Physics: CUET Mock Test - 6 - CUET MCQ

Each of the message signals has different ranges of frequencies. The type of communication system needed for a given signal depends on the bad of frequencies which are considered essential for the communication process.

The difference in mass between a nucleus and its constituent nucleons is known as the mass defect, denoted by Δm.

The formula for calculating the mass defect (Δm) is given by :

Δm = Z_mp + (A-Z)m_n - M

where:
Z is the atomic number of the nucleus,
A is the mass number of the nucleus,
mp is the mass of a proton,
mn is the mass of a neutron,
and M is the actual mass of the nucleus.

The above formula represents the difference between the mass of the individual protons and neutrons that make up the nucleus and the actual mass of the nucleus itself. The mass defect is a measure of the binding energy that holds the nucleus together.

The correct answer is option (2).

Concept:

Bohr model:

The Lyman series:

• It includes the lines emitted by transitions of the electron from an outer orbit of quantum number n₂ > 1 to the 1st orbit of quantum number n₁ = 1.
• All the energy wavelengths in the Lyman series lie in the ultraviolet band.

The Balmer series:

• It includes the lines due to transitions from an outer orbit n₂ > 2 to the orbit n₁ = 2.
• Four of the Balmer lines lie in the "visible" part of the spectrum.

Paschen series (Bohr series, n₁ = 3)

Brackett series (n₁ = 4)

Pfund series (n₁ = 5)​

Paschen series:

• When an electron in a Hydrogen atom transit from a higher energy orbit to 3rd orbit. (outer orbit n₂ = n > 3 to the orbit n₁ = 3) known as Paschen Series.
• So the empirical formula for the observed wavelengths (λ)for hydrogen (Z = 1) is
  

For λ_min, n₂ = ∞

Calculation:
Given:
The shortest wavelength in the Lyman series of the hydrogen spectrum = 912 Å.

The shortest wavelength of the Paschen series means

For λmin, n₁ = 3, n₂ = ∞

Putting the value of (1) in (2), we get,

8208 Å
The correct answer is option (1)

Concept:

• The Lyman series is a series of transitions in the hydrogen atom that result in the emission of photons in the ultraviolet part of the electromagnetic spectrum.
• The series is named after the American physicist Theodore Lyman, who first studied it in the early 1900s.

The maximum wavelength in the Lyman series occurs when the electron in the hydrogen atom undergoes a transition from the n = 2 energy level to the n = 1 energy level.
1/λ = R∞ (1/n₁² - 1/n₂²)
1/λ_max = R∞ (1/2² - 1/1²)
Putting R∞ = 1.0973731568539 × 10⁷ m^-1
​λmax = 121.6 nm
Similarly,
The minimum wavelength in the Lyman series occurs when the electron in the hydrogen atom undergoes a transition from the n = infinity energy level to the n = 1 energy level. This transition results in the emission of a photon with a wavelength of 91.2 nm.

• Therefore, the ratio of the maximum wavelength to the minimum wavelength in the Lyman series is:

Maximum wavelength/Minimum wavelength = 121.6 nm/91.2 nm = 1.333...

• So the ratio of the maximum wavelength to the minimum wavelength in the Lyman series is approximately 1.33 = 4/3.

​The correct answer is option (1)

When a light ray travels from a denser to a rarer medium, it undergoes refraction, and several properties of the light wave change. The properties that change include the velocity, wavelength, frequency, and phase.

Energy increases:
This statement is incorrect. Energy is conserved, and it remains constant as light travels from one medium to another. However, some energy may be lost due to absorption or reflection.

Frequency remains the same:
This statement is correct. The frequency of the light wave remains constant as it moves from one medium to another.

Phase changes by 90°:
This statement is incorrect. The phase change depends on the angle of incidence and the angle of refraction, and it varies for different angles. It may or may not change by 90°.

Velocity increases:
This statement is correct. The velocity of light increases as it moves from a denser to a rarer medium. This is because the refractive index of the rarer medium is less than the denser medium, and the speed of light is directly proportional to the refractive index.

Wavelength decreases:
This statement is incorrect. The wavelength of the light wave changes as it moves from one medium to another. Snell's law gives the relationship between the wavelength and the refractive index. However, the change in wavelength depends on the angle of incidence and the angle of refraction and is not a fixed value.
In conclusion, option 2 is the correct answer, as only statements (B) and (D) is correct.

