Explanation: This question asks about the correct unit used to express the force constant in simple harmonic motion. The force constant represents how strongly a system resists displacement from its equilibrium position and plays a central role in oscillatory motion.
In SHM, the restoring force acting on a particle is directly proportional to its displacement from the mean position. This relationship is expressed through Hooke’s law, where force depends on both the displacement and a proportionality constant. This constant determines how stiff or elastic the system is.
To identify the unit, we consider the relation between force and displacement. force is measured in newtons, and displacement in meters. Since the force constant links these two, its unit must relate force per unit displacement. Dimensional analysis helps here: dividing force by displacement yields the required unit.
Think of a spring—if it is stiff, even a small stretch produces a large force. If it is soft, the same stretch produces less force. The constant quantifies this behavior numerically.
In summary, the unit of the force constant comes from the ratio of force to displacement, reflecting how strongly a system restores itself to equilibrium after being disturbed.
Option b – N/m
A particle executes SHM of amplitude 2 cm. At the extreme position the force is 4 N. At a point midway between the mean and extreme position the force is :
(A) 2 N
(B) 3 N
(C) 4 N
(D) 6 N
Explanation: This question explores how the restoring force varies with position in simple harmonic motion. The particle oscillates between its mean position and extreme positions, and the force changes depending on where it is along this path.
In SHM, the restoring force is directly proportional to displacement from the mean position. At the extreme position, displacement is maximum, so the force is also maximum. As the particle moves toward the center, displacement decreases, and so does the force.
The midpoint between the mean and extreme position corresponds to half the maximum displacement. Since force depends linearly on displacement, reducing displacement to half also reduces the force proportionally. This direct proportionality is the key concept governing the behavior.
A useful analogy is stretching a spring: pulling it twice as far produces twice the restoring force. Similarly, pulling it halfway results in half the force. This proportional nature simplifies analysis in SHM.
Overall, by comparing displacements at different positions, the corresponding force can be determined using the linear relationship between force and displacement in SHM.
Option a – 2 N
A particle executes SHM of time period 8 s and amplitude 4 cm. Find the speed of a particle 2s after it passes through mean position :
(A) Zero
(B) 8 cm/s
(C) 12.56 cm/s
(D) 6.28 cm/s
Explanation: This question focuses on determining the velocity of a particle in SHM at a specific time after it crosses the equilibrium position. The motion is Periodic, and both displacement and velocity vary sinusoidally with time.
In SHM, angular frequency is related to time period, and velocity depends on both angular frequency and displacement. When the particle passes through the mean position, its velocity is maximum. As time progresses, the velocity decreases as the particle moves toward the extreme position.
After a certain time interval, the phase of motion changes, which determines the displacement and velocity. By calculating angular frequency from the given time period and using time to find phase, velocity can be evaluated using standard SHM relations.
An analogy is a pendulum: it moves fastest at the center and slows down as it rises. Similarly, after leaving the mean position, velocity gradually reduces depending on time elapsed.
Thus, using time-based phase and SHM velocity relations, the speed at the given instant can be determined systematically.
Option a – Zero
A body of Mass 0.1 kg is undergoing SHM of amplitude 1 m and period 0.2 s. What is the maximum force acting on it?
(A) 9.87 N
(B) 987 N
(C) 98.7 N
(D) 49.3 N
Explanation: This problem involves calculating the maximum restoring force experienced by a particle undergoing SHM. The force varies with displacement and is greatest at the extreme positions where displacement is maximum.
The restoring force in SHM is proportional to displacement and depends on the Mass and angular frequency of the system. Angular frequency is related to the time period, and once known, it helps determine the maximum acceleration.
Maximum force occurs when acceleration is maximum, which happens at the extreme position. Using the relation between Mass, angular frequency, and amplitude, the force can be evaluated.
Imagine a Mass attached to a spring: when stretched or compressed fully, the restoring force is strongest. At intermediate points, the force is smaller.
In summary, by combining time period, amplitude, and Mass, the maximum restoring force can be determined using SHM principles.
Option c – 98.7 N
When a Mass undergoes simple harmonic motion there is always a constant ratio between its displacement and :
Explanation: This question examines a defining characteristic of simple harmonic motion. SHM is governed by a specific relationship between displacement and another physical quantity that ensures oscillatory motion.
In SHM, acceleration is always proportional to displacement and directed toward the equilibrium position. This proportionality remains constant throughout the motion and is the fundamental condition for SHM.
The ratio between acceleration and displacement remains fixed and is related to the square of angular frequency. This ensures that motion repeats periodically and maintains a sinusoidal nature.
Consider a spring system: the further it is stretched, the stronger the restoring pull, always in a consistent proportion. This predictable relationship defines SHM behavior.
Thus, identifying the physical quantity that maintains a constant proportional relationship with displacement helps define the nature of SHM.
Option b – Acceleration
A weight suspended from a spring oscillates up and down. The acceleration of the weight is zero at
(A) Highest point
(B) Lowest point
(C) Middle point
(D) Middle of amplitude
Explanation: This question focuses on identifying the point in SHM where acceleration becomes zero. Acceleration in SHM depends on displacement and changes continuously as the particle moves.
At extreme positions, displacement is maximum, so acceleration is also maximum toward the center. As the particle moves toward the mean position, displacement decreases, and acceleration reduces.
At the equilibrium or mean position, displacement becomes zero. Since acceleration is proportional to displacement, it also becomes zero at this point.
A simple analogy is a pendulum: it experiences no restoring pull exactly at the center, but maximum pull at the ends. Similarly, acceleration behaves in the same way in SHM.
Hence, by understanding how acceleration varies with displacement, the position where acceleration becomes zero can be identified clearly.
Option c – Middle point
The necessary and sufficient condition for a motion to be simple harmonic is :
(A) Acceleration should be constant
(B) The acceleration should be proportional to the displacement from the equilibrium position
(C) The acceleration should be directed toward the equilibrium position
(D) Both ‘2’ and ‘3’
Explanation: This question asks for the precise condition that defines simple harmonic motion. Not all oscillatory motions are SHM, so a specific criterion must be satisfied.
The defining feature of SHM is that the restoring force (or acceleration) must be directly proportional to displacement and always directed toward the equilibrium position. Both magnitude and direction are essential.
If only proportionality exists without correct direction, motion won’t be oscillatory. Similarly, if direction is correct but proportionality is absent, motion won’t follow SHM behavior.
