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/q₁ = cos (ωt) ………………..A
Also, q₂ = C × V₂ = 5 × 10^-6 × 10 …………………….2
q₂ = 0.5 × 10^-4 C
From 1 and 2 ⇒ q₂/q₁ = V₂/ V₁ ⇒ q₂/q₁ = 0.5 = 1/2
From equation A, we can equate as follows ⇒ cos (ωt) = 1/2
ωt = π/3 rad ………………..3
For an LC circuit ⇒ 
ω=20000rad/s …………………….4
The current through the circuit is given as:
Current (I)=-dq/dt

Current: I = -dq/dt = q₁ωsin(ωt) = 10⁻⁴ × 20,000 × sin(π/3) = 2 × 0.866 = 1.732 A
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 curren

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
⇒
⇒

Long distance radio broadcasts use shortwave bands because only these bands can be refracted by the ionosphere.

Maxwell’s Fourth Equation
It is based on Ampere’s circuital law. To understand Maxwell’s fourth equation it is crucial to understand Ampere’s circuital law,
Consider a wire of current-carrying conductor with the current I, since there is an electric field there has to be a magnetic field vector around it. Ampere’s circuit law states that “The closed line integral of magnetic field vector is always equal to the total amount of scalar electric field enclosed within the path of any shape” which means the current flowing along the wire(which is a scalar quantity) is equal to the magnetic field vector (which is a vector quantity)

The direction of propagation of the electromagnetic wave is always perpendicular to the plane in which
So, the direction of the propagation of the wave, 
Hence, option B is the correct answer.

Electric field lines show the direction of the electric field at the point. If the electric field lines were tangential, parallel, or opposite to the equipotential surface, a tangential field will exist on the surface and work done in moving a charge on the surface is not zero.

Therefore electric field lines are always perpendicular to the equipotential surface.

Aluminium electrolytic capacitors have the Aluminium foil anode (positive terminal) which is attached and covered with a layer of Aluminium Oxide which acts as a dielectric. The whole assembly is covered using a paper separator soaked in electrolyte such as, Borax or Glycol and covered by Aluminium foil which acts as cathode ( negative electrode)

A positive charge experiences a force in the direction of the field.
If it moves opposite to the direction of the field, negative work is done by the electric field on the charge.
Potential energy of the charge increases since an external source is required to do work on the charge against the direction of the field, and the work done is stored in the charge as its potential energy.

Given, the maximum kinetic energy: K_max​=4eV
If V0​ be the stopping potential, then K_max​=eV₀​
⇒eV_0​=4eV 
⇒V₀​=4V

Photoelectric cell or photocell, device whose electrical characteristics (e.g., current, voltage, or resistance) vary when light is incident upon it. The most common type consists of two electrodes separated by a light-sensitive semiconductor material.
The photoelectric effect is the observation that many metals emit electrons when light shines upon them. Electrons emitted in this manner can be called photoelectrons.

A particle representing a quantum of light or other electromagnetic radiation. A photon carries energy proportional to the radiation frequency but has zero rest mass.

The maximum velocity of an electron emitted from the photocell is given as 3.75×10⁸ cm/s. To find the stopping potential, follow these steps:

• Use the formula for kinetic energy: KE = 0.5mv².
• Convert the velocity into metres per second: 3.75×10⁶ m/s.
• Calculate the kinetic energy of the electron: 
  KE = 0.5 × 9.11×10^-31 kg × (3.75×10⁶ m/s)².
• Convert the kinetic energy into electron volts (eV): 1 eV = 1.6×10^-19 J.
• The stopping potential, V, is given by the kinetic energy in eV, which is approximately 40 volts.

Fluorescence is the emission of light by a substance that has absorbed light or other electromagnetic radiation. It is a form of luminescence. In most cases, the emitted light has a longer wavelength, and therefore lower energy, than the absorbed radiation. The most striking example of fluorescence occurs when the absorbed radiation is in the ultraviolet region of the spectrum, and thus invisible to the human eye, while the emitted light is in the visible region, which gives the fluorescent substance a distinct color that can be seen only when exposed to UV light. Fluorescent materials cease to glow nearly immediately when the radiation source stops, unlike phosphorescent materials, which continue to emit light for some time after.

In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67×10−27) is less than the mass of incident α−particles (6.64×10−27). Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the α−particles would not bounce back if solid hydrogen is used in the α−particle scattering experiment.

Control rods are used in nuclear reactors to control the fission rate of uranium and plutonium. They are composed of chemical elements such as boron, silver, indium and cadmium that are capable of absorbing many neutrons without themselves fissioning.

To find activity of the sample --->which is the rate of disintegration.

Since radioactivity comes under 1^o kinetics.

[R]=k[A]     [A]-->amount of initial sample 10¹⁸ atoms

Given,

Half-life=2days

K=0.693/2x24x60x60 sec

R=(0.693/2x24x60x60)x10¹⁸

R≈3.5x10¹² Bq

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 magnetic susceptibility of diamagnetic materials doesn’t depend on temperature. This is often because there’s an outsize 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.

Paramagnetic materials are those when suspended freely inside the magnetic field, it slowly sets itself parallel to the direction of the magnetic field. When placed in a non-uniform magnetic field, it tends to move from weaker to a stronger magnetic field.

When suspended freely, a thin long piece of magnet comes to rest nearly in the geographical north-south direction. When placed in a non-uniform magnetic field, it tends to move from weaker to stronger magnetic field.

The magnetic dipole moment of a magnetic dipole is defined as the product of its pole strength and magnetic length. The SI unit of magnetic dipole moment is ampere (metre)² (Am²). Magnetic dipole moment is a vector quantity and is directed from south to north pole of the magnet.

Magnetic dipole moment = Pole strength × Magnetic length.
Magnetic dipole moment = 20 Am × 0.1 m
Magnetic dipole moment = 2 Am².