VITEEE: Subject Wise and Full Length VITEEE Physics Test - 10 Free Online

To find the primary voltage and current for a transformer:

• The turn ratio is 1:10, meaning the primary voltage is 1/10th of the secondary voltage.
• The secondary resistance is 200 ohms, with a current of 0.5 amps.
• Calculate the secondary voltage using Ohm's law: V = I × R.
• So, the secondary voltage is: V_s = 0.5 A × 200 Ω = 100 V.
• Using the turn ratio, the primary voltage is: V_p = V_s / 10 = 100 V / 10 = 10 V.
• Now, calculate the primary current using the formula: I_p = I_s / turn ratio.
• This gives us: I_p = 0.5 A / 10 = 0.05 A.

The primary voltage is 10 V and the primary current is 0.05 A.

An electron-volt (eV) is a unit of:

• Energy

An electron-volt is defined as the amount of energy gained by an electron when it is accelerated through an electric potential difference of one volt. This unit is commonly used in the fields of physics and chemistry to express small amounts of energy, particularly at the atomic and subatomic levels.

The capacity of a conductor is influenced by several factors:

• Size: The overall dimensions of the conductor play a crucial role in its capacity.
• Thickness: A thicker conductor can carry more current due to its increased surface area.
• Material: Different materials have varying conductivity levels, affecting the conductor's ability to transport electricity.

All these aspects work together to determine the overall capacity of a conductor.

The phenomenon of electrolysis was first described by Michael Faraday.

• Faraday made significant contributions to the understanding of electrolysis.
• He formulated the laws of electrolysis, which explain how electric current causes chemical reactions.
• These laws provide a foundation for electrochemical processes and applications.

Telecommunication through optical fibres has several key characteristics:

• Optical fibres offer extremely low transmission loss, making them efficient for long-distance communication.

• They can have a homogeneous core paired with suitable cladding for effective light transmission.

• Some optical fibres are designed with a graded refractive index, which helps to manage light propagation more effectively.

• However, unlike copper cables, optical fibres are not affected by electromagnetic interference from external sources.

To determine the specific charge of a particle accelerated through a potential difference (P.D.) of 100 volts and experiencing no deflection in uniform electric and magnetic fields:

• The particle's kinetic energy (KE) gained from the P.D. is given by: KE = eV, where e is the charge and V is the potential difference.
• Substituting the values: KE = e × 100
• In the electric field, the force on the particle is: F_E = eE
• In the magnetic field, the force is: F_B = evB
• For no deflection, these forces must be equal: eE = evB
• By cancelling e (assuming it is not zero), we get: E = vB
• Solving for v gives: v = E/B
• Substituting the values of E and B: v = (15 × 10⁶ V/m) / (5 × 10³ T) = 3 × 10³ m/s
• Now substituting v into the kinetic energy equation: KE = (1/2)mv² = eV
• Rearranging for specific charge: Specific charge = e/m = (2eV) / (mv²)
• Using v derived earlier and substituting: Specific charge = (2 × e × 100) / (m × (3 × 10³)²)
• From here, you can calculate the specific charge based on the values given and derived.

The resistance of a discharge tube can vary depending on its conditions. Here are the key points:

• Ohmic behaviour occurs when the resistance remains constant, following Ohm's law.
• Non-ohmic behaviour is observed when the resistance changes with voltage or current, meaning it does not follow Ohm's law.
• Discharge tubes are typically non-ohmic because their resistance can change significantly based on factors like temperature and voltage.
• Sometimes, under specific conditions, they may exhibit ohmic characteristics.

To find the resistance of the second coil:

• The total potential difference across both coils is the sum of the individual voltages: 3 V + 4.5 V = 7.5 V.
• Using Ohm's Law, V = I × R, where V is voltage, I is current, and R is resistance:
• The resistance of the first coil is given as 2 Ω.
• Let the resistance of the second coil be R2.
• The total resistance in series is: 2 Ω + R2.
• The total voltage across both coils can be expressed as: 7.5 V = I × (2 Ω + R2).
• From the first coil, we have: 3 V = I × 2 Ω, which gives I = 1.5 A.
• Substituting the current into the total voltage equation: 7.5 V = 1.5 A × (2 Ω + R2).
• This simplifies to: 7.5 V = 3 Ω + 1.5 A × R2.
• Rearranging gives: 1.5 A × R2 = 4.5 V.
• Solving for R2, we find: R2 = 3 Ω.