An example is a stretched spring: the restoring force always pulls it back toward the center and increases proportionally with displacement.
Therefore, both proportionality and direction together form the necessary and sufficient condition for SHM.
Option d – Both ‘2’ and ‘3’
If a hole is bored along the diameter of the Earth and the stone is dropped into the hole :
(A) The stone reaches the center of the Earth and stops there
(B) The stone reaches the other side of the Earth and stops there
(C) The stone executes SHM about the center of the Earth
(D) Stone reaches the other side of the Earth and escapes into the space
Explanation: This question considers motion under gravity inside the Earth, assuming uniform density. The gravitational force behaves differently inside compared to outside.
Inside the Earth, gravitational force decreases linearly with distance from the center. This creates a restoring force directed toward the center, proportional to displacement.
Such a force condition resembles that of SHM, where acceleration is proportional to displacement and directed toward equilibrium. Hence, the motion becomes oscillatory about the center.
Imagine a Mass attached to a spring: if displaced, it oscillates back and forth. Similarly, the stone would move through the Earth and oscillate along the diameter.
Thus, the nature of gravitational variation inside Earth leads to oscillatory motion resembling SHM.
Option c – The stone executes SHM about the center of the Earth
(C) the Speed of the particle may very or remains constant
(D) The velocity remains constant
Explanation: This question distinguishes between different types of Periodic motion and how speed behaves in such systems. Periodic motion simply means motion that repeats after equal intervals of time.
Speed behavior depends on the type of motion. In uniform circular motion, speed remains constant, while in SHM, speed varies continuously between maximum and zero.
Since Periodic motion includes various types, speed may either remain constant or change depending on the specific system. It is not restricted to one behavior.
For example, a rotating fan has constant speed, whereas a swinging pendulum speeds up and slows down during motion.
Thus, Periodic motion allows both constant and varying speed depending on the nature of the motion involved.
Option c – the Speed of the particle may very or remains constant
What determines the frequency of natural oscillations of the system?
(A) Inertia alone
(B) Elasticity alone
(C) Both inertia and elasticity
(D) Neither inertia nor elasticity
Explanation: This question focuses on factors affecting the natural frequency of a system undergoing oscillations. Frequency determines how fast the system completes oscillations.
Natural frequency depends on both inertia and elasticity. Inertia resists motion, while elasticity provides restoring force. The balance between these two determines Oscillation rate.
If inertia is large, motion is slower; if elasticity is strong, oscillations are faster. The interplay of these factors defines the system’s natural frequency.
A Mass-spring system illustrates this well: a heavier Mass oscillates slower, while a stiffer spring increases frequency.
Hence, natural frequency is governed by both resistance to motion and restoring capability of the system.
Option c – Both inertia and elasticity
The motion of a simple pendulum is SHM because
(A) It is simple to construction
(B) It oscillates in a simple way
(C) Its acceleration is proportional to the displacement from the mean position
(D) It depends on the mass of the body
Explanation: This question explores why a simple pendulum behaves like a simple harmonic oscillator under certain conditions. The motion depends on restoring forces acting on the bob.
For small angular displacements, the restoring force acting on the pendulum bob is proportional to its displacement from the mean position. This satisfies the key condition for SHM.
The approximation holds only for small angles, where sine of the angle is nearly equal to the angle itself. This makes the motion mathematically similar to SHM.
Think of gently swinging a pendulum—it oscillates smoothly and predictably, similar to a spring system.
Thus, under small-angle approximation, the proportional restoring force makes pendulum motion simple harmonic.
Option c – Its acceleration is proportional to the displacement from the mean position
Equinox means the day on which:
a. Duration of day and night are the same
b. An eclipse occurs
c. Duration of the day is greater
d. The duration of the night is greater
Explanation: This question refers to a specific astronomical event related to Earth’s motion around the Sun. Equinoxes occur at particular times of the year.
During an equinox, the Sun is positioned directly above the equator, causing nearly equal durations of day and night across the globe. This happens twice a year.
It occurs because Earth’s axis is tilted, and during equinoxes, neither hemisphere is tilted toward or away from the Sun.
A simple way to imagine it is that both hemispheres receive equal sunlight, leading to balanced day and night durations.
Thus, equinox represents a condition where day and night are approximately equal everywhere on Earth.
Option b – An eclipse occurs
The value of G depends upon:
a. The masses of bodies
b. The medium between the bodies
c. The temperature of bodies
d. None
Explanation: This question examines whether the universal gravitational constant varies with different physical conditions. The constant G appears in Newton’s law of Gravitation and governs the strength of gravitational interaction between masses.
G is considered a universal constant, meaning it has the same value everywhere in the universe regardless of the nature of the objects involved. It does not depend on the masses, distance, medium, or temperature of the bodies.
Unlike quantities such as weight or gravitational force, which change depending on conditions, G remains fixed. Its value is determined experimentally and is fundamental to gravitational Physics.
An analogy is the value of π in mathematics—it remains constant irrespective of the circle considered. Similarly, G remains unchanged across all gravitational interactions.
Therefore, G is independent of external conditions and remains constant universally.
Option d – None
The force of attraction between two unit point masses separated by unit distance is called:
a. Gravitational field strength
b. Gravitational potential
c. Acceleration due to gravity
d. Universal Gravitational constant
Explanation: This question is about identifying a fundamental physical constant defined through gravitational interaction. It refers to a specific situation involving unit masses and unit distance.
In Newton’s law of Gravitation, the force between two masses depends on their values, separation, and a constant factor. When both masses and distance are taken as unity, the resulting force directly represents that constant.
This definition helps in understanding the physical meaning of the constant—it quantifies the strength of gravitational attraction under standard conditions.
Imagine setting up a standard reference experiment where all variables are fixed at unity; the resulting force gives a direct measure of the gravitational constant.
Thus, this definition provides a conceptual way to interpret the constant used in gravitational force calculations.
Option d – Universal Gravitational constant
A 1200 kg car going 30 m/s applies its brakes and skids to rest. If the friction force between the sliding tires and the pavement is 6000 N, how far does the car skid before coming to rest:
a. 45 meters
b. 65 meters
c. 90 meters
d. 101 meters
Explanation: This problem involves motion under constant deceleration due to friction. When brakes are applied, friction opposes motion and reduces the speed of the car until it comes to rest.
