their rest position and therefore have more kinetic and potential energy during the oscillation. Pitch, on the other hand, is determined by frequency. A higher frequency sound is heard as a higher pitch. Since amplitude and frequency are independent properties, a high amplitude does not automatically imply a high pitch. Step-by-Step Solution: Step 1: Recall that in a simple harmonic wave, energy transported per unit time is proportional to amplitude squared. Step 2: Recognise that an increase in amplitude means that the wave carries more energy through the medium. Step 3: Note that pitch is an auditory sensation that depends on frequency, not on amplitude, so amplitude does not determine pitch. Step 4: Conclude that a wave with high amplitude must be associated with high energy, not necessarily with high or low pitch. Verification / Alternative check: In everyday life, when you speak more loudly, you increase the amplitude of the sound waves without changing the note you are singing. The sound becomes more energetic and intense but may have the same pitch. This confirms that amplitude controls loudness and energy, while frequency controls pitch. Why Other Options Are Wrong: Option A: High pitch is associated with high frequency, not with amplitude, so this is incorrect. Option B: Low energy is the opposite of what a high amplitude wave has, so this is clearly wrong. Option D: Low pitch again wrongly connects amplitude with pitch, so it is not correct. Common Pitfalls: Students often confuse loudness and pitch when thinking about sound. They might think that a loud sound must be high pitched. In fact, loudness relates to amplitude and pitch relates to frequency. Keeping these terms separate in your mind helps when answering wave questions in physics exams. Final Answer: A wave with high amplitude has high energy.

Question 6

An electric fuse is a safety device inserted in series with an electrical circuit. The operation of a fuse is based mainly on which effect of electric current?

Accepted Answer

Introduction / Context: Electric fuses are widely used in household wiring and electrical appliances to protect circuits from excessive current. When a fault or overload occurs, the fuse blows and cuts off the supply, preventing damage or fire. This question asks which basic effect of electric current is used in the working of a fuse. Given Data / Assumptions: A fuse is made of a thin wire or strip with low melting point material. It is connected in series with the circuit it protects. During overload, current becomes higher than the rated value. Concept / Approach: When electric current passes through a conductor, it can produce several observable effects. These include heating effect, magnetic effect, chemical effect, and in some cases light. A fuse relies on the conductor becoming hot when current exceeds a safe value. The heat generated is proportional to I^2 * R * t, where I is current, R is resistance, and t is time. If the temperature rises above the melting point of the fuse material, the fuse element melts and opens the circuit. No chemical reaction or magnetic field is required for this basic operation, so the key effect is the heating effect of current. Step-by-Step Solution: Step 1: Recall that a fuse wire is thin and has relatively high resistance so that it heats up quickly when current increases. Step 2: Understand that heat produced in a conductor is given by H proportional to I^2 * R * t, which shows that even a small increase in current can greatly increase heating. Step 3: When the temperature of the fuse wire crosses its melting point, the wire melts and breaks, stopping current flow. Step 4: Recognise that this sequence is an application of the heating effect of electric current. Verification / Alternative check: In contrast, the magnetic effect of current is used in devices such as electromagnets and electric motors. The chemical effect is used in electroplating and electrolysis. Since a fuse does not rely on magnetism or chemical reactions, the heating effect is the only reasonable choice. Why Other Options Are Wrong: Option A: The chemical effect of the current appears in electrolysis, not in a fuse. Option B: The magnetic effect is used in relays and motors, not as the main principle in a simple fuse wire. Option D: The luminous effect of the current is observed in bulbs and discharge tubes where light is emitted, which is not the primary principle behind a fuse. Common Pitfalls: Sometimes students see the small flash of light when a fuse blows and incorrectly associate it with the luminous effect. However, that flash is a side effect of intense heating and melting, not the working principle. Always focus on the fact that the fuse is designed to melt due to heating when current becomes excessive. Final Answer: The operation of an electric fuse is based on the heating effect of the current.

Question 7

In thermodynamics, the spontaneous flow of thermal energy from a hotter body to a colder body is called what?

