Explore Exams
Login Signup
Sign in Open App
JEE Exam  >  JEE Test  >  Mock Tests Main and Advanced 2026  >  JEE Main Practice Mock Test - 5 - JEE MCQ

JEE Main Practice Mock Test - 5 Free Online Test 2026


Full Mock Test & Solutions: JEE Main Practice Mock Test - 5 (75 Questions)

You can boost your JEE 2026 exam preparation with this JEE Main Practice Mock Test - 5 (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 JEE 2026 exam, helping you build speed, accuracy as per the actual exam.

Mock Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 180 minutes
  • - Total Questions: 75
  • - Analysis: Detailed Solutions & Performance Insights
  • - Sections covered: Physics - Section A, Physics - Section B, Chemistry - Section A, Mathematics - Section A

Sign up on EduRev for free and get access to these mock tests, get your All India Rank, and identify your weak areas to improve your marks & rank in the actual exam.

JEE Main Practice Mock Test - 5 - Question 1
Save

The following figure shows a network of capacitors, where the numbers indicate capacitances in microfarad. What must be the value of capacitance C, if the equivalent capacitance between points A and B is to be1 μF ?

Detailed Solution: Question 1

The series combination of 6 and 12 is equivalent to 4 and the parallel combination of 2 and 2 is also equivalent to 4. Therefore, the network can be simplified as shown in the figure.

The parallel combination of 4 and 4 is equivalent to 8 and the series combination of 8 and 4 is equivalent to 8/3. Thus, the combination in the figure above reduces to that in the figure below.

The series combination of 1 and 8 in the figure yields 8/9 as shown in the figure below.

Now, 8/3 and 8/9 are parallel and their equivalent capacitance is 32/9. Therefore, the network finally reduces to that given in the figure. Since the total capacitance between A and B is to be 1 μF, we have

JEE Main Practice Mock Test - 5 - Question 2
Save

Which of the following product of e,h,μ,G (where μ is the permeability) be taken so that the dimensions of the product are same as that of speed of light?

Detailed Solution: Question 2

Here v = eahbμcGd. Taking the dimensions


There will be 4 simultaneous equations by equating the dimensions of M, L, T and A. These are a − 2c = 0
a − b − 2c − 2d = −1, b + c − d = 0
and 2b + c + 3d = 1
Solving for ' a ',b′, ' c ' and ' d ' we get
a = −2, b = 1, c = −1, d = 0
Thus, v = e−2hμ−1G0

Clear all your doubts with EduRev
JEE Main Practice Mock Test - 5 - Question 3
Save

A circular plate of uniform thickness has a diameter of 28 cm. A circular portion of diameter 21 cm is removed from the plate as shown. O is the centre of mass of complete plate. The position of centre of mass of remaining portion will shift towards left from 'O' by

Detailed Solution: Question 3


Here πr2t is volume and mass is written in terms of density times volume, where ρ is the density of plate material and t is the thickness of the plate.

x2 = 4.5 cm

JEE Main Practice Mock Test - 5 - Question 4
Save

A spherical balloon contains air at temperature T0 and pressure P0. The balloon material is such that the instantaneous pressure inside is proportional to the square of the diameter. When the volume of the balloon doubles as a result of heat transfer, the expansion follows the law

Detailed Solution: Question 4

Pressure is directly proportional to square of the diameter of the balloon,

JEE Main Practice Mock Test - 5 - Question 5
Save

A sonometer wire resonates with a given tuning fork, forming standing waves with five antinodes between the two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by a mass M, the wire resonates with the same tuning fork, forming three antinodes for the same positions of the bridges. The value of M is

Detailed Solution: Question 5

The frequency of vibration of a string is given by

where N = number of loops (or segments) = number of antinodes. The other symbols have their usual meanings.

JEE Main Practice Mock Test - 5 - Question 6
Save

When an AC source of e.m.f. E = E0 sin(100t) is connected across a circuit, the phase difference between the e.m.f. E and the current I in the circuit is observed to be π/4, as shown in the figure. If the circuit consists possibly only of R − C or R − L or L − C in series, what will be the relation between the two elements of the circuit?

