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JEE Main Mock Test - 4 - JEE MCQ


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75 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Main Mock Test - 4

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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]

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

*Answer can only contain numeric values
JEE Main Mock Test - 4 - Question 22

Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019 m-3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to k × 10-1Ωm. Find the value of k.


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

*Answer can only contain numeric values
JEE Main Mock Test - 4 - Question 23

The resistance of the meter bridge AB in given figure is 4Ω. With a cell of emf ε = 0.5V and rheostat resistance Rh = 2 ohm, the null point is obtained at some point J. When the cell is replaced by another one of emf ε = ε2 the same null point J is found for Rh = 6Ω. The emf εis k × 10-1 V.
(Answer up to the nearest integer)


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

*Answer can only contain numeric values
JEE Main Mock Test - 4 - Question 24

When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to:


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

Min. energy required for emission of electrons,

*Answer can only contain numeric values
JEE Main Mock Test - 4 - Question 25

Calculate the Coulomb force between 2 alpha particles separated by 3.2 x 10-15m.


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

Given, Charge on an alpha particle, q1 = q2 = +2e, Distance between the particles, r = 3.2 x 10-15m
As we know, Charge on an electron, 
Now, using coulomb's law, we get, Force acting on the particles is given by,  

JEE Main Mock Test - 4 - Question 26

In KO2, the nature of oxygen species and the oxidation state of oxygen atom are (respectively)

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

In KO2, the nature of oxygen species and the oxidation state of oxygen atom are, superoxide and −1/2 respectively.
Superoxide ion is O-2.
Let X be an oxidation state of oxygen. The oxidation state of K is +1.
+ 1 + 2(X) = 0
2X = - 1
X = -1/2

JEE Main Mock Test - 4 - Question 27

Consider separate solutions of 0.500 M C2H5OH (aq), 0.100 M Mg3(PO4)(aq), 0.250 M KBr (aq) and 0.125 M Na3PO(aq) at 25oC. Which statement is true about these solutions, assuming all salts to be strong electrolytes?

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

JEE Main Mock Test - 4 - Question 28

The concentrations of fluoride, lead, nitrate and iron in a water sample from an underground lake were found to be 1000 ppb, 40 ppb, 100 ppm and 0.2 ppm, respectively. This water is unsuitable for drinking due to high concentration of

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

Fluoride - 1000 ppb = 1 ppm (which is within the permitted levels, viz. < 10 ppm)
Lead - 40 ppb (which is within the permitted levels, viz. < 50 ppb)
Iron - 0.2 ppm( is the maximum permmiscible limit)
Nitrate - 100 ppm is not suitable for drinking.
[NO3-] > 50 ppm, in drinking water can cause disease such as methemoglobinemia ('blue baby' syndrome).

JEE Main Mock Test - 4 - Question 29

The synthesis of alkyl fluorides is best accomplished by

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

JEE Main Mock Test - 4 - Question 30

The correct structure of product 'P' in the following reaction is:

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

Asn-Ser is dipeptide having following structure

JEE Main Mock Test - 4 - Question 31

The polymer obtained from the following reactions is:

Detailed Solution for JEE Main Mock Test - 4 - Question 31

JEE Main Mock Test - 4 - Question 32

When 2-butyne is treated with H2/Lindlar's catalyst, compound X is produced as the major product, and when treated with Na/liq. NH3, it produces y as the major product. Which of the following statements is correct?

Detailed Solution for JEE Main Mock Test - 4 - Question 32


Dipole moment of X is more than Y.
Boiling point of X is more than Y.

JEE Main Mock Test - 4 - Question 33

In graphite and diamond, the percentage of p-characters of the hybrid orbital in hybridisation is respectively:

Detailed Solution for JEE Main Mock Test - 4 - Question 33

JEE Main Mock Test - 4 - Question 34

Which of the following conversions involves change in both shape and hybridisation?

