VITEEE Physics Test - 5 - JEE MCQ

Photoelectric current depends on the intensity of incident light,
the number of photoelectrons emitted per second is directly proportional to the intensity of incident radiation.

Step-by-Step Solution:

Step 1: Understand the Concept

A charged particle moving in a magnetic field experiences a magnetic force. The force is given by:

Magnetic force (F) = q × v × B × sin(θ)

Where:

• q is the charge

• v is the velocity

• B is the magnetic induction (magnetic field strength)

• θ is the angle between the velocity vector and the magnetic field

If a particle is moving horizontally with uniform velocity, the net force acting on it must be zero. This means the magnetic force must be balanced by another force.

Since gravity is mentioned, and the magnetic field is uniform, the problem likely assumes that magnetic force is balancing the gravitational force, which pulls the particle downward.

So, we can set:

Magnetic force = Gravitational force

That is:

q × v × B = m × g

Now, solve for B:

B = (m × g) / (q × v)

Step 2: Plug in the Values

• m = 0.0006 kg

• g = 10 m/s²

• q = 25 × 10⁻⁹ C

• v = 1.2 × 10⁴ m/s

Now substitute:

B = (0.0006 × 10) / (25 × 10⁻⁹ × 1.2 × 10⁴)

B = 0.006 / (25 × 10⁻⁹ × 12000)

First calculate the denominator:

25 × 10⁻⁹ × 12000 = 3 × 10⁻⁴

Now divide:

B = 0.006 / 0.0003 = 20 Tesla

Final Answer:

20 T

Correct Option: (c)

Conclusion:

The magnetic field (B) must be 20 T for the magnetic force to balance the weight of the particle. The earlier solution that assumed the magnetic field was parallel to the velocity (thus giving zero force) did not consider the role of gravity. This was the mistake. In this problem, the magnetic field is perpendicular to the velocity, and it acts upward to balance gravity, allowing the particle to move horizontally with uniform velocity.

For electroplating a spoon, it is essential to place it in the correct position within the voltameter:

• The spoon should be positioned at the cathode, where it will receive the metal ions.
• At the anode, metal is oxidised and dissolved into the electrolyte.
• Placing the spoon anywhere else, such as in the middle or at arbitrary points, will not achieve the desired electroplating effect.

Optical fibres are a crucial technology for telecommunications, and understanding their characteristics is important.

• Low Transmission Loss: Optical fibres are known for having extremely low transmission losses, making them efficient for long-distance communication.
• Core and Cladding: They can feature a homogeneous core paired with suitable cladding, which helps in guiding light effectively.
• Graded Refractive Index: Optical fibres can also have a graded refractive index, allowing for improved performance in light transmission.
• Electromagnetic Interference: Unlike traditional copper cables, optical fibres are not susceptible to electromagnetic interference, providing a clearer signal.

The average power dissipated in a pure inductor of inductance L when an A.C. current is passing through it is:

The power dissipated in a pure inductor is calculated as follows:

• The current in an inductor creates a magnetic field that stores energy.
• This stored energy does not dissipate as heat, leading to no average power loss.
• As a result, the average power dissipated in a pure inductor is zero.

Let L₁ be the length and A₁ the area of crosssection of the original wire and L₂, A₂ be their values of the stretched wire
Since volume of the wire remains unchanged, we have

The solution requires an understanding of resistance in a cylindrical wire, which is influenced by its physical dimensions.

• Resistance (R) of a wire is calculated using the formula: R = ρ(L/A), where:
• ρ is the resistivity of the material.
• L is the length of the wire.
• A is the cross-sectional area.
• When the length of the wire is increased by 100%, the new length becomes 2L.
• The diameter's decrease leads to a change in the cross-sectional area, calculated as:
• If the diameter decreases, the area A = π(d/2)² also decreases.
• With both the increased length and decreased area, the new resistance can be determined:
• New Resistance R' = ρ(2L/A'), where A' is the new area.
• As a result, the resistance increases significantly due to the combined effects of lengthening and area reduction.