The friction force produces a constant deceleration, which can be calculated using Newton’s second law. Once deceleration is known, kinematic equations help determine the stopping distance.
The initial velocity is given, and final velocity becomes zero. Using these values, one of the standard equations of motion can be applied to find displacement.
A real-life example is a car stopping on a road—the stronger the braking force, the shorter the stopping distance.
Thus, by combining force, mass, and motion equations, the distance traveled before stopping can be determined.
Option c – 90 meters
The heliocentric theory was proposed by:
a. Ptolemy
b. Socrates
c. Newton
d. Copernicus
Explanation: This question refers to a major shift in our understanding of the Solar system. The heliocentric theory states that the Sun is at the center and planets revolve around it.
Before this idea, the geocentric model placed Earth at the center. The heliocentric model challenged this long-held belief and provided a more accurate explanation of planetary motion.
The theory explained observations such as retrograde motion of planets more naturally than earlier models. It marked a turning point in the scientific revolution.
An analogy is correcting a map that was previously drawn incorrectly—once corrected, all observations start making better sense.
Thus, the heliocentric theory fundamentally changed our understanding of the universe’s structure.
Explanation: This question relates to Kepler’s laws of planetary motion. The harmonic law describes a specific relationship between the time period of a planet and its distance from the Sun.
According to this law, the square of the time period of revolution is proportional to the cube of the average distance from the Sun. This relationship applies to all planets.
It helps explain why planets farther from the Sun take longer to complete an orbit. The law is fundamental in understanding orbital mechanics.
Think of runners on different tracks: those on larger tracks take more time to complete a lap, similar to planets with larger orbits.
Thus, the harmonic law provides a mathematical relationship connecting orbital time and distance.
Option d – Kepler’s 3rd law
The curl of a central force is:
a. Greater than 0
b. Equal to 0
c. Less than 0
d. Not defined
Explanation: This question deals with a property of Vector fields in Physics, specifically central forces. A central force always acts along the line joining the particle and a fixed point.
Such forces depend only on distance and are directed radially. Because of this symmetry, they do not produce any rotational tendency around that point.
Mathematically, the curl of a Vector field measures its tendency to produce rotation. For central forces, this rotational component is absent.
An example is gravitational force—it pulls objects directly toward the center without causing any twisting effect.
Hence, central forces are irrotational in nature, meaning their curl is zero.
Option b – Equal to 0
Force on current-carrying conductor kept parallel to the magnetic field is ….
(A) zero
(B) maximum
(C) infinity
(D) varying
Explanation: This question involves the magnetic force acting on a current-carrying conductor placed in a magnetic field. The force depends on current, length, magnetic field strength, and the angle between current and field.
The magnitude of force is proportional to the sine of the angle between the conductor and the magnetic field. When the conductor is parallel to the field, the angle becomes zero.
Since sine of zero is zero, the force becomes zero in this configuration. This means no magnetic force acts on the conductor when aligned with the field.
A simple analogy is pushing a door exactly along its hinge—you produce no rotation or effect.
Thus, orientation plays a crucial role, and parallel alignment results in no magnetic force.
Option a – zero
Force on the current carrying conductor of length 2m kept inclined at 30° with the magnetic field of induction 4T and carrying current 4A is :
(A) 4 N
(B) 8 N
(C) 12 N
(D) 16 N
Explanation: This question involves calculating magnetic force on a current-carrying conductor placed at an angle in a magnetic field. The force depends on current, length, magnetic field strength, and the angle between them.
The formula for magnetic force includes a sine term that accounts for orientation. When the conductor is inclined, only the perpendicular component contributes to the force.
By substituting given values into the expression and evaluating the sine of the angle, the magnitude of force can be determined.
Imagine holding a rod in flowing water: the more perpendicular it is to the flow, the greater the force it experiences.
Thus, by considering both magnitude and orientation, the force on the conductor can be evaluated.
Option d – 16 N
A wire of length 1.5 m carrying current 10A kept in uniform magnetic field 2T, experiences a force of 15N. The direction of the conductor with a magnetic field makes an angle :
(A) 30°
(B) 60°
(C) 45°
(D) 90°
Explanation: This problem requires determining the angle between a current-carrying conductor and a magnetic field using the force experienced by the conductor.
The magnetic force depends on current, length, field strength, and the sine of the angle between the conductor and field. Rearranging the formula allows solving for the angle.
By substituting known values, the sine of the angle can be calculated. Then, the angle itself is obtained using inverse trigonometric relations.
A helpful way to visualize this is adjusting the tilt of a rod in a magnetic field—changing its angle changes the force experienced.
Thus, by using the relationship between force and angle, the orientation of the conductor can be determined.
Option a – 30°
A +ve charge moves vertically upwards in a magnetic field directed towards the north. Force on charge will be towards
(A) north
(B) east
(C) west
(D) downward
Explanation: This question tests understanding of the magnetic force on a moving charge. The force depends on the velocity of the charge and the direction of the magnetic field.
The direction is determined using the right-hand rule, where thumb represents velocity and fingers represent magnetic field. The resulting force is perpendicular to both.
Since the directions are given, applying the rule gives a third perpendicular direction for the force.
A simple analogy is a three-dimensional coordinate system—each direction is mutually perpendicular to the others.
Thus, using directional rules, the force direction on the moving charge can be identified.
Option c – west
For deflection of the magnetic needle due to the current carrying a straight conductor, the axis of the needle is placed :
(A) parallel to conductor
(B) perpendicular to the conductor
(C) in any direction
(D) in any direction except perpendicular
Explanation: This question deals with the interaction between a magnetic needle and the magnetic field produced by a current-carrying conductor. The deflection depends on how the needle is oriented relative to the conductor.
A current-carrying conductor produces circular magnetic field lines around it. For the needle to experience maximum deflection, it must be aligned in such a way that the magnetic field can exert torque on it.
If the needle is parallel to the conductor, it aligns with the magnetic field lines and experiences minimal turning effect. However, when placed perpendicular, the field exerts a torque causing noticeable deflection.
This is similar to how a compass needle responds to Earth’s magnetic field—it rotates only when the field acts across it.
Thus, proper orientation of the needle relative to the conductor ensures effective deflection due to magnetic interaction.
