If the work done on the system or by the system is zero, which of the following statements for a system kept at a certain temperature is correct?
(a) Change in internal energy of the system is equal to the flow of Heat in or out of the system.
(b) Change in internal energy of the system is less than Heat transferred.
(c) Change in the internal energy of the system is more than the Heat flow.
(d) Cannot be determined.
Explanation: When a thermodynamic system undergoes a process where no mechanical work is exchanged, the energy changes inside the system depend mainly on Heat transfer. A system maintained at constant temperature involves careful balance between Heatenergy entering and leaving, along with internal Molecular activity such as vibration, translation, and interaction among particles. The internal energy of a system is a state function, meaning it depends only on the current state and not on the path taken.
In such situations, the first law of Thermodynamics becomes the central idea, linking Heat transfer and internal energy change. If mechanical work is absent, the energy exchange simplifies, and Heat becomes the only contributor influencing internal energy. The system may absorb or release Heat depending on whether it is being heated or cooled externally, but the temperature condition ensures equilibrium behavior at the macroscopic level.
The reasoning involves understanding that internal energy reflects microscopic energy changes, while Heat is energy in transit due to temperature difference. With no work involved, the relationship between these two quantities becomes direct and straightforward, governed entirely by energy conservation principles applied to thermal processes.
Overall, this scenario highlights how energy conservation operates in purely thermal interactions without mechanical involvement.
Option a – Change in internal energy of the system is equal to the flow of Heat in or out of the system
Blowing air with an open pipe is an example of ( Physics Project Class 12mcqs )
(a) isothermal process
(b) isochoric process
(c) isobaric process
(d) adiabatic process
Explanation: When air is blown through an open pipe, the gas inside does not remain isolated; instead, it continuously exchanges Matter and energy with the surroundings. The motion of air involves pressure differences that keep air flowing, and the system does not remain in a fixed enclosed state. As a result, properties like pressure and volume are not constant, and the system does not stay in equilibrium long enough for simple thermodynamic constraints like constant volume or constant pressure to fully describe it.
To analyze such a situation, one must consider how gas behaves under continuous flow conditions, where expansion and compression happen rapidly. The process also involves temperature changes due to energy transfer associated with motion and pressure variations. Since the system is not insulated, heat exchange with surroundings may also occur depending on conditions.
The classification depends on whether heat exchange is significant or suppressed. In rapid airflow situations, thermal exchange is minimal compared to mechanical energy changes, making the process closer to one where heat transfer is negligible. This leads to a model where internal energy changes are mainly due to work done during expansion and compression rather than heat flow.
Thus, the behavior of air in an open pipe reflects a fast-moving thermodynamic process dominated by mechanical effects rather than slow thermal equilibrium adjustments.
Option c – isobaric process
A cycle tire bursts suddenly. This represents an
(a) isothermal process
(b) adiabatic process
(c) isochoric process
(d) isobaric process
Explanation: A sudden burst of a cycle tyre occurs in a very short time interval, leaving almost no time for heat exchange between the air inside the tyre and the surrounding Environment. In Thermodynamics, processes are classified based on how energy is transferred and how quickly equilibrium is disturbed or restored. When a change happens extremely rapidly, the system behaves in a way where thermal interaction with surroundings becomes negligible.
During such a rapid expansion, the gas inside the tyre expands quickly due to high internal pressure. Because the process is so fast, the molecules do not have sufficient time to exchange heat with the surroundings. Instead, the internal energy of the gas is primarily used in doing work on the surrounding air and the tyre wall as it ruptures. This leads to a sharp drop in pressure and temperature inside the tyre almost instantaneously.
The key idea is that the defining feature is not the initial condition but the speed of the process. Fast, uncontrolled expansions or compressions typically fall under a category where heat exchange is assumed to be zero for simplification in Physics models. This helps in analyzing energy changes using only internal energy and work terms.
Overall, such sudden expansions are treated as highly rapid thermodynamic changes dominated by mechanical effects rather than thermal equilibrium adjustments.
Option b – adiabatic process
Who codified the first two laws of Thermodynamics and deduced that the absolute zero of temperature is -273.15°C, leading to the naming of the Kelvin temperature scale?
(a) William Crookes
(b) William Thomson
(c) Luis Alvarez
(d) Robert Hooke
Explanation: The development of Thermodynamics involved contributions from several scientists who helped formalize the relationship between heat, work, and energy. The person associated with establishing a clear theoretical framework for thermodynamic laws also worked extensively on temperature scales and the concept of absolute zero. This concept represents the theoretical limit where Molecularmotion would cease, forming the foundation of absolute temperature measurement.
In classical Physics, temperature scales were initially based on arbitrary reference points such as the freezing and boiling points of water. However, deeper studies into gas behavior and energy relationships led to the idea that there must be a lowest possible temperature. This limit was identified through extrapolation of gas laws and thermodynamic principles.
The scientist responsible for advancing these ideas also contributed to understanding energy conservation and the second law, which deals with the direction of heat flow. His work helped unify experimental observations with mathematical formulations, leading to the modern structure of Thermodynamics.
The Kelvin scale was named to honor these contributions, representing a temperature scale starting from absolute zero, which is fundamental in Physics and engineering calculations.
Overall, this reflects the historical development of Thermodynamics and the recognition of absolute temperature as a fundamental physical concept.
Option b – William Thomson
The statement that ‘heat cannot flow by itself from a body at a lower temperature to a body at a higher temperature’ is known as ( Physics Project Class 12mcqs )
Explanation: Thermodynamics describes how energy moves between systems, especially in the form of heat. One of its fundamental ideas is that heat transfer has a preferred direction in natural processes. When two bodies at different temperatures interact, energy naturally flows from the region of higher thermal energy to the region of lower thermal energy.
This directionality is not arbitrary but arises from statistical behavior of large numbers of particles. In a hotter system, particles have higher average kinetic energy, and when they interact with a cooler system, energy redistribution occurs until equilibrium is reached. The reverse process does not happen spontaneously because it would require an external input of energy to organize Molecularmotion in a highly ordered way.
This principle is one of the foundational laws that define irreversibility in nature. It explains why certain processes, like heat engines, require continuous energy input to maintain operation. It also introduces the idea of entropy, which measures disorder and the tendency of systems to evolve toward equilibrium.
Overall, this law establishes the natural direction of heat flow and forms the basis for understanding efficiency limits in thermodynamic systems.
Option c – Second Law of Thermodynamics
The greenhouse effect is the heating up of the Earth’s Atmosphere due to
(a) the ultraviolet rays
(b) gamma rays
(c) the infrared rays
(d) X-rays
Explanation: The Earth receives energy from the Sun in the form of electromagnetic radiation. When this energy reaches the Earth’s surface, part of it is absorbed and part is reflected back into the Atmosphere. The absorbed energy warms the surface, which then re-emits energy in a different form of radiation associated with its temperature.
Different types of electromagnetic radiation interact differently with atmospheric gases. Some wavelengths pass through the Atmosphere easily, while others are absorbed and trapped by certain gases. This trapping effect leads to warming of the lower Atmosphere because energy is retained instead of escaping directly into space.
The key mechanism involves radiation emitted by the Earth after absorbing Solar energy. This outgoing energy is in a lower-energy form compared to incoming Solar radiation. Certain atmospheric gases absorb this outgoing energy and re-radiate it, causing heat to remain within the atmospheric system.
This continuous absorption and re-emission cycle leads to a rise in average atmospheric temperature. The process is essential for maintaining Earth’s Climate but becomes problematic when intensified by increased concentrations of these gases.
