Physics: CUET Mock Test - 8 - CUET MCQ

Given: V₂ = 10 V; V₁ = 20 V; C = 5 μF; Inductance (I) = 0.5 mH
Initial charge on the capacitor (q₁) = C × V₁ = 5 × 10^-6 × 20 ……………….1
q₁ = 10^-4 C …………..B
The instantaneous charge on the capacitor as the capacitor discharges through the inductor ⇒ q₂
q₂ = q₁cos (ωt) ⇒ q₂/q₂ = cos (ωt) ………………..A
Also, q₂ = C × V₂ = 5 × 10^-6 × 10 …………………….2
q₂ = 0.5 × 10^-4 C
From 1 and 2 ⇒ q₂/q₂ = V₂V₁ ⇒ q₂/q₂ = 0.5 = 1/2
From equation A, we can equate as follows ⇒ cos (ωt) = 12
ωt = π/2 rad ………………..3
For an LC circuit ⇒ 
ω=20000rad/s …………………….4
The current through the circuit is given as:
Current (I)=-dq/dt
Charge decreases with respect to time, so, dq/dt obtained will be negative and this is why we add a negative sign to make a current

Concept:

The correct sequence for a generalized communication system is:

1. Information source
2. Transmitter
3. Channel
4. Receiver
5. Destination

Information source: This is the source that generates the information that needs to be communicated. It can be a human, a computer, or any other device that can produce data.

Transmitter: The transmitter processes the information from the source and converts it into a form suitable for transmission over the channel. This may involve modulation, amplification, or encoding of the signal.

Channel: The channel is the medium through which the signal travels. It can be a wired or wireless medium, such as a cable, fiber optic line, or radio frequency spectrum.

Receiver: The receiver processes the signal received from the channel and converts it back into a form suitable for the destination. This may involve demodulation, filtering, or decoding the signal.

Destination: The destination is the endpoint of communication, where the information is received and interpreted. It can be a person, a computer, or any other device that can process information.

Therefore, the correct sequence is: Information source → Transmitter → Channel → Receiver → Destination.

The correct answer is option (2)

Calculation:

Let Voltage V = V_mSin(wt)

V = 200Sin(wt)

When connected to "X",

The current flowing lags "V" by π/2

I = 5Sin(wt - π/2)

Since the current lags "V" by π/2, the element "X" would be the inductor.

Then reactance "X" would be = 200/5 = 40 ohm

The reactance "X" = 40j.

When connected to "Y" , the current 5 A is in phase with the voltage.

So, the "Y" will be a resistor.

Resistance "Y" = 200/5 = 40 ohm.

The "X" and "Y" are connected in series to the same applied voltage.

The net impedance would be = 40 + 40j

The current I' flowing, when connected in series, will be,


Peak Value of I' = 5/√2
The RMS value of I' will be =
= 5/2
= 2.5 A.
The correct answer is option (3)

Concept:

• To increase the magnification power of a refracting type telescope, we should increase the focal length of the objective lens or decrease the focal length of the eyepiece lens.
• This is because the magnification power of a telescope is determined by the ratio of the focal length of the objective lens to the focal length of the eyepiece lens.
• However, it's important to note that increasing the magnification power of a telescope beyond a certain point may not result in a clearer or better image.
• This is because factors such as atmospheric turbulence, optical aberrations, and aperture limitations can affect the clarity and resolution of the image.
• Therefore, it's important to strike a balance between magnification power and image quality.

Additional Information

Angular magnification:

Angular magnification is defined as the ratio of the angle subtended by the image of an object which is seen from the telescope to the angle subtended by the same object without the aid of the telescope and it is written as;

Angular magnification, M =

Here, fo is the focal length of an object and fE is the focal length of an eyepiece.

Angular resolution:

It is defined as the angle of resolving power of the telescope and it is written as;

Angular resolution =

Here, D is the diameter, which is the wavelength of the light which is used.

Concept:

• For a plano-convex lens, one of the surfaces is flat (Plano) and the other is curved (convex).
• The radius of curvature of the curved surface is given as 20 cm, and the refractive index of the material of the lens is 1.5.
• The focal length of the lens can be calculated using the lens maker's formula:

1/f = (n - 1) × (1/R1 - 1/R2)

where

f is the focal length of the lens,

n is the refractive index of the lens material,

R1 is the radius of curvature of the first surface (the flat surface in this case, which is infinite), and

R2 is the radius of curvature of the second surface (the curved surface in this case).