Option d – in any direction except perpendicular
The direction of the force on the current-carrying conductor in a magnetic field is given by :
(A) Fleming’s right-hand rule
(B) right-hand palm rule
(C) Flemings’s left-hand rule
(D) right-hand screw rule
Explanation: This question focuses on identifying the rule used to determine the direction of force experienced by a current-carrying conductor in a magnetic field.
The direction of force depends on three mutually perpendicular directions: current, magnetic field, and force. A specific hand rule helps relate these directions conveniently.
By aligning fingers and thumb in a particular orientation, one can determine how the conductor will move when placed in a magnetic field.
This is widely used in devices like electric motors, where conductors experience force and produce motion.
An analogy is using your hand as a 3D guide to visualize directions in space.
Thus, a standard hand rule provides a systematic way to determine force direction in such situations.
Option c – Flemings’s left-hand rule
Two charges +q and -q are attached to the two ends of a Light rod of length L. The system is given a velocity of v perpendicular to magnetic field 3. The magnetic force on the system of charges and magnitude of the force on one charge are respectively :
(A) Zero, zero
(B) Zero, qvB
(C) 2qvB, zero
(D) 2qvB, qvB
Explanation: This question explores the behavior of a system of charges moving in a magnetic field. Each charge experiences a magnetic force depending on its sign, velocity, and field direction.
The forces on positive and negative charges act in opposite directions due to their opposite nature. However, since they are attached to a rigid rod, their effects combine in a specific way.
While individual charges experience forces, the NET force on the system depends on Vector addition of these forces. In some cases, forces cancel out, while in others, they produce rotational effects.
A good analogy is two people pushing a rod in opposite directions—NET motion depends on how forces balance.
Thus, analyzing both individual and combined effects helps determine the overall force behavior of the system.
Option b – Zero, qvB
An elastic circular wire of length 1 carries a current I in an anticlockwise direction. It is placed in a uniform magnetic field B (out of paper) such that is perpendicular to the direction of B. it’s plane The wire will experience :
(A) No force
(B) A stretching force
(C) A compressive force
(D) A torque
Explanation: This question considers how a flexible current-carrying loop behaves in a magnetic field. The magnetic field interacts with current elements at every point along the wire.
Each small segment of the wire experiences a magnetic force. Due to symmetry, these forces act outward or inward depending on current direction and field orientation.
Since the wire is elastic, these distributed forces can change its shape rather than cause translation or rotation.
This is similar to a rubber band being pulled evenly in all directions—it expands uniformly.
Thus, the combined effect of magnetic forces on different segments leads to deformation of the wire instead of simple motion.
Option b – A stretching force
The force experienced by charge particles in a magnetic field is :
(A) Fleming’s force
(B) Lorentz’s force
(C) Oersted’s force
(D) Faraday’s force
Explanation: This question asks for the name of the force acting on a moving charged particle in a magnetic field. This force arises due to interaction between charge motion and magnetic field.
The force depends on charge, velocity, magnetic field strength, and the angle between velocity and field. It always acts perpendicular to both velocity and magnetic field.
This perpendicular nature causes circular or helical motion rather than linear acceleration.
A familiar example is electrons moving in circular paths inside cyclotrons or magnetic fields.
Thus, this specific force governing charged particle motion in magnetic fields has a well-defined name in Physics.
Option b – Lorentz’s force
Earth’s magnetic field always has a horizontal component except at
(A) The geographical center
(B) The magnetic poles of Earth
(C) The magnetic equator
(D) Latitude of 60⁰
Explanation: This question relates to the components of Earth’s magnetic field. The field can be resolved into horizontal and vertical components depending on location.
At most places, both components exist. However, at certain special points, one of the components becomes zero.
At the magnetic poles, the field lines are vertical, meaning there is no horizontal component present.
In contrast, at the magnetic equator, the field is horizontal and vertical component becomes zero.
A simple analogy is a Vector pointing straight up—it has no horizontal part.
Thus, understanding how Earth’s magnetic field varies with location helps identify where the horizontal component vanishes.
Option b – The magnetic poles of Earth
energy resides in a current-carrying inductor in the form of :
Explanation: This question deals with energy storage in electrical components, specifically an inductor. When current flows through an inductor, it creates a magnetic field around it.
The energy supplied to the inductor is stored in this magnetic field. As current increases, the magnetic field strengthens, storing more energy.
When current decreases, the magnetic field collapses and releases the stored energy back into the circuit.
This behavior is similar to compressing a spring, where energy is stored and later released.
Thus, inductors act as energy storage devices, with energy residing in the magnetic field generated by current.
Option a – Magnetic field
Corresponding to Maxwell, the unit of magnetic flux in the SI system is :
(A) Weber
(B) Tesla
(C) Henry
(D) Gauss
Explanation: This question involves unit conversion between different systems of measurement for magnetic flux. Maxwell is a unit used in the CGS system.
In the SI system, magnetic flux is measured using a different unit. Conversion between these units is important for consistency in calculations.
The SI unit represents magnetic flux in terms of fundamental electrical quantities.
A helpful analogy is converting centimeters to meters—different systems but representing the same physical quantity.
Thus, understanding unit systems allows proper identification of the SI equivalent of Maxwell.
Option a – Weber
If the direction of the electric current is reversed, the direction of the magnetic field will :
(A) Remain the same
(B) Reversed
(C) Make an angle
(D) Either (A) or (C)
Explanation: This question explores how magnetic field direction depends on the direction of electric current. A current-carrying conductor generates a magnetic field around it.
The direction of this field is determined using the right-hand thumb rule. If the current direction changes, the orientation of the magnetic field also changes.
This means the field lines reverse their direction while maintaining the same pattern.
An analogy is rotating a screw in the opposite direction—the motion reverses accordingly.
Thus, reversing current leads to reversal of the magnetic field direction.
Option b – Reversed
Magnetic field B due to a current element at all points on the axis of the element is :
(A) Zero
(B) Minimum
(C) Maximum
(D) Infinite
Explanation: This question examines the magnetic field produced by a small current element at points along its axis. The behavior of the field depends on geometry and direction.
For a current element, magnetic field is given by Biot–Savart law, which involves a cross product. This means the field depends on the sine of the angle between current direction and position Vector.
Along the axis, this angle becomes zero, making the sine term zero. As a result, no magnetic field is produced along that direction.
A simple analogy is pushing in the same direction as motion—no sideways effect is created.
Thus, due to geometric considerations, the magnetic field along the axis of a current element becomes zero.
