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Physics: CUET Mock Test - 4 - CUET MCQ


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30 Questions MCQ Test - Physics: CUET Mock Test - 4

Physics: CUET Mock Test - 4 for CUET 2025 is part of CUET preparation. The Physics: CUET Mock Test - 4 questions and answers have been prepared according to the CUET exam syllabus.The Physics: CUET Mock Test - 4 MCQs are made for CUET 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Physics: CUET Mock Test - 4 below.
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Physics: CUET Mock Test - 4 - Question 1
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A transformer is used to light 100 W 25 volt lamp from 250 Volt ac mains. The current in the main cable is 0.5 A. Calculate the efficiency of the transformer.

Detailed Solution for Physics: CUET Mock Test - 4 - Question 1

Given: Output power (Po) = 100 W; Ip = 0.5 A; Vp = 250 V
Input power (Pi) = Vp × Ip
Pi = 250 × 0.5 = 125 W
Efficiency = (Output power/Input power) × 100%
Efficiency = (100/125) x 100
Efficiency = 80
Therefore, the efficiency of the transformer is 80%.

Physics: CUET Mock Test - 4 - Question 2
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To convert a galvanometer into an ammeter, one should connect :

Detailed Solution for Physics: CUET Mock Test - 4 - Question 2

Concept:

  • To convert a galvanometer into an ammeter, a shunt resistor is connected in parallel with the galvanometer.
  • A shunt resistor is a low resistance wire or a coil that is connected in parallel with the galvanometer to divert most of the current away from the galvanometer.
  • The shunt resistor is designed to have a very low resistance so that most of the current flows through it instead of the galvanometer.
  • By connecting the shunt resistor in parallel with the galvanometer, the total current is split between the shunt resistor and the galvanometer.
  • The amount of current that flows through the galvanometer is then proportional to the voltage drop across it.
  • By selecting an appropriate value for the shunt resistor, the current that flows through the galvanometer can be kept within its range, and the galvanometer can be used as an ammeter to measure higher currents.

The value of the shunt resistor is selected such that the full-scale deflection of the galvanometer occurs when the maximum current to be measured flows through the ammeter. The value of the shunt resistor is calculated using the following formula:

Rs =

where:

Rs is the resistance of the shunt resistor
G is the resistance of the galvanometer
Ig is the full-scale deflection current of the galvanometer
Im is the maximum current to be measured by the ammeter
The shunt resistor should have a very low resistance compared to the resistance of the galvanometer to ensure that most of the current flows through it.

The value of the shunt resistor should also be such that it can handle the maximum current to be measured without getting damaged or changing its resistance value

The correct answer is option (3).

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Physics: CUET Mock Test - 4 - Question 3
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Given below are two statements :
Statement I: The electric field produced by a scalar source is known as electric charge.
Statement II: The magnetic field produced by a vector source is known as current element (I dl).

In the light of the above statements, choose the correct answer from the options given below:

Detailed Solution for Physics: CUET Mock Test - 4 - Question 3

  • Statement I is incorrect because electric fields are generated by electric charges, not scalar sources.

  • Statement II is correct as a current element (I dl) generates a magnetic field, which is a vector field.

  • The correct answer is option (3).

Physics: CUET Mock Test - 4 - Question 4
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Which of the following rays are used in doing LASIK (Laser - Assisted in Situ keratomileusis) eye surgery?
Detailed Solution for Physics: CUET Mock Test - 4 - Question 4

​LASIK:

  • LASIK (Laser-Assisted in Situ Keratomileusis) is a type of refractive surgery that uses a laser to correct vision problems.
  • During the LASIK procedure, a surgeon uses a laser to reshape the cornea, which is the clear front part of the eye.
  • The laser used in LASIK surgery is an ultraviolet excimer laser.
  • Therefore, ultraviolet excimer laser rays are used in doing LASIK eye surgery.

So, the Ultraviolet rays are used in doing LASIK.
The correct answer is option (1).

Physics: CUET Mock Test - 4 - Question 5
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The magnetic field of a plane electromagnetic wave is given by Bx = 2 × 10-7 sin (0.6 × 103y + 2 × 1011t) T. An expression for its electric field is :
Detailed Solution for Physics: CUET Mock Test - 4 - Question 5

Concept:

  • Electromagnetic wave: It can also be said that electromagnetic waves are the composition of oscillating electric and magnetic fields.
  • Electromagnetic waves are shown by a sinusoidal graph.
  • It consists of time-varying electric and magnetic fields which are perpendicular to each other and are also perpendicular to the direction of propagation of waves.