Overall, the phenomenon is driven by the interaction of Earth’s emitted radiation with atmospheric components that trap heat energy.
Option c – the infrared rays
Zeroth Law of Thermodynamics leads to
(a) the concept of temperature.
(b) the concept of specific heat.
(c) the concept of internal energy.
(d) None of the above
Explanation: Thermodynamics introduces several fundamental laws that define how thermal systems behave. One of the most basic principles establishes the idea of thermal equilibrium between systems. When two objects are in thermal contact and no NET heat flows between them, they are said to be in equilibrium with respect to temperature.
If a third object is in thermal equilibrium with each of the first two separately, then all three share a common thermal property. This transitive relationship forms the foundation for defining a measurable physical quantity that can be compared across different systems.
This principle allows temperature to be treated as a fundamental property that determines the direction of heat flow. It also makes it possible to construct thermometers, since temperature becomes a measurable quantity that is consistent across different systems in equilibrium.
Without this law, there would be no logical basis for comparing thermal states of different bodies or defining a universal temperature scale. It essentially provides the conceptual foundation for temperature measurement.
Overall, it establishes the basis for understanding and defining temperature as a fundamental thermodynamic property.
Option a – the concept of temperature
The first law of thermodynamics is simply the case of ( Physics Project Class 12mcqs )
(a) Charles’s Law
(b) Newton’s Law of Cooling
(c) The Law of Heat Exchange
(d) The Law of Conservation of Energy
Explanation: Energy in physical systems can neither be created nor destroyed; it can only change form. This principle is one of the most fundamental ideas in Physics and applies across mechanical, thermal, chemical, and electrical systems. In thermodynamics, this idea is expressed in terms of heat, work, and internal energy.
When energy is added to a system in the form of heat, part of it may increase the internal energy of the system, while part may be used to perform work on the surroundings. This balance ensures that total energy remains conserved during any process.
The first law of thermodynamics mathematically expresses this conservation principle by relating changes in internal energy to heat supplied and work done. It does not allow energy to appear or disappear but only to transfer between different forms.
This law is widely applicable in engines, refrigerators, and natural processes, where energy transformations are constantly occurring. It provides the foundation for analyzing all thermodynamic systems quantitatively.
Overall, it is a direct application of the universal conservation principle applied to thermal systems.
Option d – The Law of Conservation of Energy
A weightless string can bear tension up to 3.7 kgwt. A stone of Mass 0.5 kg is tied to it and rotated in a vertical circular path of radius 4 m then the maximum angular velocity of the stone :
(A) 2 rad/s
(B) 2.7 rad/s
(C) 3 rad/s
(D) 4 rad/s
Explanation: When an object moves in a vertical circular path, the tension in the string varies depending on its position. The maximum tension typically occurs at the lowest point of the motion because both weight and centripetal force act in the same direction, increasing the load on the string.
The string can withstand only a limited maximum tension before breaking. To ensure safe motion, the centripetal force required for circular motion must not exceed this limit. The centripetal force depends on Mass, angular velocity, and radius of the circular path.
As angular speed increases, the required centripetal force increases significantly because it depends on the square of angular velocity. Therefore, even a small increase in speed can greatly increase tension in the string.
At the limiting condition, the maximum allowable tension equals the sum of centripetal requirement and gravitational contribution at the lowest point. This condition determines the upper limit of angular velocity that the system can sustain safely.
Overall, the problem involves balancing Rotational Dynamics with material strength constraints of the string.
Option d – 4 rad/s
A motorcycle is going on over the bridge of radius R. The driver maintains a constant speed. As the motorcycle is ascending on the bridge, The normal force on it ( Physics Project Class 12mcqs )
(A) Increases
(B) Remains constant
(C) Decrease
(D) fluctuates
Explanation: When a vehicle moves over a curved surface like a bridge, it experiences circular motion along the vertical direction. This motion requires a centripetal force directed toward the center of curvature of the bridge. The forces acting on the motorcycle include its weight and the normal reaction from the bridge surface.
As the motorcycle ascends the curved bridge, the direction of required centripetal acceleration changes relative to gravity. At higher points on the curve, a greater portion of gravitational force contributes toward providing centripetal acceleration. This affects the contact force between the motorcycle and the surface.
The normal force is the force exerted by the surface perpendicular to contact. It adjusts depending on how much of the required centripetal force is already provided by gravity. If gravity contributes more toward centripetal motion, the surface needs to exert less normal force.
Thus, as the motorcycle climbs the bridge, the normal reaction decreases because the effective requirement for surface support reduces. This continues until the top of the bridge, where the normal force is minimum.
Overall, the change in normal force is governed by the balance between gravitational force and centripetal acceleration requirements in curved motion.
Option c – Decrease
A frictionless track ends in a circular loop of radius small body slides down the track from a 3 cm. A height h then the minimum value of ‘h’ point ‘p’ at for a body to complete a circular loop :
(A) 3 cm
(B) 7.5 cm
(C) 12.5 cm
(D) 14 cm
Explanation: When a body moves on a frictionless track and enters a vertical circular loop, it must maintain sufficient speed at the top of the loop to stay in contact with the track. The critical condition occurs at the highest point of the loop, where the normal reaction can become zero, and gravity alone provides the required centripetal force for circular motion.
Energy conservation plays a key role here. The body starts from height h, converting its gravitational potential energy into kinetic energy as it descends. At the top of the loop, part of this energy is still potential, and the rest is kinetic. The condition for just maintaining contact is that the centripetal force requirement equals the gravitational force at the top point.
This ensures that the body does not lose contact with the track. If the speed is lower than this critical value, the body will fall off before completing the loop. Therefore, the minimum height must be sufficient to provide both the energy needed to reach the top and the required kinetic energy at that point.
The analysis combines conservation of mechanical energy with circular motion conditions at the limiting case. This approach determines the threshold height needed for successful looping motion.
Option b – 7.5 cm
A vertical section of the flyover bridge is in the form of an arc of radius 19.5 m, the truck crosses the bridge without losing contact, and the center of gravity of the truck is 0.5 m above the surface. The maximum speed upto which the truck can be driven is : ( Physics Project Class 12mcqs )
(A) 7 m/s
(B) 14 m/s
(C) 21 m/s
(D) 28 m/s
Explanation: When a vehicle moves over a curved bridge, it undergoes circular motion along a vertical arc. The forces acting on it include gravitational force downward and normal reaction from the bridge surface upward. For safe motion, the vehicle must remain in contact with the bridge at all times, meaning the normal force must remain non-negative.
The critical situation occurs at the highest point of the arc, where the required centripetal force is maximum relative to the support available. The effective radius of motion is determined by the position of the center of gravity above the surface, which slightly reduces the curvature radius for the motion of the vehicle’s center of Mass.
At the limiting speed, the normal force becomes zero, and gravity alone provides the centripetal force required for circular motion. This condition allows calculation of the maximum speed the truck can maintain without losing contact with the bridge surface.
The solution uses the balance between gravitational force and centripetal force, adjusted for the effective radius of motion of the center of Mass.
Overall, it is a direct application of vertical circular motion with limiting contact condition.
Option b – 14 m/s
A stone attached to a rope of length 1 = 80 cm is rotated at a speed of 240 rpm. At the moment when the velocity is directed upward the rope breaks, to what height does the stone rise further?
(A) 10.3 m
(B) 41.2 m
(C) 20.6 m
(D) 24.9 m
Explanation: When an object is rotating in a vertical circle and the string breaks at a particular instant, the object continues to move in the direction of its instantaneous velocity due to inertia. At the moment of breaking, the stone has both tangential speed and a specific upward direction component depending on its position in the circular motion.
