VITEEE: Subject Wise and Full Length VITEEE Physics Test - 7 Free Online

To determine the value of inductive reactance in an A.C. circuit with a given phase angle:

• The phase angle between voltage and current is given as 45 degrees.
• The relationship between resistance (R), inductive reactance (X_L), and phase angle (φ) in a series circuit can be described using the tangent function:

tan(φ) = X_L / R

• For a phase angle of 45 degrees, tan(45°) = 1.
• This leads to the equation: X_L = R.

Thus, the value of inductive reactance is equal to the resistance:

• X_L = R

To determine the velocity of electrons with no deflection:

• The condition for no deflection in an electron beam is that the electric force equals the magnetic force.
• Electric force (F_E) is given by F_E = qE, where q is the charge of an electron and E is the electric field strength.
• Magnetic force (F_B) is given by F_B = qvB, where v is the velocity of the electrons and B is the magnetic field strength.
• Setting the two forces equal gives: qE = qvB.
• Cancel out q (the charge) from both sides, resulting in: E = vB.
• Rearranging this formula, we find: v = E/B.
• Substituting the values:
• E = 3 x 10³ V m^-1
• B = 5 x 10^-4 wbm^-2
• Calculating the velocity: v = (3 x 10³) / (5 x 10^-4) = 6 x 10⁶ m s^-1.

Telecommunication through optical fibres has several key characteristics:

• Low transmission loss: Optical fibres are known for their minimal loss of signal over long distances.
• Core and cladding: They can have a homogeneous core paired with suitable cladding to guide light effectively.
• Graded refractive index: Some optical fibres utilise a graded refractive index to enhance performance.
• Electromagnetic interference: Unlike traditional cables, optical fibres are not affected by electromagnetic interference from external sources.

Therefore, the statement regarding their susceptibility to electromagnetic interference is incorrect.

On removing the oil, the capacitance of the capacitor will become.
C ′ = C/K = C/2

The quantity of electricity required to liberate one gram equivalent of an element is:

• 96500 coulombs is the correct amount of electricity.
• This value is known as one Faraday, named after the scientist Michael Faraday.
• In practical terms, it indicates the total charge needed to transfer one mole of electrons.
• The equivalent of one gram of an element corresponds to its valency in electrochemical reactions.

To find the initial resistivity of the material, we can use the formula that relates resistivity change to temperature change:

• The formula is: Δρ = ρ₀ × α × ΔT
• Where:
  • Δρ is the change in resistivity (2 x 10^-8 ohm-meter)
  • ρ₀ is the initial resistivity (unknown)
  • α is the temperature coefficient of resistivity (0.0004/K)
  • ΔT is the change in temperature (50°C)
• Substituting the known values into the formula:
• 2 x 10^-8 = ρ₀ × 0.0004 × 50
• This simplifies to:
• 2 x 10^-8 = ρ₀ × 0.02
• Now, to isolate ρ₀, divide both sides by 0.02:
• ρ₀ = (2 x 10^-8) / 0.02
• Calculating this gives:
• ρ₀ = 1 x 10^-6 ohm-meter or 100 x 10^-8 ohm-meter

The initial resistivity of the material is therefore 100 x 10^-8 ohm-meter.

To determine the appropriate range and resistance combination for a microammeter, consider the following:

• The microammeter has a resistance of 100 Ω and a full-scale current of 50 μA.
• When used as a voltmeter, the total voltage range can be calculated using Ohm's law (V = I × R).
• For a 50 V range, the required series resistance is calculated as follows:
• V = 50 V
• I = 50 μA = 0.00005 A
• R_total = V / I = 50 V / 0.00005 A = 1,000,000 Ω
• Resistance needed = R_total - 100 Ω = 999,900 Ω (not feasible)
• For a 10 V range:
• V = 10 V
• I = 50 μA
• R_total = 10 V / 0.00005 A = 200,000 Ω
• Resistance needed = 200,000 Ω - 100 Ω = 199,900 Ω (not feasible)
• For a 10 mA range, the required parallel resistance can be calculated:
• I = 10 mA = 0.01 A
• R_total = V / I = 0.05 V / 0.01 A = 5 Ω
• Resistance to be added in parallel = 1 Ω or 0.1 Ω (feasible)

The resistance of a wire is determined by its length, cross-sectional area, and the material it is made from. The formula for resistance (R) is given by:

• 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.