Detailed Solution for JEE Main Mock Test - 4 - Question 34

The hybridisation in boron atom changes from sp2 to sp3 in the following conversion:
BF3 → BF-4

JEE Main Mock Test - 4 - Question 35

Which of the following lines correctly show the temperature dependence of equilibrium constant K for an exothermic reaction?

Detailed Solution for JEE Main Mock Test - 4 - Question 35


Comparing with equation of a straight line,

Since the reaction is exothermic, ΔHis negative. Therefore, slope = +ve.

Hence, option (1) is correct.

JEE Main Mock Test - 4 - Question 36

An aqueous solution contains 0.10 M H2S and 0.20 M HCI. If the equilibrium constant for the formation of HS- from H2S is 1 .0 x 10-7 and that of S2- from HS- ions is 1.2 x 10-13, then the concentration of S2- ions in aqueous solution is

Detailed Solution for JEE Main Mock Test - 4 - Question 36

In presence of external H+,

JEE Main Mock Test - 4 - Question 37

Which of the following is not a property of physical adsorption?

Detailed Solution for JEE Main Mock Test - 4 - Question 37

Physical adsorption is multilayer adsorption. Monolayer is formed in case of chemisorption.

JEE Main Mock Test - 4 - Question 38

The major product of the following reaction is:

Detailed Solution for JEE Main Mock Test - 4 - Question 38

JEE Main Mock Test - 4 - Question 39

Peroxyacetyl nitrate (PAN), an eye irritant is produced by:

Detailed Solution for JEE Main Mock Test - 4 - Question 39

Photochemical smog produce chemicals such as formaldehyde, acrolein and peroxyacetyl nitrate (PAN).

JEE Main Mock Test - 4 - Question 40

Galvanization is applying a coating of

Detailed Solution for JEE Main Mock Test - 4 - Question 40

Galvanisation or galvanization (or galvanizing as it is most commonly called) is the process of applying a protective zinc coating to iron or steel, to prevent rusting.

JEE Main Mock Test - 4 - Question 41

Which of the following atoms has the highest first ionisation energy?

Detailed Solution for JEE Main Mock Test - 4 - Question 41

JEE Main Mock Test - 4 - Question 42

Which of the following compounds is metallic and ferromagnetic?

Detailed Solution for JEE Main Mock Test - 4 - Question 42

Chromium Oxide (CrO2) is metallic compound and it shows Ferromagnetic property. It is a black synthetic magnetic solid and was once widely used in magnetic tape emulsion.

JEE Main Mock Test - 4 - Question 43

The major product formed in the following reaction is:

Detailed Solution for JEE Main Mock Test - 4 - Question 43


PCC oxidises 1° and 2° alcohols into aldehyde and ketones, respectively.

JEE Main Mock Test - 4 - Question 44

Which one of the following is not a common component of Photochemical Smog?

Detailed Solution for JEE Main Mock Test - 4 - Question 44
  • Photochemical smog is a mixture of air pollutants that have been chemically altered into further noxious compounds by exposure to sunlight.
  • The main components of photochemical smog are nitrogen oxides, Volatile Organic Compounds (VOCs), tropospheric ozone, and PAN (peroxyacetyl nitrate).
JEE Main Mock Test - 4 - Question 45

Lanthanoid contraction is caused due to:

Detailed Solution for JEE Main Mock Test - 4 - Question 45
  • The elements in the Lanthanide series are La, Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb and Lu. In Lanthanides (Ce−Lu), the atomic and ionic radii decrease steadily.
  • This steady decrease in atomic and ionic radii is known as Lanthanide Contraction. 
  • The contraction is due to the fact that f−orbitals are not capable of providing effective shielding for the valence electrons from the nuclear attraction.
*Answer can only contain numeric values
JEE Main Mock Test - 4 - Question 46

The mole fraction of glucose (C6H12O6) in an aqueous binary solution is 0.1. The mass percentage of water in it, is ________. (Nearest integer)