After the string breaks, the stone behaves like a projectile under the influence of gravity alone. Its upward component of velocity determines how high it will rise before coming to rest momentarily at the highest point of its trajectory. The conversion of kinetic energy into potential energy governs this rise in height.
The key idea is that rotational motion suddenly transitions into free motion under gravity. The angular speed is first converted into linear speed using the radius of rotation. Then only the vertical component of velocity contributes to further upward motion.
The final height depends entirely on the initial upward velocity component at the instant of release and gravitational deceleration acting on the stone.
Overall, this is a combination of circular motion and vertical projectile motion under gravity.
Option c – 20.6 m
A 4 kg ball swings in a vertical circle at the end of the chord 1 m long. What maximum speed at which it can swing if the chord can sustain maximum tension of 163.6 N? ( Physics Project Class 12mcqs )
(A) 5.57 m/s
(B) 4 m/s
(C) 10.2 m/s
(D) 31.1 m/s
Explanation: In vertical circular motion, tension in the string varies at different points of the path. The maximum tension occurs at the lowest point of the motion because both gravitational force and centripetal force act in the same direction toward the center of the circle.
The centripetal force required for circular motion depends on Mass, speed, and radius. As speed increases, the required centripetal force increases significantly, leading to higher tension in the chord. The maximum allowable speed is reached when the tension equals the sum of gravitational force and centripetal force requirement at the lowest point.
The analysis involves identifying the most critical point in the motion where stress on the string is highest. Using this condition ensures that the system remains intact throughout the circular path.
The balance between gravitational force and centripetal force determines the maximum safe velocity the system can sustain.
Overall, this is a classical application of vertical circular motion and tension constraints in Rotational Dynamics.
Option a – 5.57 m/s
The string of a pendulum of Mass m and length 1 is displaced through 90°. The minimum strength of the string to withstand the tension will be :
(A) mg
(B) 2 mg
(C) 3 mg
(D) 4 mg
Explanation: A pendulum released from a displaced position undergoes conversion of potential energy into kinetic energy as it swings downward. The tension in the string is not constant and varies depending on the position of the bob in its motion.
The maximum tension typically occurs at the lowest point of the swing, where the speed is highest. At this point, both the weight of the bob and the centripetal force requirement contribute to the tension in the string. The energy gained from the initial displacement determines the speed at the lowest point.
The analysis involves understanding how gravitational potential energy converts into kinetic energy during motion. The string must be strong enough to withstand the peak tension experienced during the swing.
At the extreme position, the bob has maximum potential energy and zero kinetic energy, while at the lowest point, kinetic energy is maximum. The combination of these effects determines the maximum force experienced by the string.
Overall, the problem combines energy conservation with circular motion dynamics to determine structural requirements of the pendulum system.
Option c – 3 mg
A small body of Mass 0.1 kg swings in a vertical circle, at the end of the chord of length 1 m. If the speed is 2 m/s when the chord makes an angle of 30° with vertical. Find the tension in the chord : (g=9.8 m/s²) ( Physics Project Class 12mcqs )
(A) 0.4 N
(B) 0.85 N
(C) 0.98 N
(D) 1.25 N
Explanation: In vertical circular motion, the tension in the string varies continuously depending on the position of the object along the circular path. At any point, the forces acting on the body include its weight acting downward and the tension acting along the string toward the center of the circle.
The required centripetal force is provided by the NET inward force, which is the difference between tension and the radial component of weight. As the angle changes, the component of gravitational force along the radial direction also changes, affecting the tension.
To analyze such motion, one must resolve forces along the radial direction and apply Newton’s second law for circular motion. The speed at that point also influences the centripetal force requirement.
This type of problem combines Vector resolution of forces with Rotational Dynamics principles. The balance between inward force requirement and gravitational component determines the tension in the string.
Overall, it is a direct application of force decomposition in non-uniform circular motion.
Option d – 1.25 N
A particle of Mass 1 kg is moving in the xy plane parallel to the y-axis, with a uniform speed of 5 m/s, 2 m away from the origin. Angular momentum of particle about origin :
(A) 10kg m² s
(B) 5kg m²/s
(C) 2.5kg m²/s
(D) 2kgm²/s
Explanation: Angular momentum measures the rotational effect of a moving particle about a reference point. It depends on both the linear momentum of the particle and its perpendicular distance from the chosen origin. When a particle moves in a straight line but does not pass through the origin, it still possesses angular momentum due to its offset path.
The magnitude of angular momentum is given by the product of Mass, velocity, and perpendicular distance from the origin to the line of motion. This distance acts as the lever arm, determining how strongly the motion contributes to rotational effect about the origin.
In this situation, the particle moves parallel to one axis at a constant distance from the origin, so its angular momentum remains constant in magnitude as long as speed and distance remain unchanged. The direction is perpendicular to the plane of motion, determined by the right-hand rule.
This concept is important in understanding how linear motion can still produce rotational effects depending on the reference point.
Overall, angular momentum here arises from linear motion offset from the origin.
Option a – 10kg m² s
A bob of Mass m attached to an inextensible string of length I is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed of rad/s about the vertical. About the point of suspension : ( Physics Project Class 12 MCQs )
(A) angular momentum changes in direction but not in magnitude
(B) angular momentum changes both in direction and magnitude
(C) angular momentum is conserved
(D) angular momentum changes in magnitude but not in direction
Explanation: When a bob rotates in a horizontal circular path, it experiences uniform circular motion around a vertical axis passing through the point of suspension. The angular momentum of the system depends on both the motion of the bob and the reference point chosen for measurement.
Since the motion is circular and steady, the magnitude of angular momentum remains constant if speed and radius remain constant. However, its direction continuously changes because angular momentum is a Vector quantity pointing perpendicular to the plane of motion.
In such motion, the forces acting are tension in the string and gravity, but the torque about the point of suspension is effectively zero for uniform motion. This leads to conservation of angular momentum in magnitude.
The changing direction arises because the position Vector of the bob keeps rotating, altering the orientation of angular momentum even though its size remains unchanged.
Overall, this represents a system where angular momentum is conserved in magnitude but continuously changes direction due to circular motion geometry.
Option a – angular momentum changes in direction but not in magnitude
If the radius of the Earth is suddenly expanded by ‘n’ times its present value without change in its mass then the length of the day is :
(A) 24 hours
(B) 24/n² hours
(C) n² 24 hours
(D) n 24 hours
Explanation: The rotation of Earth is governed by conservation of angular momentum when no external torque acts on the system. Angular momentum depends on moment of inertia and angular velocity. If the mass remains constant but the radius changes, the distribution of mass relative to the axis of rotation changes significantly.
Moment of inertia for a spherical body depends on the square of its radius. So, when the radius increases, the moment of inertia increases accordingly. To conserve angular momentum, the angular velocity must adjust inversely.
As angular velocity decreases, the time taken for one complete rotation increases. This directly affects the length of the day, which is defined by one full rotation of Earth about its axis.
Thus, an increase in radius leads to a slower rotation and a longer day. The relationship between radius and rotational period follows from conservation principles in Rotational Dynamics.
Overall, this is a direct consequence of angular momentum conservation in a rotating body undergoing change in size.
