All questions of Physics for JEE Exam

Radius of Bohr Orbit
The radius of the Bohr orbit is determined by the principal quantum number (n) of the electron in that orbit. Each orbit in the Bohr model is associated with a specific value of n.

Dependence on 1/n
- The radius of the Bohr orbit is inversely proportional to the principal quantum number (n).
- This means that as the value of n increases, the radius of the orbit decreases.
- Mathematically, the radius of the nth Bohr orbit is given by the expression r = n2h2/4π2me2, where h is the Planck constant, m is the mass of the electron, and e is the charge of the electron.

Importance of n^2
- The correct answer is option 'C', 1/n2, because the radius of the Bohr orbit depends on the square of the principal quantum number.
- The square of n (n2) appears in the formula for the radius of the Bohr orbit, indicating that the radius is directly related to n2.
- This dependence on n2 is a key feature of the Bohr model and is crucial in understanding the behavior of electrons in atoms.
In conclusion, the radius of the Bohr orbit depends on the square of the principal quantum number (n), as indicated by the formula for the radius of the orbit. Understanding this relationship is essential in grasping the fundamentals of atomic structure and electron behavior.

Series Connection of Resistors

When resistors are connected in series, they are connected end-to-end so that the current flows through each resistor in turn. The total resistance in a series circuit is equal to the sum of individual resistances. The current through each resistor is the same because there is only one path for the current to flow.

Effect on Current Across Each Resistor

The current across each resistor in a series connection remains the same. This is because the current at any point in a series circuit is the same as the current at any other point. The current is equal to the voltage divided by the total resistance of the circuit.

Since the resistors are in series, they have a cumulative effect on the total resistance of the circuit. Therefore, the voltage drop across each resistor is proportional to its individual resistance. However, the current remains the same throughout the circuit, so the voltage drop across each resistor is also proportional to its resistance.

Conclusion

In summary, when resistors are connected in series, the current across each resistor remains the same. The voltage drop across each resistor is proportional to its resistance, but the current remains constant throughout the circuit.

Unit of Magnetic Flux

The unit of magnetic flux is a measure of the total magnetic field passing through a given area. It is defined as the total amount of magnetic field passing through a given surface. The unit of magnetic flux is the Maxwell.

Maxwell

The Maxwell is named after the Scottish physicist James Clerk Maxwell. It is defined as the amount of magnetic flux that passes through a surface of one square centimeter perpendicular to a magnetic field of one Gauss. The Maxwell is equal to 10^-8 Weber.

Gauss

The Gauss is a unit of magnetic field strength. It is defined as the magnetic flux density that exerts a force of one dyne on a unit magnetic pole. One Gauss is equal to 10^-4 Tesla.

Tesla

The Tesla is the SI unit of magnetic flux density. It is defined as the magnetic flux density that exerts a force of one Newton on a unit magnetic pole. One Tesla is equal to 10^4 Gauss.

Ampere

The Ampere is the SI unit of electric current. It is defined as the amount of electric charge passing through a given point in a circuit in one second. The Ampere is not a unit of magnetic flux.

Conclusion

In conclusion, the unit of magnetic flux is the Maxwell, which is defined as the amount of magnetic flux passing through a surface of one square centimeter perpendicular to a magnetic field of one Gauss.

No

Explanation:

Electric potential at a point in an electric field is defined as the amount of work done per unit charge in bringing a positive test charge from infinity to that point.

The electric potential due to a point charge q at a distance r from the point is given by:

V = kq/r

where k is the Coulomb's constant and is equal to 1/4πε0, where ε0 is the permittivity of free space.

Since we are dealing with a point charge in air, we can substitute the value of k and ε0 and get:

V = q/4πε0r

Simplifying further, we get:

V = q/r

Therefore, the correct option is A, i.e., q/r.

Jagaga

Explanation:
When an electric dipole is placed in a non-uniform electric field, the following happens:

1. Electric field exerts a force on each charge of the dipole
2. The force on each charge is not equal in magnitude and opposite in direction. Hence, the net force on the dipole is not zero.
3. As a result, the dipole experiences a translational motion towards the direction of the stronger electric field.