Option a – Zero
The energy resides in a current-carrying conductor in the form of :
Explanation: This question explores how energy is stored when electric current flows through a conductor. Whenever current passes, it generates a magnetic field around the conductor.
The energy supplied to maintain current does not simply vanish; instead, it gets stored in the magnetic field created around the conductor. The strength of this field depends on the current and configuration of the conductor.
This stored energy can later be released when the current changes, especially in circuits containing inductive elements. The concept is fundamental in understanding electromagnetic energy storage.
A simple analogy is a stretched rubber band storing energy due to deformation. Similarly, a conductor stores energy due to the magnetic field around it.
Thus, the energy associated with a current-carrying conductor resides in the magnetic field produced by it.
Option a – Magnetic field
Current carrying conductor produces magnetic field is discovered by :
(A) Newton
(B) Oersted
(C) Joule
(D) Faraday
Explanation: This question refers to a key historical discovery in electromagnetism. It marked the first connection between Electricity and Magnetism.
Before this discovery, Electricity and Magnetism were considered separate phenomena. The observation showed that passing current through a conductor produces a magnetic effect around it.
This was demonstrated using a compass needle, which deflected when placed near a current-carrying wire. This simple experiment led to the development of electromagnetism.
An analogy is discovering that two seemingly unrelated ideas are actually connected, opening the door to new scientific understanding.
Thus, this discovery laid the foundation for the field of electromagnetism and modern electrical Technology.
Option b – Oersted
Magnetic induction can not be measured in
(A) Wb/m²
(B) T
(C) N/Am
(D) Wb/m
Explanation: This question involves identifying valid and invalid units for magnetic induction. Magnetic induction, also called magnetic field strength, has specific units derived from its physical definition.
It can be expressed in SI units such as tesla, which can also be written in equivalent forms involving force, current, and length. Some derived units are consistent with these relations.
However, not all combinations of units correctly represent magnetic induction. Units that do not match its dimensional formula are invalid.
A useful approach is dimensional analysis—checking whether the unit matches the physical quantity’s definition.
Thus, by comparing given units with standard expressions, one can identify which unit does not represent magnetic induction.
Option d – Wb/m
Which of the following is not the unit of magnetic induction?
(A) Maxwell
(B) Gauss
(C) Oersted
(D) Wb/m²
Explanation: This question tests knowledge of different unit systems and how magnetic quantities are represented. Magnetic induction has both SI and CGS units.
Common units include tesla in SI and gauss in CGS. Some older or related units may also appear in electromagnetic contexts but may not represent magnetic induction directly.
It is important to distinguish between units of magnetic field, magnetic intensity, and other related quantities, as they are often confused.
An analogy is confusing units of speed with units of acceleration—they may seem similar but represent different concepts.
Thus, identifying the incorrect unit requires understanding which units are truly associated with magnetic induction.
Option a – Maxwell
The direction of the magnetic field produced by the current-carrying conductor is given by :
(A) Fleming’s left-hand rule
(B) Fleming’s right-hand rule
(C) right-hand thumb rule
(D) right-hand palm rule
Explanation: This question is about determining the direction of the magnetic field produced by a current-carrying conductor. The field forms concentric circles around the conductor.
A specific hand rule helps determine the direction of these circular field lines based on the direction of current.
By aligning the thumb in the direction of current, the curl of the fingers indicates the direction of the magnetic field.
This is widely used in understanding electromagnetic effects in wires and coils.
A simple analogy is gripping a rod—your fingers naturally curl around it, indicating direction.
Thus, this rule provides a clear method to determine magnetic field direction around a conductor.
Explanation: This question focuses on the nature of magnetic induction as a physical quantity. Physical quantities can be scalar, Vector, or tensor depending on how they behave.
Magnetic induction has both magnitude and direction, meaning it cannot be fully described by a single number alone. Its direction is important in determining force on charges and currents.
It follows Vector addition rules and changes direction based on the source of the magnetic field.
An example is velocity, which also requires both magnitude and direction to be fully described.
Thus, magnetic induction is classified based on its directional and magnitude properties.
Explanation: This question explores the origin of magnetic fields in relation to electric charges. Charges can either be stationary or in motion.
A stationary charge produces only an Electric Field and does not generate a magnetic field. However, when charges move, they create a magnetic field around them.
This is the principle behind electric currents, where moving charges in a conductor produce magnetic effects.
An analogy is wind created by moving air—still air produces no motion, but moving air generates effects.
Thus, the production of magnetic fields is directly linked to the motion of electric charges.
Option b – moving charge
The magnetic field is nothing but :
(A) the magnetic flux per unit area
(B) the electric flux per unit area
(C) magnetic flux normally passing per unit area
(D) permeability of the medium
Explanation: This question asks for the conceptual understanding of a magnetic field. A magnetic field represents the region around a magnet or current-carrying conductor where magnetic effects can be observed.
It can be described in terms of magnetic flux passing through a unit area. The field has both magnitude and direction and is represented by field lines.
These field lines indicate how a magnetic force would act on a moving charge or magnetic material placed in that region.
A common example is iron filings around a magnet, which align along the magnetic field lines.
Thus, a magnetic field describes the influence exerted by magnetic sources in the surrounding space.
Option c – magnetic flux normally passing per unit area
The magnetic field is measured in
(A) Wb/m²
(B) Wb/Am
(C) Wb
(D) Wb/m
Explanation: This question focuses on identifying the correct unit used to measure magnetic field strength. In Physics, units are essential for expressing physical quantities consistently.
Magnetic field strength is measured using a standard SI unit, which is derived from fundamental quantities like force, current, and length.
This unit ensures uniformity across calculations and experiments in electromagnetism.
It can also be expressed in equivalent forms using derived units, making it flexible in different contexts.
An analogy is measuring distance in meters—standardization ensures clarity and consistency.
Thus, identifying the correct SI unit helps in properly representing magnetic field strength.
Option a – Wb/m²
In the interference experiment with a biprism, the distance of the slits from the screen is increased by 10% and the separation of the slits is decreased by 20%. The percentage change in fringe width is :
(A) 25%
(B) 30%
(C) 37.5%
(D) 62.5%
Explanation: This question deals with fringe width in interference experiments and how it depends on various parameters. Fringe width is directly proportional to wavelength and distance to the screen, and inversely proportional to slit separation.
When the distance between the slits and screen increases, fringe width increases proportionally. Conversely, decreasing slit separation also increases fringe width.
