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JEE Main Mock Test - 1 Free Online Test 2026


Full Mock Test & Solutions: JEE Main Mock Test - 1 (75 Questions)

You can boost your JEE 2026 exam preparation with this JEE Main Mock Test - 1 (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, Mathematics: Section B

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JEE Main Mock Test - 1 - Question 1
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The energy required to remove the electron from a singly ionized Helium atom is 2.2 times the energy required to remove an electron from Helium atom. The total energy required to ionize the Helium atom completely is:

Detailed Solution: Question 1

Option C is correct; the total energy required is 79 eV.

Let the first ionization energy be E₁ and the second ionization energy be E₂. Given E₂ = 2.2 E₁.

For the singly ionized helium ion (He+), which is hydrogen-like with atomic number Z = 2, the ionization energy from the ground state is given by 13.6 Z2 eV.

Thus E₂ = 13.6 × 22 = 13.6 × 4 = 54.4 eV.

So E₁ = E₂ / 2.2 = 54.4 / 2.2 ≈ 24.73 eV.

Total energy to ionize helium completely = E₁ + E₂ = 24.73 + 54.4 ≈ 79.13 eV, which rounds to 79 eV.

Therefore the correct option is C: 79 eV.

JEE Main Mock Test - 1 - Question 2
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A solid sphere of mass 2 kg is resting inside a cube as shown in figure The cube is moving with a velocity . Here t is time in seconds. All surfaces are smooth. The sphere is at rest with respect to the cube. What is the total force exerted by the sphere on the cube?

Detailed Solution: Question 2

Option C is correct.

Differentiate the given velocity with respect to time to get the acceleration; a = (5 i + 2 j) m s-2.

By Newton's second law the vector sum of external forces on the sphere equals m a.

Taking rightward as x and upward as y, the external forces are Nx i from the left wall, Ny j from the floor (upward), and weight mg (downward). Thus Nx i + (Ny - mg) j = m a.

With m = 2 kg, m a = (10 i + 4 j) N.

Equating components gives Nx = 10 N and Ny - mg = 4 N.

Using g = 10 m s-2 (consistent with the options), Ny = 4 + m g = 4 + 2 × 10 = 24 N.

The resultant force magnitude is √(102 + 242) = 26 N.

Hence the total force exerted by the sphere on the cube is 26 N (option C).

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JEE Main Mock Test - 1 - Question 3
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A block is resting on a piston which executes simple harmonic motion in vertical plane with a period of 2.0 s in vertical plane at an amplitude just sufficient for the block to separate from the piston. The maximum velocity of the piston is

Detailed Solution: Question 3

The situation when the block is just below the mean position is shown in Fig., the restoring forces acting on the piston cause a normal reaction F to act on the block. For the block to separate F ≥ mg i.e., mω2 A ≥ mg
(where ω= angular frequency and A = amplitude)


Now, the maximum velocity vmax at that instant = ωA

JEE Main Mock Test - 1 - Question 4
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Determine the electric field intensity at point P due to quadruple distribution shown in figure for r ≫ a

Detailed Solution: Question 4

The field at point P is superposition of field  due to each charge.


If r >> a, we can use binomial approximation :

JEE Main Mock Test - 1 - Question 5
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Two capacitors of capacitances 0.3μF and 0.6μF are connected in series across 6 volts. The ratio of energies stored in them will be

Detailed Solution: Question 5


In series combination of capacitors

JEE Main Mock Test - 1 - Question 6
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Two coherent sources of intensity ratio 100 : 1, interfere. What is the ratio of the intensity between the maxima and minima in the interference pattern?

Detailed Solution: Question 6


Intensity at minima is

∴ 
= 121 : 81
Hence the closest choice is (c).

JEE Main Mock Test - 1 - Question 7
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A block of pure silicon at 300 K has a length of 10 cm and an area of 1.0 × 10−4 m2. If a battery of emf 2 V is connected across it, what is the electron current? The mobility of electrons is 0.14 m2 V−1 s−1 and their number density is 1.5 × 1016 m−3.