Detailed Solution: Question 6

Given E = E0 sin(100t). comparing this with E = E0 sinω = 100rads−1. It follows from the figure that the current leads to the e.m.f. which is true only for R-C circuits, and not for R-L circuits. Hence the circuit does not contain an inductor. For R-C circuit, the phase difference between E and I given by tanϕ = 1/ωRC
Given ϕ = π/4. Also ω = 100rads−1. Using these values in (i), we get

JEE Main Practice Mock Test - 5 - Question 7
Save

Focal length of the plano-convex lens is 15 cm. A small object is placed at A as shown in figure. The plane surface is silvered. The image will form at

Detailed Solution: Question 7

A ray from point O at first gets refracted from the curved surface of focal length, f1 = 15 cm and then reflects from the plane mirror. Hence the plano- convex lens will become a curved mirror of focal length F.


Thus the final image will form at 12 cm to the left of lens.

JEE Main Practice Mock Test - 5 - Question 8
Save

The binding energy of deuteron  is 1.15MeV per nucleon and an alpha particle  has a binding energy of 7.1MeV per nucleon. Then in the reaction the energy Q released is

Detailed Solution: Question 8

Binding energy of  = 1.15 × number of nucleons = 1.15 × 2 = 2.3MeV. Total binding energy of reactants = 2.3 + 2.3 = 4.6MeV. Binding energy of  = 7.1 × number of nucleons = 7.1 × 4 = 28.4MeV Therefore, Q = 28.4 − 4.6 = 23.8MeV. Hence the correct choice is (c).

JEE Main Practice Mock Test - 5 - Question 9
Save

A network of nine resistors has been shown. The equivalent resistance between A and D is

Detailed Solution: Question 9

From the symmetry of circuit, B and C are at same potential, so are points E and F. The circuit can be redrawn as shown below.

Between A and B, the effective resistance is 1/2 Ω .

Between B and E, the effective resistance is 1Ω and between E and D, the effective resistance is . All the three resistances are in series and by summing, we get 2Ω , which is in parallel to 2Ω resistance. Hence, the effective resistance is 1Ω.

JEE Main Practice Mock Test - 5 - Question 10
Save

A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 radian per second. If the horizontal component of Earth's magnetic field is 0.2 x 10-4 T, then the emf developed between the two ends of the conductor is

Detailed Solution: Question 10

Induced e.m.f. =

∴ Induced e.m.f =

= 50 μV

JEE Main Practice Mock Test - 5 - Question 11
Save

A boat is moving due East in a region where Earth's magnetic field is 5 × 10-5 NA-1m-1 due North and horizontal. The boat carries a 2 m long vertical aerial. If the speed of the boat is 1.5 m/s, then the magnitude of the induced emf in the wire of the aerial is

Detailed Solution: Question 11

Here, BH = 5.0 × 10-5 N/Am, l = 2 m and v = 1.5 m/s

Induced emf = e = BHlv = 5 × 10-5 × 1.5 × 2 = 15 × 10-5 V

or, e = 0.15 mV

JEE Main Practice Mock Test - 5 - Question 12
Save

Two spherical bodies of mass M and 5M and radii R and 2R, respectively, are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body, just before collision, is

Detailed Solution: Question 12

Let the spheres collide after time t, when the smaller sphere covered distance x1 and bigger sphere covered distance x2.

The gravitational force acting between two spheres depends on the distance (x), which is a variable quantity.

The gravitational force, F(x) =

Acceleration of smaller body, a1(x) =

Acceleration of bigger body, a2(x) =

From equation of motion,

x1 = 1/2a1(x)t2 and x2 = 1/2a2(x)t2

= 5

x1 = 5x2

We know that x1 + x2 = 9R

x1 + x1 / 5 = 9R

x1 = 45R/6 = 7.5R

JEE Main Practice Mock Test - 5 - Question 13
Save

The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetic metre is 2 s. The magnet is cut along its length into three equal parts and the three parts are then tied up with their like poles together. The time period of this combination will be

Detailed Solution: Question 13

For a vibrating magnet, , where I = ml2/12, M = xl and x = Pole strength of magnet

(For three pieces together)

(x) (For three pieces together)

JEE Main Practice Mock Test - 5 - Question 14
Save

The atomic mass of 7N15 is 15.000108 amu and that of 8O16 is 15.994915 amu. The minimum energy required to remove the least tightly bound proton is (mass of a proton is 1.007825 amu)