The resistance of the second coil is 3 Ω.

To calculate the drift velocity of electrons in a copper wire, we can use the formula:

Drift velocity (v_d) = I / (n * A * e)

• I = Current = 1.344 A
• n = Number of free electrons = 8.4 x 10²² electrons/cm³
• A = Area of cross-section = 1 mm² = 1 x 10^-6 m²
• e = Charge of an electron = 1.6 x 10^-19 C

First, convert the electron density from cm³ to m³:

• n = 8.4 x 10²² electrons/cm³ = 8.4 x 10²⁸ electrons/m³

Now, substitute these values into the drift velocity formula:

• v_d = 1.344 / (8.4 x 10²⁸ * 1 x 10^-6 * 1.6 x 10^-19)

Calculating this gives:

• v_d = 1.344 / (1.344 x 10⁴) = 0.0001 m/s

Converting to mm/s:

• v_d = 0.1 mm/s

Thus, the drift velocity is approximately 0.1 mm/sec.

To calculate the effective resistance between points X and Y, follow these steps:

• Each resistor in the network has a value of 6 Ω.
• Points X and Y are connected to points A and B, respectively, through wires with negligible resistance.
• The effective resistance can be determined based on the arrangement of the resistors:
• If the resistors are in series, simply add their resistances.
• If they are in parallel, use the formula: 1/R_eff = 1/R1 + 1/R2 + ...
• Assuming a typical arrangement where two resistors are in parallel, the effective resistance is calculated as:
• R_eff = 6 Ω || 6 Ω = 6/2 = 3 Ω.

If additional resistors are involved, apply the same principles accordingly.

When yellow light strikes a surface, it does not cause the emission of electrons, while green light is effective in doing so. If red light is used, the outcome is as follows:

• No electrons are emitted from the surface.

• Red light does not possess enough energy to release electrons.

To determine the current in the transmission line after voltage transformation:

• Initial current from the generator: 100 A
• Initial voltage: 4000 V
• Transformed voltage: 240000 V

Using the principle of conservation of power:

• Power input = Power output
• Power (P) is calculated as: P = V × I
• Input power: P1 = 4000 V × 100 A = 400000 W
• Output power: P2 = 240000 V × I2

Setting input power equal to output power:

• 400000 W = 240000 V × I2
• Solving for I2 gives: I2 = 400000 W / 240000 V
• Calculating I2 results in: I2 ≈ 1.67 A

Thus, the current in the transmission line is approximately 1.67 A.

Power delivered by a source in a circuit reaches its maximum when:

• The inductive reactance (ωL) equals the capacitive reactance (ωC).
• This condition indicates that the circuit is in resonance.
• At resonance, the impedance is minimized, allowing for maximum current flow.
• It is essential for efficient energy transfer in AC circuits.

To determine the self-inductance of the coil, we can use the formula for induced electromotive force (e.m.f.) given by:

• e.m.f. = -L * (ΔI/Δt)

Where:

• L = self-inductance (in Henries)
• ΔI = change in current (in Amperes)
• Δt = time interval (in seconds)

Given the values:

• e.m.f. = 8 V
• ΔI = 2 A - 0 A = 2 A
• Δt = 0.05 s

We can rearrange the formula to find L:

• L = -e.m.f. * (Δt/ΔI)

Substituting the known values:

• L = -8 V * (0.05 s / 2 A)
• L = -8 * 0.025
• L = -0.2 H

Thus, the self-inductance of the coil is 0.2 H.

In television broadcasting, both picture and sound are transmitted simultaneously.

• The audio signal is typically frequency modulated.
• The video signal is usually amplitude modulated.

This combination allows for efficient transmission of both components, ensuring a coherent viewing experience.

The relationship between electric flux and charge inside a closed surface is described by Gauss's Law.

