VITEEE Physics Test - 3 - JEE MCQ

In A.C., the resistance of a coil behaves differently compared to D.C.

• The resistance in A.C. is generally referred to as impedance, which includes both resistance and reactance.
• As frequency increases, the reactive component of impedance often increases, leading to a rise in overall impedance.
• Therefore, the total impedance in A.C. circuits is usually higher than the D.C. resistance alone.
• This means that the effective resistance will typically increase when switching from D.C. to A.C.

To determine the required potential difference for generating X-rays with a wavelength of at least 0.2 nanometres, follow these steps:

• Calculate the energy associated with the wavelength using the formula:

  E = h * c / λ

  where:
  • h = 6.63 x 10^-34 J·s (Planck's constant)
  • c = 3 x 10⁸ m/s (speed of light)
  • λ = 0.2 x 10^-9 m (wavelength in metres)
• Substituting the values:

  E = (6.63 x 10^-34) * (3 x 10⁸) / (0.2 x 10^-9)

• Calculate E:

  E = (6.63 x 3) / 0.2 x 10^-34 x 10⁸ x 10⁹

  E = 9.945 x 10^-14 J

• Relate energy to potential difference:

  E = e * V, where e = 1.6 x 10^-19 C (elementary charge) and V is potential difference.

• Rearranging gives:

  V = E / e

• Substituting E and e:

  V = (9.945 x 10^-14) / (1.6 x 10^-19)

• Calculate V:

  V = 62156.25 V or 62.16 kV

• Convert to kilovolts:

  To convert volts to kilovolts, divide by 1000: V = 62.16 kV.

Thus, the required potential difference is approximately 62.16 kV.

To find the work function of the metal, we start with the energy of the photon and the relationship between wavelength and energy.

• The initial energy of the photon is given as 2.5 eV.
• The energy of a photon can be calculated using the equation: E = h * c / λ, where h is Planck's constant and c is the speed of light.
• When the wavelength λ is decreased by 20%, the new wavelength becomes λ' = 0.8 * λ.
• The energy of the new photon is then E' = h * c / λ', which results in an increased energy.
• Given that the velocity of emitted electrons doubles, we can relate the kinetic energy of the electrons to their velocity: K.E. = 0.5 * m * V^2.
• When velocity doubles, the new kinetic energy becomes four times the original kinetic energy.
• Using the photoelectric equation, we have: E = W + K.E., where W is the work function.
• Let the initial kinetic energy be K.E. and the new kinetic energy be 4K.E..
• Setting up the equations: 2.5 eV = W + K.E. and E' = W + 4K.E. will help solve for W.

By substituting and rearranging the equations, we can solve for the work function W. The final value leads us to determine the work function of the metal.

Current flows in semiconductors through:

• Electrons: These are negatively charged particles that move through the semiconductor material, contributing to current flow.
• Holes: These represent the absence of electrons; they act as positive charge carriers and also facilitate current flow.
• In summary, current in semiconductors is carried by both electrons and holes.

To determine the heat produced when a charged capacitor discharges, follow these steps:

• Calculate the energy stored in the capacitor using the formula: Energy (E) = 0.5 × C × V², where C is the capacitance and V is the voltage.
• In this case: C = 2 μF (2 × 10⁻⁶ F) and V = 100 V.
• Plug in the values: E = 0.5 × (2 × 10⁻⁶) × (100)².
• Calculate the energy: E = 0.5 × (2 × 10⁻⁶) × 10000 = 0.01 J.
• This energy is converted to heat when the capacitor discharges through a wire.

Thus, the heat produced is 0.01 J.

To determine the terminal potential difference of the battery, follow these steps:

• The total resistance in the circuit is the sum of the internal resistance of the battery and the external resistance.
• Calculate the total resistance:
  • Internal resistance: 1 Ω
  • External resistance: 3 Ω
  • Total resistance = 1 Ω + 3 Ω = 4 Ω
• Using Ohm's Law, the current (I) flowing through the circuit can be calculated:
  • Battery voltage (V) = 4 V
  • Current, I = V / Total resistance = 4 V / 4 Ω = 1 A
• Now, find the voltage drop across the internal resistance of the battery:
  • Voltage drop = Current x Internal resistance = 1 A x 1 Ω = 1 V
• Finally, calculate the terminal potential difference:
  • Terminal potential difference = Battery voltage - Voltage drop
  • Terminal potential difference = 4 V - 1 V = 3 V

The terminal potential difference is 3 V.