Detailed Solution for JEE Main Mock Test - 4 - Question 46

Mole fraction of glucose in aqueous solution = 0.1
Mass of glucose in the solution = 0.1 × 180 = 180 g
Mole fraction of water in aqueous solution = 1 - 0.1 = 0.9
Mass of water in the solution = 0.9 × 18 = 162 g
Total mass of the solution = 180 g + 162 g = 342 g
Mass percentage of water = (162 g/342 g) × 100 = 47.368% ≈ 47%

*Answer can only contain numeric values
JEE Main Mock Test - 4 - Question 47

For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of M/10 sulphuric acid. The unreacted acid required 20 ml of M/10 sodium hydroxide for complete neutralisation. The percentage of nitrogen in the compound is (Nearest integer)


Detailed Solution for JEE Main Mock Test - 4 - Question 47

*Answer can only contain numeric values
JEE Main Mock Test - 4 - Question 48

An element with molar mass 2.7 × 10-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 103 kg m-3, the radius of the element is approximately _______ × 10-12 m (Nearest integer).


Detailed Solution for JEE Main Mock Test - 4 - Question 48

M, molar mass of the element = 27 gm mol-1
a, edge length of the cubic unit cell = 405 pm
= 4.05 × 10-8 cm
d, density of the element = 2.7 gm/cc

= 143 × 10-12 m

*Answer can only contain numeric values
JEE Main Mock Test - 4 - Question 49

Given:
(i) 2Fe2O3(s) → 4Fe(s) + 3O2(g); G° = +1487.0 kJ mol-1
(ii) 2CO(g) + O2(g) → 2CO2(q): ΔrG° = -514.4 kJ mol-1
Free energy change, ΔfG° for the following reaction will be -_____ KJ mol-1. (Nearest integer)
2Fe2O3(s) + 6CO(g) → 4Fe(s) + 6CO2(g)


Detailed Solution for JEE Main Mock Test - 4 - Question 49

2Fe2O3(s) → 4Fe(s) + 3O2(g); ΔG° = 1487 kJ mol-1
2CO(g) + O2(g) → 2CO2(g); ΔG° = -514.4 kJ mol-1
For reaction:
2Fe2O3 + 6CO → 4Fe + 6CO2
ΔG = - 514.4 × 3 + 1487 = -56.2 kJ mol-1
≈ -56 KJ mol-1

*Answer can only contain numeric values
JEE Main Mock Test - 4 - Question 50

The number of P-O bonds in P4O6 is:


Detailed Solution for JEE Main Mock Test - 4 - Question 50

The number of P-O bonds in P4O6 is 12.

JEE Main Mock Test - 4 - Question 51

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 51

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 52

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 52

JEE Main Mock Test - 4 - Question 53

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 53

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 54

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 54

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 55

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 55

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 56

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 56


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

JEE Main Mock Test - 4 - Question 57

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 57


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

JEE Main Mock Test - 4 - Question 58

If a plane bisects the line segment joining the points (1, 2, 3) and (-3, 4, 5) at right angles, then this plane also passes through the point

Detailed Solution for JEE Main Mock Test - 4 - Question 58


Q(-1, 3, 4), where Q is the midpoint of PR.
The normal to the plane is:

Equation of plane is:
2x - y - z + d = 0
Since the plane passes through the point (-1, 3, 4), d = 9
Hence, the equation of plane is:
2x - y - z + 9 = 0
From the given options; (-3, 2, 1) satisfies it.

JEE Main Mock Test - 4 - Question 59

The value of k for which the points A (1, 0, 3), B (– 1, 3, 4), C (1, 2, 1) and D (k, 2, 5) are coplanar, is

Detailed Solution for JEE Main Mock Test - 4 - Question 59

JEE Main Mock Test - 4 - Question 60

A unit tangent vector at t = 2 on the curve x = t2 + 2, y = 4t– 5, z = 2t2 – 6t is

Detailed Solution for JEE Main Mock Test - 4 - Question 60

The position vector of any point at t is r = (2 + t2) i + (4t3 - 5) j + (2t2 - 6t) k