Since the first surface is flat, the radius of curvature R1 is infinite, and the term (1/R1) becomes zero. Therefore, the lens maker's formula simplifies to:

1/f = (n - 1) × (1/R2)------(1)

Calculation:

Substituting the values in equation (1), we get:

1/f = (1.5 - 1) × (1/20)

1/f = 0.025

f = 1/0.025

f = 40 cm

Therefore, the focal length of the plano-convex lens is 40 cm.

The correct answer is option (4)

Concept:

The height of the image formed by a convex mirror can be calculated using the mirror formula:

1/f = 1/v + 1/u

where

f is the focal length of the mirror,

v is the distance of the image from the mirror, and

u is the distance of the object from the mirror.

Calculation:

In this case, the object distance u is equal to the focal length f, so we can simplify the formula to:

1/f = 1/v + 1/f ----(1)

From sign convention u = -f

Putting u = -f in equation (1), we get,

v = f/2

This means that the image is formed at a distance of half the focal length from the mirror.

Now, to find the height of the image, we can use the magnification formula:

m = -v/u

where m is the magnification of the image. Since the image formed by a convex mirror is always virtual and upright, the magnification is negative.

Substituting the values of v and u, we get:

m = -f/2f = -1/2

This means that the image is half the size of the object, and since the object is 1 meter tall, the height of the image will be:

Height of image = m * height of object = (-1/2) * 1m = -0.5m

Therefore, the height of the image is 0.5 meters or 50 centimeters.

The correct answer is option (4)

Concept:

• Electromagnetic (EM) spectrum: The electromagnetic (EM) spectrum is the range of all types of EM radiation arranged according to frequency or wavelength.
• In order of decreasing wavelength or increasing frequency these are:
• radio waves radiation, microwave radiation, infrared, visible light, ultraviolet, X-rays, and gamma rays.

• In a vacuum, all EM radiations travel at the same speed irrespective of the wavelength or frequency of the wave.

Explanation:

• All EM radiations travel at the same speed.
• So, X-rays and UV light travels at the same speed.
• Since every EM radiations have different wavelengths according to their energy level.
• So the X-rays are different from UV light because it has a different wavelength from UV light.
• Hence the correct answer is option 2.

CONCEPT:

Dual nature of matter:

• According to de Broglie, the matter has a dual nature of wave-particle.
• The wave associated with each moving particle is called matter waves.
• ​​de Broglie wavelength associated with the particle

Where, h = Planck's constant, m = mass of a particle and v = velocity of a particle

Work-Energy Theorem:

• The work-energy theorem states that the net work done by the forces on an object is equal to the change in its kinetic energy.

​⇒ W = ΔKE

Where W = work done and ΔKE = change in kinetic energy

CALCULATION:

Given V = 2 volt, charge of an electron e = 1.6×10^-19 C and m = 9.1×10^-31 kg

• We know that if a charge of e coulombs is moved through a potential difference of V volts then the work done is given as,

⇒ W = eV -----(1)

By work-energy theorem, the work done in accelerating an electron through the electric field will be equal to the kinetic energy of the electron,

-----(2)

By equation 1 and equation 2,

-----(3)

• The de Broglie wavelength associated with the electron is given as,

-----(4)

By equation 3 and equation 4,

Since h = 6.6 × 10^-34 J-sec

⇒ λ = 8.64 × 10^-10

⇒ λ = 8.64 Å

• Hence, option 2 is correct.

Concept:

Uniform Magnetic Field:

• A uniform magnetic field is a magnetic field that has the same magnitude and direction throughout the region under consideration, thus the field lines need to be both parallel and spaced out evenly.
• It is symbolled as B.
• It is a vector quantity.

Magnetic Force:

• The magnetic force is a consequence of the electromagnetic force, one of the four fundamental forces of nature, and is caused by the motion of charges.
• Two objects containing charges with the same direction of motion have a magnetic attraction force between them.
• Similarly, objects with charges moving in opposite directions have a repulsive force between them.
• The usual way to go about finding the magnetic force is framed in terms of a fixed amount of charge q moving at constant velocity v in a uniform magnetic field B.
• The magnetic force is described by the Lorentz Force law:

In this form, it is written using the vector cross product.

Given:

Charge C or q = 0.2 C, Velocity, , Uniform magnetic Field,

Calculation:

Substitute the value in the equation, we get

So, Option 2 is correct.

Concept:

• An alternating current is an electric current that reverses direction periodically.
• The general equation of emf is given by
  • E = E₀ sin (ωt) --- (1)
  • E = E₀ cos (ωt) --- (2)
• Where, I₀ = Peak current, ω = angular frequency, t = time
• ---- (3)
• Where, T = time period.