In this scenario:

• The length of the wire is doubled: L' = 2L
• The area is halved: A' = A/2

Substituting these values into the resistance formula:

• R' = ρ(2L/(A/2))
• This simplifies to R' = ρ(4L/A)

Hence, the new resistance is:

• R' = 4R

This indicates that the resistance will increase by a factor of four. Therefore, the correct conclusion is:

• Resistance will increase 4 times.

The expected energy of electrons at absolute zero is referred to as:

• Fermi energy - This is the maximum energy level that electrons can occupy at absolute zero temperature.
• Emission energy - This term relates to energy released when an electron transitions between energy states.
• Work function - This is the minimum energy required to remove an electron from a solid.
• Potential energy - This refers to stored energy based on an object's position or state.

The correct term for the expected energy of electrons at zero Kelvin is the Fermi energy.

To calculate the efficiency of the transformer, follow these steps:

• Determine the output power: The lamp consumes 100 W.
• Calculate the input power:
  • Input voltage = 220 V
  • Main current = 0.5 A
  • Input power = Voltage × Current = 220 V × 0.5 A = 110 W
• Calculate efficiency:
  • Efficiency = (Output Power / Input Power) × 100
  • Efficiency = (100 W / 110 W) × 100 ≈ 90.91%

The transformer efficiency is approximately 90%.

The magnetic field inside a long solenoid is influenced by several factors. However, it is important to note:

• The field strength is directly proportional to the number of turns per unit length of the solenoid.
• The field also depends on the current flowing through the solenoid.
• Conversely, the radius of the solenoid does not affect the strength of the magnetic field inside it.

In summary, while the number of turns and the current are crucial, the radius of the solenoid is not a determining factor for the field strength produced within it.

The radiation types are measured by their wavelengths, with shorter wavelengths corresponding to higher energy.

Among the listed types of radiation:

• X-rays have relatively short wavelengths, typically ranging from 0.01 to 10 nanometres.
• γ-rays (gamma rays) possess even shorter wavelengths, generally less than 0.01 nanometres.
• β-rays (beta rays), which are electrons emitted from radioactive decay, do not have wavelengths in the same sense as electromagnetic waves.
• α-rays (alpha rays) consist of helium nuclei and similarly do not have a defined wavelength like electromagnetic radiation.

In conclusion, γ-rays have the shortest wavelength among these types of radiation.

The potential at any point due to a charge is given by the formula:

• V = k * (Q/r), where:
• V is the electric potential,
• k is Coulomb's constant,
• Q is the charge, and
• r is the distance from the charge to the point of interest.

In this scenario:

• The two charges are -10 C and +10 C.
• They are separated by 10 cm, so the distance from the centre to each charge is 5 cm.

Calculating the potential at the centre:

• Potential due to the -10 C charge:
• V1 = k * (-10 C) / 0.05 m
• Potential due to the +10 C charge:
• V2 = k * (10 C) / 0.05 m

Adding these two potentials:

• V = V1 + V2
• Since V1 is negative and V2 is positive, they will cancel each other out.

Thus, the total potential at the centre is:

• V = 0 V

However, since one charge is negative and the other is positive, the potential can also be considered as undefined due to the infinite potential at the location of the charges themselves.

The potential of two conducting spheres with equal charges varies based on their size.

• The potential (\(V\)) of a conducting sphere is determined by the formula: V = kQ/r, where Q is the charge, r is the radius, and k is a constant.
• Since both spheres have the same charge, the potential depends on their radii.
• The smaller sphere has a smaller radius, resulting in a higher potential.
• Conversely, the larger sphere has a lower potential due to its greater radius.

Thus, the potential is greater on the smaller sphere.