Accepted Answer

Introduction / Context: In everyday language people often confuse the words heat and temperature, but in physics they have precise meanings. This question aims to clarify that distinction by asking what we call the flow of thermal energy from a region of higher temperature to a region of lower temperature. Given Data / Assumptions: There are two bodies or regions at different temperatures. Thermal energy moves from the hotter body to the colder body. We are concerned with the energy in transit, not the energy stored. Concept / Approach: Heat in physics is defined as the energy that is transferred between systems or bodies due to a temperature difference. It flows spontaneously from higher temperature to lower temperature until thermal equilibrium is reached. Temperature, on the other hand, is a measure of the average kinetic energy of the particles in a body. Electric current, voltage, and potential difference are electrical concepts and do not describe thermal energy flow directly. Therefore, the correct term for the flow of thermal energy is heat. Step-by-Step Solution: Step 1: Identify that the situation describes energy moving because one body is hotter than another. Step 2: Recall the definition of heat as energy in transit due to temperature difference. Step 3: Note that electric current refers to flow of electric charge, not thermal energy. Step 4: Recognise that voltage and potential difference refer to electric potential energy per unit charge, not to thermal energy transfer. Verification / Alternative check: In thermodynamics, when we write the first law of thermodynamics, we use Q to denote heat added to a system, where Q represents energy transferred. Temperature is represented by T and does not itself represent an amount of energy transferred. This supports the idea that heat is the correct term for the energy flow. Why Other Options Are Wrong: Option B: Electric current is the rate of flow of electric charge in a conductor, measured in ampere. Option C: Voltage is another name for electric potential difference and relates to electric circuits. Option D: Potential difference is an electrical term and not used to describe thermal energy transfer between bodies. Common Pitfalls: Students sometimes say that a body contains heat, but strictly speaking a body contains internal energy and heat is the energy that flows between bodies. Remember that heat flows from hot to cold until temperatures become equal, and it is this flow that the term heat refers to. Final Answer: The flow of thermal energy from a hotter body to a colder body is called heat.

Question 8

A permanent magnet can exert a magnetic force on which of the following objects?

Accepted Answer

Introduction / Context: Permanent magnets and electromagnets are common sources of magnetic fields in physics and technology. Understanding how they interact with each other and with other magnetic materials is important in applications such as motors, loudspeakers, and magnetic locks. This question asks which objects a permanent magnet can affect. Given Data / Assumptions: We have at least one permanent magnet. We consider its interaction with other permanent magnets and with electromagnets. Both devices create magnetic fields in the surrounding space. Concept / Approach: A permanent magnet has a fixed magnetic field due to the alignment of magnetic domains in its material. An electromagnet generates a magnetic field when electric current flows through its coil. Magnetic forces arise whenever magnetic fields interact with magnetic materials or with other sources of magnetic field. Therefore, a permanent magnet can attract or repel another permanent magnet, and it can also interact with an electromagnet because both create magnetic fields that can influence each other. Step-by-Step Solution: Step 1: Recall that like poles of magnets repel and unlike poles attract, whether the magnets are permanent or electromagnetic. Step 2: Recognise that a permanent magnet produces a static magnetic field around it. Step 3: Understand that an electromagnet produces a magnetic field when current flows and acts like a magnet with north and south poles. Step 4: Conclude that a permanent magnet can interact with both another permanent magnet and an electromagnet, so both are affected. Verification / Alternative check: In many devices, such as electric motors, permanent magnets and electromagnets are used together. The rotor or stator may be a permanent magnet, while the other part is an electromagnet. The fact that these devices work is practical proof that permanent magnets and electromagnets can influence each other strongly. Why Other Options Are Wrong: Option A: Only other permanent magnets is too restrictive because it ignores the clear interaction with electromagnets. Option B: Only electromagnets carrying current is also incomplete because permanent magnets clearly attract or repel each other as well. Option D: Neither permanent magnets nor electromagnets is completely wrong because we know magnets do exert forces. Common Pitfalls: One mistake is to think of electromagnets and permanent magnets as completely different categories that do not interact. In reality, both produce magnetic fields and obey the same laws. Another confusion occurs if students think that only moving charges interact with magnets, but in fact magnetic fields interact regardless of whether they come from permanent alignment or from current. Final Answer: A permanent magnet can affect both permanent magnets and electromagnets.

Question 9

The solubility of a gas in a liquid, such as carbon dioxide in water, depends on external conditions. How does its solubility change with pressure and temperature?

