VITEEE Physics Test - 4 - JEE MCQ

A device that converts electrical energy into mechanical energy is known as an electric motor.

• An electric motor transforms electrical energy into motion.
• It operates on the principle of electromagnetism.
• Common applications include powering household appliances, vehicles, and industrial machines.

Moseley's law

• The frequency is directly proportional to the square of the atomic number.
• This means that as the atomic number increases, the frequency of the emitted X-rays also increases significantly.
• Moseley's law is crucial for understanding the characteristics of elements and their X-ray emissions.

Fraunhofer spectrum refers to a particular type of absorption spectrum characterised by the following features:

• Line absorption spectrum: It displays distinct dark lines against a continuous spectrum of light.
• These lines indicate the wavelengths at which light is absorbed by elements or compounds.
• Fraunhofer lines are crucial for identifying the composition of stars and other celestial bodies.

The resistance of a cell is influenced by various factors, but it does not depend on the following:

• Current drawn from the cell: The amount of current being drawn does not affect the cell's inherent resistance.
• Temperature of the electrolyte: While temperature can impact overall performance, it does not change the resistance directly.
• Concentration of electrolyte: Changes in concentration alter efficiency but not the basic resistance.
• Electromotive force (e.m.f.) of the cell: The e.m.f. is related to the cell's voltage but does not determine resistance.

Given:

f = 50 MHz

The angle of incidence (I) = 30^o

Sin i = Sin r

r = 30^o

To determine the new current flowing through the wire when its temperature is raised, follow these steps:

• The initial resistance of the wire at 20°C is 10 ohms.
• The temperature coefficient of resistivity is 5 x 10^-3 per °C.
• The current at 20°C is 30 milliamperes.
• The new temperature is 120°C.

Now, calculate the change in temperature:

• Change in temperature = 120°C - 20°C = 100°C.

Next, calculate the increase in resistance using the formula:

• New resistance = Original resistance × (1 + temperature coefficient × change in temperature)
• New resistance = 10 ohms × (1 + (5 x 10^-3 × 100))
• New resistance = 10 ohms × (1 + 0.5) = 15 ohms.

Finally, using Ohm’s Law (V = IR), where the potential difference remains constant:

• Original current = 30 mA corresponding to 10 ohms.
• Using the ratio of resistances to find the new current:
• New current = (Original current) × (Original resistance / New resistance)
• New current = 30 mA × (10 ohms / 15 ohms) = 20 milliamperes.

When a wire is stretched to double its length, its resistance changes due to several factors:

• The resistance of a wire is affected by its length, cross-sectional area, and the material's resistivity.
• When the length is doubled, the cross-sectional area is reduced, leading to an increase in resistance.
• Resistance is calculated using the formula: R = ρ(L/A), where R is resistance, ρ is resistivity, L is length, and A is cross-sectional area.
• If the length is doubled, the new resistance becomes: R' = ρ(2L/A'), where A' is the new, smaller cross-sectional area.
• Since the volume of the wire remains constant, doubling the length results in a reduction of the area by half, making A' = A/2.
• Thus, the new resistance can be expressed as: R' = ρ(2L/(A/2)) = 4ρ(L/A) = 4R.

In conclusion, when the wire is stretched to double its length, the resistance increases to four times the original resistance.

A choke coil operates based on the principle of self induction.

Here are some key points regarding choke coils:

• Self induction occurs when a changing current in a coil creates an electromotive force (EMF) that opposes the change in current.
• This property is utilised in choke coils to limit the flow of alternating current (AC).
• Choke coils are commonly used in various electrical applications, including power supplies and radio frequency circuits.

Electromagnetic waves are produced by:

• A stationary charge does not generate electromagnetic waves.
• An accelerated charge is responsible for creating electromagnetic waves.
• A uniformly moving charge produces a static electric or magnetic field, not waves.

In summary, only an accelerated charge can produce electromagnetic waves. Therefore, the correct answer is the second option.

To find the total electric flux through the surface of the cylinder, consider the following points:

• The cylinder is placed in a uniform electric field parallel to its axis.
• The electric flux is defined as the product of the electric field and the area through which it passes.
• However, for a closed surface like the cylinder, the total electric flux depends on the net charge enclosed.
• In this scenario, since the cylinder does not enclose any charge, the total enclosed charge is zero.
• According to Gauss's law, if there is no enclosed charge, the total electric flux through the surface of the cylinder is also zero.

To determine the electric potential at the midpoint between two charges:

• Consider two charges: -10 C and +10 C, separated by a distance of 10 cm.
• The midpoint is 5 cm from each charge.
• Electric potential (V) due to a point charge is calculated using the formula: V = k * Q / r, where:
• k is the electrostatic constant, approximately 8.99 x 10^9 N m²/C².
• Q is the charge in coulombs.
• r is the distance from the charge.
• Calculating the potential due to:
• The -10 C charge at 5 cm results in a negative potential.
• The +10 C charge at 5 cm results in a positive potential.
• Since the magnitudes of the charges are equal and opposite, their potentials will cancel each other out at the midpoint.
• Thus, the total potential at the centre 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

To determine the common potential of two capacitors connected in parallel, follow these steps:

• Calculate the charge on each capacitor:
• For the 10 μF capacitor charged to 250 V: Q1 = C1 x V1 = 10 μF x 250 V = 2500 μC.
• For the 5 μF capacitor charged to 100 V: Q2 = C2 x V2 = 5 μF x 100 V = 500 μC.
• Find the total charge when the capacitors are connected:
• Total charge (Q_total) = Q1 + Q2 = 2500 μC + 500 μC = 3000 μC.
• Calculate the total capacitance in parallel:
• C_total = C1 + C2 = 10 μF + 5 μF = 15 μF.
• Determine the common potential (V_common) using the formula:
• V_common = Q_total / C_total = 3000 μC / 15 μF = 200 V.