The speed of the charged particle depends on its interactions with the electric and magnetic fields.

• The particle experiences a force due to the electric field in the y-direction.
• It also feels a force from the magnetic field in the x-direction as it moves.
• The combined effects of these forces influence the particle's speed.
• The speed will vary based on the particle's position, specifically in the x and y dimensions.
• The z-coordinate does not directly affect the speed in this scenario.

Thus, the speed depends on both the x and y coordinates, but not on z.

To find the new capacitance of the capacitor:

• Initially, the capacitance \(C_1\) is 2 μF with a separation of 0.4 cm.
• When the separation is halved, it becomes 0.2 cm.
• The capacitance of a parallel plate capacitor is given by the formula:
• C = ε₀ * A / d, where:
• ε₀ is the permittivity of free space,
• A is the area of the plates, and
• d is the separation between the plates.
• When the separation \(d\) is halved, the capacitance doubles:
• New capacitance without dielectric \(C_2 = 2 * C_1 = 2 * 2 μF = 4 μF\).
• Now, the capacitor is filled with a dielectric material of value 2.8:
• The capacitance increases by the dielectric constant:
• C_new = C_2 * K, where K is the dielectric constant.
• This gives:
• C_new = 4 μF * 2.8 = 11.2 μF.

The final capacitance of the capacitor filled with the dielectric material is 11.2 μF.

The path of a particle moving under a fixed force can be described as follows:

• The particle will move in a straight line if the force remains constant in both magnitude and direction.
• If the force varies in direction or magnitude, the path could change to other forms such as a circle, parabola, or ellipse.
• In the absence of any other forces acting on the particle, a constant force will always yield a linear trajectory.

In the process of radioactive decay, β-particles are negatively charged particles emitted from the nucleus.

• These particles are primarily electrons generated through the transformation of neutrons.
• During this decay, a neutron converts into a proton, releasing an electron known as a β-particle.
• This process is a natural way for unstable nuclei to achieve stability.
• It is not related to electrons found in atomic orbits or produced by collisions.

To calculate the Q-value of the reaction p + ⁷Li → ⁴He + ⁴He:

• First, identify the atomic masses:
  • 1H: 1.007825 u
  • 7Li: 7.016004 u
  • 4He: 4.002603 u
• Calculate the total mass of the reactants:
  • Mass of reactants = mass of p + mass of ⁷Li
  • Mass of reactants = 1.007825 u + 7.016004 u = 8.023829 u
• Calculate the total mass of the products:
  • Mass of products = 2 × mass of ⁴He
  • Mass of products = 2 × 4.002603 u = 8.005206 u
• Determine the mass defect:
  • Mass defect = Mass of reactants - Mass of products
  • Mass defect = 8.023829 u - 8.005206 u = 0.018623 u
• Convert the mass defect to energy (Q-value):
  • Q-value = mass defect × 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.

The distance of closest approach for an alpha nucleus bombarding a heavy nuclear target is influenced by several factors:

• The kinetic energy of the alpha nucleus, represented as 1/2 mv², indicates that the energy depends on the mass m and the velocity v.
• The electric potential energy at the closest approach is determined by the charge of the target nucleus, noted as Ze.
• As the alpha particle approaches the target, it experiences a repulsive force due to the like charges. The distance of closest approach can be derived from the balance of kinetic and potential energy.
• This distance is inversely proportional to the charge of the target nucleus (1/Ze) and involves the velocity of the alpha particle.

In summary, the distance of closest approach is mainly influenced by the charge of the nucleus and the kinetic energy of the alpha particle. Thus, it is proportional to the terms mentioned above.

To determine the stopping potential:

• The energy of the incoming light is 2 eV.
• The work function of the metal is 0.6 eV.
• The energy available to eject electrons is calculated as:
• Energy available = Energy of light - Work function
• Energy available = 2 eV - 0.6 eV = 1.4 eV
• The stopping potential corresponds to the maximum kinetic energy of the emitted electrons, which is equal to the energy available:
• Thus, stopping potential = 1.4 V.