By applying percentage changes to both parameters, the NET effect on fringe width can be calculated.
The changes must be combined carefully, considering proportional relationships rather than simple addition.
An analogy is stretching a pattern—the more you spread it or reduce spacing, the wider it appears.
Thus, using proportional reasoning, the overall percentage change in fringe width can be determined.
Option c – 37.5%
In Young’s experiment, the angular width of a fringe formed on a distant screen is 2°. The wavelength of Light used is 6500 A°. The spacing between the slits is approximate:
(A) 1.862 mm
(B) 18.62 mm
(C) 0.1862 mm
(D) 0.01862 mm
Explanation: This question involves determining slit separation using the concept of angular fringe width in Young’s double-slit experiment. The angular fringe width depends on the wavelength of Light and the separation between the slits.
In interference, angular fringe width is directly proportional to wavelength and inversely proportional to slit separation. This means larger wavelengths produce wider fringes, while larger slit separation reduces fringe spacing.
By rearranging the relationship, slit separation can be calculated using the given angular width and wavelength. Proper unit conversion is important, especially when dealing with very small wavelengths.
An analogy is water waves passing through two openings—closer openings create wider patterns, while farther openings compress them.
Thus, by applying the relationship between angular width, wavelength, and slit separation, the required value can be obtained.
Option d – 0.01862 mm
A beam of Light of wavelength 600nm from a distant source falls on a single slit 1mm wide and the resulting Diffraction pattern is observed on a screen 2m away. The distance between the first dark fringes on either side of the central bright fringe is :
(A) 24 mm
(B) 2.4 mm
(C) 1.2 mm
(D) 12 mm
Explanation: This question focuses on Diffraction patterns produced by a single slit. When Light passes through a narrow slit, it spreads out and forms a pattern of bright and dark fringes on a screen.
The central bright fringe is the widest, and the first dark fringes appear symmetrically on both sides. The distance between these dark fringes depends on wavelength, slit width, and distance to the screen.
The formula for position of minima helps determine how far these dark fringes are located from the center. Doubling this value gives the total distance between the first dark fringes on both sides.
A simple analogy is water waves spreading after passing through a narrow gap, forming alternating regions of high and low intensity.
Thus, using Diffraction principles, the distance between the first dark fringes can be calculated.
Option b – 2.4 mm
A point is situated at 8.55 cm and 8.7 cm from two sources. The nature of illumination at that point, if the wavelength of Light is 6000 Aº, will be :
(A) Dark
(B) Bright
(C) Neither bright nor dark
(D) Both (A) and (B)
Explanation: This question examines interference based on path difference between waves arriving at a point from two sources. The nature of illumination depends on whether interference is constructive or destructive.
If the path difference is an integral multiple of wavelength, constructive interference occurs, producing brightness. If it is an odd multiple of half wavelength, destructive interference leads to darkness.
When the path difference does not exactly match these conditions, the intensity lies somewhere between bright and dark.
An analogy is two sets of water waves overlapping—depending on their phase difference, they can reinforce or cancel each other.
Thus, by comparing the path difference with wavelength, the nature of illumination at the point can be determined.
Option b – Bright
In Fraunhofer Diffraction, due to a single slit, the Light used is of wavelength 6000 A°. If the direction in which the fifth dark band occurs makes an angle of 6° 54′ with the plane of the slit, the width of the slit is :
(A) 2.5 cm
(B) 0.5 mm
(C) 0.25 mm
(D) 0.025 mm
Explanation: This question deals with determining slit width using Diffraction conditions in Fraunhofer Diffraction. Dark bands occur at specific angles where destructive interference takes place.
The condition for minima relates slit width, wavelength, and angle of Diffraction. Using this relationship, slit width can be calculated when the angle and wavelength are known.
Care must be taken to convert angular values properly into radians or use correct trigonometric values. The order of the dark band also plays a role in calculations.
An analogy is tuning a musical instrument—specific conditions produce certain notes, just as specific angles produce dark fringes.
Thus, applying the Diffraction condition allows calculation of slit width accurately.
Option d – 0.025 mm
The limit of resolution of a microscope that has a numerical aperture of the objective 0.12 will be (λ = 5800 A°) :
(A) 4 µm
(Β) 2.416 μm
(C) 3.416 μm
(D) 6 μm
Explanation: This question focuses on the resolving power of a microscope, which determines its ability to distinguish between two closely spaced objects.
The limit of resolution depends on the wavelength of Light used and the numerical aperture of the objective lens. Shorter wavelengths and larger numerical aperture improve resolution.
The formula relates these quantities inversely, meaning as numerical aperture increases, the minimum resolvable distance decreases.
A practical example is viewing fine details under a microscope—better Optics and shorter wavelengths reveal finer structures.
Thus, by applying the resolution formula, the smallest distinguishable separation can be calculated.
Option b – 2.416 μm
In a biprism experiment, the slit is illuminated with Light of wavelength 5000 A°. How many fringes will pass a point on the screen, if the path difference is altered by 0.05 mm?
(A) 10
(B) 100
(C) 5
(D) 50
Explanation: This question involves understanding how interference fringes shift when path difference changes in a biprism experiment.
Each time the path difference changes by one wavelength, one fringe moves past a given point. Therefore, the total number of fringes passing depends on how many wavelengths fit into the total path difference change.
By dividing the total change in path difference by the wavelength, the number of fringes can be determined.
An analogy is counting waves passing a fixed point—each complete wave corresponds to one fringe shift.
Thus, the number of fringes is directly related to the ratio of path difference change to wavelength.
Option b – 100
In a biprism experiment, 60 fringes are observed with sodium Light of wavelength 6000 A°. Keeping the same arrangement, if sodium light is replaced by the green light of wavelength 5000 Aº, how many fringes will be then observed?
(A) 67
(B) 69
(C) 72
(D) 74
Explanation: This question explores how the number of observed fringes depends on the wavelength of light in an interference setup.
Fringe width is directly proportional to wavelength. If wavelength decreases, fringe width decreases, meaning more fringes fit within the same space.
Thus, shorter wavelengths produce a greater number of fringes, while longer wavelengths produce fewer fringes under identical conditions.
An analogy is drawing lines closer together—you can fit more lines in the same space when spacing is reduced.
Hence, changing wavelength directly affects the number of fringes observed in an interference pattern.