Option b – 24/n² hours
A wheel is rotating with an angular frequency of 500 revolutions per minute on a shaft. The second wheel whose mass and radius are half of the first is coupled with the first wheel then the angular speed of rotation becomes : ( Physics Project Class 12 MCQs )
(A) 500 rpm
(B) 450 rpm
(C) 250 rpm
(D) 125 rpm
Explanation: When two rotating bodies are coupled, they tend to reach a common angular velocity due to internal friction or mechanical linkage. In such systems, angular momentum is conserved if no external torque acts on the system.
The total moment of inertia of the system depends on mass distribution and radius of rotation. When a second wheel with different mass and radius is added, it changes the overall moment of inertia, affecting the final angular velocity after coupling.
Since angular momentum is conserved, an increase in moment of inertia leads to a decrease in angular velocity, and vice versa. The final angular speed depends on the balance between initial rotational energy distribution and combined inertia of both wheels.
The interaction continues until both wheels rotate together with a common angular speed.
Overall, this problem illustrates conservation of angular momentum in coupled rotational systems.
Option b – 450 rpm
A particle of mass m is rotating in a plane in a circular path of radius Y’. Its angular momentum is L. The centripetal force acting on the particle is :
(A) L²/mr³
(B) L²m/r
(C) L²/mr²
(D) L²m/r²
Explanation: In uniform circular motion, a particle moves along a circular path while continuously changing its direction of velocity. This change in direction requires a centripetal force directed toward the center of the circle. Angular momentum plays a key role in describing rotational motion and is related to both linear momentum and the perpendicular distance from the axis of rotation.
Angular momentum depends on the product of mass, velocity, and radius of the circular path. When expressed in terms of angular quantities, velocity can be related to angular speed, allowing angular momentum to represent the rotational inertia of the moving particle. The centripetal force is responsible for maintaining circular motion and depends on the square of velocity and inversely on radius.
By connecting angular momentum with linear velocity, the centripetal force can be expressed in terms of rotational quantities. This shows how rotational and translational descriptions of motion are interconnected.
The key idea is that angular momentum provides an alternative way of describing circular motion, while centripetal force ensures the continuous inward acceleration required to maintain that motion.
Overall, this highlights the relationship between Rotational Dynamics and the force required to sustain circular motion.
Option a – L²/mr³
If the ice of polar caps melts the duration of the day will : ( Physics Project Class 12 MCQs )
(A) Increase
(B) 20 gm cm²
(C) Remains the same
(D) cannot be predicted
Explanation: The duration of a day depends on the Earth’s angular velocity of rotation. Any redistribution of mass on Earth can affect its moment of inertia, which in turn influences rotational speed due to conservation of angular momentum.
When polar ice melts, water redistributes from regions near the poles toward lower latitudes. This increases the average distance of mass from the Earth’s rotation axis, thereby increasing the Earth’s moment of inertia.
Since angular momentum remains conserved in the absence of external torque, an increase in moment of inertia leads to a decrease in angular velocity. A slower angular velocity means Earth takes more time to complete one full rotation.
Thus, changes in mass distribution can directly influence the length of a day through Rotational Dynamics principles.
Overall, this is an application of angular momentum conservation in a real-world geophysical system.
Option a – Increase
A particle performs UCM with an angular momentum of L. If the frequency of the particle in motion is doubled and Its KE is halved, the angular momentum becomes :
(A) L/4
(B) L/2
(C) L
(D) 4L
Explanation: In uniform circular motion, angular momentum depends on mass, radius, and angular velocity. Frequency is directly related to angular velocity, meaning changes in frequency affect rotational motion significantly. Kinetic energy in circular motion depends on the square of velocity, which is also linked to angular velocity.
When frequency increases, angular velocity increases proportionally. However, if kinetic energy simultaneously decreases, it implies a change in either radius or effective velocity contributing to motion. These two conditions together affect the system’s rotational characteristics.
Angular momentum is directly proportional to both moment of inertia and angular velocity. Any change in these quantities alters angular momentum. Since frequency and kinetic energy change in different ways, their combined effect determines the final angular momentum.
The key idea is that angular momentum is sensitive to both rotational speed and mass distribution, and cannot be determined by a single parameter alone.
Overall, this problem focuses on how multiple simultaneous changes in rotational parameters influence angular momentum.
Option a – L/4
A body starts from rest and completes 20 revolutions in 5 minutes then its number of revolutions in the next 5 minutes : ( Physics Project Class 12 MCQs )
(A) 20 revolutions
(B) 40 revolutions
(C) 60 revolutions
(D) 80 revolutions
Explanation: When a body starts from rest and undergoes rotational motion with uniform angular acceleration, its angular displacement increases with the square of time. This means the number of revolutions is not directly proportional to time but follows a quadratic relationship.
In uniformly accelerated rotational motion, the angular displacement is proportional to the square of time elapsed. Therefore, if the number of revolutions in the first interval is known, the total displacement after a longer time can be determined using this time-squared relationship.
The motion starts from rest, so initial angular velocity is zero. As time progresses, angular velocity increases steadily due to constant angular acceleration. This leads to a non-linear increase in revolutions over equal time intervals.
By comparing successive time intervals, the additional revolutions in the next interval can be determined using proportional relationships derived from the time-squared dependence.
Overall, this problem highlights the nature of uniformly accelerated rotational motion starting from rest.
Option c – 60 revolutions
A flywheel of MI 15 kgm² slows down from 90 rad/ s to 40 rad/s in 20 seconds by a constant torque than the number of revolutions during this time interval :
(A) 100 rev
(B) 207 rev
(C) 215 rev
(D) 1300 rev
Explanation: A flywheel undergoing uniform angular deceleration experiences a steady decrease in angular velocity over time. The angular displacement during such motion can be determined using average angular velocity multiplied by time, similar to linear kinematics.
Since the angular velocity decreases uniformly, the average angular velocity is taken as the mean of initial and final angular velocities. This average value represents the effective rotational speed over the given time interval.
The total angular displacement is then obtained by multiplying this average angular velocity with the time duration. This gives the total rotation in radians, which can be converted into revolutions.
The moment of inertia is not directly needed for displacement calculation but is relevant for torque and energy considerations. The key idea is the kinematic behavior of rotational motion under constant angular acceleration.
Overall, this problem applies rotational kinematics to determine angular displacement during uniform deceleration.
Option b – 207 rev
A 0.5 kW motor acts for 10 seconds on an initially non-rotating wheel with a moment of inertia of 1 kg m². What is the angular velocity developed in the wheel neglecting friction? ( Physics Project Class 12 MCQs )
(A) 50 rad/s
(B) 120 rad/s
(C) 70 rad/s
(D) 160 rad/s
Explanation: A motor supplying constant power transfers energy into a rotating system over time. power represents the rate at which work is done, so multiplying power by time gives the total energy supplied to the system.
This energy is converted into rotational kinetic energy of the wheel. Rotational kinetic energy depends on moment of inertia and square of angular velocity. As energy increases, angular velocity increases accordingly.
Since the wheel starts from rest, all supplied energy goes into increasing rotational motion. The relationship between energy and angular velocity allows determination of final speed using energy conservation principles.
The process assumes no energy losses due to friction or other resistive forces, meaning all input energy contributes to rotation.
Overall, this is an application of energy transfer into rotational motion using power and kinetic energy relations.
Option c – 70 rad/s
A wheel of the moment of inertia 50 kg m² is rotating at a uniform angular velocity of 5 rad/ s. The torque required to stop it in 2 s has a magnitude :
(A) 125 Nm
(B) 250 Nm
(C) 500 Nm
(D) 1000 Nm
Explanation: When a rotating object is brought to rest, it undergoes angular deceleration due to an applied torque. Torque is directly related to the rate of change of angular momentum, which depends on moment of inertia and angular acceleration.