However, the electric field also exerts a torque on the dipole. This torque tends to align the dipole with the direction of the electric field.

The magnitude of the torque is given by the formula: τ = pEsinθ
where p is the dipole moment, E is the electric field, and θ is the angle between the dipole moment and the electric field.

The torque tends to rotate the dipole until it aligns with the direction of the electric field. This is because the torque is maximum when θ = 90° and zero when θ = 0° or 180°.

Thus, when an electric dipole is placed in a non-uniform electric field, both force and torque act on it. The force tends to translate the dipole, while the torque tends to rotate it. Hence, the correct option is B.

SI Unit of Magnetic Flux

The SI unit of magnetic flux is Weber.

Explanation

Magnetic flux is the measure of the total magnetic field that passes through a given area. It is defined as the total number of magnetic field lines passing through a given surface area. The SI unit of magnetic flux is Weber (Wb).

- Weber (Wb): It is defined as the magnetic flux that passes through an area of one square meter when a magnetic field of one Tesla is perpendicular to the surface. One Weber is equal to one Tesla-meter squared (T-m2).

Other Units

- Maxwell (Mx): It is the CGS unit of magnetic flux, defined as the magnetic flux that passes through an area of one square centimeter when a magnetic field of one Gauss is perpendicular to the surface. One Weber is equal to 10^8 Maxwell.
- Tesla meter squared (T-m2): It is the SI unit of magnetic flux density multiplied by area. One Weber is equal to one Tesla-meter squared (T-m2).

Conclusion

The SI unit of magnetic flux is Weber (Wb), which is defined as the magnetic flux that passes through an area of one square meter when a magnetic field of one Tesla is perpendicular to the surface.

The cauchy integral formula states that the values of a holomorphic function inside a disk are determined by the values of that function on the boundary of the disk .
Cauchy integral formula may be used to obtain experience for the derivatives of F (z).

Explanation:

When a lens is immersed in a transparent liquid, it can become invisible under the condition of having the same refractive index as the surrounding liquid.

Refractive Index:
The refractive index is a measure of how much a material bends or refracts light. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. The refractive index determines how much light rays are bent when they pass from one medium to another.

Immersion in a Transparent Liquid:
When a lens is immersed in a transparent liquid, such as water or oil, the light rays entering the lens from the surrounding medium will experience refraction at the lens-medium interface. The direction of the light rays changes due to the change in the refractive index at the interface.

Same Refractive Index:
If the refractive index of the lens material is the same as that of the surrounding liquid, then there is no change in the direction of the light rays at the lens-medium interface. This means that the light rays passing through the lens will continue to travel in a straight line without any deviation.

Consequence:
As a result, when the refractive index of the lens is the same as that of the surrounding liquid, the lens becomes invisible. Light passes through the lens without any refraction or bending, making it indistinguishable from the surrounding medium.

Example:
For example, if a lens made of glass (with a refractive index of 1.5) is immersed in a liquid with the same refractive index of 1.5, the lens will appear invisible when viewed from the outside. The light rays will pass through the lens without any noticeable change in their direction.

Conclusion:
In conclusion, when a lens is immersed in a transparent liquid and it becomes invisible, it is because the refractive index of the lens is the same as that of the surrounding liquid. This condition ensures that light rays passing through the lens do not experience any refraction or bending, making the lens indistinguishable from the surrounding medium.

The expression for Curie is Ci.

Calculation of Peak Reverse Voltage in Full-Wave Rectifier

Full-Wave Rectifier: A full-wave rectifier is an electronic circuit that converts the complete AC signal into a pulsating DC signal. It is made up of two diodes that are connected to the AC source in either a bridge or center-tapped configuration.

Peak Reverse Voltage (PIV): The maximum voltage that appears across the diode when it is in reverse bias is known as the Peak Reverse Voltage.

