Physics: CUET Mock Test - 5 - CUET MCQ

Given: E_rms = 300 V
The required equation is → P.I.V. = 2 × E₀ or P.I.V. = 2√2 × E_rms
P.I.V. = 2√2 × 300 V
P.I.V. = 848.52 V ≈ 849 V.

Concept:

The photoelectric effect:

• The photoelectric effect is the phenomenon where electrons are emitted from the surface of a metal when radiation of appropriate frequency, such as visible or ultraviolet light, is allowed to incident on it.
• This effect was first observed by Heinrich Hertz in 1887 and was later explained by Albert Einstein in 1905.
• In the photoelectric effect, a photon of sufficient energy strikes the metal's surface, transferring its energy to an electron in the metal.
• If the photon's energy is greater than the metal's work function, which is the minimum energy required to remove an electron from the metal, the electron can be ejected from the metal's surface.

Application:

The photoelectric effect has many practical applications, including photovoltaic cells used in solar panels, photoelectric detectors used in cameras, and photocells used in automatic doors and burglar alarms.

Additional InformationNuclear fission:

• Nuclear fission is the process where the nucleus of an atom is split into smaller nuclei, releasing a large amount of energy.
• This process is used in nuclear power plants to generate electricity.

Compton effect:

• The Compton effect is the phenomenon where a photon loses some of its energy when it collides with an electron, resulting in the photon's wavelength increasing.
• This effect is used in medical imaging to produce X-ray images of the body.

Thermionic emissions:

• Thermionic emissions are the emissions of electrons from heated material.
• This process is used in electron guns to produce the electrons used in cathode ray tubes and vacuum tubes.

The correct answer is option (3)

Concept:

De Broglie wavelength:

• The De Broglie wavelength is given by the formula λ = h/p---(1)

where h is Planck's constant and p is the momentum of the particle.

• Since all the particles have the same kinetic energy, we can say that they all have the same momentum, as momentum is given by p = √(2mK), where m is the mass of the particle and K is the kinetic energy.

Now, comparing the masses of the particles

We can see that the α particle (which consists of two protons and two neutrons) has the largest mass, followed by the deuteron (which consists of a proton and a neutron), then the proton, and finally the electron with the smallest mass.

Since for the same kinetic energy K, the de Broglie wavelength is inversely proportional to the square root of the mass.

Out of the given particles, the alpha particle has the highest mass and would possess De Broglie wavelength.

The correct answer is option (3)

Concept:
The ratio of radii of two nuclei can be approximated by the following formula:

Where r₁ and r₂ are the radii of the two nuclei, and A₁ and A₂ are their atomic mass numbers.
Using this formula, we can calculate the ratio of radii of the two nuclei with atomic mass numbers 27 and 8 respectively as follows:

Therefore, the ratio of radii of the two nuclei is 1.5 i.e. 3/2.
The correct answer is option (1)

Important Points

• This formula is based on the assumption that the nuclear density is constant for both nuclei.
• It is derived from the fact that the volume of a nucleus is proportional to its mass number, and the radius of a nucleus is proportional to the cube root of its volume.
• However, it should be noted that this formula is only an approximation and may not hold true for all nuclei.

Calculation:

The amount of substance left after 12 years can be found using the formula for radioactive decay:

N(t) = No * (1/2)​(t/t_1/2)

where N(t) is the amount of substance remaining after time t, No is the original amount of substance, t_1/2 is the half-life, and t is the time elapsed.

In this case, No (since No is the original mass), t1/2 = 4 years, and t = 12 years.

Substituting these values into the formula gives:

N(12) = No * (1/2)^12/4
N(12) = No * (1/2)³
N(12) = N_o/8
N(12) = 0.125 * No

Therefore, after 12 years, the amount of substance remaining is 0.125 times the original mass or 12.5% of the original mass.

N = N0/8

The correct answer is option (3)

Concept:

Uranium:

• Uranium is a radioactive element commonly used as fuel in nuclear reactors. It can undergo nuclear fission, releasing a large amount of energy.
• The rate of the fission reaction can be controlled by adjusting the concentration of Uranium-235, a fissile isotope of uranium, in the fuel rods.
• It is used for fission reactions.

Moderator:

• In a nuclear reactor, the fission reaction produces fast-moving neutrons that need to be slowed down in order to interact with the fuel and sustain the reaction.
• The moderator is a material, usually water or graphite, that is used to slow down the neutrons to a speed at which they can interact with the fuel and produce more neutrons.
• It Slows down the fast-moving neutrons

Control rod:

• Control rods are devices made of neutron-absorbing materials such as boron or cadmium that can be inserted or removed from the reactor core to control the rate of the fission reaction.
• By absorbing neutrons, the control rods reduce the number of neutrons available to interact with the fuel, slowing down or stopping the reaction.
• The reaction rate can be controlled by it.

Coolant:

• The coolant is a fluid, typically water or gas, that circulates through the reactor core to transfer heat from the fuel rods to a steam turbine.
• The heat generated by the fission reaction is used to produce steam, which drives the turbine and generates electricity.
• The coolant also helps to regulate the temperature of the core and prevent it from overheating.
• It transfers heat from the core to turbine.

​So, the answer would be (A) - (III), (B) - (II), (C) - (I), (D) - (IV)

The correct answer is option (3).

