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JEE Main Mock Test - 4 - JEE MCQ


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30 Questions MCQ Test Mock Tests for JEE Main and Advanced 2025 - JEE Main Mock Test - 4

JEE Main Mock Test - 4 for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2025 preparation. The JEE Main Mock Test - 4 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Mock Test - 4 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Mock Test - 4 below.
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JEE Main Mock Test - 4 - Question 1

Two radioactive substances A and B have decay constants 5λ and λ, respectively. At t = 0, a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become (1/e)will be:

Detailed Solution for JEE Main Mock Test - 4 - Question 1

JEE Main Mock Test - 4 - Question 2

The magnetic field and number of turns of the coil of an electric generator is doubled then the magnetic flux of the coil will:

Detailed Solution for JEE Main Mock Test - 4 - Question 2

Given: N = 2N1 and B = 2B1
The magnetic flux through the electric generator when the magnetic field is B, current flowing is A and the number of turns is N
φ = N B A cos θ ....(1)
The magnetic flux through the electric generator when the magnetic field and number of turns of the coil of an electric generator is doubled
φ1 = N1 B1 A cosθ   ...(2)
⇒ φ= (2N)(2B) A cos θ = 4 N B A cos θ = 4φ [∵φ = NBA cos θ]

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JEE Main Mock Test - 4 - Question 3

If λ1 and λ2  are the wavelengths of the first members of Lyman and Paschen series, respectively, then λ1 : λ2 is:

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

For the wavelength of the first member of Lyman series:

For the wavelength of the first member of Paschen series:

JEE Main Mock Test - 4 - Question 4

Three rods of Copper, Brass and Steel are welded together to from a Y –shaped structure. Area of cross – section of each rod = 4 cm2. End of copper rod is maintained at 100°C where as ends of brass and steel are kept at 0°C. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is

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


JEE Main Mock Test - 4 - Question 5

In an experiment for determination of refractive index of glass of a prism by i – δ plot, it was found that a ray incident at an angle of 35° suffers a deviation of 40° and that it emerges at an angle of 79°. In that case, which of the following is closest to the maximum possible value of the refractive index?

Detailed Solution for JEE Main Mock Test - 4 - Question 5


1.5 is the nearest option.

JEE Main Mock Test - 4 - Question 6

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10-12 m, the minimum electron energy required is close to:

Detailed Solution for JEE Main Mock Test - 4 - Question 6

JEE Main Mock Test - 4 - Question 7

Unpolarized light of intensity I is incident on a system of two polarizers, A followed by B. The intensity of emergent light is I/2. If a third polarizer C is placed between A and B, the intensity of emergent light is reduced to I/3. The angle between polarizers A and C is θ. Then,

Detailed Solution for JEE Main Mock Test - 4 - Question 7

For the system of polarizers A and B, the intensity of the emergent light is I/2. So, A and B have same alignment of transmission axis.
Let's assume that C is introduced at θ angle w.r.t. A.
Then,
Output Intensity = I/2
So, we have

JEE Main Mock Test - 4 - Question 8

Assume that an electric field  exists in space. Then the potential difference VA – VO, where VO is the potential at the origin and VA the potential at x = 2 m is:

Detailed Solution for JEE Main Mock Test - 4 - Question 8

JEE Main Mock Test - 4 - Question 9

In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery and a high resistance of 11 kΩ are used. The figure of merit of the galvanometer is 60 μA/division. In the absence of shunt resistance, the galvanometer produces a deflection of θ = 9 divisions, when current flows in the circuit. The value of the shunt resistance that can cause the deflection of θ/2 is closest to:

Detailed Solution for JEE Main Mock Test - 4 - Question 9


For deflection of θ/2, current also reduces to I/2 with shunt resistance S.
Hence,

JEE Main Mock Test - 4 - Question 10

A current of 1 A is flowing on the sides of an equilateral triangle of side 4.5 × 10-2 m. The magnetic field at the centroid of the triangle will be:

Detailed Solution for JEE Main Mock Test - 4 - Question 10


Or 4 × 10-5 Wb/m2

JEE Main Mock Test - 4 - Question 11

A rocket is fired from the earth towards the sun. At what distance from the earth's centre is the gravitational force on the rocket zero ? Mass of the sun = 2 × 1030 kg, mass of the earth 6 × 1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 1011 m).

Detailed Solution for JEE Main Mock Test - 4 - Question 11

As the Rocket is fired from earth (E) towards the sun (s) at point (P) the distance from center of the earth.

The Force acts on the rocket are: These two gravitational forces are in opposite directions.

As we know gravitational force between two objects which are at distance R from each other. 
G = universal gravitational constant, m1m2 = mass of the objects, R = Distance between them
By the gravitational Force
Force on Rocket due to Earth = Force on Rocket due to sun

By simplification:


JEE Main Mock Test - 4 - Question 12

A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

Detailed Solution for JEE Main Mock Test - 4 - Question 12

The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.
As we know that, the value of the ratio of specific heat for standard gas = 1.40.