To stop the wheel in a given time, a constant angular deceleration is assumed. This deceleration is determined by the change in angular velocity over the stopping time.
Once angular acceleration is known, torque can be calculated using the relationship between torque, moment of inertia, and angular acceleration. A larger moment of inertia or higher initial angular velocity requires a greater opposing torque to bring the system to rest.
The key idea is that torque is responsible for changing rotational motion, just as force changes linear motion.
Overall, this problem uses Rotational Dynamics to determine the braking torque required to stop a rotating system.
Option a – 125 Nm
A motor running at a rate of 1200 rpm can supply a torque of 80 Nm. What power does it develop?
(A) 1.6 π kW
(B) 3.2 π kW
(C) 0.8 π kW
(D) 4.8 π kW
Explanation: power in rotational systems is the rate at which work is done by a torque acting on a rotating object. It depends on both the magnitude of torque and the angular velocity of rotation.
Angular velocity can be obtained by converting revolutions per minute into radians per second. Once angular velocity is known, power is calculated as the product of torque and angular velocity.
A higher torque or higher rotational speed results in greater power output. This relationship is fundamental in understanding how engines and motors deliver energy to mechanical systems.
The concept shows how rotational motion translates into useful mechanical work in practical applications.
Overall, this problem connects torque, angular velocity, and power in rotational mechanics.
Option b – 3.2 π kW
Starting from rest a fan takes 5 seconds to attain the maximum speed of 400 rpm. Assuming constant acceleration, the time taken by the fan in attaining half the maximum speed is :
(A) 20 s
(B) 10 s
(C) 2.5 s
(D) 2.0 s
Explanation: When a rotating system starts from rest with constant angular acceleration, its angular velocity increases linearly with time. This means speed is directly proportional to time under uniform acceleration conditions.
If the system reaches a certain maximum speed in a known time, any intermediate speed can be related proportionally to the time taken to reach it. This linear relationship simplifies analysis of rotational motion under constant acceleration.
Half of the final speed corresponds to half of the time required to reach the full speed when starting from rest.
The key idea is the proportional relationship between angular velocity and time in uniformly accelerated rotational motion.
Overall, this problem demonstrates linear growth of angular velocity under constant acceleration.
Option c – 2.5 s
A motor of an engine is rotating about its axis with an angular velocity of 100 rev/min. It comes to rest in 15 seconds after being switched off. A number of revolutions made by the motor before coming to rest :
(A) 25 rev
(B) 25 π rev
(C) 12.5 rev
(D) 50 π rev
Explanation: When a rotating object slows down uniformly, its angular velocity decreases at a constant rate until it reaches zero. The total angular displacement during this deceleration can be found using the average angular velocity over the time interval.
Since the deceleration is uniform, the average angular velocity is the mean of initial and final angular velocities. Multiplying this average value by time gives total rotation in radians, which can be converted into revolutions.
The process assumes no sudden changes in torque, meaning the slowing down is smooth and steady.
Overall, this is a straightforward application of rotational kinematics under uniform deceleration.
Option c – 12.5 rev
The ratio of the accelerations for a Solid sphere (mass m and radius R) rolling down an incline of angle ‘0’ without slipping and slipping down the incline without rolling is :
(A) 5 : 7
(B) 2 : 5
(C) 2 : 3
(D) 7 : 5
Explanation: When a body moves down an inclined plane, its acceleration depends on how the motion occurs—pure rolling or slipping. In pure rolling motion, part of the gravitational energy is used in rotational motion as well as translational motion, so the effective linear acceleration is reduced compared to free sliding. In slipping motion, there is no rotational constraint, so the body accelerates purely due to gravity components along the incline.
For a Solid sphere, rotational inertia plays a significant role in reducing acceleration during rolling. The moment of inertia of the sphere causes energy to be shared between rotation and translation, making it move slower than a body that is simply sliding without rotation.
In contrast, when slipping occurs without rolling, no rotational energy is involved, and the full component of gravitational force along the incline contributes to linear acceleration. This makes slipping motion faster than rolling motion for the same object.
The comparison between these two accelerations highlights how rotational inertia influences motion on inclined planes and demonstrates the difference between constrained and unconstrained motion.
Overall, this is a comparison between Rotational Dynamics and pure translational motion on an incline.
Option a – 5 : 7
A body rolls down an inclined plane. If its kinetic energy of rotational motion is 40% of its kinetic energy of translational then the body is :
Explanation: In rolling motion, a body possesses both translational and rotational kinetic energy. The distribution between these two depends on the shape and moment of inertia of the object. The rotational kinetic energy is associated with spinning about its axis, while translational kinetic energy is associated with motion of its center of mass.
The ratio of rotational to translational kinetic energy is directly related to the moment of inertia factor of the body. Different shapes like Solid spheres, hollow spheres, rings, and discs have distinct distributions of mass, leading to different energy ratios during rolling.
When rotational kinetic energy is a specific fraction of translational kinetic energy, it indicates a characteristic value of the moment of inertia. This allows identification of the type of body based on standard rolling motion results.
The key idea is that energy distribution in rolling motion uniquely depends on geometry and mass distribution.
Overall, this problem uses energy partition in rolling motion to identify the nature of the rolling object.
The Solid cylinder is raised to a certain height on the inclined plane and then rolls down with a velocity of 7 m/s. What is the height to which the cylinder is raised?
(A) 1.2 m
(B) 4.9 m
(C) 3 m
(D) 3.75 m
Explanation: When a rigid body rolls without slipping, its gravitational potential energy is converted into both translational and rotational kinetic energy. Unlike pure sliding, part of the energy goes into rotation, so the final speed is lower for the same height.
At the starting point, the body has only potential energy due to its height. As it rolls down, this energy is conserved and redistributed into motion. At the bottom, the total energy is split between linear motion of the center of mass and rotational motion about its axis.
The moment of inertia of the cylinder determines how much energy is stored in rotation. A Solid cylinder has a fixed relationship between mass distribution and radius, which influences the final speed achieved for a given height.
Using energy conservation principles, the height can be determined by equating initial potential energy to total kinetic energy at the bottom.
Overall, this is a direct application of energy conservation in rolling motion.
Option d – 3.75 m
An inclined plane makes an angle of 30° with the horizontal. A Solid sphere rolling down this inclined plane from rest without slipping has linear acceleration equal to
(A) 3g
(B) 5g
(C) 5g/14
(D) 5g/7
Explanation: When a rigid body rolls down an inclined plane without slipping, its acceleration depends on both gravitational force and rotational inertia. Unlike a sliding object, part of the gravitational energy is used to produce rotational motion, reducing the NET linear acceleration.
The slope angle determines the component of gravitational force acting along the incline. The presence of rotational motion introduces an effective resistance to acceleration because energy is shared between translation and rotation.
Different shapes have different moments of inertia, which affect how much of the gravitational force contributes to linear motion. A Solid sphere has a specific inertia factor that leads to a characteristic acceleration value lower than that of a sliding object.
The acceleration remains constant along the incline if there is no slipping and friction is sufficient to maintain rolling.
Overall, this problem highlights the combined effect of gravity and rotational inertia in rolling motion.
Option c – 5g/14
A Solid sphere of mass 2 kg rolls up a 30° incline with an initial speed of 10 m/s. The maximum height reached by the sphere is : (g = 10 m/s²)
(A) 3.5 m
(B) 7 m
(C) 10.5 m
(D) 14 m
Explanation: When a rolling body moves upward on an incline, its kinetic energy gradually converts into gravitational potential energy. Since the body is rolling without slipping, both translational and rotational kinetic energies decrease simultaneously as it climbs.