Concept:

The intensity of a sinusoidal plane electromagnetic wave defined as the average value of poynting vector taken over one cycle (in free space)

I = P_av/(4πr²) = E_m²/2μ₀c

Where,

Pav → Average power

Em → Maximum value of electric field

r → Distance

μ0 → permeability of free space = 4π × 10^-7 m kg s^-2 A^-2

c → Speed of light = 3 × 10⁸ m/s

Calculation:

Given

Pav = 1000 W

r = 3.5 m

μ0 = 4π × 10-7 m kg s-2 A-2

c = 3 × 108 m/s

Pav/(4πr2) = Em2/2μ0c

⇒ Em = √[2μ0cPav/(4πr2)]

∴ Em = √[4π × 10-7 ×3 × 108× 1000/(2π3.52)]

=69.98 ≈ 70 V/m

The correct answer is Both I and II.

Key Points

• When a ray of light passes from one medium to another, it suffers a change in direction at the boundary of two mediums is called refraction.
  • The change in direction because the speed of light travels at different speeds in different mediums.
  • Refractive Index of a medium defines that it is denser or rarer.
  • When a light ray travels from a lighter to a denser medium, the light ray bends towards the normal.
  • When a light ray travels from a denser to a lighter medium the light ray bends away from the normal.
• Snell’s law of refraction
  • “The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given color and for the given pair of media”.
• The refractive index of water is 1.33.
  • It means that light travels 1.333 times faster in a vacuum than it does in water.

Concept:
According to wave-particle duality, the de-Broglie wavelength in Quantum Mechanics determines the probability density of finding the object in a space
λ = h/p

Explanation:

⇒ = 1 (As we have the same de-Broglie wavelength)

⇒ p₁ = p₂

Since m₁ > m₂
Thus E₁ < E₂.
The correct answers are option (3).

The correct answer is option 3):(1500 C)

Concept:

If Q charge flows through the conductor for ‘t’ seconds, then the current given by that conductor is

I=

Q = I × t

I = current

t = time

Calculation:

Given

I = 5 A

t = 5 min = 300 sec

Q = 5× 300

= 1500 C

Concept:

• The refractive index, also called the index of refraction, measures the bending of a ray of light when passing from one medium into another.
• Also defined as the ratio of the speed of light in a vacuum to the speed of light in a medium.
• The refractive index of a medium can be calculated using the following formula:
• n = c/v
• where n is the refractive index of the medium, c is the velocity of light in vacuum, and v is the velocity of light in the medium.
• Lens maker formula.
• 

Explanation:

Let the refractive index of the medium is n₁.

• The refractive index of the liquid must be equal to 1.47 in order to make the lens disappear.
• This means n₁ = n.
• From lens maker formula.
• This gives 1/f = 0 or f → ∞.
• The lens in the liquid will act like a plane sheet of glass.
• The liquid could be glycerine.

By thin lens formula,
1/f = 1/v - 1/u
1/f = 1/(-50) - 1/(-25)
f = +50 cm.

Original fringe width is given as → β = Dλ/d.
So, the new fringe width is → β’ = 
β’ = 4β
Therefore, the new fringe width is 4 β.

Angular magnification is maximum when the final image is formed at the near point.

u = -7.1 cm.
Therefore, the object should be placed 7.1 cm before the lens to avoid angular magnification.

Fringe width is expressed as:
β = Dλ/d.
So, as the separation (d) between the two slits decreases, the fringe width increases. They are inversely proportional to each other.

Given: I₀ = 0.092 A
The required equation is → I_rms = I_o/√2
I_rms = 0.092/1.414
I_rms = 0.065 A

Given: f = +10 cm; v = -30 cm
Required equations →
1/f = 1/v - 1/u
m = (size of the image/size of the object) = v/u
1/u = 1/v - 1/f = 1/(-30) - 1/10 = 
u = (-15)/2 = -7.5cm
m =v/u = -(30/(-7.5)) = -300/-75 = +4
Therefore, the object should be placed at a distance of 7.5 cm from the lens to get the image at a distance of 30 cm from the lens. It is four times enlarged and is erect.

N = N₀e^{-(λ1 + λ2) t}.
4 = e^{(λ1 + λ2) t}.
t =  = 1080 year.

The sudden increase in current in a Zener diode is due to the rupture of the many covalent bonds present. Therefore, the Zener diode should be connected in reverse bias.

A rectifier is based on the fact that a forward bias p-n junction conducts and a reverse bias p-n junction does not conduct electricity. The rectifier is used to convert alternating current voltage (ac) to direct current voltage (dc).

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.

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.

In Amplitude modulation, the modulating wave is superimposed on a high-frequency carrier wave such that the amplitude of the modulated wave varies as the amplitude of the modulating wave.

Protons are not emitted by radioactive substances during their decay. They are positively charged subatomic particles found in the atomic nucleus. The others can be emitted by radioactive substances during their decay.

Semiconductor diode can be used as a rectifier. The function of a rectifier is that it converts an alternating current into direct current by allowing the current to pass through in one direction.

Rectifier is a device that does the process of rectification. This means that rectifiers straighten the direction of the current flowing through it. There are mainly 2 types of rectifiers, namely, full-wave rectifiers and half-wave rectifiers.

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.

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.

AGC stands for automatic gain control. Automatic gain control changes (AGC) the overall gain of a receiver automatically so that the strength of the received signal remains almost constant.