Therefore, the change in the resistance of the wire will be substantial, indicating a marked increase in overall resistance. The exact percentage can be evaluated based on the area reduced, often leading to a greater than 200% increase in resistance.

To calculate the electric current generated by an electron revolving in a circular path, follow these steps:

• First, determine the current (I) using the formula: I = n × q, where:
  • n is the number of revolutions per second (frequency),
  • q is the charge of the electron.
• Given that the frequency (n) is 5 x 10¹⁵ cycles/sec and the charge (q) is 1.6 x 10^-19 C, substitute these values into the formula:
• Calculate the current: I = (5 x 10¹⁵) × (1.6 x 10^-19).
• This results in: I = 8 x 10^-4 A or 0.8 mA.

The electric current in the circle is therefore 0.8 mA.

To find the temperature at which the resistance of a copper wire becomes three times its resistance at 0°C, we can use the formula for resistance change with temperature:

• R = R₀ (1 + αT)
• Where:
• R = resistance at temperature T
• R₀ = resistance at 0°C
• α = temperature coefficient of resistance (4 x 10^-3 per °C)
• T = temperature in °C
• Given that R = 3R₀, we can set up the equation:

3R₀ = R₀(1 + αT)

• Dividing both sides by R₀ gives:
• 3 = 1 + αT
• Now, isolate T:

3 - 1 = αT

2 = αT

T = 2/α

• Substituting α:
• T = 2 / (4 x 10^-3)
• T = 500°C

The temperature at which the resistance becomes three times its value at 0°C is 500°C.

To determine which particle exhibits the maximum de-Broglie's wavelength when moving at the same velocity, consider the following:

• De-Broglie's wavelength is inversely related to the momentum of a particle.
• Momentum is defined as the product of mass and velocity.
• Among the given particles, the mass varies:
• Neutrons have a mass of approximately 1 atomic mass unit (amu).
• Protons also have a mass near 1 amu.
• Beta particles (electrons) have a significantly smaller mass (around 0.0005 amu).
• Alpha particles, consisting of two protons and two neutrons, have a larger mass (approximately 4 amu).
• Since de-Broglie's wavelength increases as mass decreases, the particle with the smallest mass will have the largest wavelength.
• Thus, the β-particle (electron) will have the maximum de-Broglie's wavelength as it has the least mass.

The core of a transformer is laminated to:

• Reduce energy losses caused by eddy currents.

• Improve efficiency by limiting the flow of these currents.

• Enhance the overall performance of the transformer.

The mutual inductance of two coils depends on several factors:

• Relative position and orientation: The way the coils are positioned and aligned affects the magnetic coupling between them.
• Material properties: The type of materials used in the coils can influence their inductive behaviour.
• Changing currents: The rates at which the currents in both coils change play a crucial role in determining the mutual inductance.

However, the actual currents flowing through the coils do not directly affect the mutual inductance itself.

Which of the following waves have maximum wavelength?

The wave with the maximum wavelength among the options listed is:

• Radio waves

Here's a brief overview of each type of wave:

• X-rays: Have short wavelengths, typically in the range of 0.01 to 10 nanometres.
• IR-rays: Infrared rays have longer wavelengths than X-rays, ranging from about 700 nanometres to 1 millimetre.
• UV-rays: Ultraviolet rays fall between X-rays and visible light, with wavelengths from 10 to 400 nanometres.
• Radio waves: These have the longest wavelengths, ranging from 1 millimetre to thousands of kilometres.

Thus, among the listed options, radio waves possess the maximum wavelength.

The question involves calculating the total electric flux from a cube containing dipoles.

The total electric flux, denoted by Φ, through a closed surface can be determined using Gauss's Law, which states:

• The total electric flux through a closed surface is proportional to the net charge enclosed within that surface.
• For dipoles, which consist of equal and opposite charges, the net charge is zero.
• Since there are eight dipoles, each having a charge of magnitude e, the total enclosed charge remains zero.

