ICSE Physics Class 10 Goyal Brothers Solutions

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    ICSE Physics Class 10 Goyal Brothers Objective with Solutions for Students

    Consider the following physical quantities: energy, power, pressure, impulse, temperature, gravitational potential. Which of the above is/are the Vector quantity/quantities?

    (a) Impulse only

    (b) Impulse and pressure only

    (c) Impulse, temperature, and pressure

    (d) Gravitational potential

    Explanation:
    This question asks us to identify which of the listed physical quantities possess both magnitude and direction. In Physics, quantities are broadly classified as scalar or Vector depending on how they are defined and measured.

    Scalar quantities are described only by magnitude. Examples include Mass, temperature, energy, and time. These quantities do not require any directional information to fully specify their values. Vector quantities, on the other hand, require both magnitude and direction. Common examples include displacement, velocity, acceleration, momentum, and force. When analyzing a list of quantities, the key step is to check whether direction is an inherent part of the quantity’s definition.

    To approach the problem, examine each quantity individually. Some quantities describe the rate at which energy is transferred or stored in a system, while others describe conditions such as thermal state or potential energy per unit Mass in a field. These typically depend only on magnitude. However, quantities that arise directly from force or momentum interactions may carry directional properties because the underlying physical processes involve motion or directional change.

    In Physics, impulse is associated with the change in momentum of an object during a short time interval. Momentum itself is a directional quantity because it depends on velocity, which has direction. Therefore, whenever a physical quantity is derived from directional motion or force, it may inherit Vector characteristics.

    A useful way to think about this is to imagine pushing an object in a particular direction. The change in motion produced during that push occurs along that same direction, showing how certain quantities depend on direction.

    Overall, identifying Vector quantities requires recognizing which physical quantities fundamentally depend on directional motion or forces rather than just magnitude.

    Option a – Impulse only

    A piece of stone tied to a string is made to revolve in a circular orbit of radius r with the other end of the string as the center. If the string breaks, the stone will:

    (a) move away from the center

    (b) move towards the center

    (c) move along a tangent

    (d) stop

    Explanation:
    The situation described involves a stone moving in a circular path while attached to a string. Circular motion occurs when an object continuously changes direction while maintaining motion along a curved path. Understanding what happens when the string breaks requires knowledge of centripetal force and Newton’s laws of motion.

    For an object to move in a circle, there must be a force constantly directed toward the center of the circle. This inward force is called centripetal force. In this case, the tension in the string provides that force, pulling the stone toward the center and preventing it from moving away in a straight line. Because of this continuous inward pull, the stone’s direction of motion keeps changing, producing circular motion.

    At any instant during circular motion, the velocity of the object is directed along the tangent to the circle. This means that although the object follows a curved path, its instantaneous velocity always points along the straight line touching the circle at that point.

    If the string suddenly breaks, the centripetal force disappears immediately. Without this inward force, there is nothing left to continuously change the direction of motion. According to Newton’s First Law of Motion, an object continues moving with the same velocity unless acted upon by an external force.

    A common real-life example is swinging a ball attached to a string and then releasing it. When released, the ball moves in the direction it was heading at that instant rather than continuing in a circle.

    Thus, the motion after the string breaks is determined by the direction of the stone’s instantaneous velocity at that moment.

    Option c – move along a tangent

    Which one of the following statements is not true?

    (a) The gravitational force of Earth acting on a body of Mass 1 kg is 9.8 newton

    (b) The force acting on an object of Mass 5 kg moving with a uniform velocity of 10 m/s on a frictionless surface is zero

    (c) The SI unit of weight is kg

    (d) The momentum of a man having a Mass of 100 kg walking with a uniform velocity of 2 m/s is 200 newton-seconds

    Explanation:
    This question requires identifying the statement that does not correctly represent a physical concept. Such Questions test conceptual understanding of basic mechanics, especially force, momentum, and physical units.

    In mechanics, force is defined by Newton’s Second Law as the product of Mass and acceleration. The gravitational force acting on a body near the Earth’s surface is equal to its weight, which can be calculated using the relation
    𝐹
    =
    𝑚
    𝑔
    F=mg, where
    𝑔
    g represents the acceleration due to gravity. The standard value of
    𝑔
    g near the Earth’s surface is approximately 9.8 m/s2.

    Another important principle is that if an object moves with uniform velocity, its acceleration is zero. Since force is proportional to acceleration, the NET force acting on such an object must also be zero. This idea follows directly from Newton’s laws.

    Momentum is another important physical quantity defined as the product of Mass and velocity. Its unit is kg·m/s, which is equivalent to newton-seconds (N·s). Momentum represents the quantity of motion possessed by an object and depends on both Mass and velocity.

    In addition to understanding physical definitions, it is also important to know the correct SI units of different quantities. Each physical quantity has a specific unit assigned in the International System of Units. Confusing units with other quantities can lead to incorrect statements.

    Therefore, when solving this type of question, carefully evaluate each statement based on physical definitions, units, and fundamental laws of motion before identifying which one is incorrect.

    Option c – The SI unit of weight is kg

    Which one of the following statements is correct?