The maximum current that flows through a fuse wire before it blows out varies with its radius.

The relationship between the maximum current and the radius can be summarised as follows:

• The maximum current is directly proportional to the radius of the fuse wire.
• This means that as the radius increases, the maximum current capacity also increases.
• Therefore, the formula can be expressed simply as the radius itself.

A non-ohmic resistor is a component that does not follow Ohm's Law, meaning its resistance varies with voltage and current.

Examples of non-ohmic resistors include:

• Diodes, which allow current to flow in one direction only and have a non-linear current-voltage relationship.
• Some types of thermistors, which change resistance significantly with temperature.

In contrast, materials like copper and aluminium are considered ohmic, as they maintain a constant resistance under varying conditions. Therefore, the correct choice demonstrates a clear distinction between ohmic and non-ohmic behaviour.

Optical fibres have several key characteristics that make them suitable for telecommunications:

• Low transmission loss: Optical fibres are known for their extremely low loss during signal transmission, allowing for long-distance communication without significant degradation.
• Core and cladding: They can have a homogeneous core paired with a suitable cladding to facilitate light transmission.
• Graded refractive index: Some optical fibres are designed with a graded refractive index, which helps in reducing signal distortion over long distances.
• Electromagnetic interference: Unlike traditional cables, optical fibres are not affected by electromagnetic interference from external sources.

When a proton is accelerated with a 1 volt potential difference, its kinetic energy can be determined as follows:

• The energy gained by a proton when accelerated through a potential difference is equal to the potential difference itself. Thus, for a 1 volt potential difference, the kinetic energy becomes:
• Kinetic Energy = 1 eV
• This is because the energy in electron volts (eV) directly correlates to the voltage applied.

Therefore, when a proton is accelerated through a 1 volt potential difference, its kinetic energy is 1 eV.

What is increased in a step-down transformer?

A step-down transformer is designed to reduce voltage while increasing current. Here’s how it works:

• Voltage reduction: The primary voltage is higher than the secondary voltage.
• Current increase: As voltage decreases, the current increases proportionally to maintain power levels.
• Power conservation: The transformer ensures that power input equals output, minus losses.

In summary, a step-down transformer increases current while decreasing voltage. This relationship is essential for efficient energy transfer.

To determine the coefficient of mutual induction for the secondary coil, we can use the relationship between induced e.m.f., mutual inductance, and the rate of change of current.

• The induced e.m.f. in the secondary coil is given as 5 mV.
• The rate of change of current is 20 A/s.
• Using the formula for mutual inductance: e = -M (di/dt), where:
  • e = induced e.m.f. (in volts)
  • M = mutual inductance (in henries)
  • di/dt = rate of change of current (in A/s)
• Rearranging gives: M = -e / (di/dt).
• Substituting the values:
  • M = 5 mV / 20 A/s
  • M = 0.005 V / 20 A/s
  • M = 0.00025 H or 0.25 mH.

The coefficient of mutual induction for the secondary coil is therefore 0.25 mH.

Which of the following rays are not electromagnetic waves?

Among the options listed, the following are not electromagnetic waves:

• β-rays - These are charged particles, specifically electrons, and do not fall under the category of electromagnetic radiation.

The other options are indeed forms of electromagnetic waves:

• X-rays - High-energy electromagnetic radiation used in medical imaging.
• γ-rays - Extremely high-frequency electromagnetic radiation emitted from radioactive decay.
• Heat rays - Typically refer to infrared radiation, which is also an electromagnetic wave.

