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JEE Main Mock Test - 3 - Question 1

In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and D are 1%, 2%, 3% and 4%, respectively. Then the maximum percentage of error in the measurement X, where  will be

Detailed Solution for JEE Main Mock Test - 3 - Question 1


JEE Main Mock Test - 3 - Question 2

In an electromagnetic wave in free space the root mean square value of the electric field is Erms = 6 V/m.
The peak value of the magnetic field is:-

Detailed Solution for JEE Main Mock Test - 3 - Question 2

JEE Main Mock Test - 3 - Question 3

The position of the direct image obtained at O, when a monochromatic beam of light is passed through a plane transmission grating at normal incidence is shown in figure. The diffracted images A, B and C correspond to the first, second and third order diffraction. When the source is replaced by another source of shorter wavelength

Detailed Solution for JEE Main Mock Test - 3 - Question 3

JEE Main Mock Test - 3 - Question 4

A 2.0 cm object is placed 15 cm in front of a concave mirror of focal length 10 cm. What is the size and nature of the image?

Detailed Solution for JEE Main Mock Test - 3 - Question 4

JEE Main Mock Test - 3 - Question 5

The engine of a motorboat is switched off when its speed is v0. If motorboat experiences variable acceleration as dv/dt = −βv3, where β is a positive constant and v instantaneous speed, then the speed of motorboat after t second (t = 0 when switch is off) is

Detailed Solution for JEE Main Mock Test - 3 - Question 5

JEE Main Mock Test - 3 - Question 6

In a photodiode, the conductivity increases when the material is exposed to light. It is found that conductivity changes only if the wavelength is less than 600 nm. What is the bandgap?

Detailed Solution for JEE Main Mock Test - 3 - Question 6

Given,
The maximum wavelength of light, which increases the conductivity, λ = 600 nm
Hence the energy of this photon must be equal to band gap (Eg).

JEE Main Mock Test - 3 - Question 7

If the magnetic dipole moment of an atom of diamagnetic material, paramagnetic material and ferromagnetic material is denoted by μd, μp and μf respectively, then

Detailed Solution for JEE Main Mock Test - 3 - Question 7

The magnetic dipole moment of a diamagnetic material is zero as each of its pair of electrons has opposite spins, i.e., μd = 0. Paramagnetic substances have dipole moment > 0, i.e., μp ≠ 0, because of the excess of electrons in its molecules spinning in the same direction.
Ferromagnetic substances are very strong magnets and they also have a permanent magnetic moment, i.e., μf ≠ 0.

JEE Main Mock Test - 3 - Question 8

Which of the following will pair up to produce stationary wave?
(1) Z1 = A cos(kx − ωt)
(2) Z2 = A cos(kx + ωt)
(3) Z3 = A cos(ky − ωt)
(4) Z4 = A cos(kz + ωt)

Detailed Solution for JEE Main Mock Test - 3 - Question 8

Stationary waves are produced by two identical waves travelling in opposite direction in same plane.
∴ for stationary wave
Z1 = A cos(kx − ωt)
Z2 = A cos(kx + ωt)

JEE Main Mock Test - 3 - Question 9

A particle is undergoing SHM. Displacement vs time curve is as shown in the figure.

The correct statement is

Detailed Solution for JEE Main Mock Test - 3 - Question 9

The phase of a particle executing SHM is defined as the state of a particle as regard to its position and direction of motion at any instant of time. In the given curve, phase is same when t = 1 s and t = 5 s. Also phase is same when t = 2 s and t = 6 s.

JEE Main Mock Test - 3 - Question 10

Match List - I (Electromagnetic wave type) with List - II (Its association/application) and select the correct option from the choices given below the lists :

Detailed Solution for JEE Main Mock Test - 3 - Question 10

Infrared - Muscular treatment
Radio wave - Broadcasting
X Rays - To detect fracture
Uv Rays → Absorbed ozone layer

JEE Main Mock Test - 3 - Question 11

Figure shows a cyclic process ABCDBEA performed on an ideal cycle. If PA = 2 atm, PB = 5 atm and PC = 6 atm, VE – VA = 20 litre, find the work done (in KJ) by the gas in the complete process. (1 atm. pressure = 1 × 105 Pa)

Detailed Solution for JEE Main Mock Test - 3 - Question 11

Evidently, the entire cycle can be visualised as made up of two cycles, i.e., cycle ABEA and cycle BCDB.
W1 = area of the loop ABEA in the P-V diagram

Work done in the cycle BCDB, by the gas
W2 = –(area of loop BCDB)
Now,evidently, triangles ABE and BCD are similar, the corresponding angles being equal
(viz., ∠EBA =∠DBC =∠EAB =∠BCD, etc.)