According to Gauss's Law:

• The total electric flux, Φ, through a closed surface is proportional to the charge, Q, enclosed within that surface.
• Mathematically, this is expressed as: Φ = Q/ε₀, where ε₀ is the permittivity of free space.

To determine the enclosed charge based on the incoming and outgoing electric flux:

• If φ₁ is the flux entering the surface and φ₂ is the flux leaving, the net flux is φ₁ - φ₂.
• The enclosed charge can thus be calculated as: Q = ε₀(φ₁ - φ₂).

This shows that the charge inside the surface depends on the difference between the incoming and outgoing flux. A positive result indicates a net inward flux, suggesting a net positive charge inside the surface.

To calculate the potential energy between two equal positive charges:

• We use the formula for electric potential energy: U = k × (q₁ × q₂) / r, where:
  • U is the potential energy,
  • k is Coulomb's constant (approximately 8.99 x 10⁹ N m²/C²),
  • q₁ and q₂ are the charges (1 μC each),
  • r is the distance between the charges (1 m).
• Substituting the values:
  • q₁ = 1 x 10^-6 C,
  • q₂ = 1 x 10^-6 C,
  • r = 1 m.
• The calculation becomes: U = (8.99 x 10⁹) × (1 x 10^-6) × (1 x 10^-6) / 1.
• This simplifies to: U = 8.99 x 10^-3 J.

The potential energy of two equal positive charges 1 μC each, held 1 m apart in air, is approximately 9 x 10^-3 J.

To determine the work done by the electric force when expanding a spherical shell of charge:

• The electric field E due to a charged spherical shell outside the shell (at distance r >a) is given by:
• E = kQ/r², where k is Coulomb's constant.

Work done (W) during the expansion from radius a to b can be calculated as:

• W = ∫_a^b F·dr, where F = Q·E.
• Substituting for E: W = ∫_a^b (Q·(kQ/r²)) dr.
• Simplifying gives: W = kQ² ∫_a^b (1/r²) dr.
• This integral evaluates to: W = kQ²[-1/r]_a^b = kQ²(1/a - 1/b).

Thus, the work done by the electric force during the expansion is:

• W = kQ²(1/a - 1/b).

To find the electric potential at a specific point given the work done and the charge:

• The formula for electric potential \( V \) is given by: V = W/Q
• Where:
  • W = Work done (in joules)
  • Q = Charge (in coulombs)
• In this case:
  • Work done, \( W = 100 \, \text{J} \)
  • Charge, \( Q = -5 \, \text{C} \)
• Substituting the values into the formula: V = 100 \, \text{J} / -5 \, \text{C} = -20 \, \text{V}

The electric potential at this point is -20 V.

Correct option: B.

Let the plate area be A and the separation be d. The capacitance without dielectric is C₀ = ε₀ A / d.

Each dielectric occupies half the area, so each region has area A/2.

The capacitance of the region with dielectric K₁ is C₁ = ε₀ K₁ (A/2) / d = (C₀ K₁)/2.

The capacitance of the region with dielectric K₂ is C₂ = ε₀ K₂ (A/2) / d = (C₀ K₂)/2.

Since the two regions are side-by-side between the same plates, they are connected in parallel, so total capacitance is the sum: C = C₁ + C₂ = C₀ (K₁ + K₂) / 2.

Therefore the correct choice is option B.

To determine which particle will describe the smallest circle when projected with the same velocity perpendicular to a magnetic field, consider the following factors:

• Charge: The charge of the particle affects its motion in a magnetic field. A charged particle experiences a force that causes it to move in a circular path.
• Mass: The mass of the particle also plays a crucial role. Lighter particles will curve more sharply than heavier ones when subjected to the same force.
• Radius of Circular Motion: The radius of the circle is inversely proportional to the particle's mass. Therefore, the smaller the mass, the smaller the radius.

Given these factors:

• The electron, being the lightest among the options, has the smallest mass.
• As a result, it will describe the smallest circle when projected under the same conditions.

The nuclear size is measured in the units of:

• Fermi is the most common unit for measuring nuclear size.
• Another unit that may be encountered is Angstroms, but it is less typical for nuclear dimensions.
• Centimetres and microns are not appropriate for this context.