Given:
Alternating Voltage (V) = 300 V

Formula:
PIV = 2V

Calculation:
PIV = 2V
PIV = 2 * 300
PIV = 600 V (This is the maximum voltage that appears across the diode)

However, the diode is rated for a maximum reverse voltage of 849 V. Therefore, the PIV will be equal to the maximum reverse voltage rating of the diode, which is 849 V.

Answer:
The value of Peak Reverse Voltage (PIV) in this case is 849 V, which is option A.

Negative Charge and Electric Field Direction

Electric Field Direction of a Negative Charge

The electric field is a vector quantity that describes the direction and magnitude of the electric force that a charge would experience if it were placed in the field. The direction of the electric field created by a negative charge is towards the charge.

Explanation

When a negative charge is placed at a point in space, it creates an electric field around it. The electric field lines radiate outwards from the charge and are directed towards it. This is because the electric field is created by the negative charge, and any positive charge placed in the field would experience a force towards the negative charge.

Example

For example, consider a negative charge of -2 μC located at a point in space. The electric field created by this charge would be directed towards the charge, as shown in the figure below.



Conclusion

In summary, the direction of the electric field created by a negative charge is towards the charge. This is because the electric field is created by the negative charge, and any positive charge placed in the field would experience a force towards the negative charge.

Van de Graff Generator and Potential Difference

What is a Van de Graff Generator?
A Van de Graff generator is an electrostatic generator which uses a moving belt to accumulate electric charge on a hollow metal globe on the top of the machine. The charge is then stored on the globe and can be used for various experiments and demonstrations.

How does it work?
The Van de Graff generator works by a process called triboelectric effect. In this process, when two materials come in contact, electrons are exchanged between them. This creates a charge imbalance, which can be used to generate electricity. The belt of the Van de Graff generator is made from a material that has a high affinity for electrons, such as rubber. When the belt moves, it rubs against a metal comb, which strips electrons from the belt. These electrons are then transferred to the globe, creating a charge on it.

What is Potential Difference?
Potential difference is the difference in voltage between two points in an electric circuit. It is measured in volts (V) and represents the amount of energy that is required to move a unit charge from one point to another.

Order of Potential Difference built up by the Van de Graff Generator
The potential difference built up by the Van de Graff generator can be quite high, due to the accumulation of charge on the globe. The correct answer is option B, potential difference of the order of several million volts. In fact, some Van de Graff generators can produce potential differences of up to 30 million volts. This high potential difference makes the Van de Graff generator useful for experiments and demonstrations involving high voltage electricity, such as lightning bolts or electric discharges.

Interference pattern and phase difference

Interference is a phenomenon that occurs when two or more waves meet at a single point in space. When these waves have the same frequency, amplitude, and are in phase, they create an interference pattern that is characterized by alternating constructive and destructive interference. The phase difference between the two sources plays a crucial role in determining the nature of the interference pattern.

Continuous phase difference variation

If the phase difference between the two sources varies continuously, it implies that the sources are not coherent. This is because a coherent source has a constant phase difference, and any variation in the phase difference will result in a change in the interference pattern. When the phase difference is continuously varying, the interference pattern will disappear completely.

Reason for the disappearance of the interference pattern

The reason for the disappearance of the interference pattern is that the waves from the two sources are not synchronized. They do not have a fixed phase relationship, and as a result, they interfere randomly with each other. This means that the constructive and destructive interference cancel out each other, resulting in a uniform distribution of energy throughout the space.

Conclusion

In summary, the interference pattern created by two sources depends on the phase difference between them. When the phase difference varies continuously, the interference pattern disappears completely. This is because the waves from the two sources are not coherent, and they interfere randomly with each other, resulting in a uniform distribution of energy throughout the space.

"In Young" can refer to several things:

1. A common Korean given name for females. It means "kind and beautiful."

2. A South Korean actress named Kim In-young, who is known for her roles in the films "I Saw the Devil" and "The Throne."

3. A South Korean singer and former member of the girl group Girls' Day, whose full name is Park In-young.

Without further context, it is difficult to determine the specific meaning of "In Young."