∴ W2 = –0.33 kJ
∴ Total work done by the gas
ΔW = W1 + W2 = 3 kJ – 0.33 kJ = 2.67 kJ

JEE Main Mock Test - 3 - Question 12

If first excitation potential of a hydrogen like atom is V electron volt, then the ionization energy of this atom will be:

Detailed Solution for JEE Main Mock Test - 3 - Question 12


JEE Main Mock Test - 3 - Question 13

Let ω is the angular velocity of the earth's rotation about its own axis and acceleration due to gravity on the earth's surface at the poles is g. An object weighed by a spring balance gives the same reading at the equator, as well as at a height h above the poles (h ≪ R). What is the value of h ?

Detailed Solution for JEE Main Mock Test - 3 - Question 13

The value of acceleration due to gravity at the equator is g= g − Rω2 and the acceleration due to gravity at a height h above the pole is . It is given that

*Answer can only contain numeric values
JEE Main Mock Test - 3 - Question 14

A charge particle having charge Q is placed at the centre of hollow hemispherical sphere whose a portion is removed as shown. If the radius of the hollow sphere is R, then flux through the remaining surface is equal to Q/xε0. The value of x is?


Detailed Solution for JEE Main Mock Test - 3 - Question 14

*Answer can only contain numeric values
JEE Main Mock Test - 3 - Question 15

A tuning fork and uniform string of length 100 cm give 4 beats /s. The string is made shorter without any change in tension and the mode of oscillation, until its frequency becomes equal to that of the fork. If now the length of the string is 99 cm, and the frequency of the fork is 100nHz, find the value of n.


Detailed Solution for JEE Main Mock Test - 3 - Question 15


Let the freq. of tuning fork be f
Earlier the freq. of the string is f−4 & later it is equal to f

*Answer can only contain numeric values
JEE Main Mock Test - 3 - Question 16

An ionization chamber with parallel conducting plates as anode and cathode has 5 × 107 electrons and the same number of singly-charged positive ions per cm3. The electrons are moving at 0.4 m s−1. The current density from anode to cathode is 4 μA m−2. The velocity of positive ions moving towards cathode is V, find 100V in m/s.


Detailed Solution for JEE Main Mock Test - 3 - Question 16

In general, the current is,
i = neAvd
Current density is,
J = i/A = nevd 
For electrons, the current density is,

The total current density for the material is,

*Answer can only contain numeric values
JEE Main Mock Test - 3 - Question 17

Three identical hollow sphere, each of mass 2kg and radius 10cm are placed as shown in diagram.Consider an axis XX′ which is touching to two shells and passing through diameter of third shell. Calculate Moment of inertia of the system consisting of these three spherical shells XX′ axis in g m2?


Detailed Solution for JEE Main Mock Test - 3 - Question 17


Using parallel axes theorem,

*Answer can only contain numeric values
JEE Main Mock Test - 3 - Question 18

A uniform rod of length l is acted upon by a force F in a gravity-free region, as shown in the figure. If the area of cross-section of the rod is A and it's Young's modulus is Y, then the elastic potential energy stored in the rod due to elongation is . Then find the value of n?


Detailed Solution for JEE Main Mock Test - 3 - Question 18


JEE Main Mock Test - 3 - Question 19

Identify the pair of elements with the highest and lowest electronegativity respectively

Detailed Solution for JEE Main Mock Test - 3 - Question 19

Fluorine (F) is the element with highest electronegativity, lying at the extreme right in the second period.
Francium (Fr) is the most electropositive or least electronegative element, lying at the bottom left of the periodic table.

JEE Main Mock Test - 3 - Question 20

Which of the following facts regarding boron and silicon is not true?

Detailed Solution for JEE Main Mock Test - 3 - Question 20

The halides of B and Al are hydrolysed.

JEE Main Mock Test - 3 - Question 21

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 mA current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 F = 96500C mol−1)

Detailed Solution for JEE Main Mock Test - 3 - Question 21

2H2O + 2e− → H+ 2OH
For 0.01 mole H2 0.02 mole of electrons are consumed
charge required = 0.02 × 96500 C = i × t

JEE Main Mock Test - 3 - Question 22

An oxide of nitrogen is reddish brown and paramagnetic at room temperature but it decolourises and also loses its paramagnetism on freezing it. The oxide at room temperature is

Detailed Solution for JEE Main Mock Test - 3 - Question 22

2Pb(NO3)→ 2PbO + 4NO+ O2
i.e. can be prepared by heating heavy metal nitrate of Pb(NO3)2. It is a red brown gas but on cooling below 273 K converts to colourless liquid and retains as dimer. In NO2 molecules 17 valence e−are there (12 that of O & 5 that of N) i.e. one e−unpaired results to paramagnetic nature. So this is an odd emolecule on being dimer with even ebecomes stable & loses paramagnetic characteristics.

JEE Main Mock Test - 3 - Question 23

T is a point on the tangent to a parabola y2 = 4ax at its point P. TL and TN are the perpendiculars on the focal radius SP and the directrix of the parabola respectively. Then

Detailed Solution for JEE Main Mock Test - 3 - Question 23

Let the equation of the parabola be y² = 4ax.