The correct answer is option (2)

The correct answer is option (4)

Concept:

Forward Bias:

• When forward biased, the applied voltage V of the battery mostly drops across the depletion region and the voltage drops across the p-side and n-side of the p-n junction is negligibly small.
• In forward biasing the forward voltage opposes the potential barrier Vbi. As a result, the potential barrier height is reduced and the width of the depletion layer decreases.
• As forward voltage is increased, at a particular value the depletion region becomes very much narrow such that a large number of majority charge carriers can cross the junction.
  ​

Explanation:

• When a forward bias is applied to a p-n junction diode, the potential barrier at the junction is reduced.
• This allows current to flow easily across the junction. Electrons from the N-type material and holes from the P-type material move across the junction and recombine, releasing energy in the form of light or heat.
• As a result, the potential barrier height is reduced and the width of the depletion layer decreases.
• The current flows from the P-type material to the N-type material, and the diode has a low resistance to the flow of current. In essence, the forward bias allows the diode to conduct electricity, which is why it is commonly used as a rectifier to convert AC to DC.

Additional Information

Reversed Bias:

• When reverse biased, more charge carriers are depleted, resulting in the widening of the depletion region.
• This increases the opposing electric field for the diffusion carriers and does not allow them to cross the junction, offering a high resistance

This can also be understood with the VI characteristic of a p-n junction diode:

CONCEPT:

• Lens: The transparent curved surface which is used to refract the light and make an image of any object placed in front of it is called a lens.
  • Convex lens: ​A lens having two spherical surfaces, bulging outwards is called a double convex lens (or simply convex lens).
  • It is thicker in the middle as compared to the edges.
  • Convex lenses converge light rays and hence, convex lenses are also called converging lenses.

Explanation:

• When half of the lens is covered by the black paper then the light from the object will fall on the rest half part of the lens and form a complete image of the object. So option 1 is not correct.
• As only half of the total light rays will fall on the lens due to which the intensity of the image will be very less (diminished). Hence option 2 is correct.
• As the image will be formed the same as that in the full lens. If it was inverted initially then now also will be inverted and if it was erect previously then now also it will be erect. Hence option 3 is wrong.
• There will be no change in the size of the image. It will be the same as that in the case of a full lens. So option 4 is wrong.

• When two plane mirrors placed at 90° to each other then, the incident ray and the reflected ray are antiparallel to each other.
• Angle between the incident ray and reflected ray is 180°
• As shown in the figure are two plane mirrors XY and YZ (XY ⊥ YZ) joined at their edge.
• Also shown is a light ray falling on one of the mirrors and reflected back parallel to its original path as a result of this arrangement.

• The two mirrors are now rotated by an angle θ to their new position X'YZ', as shown.
• As a result, the new reflected ray is at an angle α from the original reflected ray.
• Then α = 0°, because, after rotation of the mirror combination, reflected rays are parallel to the incident rays,
• So, the new reflected ray is parallel to the original reflected ray.

Additional Information

• When two plane mirrors are placed at an angle other than 90 degrees to each other, the incident ray and reflected ray are no longer antiparallel, and their relationship is determined by the angles of incidence and reflection for each mirror.
• The angles formed by the reflected rays can be different from each other and will depend on the angle between the mirrors.

Concept:

• The Gold Foil Experiment: A very thin sheet of gold foil was bombarded with alpha particles.
  • The experiment result of the Bombardment of gold foil with alpha particles showed that a very small part of alpha particles was deflected.
  • This experiment showed that the atom consists of a small and dense positively charged interior surrounded by a cloud of negative charge electrons.

Figure (A) The experimental setup for Rutherford's gold foil experiment
Figure (B) Rutherford found that a small percentage of alpha particles were deflected at large angles

Explanation:

• In Rutherford's Gold foil, alpha-particle scattering experiment, Very thin sheets of gold foil were bombarded with fast-moving alpha particles.
• Since alpha particles are positive charge ions, He used them to know the exact position of positive charge inside the atom.
• He found a very small, dense, positively-charged nucleus at its center (bottom).
• So the correct answer is option 1.

Given:

Number of electrons

Charge of electron beam

Area

Time

Concept:

The total charge crossing a perpendicular cross-section in one second is

• q=ne
• Current
• Current density,

Explanation:

The total charge crossing a perpendicular cross-section in one second is

q=ne

Hence, the correct answer is .