    (a) Acceleration due to gravity decreases with the increase in altitude

    (b) Acceleration due to gravity increases with the increase of depth (assuming Earth to be a sphere of uniform density)

    (c) Acceleration due to gravity decreases with the increase of latitude

    (d) Acceleration due to gravity is independent of the Mass of the Earth

    Explanation:
    This question focuses on how the acceleration due to gravity varies under different conditions. The acceleration due to gravity, commonly denoted by
    𝑔
    g, is the acceleration experienced by objects due to the Earth’s gravitational attraction.

    Near the Earth’s surface, the value of
    𝑔
    g is approximately 9.8 m/s2. However, this value is not perfectly constant everywhere. It varies slightly depending on factors such as altitude, depth below the Earth’s surface, and latitude. These variations occur because gravitational attraction depends on the distance from the Earth’s center and the distribution of Mass within the planet.

    When an object moves to higher altitudes above the Earth’s surface, the distance between the object and the Earth’s center increases. According to the law of Gravitation, gravitational force decreases with increasing distance. As a result, the value of
    𝑔
    g changes gradually with altitude.

    Similarly, when moving inside the Earth, the gravitational effect changes because only the mass lying at smaller radii contributes effectively to the gravitational pull. This causes variations in the acceleration due to gravity with depth.

    Latitude also influences the value of
    𝑔
    g. Because the Earth rotates and is slightly flattened at the poles, gravitational acceleration differs slightly between the equator and the poles.

    To determine which statement is correct, one must analyze how these physical factors influence gravitational acceleration and compare them with the statements provided in the options.

    Option a – Acceleration due to gravity decreases with the increase in altitude

    A barometer kept in an elevator reads 760 mm when the elevator is at rest. When the elevator moves in an upward direction with increasing speed, the reading of the barometer will be:

    (a) 760 mm

    (b) < 760 mm

    (c) > 760 mm

    (d) Zero

    Explanation:
    A barometer measures atmospheric pressure by balancing the pressure of air against a column of mercury. The height of the mercury column represents the atmospheric pressure acting on the surface of the liquid.

    When the barometer is placed in an elevator at rest, the reading corresponds to the normal atmospheric pressure at that location. However, when the elevator starts accelerating, the effective forces acting on objects inside the elevator change due to the acceleration of the system.

    In an accelerating frame such as a moving elevator, objects behave as if an additional force is acting on them. When the elevator accelerates upward, the effective weight of objects inside the elevator increases because the upward acceleration adds to the gravitational effect. This means liquids inside measuring instruments experience greater effective pressure.

    Since the barometer operates based on pressure balance, any change in effective gravitational influence on the mercury column can alter the height of the column. The apparent weight of the mercury increases in an upward accelerating frame, which influences the pressure balance inside the instrument.

    A similar effect can be observed when standing on a weighing machine inside an accelerating elevator. If the elevator accelerates upward, the scale reading becomes larger because the apparent weight increases.

    Therefore, the reading of the barometer changes depending on the acceleration of the elevator and the resulting change in effective forces acting on the mercury column.

    Option b – < 760 mm

    In wind power, which form of energy is converted into electrical energy?

    (a) Kinetic energy

    (b) Potential energy

    (c) Solar energy

    (d) Radiant energy

    Explanation:
    Wind power generation involves converting the energy of moving air into electrical energy using wind turbines. Wind is moving air caused by pressure and temperature differences, giving it kinetic energy, expressed as ½ m v2, where m is mass and v is velocity.

    As wind pushes the turbine blades, it rotates them, converting kinetic energy into mechanical energy in the form of rotation. The turbine shaft then drives a generator, where electromagnetic induction converts mechanical energy into electrical energy.

    An analogy is a bicycle dynamo: as the wheel rotates, it powers a small generator to Light a bulb. Wind turbines use a similar principle on a larger scale.

    Thus, the motion of air is harnessed and transformed through a sequence of energy conversions to produce Electricity.

    Option a – Kinetic energy

    The specific Heat of a gas can have

    (a) any positive value ranging from zero to infinity.

    (b) always negative value.

    (c) any positive value ranging from five to infinity.

    (d) zero value.

    Explanation:
    Specific Heat is the Heat energy required to raise the temperature of a unit mass of a substance by 1°C or 1 K. For gases, specific Heat depends on the heating conditions, either constant volume (Cv) or constant pressure (Cp).

    At constant pressure, part of the Heat supplied does work during expansion, so Cp > Cv. Specific Heat is always positive for gases because heating increases kinetic energy of particles. Molecular structure, degrees of freedom, and thermodynamic constraints affect the magnitude of specific Heat.

    This concept helps in understanding gas behavior in engines, refrigerators, and thermodynamic cycles. Knowing specific Heat allows calculation of temperature change when energy is supplied.

    Overall, gases can have a range of positive specific Heat values depending on the type and condition of heating.

    Option a – any positive value ranging from zero to infinity

    The thermal capacity of a body depends on the

    (a) mass of the body only.

    (b) mass and shape of the body only.

    (c) density of the body.

    (d) mass, shape, and temperature of the body.

    Explanation:
    Thermal capacity, or Heat capacity, is the amount of heat required to raise the temperature of a body by 1°C. It depends on the mass of the body and the specific heat capacity (c) of the material, given by C = m × c.

    Material type affects how energy is stored, as different substances absorb heat differently. Shape generally does not directly affect heat capacity, though larger or more complex objects can store more energy due to mass. Temperature also indirectly affects capacity if specific heat varies with temperature.

    For example, water has a higher heat capacity than Metals, so it heats up more slowly. Understanding heat capacity is crucial in thermal management and engineering applications.