JEE Main Mock Test - 4 - Question 61

For each t ∈ R, let [t] be the greatest integer less than or equal to t. Then,

Detailed Solution for JEE Main Mock Test - 4 - Question 61

JEE Main Mock Test - 4 - Question 62

Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row consists of two balls and so on. If 99 more identical balls are added to the total number of balls used in forming the equilateral triangle, then all these balls can be arranged in a square, each of whose sides contains exactly 2 balls less than the number of balls that each side of the triangle contains. Then, the number of balls used to form the equilateral triangle is:

Detailed Solution for JEE Main Mock Test - 4 - Question 62

Let n be the total number of balls required to formed equilateral triangle, then

n2 + n + 198 = 2(n2 + 4 - 4n)
n2 - 9n - 190 = 0
n2 - 19n + 10 - 190 = 0
n(n - 19) + 10(1 - 19) = 0
n = 19
Number of balls = (19)(20)/2 = 190

JEE Main Mock Test - 4 - Question 63

Consider the following three statements:

P : 5 is a prime number.
Q : 7 is a factor of 192
R : L.C.M. of 5 and 7 is 35.

Then the truth value of which one of the following statements is true?

Detailed Solution for JEE Main Mock Test - 4 - Question 63

JEE Main Mock Test - 4 - Question 64

The differential equation of the family of curves x2 = 4b(y + b), b ∈ R is

Detailed Solution for JEE Main Mock Test - 4 - Question 64

x= 4b(y + b) .... (i)
x2 = 4by + 4b2
2x = 4by' + 0

⇒ b = x / (2y')
Put the value of b in equation (i) we get
x2 = 4(x / (2y'))(y + (x/(2y')))
x(y')2 = x + 2yy'

JEE Main Mock Test - 4 - Question 65

If m is the minimum value of k for which the function f(x) = x is increasing in the interval [0, 3] and M is the maximum value of f in [0, 3] when k = m, then the ordered pair (m, M) is equal to

Detailed Solution for JEE Main Mock Test - 4 - Question 65

JEE Main Mock Test - 4 - Question 66

Let Then, in the interval (-2, 2), g is:

Detailed Solution for JEE Main Mock Test - 4 - Question 66


JEE Main Mock Test - 4 - Question 67

Find the modulus and argument of the complex number .

Detailed Solution for JEE Main Mock Test - 4 - Question 67

JEE Main Mock Test - 4 - Question 68

Which of the following is correct?

Detailed Solution for JEE Main Mock Test - 4 - Question 68

We know that to every square matrix, A = [aij] of order n. We can associate a number called the determinant of square matrix A, where aij = (i, j)th element of A. Thus, the determinant is a number associated with a square matrix.

JEE Main Mock Test - 4 - Question 69

A line with positive direction cosines passes through the point P(2, - 1, 2) and makes equal angles with the coordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ is equal to

Detailed Solution for JEE Main Mock Test - 4 - Question 69

JEE Main Mock Test - 4 - Question 70

The curve amongst the family of curves, represented by the differential equation. (x2 - y2)dx + 2xy dy = 0 which passes through (1, 1) is:

Detailed Solution for JEE Main Mock Test - 4 - Question 70

*Answer can only contain numeric values
JEE Main Mock Test - 4 - Question 71

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


Detailed Solution for JEE Main Mock Test - 4 - Question 71


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 72

The value of . Find the value of m.


Detailed Solution for JEE Main Mock Test - 4 - Question 72

*Answer can only contain numeric values
JEE Main Mock Test - 4 - Question 73

If the third term in the binomial expansion of (1 + x log2x)5 equals 2560, and the fractional possible value of x is k/4. Find the value of k. 


Detailed Solution for JEE Main Mock Test - 4 - Question 73

*Answer can only contain numeric values
JEE Main Mock Test - 4 - Question 74

where x and y are real numbers, then y - x is equal to _____.


Detailed Solution for JEE Main Mock Test - 4 - Question 74

*Answer can only contain numeric values
JEE Main Mock Test - 4 - Question 75

The value of the integral  (where [x] denotes the greatest integer less than or equal to x) is ___.


Detailed Solution for JEE Main Mock Test - 4 - Question 75

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