The situation is shown in adjacent figure, P → is parallel to E →
∴ U = - pE cos 0 = - pE = Minimum
and the net force on the dipole is zero

The first atomic reactor was built by Enrico Fermi in 1942.

• Fermi was a renowned physicist.
• He led the team that created the first controlled nuclear chain reaction.
• This milestone was achieved at the University of Chicago.
• The reactor was named Chicago Pile-1.
• This event marked a significant advancement in nuclear physics.

The mass defect in a nuclear reaction is 0.3 grams. To calculate the energy liberated, we can use Einstein's equation, E=mc².

• Mass defect (m): 0.3 grams, which is equivalent to 0.0003 kg.
• Speed of light (c): 3 x 10⁸ m/s.
• Using the formula, E = mc²:
• E = 0.0003 kg x (3 x 10⁸ m/s)²
• E = 0.0003 kg x 9 x 10¹⁶ m²/s²
• E = 2.7 x 10¹³ joules.
• Convert joules to kilowatt hours:
• 1 kilowatt hour = 3.6 x 10⁶ joules.
• E (in kWh) = 2.7 x 10¹³ joules / 3.6 x 10⁶ joules/kWh.
• E = 7.5 x 10⁶ kWh.

The total energy liberated in this nuclear reaction is 7.5 x 10⁶ kilowatt hours.

The photoelectric effect occurs when light or radiation strikes a metal surface and causes the emission of electrons. However, this effect is dependent on the energy of the incident radiation.

• The energy of the incident rays must be above a certain threshold to release electrons.
• Ultraviolet rays typically have enough energy to trigger the photoelectric effect.
• However, if the energy is insufficient, such as that from infrared rays or radiowaves, the effect will not occur.
• On the other hand, X-rays possess high energy and can also cause the photoelectric effect.
• In summary, the type of radiation significantly influences whether the photoelectric effect happens.

The photoelectrons will be emitted because wavelength of incident radiation is less than threshold wavelength (λ < λ₀)

To find the refractive index of the glass plate when the reflected and refracted rays are perpendicular, follow these steps:

• Let the angle of incidence be 60°.
• According to the problem, the reflected ray and the refracted ray are mutually perpendicular.
• This means the angle of refraction (r) plus the angle of reflection (which is equal to the angle of incidence) sums to 90°.
• Thus, we have the equation: 60° + r = 90°, which implies r = 30°.
• Using Snell's Law: n₁ sin(θ₁) = n₂ sin(θ₂), where:
• n₁ is the refractive index of air (approximately 1),
• θ₁ is the angle of incidence (60°),
• n₂ is the refractive index of glass,
• θ₂ is the angle of refraction (30°).
• Substituting into Snell's Law gives: 1 * sin(60°) = n * sin(30°).
• Since sin(60°) = √3/2 and sin(30°) = 1/2, we can rewrite the equation as:
• √3/2 = n * 1/2.
• Solving for n yields n = √3.

Therefore, the refractive index of the glass plate is √3.

Intrinsic semiconductors are electrically neutral, whereas extrinsic semiconductors contain impurities that introduce additional charge carriers.

When impurities are added to an intrinsic semiconductor, the following occurs:

• If a donor impurity is added (such as phosphorus in silicon), the semiconductor becomes negatively charged due to the excess electrons.
• If an acceptor impurity is introduced (like boron in silicon), it results in a positively charged semiconductor due to the presence of holes, which act as positive charge carriers.

Thus, an extrinsic semiconductor can be:

• Positively charged if it has more holes.
• Negatively charged if it has more free electrons.

In summary, the charge of an extrinsic semiconductor depends on the type of impurity added.

The depletion layer is primarily composed of:

• Electrons: These are negatively charged particles that are typically depleted in this region.
• Protons: These are positively charged particles, but they do not contribute significantly to the depletion layer.
• Mobile ions: These ions can move freely and are not generally found in the depletion layer.
• Immobile ions: These ions remain fixed in place and contribute to the charge distribution in the depletion layer.

In summary, the depletion layer consists mainly of immobile ions which create an electric field that influences the behaviour of charge carriers in semiconductor devices.