  • If the electric field oscillated in the x-axis, then the magnetic field will oscillate in the y-axis and the direction of propagation of the electromagnetic wave is in the z-axis.
  • The speed of electromagnetic waves equals the speed of light in air.
  • The relation between the magnitude of the electric field and the magnetic field is given as E0 = B0c

Explanation:
Given,
The magnetic field in a plane electromagnetic wave, Bx = 2 × 10-7 sin (0.6 × 103y + 2 × 1011t) T.
Take, the speed of light, c = 3 ×108 ms-1
The relation between the magnitude of the electric field and the magnetic field is given as E0 = B0 c
E0 = 2 ×10-7 × 3 ×108
E0 = 60 V/m
So, the electric field in a plane electromagnetic wave is Ez = 60 sin (0.6 × 103y + 2 × 1011t) V/m.
The correct answer is option (4)

Physics: CUET Mock Test - 4 - Question 6
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Number of photoelectrons emitted per second is proportional to:
Detailed Solution for Physics: CUET Mock Test - 4 - Question 6

Concept:

Keeping the frequency constant of incident radiation,

  • The number of photoelectrons emitted per second is proportional to the intensity of incident light .This relationship is known as the photoelectric effect.
  • According to the photoelectric effect, when light of a certain frequency is incident on a metal surface, electrons are emitted from the metal surface.
  • The number of electrons emitted is proportional to the intensity of the incident light.

This relationship is expressed by the equation:

I = K *I0

Where I is the number of photoelectrons emitted per second, I0 is the intensity of the incident light, and K is a proportionality constant known as the photoelectric constant.

This equation shows that the number of photoelectrons emitted per second increases with increasing intensity of the incident light. However, the maximum kinetic energy of the photoelectrons is only dependent on the frequency of the incident light, and is independent of the intensity of the light.

The correct answer is option (1)

Physics: CUET Mock Test - 4 - Question 7
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The atomic hydrogen emits a line spectrum consisting of various series. Which series of hydrogen atomic spectra is lying in the visible region?
Detailed Solution for Physics: CUET Mock Test - 4 - Question 7

CONCEPT
→ When an electron comes down from a higher state (n > 2) to the n = 2 state, the e-m radiation emitted is called Balmer lines.
→ Visible spectrum lies in the range 380 to 700 nm. Only Balmer lines emit electromagnetic radiation that lies in the visible spectrum.

So, the correct answer is option (2).

Physics: CUET Mock Test - 4 - Question 8
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What will be the value of equivalent resistance when the resistors of 3 ohm and 6 ohm are parallel
Detailed Solution for Physics: CUET Mock Test - 4 - Question 8

The correct answer is 2Ω.
When resistors are connected in parallel, the formula for calculating the equivalent resistance is given by -

Given:
R1 = 3 and R2 = 6
Substituting given values and solving the above equation


R = 2Ω

Thus, the value of equivalent resistance when the resistors of 3 ohm and 6 ohm are parallel will be 2Ω.

Physics: CUET Mock Test - 4 - Question 9
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In the Davisson and Germer experiment, the velocity of electron emitted from the electron gun can be increased by
Detailed Solution for Physics: CUET Mock Test - 4 - Question 9

CONCEPT:

  • Davisson Germer experiment was conducted in 1927 to study the wave nature of the electron.
  • The electrons are emitted by an electron gun and are accelerated by applying a suitable potential difference between anode and cathode.
  • The finely collimated beam is directed towards the Ni crystal.
  • The scattered electrons are collected using a movable detector and a graph is plotted between applied potential and angle of reflection.
  • Analysis of those graphs showed the wave nature of the electron.