Detailed Solution: Question 7

Given μe = 0.14 m2 V−1 s−1,ne = 1.5 × 1016 m−3, l = 10 cm = 0.1 m, A = 1.0 × 10−4 m2 and V = 2 volts.
The electric field in the block is

The drift speed of electrons is
ve = μeE = 0.14 × 20 = 2.8 ms−1
∴ Electron current Ie is

JEE Main Mock Test - 1 - Question 8
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Assertion: The speed of sound in a gas is not affected by change in pressure provided the temperature of the gas remains constant.
Reason: The speed of sound is inversely proportional to the square root of the density of the gas.

Detailed Solution: Question 8

Assertion and Reason are both correct and reason is not the correct explanation of assertion.
“At constant T, P and ρ  change proportionally, keeping v constant.”

JEE Main Mock Test - 1 - Question 9
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Two identical blocks A and B are placed on two inclined planes as shown in diagram. Neglect air resistance and other friction

Read the following statements and choose the correct options.
Statements I : Kinetic energy of 'A' on sliding to J will be greater than the kinetic energy of B on falling to M.
Statements II : Acceleration of ' A ' will be greater than acceleration of ' B ' when both are released to slide on inclined plane
Statements III : Work done by external agent to move block slowly from position B to O is negative

Detailed Solution: Question 9

Statement I : Work done by gravity is same for motion from A to J and B to M for equal mass. So K.E. will be equal.
Statement II : Acceleration = g sinθ

JEE Main Mock Test - 1 - Question 10
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The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the volume of the sheet to correct significant figures.

Detailed Solution: Question 10

Length , l = 4,234 m
Breadth, b = 1.005 m
Thickness , t = 2.01 x 10-2 m
Volume = l x b x t
⇒ V = 4.234 x 1.005 x 0.0201 = 0.0855289 = 0.0855 m3 (significant figure = 3)

JEE Main Mock Test - 1 - Question 11
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A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in the figure. When the string is cut, the initial angular acceleration of the rod is:
[2013]

Detailed Solution: Question 11

Weight of the rod will produce the torque,
τ = mg x L / 2 = I α = mL2 / 3 α (∵ Irod = ML2 / 3)
Hence, angular acceleration, α = 3g / 2L

JEE Main Mock Test - 1 - Question 12
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A particle is projected with a velocity v, so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is   

Detailed Solution: Question 12

JEE Main Mock Test - 1 - Question 13
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A body starts from rest, the ratio of distances travelled by the body during 3rd and 4thseconds is :

Detailed Solution: Question 13

The velocity after 2 sec = 2a and velocity after 3 sec = 3a
For some constant acceleration a,
Now distance travelled in third second, s3 = 2a + 1 / 2 a
= 5/2 a
Similarly distance travelled in fourth second, s4= 3a + 1 / 2 a
= 7/2 a
Hence the required ratio is 5/7

JEE Main Mock Test - 1 - Question 14
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A ray of light is incident at an angle of 60° on one face of a prism of angle 30°. The emergent ray of light makes an angle of 30° with incident ray. The angle made by the emergent ray with second face of prism will be:

Detailed Solution: Question 14

For the prism, angle of deviation is given by:

Hence, emergent ray will be perpendicular to the second face or we say that the angle made by the emergent ray is 90°.

JEE Main Mock Test - 1 - Question 15
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The diagram of a logic circuit is given below. The output F of the circuit is given by:

Detailed Solution: Question 15

By law of distribution of Boolean Algebra, we have A+ (B. C) = (A+B). (A + C)
Step 1: Write outputs for the OR gates with inputs W and X, and W and Y.
Input for first OR gate are W and X. The output for this OR gate is, Y1 = W + X
Input for second OR gate are W and Y. The output for this OR gate is, Y2 = W + Y
Step 2: Write output for AND gate whose inputs are the output of OR gate.
F = Y1 • Y2 ⇒ F = (W + X). (W + Y)
By using Law of distribution of Boolean Algebra, A + (B. C) = (A+B). (A + C)
Therefore, F = W + (X. Y)

JEE Main Mock Test - 1 - Question 16
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In the circuit shown in the Figure, cell is ideal and R2 = 100Ω. A voltmeter of internal resistance 200Ω reads V12 = 4 V and V23 = 6 V between the pair of points 1 - 2 and 2 - 3 respectively. What will be the reading of the voltmeter between the points 1 - 3. 