To determine the ratio of electric flux through two concentric spheres:

• Consider a sphere S₁ with radius r₁ that encloses a total charge Q.
• There is another concentric sphere S₂ with radius r₂ where r₂ is greater than r₁.
• No additional charges exist in the region between the two spheres.
• According to Gauss's law, the electric flux Φ through a closed surface is directly proportional to the charge enclosed by the surface.
• For both spheres, the charge enclosed is the same, which is Q.
• Thus, the electric flux through both spheres is equal:
• Φ₁ = Φ₂ (where Φ₁ is the flux through S₁ and Φ₂ is the flux through S₂)
• Therefore, the ratio of electric flux through S₁ and S₂ is:
• Φ₁/Φ₂ = 1

An electron carries a charge of e coulombs and passes through a potential difference of V volts. Its energy in joules can be calculated as follows:

• The energy gained by the electron when it moves through a potential difference is given by the formula: Energy = Charge × Voltage.
• In this case, the charge of the electron is e coulombs and the potential difference is V volts.
• Therefore, the energy in joules is: Energy = e × V.

This means that the energy of the electron after passing through the potential difference is directly proportional to both its charge and the voltage it encounters.

When an electric dipole is placed in a uniform electric field, its potential energy depends on the orientation of the dipole moment relative to the field.

The potential energy (U) of a dipole in an electric field is given by:

• If the dipole moment (p) is aligned with the field (0°), the potential energy is at a minimum.
• If the dipole moment is opposite to the field (180°), the potential energy is at a maximum.
• At an angle of 90°, the potential energy is zero, indicating a neutral position.

To achieve minimum potential energy, the dipole moment should align with the electric field. Therefore, the angle between the dipole moment and the electric field should be:

• 0° for minimum potential energy.

C is inversely proportional to d, so:
C ∝ 1/d

This implies:
C₂ / C₁ = d₁ / d₂

Now using the given values:
10 / C₂ = 4 / 8

Solving for C2:
Cross-multiplying:
10 × 8 = 4 × C₂

So:
80 = 4 × C₂
Divide both sides by 4:
C₂ = 20 μF

The cyclotron is a type of particle accelerator that can accelerate particles with a significant mass, such as protons, deuterons, and alpha particles. However, it cannot effectively accelerate electrons. This limitation is due to:

• Mass: Electrons have a very low mass compared to other particles. This causes them to reach relativistic speeds quickly, making it difficult to maintain the synchronisation needed for acceleration in a cyclotron.
• Magnetic field: The cyclotron relies on a magnetic field to keep particles moving in a circular path. Because electrons are so light, they experience a much larger deflection in the magnetic field, complicating control.

As a result, other types of accelerators, like linear accelerators, are more suited for accelerating electrons.

The difference between the mass of a nucleus and the total mass of its constituents is known as the:

• Mass defect - This term describes the discrepancy between the mass of an atomic nucleus and the combined mass of its individual protons and neutrons.
• This difference occurs due to the binding energy that holds the nucleus together, which results in a lower mass.
• The mass defect is crucial for understanding nuclear stability and energy release in nuclear reactions.

To find the velocity of the body at point P, we can apply the principle of conservation of energy, which states that the total energy remains constant.

The energy at point P consists of kinetic and potential energy:

• Kinetic Energy (KE): KE = 0.5 * m * v²
• Potential Energy (PE): PE = q * V

At point Q, the total energy can be expressed as:

• KE_Q = 0.5 * m * v_Q²
• PE_Q = q * V_Q

Given:

• Mass (m) = 1 g = 0.001 kg
• Charge (q) = 10^-8 C
• Potential at P (V_P) = 600 V
• Potential at Q (V_Q) = 0 V
• Velocity at Q (v_Q) = 20 cm/s = 0.2 m/s

Calculate the energy at Q:

• KE_Q = 0.5 * 0.001 * (0.2)² = 0.00002 J
• PE_Q = 10^-8 * 0 = 0 J

At point Q, the total energy is:

• Total Energy_Q = KE_Q + PE_Q = 0.00002 J

Now, at point P:

• PE_P = 10^-8 * 600 = 0.000006 J

Using conservation of energy:

• Total Energy_P = KE_P + PE_P
• 0.00002 J = KE_P + 0.000006 J
• KE_P = 0.00002 - 0.000006 = 0.000014 J

Now, solve for the velocity at P:

• 0.000014 = 0.5 * 0.001 * v_P²
• v_P² = (0.000014 * 2)

Radioactive decay emits various particles, but not all. Here’s a breakdown:

• Protons: These are typically not emitted during decay processes.
• Neutrinos: Often released, especially in beta decay.
• Helium nuclei: Commonly emitted in alpha decay.
• Electrons: Usually emitted during beta decay.