EXPLANATION:

  • The velocity of an electron accelerated through a potential is given by


Where e = Charge of electron and Vo = potential, m = mass of the particle

  • From the above equation it is clear that

⇒ V ∝ √Vo

  • The velocity of the electron is directly proportional to the square root of the potential difference between the anode and filament. Hence, option 2 is the answer
Physics: CUET Mock Test - 4 - Question 10
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The self-inductance of a coil is 100 mH. The magnetic flux linked with each turn of the coil ,while a current of 8 mA passes through it is found to be :
Detailed Solution for Physics: CUET Mock Test - 4 - Question 10

Concept:

​Magnetic flux:

  • It is defined as the number of magnetic field lines passing through a given closed surface.
  • It provides the measurement of the total magnetic field that passes through a given surface area.
  • Formula, ϕ = BA cosθ , where B = magnetic field, A = area, θ = angle between the area and magnetic field
  • The relation between the magnetic flux and current is given as, ϕ = LI, where L = self-inductance, I = current​

Calculation:
Given,
self inductance, L = 100 mH, current, I = 8 mA
The magnetic flux linked with each turn of the coil is given as,
ϕ = LI
ϕ = 100 × 10-3 × 8 × 10-3
ϕ = 8 × 10-4 Wb
Hence, the magnetic flux linked with each turn of the coil is 8 × 10-4 Wb.

Physics: CUET Mock Test - 4 - Question 11
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How many main types of rectifiers are there?
Detailed Solution for Physics: CUET Mock Test - 4 - Question 11

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.

Physics: CUET Mock Test - 4 - Question 12
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The electron emitted in β – radiation originates from where?
Detailed Solution for Physics: CUET Mock Test - 4 - Question 12

In β-emission, a neutron of nucleus decays into a proton and a β-particle.
The reaction given as:
0n1 → 1H2 + -1e0.

Physics: CUET Mock Test - 4 - Question 13
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What is the ripple factor for a half-wave rectifier?
Detailed Solution for Physics: CUET Mock Test - 4 - Question 13

For a half-wave rectifier,

r = 1.21

Physics: CUET Mock Test - 4 - Question 14
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The frequencies for transmitting music is which of the following?
Detailed Solution for Physics: CUET Mock Test - 4 - Question 14

To transmit music, an approximate bandwidth of 20 kHz is required because of the high frequencies produced by the musical instruments. The audible range of frequencies extends from 20 Hz to 20 kHz.

Physics: CUET Mock Test - 4 - Question 15
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Which of the following is mainly sued in the production of Integrated Circuits?
Detailed Solution for Physics: CUET Mock Test - 4 - Question 15

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.

Physics: CUET Mock Test - 4 - Question 16
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According to Kirchhoff’s Loop Rule
Detailed Solution for Physics: CUET Mock Test - 4 - Question 16

Kirchhoff’s loop rule is based on the principle of conservation of energy. The work done in transporting a charge in a closed loop is zero. The algebraic sum ( since potential differences can be both positive and negative) of potential differences around any closed loop is always zero.

Physics: CUET Mock Test - 4 - Question 17
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The Wheatstone bridge is balanced for four resistors R1,R2,R3 and R4 with a cell of emf 1.46 V. The cell is now replaced by another cell of emf 1.08 V. To obtain the balance again
Detailed Solution for Physics: CUET Mock Test - 4 - Question 17

The balance point of the Wheatstone’s bridge is determined by the ratio of the resistances. The change in the emf of the external battery will have no effect on the balance point.
 

Explanation:


  • Initial Balanced Wheatstone Bridge: In the initial balanced Wheatstone bridge configuration, the emf of the cell is 1.46 V and all four resistors R1, R2, R3, and R4 are set to specific values to achieve balance.

  • Replacement of Cell: When the cell is replaced by another cell with an emf of 1.08 V, the balance of the Wheatstone bridge is disrupted.

  • Requirement for Rebalancing: In order to rebalance the Wheatstone bridge with the new cell of emf 1.08 V, no resistance needs to be changed.

  • Reasoning: The balance of the Wheatstone bridge is determined by the ratio of the resistances in the bridge arms and not by the absolute values of the resistances. As long as the ratio of the resistances remains the same, the balance will be maintained regardless of the emf of the cell.

  • Conclusion: Therefore, in this scenario, no resistance needs to be changed to obtain the balance again with the new cell of emf 1.08 V.


    •  
Physics: CUET Mock Test - 4 - Question 18
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Two cells of emf 1.25V , 0.75V and each of internal resistance 1Ω are connected in parallel. The effective emf will be
Detailed Solution for Physics: CUET Mock Test - 4 - Question 18

We want the Voltage difference VAB.
let A and B be open and not connected to any thing.
There is a current that flows from cell of larger emf to the cell of small emf.
Call that current as  I Amperes.