Detailed Solution: Question 14

Δm = mp + mnitrogen − moxygen

Energy required = Δm × 931.5 MeV

= (1.007825 + 15.000108 − 15.994915) × 931.5 MeV

≈ 12.13 MeV

JEE Main Practice Mock Test - 5 - Question 15
Save

A wire carrying current I is tied between points P and Q and is in the shape of a circular arch of radius R due to a uniform magnetic field B (perpendicular to the plane of the paper, shown by xxx) in the vicinity of wire. If the wire subtends an angle 2θo at the centre of the circle (of which it forms an arch) then the tension in the wire is:

Detailed Solution: Question 15

Consider any small element of arc of angle dθ and length dl . Then magnetic force df = Idl.B outwards

For equilibrium of element

For small angle dθ/2, the value of sindθ/2 = dθ/2

JEE Main Practice Mock Test - 5 - Question 16
Save

Statement -1 : Two stones are simultaneously projected from level ground from same point with same speed but different angles with horizontal. Both stones move in same vertical plane. Then the two stones may collide in mid air.

Statement - 2 For two stones projected simultaneously from same point with same speed at different angles with horizontal, their trajectories may intersect at some point.

Detailed Solution: Question 16

Both the stones cannot meet ( collide) because their horizontal component of velocities are different. Hence statement I is false.

JEE Main Practice Mock Test - 5 - Question 17
Save

A ring of mass m, radius r having charge q uniformly distributed over it and free to rotate about it's own axis is placed in a region having a magnetic field B parallel to its axis. If the magnetic field is suddenly switched off, the angular velocity acquired by the ring is

Detailed Solution: Question 17

⇒

JEE Main Practice Mock Test - 5 - Question 18
Save

If the distance between the Earth and the Sun were half its present value, the number of days in a year would have been

Detailed Solution: Question 18

Therefore,

= 129 days

JEE Main Practice Mock Test - 5 - Question 19
Save

A pump motor is used to deliver water at a certain rate from a given pipe. To obtain n times water from the same pipe in the same time the amount by which the power of the motor should be increased

Detailed Solution: Question 19

The volume delivered per second = Av

Mass delivered per second = Avd

Momentum delivered per second = Av2d = Force

Power = Force × Velocity = Av3d

i.e., Power ∝ v3

JEE Main Practice Mock Test - 5 - Question 20
Save

A cylinder weighing 450 N with a radius of 30 cm is held fixed on an incline that is rotating at 0.5 rad s−1. The cylinder is released when the incline is at position θ equal to 30°. If the cylinder is at 6 m from the bottom A at the instant of release, what is the initial acceleration of the centre of the cylinder relative to the incline, if there is no slipping? (g = 10 m s−2 )

Detailed Solution: Question 20

From the free body diagram of the cylinder, the torque about the point of contact O.

(R = 0.3 m = radius of circle, r = 6 m)

*Answer can only contain numeric values
JEE Main Practice Mock Test - 5 - Question 21
Save

Force of 4 N is applied on a body of mass 20 kg. Find the work done in Joules in 3rd second.


Detailed Solution: Question 21

Acceleration

(Displacement in nth second is given by Snth = u + a/2(2n−1) )

Distance covered by body in 3rd second

*Answer can only contain numeric values
JEE Main Practice Mock Test - 5 - Question 22
Save

Three point charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown in the figure. If the net electrostatic energy of the configuration is zero, then what is the value of 10×∣Q/q∣? [Take √2 =1.4]


Detailed Solution: Question 22

Net electrostatic energy of the configuration will be

Putting U = 0, we get

*Answer can only contain numeric values
JEE Main Practice Mock Test - 5 - Question 23
Save

A wire of length L and 3 identical cells, of negligible internal resistance, are connected in series. Due to the current, the temperature of the wire is raised by ΔT in a time t. A number of N similar cells are now connected in series with a wire of the same material and area of cross-section but of length 2L. The temperature of the wire is raised by the same amount ΔT in the same time. Find the value of N.


