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CUET Exam  >  CUET Test  >   UG Mock Test Series 2026  >  Physics: CUET Mock Test - 4 - CUET MCQ

CUET UG Physics: Mock Test - 4 Free Online Test 2026


Full Mock Test & Solutions: Physics: CUET Mock Test - 4 (50 Questions)

You can boost your CUET 2026 exam preparation with this Physics: CUET Mock Test - 4 (available with detailed solutions).. This mock test has been designed with the analysis of important topics, recent trends of the exam, and previous year questions of the last 3-years. All the questions have been designed to mirror the official pattern of CUET 2026 exam, helping you build speed, accuracy as per the actual exam.

Mock Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 60 minutes
  • - Total Questions: 50
  • - Analysis: Detailed Solutions & Performance Insights

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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: 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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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: Question 2

  • 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).

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Physics: CUET Mock Test - 4 - Question 3
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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: Question 3

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 4
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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: Question 4

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 5
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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: Question 5

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 6
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The electron emitted in β – radiation originates from where?

Detailed Solution: Question 6

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 7
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What is the ripple factor for a half-wave rectifier?

Detailed Solution: Question 7

For a half-wave rectifier,

r = 1.21

Physics: CUET Mock Test - 4 - Question 8
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The frequencies for transmitting music is which of the following?

Detailed Solution: Question 8

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 9
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Which of the following is mainly sued in the production of Integrated Circuits?

Detailed Solution: Question 9

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 10
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According to Kirchhoff’s Loop Rule

Detailed Solution: Question 10

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 11
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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: Question 11

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 12
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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: Question 12

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 13
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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: Question 13

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 14
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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: Question 14

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 15
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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: Question 15

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 16
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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: Question 16

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 17
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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: Question 17

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 18
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Capacitors and capacitance:
(A) Capacitance depends on area and separation.
(B) Increases with dielectric insertion.
(C) Unit is farad.
(D) Is independent of plate material.

Detailed Solution: Question 18

The capacitance of a parallel plate capacitor is C = ε₀A/d (depends on area A and separation d). Inserting a dielectric with dielectric constant K increases capacitance by a factor of K (C = Kε₀A/d). The SI unit of capacitance is the farad (F). For ideal capacitors, capacitance depends only on geometry and the medium between plates, not on the conducting material of the plates.

Physics: CUET Mock Test - 4 - Question 19
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Combination of capacitors in series and parallel:
(A) In series, total C is less than individual.
(B) In parallel, total C is sum.
(C) Charge is same in series.
(D) Voltage is same in series.

Detailed Solution: Question 19

In series combination, the equivalent capacitance is less than the smallest individual capacitance (1/Ceq = Σ1/Ci). In parallel, capacitances simply add: Ceq = ΣCi. In series connection, the same charge appears on each capacitor. Voltage divides across capacitors in series (same voltage across each in parallel).

Physics: CUET Mock Test - 4 - Question 20
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Dielectrics and polarization:
(A) Dielectrics reduce the electric field inside.
(B) Polarization is induced dipole moment per unit volume.
(C) Dielectric constant K > 1 for dielectrics.
(D) Polarization occurs only in conductors.

Detailed Solution: Question 20

When a dielectric is inserted in an electric field, polarization produces an opposing field, reducing the net electric field inside the material. Polarization P is defined as the induced dipole moment per unit volume. The relative permittivity (dielectric constant) K is always greater than 1 for dielectrics. Polarization is a characteristic property of dielectrics/insulators, not conductors.

Physics: CUET Mock Test - 4 - Question 21
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Two concentric shells of uniform density of mass M1 and M2 are situated as shown in the figure. The forces experienced by a particle of mass m when placed at positions A, B and C respectively are (given OA = p, OB = q and OC = r)

Detailed Solution: Question 21

We know that attraction at an external point due to spherical shell of mass M is GM/r2 while at an internal point is zero. So, for particle at point C
OA = G (M1 + M2)m/p2
OB = G M1m/q2
OC = 0
 

Physics: CUET Mock Test - 4 - Question 22
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Power versus time graph for a given force is given below. Work done by the force upto time t(<t0).

Detailed Solution: Question 22

The work done by force from time t = 0 to t = t sec. is given by shaded area in graph below.
Hence as t increases, this area increases. 

∴ Work done by force keeps on increasing

Physics: CUET Mock Test - 4 - Question 23
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Four similar particles of mass m are orbiting in a circle of radius r is the same direction and same speed because of their mutual gravitational attractive force as shown in the figure. Speed of a particle is given by

Detailed Solution: Question 23

Net force is towards centre

 

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: 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: 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: 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: 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: 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: 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: 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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This mock test contains 50 questions designed to replicate the pattern, difficulty level and time pressure of the actual CUET exam. It is part of CUET UG Mock Test Series 2026 on EduRev, covering all the topics that are being asked as per the current CUET syllabus. Attempting this test under timed conditions will help you assess your exam readiness, identify subject-wise weak areas, and improve your speed and accuracy before the actual exam.

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