In summary, among the options listed, protons are not emitted by radioactive substances during their decay.

A heavy nucleus at rest breaks into two fragments which fly off with velocities 8:1. The ratio of radii of the fragments is

To determine the ratio of the radii of the fragments after the nucleus breaks apart, we can use the principles of conservation of momentum and the relationship between mass, velocity, and radius in nuclear physics.

• The fragments move with velocities in the ratio of 8:1.
• According to conservation of momentum, the momentum before the break must equal the total momentum after.
• If we let the masses of the fragments be m1 and m2, their velocities are v1 and v2, respectively.
• The relationship between mass and radius is based on the volume of each fragment, which is proportional to the cube of the radius.
• Therefore, if we denote the radii of the fragments as r1 and r2, we can express the masses in terms of the radii: m1 ∝ r1³ and m2 ∝ r2³.
• Using the given velocity ratio (8:1) and the relationship of mass and radius, we can derive the ratio of the radii.

From this analysis, we find that the ratio of the radii is 1:2, which indicates that one fragment is larger than the other in a specific ratio. The final answer can be determined through calculation based on the ratios of mass and velocity. Thus, we conclude that:

• The ratio of the radii of the fragments is 1:2.

Stopping potential for photoelectrons

The stopping potential in the photoelectric effect is influenced by several factors:

• Frequency of light: It plays a crucial role. Higher frequencies can yield photoelectrons with greater energy.
• Cathode material: The specific material affects the work function, thus influencing the stopping potential.
• Intensity of light: This does not directly impact the stopping potential, but rather the number of emitted electrons.

In summary, the stopping potential is determined by both the frequency of the incident light and the nature of the cathode material.

The photoelectric effect involves the emission of electrons from a metallic surface when it is exposed to light photons. When photons of energy hv strike a surface with a work function of hv₀, electrons are released.

• The kinetic energy of the emitted electrons can vary.
• The maximum kinetic energy of the most energetic electrons is given by the equation hv - hv₀.
• Not all emitted electrons will have the same kinetic energy; they exhibit a distribution.
• The most energetic electrons will have their kinetic energy at the maximum calculated value, hv - hv₀.
• The energy of the photons must exceed the work function for electrons to be emitted.

The magnifying power of a simple microscope, M, is expressed in various formulas.

• M = 1 + F/D: This equation relates the magnifying power to the focal length (F) and the distance (D) from the lens to the object.
• M = ν/μ: This formula uses the near point (ν) and the distance of distinct vision (μ), representing magnification in terms of vision.
• M = μ/ν: This is another variation, which inversely relates the near point and distance of distinct vision.
• M = 1 + D/F: This equation also connects the magnification with the distance (D) and focal length (F).

Each formula provides insight into how a simple microscope enhances the size of an object viewed through it.

The region near the junction of a p-n diode where there are no charge carriers is called

• Barrier: This term refers to the potential energy barrier formed at the junction.
• Depletion layer: This is the correct term for the area devoid of charge carriers.
• Base: This is not applicable in this context.
• None: This option does not provide a valid answer.

In a properly biased transistor:

• The emitter-base depletion layer is typically small.
• In contrast, the base-collector depletion layer is generally large.
• This is due to the different doping levels and the applied biasing conditions.
• The small emitter-base depletion layer allows for efficient charge injection from the emitter into the base.
• The large base-collector depletion layer helps in maintaining the transistor's active region for amplification.

P-type semiconductors contain a deficiency of electrons, which creates "holes" that can carry positive charge. Here are some key points about them:

• P-type semiconductors are formed by adding acceptor impurities to a pure semiconductor.
• This process results in an absence of electrons, leading to positively charged holes.
• The presence of these holes allows for the conduction of electricity.
• P-type materials are essential in various electronic devices, particularly in diodes and transistors.