For an adiabatic process, we have: 
The final volume is compressed to half of its initial volume.

JEE Main Mock Test - 4 - Question 13

The figure shows a 2.0 V potentiometer used for the determination of the internal resistance of a 1.5 V cell. The balance point of the cell in the open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm of the potentiometer wire. Determine the internal resistance of the cell.

Detailed Solution for JEE Main Mock Test - 4 - Question 13

The internal resistance of the cell = r
Balance point of the cell in open circuit, I1 = 76.3cm
An external resistance (R.) is connected to the circuit with R = 9.5Ω
New balance point of the circuit, I2 = 64.8cm
Current flowing through the circuit = 1
Using the relation connecting resistance and emf is,

JEE Main Mock Test - 4 - Question 14

What speed should a galaxy move with respect to us so that the sodium line at 589.0 nm is observed at 589.6 nm?

Detailed Solution for JEE Main Mock Test - 4 - Question 14

Given: Wavelength of sodium line (λ1) = 589nm Wavelength at which sodium line is observed (λ2) = 589.6nm
Change in wavelengths is given by Δλ = λ2 - λ
Substituting the values in above equation we get, Δλ 589.6 - 589 = 0.6nm
Velocity in terms of wavelength is given by,

Substituting the values in above equation we get, 

JEE Main Mock Test - 4 - Question 15

Light from a point source in the air falls on a spherical glass surface (μ=1.5 and radius of curvature 20cm). The distance of the light source from the glass surface is 100cm. The position where the image is formed is:

Detailed Solution for JEE Main Mock Test - 4 - Question 15

As per the given criteria, Refractive index of air, μ1 = 1, Refractive index of glass, μ2 = 1.5, Radius of curvature, R = 20 cm, Object distance, u = -100 cm
We know that,

JEE Main Mock Test - 4 - Question 16

One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle ( directed towards the centre ) is:

Detailed Solution for JEE Main Mock Test - 4 - Question 16

When a particle connected to a string revolves in a circular path around a center, the centripetal force is provided by the tension produced in the string. Thus, in the given case, the net force on the particle is the tension T, i,e F = T = mv2/1 Where F is the net force acting on the particle.

JEE Main Mock Test - 4 - Question 17

A metal block of area 0.10 m2 is connected to a 0.01 kg mass via a string that passes over a massless and frictionless 0.01 kg pulley as shown in the figure. A liquid with a film thickness of 0.3 mm is placed between the block and the table. When released the block moves to the right with a constant speed of 0.085 ms-1. The coefficient of viscosity of the liquid is: (Take g = 10 ms-2)

Detailed Solution for JEE Main Mock Test - 4 - Question 17

Here, m = 0.01 kg, l = 0.3 mm = 0.3 × 10-3 m, g = 10 ms-2 V= 0.085 ms-1, A = 0.1 m2
The metal block moves to the right because of the tension in the string. The tension T is equal in magnitude to the weight of the suspended mass m. 
Thus, the shear force F is: F = T = mg = 0.010 kg x 9.8 ms-2 = 9.8 x 10-2 N
Shear stress on the fluid =  

JEE Main Mock Test - 4 - Question 18

Assume that the light of wavelength is 6000Å coming from a star. What is the time resolution of a telescope whose objective has a diameter of 100 inch?

Detailed Solution for JEE Main Mock Test - 4 - Question 18

Resolving power (R.P.) of the astronomical telescope is its ability to form separate images of two neighbouring astronomical objects like stars etc.

where D is diameter of objective lens and λ is wave length of light used.

JEE Main Mock Test - 4 - Question 19

The magnetic field in a plane electromagnetic wave is given by By = (2 x 10-7) T Sin (0.5 x 103x + 1.5 x 1011t). This electromagnetic wave is:

Detailed Solution for JEE Main Mock Test - 4 - Question 19

The magnetic field in a plane of the electromagnetic wave is given by, B = 2 x 10-7 sin (0.5 × 103x + 1.5 × 1011t)
Comparing this equation with the standard wave equation: By = B0 sin[Kx + ωt]
From the above equation, K = 0.5 x 103 = propagation constant

The wavelength range of microwaves is 10-3 m to 0.3 m. The wavelength of this wave lies between 10-3m to 0.3 m, so the equation represents microwaves.

JEE Main Mock Test - 4 - Question 20

A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.