The longest wavelength of light that can cause photoelectron emission from a substance with a work function of 4.0 eV can be calculated using the energy-wavelength relationship.

• The energy of a photon is given by the equation: E = hc/λ, where:
• E is the energy in electron volts (eV).
• h is Planck's constant (approximately 4.14 x 10^-15 eV·s).
• c is the speed of light (approximately 3.0 x 10⁸ m/s).
• λ is the wavelength in metres.
• To find the longest wavelength that can cause emission:
• Set E equal to 4.0 eV.
• Rearrange to find λ: λ = hc/E.
• Substituting values gives λ ≈ 310 nm.
• Thus, the longest wavelength for photoelectron emission is approximately 310 nm.

Correct option: A - 100 Cd.

For a point source, surface illumination E varies as E ∝ I/r², where I is luminous intensity and r is the distance from the source.

The required exposure (illumination × time) for the same photographic print is constant, so E₁ × t₁ = E₂ × t₂.

Compute E₁ = 50/1² = 50 and E₂ = I/2² = I/4.

Substitute into the exposure equality: 50 × 10 = (I/4) × 20.

Simplify: 500 = 5I, hence I = 100 Cd.

Therefore the correct answer is A (100 Cd).

To determine the number of images formed by two inclined mirrors:

• Use the formula: n = 360° / θ - 1, where θ is the angle between the mirrors.
• Here, θ = 72°.
• Calculate the value: n = 360° / 72° - 1.
• This simplifies to n = 5 - 1.
• Thus, the total number of images formed is 4.

In conclusion, the number of images created by the mirrors is 4.

The minimum potential difference between the base and emitter needed to turn on a silicon transistor is typically around:

• 0.6 V to 0.7 V for standard operation.
• This range allows the transistor to enter the active region.
• Higher voltages may be used, but they are generally not necessary.
• Ensure that the transistor's specifications are checked for precise values.

The energy gap in a metal is:

• The energy gap is generally zero in metals.
• This means that electrons can move freely, allowing for high electrical conductivity.
• In contrast to insulators and semiconductors, metals do not have a significant energy gap.
• The properties of the metal can influence its behaviour, but typically, the energy gap remains minimal.

The minimum frequency of radiation that can be absorbed by a material with a band gap of 2.0 eV can be calculated using the formula:

• Frequency (ν) = Energy (E) / Planck's constant (h)
• Planck's constant (h) is approximately 6.626 x 10^-34 J·s.
• First, convert the energy from electron volts to joules:
• 2.0 eV = 2.0 x 1.602 x 10^-19 J = 3.204 x 10^-19 J.
• Now, calculate the frequency:
• ν = (3.204 x 10^-19 J) / (6.626 x 10^-34 J·s)
• This results in a frequency of approximately 4.83 x 10¹⁴ Hz.
• Therefore, the minimum frequency of the radiation that can be absorbed is nearly 5 x 10¹⁴ Hz.

In a P-type semiconductor, electrical conduction occurs primarily through:

• Holes: These are the primary charge carriers in P-type materials.
• Electrons: Although present, their number is significantly lower compared to holes.
• Charge Carrier Dynamics: The conduction process relies on the movement of holes, which are created when electrons are absent.

In summary, P-type semiconductors conduct electricity mainly because of a large number of holes, with electrons playing a minor role. The behaviour of these charge carriers is crucial for the operation of semiconductor devices.

In intrinsic semiconductors at room temperature, the number of electrons and holes is:

• The number of electrons is equal to the number of holes.
• This equality occurs because each electron that gains enough energy to become free leaves behind a hole.
• Both carriers are generated from the same source, resulting in a balance.
• Thus, in an intrinsic semiconductor, the population of electrons and holes is equal.

The p-n junction diode is primarily used as:

• Rectifier: Converts alternating current (AC) to direct current (DC).
• Amplifier: Can be used in specific configurations to boost signals.
• Oscillator: Can generate oscillating signals in certain circuits.
• Modulator: Can modulate signals in communication systems.