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JEE Main Mock Test - 1 - JEE MCQ


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30 Questions MCQ Test Mock Tests for JEE Main and Advanced 2025 - JEE Main Mock Test - 1

JEE Main Mock Test - 1 for JEE 2025 is part of Mock Tests for JEE Main and Advanced 2025 preparation. The JEE Main Mock Test - 1 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Mock Test - 1 MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Mock Test - 1 below.
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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 for JEE Main Mock Test - 1 - Question 1

On removing an electron, the Helium atom becomes a single electron atom.
Applying Bohr's model, the energy of the single electron is given by 
(where Z is atomic number and n is number of the shell in which the electron resides).
Thus, the energy required is:

Energy required to remove the electron from singly ionized Helium atom = 54.4 eV

Energy required to remove the electron from Helium atom = x eV

So, 54.4 eV = 2.2x

x = 24.73 eV

Total energy required to ionize the Helium atom completely = 54.4 + 24.73 = 79.13 eV

Hence, the nearest option is 'C'.

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 for JEE Main Mock Test - 1 - Question 2

The velocity of the sphere is same as that of the cube, which is given as 
Hence, acceleration of the sphere: 

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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 plain 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 for JEE Main Mock Test - 1 - 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 for JEE Main Mock Test - 1 - 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 for JEE Main Mock Test - 1 - 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 for JEE Main Mock Test - 1 - Question 6


Intensity at minima is

∴ 
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 for JEE Main Mock Test - 1 - 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 for JEE Main Mock Test - 1 - Question 8

Assertion and Reason are both correct and reason is not the correct explanation of assertion.

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 for JEE Main Mock Test - 1 - 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 for JEE Main Mock Test - 1 - 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 for JEE Main Mock Test - 1 - 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 for JEE Main Mock Test - 1 - 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 for JEE Main Mock Test - 1 - 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 for JEE Main Mock Test - 1 - 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 for JEE Main Mock Test - 1 - 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 for JEE Main Mock Test - 1 - 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 into its constituent nucleons is:

(Mass of 119.902199amu, mass of proton = 1.007825amu and mass of neutron = 1.008665amu)
Detailed Solution for JEE Main Mock Test - 1 - Question 17

Given, Z 50, A - Z = 120 - 50 = 70
Δm = Z.mp + (A - Z) mn - M= [50 × 1.007825 +70 × 1.008665 - 119.902199] = 1.095601 MeV
E = 1.095601 × 931.478 MeV = 1020.53 MeV ≈ 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 for JEE Main Mock Test - 1 - 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] 

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 for JEE Main Mock Test - 1 - 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 for JEE Main Mock Test - 1 - Question 20

N = qvB

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:
Detailed Solution for JEE Main Mock Test - 1 - 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 for JEE Main Mock Test - 1 - 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 for JEE Main Mock Test - 1 - 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 for JEE Main Mock Test - 1 - 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 for JEE Main Mock Test - 1 - 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 lobs 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 for JEE Main Mock Test - 1 - 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 for JEE Main Mock Test - 1 - Question 27


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