At the highest point, the velocity becomes zero, meaning all kinetic energy has been converted into potential energy. The presence of rotational motion means that both forms of kinetic energy must be considered in the energy conversion process.
The moment of inertia of a solid sphere determines the ratio of rotational to translational kinetic energy, which affects how much total energy is available for gaining height.
Energy conservation is the key principle used to relate initial speed to final height reached against gravity.
Overall, this is a rolling motion energy conversion problem involving both translational and rotational energy.
Option b – 7 m
A solid sphere rolls down an inclined plane the percentage of total energy which is rotational K.E. is :
(A) 28%
(B) 72%
(C) 100%
(D) 75%
Explanation: In rolling motion without slipping, the total kinetic energy of a rigid body is divided into translational kinetic energy of its center of mass and rotational kinetic energy about its axis. The proportion between these two depends on the moment of inertia of the body.
For a solid sphere, the mass distribution is such that its rotational inertia is relatively small compared to other shapes like rings or hollow spheres. This results in a specific fixed ratio of rotational to total kinetic energy during pure rolling motion.
The distribution of energy remains constant for a given shape because both translational and rotational speeds are linked through the no-slip condition. This makes energy partition a characteristic property of the object.
Overall, the percentage of rotational energy is determined entirely by the geometry and mass distribution of the rolling body.
Option a – 28%
A solid sphere rolls down an inclined plane, and the percentage of total energy which is translational K.E. is :
(A) 28%
(B) 72%
(C) 100%
(D) 25%
Explanation: During pure rolling motion, the total kinetic energy is shared between translational motion of the center of mass and rotational motion about the axis. The no-slip condition ensures a fixed relationship between angular velocity and linear velocity.
For a solid sphere, the translational component forms a specific fraction of the total energy due to its particular moment of inertia. The relatively small rotational inertia means most of the energy remains in translational motion compared to rotation.
This fixed ratio arises because both types of motion are linked through geometry and constraints of rolling. As a result, the energy distribution remains constant regardless of speed.
Overall, translational kinetic energy dominates the total energy in a rolling solid sphere due to its mass distribution.
Option b – 72%
The speed of a solid sphere after rolling from a height of 14 m horizontally is (g = 9.8 m/s²) :
(A) 14 m/s
(B) 7 m/s
(C) 9.8 m/s
(D) 10 m/s
Explanation: When a rigid body rolls without slipping from a height, gravitational potential energy is converted into both translational and rotational kinetic energy. The final speed depends on how the energy is shared between these two forms.
A solid sphere has a specific moment of inertia that determines the fraction of energy used for rotation. Because of this, the final linear speed is less than that of a freely sliding object.
Energy conservation is applied by equating initial potential energy with total kinetic energy at the bottom of the incline. The rolling condition ensures a fixed relationship between angular and linear velocity, allowing the speed to be determined.
Overall, this is a direct application of energy conservation in rolling motion systems.
Option a – 14 m/s
A small spherical liquid drop is moving in a viscous medium. The viscous force does not depend on
(A) the nature of the medium
(B) the density of the medium
(C) the instantaneous speed of the spherical drop
(D) the radius of the spherical drop
Explanation: When an object moves through a viscous Fluid, it experiences a resistive force due to internal friction between Fluid layers. This force depends on several factors such as the viscosity of the Fluid, the size and shape of the object, and its velocity.
For a spherical object, the viscous drag is influenced by its radius, the viscosity coefficient of the medium, and the relative speed of motion. The nature of the Fluid also affects how strongly it resists motion.
However, certain properties of the object do not influence viscous force directly. Instead, only those parameters related to motion and Fluid interaction determine the magnitude of resistance.
The key idea is that viscous force is governed by Fluid dynamics and not by unrelated physical properties that do not affect flow behavior.
Overall, this highlights the dependence of viscous drag on motion-related and Fluid-related parameters only.
Option b – the density of the medium
If the temperature rises, the coefficient of viscosity of a liquid :
(A) decreases
(B) increases
(C) remains unchanged
(D) increases for some liquids and decreases for others
Explanation: Viscosity is a measure of a Fluid’s resistance to flow, arising from internal friction between its layers. In liquids, this resistance is mainly due to intermolecular forces that hold molecules together and oppose their relative motion.
As temperature increases, the kinetic energy of liquid molecules increases, allowing them to move more freely. This reduces the effectiveness of intermolecular attractions, making it easier for layers of liquid to slide past one another.
Because of this reduced intermolecular interaction, the internal resistance to flow decreases. Therefore, liquids generally become less viscous at higher temperatures.
This behavior is opposite to gases, where viscosity typically increases with temperature due to increased Molecular collisions.
Overall, temperature affects Molecular motion and directly influences the ease of flow in liquids.
Option a – decreases
A metal plate 48 cm² in area rests horizontally on a layer of oil 1 mm thick. A force of 0.25 N applied to the plate horizontally keeps it moving with a uniform speed of 2.5 cm/s. The coefficient of viscosity of the oil is :
(A) 1.083 Ns/m²
(B) 3.083 Ns/m²
(C) 2.083 Ns/m²
(D) 4.083 Ns/m²
Explanation: When a thin layer of liquid separates a moving surface from a fixed Base, the motion experiences viscous drag due to internal friction between adjacent Fluid layers. Each layer of the Fluid tends to move at a different speed, creating a velocity gradient across the thickness of the liquid film.
The resisting force depends on the area of contact, the velocity difference, the thickness of the fluid layer, and the coefficient of viscosity. For steady motion at constant speed, the applied force balances the viscous force exactly, meaning there is no NET acceleration.
The velocity gradient is assumed linear for a thin fluid layer, which allows a direct relationship between force and viscosity. Larger area or higher speed increases resistance, while greater thickness reduces it because layers are farther apart and interact less strongly.
This type of motion is commonly analyzed using basic fluid mechanics assumptions for laminar flow between parallel surfaces.
Overall, this problem applies the concept of viscous drag in a thin liquid film under steady motion conditions.
Option c – 2.083 Ns/m²
A plate of area 100 cm² is lying on the upper surface of a 3 mm thick oil film. If the coefficient of viscosity of the oil is 15.5 poise then the horizontal force required to move the plate with a velocity of 3 cm/s will be
(A) 0.155 N
(B) 15.5 N
(C) 1.55 N
(D) 155 N
Explanation: When a solid surface moves over a viscous liquid layer, the liquid resists motion due to internal friction between its layers. This resistance arises because different layers of the liquid move at different velocities, creating a velocity gradient across the thickness of the film.
The force required to maintain uniform motion depends on the viscosity of the liquid, the area of contact, the speed of motion, and the thickness of the fluid layer. A thicker layer reduces the gradient, while higher viscosity increases resistance significantly.
The motion is assumed to be steady and laminar, meaning the flow is smooth and orderly without turbulence. In such conditions, the velocity profile across the liquid film is taken as linear, simplifying the relationship between force and motion.
The applied force balances the viscous drag when the plate moves with constant speed, ensuring no acceleration occurs.
Overall, this is an application of viscous flow between parallel surfaces under steady-state conditions.
Option a – 0.155 N
Stoke’s law is applicable only to :
(A) púre liquids
(B) solutions
(C) non-viscous liquids
(D) viscous liquids
Explanation: When a small spherical object moves through a fluid, it experiences a resistive force due to viscosity. This force depends on the size of the object, the velocity of motion, and the viscosity of the surrounding fluid. However, the validity of this relationship depends on the nature of the flow around the object.