1.25 V - 0.75 V - I * R2 - I * R1 = 0

I = 0.5 / (R1+R2)
VAB =  -0.75 - I * R2 =  - 0.75 - 0.5 * R2 / (R1+R2)

= - (0.75 R1 + 1.25 R2) / (R1+R2)

= - ().75 * 1 + 1.25 * 1) / (1 + 1)  volts

=  - 1 volts

Physics: CUET Mock Test - 4 - Question 19
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 If the length of the filament of a heater is reduced by 10% the power of the heater will
Detailed Solution for Physics: CUET Mock Test - 4 - Question 19

Power P= V2/R​
If length is reduced by 10% then, new resistance of filament will be R′.
R′ = R−10% of R
R′ = 0.9R
Now new power of heater is P2​
P2​ = V2​​/R′ = V2/0.9R ​ ​= 1.1 P
% increase powe r = 11%

Physics: CUET Mock Test - 4 - Question 20
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Two point charges (-q) and (+4q) are placed at separation 'r'. Where should a third charge be placed so that entire system of charges becomes in equilibrium?
Detailed Solution for Physics: CUET Mock Test - 4 - Question 20

Calculation:

Let a charge "Q" is placed at a distance of "x" from (-q).

Let the force exerted by "-q" be F1 and "4q" be F2 on "Q".

To make the system in equilibrium, the net force acting on "Q" should be zero.

i.e., F1 + F2 = 0.-----(1)

F1 = K(-q)Q/x2

F2 = K(4q)Q/(r + x)2

Putting the value of F1 and F2 in equation (1), we get,

K(-q)Q/x2 + K(4q)Q/(r + x)2 = 0

(x + r)/x = 2

x = r

The charge "Q" is placed at a distance of "r" from charge "-q" on the extreme side of (-q).

The correct answer is option (1).

Physics: CUET Mock Test - 4 - Question 21
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The variation of electric field with respect to distance from centre of a charged conducting spherical shell of radius R is given by :
Detailed Solution for Physics: CUET Mock Test - 4 - Question 21

Concept:

  • For a conducting spherical shell of radius R that carries a charge Q, the electric field inside the shell is zero.
  • And outside the shell, the electric field behaves as if all the charge Q is concentrated at the center of the shell.
  • Thus, the electric field outside the shell, at a distance r from the center, is given by Coulomb's law:

E = kQ/r2
where k is the Coulomb constant.

Therefore, the electric field outside the shell varies inversely with the square of the distance from the center of the shell.
That is, as the distance from the center of the shell increases, the electric field strength decreases rapidly.
So, the plot of electric field w.r.t the distance "r" would be,

The correct answer is option "2"

Physics: CUET Mock Test - 4 - Question 22
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A conducting sphere is charged. If the electric field at a distance 20 cm from the centre of the sphere is 1.2 × 103 NC-1 and points radially inwards, the net charge on the sphere is:
Detailed Solution for Physics: CUET Mock Test - 4 - Question 22

Concept:

We know that for a conducting sphere, the electric field outside the sphere is the same as the field due to a point charge located at the center of the sphere. So, we can use the formula for the electric field due to a point charge to find the net charge on the sphere.

The electric field due to a point charge q at a distance r from it is given by:

E = k*q/r2

where k is the Coulomb constant.

In this case, the electric field at a distance of 20 cm from the center of the sphere is given as 1.2 × 103 NC-1, and it points radially inwards.

This means that the point charge at the center of the sphere is negative.

So, we have:

E = k*q/r2

= q = ------(1)

Calculation:

Given that

E = 1.2 × 103 NC-1

r = 20 cm = 0.2 m.

k = 9 × 109

Substituting values of k, r, E in the equation we get,

q =

=

= 5.33 ×10-9 C

Since the point charge at the center of the sphere is negative.

q = - 5.3 ×10-9 C

The correct answer is option (4)

Physics: CUET Mock Test - 4 - Question 23
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A parallel plate capacitor having cross-sectional area 'A' and separated by distance 'd' is filled by copper plate of thickness b. It's capacitance is :
Detailed Solution for Physics: CUET Mock Test - 4 - Question 23

Concept:
The capacitance of a parallel plate capacitor is given by the formula:

C = εA/d
where
C is the capacitance,
ε is the permittivity of the medium between the plates,
A is the area of each plate, and d is the distance between the plates.