As a result, the total electric flux coming out of the cube is:

• Φ = 0

The potential at the midpoint between two charges can be calculated as follows:

• The two charges are -10 C and +10 C, placed 10 cm apart.
• The midpoint is 5 cm from each charge.
• Electric potential (V) due to a point charge is given by the formula: V = k * (q / r), where:
  • k is Coulomb's constant.
  • q is the charge.
  • r is the distance from the charge.
• Calculating the potential from each charge at the midpoint:
  • For -10 C: V1 = k * (-10 C / 0.05 m)
  • For +10 C: V2 = k * (10 C / 0.05 m)
• Since the potentials due to each charge are equal in magnitude but opposite in sign, they will cancel each other out:
• Total potential = V1 + V2 = 0

The potential at the midpoint between the charges is zero.

To calculate the potential energy of the system:

• The system consists of two equal charges, q, separated by a distance of 2a, with a third charge of -2q positioned at the midpoint.
• The potential energy between two point charges is given by the formula: U = k * (q1 * q2) / r, where k is the Coulomb's constant, q1 and q2 are the charges, and r is the distance between them.
• Calculate the potential energy contributions from each pair of charges:
• Between the two charges q:

  U1 = k * (q * q) / (2a) = k * q² / (2a).

• Between one charge q and the charge -2q:

  U2 = k * (q * -2q) / (a) = -2k * q² / a.

• For the other charge q and the charge -2q:

  U3 = k * (q * -2q) / (a) = -2k * q² / a.

• Now, sum all contributions to find the total potential energy:

  U_total = U1 + U2 + U3 = k * q² / (2a) - 2k * q² / a - 2k * q² / a.

• This simplifies to:

  U_total = k * q² / (2a) - 4k * q² / a = -7k * q² / (2a).

• Therefore, the final result for the potential energy is:

  U_total = -7q² / (8π ∈₀ a).

When an uncharged capacitor is connected to a battery, energy is involved in the charging process. The following points outline the energy loss as heat:

• The total energy supplied by the battery is represented as Q V, where Q is the charge and V is the voltage.
• However, not all of this energy is stored in the capacitor; some is lost as heat due to resistance in the circuit.
• The energy actually stored in the capacitor is 1/2 Q V.
• Thus, the energy lost as heat during charging can be determined by the difference between the energy supplied and the energy stored.

In summary, the energy lost as heat is equal to:1/2 Q V.

An electric dipole in a non-uniform electric field experiences:

• Torque: The dipole will be subjected to a rotational force due to varying field strengths across its ends.
• Net Force: The dipole can also experience a net force because the electric field strength acting on each end is different.

In summary, an electric dipole in a non-uniform electric field will experience both a torque and a net force. This is a result of the differences in electric field strength affecting its positive and negative charges.

In the process of β⁺ decay, several key changes occur within the nucleus:

• Proton conversion: A proton in the nucleus is transformed into a neutron.
• Positron emission: This transformation releases a positron, which is the antimatter counterpart of an electron.
• Neutrino production: Alongside the positron, a neutrino is also emitted, which carries away some energy.
• Nuclear composition change: The atomic number of the element decreases by one, while the mass number remains unchanged.

This decay process is significant in understanding nuclear reactions and the stability of atomic nuclei.

A current-carrying loop in a uniform magnetic field will achieve equilibrium when it is inclined at:

• The loop will be in a stable position when its plane is aligned with the magnetic field.
• This occurs at an angle of zero degrees to the field direction.
• At this position, the torque acting on the loop is minimised, leading to stability.
• Other angles, such as 45 degrees or 90 degrees, will not provide the same stability.

Fast neutrons are generally more effective in inducing nuclear fission compared to other particles. Here's why:

• Neutron interactions: Neutrons can easily penetrate the nucleus without being repelled by the positive charge, unlike protons.
• Energy levels: Fast neutrons have sufficient energy to overcome the nuclear forces holding the nucleus together.
• Fission probability: Slow neutrons, or thermal neutrons, are also effective due to their ability to be captured by the nucleus, leading to fission.
• Protons: Fast and slow protons are less effective as they face repulsion from the nucleus, making fission less likely.