Calculating Electric Potential Due to a Point Charge

Given:
Charge (q) = 10C
Distance (r) = 8cm

Formula:
Electric potential (V) = k(q/r)

where k is the Coulomb's constant which is equal to 9 x 10^9 Nm^2/C^2.

Calculation:
Electric potential (V) = k(q/r)
V = (9 x 10^9 Nm^2/C^2) x (10C/0.08m)
V = 1.125 x 10^12 V

Answer:
The electric potential due to a point charge of 10C at a distance of 8cm away from the charge is 1.125 x 10^12 V. Therefore, the correct option is (b).

The correct answer is option 'D' - FM radio. FM radio is a suitable communication channel for the transmission of a message signal with a bandwidth of 200kHz. Here's an explanation of why FM radio is the best choice:

Frequency Modulation (FM) Radio:
- Frequency modulation is a technique where the frequency of the carrier signal is varied in proportion to the message signal.
- In FM radio, the audio signal (message signal) is used to modulate the frequency of the carrier signal.
- FM radio offers high-quality sound reproduction and can transmit a wide range of audio frequencies.

Explanation:
1. Bandwidth Requirement:
- The bandwidth of a communication channel represents the range of frequencies it can transmit.
- In this case, the message signal has a bandwidth of 200kHz, which means it contains frequencies ranging from 0Hz to 200kHz.
- To transmit the complete message signal without distortion, the communication channel must have a bandwidth equal to or greater than 200kHz.

2. TV Transmission:
- TV transmission typically requires a much higher bandwidth than FM radio.
- TV signals carry both video and audio information, resulting in a larger bandwidth requirement.
- TV transmission channels typically have a bandwidth of several MHz, making it unsuitable for a 200kHz message signal.

3. Optical Fiber:
- Optical fiber is a high-speed transmission medium that uses light signals to transmit information.
- Optical fiber can provide extremely high bandwidth and is commonly used for long-distance data transmission.
- However, the bandwidth of optical fiber is typically in the range of several GHz, far exceeding the 200kHz bandwidth requirement of the message signal.
- Therefore, using optical fiber for transmitting a 200kHz message signal would be inefficient and unnecessary.

4. AM Radio:
- AM radio (Amplitude Modulation) is another commonly used communication channel.
- In AM radio, the amplitude of the carrier signal is varied in proportion to the message signal.
- While AM radio can transmit audio signals, its bandwidth is typically limited to a few kHz, making it unsuitable for a 200kHz message signal.

5. FM Radio:
- FM radio (Frequency Modulation) is well-suited for transmitting a message signal with a 200kHz bandwidth.
- FM radio can provide a wide frequency range, allowing for the transmission of high-quality audio signals.
- The bandwidth of FM radio stations typically ranges from 150kHz to 200kHz, which matches the bandwidth requirement of the message signal.
- Therefore, FM radio is the best choice among the given options for transmitting a message signal with a bandwidth of 200kHz.

Given:
Charge, q = 4 × 10⁻³ C
Electric potential, V = 2 × 10² V

To find: Work done, W

Formula:
Work done = q × ∆V
where, ∆V = change in electric potential

Calculation:
Work done = q × ∆V
∆V = Vf – Vi
where, Vf = final potential = 2 × 10² V (given)
Vi = initial potential = 0 V (at infinity)
∆V = 2 × 10² V – 0 V = 2 × 10² V

Work done = q × ∆V
= 4 × 10⁻³ C × 2 × 10² V
= 0.8 J

Therefore, the amount of work done to bring a charge of 4 × 10⁻³C charge from infinity to a point whose electric potential is 2 × 10²V is 0.8 J. Hence, option (a) is the correct answer.

Solution:

Given, charges at each corner of the square = 20 esu

Side length of the square = 10 cm

Electric field intensity at the center of the square is to be found.