Calculation:

For a positive half cycle, time period t = T/2

Then average e.m.f. during the positive half cycle of an A.C is

E_av (half cycle) = E/t

From equation 1 and 3
⇒
⇒

Concept:

For transmitting a signal, we need an antenna or an aerial.
→The antenna should have a size comparable to the wavelength of the signal (at least λ/4 in dimension) so that the antenna properly senses the time variation of the signal.(where λ is wavelength)
→We can obtain transmission with a reasonable antenna length if the transmission frequency is high (for example, if v is 1 MHz, then λ is 300 m).
→Therefore, there is a need of translating the information contained in our original low-frequency baseband signal into high or radio frequencies before transmission.
→The minimum size of the antenna required to send the electromagnetic wave signal is given as,

length of antenna = λ/4

Explanation:

Given:

Length of antenna =100 m

Height of building = 500 m

The wavelength of wave is

λ = 4 (length of antenna) = 4 (100) = 400 m

Hence, the option (1) is the correct answer.

The most common application of LC oscillators is radio transmitters and receivers. LC oscillators have good phase noise characteristics as well as offer ease of implementation. Due to these factors, LC oscillators are most commonly used in radio-frequency circuits.

When we are connecting a charged capacitor to an inductor, the electric current in the circuit and charge on the capacitor undergoes LC oscillations. All the others are incorrect statements regarding LC Oscillations.

The self-inductance of a coil is said to be one henry if an induced emf of one volt is set up in it when the current in it changes at the rate of one ampere per second. Self inductance is defined as the induction of a voltage in a current-carrying wire when the current in the wire itself is changing.

Self-inductance is directly proportional to the number of turns and the area of cross-section. The self-inductance of a solenoid increases μ_r times if it is wound over an iron core of relative permeability μ_r.

The mutual inductance of two coils is the property of their combination. It does not matter which one of them functions as the primary or the secondary coil. Hence, mutual inductance does not depend on reciprocity.

NΦ = MI
200 × 500 = M × 10
M = 10⁴ H

Self-inductance = Magnetic flux / Current
Self inductance = 10×10^-6 / 5 × 10^-3
Self inductance = 2 × 10^-3 H
Self inductance = 2mH

Self-inductance = Magnetic flux / Current
Self-inductance = 50×10⁻³/5
Self-inductance = 1 × 10^-2 Wb

Mutual Inductance is the interaction of one coils magnetic field on another coil as it induces a voltage in the adjacent coil. Unit of Mutual inductance = V / As⁻¹.
SI unit = VsA^-1.

Self-inductance = 

Magnetic flu x Current

Self-inductance = [M L² T^-2 A^-2]

Magnetization is also termed as magnetic polarization. Magnetic polarization is basically a balance between the magnetic flux density in a space which is devoid of matter and the magnetic flux density in a space with matter, i.e. in a material.

The statement magnetic intensity and intensity of magnetization are the same is the false one. They are not the same. When a magnet is entering a magnetic field, then the poles of the magnet experiences certain forces. Magnetic intensity refers to the measure of these forces. But, the intensity of magnetization explains the change in the magnetic moment of a magnet as a function of volume. All the other statements are valid.

Diamagnetic materials are those which when placed in a magnetizing field are feebly magnetized in a direction opposite to that of the magnetizing field. Therefore, it is feebly repelled by a magnet. Example: zinc, gold, etc.

The relative permeability of ferromagnetic materials is much larger than 1. Ferromagnetic substances are those, which when placed in a magnetic field are strongly magnetized in the direction of the magnetizing field, i.e., it is strongly attracted by a magnet.

Diamagnetic materials have a small and negative value of magnetic susceptibility. Paramagnetic materials have a small and positive value of magnetic susceptibility and ferromagnetic materials have a large and positive value of magnetic susceptibility.

Iron is a ferromagnetic material. Ferromagnetism decreases with a rise in temperature, and above a certain temperature called “Curie’s temperature,” ferromagnets become paramagnets. For iron, it is approximately, 800 °C.

Permanent magnets are made up of materials having high retentivity, high coercivity, and high permeability such as steel. Other suitable materials for making permanent magnets are alnico (an alloy of iron, aluminum, nickel, copper, and cobalt), cobalt steel, and ticonal. That leaves zinc as the one that cannot be used to make permanent magnets.

The magnetic susceptibility of diamagnetic materials doesn’t depend upon temperature. This is often because there’s an outsized barrier between the bottom state and therefore the excited states of diamagnetic substances. Hence, it remains constant during the change of temperature. All the other statements are false.