Detailed Solution: Question 23

In the first case,

(3E)2 / Rt = msΔT ... (i)

When the length of the wire is doubled, resistance and mass both are doubled. Therefore, in the second case,

(NE)2 / 2R x t = (2m)sΔT ... (ii)

Dividing Eq.(ii) by (i) we get,

N2 / 18 = 2

or N2 = 36

or N = 6

*Answer can only contain numeric values
JEE Main Practice Mock Test - 5 - Question 24
Save

Ultraviolet light of wavelength 300 nm and intensity 1.0 Wm-2 falls on the surface of a photosensitive material. If one percent of the incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of 1.0 cm2 of the surface is nearly p × 1012 s-1. The value of p is (Nearest integer)


Detailed Solution: Question 24

Power is given by:

Intensity of light, I = Power per unit area

Hence, the number of photons emitted in area of 1cm2 per second is given by:

As only 1 percent of photons produce photoelectrons, the number of electrons emitted per sec is:

ne = 1.5 × 1012 s-1

ne ≈ 2 × 1012 s-1

JEE Main Practice Mock Test - 5 - Question 25
Save

At 780 K and 10 atmosphere pressure the equilibrium constant for the reaction 2 A(g) ⇌ B(g) + C(g) is 3.52. At the same temperature and 7.04 atmosphere pressure, the equilibrium constant for the same reaction is

Detailed Solution: Question 25

The value of the equilibrium constant is dependent on temperature but is independent of the pressure. Thus, the value of the equilibrium constant will remain 3.52.

JEE Main Practice Mock Test - 5 - Question 26
Save

For which of the following Kp may be equal to 0.5 atm

Detailed Solution: Question 26

For Kp = 0.5 atm
Δn = 1 (since the unit is atm)
and PCl5 ⇌ PCl3 + Cl2
Δn = 1

JEE Main Practice Mock Test - 5 - Question 27
Save

Which of the following is not involved in the formation of photochemical smog?

Detailed Solution: Question 27

Photochemical smog is formed by the combination of smoke/dust/fog articles with the oxides of Nitrogen (Like NO) and Hydrocarbons in the presence of sunlight. It is brown in colour ad hazy.
Ozone (O3) contributes towards photochemical smog as it acts as an initiator of various reactions that lead to the formation of photochemical smog.
On the other hand, SO2 does not contribute in any way towards photochemical smog.

JEE Main Practice Mock Test - 5 - Question 28
Save

The order of reactivity of hydrogen halides with an alcohol is

Detailed Solution: Question 28

The order of reactivity is HCl < HBr < HI.

JEE Main Practice Mock Test - 5 - Question 29
Save

The value of a for which one root of the equation x2 −(a + 1)x + a2 + a − 8 = 0 exceeds 2 and other is less than 2, are given by

Detailed Solution: Question 29

As the roots are real and distinct
(a + 1)2 − 4(a2 + a − 8) > 0
3a2 + 2a − 33 < 0

Also, for

From (1) and (2), we get −2 < a < 3.

JEE Main Practice Mock Test - 5 - Question 30
Save

A parabola is drawn through two given points A(2, 0) and B(−2, 0) such that its directrix always touch the circle x2 + y2 = 16, then locus of focus of the parabola is

Detailed Solution: Question 30

Let focus be S(h, k), then
(h − 2)2 + k2 = 4(cosθ − 2)2 …(i)
(h + 2)2 + k2 = 4(cosθ + 2)2 …(ii)
⇒ cosθ = h/4

View more questions
360 docs|100 tests
Join Course
Information about JEE Main Practice Mock Test - 5 Page
In this test you can find the Exam questions for JEE Main Practice Mock Test - 5 solved & explained in the simplest way possible. Besides giving Questions and answers for JEE Main Practice Mock Test - 5, EduRev gives you an ample number of Online tests for practice
Download as PDF

About this JEE Mock Test

This mock test contains 75 questions designed to replicate the pattern, difficulty level and time pressure of the actual JEE exam. It is part of Mock Tests for JEE Main and Advanced 2026 on EduRev, covering all the topics that are being asked as per the current JEE 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.

Explore more JEE Resources

If you found this JEE Main Practice Mock Test - 5 helpful, you can further enhance your JEE preparation with complete mock tests, previous year questions & practice tests available in Mock Tests for JEE Main and Advanced 2026 course on EduRev. The course offers:
  • Video Lectures: covering all the topics of Mock Tests for JEE Main and Advanced 2026
  • Notes for Revision: offering revision notes, important questions, summaries, flashcards, cheatsheets & more for JEE preparation
  • MCQ Test: variety of questions as per the pattern of JEE exam
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup