VITEEE: Subject Wise and Full Length VITEEE Physics Test - 8 Free Online

V = 50 volt; I = 12 ampere
Total power consumed = 50x12 W = 600 W
Since efficiency of motor is 30%, therefore, 70% of power is lost in Joule heating.
If Q is the heat produced, then
Q = 70% of 600 W or Q = 70/100 x 600 W = 420 W
Also, Q = I²R
∴ 420 = (12)²R or R = 420/(12 x 12(    Ω = 2.92 Ω

To determine the ratio of the maximum speeds of electrons emitted by photons with energies of 1 eV and 2.5 eV, we need to follow these steps:

• Calculate the energy available for the electrons after overcoming the work function of 0.5 eV.
• For the 1 eV photon:
  • Net energy = 1 eV - 0.5 eV = 0.5 eV
  • Using the kinetic energy formula: K.E. = 0.5 * m * v², we find v.
• For the 2.5 eV photon:
  • Net energy = 2.5 eV - 0.5 eV = 2.0 eV
  • Again using the kinetic energy formula, we find the speed.
• Calculate the maximum speeds of the electrons:
  • For 1 eV: v₁ ∝ √(0.5)
  • For 2.5 eV: v₂ ∝ √(2.0)
• Now, find the ratio of the speeds:
  • Speed ratio = v₂ / v₁ = √(2.0) / √(0.5) = √(4) = 2

The ratio of the maximum speeds of the emitted electrons is therefore 2 : 1.

To determine the new resistance after melting and recasting a wire:

• The original resistance of the wire is denoted as R.
• When the wire is melted and recasted to half its length, the new length becomes l/2, where l is the original length.
• Resistance is calculated using the formula: R = ρl/A, where ρ is the resistivity and A is the cross-sectional area.
• Since the volume of the wire remains constant, when the length is halved, the cross-sectional area must double to maintain the same volume. Therefore, A' = 2A.
• The new resistance, R', can be expressed as: R' = ρ(l/2)/(2A).
• This simplifies to: R' = ρl/(4A) = R/4.

Hence, the new resistance of the wire is R/4.

The rate at which the weights of the cathodes in the copper and silver voltmeters increase is influenced by several factors related to the materials used.

• The rate of weight increase is directly proportional to the atomic mass of each metal.
• The relationship is based on the principle that heavier atoms deposit more mass during electrolysis.
• Therefore, comparing copper and silver, the weight gain will depend on their respective atomic masses.

The energy per unit volume for a capacitor can be expressed using different formulas. Here are the relevant points:

• The energy density, or energy per unit volume, is important for understanding the capacitor's performance.
• For a capacitor with area A and separation d, kept at a potential difference V, the energy density can be derived from the capacitor's characteristics.
• Key formulas for energy density include:
• 1/2 ε₀ V²/d²
• 1/4 ε₀ V² d²
• 1/2 CV²
• Q²/2C
• Each formula highlights different aspects of a capacitor's energy storage capabilities.

To determine the different combinations of three equal resistors, consider the following:

• Series Connection: All three resistors connected in a single path.
• Parallel Connection: All three resistors connected alongside each other.
• Two in Series, One in Parallel: Two resistors connected in series, then connected in parallel with the third.
• Two in Parallel, One in Series: Two resistors connected in parallel, then connected in series with the third.
• Single Resistor Configurations: Each resistor can also be used individually.

This results in a total of five distinct configurations. The combinations reflect different ways to arrange the resistors to achieve varying total resistance values.

To determine the kinetic energy of a proton accelerated by a potential difference of 1 kV:

• The kinetic energy (KE) gained by a charged particle is calculated using the formula: KE = qV, where q is the charge and V is the potential difference.
• The charge of a proton is the same as that of an electron, approximately 1.6 x 10^-19 C.
• For a potential difference of 1 kV, the energy gained by the proton is:
• KE = (1.6 x 10^-19 C) x (1000 V) = 1.6 x 10^-16 J
• To convert this energy into electronvolts (eV), we note that 1 eV = 1.6 x 10^-19 J.
• Thus, the kinetic energy in eV is:
• KE = 1.6 x 10^-16 J / (1.6 x 10^-19 J/eV) = 1000 eV = 1 keV

Therefore, the kinetic energy of the proton is 1 keV.

In an L-R circuit, the time constant is defined as the time it takes for the current to rise from zero to approximately 63% of its maximum value.

• The time constant is represented by the symbol τ.
• It is calculated using the formula: τ = L/R, where L is inductance and R is resistance.
• At one time constant (τ), the current reaches about 0.63 I₀, where I₀ is the maximum current.
• This growth reflects the exponential nature of current increase in an L-R circuit.

Lenz's law arises from the principle of conservation of energy. This law states that the direction of an induced current is such that it opposes the change in magnetic flux that produced it. Here are key points to understand:

• Opposition to change: The induced current will always act to counteract the change in the magnetic environment.
• Energy conservation: The law ensures energy is conserved in electromagnetic systems, preventing the creation of energy from nothing.
• Practical examples: Applications can be seen in electric generators and inductors, where changes in magnetic fields induce currents.

This principle illustrates the fundamental link between electromagnetism and energy conservation.

To find the potential difference, we can use the formula for electrical work:

• W = Work done
• Q = Charge
• V = Potential difference

The relationship between these is given by:

W = QV

From this, we can rearrange the formula to find V:

V = W/Q

Substituting the values:

• W = 2 joules
• Q = 20 coulombs

Now, we calculate:

V = 2/20 = 0.1 volts

To find the electric potential at point P due to an electric dipole, consider the following:

• The dipole consists of two equal charges, q, of opposite signs, separated by a distance of 2a.
• The dipole moment, p, is defined as p = q × 2a.
• Point P is located at a distance r from the dipole's centre, with the line connecting the dipole to point P making an angle θ with the dipole's axis.
• The electric potential (V) at point P due to the dipole can be expressed using the formula:

V = p cos θ / (4 π ε₀ r²)

• This formula indicates that the potential decreases with the square of the distance from the dipole.
• The term cos θ shows that the potential varies depending on the angle relative to the dipole's axis.

The capacitance of a parallel plate capacitor increases to 4/3 of its original value when a dielectric slab of thickness t = d/2 is inserted between the plates. We need to find the dielectric constant of the slab.

To determine the dielectric constant, follow these steps:

• Let the original capacitance be C₀ and the new capacitance, after inserting the dielectric, be C.
• The new capacitance is given by C = 4/3 C₀.
• The capacitance of a parallel plate capacitor without a dielectric is calculated using the formula:
• C₀ = 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.
• With the dielectric inserted, the capacitance becomes:
• C = (ε₀ A)/(d - t) + (ε_r ε₀ A)/t, where:
• ε_r is the dielectric constant,
• t = d/2
• Substituting t into the formula gives:
• C = (ε₀ A)/(d - d/2) + (ε_r ε₀ A)/(d/2).
• This simplifies to:
• C = (2ε₀ A/d) + (2ε_r ε₀ A/d).
• Factoring out common terms results in:
• C = (2ε₀ A/d)(1 + ε_r).
• Setting this equal to 4/3 C₀ allows us to solve for ε_r.
• Through calculations, it can be determined that:
• ε_r = 2.

B - 8 J

Let the capacitance of each capacitor be C. For five equal capacitors in series, C_series = C/5.

Given C_series = 4 μF, so C = 5 × 4 μF = 20 μF.

When connected in parallel, total capacitance is C_total = 5 × C = 100 μF.

The energy stored by a capacitor is given by U = 1/2 C V².

Substituting C_total = 100 × 10^-6 F and V = 400 V:

U = 1/2 × 100 × 10^-6 × (400)² = 0.5 × 100 × 10^-6 × 160000 = 8 J

Therefore the total energy stored is 8 J, so option B is correct.

To find the point where the electric potential is zero between two charges:

• Let the distance from the +4 μC charge to the point where potential is zero be x.
• The distance from the -6 μC charge to this point will be (20 - x) cm.
• The electric potential (V) due to a point charge is given by the formula:

V = k * (q/r), where k is the Coulomb's constant, q is the charge, and r is the distance from the charge.

• For the +4 μC charge, the potential is V1 = k * (4 × 10^-6) / x.
• For the -6 μC charge, the potential is V2 = k * (-6 × 10^-6) / (20 - x).

Setting the total potential to zero:

• V1 + V2 = 0
• k * (4 × 10^-6) / x + k * (-6 × 10^-6) / (20 - x) = 0
• Canceling k and rearranging gives:
• 4/(20 - x) = 6/x

Cross-multiplying leads to:

• 4x = 6(20 - x)
• 4x = 120 - 6x
• 10x = 120
• x = 12 cm.

The point where the electric potential is zero is 12 cm from the +4 μC charge.

2.5 mA
Potential gradient in the wire
= 2 2 = 1V/m
P.D. across a balancing length of 5 mm
= 5 x 10^-3 x 1
= 5 x 10^-3 V
∴ Current in the second circuit
= 5 x 10^-3/2 = 2.5 x 10^-3 A
= 25 x 10^-4 A

In β ⁺ decay, the following changes occur within the nucleus:

• Positron emission: A proton is transformed into a neutron, resulting in the release of a positron.
• Neutrino production: Alongside the positron, a neutrino is also emitted.
• Nuclear composition: The atomic number decreases by one, while the mass number remains unchanged.
• Energy release: This decay process releases energy, contributing to the overall stability of the nucleus.

When a uranium atom undergoes fission, it splits into two or more smaller nuclei along with additional particles. This process leads to a change in mass. Here's what happens:

• The total weight of the resulting materials is less than that of the original uranium atom.
• This mass loss is due to the conversion of some mass into energy, as described by Einstein's equation, E=mc².
• The energy released during fission is substantial, contributing to the overall efficiency of nuclear reactions.

In summary, the mass of the fission products is always less than the mass of the original uranium atom, confirming that mass is converted to energy during the reaction.

The stopping potential of photoelectrons varies depending on the type of radiation used.

• Ultraviolet radiation has a higher energy compared to infrared radiation.
• Higher energy results in a greater maximum kinetic energy for photoelectrons, leading to a higher stopping potential.
• In contrast, infrared radiation has lower energy, producing photoelectrons with less kinetic energy.
• This means the stopping potential will be lower when using infrared.

Therefore, the stopping potential is more when ultraviolet radiation is used compared to infrared.

The distance of closest approach for an alpha nucleus can be understood through the principles of electrostatics and kinetic energy.

• The alpha nucleus possesses kinetic energy expressed as 1/2 mv².
• As it approaches a heavy nuclear target with charge Ze, the kinetic energy will convert into potential energy due to the electrostatic force between the charges.
• At the point of closest approach, the kinetic energy is fully converted to potential energy:
• Potential energy at closest approach = k * Ze * (1/r), where r is the distance of closest approach and k is Coulomb's constant.
• Thus, we can derive that the distance of closest approach, r, is proportional to:
• 1. The kinetic energy of the alpha particle (which depends on v²) and
• 2. The inverse of the nuclear charge Ze.
• Therefore, the distance of closest approach is inversely related to Ze and directly related to 1/v².

Correct option: B. I

For parallel sunlight, the power collected by a lens equals the solar irradiance multiplied by the area of the lens; for lenses of the same diameter this collected power is the same regardless of focal length.

The lens forms an image of the Sun at its focal plane. The linear size of that image is proportional to the focal length (f), so the image area ∝ f². Thus the irradiance at the image (power per unit area) varies as 1/f².

Halving the focal length from 30 cm to 15 cm reduces the image area by a factor (30/15)² = 4, so the image irradiance increases by a factor 4. However, the total power on the image = irradiance × image area remains equal to the power collected by the lens, which is unchanged for equal lens diameters.