Stoke’s law applies when the motion occurs at very low speeds, ensuring smooth and orderly flow of fluid layers around the object. In such cases, inertial effects are negligible compared to viscous forces, and the flow remains laminar rather than turbulent.
The shape of the object also matters, as the law is derived specifically for spherical bodies where symmetry simplifies the flow pattern. Deviations from these conditions lead to more complex drag behavior that is not described by this law.
Overall, the law is valid under conditions where viscous forces dominate and fluid motion remains smooth and predictable around a spherical object.
Option d – viscous liquids
The SI unit of the coefficient of viscosity is :
(A) m/kg-s
(B) kg/m-s²
(C) m-s/kg²
(D) kg/m-s
Explanation: Viscosity is a physical property of fluids that measures their resistance to gradual deformation or flow. It arises due to internal friction between adjacent layers of fluid moving at different velocities. The coefficient of viscosity quantifies this resistance in terms of force required to maintain a velocity gradient in a fluid.
From the definition, viscosity relates shear stress to velocity gradient. Shear stress is force per unit area, and velocity gradient represents change in velocity per unit distance. Combining these physical quantities leads to a derived SI unit based on fundamental mechanical units.
The unit reflects how much force is needed to move a fluid layer relative to another layer over a given area and distance.
Overall, the SI unit is derived from force, length, and time considerations in fluid motion.
Option d – kg/m-s
The coefficient of viscosity of a liquid does not depend upon
(A) the density of the liquid
(B) the temperature of the liquid
(C) the pressure of the liquid
(D) the nature of liquid
Explanation: The coefficient of viscosity is a property that describes how strongly a liquid resists internal flow. It depends on the nature of intermolecular forces within the liquid, as well as external conditions that affect Molecular motion.
Temperature has a significant effect because it changes Molecular kinetic energy, altering the strength of interactions between layers. Pressure can also influence viscosity in some cases, especially at high values, by affecting Molecular spacing.
However, certain properties of the liquid, such as how much mass is contained in a given volume, do not directly determine viscosity. Instead, viscosity is more closely related to Molecular structure and interaction strength rather than bulk density alone.
The key idea is that viscosity reflects internal Molecular behavior rather than purely macroscopic mass distribution.
Overall, it depends on molecular interaction characteristics rather than all physical properties of the fluid.
Option a – the density of the liquid
1 centipoise is equal to :
(A) 0.001 kg/m-s
(B) 1 kg/m-s
(C) 0.1 kg/m-s
(D) 1000 kg/m-s
Explanation: Viscosity can be expressed in different unit systems depending on the scale and application. The poise is a unit used in the CGS system to measure viscosity, while the SI system uses a different derived unit based on force, length, and time.
Since centipoise is a fractional unit of poise, it represents one-hundredth of a poise. Converting between these units requires understanding the relationship between CGS and SI systems and how fundamental mechanical units translate between them.
The conversion reflects how small-scale fluid resistance is quantified in practical measurements, especially in laboratory and engineering applications.
Overall, it is a straightforward unit conversion within the CGS system of viscosity measurement.
Option a – 0.001 kg/m-s
One poise is equivalent to :
(A) 0.1 Pa-s
(B) 0.001 Pa-s
(C) 0.01 Pa-s
(D) 0.0001 Pa-s
Explanation: The poise is a unit of dynamic viscosity in the CGS system, representing resistance to flow in fluids. It is defined based on force applied over an area and the resulting velocity gradient in a fluid layer.
To relate it to the SI system, it must be expressed in terms of standard mechanical units such as kilogram, meter, and second. This conversion helps unify fluid mechanics calculations across different measurement systems.
Viscosity expressed in SI units is commonly used in scientific and engineering contexts because it aligns with the standard international system.
Overall, this is a unit conversion between CGS and SI systems for viscosity.
Option a – 0.1 Pa-s
Two sound waves of wavelengths 0.87 m and 0.885 m produce 7 beats per second. The velocity of sound in the air will be near about :
(A) 359 m/s
(B) 340 m/s
(C) 280 m/s
(D) 320 m/s
Explanation: Beat frequency occurs when two sound waves of slightly different frequencies interfere with each other. The frequency of each wave is related to its wavelength and the speed of sound through the medium.
Since both waves travel through the same medium, their speeds are equal, and frequency differences arise solely due to differences in wavelength. The relationship between frequency, wavelength, and wave speed allows determination of sound velocity using the beat frequency condition.
The beat frequency is equal to the absolute difference between the two frequencies. By expressing frequencies in terms of wave speed and wavelength, an equation involving a single unknown (speed of sound) can be formed.
This method is commonly used in wave mechanics to determine wave speed experimentally.
Overall, this problem applies wave interference and frequency-wavelength relationships to determine sound speed.
Option a – 359 m/s
Eleven tuning forks are arranged in ascending order of frequencies. Each fork produces 8 beats per second with the other. The last fork is an octave of the first. The frequency of the 10th fork is :
(A) 80 Hz
(B) 152 Hz
(C) 172 Hz
(D) 184 Hz
Explanation: When multiple tuning forks are arranged in increasing order of frequency, the beat frequency between adjacent forks remains constant if the difference between successive frequencies is uniform. This creates an arithmetic progression of frequencies.
An octave relationship means the highest frequency is twice the lowest, providing a fixed boundary condition for the sequence. Using this condition, the common difference between frequencies can be determined.
Once the pattern is established, the frequency of any specific tuning fork in the sequence can be found using the arithmetic progression formula.
Beat frequency helps define the spacing between adjacent frequencies, while the octave condition fixes the overall range.
Overall, this is a combination of wave interference and arithmetic progression in frequency distribution.
Option b – 152 Hz
A tuning fork C sounded together with a tuning fork D of frequency 256 Hz emits 2 beats. On loading the tuning fork D the number of beats, heard is 1 per second. The frequency of tuning fork C is :
(A) 257 Hz
(B) 256 Hz
(C) 258 Hz
(D) 254 Hz
Explanation: Beat frequency arises when two sound sources have slightly different frequencies. The difference between their frequencies determines how many beats are heard per second.
When one tuning fork is loaded, its frequency changes due to added mass, usually decreasing in value. This changes the beat frequency observed between the two forks.
By comparing beat frequencies before and after loading, one can determine whether the original frequency was higher or lower than the known reference. This comparison helps identify the correct frequency relationship.
The key idea is that beat frequency depends only on the absolute difference between frequencies, and changes in one source shift this difference in a predictable way.
Overall, this problem uses beat frequency changes to infer unknown frequency values.
Option d – 254 Hz
Two bodies have frequencies of 252 and 256 vibrations per second respectively. The beat frequency produced when they vibrate together is :
(A) 16
(B) 12
(C) 20
(D) 4
Explanation: When two sound waves of slightly different frequencies are produced together, they interfere with each other and create a Periodic variation in intensity known as beats. This happens because the waves alternately reinforce and cancel each other depending on their phase relationship.
The number of beats heard per second depends on how quickly the phase difference between the two waves changes. If the frequencies are very close, the beats are slow and easily noticeable. If the difference is larger, the beats occur more rapidly.
Beat frequency is determined purely by the absolute difference between the two frequencies, regardless of which one is higher or lower. This makes it a simple but powerful tool for comparing close frequencies in sound waves.
The phenomenon is widely used in tuning musical instruments, where small differences in frequency are detected through the presence of beats.
Overall, this problem applies the basic concept of wave interference and frequency difference in sound waves.