Option c – 72
For the same angle of incidence, the angles of refraction in media ‘P’, ‘Q’, ‘R’, and ‘S’ are 50°, 40°, 30°, and 20° respectively. The speed of light is minimum in medium :
(A) P
(B) Q
(C) R
(D) S
Explanation: This question uses refraction to determine the relative speed of light in different media. According to Snell’s law, the refractive index is related to the angle of refraction.
A smaller angle of refraction indicates a higher refractive index, which corresponds to lower speed of light in that medium.
Thus, by comparing angles of refraction for different media, one can determine where light travels slowest.
An analogy is a vehicle slowing down more in rough terrain compared to smooth roads.
Therefore, the medium with the smallest refracted angle will have the minimum speed of light.
Option d – S
Two points separated by a distance of 0.1mm can just be inspected in a microscope when the light of wavelength 6000 A is used. If the light of wavelength 4800 A° is used, the limit of resolution will become :
(A) 0.8 mm
(B) 0.15 mm
(C) 0.08 mm
(D) 0.85 mm
Explanation: This question examines how resolution depends on wavelength in optical instruments. The limit of resolution is directly proportional to wavelength.
When wavelength decreases, the resolving power improves, allowing finer details to be distinguished. Conversely, larger wavelengths reduce resolution.
This is why electron microscopes, which use much shorter wavelengths, can achieve far higher resolution than optical microscopes.
An analogy is using a finer brush to draw details—smaller tools allow more precision.
Thus, resolution improves with decreasing wavelength and worsens with increasing wavelength.
Option c – 0.08 mm
Two interfering beams have intensities in the ratio of 9 : 4. Then, the ratio of maximum to minimum intensity in the interference pattern is :
(A) 9 : 4
(B) 6 : 4
(C) 25 : 1
(D) 5 : 1
Explanation: This question deals with intensity variation in interference patterns. When two waves interfere, the resulting intensity depends on their amplitudes or intensities.
Maximum intensity occurs when waves are in phase, while minimum intensity occurs when they are out of phase. The ratio depends on the square roots of the individual intensities.
Using the relationship between amplitude and intensity, the ratio can be derived mathematically.
An analogy is combining two sound waves—louder and softer sounds combine differently depending on their phases.
Thus, using interference principles, the ratio of maximum to minimum intensity can be determined.
Option c – 25 : 1
In Y. E., to obtain interference fringes, the light used consists of two wavelengths 5200 A and 6500 A°. The distance between the two slits is 2 mm and the distance between the plane of the slits and the screen is 2 m. What is the least distance from the central bright band, where the bright bands due to the two wavelengths overlap?
(A) 2.6 mm
(B) 1.56 mm
(C) 5.3 mm
(D) 1.6 mm
Explanation: This question explores the condition under which bright fringes from two different wavelengths overlap in Young’s double-slit experiment. Each wavelength produces its own interference pattern with different fringe spacing.
Bright fringes occur when path difference equals an integer multiple of wavelength. For two wavelengths to coincide, their respective fringe positions must match at some point.
This happens when a common multiple of the two wavelengths satisfies the interference condition. The least such position gives the nearest point of overlap from the central maximum.
An analogy is two repeating patterns aligning at certain intervals, like beats in music syncing periodically.
Thus, by finding a common condition for both wavelengths, the position of coinciding bright fringes can be determined.
Option a – 2.6 mm
In a biprism experiment, the slit and eyepiece are 10 cm and 80 cm away from the biprism when a convex lens was interposed at 40 cm from the slit, the separation of two magnified images of the slit was found to be 4 mm, and the distance between two virtual images will be :
(A) 1.6 mm
(B) 2 mm
(C) 3.2 mm
(D) 5 mm
Explanation: This question involves determining the separation between two virtual coherent sources formed in a biprism experiment. These sources are responsible for producing interference fringes.
The separation depends on the geometry of the setup and magnification produced by an inserted lens. measurements taken after magnification must be related back to the original separation.
Using lens formula and magnification relations, the actual distance between the virtual images can be calculated from the observed value.
An analogy is viewing an object through a magnifying glass—the observed size differs from the actual size and must be corrected.
Thus, by accounting for magnification, the true separation between virtual sources can be obtained.
Option c – 3.2 mm
A ray of light is incident on a glass slab with a refractive index of 1.5 making an angle of 40° with the surface. The angle of refraction in glass is
(A) 30° 42′
(B) 35° 42′
(C) 19° 28′
(D) 35° 36′
Explanation: This question applies Snell’s law to determine the angle of refraction when light enters a denser medium. The angle given with the surface must first be converted to the angle with the normal.
Snell’s law relates the sine of the angle of incidence and refraction through the refractive indices of the two media. Once the correct angle of incidence is obtained, the refracted angle can be calculated.
Careful attention to geometry is essential, especially when angles are given relative to the surface rather than the normal.
An analogy is a car entering a muddy region—it slows and changes direction depending on entry angle.
Thus, by applying Snell’s law correctly, the angle of refraction can be determined.
Option a – 30° 42′
A ray of light is incident on the surface of a glass plate of refractive index 1.55 at the polarising angle. The angle of refraction is
(A) 32° 49′
(B) 35° 16′
(C) 159 49′
(D) 67°
Explanation: This question involves Brewster’s law, which relates the polarizing angle to the refractive index of a medium. At this angle, reflected light becomes completely polarized.
According to the law, the tangent of the polarizing angle equals the refractive index. Additionally, at this condition, the reflected and refracted rays are perpendicular to each other.
Using this geometric relationship, the angle of refraction can be determined from the polarizing angle.
An analogy is light bouncing off water at a specific angle that reduces glare, commonly seen with polarized sunglasses.
Thus, using Brewster’s condition, the refracted angle can be found through geometric relations.
Option a – 32° 49′
When a source of light is receding away from an observer, then the spectral lines will get displaced towards
(A) violet region
(B) blue region
(C) ultraviolet region
(D) red region
Explanation: This question refers to the Doppler effect for light. When a source of light moves relative to an observer, the observed wavelength changes.
If the source moves away, the wavelength increases, shifting toward longer wavelengths. This shift is known as redshift in the visible Spectrum.
Conversely, if the source approaches, wavelengths decrease, leading to blueshift.
An everyday analogy is the change in pitch of a passing siren—frequency decreases as it moves away.
Thus, motion of the source relative to the observer determines the direction of spectral shift.