    Thus, thermal capacity depends mainly on the mass and the specific heat of the substance.

    Option d – mass, shape, and temperature of the body

    Which one of the following statements regarding the expansion of materials due to heating is not correct?

    (a) As the ice melts, it expands uniformly up to 4°C.

    (b) Mercury thermometer works using the principle of expansion due to heating.

    (c) A small gap is kept between two rails to allow for expansion due to heating.

    (d) The length of a metallic wire increases when its temperature is increased.

    Explanation:
    Thermal expansion refers to the increase in size of a material when heated due to increased atomic vibrations. Solids expand linearly, liquids expand volumetrically, and gases expand significantly when heated.

    Some substances, like water, show anomalous expansion: it contracts between 0°C and 4°C, reaching maximum density at 4°C. Mercury in thermometers expands uniformly with temperature. Rails and bridges are built with gaps to allow for expansion and prevent structural damage.

    To determine an incorrect statement, compare each claim against known physical principles of expansion and special behaviors, such as water’s unusual contraction near freezing.

    Thermal expansion explains practical design considerations and why temperature changes affect materials differently.

    Option a – As the ice melts, it expands uniformly up to 4°C

    When water is heated from 0°C to 10°C, its volume

    (a) increases

    (b) decreases

    (c) does not change

    (d) first decreases and then increases

    Explanation:
    Water shows anomalous expansion near freezing. Between 0°C and 4°C, heating reduces water’s volume as hydrogen bonds rearrange molecules closer together. Density increases up to 4°C.

    Above 4°C, water behaves normally: heating increases Molecular motion, causing expansion. This behavior is vital in lakes and ponds: ice forms on the surface, insulating aquatic life below.

    Thus, water’s volume decreases initially and then increases with temperature between 0°C and 10°C.

    Option d – first decreases and then increases

    Gases have

    (a) only aerial expansion

    (b) only volume expansion

    (c) no definite shape

    (d) Both (b) and (c)

    Explanation:
    Gases do not have a definite shape; they expand to fill their container. This expansion is primarily volumetric and aerial, meaning the gas spreads in all directions.

    Unlike Solids or liquids, the particles in gases move freely and randomly. This is why gases conform to the shape and volume of containers. Understanding gas behavior is fundamental in Thermodynamics and explains phenomena like diffusion and pressure variations.

    Thus, gases have no fixed shape and exhibit volume expansion.

    Option d – Both (b) and (c)

    Water is used in a hot water bag because

    (a) it is easily available.

    (b) it has high specific gravity.

    (c) it has high specific heat.

    (d) it is a liquid substance.

    Explanation:
    Water has a high specific heat capacity, meaning it can store and release a large amount of heat with minimal temperature change. This property makes it ideal for retaining warmth in hot water bags.

    It is also easily available and a liquid, making it convenient to pour into flexible bags. The high heat storage ensures gradual, long-lasting heat release for therapeutic purposes.

    Thus, water is used in hot water bags primarily because of its high specific heat.

    Option c – it has high specific heat

    The majority of charge carriers in a p-type semiconductor are

    (a) free electrons

    (b) conduction electrons

    (c) ions

    (d) holes

    Explanation:
    A p-type semiconductor is created by doping a pure semiconductor with an element that has fewer valence electrons (e.g., aluminum in silicon). This creates holes, which are locations where electrons are missing.

    These holes act as the primary charge carriers, moving through the crystal lattice as electrons fill vacancies. Minority carriers are electrons, but the flow of holes dominates electrical conduction in p-type materials.

    Thus, in p-type semiconductors, holes are the majority carriers.

    Option d – holes

    In an n-type semiconductor

    (a) electrons are the only carriers.

    (b) holes are the only carriers.

    (c) electrons are the majority carriers, and holes are the minority carriers.

    (d) None of the above

    Explanation:
    An n-type semiconductor is produced by doping a semiconductor with elements having extra valence electrons (e.g., phosphorus in silicon). The extra electrons become majority carriers, moving freely through the lattice.

    Holes exist as minority carriers but contribute less to conduction. This contrasts with p-type semiconductors, where holes dominate. Understanding carrier types is essential for designing diodes, transistors, and integrated circuits.

    Thus, electrons are the majority carriers, and holes are the minority carriers in n-type semiconductors.

    Option c – electrons are the majority carriers, and holes are the minority carriers

    Which of the following is not a semiconductor?

    (a) Silicon

    (b) Krypton

    (c) Selenium

    (d) Germanium

    Explanation:
    Semiconductors are materials with electrical conductivity between that of a conductor and an insulator. Common semiconductors include silicon, germanium, and selenium. Their conductivity can be modified by doping.

    Elements like krypton are noble gases with complete electron shells and do not conduct Electricity under normal conditions. Identifying non-semiconductors requires knowledge of elemental properties and electrical behavior.

    Thus, the material not showing semiconductor properties is easily recognized based on its electronic configuration.

    Option b – Krypton

    Which one of the following is not a semiconductor?

    (a) Silicon

    (b) Germanium

    (c) Quartz

    (d) Gallium arsenide

    Explanation:
    Semiconductors like silicon, germanium, and gallium arsenide have conductivity between Metals and insulators, and their properties can be altered by doping.