Let's consider a square with side length 20 cm such that the given square is one of its four quadrants, as shown below:


Charge in each quadrant = 20 esu

Charge in the entire square = 4 × 20 esu = 80 esu

Distance of the center of the square from any corner = side length / √2 = 10 / √2 cm

By symmetry, the electric field due to the charges in each quadrant at the center of the square will be equal in magnitude and direction, and they will cancel out each other. Therefore, the net electric field at the center of the square will be zero.

Hence, the correct option is A) 0.

Explanation:
When an electric dipole is placed in a uniform electric field, it experiences a torque but no net force. This is because the electric field exerts equal and opposite forces on the charges of the dipole, resulting in a net force of zero. However, the two forces produce a torque on the dipole, causing it to rotate until it aligns with the electric field.

To understand this concept better, let us consider the following points:

1. Electric Dipole: An electric dipole consists of two equal and opposite charges separated by a small distance. The direction of the dipole is from the negative charge to the positive charge.

2. Uniform Electric Field: A uniform electric field is one in which the direction and magnitude of the electric field are the same at all points.

3. Torque: Torque is the rotational equivalent of force. It is the product of the force and the perpendicular distance from the pivot point to the line of action of the force.

4. Force: Force is a push or pull exerted on an object due to the interaction with another object.

Now, let us consider an electric dipole placed in a uniform electric field:

• Since the electric field is uniform, the force on the positive charge and the negative charge of the dipole will be equal in magnitude but opposite in direction.

• As a result, there will be no net force on the dipole, since the two forces cancel each other out.

• However, there will be a torque on the dipole due to the two equal and opposite forces acting at a distance from each other.

• The torque will cause the dipole to rotate until it aligns with the direction of the electric field.

• Once the dipole is aligned with the electric field, there will be no torque acting on it, and it will remain in that position.

Therefore, we can conclude that when an electric dipole is placed in a uniform electric field, it experiences a torque but no net force.

Stopping Potential When Frequency of Incident Radiation is Equal to Threshold Frequency

When the frequency of the incident radiation is equal to the threshold frequency, the stopping potential will be zero.

Explanation:

- The threshold frequency is the minimum frequency of the incident radiation required to emit electrons from a metal surface.
- The photoelectric effect states that electrons will be emitted from a metal surface when it is exposed to incident radiation of a frequency greater than or equal to the threshold frequency.
- The stopping potential is the minimum voltage required to stop the emitted electrons from reaching the anode in a photoelectric setup.
- When the frequency of the incident radiation is equal to the threshold frequency, the emitted electrons will have zero kinetic energy.
- Therefore, the stopping potential required to stop these electrons will be zero.
- This is because the electrons do not have enough energy to reach the anode, so no voltage is required to stop them.

Conclusion:

In conclusion, the stopping potential will be zero when the frequency of the incident radiation is equal to the threshold frequency. This is because the emitted electrons do not have enough energy to reach the anode, so no voltage is required to stop them.

MH (millihenry)
b)10 mH
c)100 mH
d)500 mH

We can use the formula for self-inductance:

L = Φ/I

where L is the self-inductance in henries, Φ is the magnetic flux in webers, and I is the current in amperes.

Converting the given magnetic flux to webers, we have:

Φ = 50 milliwebers = 0.05 webers

Substituting the values into the formula, we get:

L = Φ/I = 0.05 webers/5 A = 0.01 henries

Converting the result to millihenries, we have:

L = 0.01 H = 10 mH

Therefore, the answer is b) 10 mH.

To find the motional emf in the bar, we need to use Faraday's law of electromagnetic induction. According to this law, the emf induced in a conductor is equal to the rate of change of magnetic flux through the conductor.