Detailed Solution: Question 16


Let emf of the cell be E. Current through the voltmeter (when connected between 1-2) is 
Current through R1 = 4/R1
∴ Current through 
∴ Potential difference across 
∴
When the voltmeter is connected between 2 - 3
Current through voltmeter = 6/200 A
Current through R2 = 6/100 A
∴ Current through

∴
From (1) and (2) 
Put this in 
When connected across 1-3, the voltmeter will read E = 12 V.

JEE Main Mock Test - 1 - Question 17
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The energy required to separate the typical middle mass nucleus 12050Sn into its constituent nucleons is:

(Mass of 12050Sn = 119.902199 amu, mass of proton = 1.007825 amu, and mass of neutron = 1.008665 amu)

Detailed Solution: Question 17

Given, Z = 50, A = 120, so number of neutrons = A - Z = 120 - 50 = 70

Mass defect, Δm = (Z × mp) + ((A - Z) × mn) - M

= (50 × 1.007825) + (70 × 1.008665) - 119.902199

= 50.39125 + 70.60655 - 119.902199 = 1.095601 amu

Energy released (binding energy) E = Δm × 931.478 MeV/amu

= 1.095601 × 931.478 = 1020.53 MeV ≈ 1021 MeV

Therefore, the energy required to separate the nucleus into its constituent nucleons is approximately 1021 MeV.

JEE Main Mock Test - 1 - Question 18
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The astronomical phenomenon when the planet Venus passes directly between the Sun and the earth is known as Venus transit. For two separate persons standing on the earth at points M and N, the Venus appears as black dots at points M' and N' on the Sun. The orbital period of Venus is close to 220 days. Assuming that both earth and Venus revolve on circular paths and taking distance MN = 1000 km, calculate the distance M'N' on the surface of the Sun.
[Take (2.75)1/3 = 1.4]

Detailed Solution: Question 18

Let radius of circular orbit of the Earth and Venus be re and rv respectively (rc/rv) = (365/220)2 [Kepler's third law] 

⇒ rc/rv = (2.75)1/3 = 1.4
From the drawing given in the problem M'N'/MN = N'V/NV

JEE Main Mock Test - 1 - Question 19
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A straight wire of length L and radius a has a current I. A particle of mass m and charge q approaches the wire moving at a velocity v in a direction anti parallel to the current. The line of motion of the particle is at a distance r from the axis of the wire. Assume that r is slightly larger than a so that the magnetic field seen by the particle is similar to that caused by a long wire. Neglect end effects and assume that speed of the particle is high so that it crosses the wire quickly and suffers a small deflection θ in its path. Calculate θ.

Detailed Solution: Question 19




Force on the particle is 
This force is always perpendicular to the velocity. Since deflection is small, the force is nearly in (↑) direction always.
Impulse is: 

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 20
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In gravity free space, a bead of charge 1μC and mass 3mg is threaded on a rough rod of friction coefficient μ = 0.3. A magnetic field of magnitude 0.2 T exists perpendicular to the rod. The bead is projected along the rod with a speed of 4 m/s. How much distance (in m) will the bead cover before coming to rest?


Detailed Solution: Question 20

N = qvB

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 21
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A water barrel having water up to depth 'd' is placed on a table of height 'h'. A small hole is made on the wall of the barrel at its bottom. If the stream of water coming out of the hole falls on the ground at a horizontal distance 'R' from the barrel, then the value of 'd' is R^2/kh. what is k?