Accepted Answer

Introduction / Context: The solubility of gases in liquids is important in many everyday situations, such as the carbonation of soft drinks, dissolved oxygen in water for aquatic life, and industrial absorption processes. This question examines the combined effect of pressure and temperature on gas solubility in a liquid. Given Data / Assumptions: We consider a gas dissolved in a liquid, for example carbon dioxide in water. Pressure of the gas above the liquid can be varied. Temperature of the system can also be changed. We assume no chemical reaction is changing the gas into another species. Concept / Approach: Henry law states that at constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid. Thus, higher pressure pushes more gas molecules into solution. Temperature has the opposite effect; as temperature rises, gas molecules gain kinetic energy and escape more easily from the liquid, reducing solubility. That is why warm soda loses its fizz quickly. Therefore, we must look for the option that combines increased solubility with increasing pressure and decreased solubility with increasing temperature. Step-by-Step Solution: Step 1: Apply Henry law and conclude that solubility is proportional to pressure, so increasing pressure increases solubility. Step 2: Consider the effect of temperature. Higher temperature gives gas molecules more energy, so they leave the liquid phase more readily. Step 3: Hence, increasing temperature generally decreases the solubility of gases in liquids. Step 4: Combine these results. Solubility increases with increasing pressure and decreases with increasing temperature, which matches option C. Verification / Alternative check: Think about a sealed cold drink bottle. It is filled under high pressure, which allows much more carbon dioxide to dissolve. When you open the bottle and release pressure, gas comes out as bubbles. Also, if the drink is warmed, it goes flat faster, showing that gas solubility is lower at higher temperature. Why Other Options Are Wrong: Option A: It increases with increasing pressure only mentions pressure and ignores the temperature effect, so it is incomplete. Option B: It decreases with increasing temperature only mentions temperature and neglects the known effect of pressure. Option D: It is independent of both pressure and temperature contradicts experimental evidence and basic gas solubility theory. Common Pitfalls: Students may remember only one factor, such as pressure, and forget temperature. Others might confuse gas solubility with the behaviour of solids, where solubility often increases with temperature. It is important to distinguish that for gases, higher temperature usually reduces solubility in liquids. Final Answer: The correct combined behaviour is that gas solubility increases with increasing pressure and decreases with increasing temperature.

Question 10

To make a simple electromagnet in a school laboratory, which of the following items are required?

Accepted Answer

Introduction / Context: Electromagnets are magnets produced by the flow of electric current through a coil. They are widely used in electric bells, relays, cranes, and many other devices. In school experiments, students build a simple electromagnet to understand how electricity and magnetism are related. This question asks which components are needed to construct such an electromagnet. Given Data / Assumptions: We want a simple electromagnet that works on direct current. A coil of wire carrying current must be wound around a magnetic core. A suitable power source and core material are available in the options. Concept / Approach: When electric current passes through a conductor, it produces a magnetic field around it. Winding the conductor into a coil concentrates the field. Placing a soft iron core inside the coil further strengthens the field because the core becomes magnetised. To maintain current in the coil, a direct current source, such as dry cell batteries, is required. Therefore, we need three things together: a power source, an insulated copper wire to form the coil, and a soft iron core such as a nail. Step-by-Step Solution: Step 1: Identify that an electromagnet requires a current carrying coil and a core. Step 2: Recognise that dry cell batteries supply the necessary direct current. Step 3: The thin insulated copper wire is used to make many turns of a coil around the core, increasing the magnetic field strength. Step 4: The iron nail or rod acts as a soft iron core that becomes strongly magnetised when current flows through the surrounding coil. Verification / Alternative check: A typical classroom experiment uses exactly these components. Students wrap insulated copper wire around an iron nail, connect the ends of the wire to a battery, and then observe that the nail can attract small paper clips, showing that an electromagnet has been created. Why Other Options Are Wrong: Option A: A power source alone cannot form a magnet without a coil and core. Option B: A coil of wire alone, without current, does not create a meaningful magnetic field. Option C: A soft iron nail alone has only weak magnetism and does not become a strong magnet without a surrounding current carrying coil. Common Pitfalls: Some learners think that any piece of iron connected to a battery becomes an electromagnet, but without many turns of wire, the field is very weak. Others forget the need for insulation on the wire so that turns do not short circuit. Remember that a complete electromagnet setup requires all the components listed together. Final Answer: To make a simple electromagnet you need all of the above components together.

Question 11

In driving and vehicle dynamics, skids that occur when the brakes are applied so hard that the front or rear wheels lose traction with the road surface are called what type of skids?

Accepted Answer

Introduction / Context: Safe driving requires good understanding of how tyres interact with the road surface. When friction is lost due to sudden or excessive actions, a vehicle can skid. This question focuses on the specific situation in which brakes are applied so hard that the front or rear wheels lock and lose traction, leading to a skid. Given Data / Assumptions: The driver applies the brakes very hard. The tyres lose grip and begin to slide instead of roll. We are asked to identify the type of skid caused mainly by braking. Concept / Approach: A skid occurs when the frictional force between tyre and road is not sufficient to support the demanded acceleration or deceleration. Different types of skids are classified by their causes, such as cornering skids during turning, power skids due to excessive acceleration, and braking skids caused by harsh braking. When brakes are applied so strongly that the wheels lock up, the tyres slide over the road rather than roll, and the skid is termed a braking skid. Step-by-Step Solution: Step 1: Recognise that the scenario describes intense braking that causes wheels to lose rolling motion. Step 2: Connect this description with the definition of a braking skid, which is directly caused by excessive braking force. Step 3: Cornering skids are primarily related to turning at excessive speed rather than straight line braking. Step 4: Blowout skids result from sudden tyre failure, and power skids relate to over acceleration, so neither matches the given condition. Verification / Alternative check: Driver training manuals often describe four basic skids: braking, power, cornering, and blowout. The description given in the question, where brakes are applied so hard that wheels lose traction, matches exactly the definition of a braking skid. Anti lock braking systems in modern vehicles are designed specifically to prevent this kind of skid by modulating brake pressure. Why Other Options Are Wrong: Option A: Cornering skids happen when a vehicle is turned too sharply or too fast, causing sideways sliding, not necessarily due to harsh braking. Option B: Blowout skids are associated with sudden tyre bursts or blowouts, which is not mentioned here. Option C: Power skids occur when too much power is applied to the drive wheels, causing them to spin and lose traction, again a different situation. Common Pitfalls: Students sometimes choose cornering skids because they associate skids in general with turning. It is important to read the cause used in the question. The wording clearly emphasises brakes applied so hard that traction is lost, which points to braking skids. In examinations, pay close attention to the described cause of the skid. Final Answer: These skids are specifically called braking skids.