Calculation:
When a copper plate of thickness b is placed between the plates of the capacitor, the distance between the plates effectively reduces to d' = d - b, where b is the thickness of the copper plate. The permittivity of copper is very close to that of vacuum, so we can assume that the permittivity between the plates remains the same.

Therefore, the new capacitance of the capacitor is given by:
C' = εA/d'
C' = εA/(d - b)
C' =
The correct answer is option (2)

Physics: CUET Mock Test - 4 - Question 24
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The power in an AC circuit contains an inductor of 30 mH, a capacitor of 300 μF, a resistor of 70 Ω, and an AC source of 24 V, 60 Hz. Calculate the energy dissipated in the circuit in 1000 s.
Detailed Solution for Physics: CUET Mock Test - 4 - Question 24

We know that, Pav = Vrms Irms cos⁡Φ ……1 and cosΦ;= R/Z ....2
In LCR, cosΦ = VR/V = IR/IZ and Irms = Vrms/Z ....3
Substituting 2 and 3 in 1


Given: Vrms = 24 V; Resistance (R) = 70 Ω; Inductance (I) = 30 mH = 30 × 10-3 H; Capacitance (C) = 300 μF
XL = 2πυL = 2π (60)(30 × 10-3)
XL= 11.304 Ω
XC = (1/(2πvC)) = (1/(2π(60)(300×10−6)))
XC = 8.846 Ω
For series LCR circuit → Z = 

Z = 70.04 ≈ 70 Ω
So, the energy used in 1000 seconds is → Pavt =

Therefore, the energy dissipated in the circuit at 1000 seconds is 8.22 × 103J.

Physics: CUET Mock Test - 4 - Question 25
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In a step-down transformer, the number of turns in the secondary coil is 20 and the number of turns in the primary coil is 100. If the voltage applied to the primary coil is 120 V, then what is the voltage output from the secondary coil?
Detailed Solution for Physics: CUET Mock Test - 4 - Question 25

Given: Ns = 20; Np = 100; Vp = 120 V
For a step-down transformer → Ns/Np = Vs/Vp

Vs = 24 V
Therefore, the voltage output from the secondary coil is 24 V.

Physics: CUET Mock Test - 4 - Question 26
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In Young’s double-slit experiment with monochromatic light, how is fringe width affected, if the screen is moved closer to the slits?
Detailed Solution for Physics: CUET Mock Test - 4 - Question 26

In Young's double-slit experiment, the fringe width (β) is given by the formula:

β = λD / d

Where:

  • λ is the wavelength of the light,

  • D is the distance between the slits and the screen,

  • d is the distance between the two slits.

If the screen is moved closer to the slits, the distance D decreases. Since fringe width β is directly proportional to D, reducing the distance D will result in a decrease in the fringe width.

Physics: CUET Mock Test - 4 - Question 27
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Identify the expression for Bohr’s second postulate.
Detailed Solution for Physics: CUET Mock Test - 4 - Question 27

Bohr’s second postulate states that the angular momentum (L) of an electron in orbit is quantized: L = nh/2π, where n is an integer and h is Planck’s constant. Thus, the correct expression is L = nh/2π.

Physics: CUET Mock Test - 4 - Question 28
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Which among the following is a portion of a transparent refracting medium bound by one spherical surface and the other plane surface?
Detailed Solution for Physics: CUET Mock Test - 4 - Question 28

Lens is a portion of a transparent refracting medium bound by two spherical surfaces or one spherical surface and the other plane surface. Lenses are divided into two classes, namely, convex lens or converging lens, and concave lens or diverging lens.

Physics: CUET Mock Test - 4 - Question 29
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How much intensity of the image is increased if the diameter of the objective of a telescope is doubled?
Detailed Solution for Physics: CUET Mock Test - 4 - Question 29

On doubling the diameter, the area of the objective increases four times. Its light-gathering power increases four times. The brightness of the image also increases four times. Therefore, the intensity of the image increases by four.

Physics: CUET Mock Test - 4 - Question 30
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The wavelength of light coming from a sodium source is 589 nm. What will be its wavelength in the water?
Detailed Solution for Physics: CUET Mock Test - 4 - Question 30

λ = 589 nm, μ = 1.33.
Wavelength in water, λW = λ/μ
λW = 589/1.33
λW = 443 nm.

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