Detailed Solution: Question 21


By Bernoulli's equation: mgd = 1/2 mv2
v = √2gd ........... (1)
By Projectile motion equation: 1/2 gt2 = h

Therefore, 
So, d = R2/4h

JEE Main Mock Test - 1 - Question 22
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A reaction mixture containing H2, N2 and NH3 has partial pressures 2 atm,1 atm and 3 atm respectively at 725 K. If the value of Kp for the reaction, N2(g) + 3H2(g) ⇌ 2NH3(g) is 4.28 × 10−5 atm−2 at 725 K, in which direction the net reaction will go ?

Detailed Solution: Question 22


= 1.125 atm−2
Since value of Qp is larger than Kp (4.28 × 10−5 atm−2), it indicates net reaction will proceed in backward direction.

JEE Main Mock Test - 1 - Question 23
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If SO2 content in the atmosphere is 0.12ppm by volume, pH of rain water is (assume 100% ionisation of acid rain as monobasic acid)

Detailed Solution: Question 23

0.12ppm = 0.12 g in 106 mL
[H+] = [SO2]

pH = 5.7

JEE Main Mock Test - 1 - Question 24
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Consider the following reaction

Which of the following statements are correct ?
a. I is the major product of the reaction
b. II is the major product of the reaction
c. Formation of I is in accordance with Saytzeff's rule
d. II is more stable because it is more substituted

Detailed Solution: Question 24

JEE Main Mock Test - 1 - Question 25
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Which of the following statements are correct?
1. Electron density in xy plane in  orbital is zero.
2. Electron density in xy plane in  orbital is zero.
3. 2s orbital has only one spherical node in it.
4. For 2px orbital yz is the nodal plane.

Detailed Solution: Question 25

The four lobes of  orbital are lying along x and y axes, so, density in XY plane can't be zero. 
The two lobes of  orbital are lying along z axis, and contain a ring of negative charge surrounding the nucleus in xy plane. So, the density in XY plane is non-zero.
2s orbitals have one spherical node, where the electron density is zero.
In 2px orbital, both the lobes lie along x axis. Hence,the density in yz plane is zero, thus it is the nodal plane.

JEE Main Mock Test - 1 - Question 26
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Assertion (A): Mg2+ and Al3+ are isoelectronic but the magnitude of ionic radius of Al3+ is less than that in Mg2+.
Reason (R): The effective nuclear charge on the outermost electrons in Al3+ is greater than that in Mg2+.
The correct option among the following is

Detailed Solution: Question 26

Higher the electrostatic attraction between nucleus and valence electron, smaller will be the size of the atom/ion.
Electrostatic attraction is given by Zeff = Z−S where, Z= the number of protons in the nucleus of an atom or ion (the atomic number) and S= shielding from core electrons.
In case of Al3+ and Mg2+, they are isoelectronic species and thus have same number of electrons as 1s2,2s2,2p6 isoelectric series of atoms and ions with different numbers of protons (and thus different nuclear attraction, gives) the relative ionic sizes of each atom or ion with respect to atomic number.
Atomic number, ZAl = 13 and ZMg = 12.
Thus, Al3+ has lower ionic radii than Mg2+ due to higher nuclear charge.
Hence, A is true, R is true and R is the correct explanation for A.

JEE Main Mock Test - 1 - Question 27
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The values of  satisfying the equation 

Detailed Solution: Question 27


JEE Main Mock Test - 1 - Question 28
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The number of solution of equation, is/are

Detailed Solution: Question 28

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 29
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If 2061753 is divided by 17, then find the remainder.


Detailed Solution: Question 29




= 2 × (16)438
= 2 × (17−1)438
= 2(17k + 1)
= 34k + 2
Rem = 2

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 30
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If 
and F = [40] are matrices and α, β are roots of equation AB5C10D5E = F then the value of (1 − α)(1 − β) is


Detailed Solution: Question 30


Here, BC = CB = I
∴ B5C10B5 = I10 = I
So, we get

⇒1(x2 − 5x + 20) + 25(x + 2) = 40 ⇒ x2 + 20x + 30 = 0
∴ (1 − α)(1 − β) = 1 − (α + β) + αβ = 1 − (−20) + 30 = 51

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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.

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