Question 12

Find the electric current in amperes when an electric charge of 1200 coulomb passes through a conductor uniformly in 6 seconds.

Accepted Answer

Introduction / Context: Electric current measures how much electric charge flows through a conductor per unit time. This is a fundamental idea in basic electricity and circuits. The question provides a total charge and a time interval and asks you to calculate the current in amperes using the definition formula. Given Data / Assumptions: Total charge Q flowing through the conductor is 1200 coulomb. Time t taken for this charge to pass is 6 seconds. The current is steady and uniform during this time. We need the current I in ampere. Concept / Approach: By definition, electric current I is the rate of flow of charge. Mathematically, I = Q / t, where Q is the total charge that flows in time t. Substituting the provided values directly into this formula gives the answer. The units are consistent: coulomb divided by second gives ampere, the SI unit of current. Step-by-Step Solution: Step 1: Write the basic formula for current: I = Q / t. Step 2: Substitute Q = 1200 coulomb and t = 6 second into the formula. Step 3: Perform the division: I = 1200 / 6. Step 4: Calculate 1200 / 6 = 200, so I = 200 ampere. Verification / Alternative check: We can perform a rough check. If 600 coulomb flowed in 6 seconds, the current would be 100 ampere. Because the actual charge is 1200 coulomb, which is twice 600, the current should be twice 100 ampere, that is 200 ampere. This matches our detailed calculation and confirms the result. Why Other Options Are Wrong: Option A: 7200 A would correspond to 1200 coulomb in only 0.166 second, not 6 seconds, so it is not correct for this data. Option C: 100 A would be correct if Q were 600 coulomb instead of 1200 coulomb, so it is too small. Option D: 20 A is an order of magnitude smaller than the correct value and does not satisfy the formula I = Q / t when Q = 1200 coulomb and t = 6 second. Common Pitfalls: Students sometimes confuse the formula and use I = t / Q by mistake or forget to divide properly. Another common error is mishandling units, but here the units are already in SI form, coulomb and second. Always remember that current is charge per unit time, so the charge goes in the numerator and time in the denominator. Final Answer: The electric current is 200 A.

Question 13

An object with a mass of 22 kilogram is moving with a velocity of 5 metre per second. What is the kinetic energy possessed by the object?

Accepted Answer

Introduction / Context: Kinetic energy is the energy that a body possesses due to its motion. It depends on both the mass of the object and its speed. This question checks if you can apply the standard kinetic energy formula to find the energy of a moving object when its mass and velocity are given. Given Data / Assumptions: Mass m of the object is 22 kilogram. Velocity v of the object is 5 metre per second. The motion is non relativistic, so classical mechanics applies. We are asked to calculate kinetic energy in joule. Concept / Approach: The formula for kinetic energy of a body of mass m moving with speed v is KE = (1 / 2) * m * v^2. Here, m is in kilogram and v is in metre per second, so the result will be in joule because 1 joule equals 1 kilogram metre^2 per second^2. We simply substitute the given values and simplify the arithmetic. Step-by-Step Solution: Step 1: Write the formula for kinetic energy: KE = (1 / 2) * m * v^2. Step 2: Substitute m = 22 kilogram and v = 5 metre per second. Step 3: First compute v^2 = 5^2 = 25. Step 4: Compute KE = (1 / 2) * 22 * 25. Half of 22 is 11, so KE = 11 * 25 = 275 joule. Verification / Alternative check: Another way is to multiply directly: 22 * 25 = 550, then divide by 2 to get 275. Both approaches yield the same result. Dimensional analysis also confirms that kilogram times metre^2 per second^2 gives joule, so the units are consistent. Why Other Options Are Wrong: Option B: 110 joule might result from taking half of the mass but forgetting to multiply by v^2 correctly; it is too small compared to the correct value. Option C: 550 joule corresponds to m * v^2 without dividing by 2, which means the factor 1 / 2 in the formula was omitted. Option D: 2750 joule is ten times larger than the correct answer and does not arise from any correct interpretation of the given numbers. Common Pitfalls: Common mistakes include forgetting to square the velocity, misplacing the factor 1 / 2, or doing incorrect multiplication. Always compute v^2 first, then multiply by mass, then multiply by 1 / 2. Careful stepwise calculation helps prevent arithmetic errors during examinations. Final Answer: The kinetic energy of the object is 275 joule.