1. Calculate the magnetic flux:
The magnetic flux through the bar can be calculated using the formula:
ϕ = BAcosθ

Given:
B = 50 T (magnetic field magnitude)
A = length × width = 2m × 1m = 2m² (area of the bar)
θ = angle between the magnetic field and the normal to the bar = 0° (since the bar is falling vertically)

Substituting the values, we get:
ϕ = (50 T)(2m²)(cos0°)
ϕ = 100 Tm²

2. Calculate the rate of change of magnetic flux:
Since the bar is falling, the rate of change of magnetic flux can be calculated by dividing the change in magnetic flux by the change in time:
dϕ/dt = Δϕ/Δt

Given:
Δϕ = change in magnetic flux = 100 Tm²
Δt = change in time = 40 m

Substituting the values, we get:
dϕ/dt = (100 Tm²)/(40 m)
dϕ/dt = 2.5 T/s

3. Calculate the motional emf:
Using Faraday's law, the motional emf can be calculated by multiplying the rate of change of magnetic flux by the number of turns in the bar. Since the bar is a single loop, the number of turns is 1:
ε = -N(dϕ/dt)

Given:
N = number of turns = 1
dϕ/dt = 2.5 T/s

Substituting the values, we get:
ε = -(1)(2.5 T/s)
ε = -2.5 V

Note: The negative sign indicates the direction of the induced current. Since the bar is falling, the induced current will flow in the opposite direction.

4. Convert the motional emf to a positive value:
Since the magnitude of the motional emf is always positive, we can convert it to a positive value by taking the absolute value:
|ε| = 2.5 V

Therefore, the motional emf in the bar when it has fallen 40 meters is 2.5 V. None of the given options match the correct answer.

Instantaneous Power Varies in Both Magnitude and Sign Over a Cycle

Explanation:

- Alternating current (AC) power is a type of electrical power where the direction and magnitude of the current flow changes periodically.
- The power consumed by an AC circuit is not constant but varies with time due to changes in voltage and current.
- Instantaneous power is the power consumed or delivered by a circuit at any instant of time and is given by the product of voltage and current at that instant.
- Instantaneous power is a function of time and varies continuously over a cycle.
- In an AC circuit, the voltage and current are sinusoidal and vary with time according to a sine wave.
- The product of voltage and current also varies with time and follows a wave-like pattern.
- The sign of instantaneous power depends on the phase relationship between voltage and current. If voltage and current are in phase, the power is positive, indicating that the circuit is consuming power. If voltage and current are out of phase, the power is negative, indicating that the circuit is delivering power.
- The magnitude of instantaneous power depends on the amplitude of voltage and current at that instant.
- Therefore, instantaneous power varies in both magnitude and sign over a cycle in an AC circuit.

Conclusion:

- Apparent power is the product of root mean square (RMS) voltage and current and is a constant value for a given circuit.
- Effective power or real power is the average power consumed by a circuit over a cycle and is a positive value.
- Average power is the time average of instantaneous power over a cycle and is equal to effective power.
- Instantaneous power is the power consumed or delivered by a circuit at any instant of time and varies in both magnitude and sign over a cycle.

There are different types of charges in physics, but assuming you are referring to electric charges, here are three examples:

1) A positively charged proton has a charge of +1.6 x 10^-19 Coulombs. This charge is fundamental to the structure of atoms, as it attracts negatively charged electrons to form the nucleus of an atom.

2) An negatively charged electron has a charge of -1.6 x 10^-19 Coulombs. Electrons are responsible for the flow of electricity and the behavior of materials in electrical circuits.

3) A charged balloon that has been rubbed with a woolen cloth can acquire a static electric charge due to the transfer of electrons from one surface to another. This type of charge is called triboelectric charging and can be observed in everyday life, such as when hair stands up after rubbing a balloon on it.

We can use the formula for magnetization:

M = NI/L

where N is the number of turns per unit length, I is the current, and L is the length of the solenoid.

First, we need to find the magnetic field inside the solenoid. We can use the formula:

B = μ0μrIN

where μ0 is the permeability of free space (4π x 10^-7 Tm/A), μr is the relative permeability of the material, I is the current, and N is the number of turns per unit length.

Plugging in the values, we get:

B = (4π x 10^-7 Tm/A)(200)(2 A)(200 turns/m) = 0.50 T

Now we can calculate the magnetization:

M = (200 turns/m)(2 A)/1 m = 400 A/m

Therefore, the magnetization of the material is 400 A/m.