    Quartz, however, is an insulator under normal conditions. It has a rigid crystal lattice with very few free electrons, so it does not conduct Electricity efficiently. Understanding the distinction helps in selecting materials for electronics and circuit design.

    Thus, quartz is not considered a semiconductor.

    Option c – Quartz

    Silicon is which type of material?

    (a) Semiconductor

    (b) Insulator

    (c) Conductor

    (d) No option is correct

    Explanation:
    Silicon is a widely used material in electronics due to its semiconducting properties. Its conductivity is moderate and can be enhanced or reduced by introducing impurities, a process called doping.

    Silicon forms the backbone of modern devices such as diodes, transistors, and integrated circuits. Its semiconducting nature allows it to control current flow efficiently, making it crucial for digital and analog electronics.

    Thus, silicon is classified as a semiconductor.

    Option a – Semiconductor

    Which of the following is widely used in making semiconductor chips?

    (a) Radium

    (b) Sodium

    (c) Germanium

    (d) Sulphur

    Explanation:
    Semiconductor chips require materials with controlled electrical properties. Germanium and silicon are classic semiconductors, but silicon is preferred due to its stability, abundance, and ease of forming high-quality crystals.

    The manufacturing process involves slicing silicon wafers, doping specific regions, and creating circuits through photolithography. Choosing the right semiconductor ensures device reliability and performance.

    Thus, silicon is the primary material for semiconductor chips.

    Option c – Germanium

    When a piece of pure silicon is doped with aluminum, then

    (a) the conductivity of the doped silicon piece will remain the same.

    (b) the doped silicon piece will become n-type.

    (c) the doped silicon piece will become p-type.

    (d) the resistivity of the doped silicon piece will increase.

    Explanation:
    Doping silicon with aluminum, which has fewer valence electrons than silicon, creates holes in the crystal lattice. These holes act as positive charge carriers, making the material p-type.

    Electrical conduction in the doped silicon occurs primarily through the movement of these holes. This principle is foundational in constructing p-n junctions and designing electronic devices.

    Thus, the doped silicon becomes p-type.

    Option c – the doped silicon piece will become p-type

    Avalanche breakdown is due to

    (a) collision of minority charge carrier.

    (b) increase in depletion layer thickness.

    (c) decrease in depletion layer thickness.

    (d) None of the above

    Explanation:
    Avalanche breakdown occurs in a p-n junction under high reverse bias. Minority carriers accelerate due to the Electric Field and collide with atoms in the lattice, generating additional electron-hole pairs.

    This chain reaction rapidly increases current, potentially damaging the diode if uncontrolled. It is utilized in avalanche diodes designed to handle such conditions safely.

    Thus, avalanche breakdown is caused by collisions of charge carriers in strong reverse bias.

    Option b – increase in depletion layer thickness

    The p-n junction diode is used as

    (a) An amplifier

    (b) A rectifier

    (c) An oscillator

    (d) A modulator

    Explanation:
    A p-n junction diode allows current to flow primarily in one direction, making it ideal as a rectifier. Rectifiers convert Alternating Current (AC) into direct current (DC).

    In the forward-biased condition, electrons and holes move across the junction, allowing current flow. In reverse bias, the junction blocks current, preventing backward flow. This property is essential for power supplies and electronics circuits.

    Thus, the diode functions as a rectifier.

    Option b – A rectifier

    The full form of LED is

    (a) Light Emitting Diode

    (b) Light Emitting Device

    (c) Light Enhancing Device

    (d) Light Enhancing Diode

    Explanation:
    LED stands for Light Emitting Diode, a semiconductor device that emits Light when a forward current passes through it.

    Electrons recombine with holes in the semiconductor, releasing energy in the form of photons. LEDs are energy-efficient, have long lifespans, and are widely used in displays, indicators, and lighting.

    Thus, the full form of LED is Light Emitting Diode.

    Option a – Light Emitting Diode

    The reverse biasing in a p-n junction diode

    (a) decreases the potential barrier.

    (b) increases the potential barrier.

    (c) increases the number of minority charge carriers.

    (d) increases the number of majority charge carriers.

    Explanation:
    In reverse bias, the p-n junction diode’s potential barrier increases, preventing majority carriers from crossing the junction.

    Minority carriers may still contribute to a very small leakage current. The widening of the depletion region under reverse bias ensures the diode blocks current flow efficiently, a principle used in protection circuits and rectifiers.

    Thus, reverse bias increases the potential barrier of the diode.

    Option b – increases the potential barrier

    In a p-n junction, avalanche current flows in the circuit when biasing is

    (a) forward

    (b) reverse

    (c) zero

    (d) excess

    Explanation:
    Avalanche current occurs under reverse bias when the applied voltage is very high. Minority carriers accelerate, collide with lattice atoms, and generate more carriers, causing a rapid increase in current.

    This principle is exploited in avalanche diodes for voltage regulation and surge protection. Understanding the biasing condition is critical to predicting and controlling avalanche breakdown.

    Thus, avalanche current flows under reverse bias.

    Option b – reverse

    In thermocouples, current flow due to

    (A) Protons

    (B) Electrons

    (C) +ve ions

    (D) -ve ions

    Explanation:
    In a thermocouple, current flows because of electrons moving between two dissimilar Metals when there is a temperature difference between junctions.

    The Seebeck effect causes electrons to flow from the hot junction to the cold junction, creating an electromotive force (emf) and measurable current. This principle underpins temperature measurement in thermocouples.