Question 14

A body of mass 4.0 kilogram is initially at rest. Under the action of a constant unbalanced force it gains a speed of 5 metre per second. How much work is done by the force on the body?

Accepted Answer

Introduction / Context: Work and kinetic energy are closely related in mechanics. When a net force acts on a body and changes its speed, the work done by that force is equal to the change in the kinetic energy of the body. This question uses that relationship to find the work done when a body is accelerated from rest to a certain speed. Given Data / Assumptions: Mass m of the body is 4.0 kilogram. Initial velocity u is 0 metre per second because the body is at rest. Final velocity v is 5 metre per second. The applied force is constant and acts along the direction of motion. Concept / Approach: The work energy theorem states that the net work done on an object equals the change in its kinetic energy. Mathematically, W = KE final minus KE initial. Kinetic energy is given by KE = (1 / 2) * m * v^2. Using the initial and final velocities, we can calculate the change in kinetic energy and thus the work done by the force. Step-by-Step Solution: Step 1: Compute initial kinetic energy: KE initial = (1 / 2) * m * u^2. Since u = 0, KE initial = 0. Step 2: Compute final kinetic energy: KE final = (1 / 2) * m * v^2 = (1 / 2) * 4 * 5^2. Step 3: Calculate v^2 = 5^2 = 25, so KE final = (1 / 2) * 4 * 25 = 2 * 25 = 50 joule. Step 4: Use work energy theorem: W = KE final minus KE initial = 50 minus 0 = 50 joule. So the work done is 50 joule. Verification / Alternative check: The answer can be checked by estimating. A 4 kilogram mass moving at 5 metre per second is not extremely fast, so an energy of tens of joule is reasonable. The calculation 0.5 * 4 * 25 naturally gives 50, so the arithmetic is straightforward and consistent. Why Other Options Are Wrong: Option B: 30 joule does not match the computed change in kinetic energy and likely arises from a random mistake. Option C: 20 joule could result from using an incorrect formula such as m * v instead of (1 / 2) * m * v^2. Option D: 40 joule might come from miscalculating 0.5 * 4 * 5^2 as 40 instead of 50, showing a multiplication error. Common Pitfalls: Students sometimes forget that work depends on change in kinetic energy, not just on final energy. Others may incorrectly square the velocity or misplace decimal points. Keeping the formula W = (1 / 2) * m * v^2 minus (1 / 2) * m * u^2 in mind and performing each step carefully helps avoid such mistakes. Final Answer: The work done by the force on the body is 50 joule.

Question 15

To convert a temperature measured on the Celsius scale into the Kelvin scale, what mathematical operation must you perform with the number 273 on the given Celsius temperature?

Accepted Answer

Introduction / Context: Temperature can be expressed in different scales, the most common being Celsius and Kelvin. The Kelvin scale is the SI unit of temperature and is widely used in scientific calculations. Understanding how to convert between Celsius and Kelvin is basic but essential knowledge in physics and chemistry. Given Data / Assumptions: We have a temperature value on the Celsius scale, denoted by degree Celsius. We want the corresponding value on the Kelvin scale. We use the simple linear relationship between the two scales. Concept / Approach: The Celsius and Kelvin scales have the same size degree, but their zero points differ. Absolute zero, the lowest possible temperature, is 0 kelvin, which corresponds to approximately minus 273 degree Celsius. Therefore, the relation is K = degree Celsius + 273. For most school level problems, the rounding to 273 is sufficient, though more precisely it is 273.15. The key point is that you add 273 to convert Celsius to Kelvin. Step-by-Step Solution: Step 1: Recall the relation between the scales: K = degree Celsius + 273. Step 2: Identify that we want to convert from Celsius to Kelvin, so Kelvin is the subject of the formula. Step 3: To obtain K, we add 273 to the Celsius value, we do not subtract, multiply, or divide by 273. Step 4: Conclude that the correct operation is to add 273 to the given Celsius temperature. Verification / Alternative check: Take a familiar example. The freezing point of water is 0 degree Celsius. On the Kelvin scale, this temperature is 273 kelvin. Since we know that 0 plus 273 equals 273, the formula K = degree Celsius + 273 is confirmed. Similarly, 100 degree Celsius corresponds to about 373 kelvin, again adding 273. Why Other Options Are Wrong: Option A: Subtract 273 would give negative values for many ordinary temperatures and does not match the known conversion. Option B: Multiply by 273 would produce extremely large values that do not correspond to known Kelvin temperatures. Option D: Divide by 273 would drastically reduce the temperature value and has no physical meaning in this context. Common Pitfalls: Some students confuse the difference between degrees and the actual scale shift. They might remember the number 273 but use it in the wrong operation. To avoid this, remember a simple reference: 0 degree Celsius is 273 kelvin, so you must add 273 to move from Celsius to Kelvin. Final Answer: To convert Celsius to Kelvin you must add 273 to the Celsius temperature.