Option d – 4
When a tuning fork of frequency 341 Hz is sounded with another tuning fork, six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first tuning fork, the number of beats is two per second. The natural frequency of the second tuning fork is :
(A) 335 Hz
(B) 339 Hz
(C) 343 Hz
(D) 347 Hz
Explanation: Beat frequency arises due to interference between two sound sources having slightly different frequencies. The difference in their frequencies determines the number of beats per second. When one tuning fork is altered by adding wax, its frequency decreases, changing the beat frequency observed.
The key idea is that loading a tuning fork reduces its frequency, while the other remains unchanged. By comparing the two observed beat frequencies before and after loading, the original unknown frequency can be identified.
Two possible frequency cases exist depending on whether the unknown frequency is higher or lower than the known reference. The change in beat frequency after loading helps eliminate ambiguity and determine the correct original value.
This method relies on understanding how frequency changes affect interference patterns between sound waves.
Overall, this is an application of beat frequency variation to determine an unknown sound frequency.
Option d – 347 Hz
Two tuning forks A and B produce 8 beats per second when sounded together. When B is slightly loaded with wax the beats are reduced to 4 per second. If the frequency of A is 512 Hz, the frequency of B is :
(A) 508 Hz
(B) 516 Hz
(C) 504 Hz
(D) 520 Hz
Explanation: Beats occur when two sound waves of slightly different frequencies interfere. The beat frequency depends on the absolute difference between the two frequencies. When one tuning fork is loaded, its frequency decreases, changing this difference.
Initially, the two frequencies differ by a certain amount producing a known beat frequency. After loading, the frequency of one fork shifts downward, reducing the difference between the two frequencies.
By analyzing how the beat frequency changes before and after loading, one can determine whether the unknown frequency was initially higher or lower than the known one.
The key principle is that the change in beat frequency direction reveals the relative ordering of frequencies, allowing the original unknown frequency to be identified.
Overall, this problem uses variation in beat frequency to determine an unknown sound frequency.
Option d – 520 Hz
A source of the sound of frequency 450 cycle/sec is moving, towards a stationary observer with 34 m/s. If the speed of sound is 340 m/s, then the apparent frequency will be :
(A) 410 cycle/sec
(B) 500 cycle/sec
(C) 550 cycle/sec
(D) 450 cycle/sec
Explanation: When a sound source moves relative to an observer, the observed frequency changes due to the Doppler effect. As the source approaches, the wavefronts get compressed, reducing the effective wavelength in the direction of motion. This causes the observer to detect a higher frequency than the actual emitted frequency.
The change in frequency depends on the speed of sound in the medium and the speed of the source. A faster approaching source leads to greater compression of wavefronts and a larger increase in observed frequency.
The principle is widely used in applications such as radar, astronomy, and motion detection, where frequency shifts indicate relative motion between source and observer.
The key idea is that motion alters the spacing of wavefronts, which directly affects perceived frequency.
Overall, this is an application of the Doppler effect for a moving sound source approaching a stationary observer.
Option b – 500 cycle/sec
Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce 4 beats per second. The highest frequency is twice the lowest. The possible highest and lowest frequencies are :
(A) 80 Hz and 40 Hz
(B) 100 Hz and 50 Hz
(C) 44 Hz and 22 Hz
(D) 72 Hz and 36 Hz
Explanation: When tuning forks are arranged in increasing order with a constant beat frequency between adjacent forks, the frequency differences between successive forks remain uniform. This forms an arithmetic progression in frequency values.
The beat frequency between adjacent forks determines the common difference of this progression. With a fixed number of forks and a known total range (highest frequency being twice the lowest), the full sequence can be constructed.
The arithmetic nature of the frequencies ensures that each step increases by a constant amount, allowing systematic determination of all frequencies.
The relationship between beat frequency and frequency difference is crucial in building the sequence.
Overall, this is a combination of arithmetic progression and wave interference concepts in sound frequency arrangements.
Option d – 72 Hz and 36 Hz
Two tuning forks have frequencies of 450 Hz and 454 Hz respectively. On sounding these forks together, the time interval between successive maximum intensities will be :
(A) 1/4 second
(B) 1/2 second
(C) 1 second
(D) 2 second
Explanation: When two sound waves of slightly different frequencies interfere, they produce beats, which are Periodic variations in intensity. The time interval between successive maxima corresponds to the time taken for one complete cycle of constructive interference.
This interval is inversely related to the difference between the two frequencies. A smaller difference results in slower beats and larger time intervals between maxima, while a larger difference leads to faster variations.
The phenomenon arises due to continuous phase change between the two waves, causing alternating reinforcement and cancellation of sound intensity.
This concept is commonly used in acoustics to analyze frequency differences in sound waves.
Overall, this is an application of beat frequency and wave interference in time-domain analysis.
Option a – 1/4 second
A tuning fork gives 5 beats with another tuning fork of frequency 100 Hz. When the first tuning fork is loaded with wax, then the number of beats remains unchanged, then what will be the frequency of the first tuning fork?
(A) 95 Hz
(B) 100 Hz
(C) 105 Hz
(D) 110 Hz
Explanation: Beat frequency depends on the absolute difference between two sound frequencies. When one tuning fork is loaded, its frequency decreases. If the beat frequency remains unchanged after loading, it indicates a special relationship between the two frequencies.
This suggests that the original frequency difference remains constant even after one frequency is reduced, implying a symmetric condition where the unknown frequency lies exactly at a midpoint-like relation relative to the change caused by loading.
By analyzing how loading affects frequency and comparing beat patterns before and after, the correct frequency relationship can be determined.
The key idea is that unchanged beats despite frequency modification indicate a balanced frequency configuration.
Overall, this is a beat frequency consistency problem used to infer an unknown tuning fork frequency.
Option c – 105 Hz
Two tuning forks of frequency 256 and 258 vibrations per second are sounded together, then the time interval between consecutive maxima heard by the observer is :
(A) 2 sec
(B) 0.5 sec
(C) 250 sec
(D) 252 sec
Explanation: When two sound waves of nearly equal frequencies interfere, they produce beats characterized by alternating loud and soft sound intensity. The time interval between consecutive maxima corresponds to the period of the beat phenomenon.
This interval is determined by the difference between the two frequencies. A smaller difference leads to a longer time between successive maxima, while a larger difference shortens it.
The Periodic reinforcement of sound occurs when the waves come back into phase after a full cycle of relative phase change.
This is a direct consequence of wave superposition and interference in sound waves.
Overall, this is an application of beat frequency and Periodic wave interference.
Option b – 0.5 sec
Two tuning forks A and B vibrating simultaneously produce 5 beats. The frequency of B is 512 Hz. It is seen that if one arm of A is filed, then the number of beats increases. The frequency of A will be :
(A) 502 Hz
(B) 507 Hz
(C) 517 Hz
(D) 522 Hz
Explanation: Beat frequency depends on the difference between two sound frequencies. When a tuning fork is filed, its frequency increases due to reduction in mass, which changes its vibration characteristics.
Initially, the two forks produce a certain number of beats based on their frequency difference. After modifying one fork, the change in beat frequency reveals whether the original frequency was higher or lower than the reference fork.
If beats increase after filing, it indicates that the frequency difference has increased, helping determine the relative ordering of the original frequencies.
The key idea is that changes in physical structure of a tuning fork directly affect its frequency, which in turn affects beat patterns.
Overall, this is a frequency comparison problem using beat variation under modification.
Option c – 517 Hz
We covered all the physics project class 12 ISC mcqs above in this post for free so that you can practice well for the exam.
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