    Thus, electron flow is responsible for current in thermocouples.

    Option b – Electrons

    If the temperature of the hot junction is increased by keeping the cold junction at the same temperature, thermoemf :

    (A) Always increases

    (B) Always decreases

    (C) May increases or decrease

    (D) Always remains constant

    Explanation:
    Thermoelectric emf in a thermocouple depends on the temperature difference between the hot and cold junctions. Increasing the hot junction temperature increases this difference, enhancing electron flow.

    The Seebeck effect causes electrons to migrate from the hot side to the cold side, producing a voltage. The magnitude of thermo emf is proportional to the temperature gradient, though it also depends on the metal pair used in the thermocouple.

    Thus, altering the hot junction temperature while keeping the cold junction constant changes the thermo emf.

    Option c – May increases or decrease

    At inversion temperature, emf across thermocouple becomes :

    (A) Maximum

    (B) Minimum

    (C) Zero

    (D) Maximum and then decreases

    Explanation:
    The inversion temperature is the temperature at which the thermo emf of a thermocouple becomes zero. Beyond this point, increasing the hot junction temperature does not produce an emf.

    This occurs because the NET movement of electrons due to the Seebeck effect cancels out at this temperature. Different metal pairs have different inversion temperatures. Understanding this is important in precision temperature measurements and high-temperature applications.

    Thus, emf becomes zero at the inversion temperature.

    Option c – Zero

    Using a thermocouple :

    (A) Electrical energy is converted into heat energy

    (B) Heat energy is converted into electrical energy

    (C) Chemical energy is converted into electric energy

    (D) Electrical energy is converted into chemical energy

    Explanation:
    Thermocouples convert heat energy into electrical energy using two dissimilar Metals joined at junctions. The Seebeck effect generates a voltage proportional to the temperature difference.

    This allows temperature measurement without needing an external power source. Thermocouples are widely used in industrial furnaces, engines, and scientific instruments due to their accuracy and durability.

    Thus, thermocouples convert thermal energy into electrical energy.

    Option b – Heat energy is converted into electrical energy

    Thermocouples cannot be used as

    (A) Thermometer

    (B) Voltmeter

    (C) Refrigerator

    (D) A. C. power generator

    Explanation:
    Thermocouples are designed to measure temperature or generate electrical signals from heat. They cannot act as refrigerators because cooling requires transferring heat against a gradient using external energy.

    While they can detect temperature changes, they cannot pump heat to lower a junction’s temperature like a refrigeration system. Understanding the operational limits ensures correct application in instruments.

    Thus, thermocouples cannot function as refrigerators.

    Option d – A. C. power generator

    If the temperature of the cold junction is kept constant and the temperature of the hot junction is continuously increased then :

    (A) Thermo emf increases continuously

    (B) Thermo emf decreases continuously

    (C) Thermo emf first increases and then decreases

    (D) Change in emf depends upon the metal pair used in the thermocouple

    Explanation:
    Thermo emf in a thermocouple depends on the temperature difference between the junctions. Continuously increasing the hot junction while keeping the cold junction constant increases the emf proportionally.

    However, the exact emf change can depend on the metal pair and its properties, as some Metals have non-linear behavior at high temperatures. This principle is critical for designing accurate thermocouple measurement systems.

    Thus, thermo emf generally increases with a larger temperature difference.

    Option c – Thermo emf first increases and then decreases

    At neutral temperature, emf across thermo couple becomes :

    (A) Maximum

    (B) Minimum

    (C) Zero

    (D) Zero and then increases

    Explanation:
    Neutral temperature is the point at which the thermo emf remains unchanged even if the reference junction’s temperature varies.

    This occurs because at neutral temperature, the effects of cold junction changes cancel out in the Seebeck effect. Determining neutral temperature ensures accurate temperature readings when the reference junction cannot be perfectly maintained.

    Thus, emf remains stable at neutral temperature.

    Option a – Maximum

    Neutral temperature is the temperature of the hot junction at which :

    (A) Thermo emf is minimum

    (B) Thermo emf is maximum

    (C) Thermo emf changes its direction

    (D) Seebeck effect is observed

    Explanation:
    Neutral temperature is defined as the hot junction temperature where thermo emf is independent of the cold junction’s temperature.

    At this point, the variation in cold junction temperature does not affect the output voltage, simplifying calibration and improving measurement accuracy. Different metal pairs have specific neutral temperatures.

    Thus, neutral temperature ensures emf stability despite cold junction variations.

    Option b – Thermo emf is maximum

    The neutral temperature of the thermocouple is dependent upon :

    (A) the Metals used in the thermocouple

    (B) the Temperature of the cold junction

    (C) the Temperature of the hot junction

    (D) the Inversion temperature

    Explanation:
    Neutral temperature depends primarily on the Metals used in the thermocouple. Different metal combinations exhibit different Seebeck coefficients, affecting the temperature at which emf becomes insensitive to the reference junction.

    The cold junction temperature and surrounding conditions do not determine neutral temperature; only the material properties Matter. This is essential for selecting appropriate metal pairs for high-accuracy temperature measurements.

    Thus, neutral temperature depends on the thermocouple Metals.