Question 16

For a given conducting material at a fixed temperature, the electrical resistance of a conductor is inversely proportional to which of the following quantities?

Accepted Answer

Introduction / Context: Resistance is a property of a conductor that opposes the flow of electric current. It depends on the material as well as on the geometry of the conductor. This question checks whether you remember how resistance varies with length and cross sectional area for a uniform conductor made of a given material. Given Data / Assumptions: The conductor is made of a fixed material, so its resistivity is constant. Temperature is held constant, so resistivity does not change with heating. We examine the relationship between resistance and geometric factors such as length and area of cross section. Concept / Approach: The fundamental relation for resistance of a uniform conductor is R = rho * L / A, where R is resistance, rho is resistivity of the material, L is length, and A is area of cross section. From this expression, for constant rho and constant temperature, resistance is directly proportional to length and inversely proportional to area of cross section. Thus, if area is increased, resistance decreases, and if area is reduced, resistance increases. Step-by-Step Solution: Step 1: Write the formula R = rho * L / A. Step 2: Note that rho and temperature are constant according to the question. Step 3: Recognise that R is proportional to L and inversely proportional to A, because rho is constant. Step 4: Conclude that the quantity with inverse relationship to resistance is the area of cross section of the conductor. Verification / Alternative check: In practical wiring, thick wires are used for high current because they have large cross sectional area and therefore lower resistance. Thin wires, which have small area, have higher resistance. This real world observation supports the inverse relationship between resistance and area of cross section. Why Other Options Are Wrong: Option A: Length is directly proportional to resistance. Doubling length doubles resistance, so it is not an inverse relation. Option B: Resistivity is a material property, not a geometric factor. For a given material it is constant and does not have an inverse relation to resistance in this context. Option C: Temperature is held constant in the question. In many materials resistance increases with temperature, but this is not the simple inverse proportionality being asked about. Common Pitfalls: Sometimes students mistakenly think that both length and area have the same kind of relation with resistance. It is essential to remember that length appears in the numerator of R = rho * L / A, while area is in the denominator. Visualising a long thin wire versus a short thick wire can help: a long thin wire has high resistance, whereas a short thick wire has low resistance. Final Answer: Resistance is inversely proportional to its area of cross section.

Question 17

According to Newtonian mechanics, which of the following physical quantities never occurs singly in nature but always appears as an interaction between two bodies?

Accepted Answer

Introduction / Context: In classical mechanics, forces describe interactions between bodies. Newton third law states that every action has an equal and opposite reaction. This question asks which quantity from the list is never isolated in nature but always appears as part of such an interaction pair. Given Data / Assumptions: We consider the standard definitions of inertia, momentum, force, and velocity. Newton laws of motion apply. We focus on whether the quantity can be associated with a single body alone or must involve at least two interacting bodies. Concept / Approach: Force is defined as an interaction that can change the motion of a body. Newton third law tells us that forces always come in pairs: if body A exerts a force on body B, then body B exerts an equal and opposite force on body A. Thus the concept of force inherently involves at least two bodies. Inertia, momentum, and velocity, on the other hand, are properties that can be assigned to a single body without reference to another body. Step-by-Step Solution: Step 1: Recall Newton third law, which describes action and reaction pairs of forces between two bodies. Step 2: Recognise that a force cannot exist as a single isolated entity; it always involves one body acting on another. Step 3: Note that inertia is the tendency of a body to resist change in its state of motion and is simply a property of that body. Step 4: Momentum (mass times velocity) and velocity itself are also properties that can be defined for a single body, not necessarily tied to a pair of bodies. Verification / Alternative check: Consider an object at rest on a table. The table exerts an upward normal force on the object, while the object exerts an equal downward force on the table. These are two forces forming an interaction pair. There is no way to have one of these forces without the corresponding other force, which illustrates that a single, isolated force is not observed in nature. Why Other Options Are Wrong: Option A: Inertia is a scalar property of a body related to its mass and does not require another body to be defined. Option B: Momentum is defined as mass times velocity and can be specified for a single particle or body. Option D: Velocity is a vector that gives the speed and direction of motion of a body and does not require a second body to exist. Common Pitfalls: Some students think of force as something that a single body possesses, like mass. In reality, force describes the interaction between bodies. Remembering the phrase forces always occur in action reaction pairs can help keep this idea clear. It is a core part of Newtonian mechanics and frequently tested in conceptual questions. Final Answer: The quantity that never occurs singly in nature is force.