The photoelectric current is proportional to the total number of incident photons per second (i.e., the total power absorbed) provided the entire image falls on the active metal surface. Since the collected power is the same, the photoelectric current remains I.

If the metal surface were smaller than the solar image so that only part of the collected light is intercepted, the current could change; under the usual assumption that the detector intercepts the whole image, the current is unchanged.

To calculate the total luminous flux emitted from a unidirectional bulb, follow these steps:

• Luminous intensity is measured in candela (cd), which represents the power emitted by a light source in a particular direction.
• The formula for total luminous flux (in lumens) is: Flux (lumens) = Luminous Intensity (candela) × Solid Angle (steradians).
• For a unidirectional bulb, the solid angle is typically considered to be 2π steradians.
• Thus, total luminous flux can be calculated as: 100 candela × 2π steradians.
• Calculating gives: 100 × 6.2832 ≈ 628.32 lumens.

This demonstrates how to derive the luminous flux from a bulb with a given intensity. The values may vary based on specific configurations and assumptions regarding the emission pattern.

A concave lens with a focal length of 20 cm, positioned in contact with a plane mirror, functions as a particular type of mirror.

• The effective focal length of the system can be calculated using the formula for combined lenses and mirrors.
• When a concave lens is used with a plane mirror, it creates a virtual image.
• The formula for the combined focal length (F) of a lens and mirror in contact is:
• F = (f_lens * f_mirror) / (f_lens + f_mirror)
• In this case, the focal length of the plane mirror is considered infinite.
• Thus, the formula simplifies, leading to a focal length equivalent to that of the lens alone.
• However, the effective focal length becomes negative, indicating a concave configuration.
• The resultant focal length for the mirror-lens system is:
• -10 cm
• This corresponds to a concave mirror with a focal length of 10 cm.

In the context of semiconductors, the following statement is incorrect:

• Resistivity falls between that of conductors and insulators.
• The temperature coefficient of resistance is typically positive.
• Doping indeed increases conductivity.
• At absolute zero, semiconductors behave like insulators.

The Fermi level in an intrinsic semiconductor is positioned approximately midway between the conduction band and the valence band.

Key points include:

• The Fermi level represents the energy level at which the probability of finding an electron is 50% at absolute zero temperature.
• In an intrinsic semiconductor, this level indicates a balance between the number of electrons and holes.
• This positioning allows for efficient electrical conductivity when external energy is applied.

Indium impurity in Germanium creates a

• p - type semiconductor.

When indium is introduced as an impurity in germanium, it contributes to the formation of holes, which are essential for electrical conduction. This process results in the transformation of germanium into a p - type semiconductor, characterised by an abundance of positive charge carriers (holes). In summary:

• Indium adds holes to germanium.
• It increases positive charge carriers.
• This defines the material as a p - type semiconductor.

When heating and cooling platinum and silicon, their resistances behave differently:

• Resistance of platinum: Increases with temperature.
• Resistance of silicon: Generally decreases with increasing temperature and may increase as it cools, depending on the conditions.

Therefore, during the cooling process:

• The resistance of platinum will increase.
• The resistance of silicon may either decrease or increase, depending on its state and the cooling rate.

When silicon is doped with certain impurity atoms, it can become an n-type semiconductor. The following points explain how this process works:

• Phosphorus is a common dopant that adds extra electrons. This results in an increase in negative charge carriers, creating an n-type semiconductor.
• Other elements, such as aluminium and magnesium, do not contribute free electrons; instead, they create 'holes' which lead to p-type semiconductors.
• In summary, only phosphorus is effective for producing n-type semiconductors by donating additional electrons.

In a P-N junction diode:

• The current during reverse bias is typically very low.
• While reverse bias current remains minimal, forward bias current is not influenced by the voltage.
• Reverse bias current can be affected by the applied voltage.
• In forward bias conditions, the current is significantly higher compared to reverse bias.