    Option a – the Metals used in the thermocouple

    The inversion temperature of the thermocouple depends upon

    (A) Neutral temperature

    (B) the Temperature of the cold junction and material used in the thermocouple

    (C) Experimental arrangement

    (D) the Temperature of the surrounding air

    Explanation:
    The inversion temperature is the temperature at which the thermocouple emf becomes zero. It depends on both the cold junction temperature and the metal pair used.

    At this temperature, the NET electron flow due to the Seebeck effect cancels out. This is significant in high-temperature thermocouple applications, ensuring the device operates within the usable temperature range without signal reversal.

    Thus, inversion temperature depends on the cold junction and thermocouple materials.

    Option b – the Temperature of the cold junction and material used in the thermocouple

    Which thermocouple form the following will produce maximum thermo emf if the junctions are kept at the same temperature difference? :

    (A) Copper – bismuth

    (B) Antimony – bismuth

    (C) Iron-nickel

    (D) Copper – iron

    Explanation:
    Thermo emf magnitude depends on the Seebeck coefficients of the metal pair. Pairs like antimony-bismuth produce larger emf than others like copper-iron or copper-bismuth because their electron Transport properties create a stronger voltage per degree of temperature difference.

    Selecting the optimal metal pair ensures maximum sensitivity and voltage output for temperature measurements over a given range.

    Thus, the metal pair with the highest Seebeck coefficient gives maximum thermo emf.

    Option b – Antimony – bismuth

    When the number of thermocouples is joined in series the NET thermo emf :

    (A) Is decreased

    (B) Is increased

    (C) Becomes zero

    (D) Remains the same

    Explanation:
    Connecting thermocouples in series, known as a thermopile, increases the NET thermo emf because the individual voltages of each thermocouple add algebraically.

    This principle allows for generating higher voltages from small temperature differences and is useful in power generation or precision temperature measurement. The total emf is proportional to the number of junctions and the temperature gradient of each.

    Thus, joining thermocouples in series increases the NET emf.

    Option b – Is increased

    The ionization energy of a hydrogen Atom is 13.6 eV. According to Bohr’s theory, the energy for a transition between the 3rd and 4th orbits is

    a. 1.51 eV

    b. 3.40 eV

    c. 0.66 eV

    d. 2.88 eV

    Explanation:
    Bohr’s theory states that the energy levels of hydrogen are quantized and given by En = −13.6/n2 eV, where n is the orbit number.

    The energy difference between two levels, ΔE = E4 − E3, corresponds to the photon emitted or absorbed during the transition. Calculating ΔE involves substituting n = 3 and n = 4 into the formula.

    This principle underpins hydrogen’s emission and absorption spectra, used in spectroscopy and understanding Atomic Structure.

    Thus, the energy for a transition can be calculated using Bohr’s quantized energy formula.

    Option c – 0.66 eV

    What is the energy of Light of wavelength 4500 Å?

    a. 3 eV

    b. 2.75 eV

    c. 2.55 eV

    d. 2.32 eV

    Explanation:
    The energy of light is given by E = hc/λ, where h is Planck’s constant, c is the speed of light, and λ is the wavelength.

    For visible light with wavelength 4500 Å (1 Å = 10−10 m), substituting constants allows calculation of the photon energy in electron volts (eV). Shorter wavelengths correspond to higher energy photons, which is why violet light is more energetic than red.

    This principle is fundamental in photoelectric effect, spectroscopy, and understanding light-Matter interactions.

    Thus, photon energy can be calculated directly from its wavelength using E = hc/λ.

    Option b – 2.75 eV

    The work function of a metal is 2.5 eV. If radiation of wavelength 4000 Å falls on it, what will be the energy of emitted photoelectrons?

    a. 0.0063 eV

    b. 0.0603 eV

    c. 0.603 eV

    d. 6.3 eV

    Explanation:
    Photoelectric effect states that electrons are emitted when incident photon energy exceeds the metal’s work function (φ). Photon energy is E = hc/λ.

    The kinetic energy of the emitted electrons is KE = E − φ. Converting the wavelength into meters and using Planck’s constant and speed of light allows calculating the energy of the photoelectrons. This demonstrates energy conservation between photons and emitted electrons.

    Thus, the emitted electron energy depends on photon energy minus the metal’s work function.

    Option c – 0.603 eV

    The longest wavelength that will cause photoelectric emission from a sodium surface is 5830 Å. What is its work function?

    a. 2.12 eV

    b. 1.21 eV

    c. 3.12 eV

    d. 2 eV

    Explanation:
    The threshold wavelength corresponds to the minimum photon energy required to emit electrons. work function φ = hc/λ, where h is Planck’s constant, c is speed of light, and λ is the threshold wavelength.

    Electrons will not be emitted if the photon energy is less than φ. This is a direct application of the photoelectric effect, linking wavelength and energy.

    Thus, the work function is determined from the threshold wavelength using φ = hc/λ.

    Option a – 2.12 eV

    The work function of a metal is 2.4 eV, and the stopping potential is 1.2 volts. What is the energy of incident photons?

    a. 2.6 volts

    b. 3.6 volts

    c. 1-volt

    d. 3 volts

    Explanation:
    Photoelectric effect gives KE = eVstop = Ephoton − φ, where φ is the work function and Vstop is the stopping potential.

    Rearranging, Ephoton = φ + eVstop. Converting φ and Vstop into electron volts allows calculation of incident photon energy. This illustrates energy conservation during electron emission.

    Thus, the incident photon energy equals work function plus the stopping potential energy.