Detailed Solution for JEE Main Mock Test - 4 - Question 20

*Answer can only contain numeric values
JEE Main Mock Test - 4 - Question 21

Ice at -20°C is added to 50 g of water at 40°C. When the temperature of the mixture reaches 0°C, it is found that 20 g of ice is still unmelted. The amount of ice (in nearest integer) added to the water was _____g.
(Specific heat of water = 4.2 J/g/°C
Specific heat of Ice = 2.1 J/g/°C
Heat of fusion of water at 0°C = 334 J/g)


Detailed Solution for JEE Main Mock Test - 4 - Question 21

Let amount of ice is m gm.
According to principal of calorimeter heat taken by ice = heat given by water
∴ 20 × 2.1 ×  m + (m - 20) × 334
= 50 ×  4.2 ×  40
376 m = 8400 + 6680
m = 40

JEE Main Mock Test - 4 - Question 22

If the standard deviation of x1, x2, …, xn is 3.5, then the standard deviation of – 2x1 – 3, – 2x2 – 3, …, – 2xn – 3 is:

Detailed Solution for JEE Main Mock Test - 4 - Question 22

The standard deviation of a set remains unchanged if each data is increased or decreased by a constant.
However, it changes similarly when data is multiplied or divided by a constant.
∴ The SD for the new data set will be = −2 × 3.5 = −7

JEE Main Mock Test - 4 - Question 23

Consider three observations a, b and c such that b = a + c. If the standard deviation of a + 2, b + 2, c + 2 is d, then which of the following is true?

Detailed Solution for JEE Main Mock Test - 4 - Question 23

JEE Main Mock Test - 4 - Question 24

In how many different ways can the letters of the word 'GEOGRAPHY' be arranged such that the vowels must always come together?

Detailed Solution for JEE Main Mock Test - 4 - Question 24

Given:

The given number is 'GEOGRAPHY'

Calculation:

The word 'GEOGRAPHY' has 9 letters. It has the vowels E, O, A in it, and these 3 vowels must always come together. Hence these 3 vowels can be grouped and considered as a single letter. That is, GGRPHY(EOA).

Let 7 letters in this word but in these 7 letters, 'G' occurs 2 times, but the rest of the letters are different.

Now,

The number of ways to arrange these letters = 7!/2!

⇒ 7 × 6 × 5 × 4 × 3 = 2520

In the 3 vowels(EOA), all vowels are different

The number of ways to arrange these vowels = 3!

⇒ 3 × 2 × 1 = 6

Now, 

The required number of ways = 2520 × 6 

⇒ 15120

∴ The required number of ways is 15120.

JEE Main Mock Test - 4 - Question 25

The equations ax + 9y = 1 and 9y - x - 1 = 0 represent the same line if a =

Detailed Solution for JEE Main Mock Test - 4 - Question 25

Given:

Equation1 = ax + 9y = 1

Equation2 = 9y - x - 1 = 0 

Concept used:

If linear equations are a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. Here, the equations have an infinite number of solutions, if

a1/a2 = b1/b2 = c1/c2

Calculation:

We have equations,

ax + 9y = 1 

⇒ ax + 9y - 1 = 0

and, 9y - x - 1 = 0

⇒ x - 9y + 1 = 0

Here, a1 = a, b1 = 9, c1 = -1

and, a2 = 1, b2 = -9, c2 = 1

As we know that 

a1/a2 = b1/b2 = c1/c2

⇒ a/1 = 9/-9 = -1/1

⇒ a = -1 = -1

⇒ a = -1

∴ The value of a is -1.

JEE Main Mock Test - 4 - Question 26

Let R be a relation defined on the set of N natural numbers as R = {(x, y): y is a factor of x, x, y∈ N} then,

Detailed Solution for JEE Main Mock Test - 4 - Question 26

R = {(x, y): y is a factor of x, x, y∈ N}
As we know that 2 is a factor of 4
So, according to the options (4, 2) ϵ R

JEE Main Mock Test - 4 - Question 27

If y = y(x) is the solution of the differential equation e such that y(0) = 0, then y(1) is equal to:

Detailed Solution for JEE Main Mock Test - 4 - Question 27


(ey) = xex + exc
At y(0) = 0,
c = 1
y = x + ln(x + 1)
y(1) = 1 + ln2

JEE Main Mock Test - 4 - Question 28

The mean and variance of a binomial distribution are 4 and 2, respectively. What is the probability of two successes?

Detailed Solution for JEE Main Mock Test - 4 - Question 28


Hence, P(X = 2) = 28/256 = 7/64

*Answer can only contain numeric values
JEE Main Mock Test - 4 - Question 29

Let A = R and A4 = [aij]. If a11 = 109, then a22 is equal to _______.


Detailed Solution for JEE Main Mock Test - 4 - Question 29


Given: (x2 + 1)2 + x2 = 109
Let x2 + 1 = t
t2 + t - 1 = 109
 (t - 10)(t + 11) = 0
 t = 10 = x2 + 1 = a22

*Answer can only contain numeric values
JEE Main Mock Test - 4 - Question 30

The value of . Find the value of m.


Detailed Solution for JEE Main Mock Test - 4 - Question 30

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