Physics: CUET Mock Test - 6 - CUET MCQ

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 intial energy stored in the capacitor C of potential difference V is
E = (1/2) CV² 
When the p.d increased by ΔV, the energy stored will be ΔE.

The capacitance of a spherical conductor of radius R is 

When a charged particle moves in an electric field, its electric potential energy decreases, its kinetic energy will increase i.e. the total energy is conserved.

The initial energy of the capacitor of capacitance C and charge Q₁,  

When the charge increases to Q₂ the change in the energy of the capacitor, 

Given percentage increase of energy,

Since, Q₂ - Q₁ = 2
⇒ Q₁ = 20 C

With a known resistance in one of the gaps, the meter bridge is used to determine the value of unknown resistance by the formula. 

Resistors 20Ω,100Ω and 25Ω will be in parallel. Their equivalent is 10Ω.


p.d. across 10Ω,10I = 10 × 10 = 100 V
This will be the voltmeter reading. Also, this will be the p.d. across each of 20Ω,100Ω and 25Ω resistors.
Ammeter reading = current through 25Ω=100/25=4 A.

So, 400 = R₀[1 + 0.005 × 2000]

• A potentiometer is considered to be sensitive if the potential gradient dV/dl is low.
• Such a potentiometer can measure very small changes in potential difference.
• Increasing the length of the potentiometer wire decreases the potential gradient. Its sensitivity increases. Increasing potential gradient decreases the sensitivity.
• Increasing the emf of the primary cell and by decreasing the length, potential gradient increases.

Given, i₁​=i₂​=i
∴F=μ_0​i²l​/2πb
Hence, force per unit length is F=μ₀​i²​/2πb

The permeability of free space, μ0, is numerically equal to 4π x 10-7. For air and other non-magnetic materials, the absolute permeability is the same constant. For magnetic materials, absolute permeability is not a fixed constant but varies non-linearly with the flux density.

The long straight wire and side AB  carry current in the same direction, hence will attract each other.
The long straight wire and side CD carry current in the opposite direction, hence will repel each other.
Force on side BC  will be equal and opposite to force on side DA.
Since CD  is farther from the wire than AB,  the force of attraction on  AB  will exceed the force of repulsion on CD.
Hence, there will be a net force of attraction on the loop ABCD and it will move towards the wire.

Explanation:A will become positive with respect to B because current  indused is in clockwise direction

e = NBAω = 20 x 3 x 10^-2 x (3.14 x 0.08 x 0.08) x 50 = 0.603C

average power loss due to joule heating

Explanation:

- When XL > XC, the power factor is not equal to 1.
- The power factor is the ratio of the resistance (R) to the impedance (Z) in a circuit.
- When the inductive reactance (XL) is greater than the capacitive reactance (XC), the power factor is not equal to 1.
- This is because the presence of more inductive reactance causes the current to lag behind the voltage, leading to a phase difference.
- In such cases, the power factor is less than 1, indicating that the circuit is not purely resistive and has reactive components affecting the power flow.

Hotwire instruments are based on the heat I_rms²​Rt and/or power I_rms²​R producing property of current. Hence it can measure both ac and dc current as both produce heat when passed through a conductor.

The electromagnetic waves are produced by vibration of an electric charge. This creates a wave which is both as electric and magnetic components. This way oscillates in perpendicular planes with respect to each other and in a phase.This creates all electromagnetic waves begin with an oscillating charged particle which creates an oscillating electric and magnetic field.

Magnetic field part of a harmonic electromagnetic wave in vacuum
,B₀​=510×10⁻⁹T
Speed of light,
C=3×10⁸m/s
E=cB_o​=153N/C

TV signals being of high frequency are not reflected by the ionosphere. Therefore, to reflect these signals, satellites are needed. That is why, satellites are used for long distance TV transmission.
 Most long-distance shortwave (high frequency) radio communication—between 3 and 30 MHz—is a result of skywave propagation.
This 3-30 MHz is a range of frequencies which are used in sky waves propagation so that the ionosphere is capable of reflecting it.

This is because for maximum diffraction the slit width always equals to the wavelength of light.