    Option b – 3.6 volts

    1000 small water drops, each of radius r and charge q, coalesce to form one spherical drop. The potential of the big drop is larger than that of the smaller ones by a factor

    a. 10

    b. 100

    c. 10,000

    d. 1000

    Explanation:
    The potential V of a sphere is V = Q/R, where Q is charge and R is radius. When 1000 drops combine, the total volume is conserved: R3 = 1000 r3, giving R = 10 r. Total charge Q = 1000 q.

    Thus, Vbig = Q/R = 1000 q / 10 r = 100 (q/r). The factor increase compared to a small drop is 100. This principle applies to charged droplets and Electrostatics experiments.

    Thus, coalescence increases potential by a factor of 100.

    Option b – 100

    A parallel plate Capacitor is given a charge Q. If the separation between the plates is doubled, its capacity will be

    a. halved

    b. doubled

    c. remains the same

    d. None of the above

    Explanation:
    Capacitance of a parallel plate Capacitor is C = ε₀A/d, where d is the plate separation. Doubling d reduces capacitance by half.

    Stored charge remains constant if the Capacitor is isolated, but voltage across the plates changes. This relation is essential in designing Capacitors for electronics and energy storage.

    Thus, increasing separation decreases capacitance inversely.

    Option a – halved

    A parallel plate Capacitor is charged, and the charging battery is then disconnected. If the plates of the Capacitor are moved further apart by means of insulating handles

    a. The voltage across the plates increases

    b. The stored capacitance decreases

    c. The stored capacitance increases

    d. remains constant

    Explanation:
    When the battery is disconnected, charge Q is constant. Capacitance decreases if plate separation increases: C = ε₀A/d.

    Voltage V = Q/C therefore increases as capacitance decreases. The energy stored U = ½ Q²/C also increases, showing mechanical work done to separate plates converts into electrical energy.

    Thus, voltage increases when the plate separation increases with an isolated Capacitor.

    Option a – The voltage across the plates increases

    The work required to bring the dipole from infinity to the point under consideration, in such a way that +q and -q are at all times on the same equipotential surface is known as

    a. potential energy

    b. kinetic energy

    c. static energy

    d. gravitational energy

    Explanation:
    Work done on a dipole in an Electric Field, keeping the charges on equipotential surfaces, corresponds to the potential energy of the dipole: U = −p·E cosθ, where p is dipole moment and θ is angle with the field.

    No work is done in moving along equipotential surfaces, only orientation matters. This is fundamental in understanding Molecular interactions, torque on dipoles, and energy storage in electric fields.

    Thus, potential energy quantifies the work associated with dipole orientation in an Electric Field.

    Option a – potential energy

    The theory of dielectric was begun by Faraday and subsequently developed by

    a. Kelvin

    b. Boltzmann

    c. Maxwell theory

    d. Gauss’s theory

    Explanation:
    Faraday initiated the study of dielectrics, materials that do not conduct Electricity but can be polarized in an Electric Field. Maxwell later formalized this through Maxwell’s equations, which describe electric and magnetic fields and how dielectrics respond.

    Dielectrics increase capacitance by reducing the effective Electric Field between Capacitor plates. Understanding this theory is crucial in designing Capacitors, insulators, and electrical circuits where energy storage and insulation are required.

    Thus, the dielectric theory began with Faraday and was developed by Maxwell.

    Option c – Maxwell theory

    The Electric Field that is responsible for polarizing a Molecule of the dielectric may be polar or non-polar is called the

    a. magnetic field

    b. Molecular field

    c. potential field

    d. gravitational field

    Explanation:
    The Molecular field is the effective Electric Field acting on molecules in a dielectric. It can polarize both polar and non-polar molecules by shifting charges or inducing dipoles.

    This field explains why dielectrics increase capacitance and how polarization occurs at a microscopic level. The Molecular field concept is foundational in understanding dielectric behavior and electric insulation.

    Thus, the polarizing field in a dielectric is called the Molecular field.

    Option b – Molecular field

    Which statement is not associated with nuclear fission?

    (a) Neutron-induced nuclear reaction.

    (b) Energy generation in stars.

    (c) The source of energy in a nuclear reactor that produces Electricity.

    (d) Enormous energy is released in an Atom bomb.

    Explanation:
    Nuclear fission involves splitting a heavy nucleus into smaller fragments, releasing energy. It is used in nuclear reactors and atomic bombs.

    However, energy generation in stars primarily occurs through nuclear fusion, where lighter nuclei combine. Understanding the distinction between fission and fusion is critical for energy applications, safety, and astrophysics.

    Thus, energy generation in stars is unrelated to nuclear fission.

    Option b – Energy generation in stars

    The basic scientific principle behind a nuclear reactor is

    (a) nuclear fusion.

    (b) controlled nuclear fusion.

    (c) uncontrolled nuclear fission.

    (d) controlled nuclear fission.

    Explanation:
    A nuclear reactor operates on controlled nuclear fission, where a chain reaction splits heavy nuclei (like uranium-235) while maintaining a steady energy release.

    Control rods absorb excess neutrons to regulate the reaction. This principle allows continuous Electricity generation in reactors safely, unlike uncontrolled fission in an atomic bomb.

    Thus, controlled nuclear fission is the core principle of nuclear reactors.