In conclusion, while both fast and slow neutrons can induce fission, fast neutrons generally have a higher probability of success.

To calculate the Q-value of the reaction:

• Identify the atomic masses involved in the reaction:
• Proton (p): 1.007825 u
• Lithium-7 (7Li): 7.016004 u
• Helium-4 (4He): 4.002603 u
• Write the reaction equation:

p + ⁷Li → 2⁴He

• Calculate the total mass of reactants:

Total mass = mass of p + mass of 7Li = 1.007825 u + 7.016004 u = 8.023829 u

• Calculate the total mass of products:

Total mass = 2 × mass of 4He = 2 × 4.002603 u = 8.005206 u

• Determine the mass difference:

Mass difference = mass of reactants - mass of products = 8.023829 u - 8.005206 u = 0.018623 u

• Convert mass difference to energy using the conversion factor:

Q-value (in MeV) = mass difference (u) × 931.5 MeV/u

Q-value = 0.018623 u × 931.5 MeV/u ≈ 17.35 MeV

The Q-value of the reaction is approximately 17.35 MeV.

To determine the threshold wavelength of a metal with a work function of 3 eV:

• The threshold wavelength can be calculated using the formula:
• λ = (hc) / φ
• Where:
• h = Planck's constant (6.626 x 10^-34 J·s)
• c = Speed of light (3 x 10⁸ m/s)
• φ = Work function in joules (1 eV = 1.6 x 10^-19 J)
• Convert work function to joules:
• 3 eV = 3 x 1.6 x 10^-19 J = 4.8 x 10^-19 J
• Substituting the values:
• λ = (6.626 x 10^-34 J·s * 3 x 10⁸ m/s) / (4.8 x 10^-19 J)
• Calculating gives:
• λ ≈ 4.13 x 10^-7 m or 4133 Å

Conclusion: The threshold wavelength is approximately 4133 Å.

To find the new wavelength of light in glass:

• The refractive index (n) of the glass is 1.5.
• The original wavelength of light in air is 6000 Å.
• The relationship between wavelength in air (λ₀) and wavelength in glass (λ) is given by:
• λ = λ₀ / n
• Substituting the values:
• λ = 6000 Å / 1.5
• This calculation results in:
• λ = 4000 Å

The wavelength of light in the glass is therefore 4000 Å.

The spectrum from a sodium vapour lamp is classified as an:

• Emission spectrum

The key characteristics of an emission spectrum include:

• It is produced when atoms or molecules emit light.
• The spectrum consists of bright lines or bands at specific wavelengths.
• Each element has a unique emission spectrum, aiding in its identification.

In contrast, other types of spectra include:

• Continuous spectrum: Displays a seamless range of wavelengths without gaps.
• Absorption spectrum: Shows dark lines where light has been absorbed by atoms or molecules.
• Band spectrum: Composed of closely spaced lines, often seen in molecular emissions.

The minimum potential difference required to turn a silicon transistor 'ON' is typically around:

• 0.6 to 0.7 V for the base-emitter junction.
• This voltage is necessary to overcome the barrier for conduction.
• Values may vary slightly depending on specific transistor designs.

The potential barrier in the depletion layer is caused by:

• Interaction between electrons and holes in the semiconductor material.
• Formation of a region devoid of mobile charge carriers.
• Presence of ions that create an electric field, establishing the barrier.

This barrier affects the movement of charge carriers and impacts the overall behaviour of the semiconductor device.

The intrinsic semiconductor behaves as an insulator at

• The temperature at which intrinsic semiconductors become insulators is typically close to absolute zero, around 0 K.

• At higher temperatures, more charge carriers are available, enabling conductivity.

• As the temperature decreases, the excitation of electrons becomes less likely, leading to insulating behaviour.