Question 18

When you look into a mirror and observe that your image appears upright but smaller than your actual size, what type of mirror are you using?

Accepted Answer

Introduction / Context: Spherical mirrors are widely used in daily life, for example as rear view mirrors in vehicles or in shaving mirrors. They can produce images that are larger, smaller, real, or virtual depending on their shape and the object position. This question focuses on identifying which type of mirror forms an upright image that is smaller than the object. Given Data / Assumptions: The image seen in the mirror is upright, not inverted. The image is smaller than the actual object. No special distances are specified, so we consider typical use where the object is in front of the mirror beyond its focal region in a normal viewing situation. Concept / Approach: A plane mirror always forms an image that is the same size as the object and upright. A concave mirror can form enlarged upright images when the object is close, or real inverted images when the object is beyond the focal point. A convex mirror, which bulges outward, always forms images that are virtual, upright, and diminished in size, regardless of the exact object distance. This property is why convex mirrors are used as rear view and side mirrors in vehicles, as they give a wider field of view. Step-by-Step Solution: Step 1: Note that the image is upright, so the mirror is not forming a real inverted image as a concave mirror would do for objects beyond the focal length. Step 2: Observe that the image is smaller than the object, which rules out a plane mirror, since plane mirrors give same size images. Step 3: Recognise that a convex mirror always produces a virtual, upright, and diminished image for any object position in front of it. Step 4: Conclude that the mirror described must be a convex mirror. Verification / Alternative check: The phrase objects in mirror are closer than they appear seen on vehicle mirrors is usually associated with convex mirrors. These mirrors show a wide scene but with reduced image size. This matches the description of an upright but smaller image. Concave mirrors used as makeup mirrors give enlarged images when the face is within the focal length, not smaller ones. Why Other Options Are Wrong: Option A: A concave mirror can give either enlarged upright images or inverted images depending on object distance, but in common arrangements it does not give a smaller upright image. Option C: A plano concave mirror has one flat and one concave surface and would behave similarly to a concave mirror for reflection, not matching the described image size. Option D: A plane mirror always produces an image of the same size as the object, so it cannot create a smaller image as given in the question. Common Pitfalls: Students sometimes confuse concave and convex mirrors because of their similar names. A useful memory aid is that convex mirrors spread out light rays and tend to make images appear smaller and farther away, which is what we see in vehicle rear view mirrors. Concave mirrors can concentrate light and are used where magnified images are needed. Final Answer: The mirror that produces an upright, smaller image is a convex mirror.

Question 19

In geometrical optics, what is the focal length of a concave mirror that has a radius of curvature of 20.0 cm?

Accepted Answer

Introduction / Context: Concave mirrors are widely used in devices such as shaving mirrors, head lights, reflecting telescopes, and many optical instruments. A key parameter that describes any spherical mirror is its focal length, which tells us how it converges or diverges light rays coming from infinity. In this question from basic geometrical optics, we are given the radius of curvature of a concave mirror and are asked to determine its focal length. Understanding the relationship between radius of curvature and focal length is very important for quick numerical calculations in physics examinations and practical applications in optical design and laboratory work. Given Data / Assumptions: • The mirror is a concave spherical mirror. • Radius of curvature R = 20.0 cm. • We assume the standard sign convention for spherical mirrors is used in theory, but for this simple magnitude question only the absolute value is needed. • We are asked for the focal length f in centimetres. Concept / Approach: In geometrical optics, a spherical mirror has a simple relationship between its focal length f and its radius of curvature R. The focus is located at the midpoint between the pole of the mirror and its center of curvature. For both concave and convex spherical mirrors, the magnitude of the focal length is given by f = R / 2. For a concave mirror, the focal length is treated as negative in sign convention, but when a question asks for focal length without focusing on sign, usually only the magnitude is expected in the answer. We therefore focus on the simple proportional relationship and compute half of the given radius of curvature. Step-by-Step Solution: Step 1: Recall the mirror formula relating focal length and radius of curvature: f = R / 2. Step 2: Substitute the given radius of curvature R = 20.0 cm into the formula. Step 3: Calculate f = 20.0 cm / 2 = 10.0 cm. Step 4: Report the focal length as 10 cm in magnitude, which is the required answer for this conceptual numerical question. Verification / Alternative check: One way to remember this relationship is geometric. For a spherical mirror, light rays coming from infinity and parallel to the principal axis reflect through a point that is midway between the mirror surface and the center of curvature. That midpoint is, by definition, the focal point. Since the center of curvature is at a distance R from the pole, the midpoint is at R / 2. This geometric interpretation is independent of the sign convention and simply confirms that the focal length in magnitude is half of the radius of curvature. You can also recall that textbook ray diagrams for spherical mirrors always show the focus halfway to the center of curvature, which is another confirmation. Why Other Options Are Wrong: Option a (5 cm): This would correspond to f = R / 4, which has no standard basis for spherical mirrors. Option c (15 cm): This value is three quarters of the radius and does not match the required half relationship. Option d (20 cm): This equals the full radius of curvature and would only be correct if focal length equalled radius, which is not true. Option e (25 cm): This is larger than the radius and has no physical meaning in the context of the standard mirror formula. Common Pitfalls: Students sometimes confuse focal length with radius of curvature and choose R instead of R / 2. Another frequent mistake is mixing sign convention with magnitude and worrying about negative signs when the question clearly asks only for the numerical value. Some learners also confuse formulas for lenses and mirrors, but the relationship f = R / 2 for spherical mirrors is a very direct one and should be memorised as a basic result. Always read the question carefully to see whether it is asking for magnitude or for a sign sensitive value in a coordinate system. Final Answer: The focal length of the concave mirror is 10 cm.