The common potential across the capacitors is 200 V.

Substances that share the same chemical properties but have different atomic weights are known as:

• Isotopes - These are variants of a chemical element.
• They have the same number of protons.
• However, they differ in the number of neutrons.
• This variation affects their atomic weights.

Isotopes can be stable or radioactive, impacting their behaviour in chemical reactions and applications.

The magnetic effect of current was first discovered by Hans Christian Oersted.

• He found that an electric current creates a magnetic field.
• This discovery was significant in linking electricity and magnetism.
• Oersted’s work laid the foundation for further studies by scientists such as Ampère and Faraday.

Number of neutrons in C¹² and C¹⁴ are:

The number of neutrons in the isotopes of carbon can be determined by the following:

• C¹²: This isotope has a total of 12 nucleons (protons + neutrons). Since carbon has 6 protons, it contains:
• 12 - 6 = 6 neutrons
• C¹⁴: This isotope has a total of 14 nucleons. With 6 protons, it contains:
• 14 - 6 = 8 neutrons

Thus, the number of neutrons in C¹² is 6 and in C¹⁴ is 8.

In a nuclear reactor, heavy water serves as a vital component.

• It functions primarily as a moderator, slowing down neutrons to sustain the nuclear reaction.
• Heavy water is effective in maintaining the reaction with certain types of fuel.
• Unlike other materials, it does not absorb neutrons, which enhances its role in the reactor.
• This property enables the reactor to operate efficiently with lower enrichment levels of uranium.

To determine the longest wavelength of light that can cause photoelectron emission, we can use the relationship between energy and wavelength.

The energy of a photon can be calculated using the formula:

• E = hc/λ

Where:

• E is the energy in electron volts (eV),
• h is Planck's constant (approximately 4.135 x 10^-15 eV·s),
• c is the speed of light (approximately 3.00 x 10⁸ m/s),
• λ is the wavelength in metres.

To find the longest wavelength that can cause emission:

• We know the work function of the substance is 4.0 eV.
• Set E = 4.0 eV and rearrange the formula to solve for λ:
• λ = hc/E.

Substituting the values:

• λ = (4.135 x 10^-15 eV·s × 3.00 x 10⁸ m/s) / 4.0 eV.
• This calculation gives λ ≈ 310 nm.

Thus, the longest wavelength of light that can cause photoelectron emission from this substance is approximately:

• 310 nm.

To find the refractive index of a glass plate when the angle between the reflected and refracted rays is a right angle, we can use Snell's law.

• According to 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 (θ),
• n2 is the refractive index of glass (μ),
• θ2 is the angle of refraction.

When the angle between the reflected and refracted rays is a right angle, we have:

• θ + θ2 = 90°
• This implies θ2 = 90° - θ.

Substituting θ2 into Snell's law gives:

• sin(θ) = μ sin(90° - θ)
• Since sin(90° - θ) = cos(θ), we further have:
• sin(θ) = μ cos(θ).

Rearranging this equation leads to:

• μ = sin(θ) / cos(θ)
• Thus, μ = tan(θ).

Hence, the refractive index of the glass plate is given by:

• μ = tan(θ).

When Arsenic is added as an impurity to Silicon, the resulting material becomes an n-type semiconductor.

• Arsenic is a donor impurity that introduces extra electrons.
• This increases the conductivity of Silicon.
• In n-type semiconductors, the majority charge carriers are electrons.

The potential barrier in the depletion layer is primarily caused by the presence of charged ions.

• The depletion layer is formed when p-type and n-type semiconductor materials come into contact.
• When these materials meet, electrons from the n-type region combine with holes in the p-type region.
• This process leads to a region devoid of mobile charge carriers, known as the depletion layer.
• In this layer, fixed, charged ions remain, creating an electric field and resulting in a potential barrier.

In the minimum deviation position of a prism:

• The relationship between the refractive index μ and the angle of minimum deviation A can be expressed with the formula:
• μ = sin((A + D)/2) / sin(A/2)
• Here, D is the angle of the prism.
• In the minimum deviation condition, the prism's angles are balanced, leading to a simplified equation.
• From this equation, we can derive the angle A in terms of μ:

A = 2 * sin^-1(μ/2)

The dynamic plate resistance can be calculated using the formula:

• Dynamic Plate Resistance (r) = ΔV / ΔI

Where:

• ΔV is the change in voltage.
• ΔI is the change in current.

In this case:

• ΔV = 150 V - 100 V = 50 V
• ΔI = 12 mA - 7.5 mA = 4.5 mA

Now, converting the current change into amperes:

• 4.5 mA = 0.0045 A

Substituting the values into the formula gives:

• r = 50 V / 0.0045 A
• r = 11111.11 Ω or approximately 11.1 KΩ

Thus, the dynamic plate resistance is approximately 11.1 KΩ.

When arsenic is added as an impurity to silicon, the resulting material is an N-type semiconductor.

• Arsenic is a donor impurity in silicon.

• It introduces extra electrons, enhancing conductivity.

• This process creates an N-type semiconductor, characterised by negative charge carriers.

• In contrast, impurities like boron would create a P-type semiconductor.

In a good conductor, the energy gap between the conduction band and the valence band is:

• The energy gap is typically zero or very small.
• This allows for easy movement of electrons.
• As a result, conductors can efficiently carry electric current.

To create a P-type semiconductor, the impurity added to pure germanium is:

• Aluminium is typically used as the dopant.
• P-type semiconductors are formed by adding elements that have fewer valence electrons than the semiconductor material.
• These impurities create 'holes' in the electron structure, allowing positive charge carriers.
• Other common dopants for P-type semiconductors include indium and gallium.