Point P be (at², 2at) and the point T be (h, k).

Equation of tangent at P is
ty = x + at².

It passes through T(h, k):
tk = h + at² .....(i)

Slope of SP = (2at - 0) / (at² - a) = 2t / (t² - 1).

TL is perpendicular to SP.
Then equation of TL is
2ty + (t² - 1)x - 2kt - (t² - 1)h = 0 .....(ii)

SL = perpendicular distance of S(a, 0) from (ii).

SL = (a + h) ........(iii)

Equation of directrix is x = -a ........(iv)

TN = perpendicular distance of T(h, k) from (iv)
TN = (h(1) + k(0) + a) / √(1² + 0²)

TN = h + a ........(v)

From (iii) and (v)
SL = TN

JEE Main Mock Test - 3 - Question 24

The general solution of sin2θ sec θ + √3 tan θ = 0 is

Detailed Solution for JEE Main Mock Test - 3 - Question 24

JEE Main Mock Test - 3 - Question 25

The fundamental period of the function f(x) = e{4{x}+3}, where {⋅} denotes the fractional part function is

Detailed Solution for JEE Main Mock Test - 3 - Question 25

f(x) = e{4{x}+3}=e{4(x}}=e{4x}
because {4{x} + 3} = {4{x}} and 4{x} = 4x − 4[x]⟹{4{x}} = {4x}

JEE Main Mock Test - 3 - Question 26

In the figure, AB, DE and GF are parallel to each other and AD, BG and EF are parallel to each other. If CD : CE = CG : CB = 2 : 1, then the value of area (△AEG): area (ΔABD) is equal to

Detailed Solution for JEE Main Mock Test - 3 - Question 26


= 7/2

JEE Main Mock Test - 3 - Question 27

Let p, q, r be three statements, then (p → (q → r)) ↔ ((p ∧ q) → r), is a

Detailed Solution for JEE Main Mock Test - 3 - Question 27

p → (q → r) ≡ ∼p∨(q → r)
(q → r) ≡ ∼p∨(q → r)
≡ ∼p∨(∼q∨r)
≡ [(∼p)∨(∼q)]∨r
≡ ∼(p∧q)∨r
≡ p∧q → r
∴ given statement is a tautology.

*Answer can only contain numeric values
JEE Main Mock Test - 3 - Question 28

A bag contains six balls of different colours. Two balls are drawn in succession with replacement. The probability that both the balls are of the same colour is p. Next four balls are drawn in succession with replacement and the probability that exactly three balls are of the same colour is q. If p: q = m : n, where m and n are coprime, then m + n is equal to:


Detailed Solution for JEE Main Mock Test - 3 - Question 28

p = 6 / 36 = 1 / 6

q = (6C₁ × 5C₁ × 4! / 3!) / 6⁴ = 120 / 1296 = 5 / 54

p / q = (1 / 6) / (5 / 54) = (54 / 6 × 5) = 9 / 5 = m / n

m + n = 14

*Answer can only contain numeric values
JEE Main Mock Test - 3 - Question 29

If the mean and variance of four numbers 3, 7, x and y(x > y) be 5 and 10 respectively, then the mean of four numbers 3 + 2x, 7 + 2y, x + y and x - y is ______.


Detailed Solution for JEE Main Mock Test - 3 - Question 29

5 = (3 + 7 + x + y) / 4 ⇒ x + y = 10

Var(x) = 10 = (3² + 7² + x² + y²) / 4 - 25

140 = 49 + 9 + x² + y²
x² + y² = 82
x + y = 10

⇒ (x, y) = (9, 1)

Four numbers are 21, 9, 10, 8

Mean = 48 / 4 = 12

*Answer can only contain numeric values
JEE Main Mock Test - 3 - Question 30

Let S be the set of values of λ, for which the system of equations 
6λx - 3y + 3z = 4λ2,
2x - 6λy + 4z = 1,
3x + 2y + 3λz = λ has no solution. Then is equal to ___________.


Detailed Solution for JEE Main Mock Test - 3 - Question 30

Given that S be the set of values of λ for which given system of equations has no solution.

Therefore for the given set of equations

The given equation simplifies as follows:

6λ(18λ² - 8) + 3(6λ - 12) + 3(4 - 18λ) = 0
⇒ 18λ³ - 14λ - 4 = 0
⇒ (λ - 1)(3λ + 1)(3λ + 2) = 0

Thus, the values of λ are:
λ = 1, -1/3, -2/3

Also for each values of  we have

which implies that, for each values of λ, the given system of equations has no solution.
Therefore S∈ {1, -1/3, -2/3} and 

= 12 (|1| + |-1/3| + |-2/3|)

= 12 (1 + 1/3 + 2/3) = 12 (6/3) = 24

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