Question 20

A body of mass 2 kg is thrown vertically upward with an initial velocity of 20 m/s. Taking g = 10 m/s^2, what will be its kinetic energy after 2 seconds of upward motion?

Accepted Answer

Introduction / Context: Projectile motion and vertical throws are common topics in basic mechanics. When a body is thrown straight up, gravity slows it down until its velocity becomes zero at the highest point, and then it falls back. This question connects kinematics with kinetic energy by asking what the kinetic energy of a body is after a certain time during its ascent. Understanding how velocity changes under constant acceleration due to gravity and how kinetic energy depends on velocity is very important for many physics problems in examinations. Given Data / Assumptions: • Mass m = 2 kg. • Initial velocity u = 20 m/s upward. • Acceleration due to gravity g = 10 m/s^2 acting downward. • Time of motion considered t = 2 s after the body is thrown. • Air resistance is neglected and motion is vertical. Concept / Approach: The motion is uniformly accelerated under gravity. For an object thrown upward, its velocity decreases linearly with time according to the equation v = u - g * t, where v is the velocity at time t. Once we find the velocity after 2 seconds, we can calculate the kinetic energy using the standard expression for translational kinetic energy, KE = (1 / 2) * m * v^2. If the velocity becomes zero at that instant, the kinetic energy will also become zero, because kinetic energy depends on the square of speed. Step-by-Step Solution: Step 1: Use the kinematic relation v = u - g * t for upward motion. Step 2: Substitute u = 20 m/s, g = 10 m/s^2, and t = 2 s. Step 3: Compute v = 20 - 10 * 2 = 20 - 20 = 0 m/s. Step 4: Use the kinetic energy formula KE = (1 / 2) * m * v^2. Step 5: Substitute m = 2 kg and v = 0 m/s to get KE = (1 / 2) * 2 * 0^2 = 0 J. Step 6: Therefore, the kinetic energy after 2 seconds is 0 J. Verification / Alternative check: We can also reason physically. The time to reach the highest point is t_top = u / g = 20 / 10 = 2 s. This means that exactly after 2 seconds, the body reaches its maximum height, and at that instant its velocity becomes zero before it starts descending. At the highest point in a vertical throw, all the initial kinetic energy has been converted into gravitational potential energy. Since kinetic energy is zero when velocity is zero, the answer of 0 J is fully consistent with the energy viewpoint as well. Why Other Options Are Wrong: Option b (100 J): This would correspond to a non zero velocity and does not match the correct v = 0 m/s at t = 2 s. Option c (200 J): This value is equal to the initial kinetic energy (1 / 2 * 2 * 20^2 = 400 J) divided by 2, but there is no reason for the energy to be half at this particular time. Option d (400 J): This is actually the initial kinetic energy at t = 0 s, not the kinetic energy after 2 seconds. Option e (20 J): This is an arbitrary small value that does not follow from the kinematic or energy calculations. Common Pitfalls: A common error is to forget the sign in the kinematic equation and write v = u + g * t for upward motion, which would give an incorrect non zero velocity. Some learners also directly use displacement formulas instead of focusing on velocity, which can lead to confusion. It is also easy to mistakenly assume that kinetic energy decreases linearly with time, but in reality it depends on the square of velocity. Always calculate the velocity at the required time using correct sign conventions and then substitute into the energy formula. Final Answer: The kinetic energy of the body after 2 seconds is 0 J.

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