The correct answer is option 4 i.e., Marie Curie discovered Radium.​

The frequencies for transmitting music is which of the following?
a) Low
b) High
c) Moderate
d) Very high

When the bandwidth is large enough to accommodate a few harmonics, the information is not lost and the rectangular signal is more or less recovered. This is so because the higher the harmonic, the less is its contribution to the waveform. Therefore, the value of the harmonic is inversely proportional to its contribution to the waveform.

For speech signals, the frequency range of 300 Hz to 3100 Hz is considered adequate. Therefore, speech signals require a bandwidth of 2800 Hz (3100 Hz – 300 Hz) for commercial telephonic communication.

SiO₂ acts as a diffusion mask permitting selective diffusions into the silicon wafer through the window etched into oxide. In other words, it is used for creating a protective SiO₂ layer on the wafer surface of the IC. Therefore, SiO₂ acts as an insulating component for integrated circuits.

An Integrated circuit (IC) is an assembly of electronic components, fabricated as a single unit, in which miniaturized active devices and passive devices and their interconnections are built upon a thin substrate of semiconductor material. There are about 600 different types of Integrated circuits.

Silicon is the one mainly used in the production of Integrated circuits. This is because silicon possesses many characteristics that are ideal for ICs. Silicon is used because it can be used as either an insulator or a semiconductor. This property is crucial for the manufacture of Integrated Circuits.

The maximum power rating possible for a microcircuit is 10 Watts. Integrated Circuits can’t be repaired because the individual components inside the IC are too small. As a result, the facility rating for many of the IC’s doesn’t exceed quite 10 watts. It’s impossible to manufacture high power IC’s and this is often one among the most important disadvantages of Integrated Circuits.

ICs are both small in size as well as in weight. As the fabrication process is used for the integration of active and passive components on a silicon integrated circuit, the IC becomes a lot smaller. Also, due to its small size, the weight of the IC reduces, when compared to a discrete circuit.

Capacitors are not easy to fabricate on an Integrated Circuit. In fact, they are the most difficult ones to fabricate. This is because capacitors require a lot of attributes such as electromagnetic compatibility, microfabrication techniques, new magnetic materials, etc. and to integrate them, it costs a lot. Therefore, fabricating capacitors is a challenging task.

A transistor has three doped regions forming two p-n junctions between them. There are two types of transistors, namely n-p-n transistors and p-n-p transistors. In n-p-n transistors, the two segments of the n-type semiconductor are separated by a segment of p-type semiconductor, and in a p-n-p transistor, it’s just the opposite scenario.

The collector is the section on the other side of the transistor that collects the charge carriers supplied by the emitter. It is moderately doped but large in size and is always kept in reverse bias with respect to the base.

In an n-p-n transistor, as the base is very thin and lightly doped, a very few electrons from the emitter combine with the holes of the base, giving rise to base current and the electrons finally collected by the positive terminal of the battery gives rise to collector current. This base current is a small fraction of collector current depending on the shape of a transistor, thickness of the base, doping levels, and bias voltage.

The expression for current gain is given as:
β = I₀ / I_B
I_C = 50I_B
Since, I_E = I_B+I_C 
⇒ I_E = I_B + 50I_B
I_E = 51I_B
I_B = IE51 = 5.551 = 0.107 mA

The expression of voltage gain is as follows:
A_V = αR_L / R_e
Given: α = 0.90, R_L = 5.0 kΩ = 5000 Ω, R_e = 50 Ω
A_V = 0.90 × 5000 / 50
A_V = 90

Power gain is defined as the ratio of output power to input power. It can also be determined as the product of current gain and voltage gain.
Given: α = 0.80, voltage gain (A_V) = 95
The required equation
⇒ A_P = α × A_V
A_P = 0.80×95
A_P = 76

I_E = I_B + I_C
I_E / I_C = I_B / I_C + 1
1α = 1 / β + 1
α = β / 1 + β
Therefore, β = α / 1−α

Emitter is the section or region on one side of the transistor that supplies charge carriers. It is heavily doped and is always kept forward biased with respect to the base so that it can supply a large number of charge carriers to the base.

The radius of an electron orbit in a hydrogen atom is of the order of 10^-11 m. It is equal to the most probable distance between the nucleus and the electron in a hydrogen atom in its ground state.

Increase in angular momentum = h /2π.
h /2π = 6.6 × 10⁻³⁴ /2 × 3.14
h /2π = 1.05 × 10^-34 Js.