Option d – red region
Two mirrors are placed perpendicular to each other. A ray strikes the first mirror and after reflection from the first mirror it falls on the second mirror. The ray after reflection from the second mirror will emerge :
(A) perpendicular to the original ray
(B) parallel to the original ray
(C) at 45° to the original ray
(D) at 60° to the original ray
Explanation: This question involves successive reflections from two mirrors placed at right angles. Each reflection changes the direction of the ray according to the law of reflection.
After the first reflection, the ray changes direction based on the mirror orientation. The second reflection further alters its path.
When mirrors are perpendicular, the combined effect of two reflections results in a specific predictable direction relative to the original ray.
A useful analogy is bouncing a ball off two walls at right angles—it returns along a particular path depending on angles.
Thus, analyzing both reflections step by step reveals the final direction of the ray.
Option b – parallel to the original ray
Two Nicol prisms are placed with their principal planes at 60°. What percentage of light incident on the combination is transmitted?
(A) 10 %
(B) 25 %
(C) 12.5 %
(D) 70 %
Explanation: This question involves polarization of light and Malus’ law. When polarized light passes through an analyzer, its intensity depends on the angle between the transmission axes.
According to Malus’ law, transmitted intensity is proportional to cos2θ, where θ is the angle between the axes of the two prisms.
By substituting the given angle, the fraction of transmitted light can be calculated.
An analogy is passing light through two window blinds—alignment affects how much light gets through.
Thus, using Malus’ law, the percentage of transmitted light can be determined.
Option c – 12.5 %
Air has a refractive index of 1.0003. The thickness of the air column which will have one more wavelength of yellow light (6000 A° ) than in the same thickness of vacuum is :
(A) 1 mm
(B) 3 mm
(C) 2 mm
(D) 4 mm
Explanation: This question compares the optical path length in air and vacuum. Light travels slightly slower in air than in vacuum due to refractive index differences.
The number of wavelengths present in a given physical length depends on wavelength in that medium. A change in medium alters wavelength and hence the number of waves.
The condition for one extra wavelength involves comparing optical path lengths and refractive index differences.
An analogy is fitting more waves into a stretched or compressed medium.
Thus, by relating refractive index and wavelength, the required thickness can be calculated.
Option c – 2 mm
An air bubble inside a glass slab appears to be 6 when viewed from one side and 4 cm deep when viewed from the opposite side. If the refractive deep glass is 1.5, then the thickness of the glass index of the slab is :
(A) 9 cm
(B) 15 cm
(C) 6 cm
(D) 3 cm
Explanation: This question deals with apparent depth and refraction through a glass slab. Due to refraction, objects inside a medium appear at different depths when viewed from different sides.
The apparent depth depends on the refractive index and actual thickness of the slab. Observations from both sides provide two apparent depths.
By relating real depth and apparent depth using refractive index, equations can be formed to determine the actual thickness.
An analogy is looking at an object in water—it appears closer than it actually is.
Thus, using refraction principles from both sides, the thickness of the slab can be determined.
Option b – 15 cm
Light of a certain wavelength is incident normally on a glass plate of thickness 6 cm and refractive index of 1.5. The light takes the same time to travel from the source to the incident surface as it takes to travel through the slab. What is the distance of the source from an incident surface :
(A) 8 cm
(B) 9 cm
(C) 10 cm
(D) 11 cm
Explanation: This question involves comparing travel time of light in air and in a denser medium. Light travels slower in a medium than in air due to refractive index.
Time taken depends on distance and speed. Since speed changes in different media, equal time implies different distances traveled.
Using the relation between speed, refractive index, and distance, the required distance of the source can be calculated.
An analogy is walking on smooth ground versus sand—slower speed in sand means shorter distance in the same time.
Thus, by equating travel times in two media, the unknown distance can be determined.
Option b – 9 cm
A ray of light is incident on a glass block such that the reflected ray is completely polarised. The refracted beam suffers a deviation of 26° from its original path in the air. What is the refractive index for glass?
(A) 1.33
(B) 1.6
(C) 2.4
(D) 1.875
Explanation: This question involves polarization of light at a specific angle known as Brewster’s angle. At this angle, the reflected light becomes completely plane-polarized.
Brewster’s law states that the tangent of the polarizing angle is equal to the refractive index of the medium. Also, at this condition, the reflected and refracted rays are perpendicular to each other.
By using the given deviation or angle information, the polarizing angle can be identified. Once this angle is known, the refractive index is calculated using the tangent relationship.
An analogy is glare reduction from water surfaces—light reflecting at certain angles becomes polarized, which is why polarized sunglasses are effective.
Thus, by applying Brewster’s law and geometric relations, the refractive index of glass can be determined.
Option b – 1.6
The refractive index of glass and that of water are 3/2 and 4/3 respectively. By what fraction would the wavelength of light change if it goes from glass to water :
(A) 1.25
(B) 0.125
(C) 3.25
(D) 1.43
Explanation: This question examines how the wavelength of light changes when it travels between two media of different refractive indices. The speed of light varies depending on the medium.
Frequency of light remains constant during refraction, but wavelength changes because speed changes. Since wavelength is directly proportional to speed, a change in medium alters wavelength accordingly.
When moving from a denser medium to a less dense one, the speed increases, and hence the wavelength increases.
An analogy is waves traveling from shallow to deeper water—they spread out and increase in wavelength.
Thus, by comparing refractive indices of the two media, the fractional change in wavelength can be determined.
Option b – 0.125
Light of a certain wavelength is incident on a glass slab of refractive index 1.6 making an angle of incidence 50°. If the thickness of the glass is 3 cm, what is the distance it will cover in glass :
(A) 4 cm
(B) 3 cm
(C) 4.619 cm
(D) 3.416 cm
Explanation: This question deals with the path of a light ray inside a glass slab. When light enters at an angle, it bends due to refraction and travels a longer path inside the medium.
The actual distance traveled depends on the thickness of the slab and the angle of refraction inside the glass. This angle is determined using Snell’s law.
Once the refracted angle is known, trigonometric relations can be used to find the path length within the slab.
An analogy is walking diagonally across a room—you cover more distance than walking straight across.
Thus, by combining refraction laws and geometry, the distance traveled by the ray inside the slab can be calculated.
Option d – 3.416 cm
We covered all the Class 11ISCPhysics solutions above in this post for free so that you can practice well for the exam.
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