    Option d – controlled nuclear fission

    Heavy water is used as a moderator in a nuclear reactor. The function of the moderator is

    (a) to control the energy released in the reactor.

    (b) to absorb neutrons and stop the chain reaction.

    (c) to cool the reactor.

    (d) to slow down the neutrons to thermal energy

    Explanation:
    Moderators slow down fast neutrons produced in fission to thermal energies, increasing the probability of further fission in fuel nuclei. Heavy water (D₂O) is effective because it slows neutrons without capturing them.

    By controlling neutron speed, reactors maintain a sustained chain reaction efficiently. Understanding moderator choice is crucial for reactor design and safety.

    Thus, the moderator slows neutrons to enhance fission efficiency.

    Option d – to slow down the neutrons to thermal energy

    Heavy water is manufactured in India at

    (a) Trombay

    (b) Assam

    (c) Delhi

    (d) Bhilai

    Explanation:
    Heavy water (D₂O) production in India occurs at Trombay using electrolysis and chemical exchange processes.

    It is critical for nuclear reactors, serving as a neutron moderator. Knowledge of production sites ensures awareness of nuclear infrastructure and industrial capabilities.

    Thus, Trombay is the heavy water production facility in India.

    Option a – Trombay

    The mass-energy relation is the outcome of

    (a) quantum theory.

    (b) general theory of relativity.

    (c) field theory of relativity.

    (d) special theory of relativity.

    Explanation:
    Einstein’s special theory of relativity introduced the mass-energy equivalence: E = mc². It shows mass can be converted into energy and vice versa.

    This principle underlies nuclear reactions, particle Physics, and energy generation, linking mass loss in fission or fusion to released energy. It fundamentally changed classical Physics and our understanding of energy.

    Thus, mass-energy equivalence arises from special relativity.

    Option d – special theory of relativity

    The binding energy of the nucleus is a measure of its

    (a) charge

    (b) mass

    (c) momentum

    (d) stability

    Explanation:
    Binding energy is the energy required to disassemble a nucleus into its constituent protons and neutrons. It reflects the stability of the nucleus.

    Higher binding energy per nucleon indicates a more stable nucleus. Understanding this concept is key in nuclear reactions, including fission and fusion, as energy release depends on changes in nuclear binding energy.

    Thus, nuclear binding energy quantifies stability.

    Option d – stability

    A motor and engine are rotating about its axis with an angular velocity of 100 rad/min. It comes to rest in 15 s after beings switched off. Assuming constant retardation, calculate the number of revolutions made by it before coming to rest :

    (A) 10.5 rev

    (B) 5.5 rev

    (C) 24.5 rev

    (D) 12.5 rev

    Explanation:
    Angular motion under constant retardation can be analyzed using kinematic equations: θ = ω₀t − ½αt², where ω₀ is initial angular velocity and α is angular deceleration.

    Converting rad/min to rad/s allows calculating the total angular displacement in radians, then converting to revolutions by dividing by 2π. This method is analogous to linear deceleration in straight-line motion.

    Thus, the total revolutions are determined using rotational kinematics.

    Option d – 12.5 rev

    Starting from rest a fan rotates and takes 5 s to attain the maximum speed of 400 rpm. Assuming constant acceleration, find the time taken by the fan in attaining half the maximum speed :

    (A) 2.5 s

    (B) 1.5 s

    (C) 4.5 s

    (D) 5.2 s

    Explanation:
    For uniform angular acceleration, the angular velocity ω = αt. Half of maximum speed is reached when ω = ωmax/2, giving t = T/2.

    This parallels linear motion where distance covered under constant acceleration depends on initial velocity and time. Converting rpm to rad/s ensures correct calculation in standard units.

    Thus, time to reach half the speed is half the total acceleration time.

    Option a – 2.5 s

    A wheel of radius 10 cm can rotate freely about its center. A string is wrapped over its rim and is pulled by a force of 5 N it is found that the torque produces an angular acceleration of 2 rad/s² in the wheel. Calculate the moment of inertia of the wheel :

    (A) 1.25 kg m²

    (B) 0.5 kg m²

    (C) 0.25 kg m²

    (D) 0.1 kg m²

    Explanation:
    The torque τ applied to a wheel produces angular acceleration α according to τ = Iα, where I is the moment of inertia.

    Here, τ = F × r, with F being the applied force and r the wheel radius. By substituting τ and α, the moment of inertia can be calculated. This is analogous to Newton’s second law (F = ma) in rotational motion.

    Thus, the moment of inertia quantifies the wheel’s resistance to angular acceleration.

    Option c – 0.25 kg m²

    Three particles of each of mass m are situated at the vertices of an equilateral triangle ABC of side L. Find the MI of the system about line AX perpendicular to AB in plane ABC :

    (A) ML²/5

    (B) 2 ML²/5

    (C) 5 ML²/4

    (D) 4 ML²/5

    Explanation:
    Moment of inertia depends on mass distribution relative to the axis of rotation. Using the perpendicular axis theorem and symmetry, the contribution of each particle to the total I is calculated.

    For particles on a triangle, distances from the axis are used to sum I = Σ m r². This method generalizes to composite systems and rigid body dynamics.

    Thus, the total moment of inertia is the sum of each particle’s m r² relative to the axis.

    Option c – 5 ML²/4

    We covered all the ICSE Physics Class 10 Goyal Brothers solutions mcqs above in this post for free so that you can practice well for the exam.

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