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JEE Main Mock Test - 1 - JEE MCQ


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75 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Main Mock Test - 1

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JEE Main Mock Test - 1 - Question 1

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

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 2

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 3

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 3

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 4

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 4

JEE Main Mock Test - 1 - Question 5

An object is said to be in uniform motion in a straight line if its displacement

Detailed Solution for JEE Main Mock Test - 1 - Question 5

Explanation:Uniform motion is the kind of motion in which a body covers equal displacement in equal intervals of time. It does not matter how small the time intervals are, as long as the displacements covered are equal.

If a body is involved in rectilinear motion and the motion is uniform, then the acceleration of the body must be zero.

JEE Main Mock Test - 1 - Question 6

Two cars are moving in same direction with speed of 30 kmph. They are separated by a distance of 5 km. What is the speed of a car moving in opposite direction if it meets the two cars at an interval of 4 min?   

Detailed Solution for JEE Main Mock Test - 1 - Question 6

Lets say the speed of second car be v. Now with respect to car 1, speed of car 2 is v - 30. Now if the cross each other at just after 4 min then the distance traveled by car 2 relative to car 1 is (v -30) x 4/60 and we know the distance travelled relatively is 5km
Thus we get 
(v -30) x 4/60 = 5
V = 300/4 - 30
= 45 km/h

JEE Main Mock Test - 1 - Question 7

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 7

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 + ½ a
= 5/2 a
Similarly distance travelled in fourth second, s4= 3a + ½ a
= 7/2 a
Hence the reqiuired ratio is 5/7

JEE Main Mock Test - 1 - Question 8

The magnetic field associated with a light wave is given, at the origin, by B = B0 [sin(3.14 × 107)ct + sin(6.28 × 107)ct]. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons ?
(c = 3 × 10ms-1, h = 6.6 × 10-34 J-s)

Detailed Solution for JEE Main Mock Test - 1 - Question 8

B = B0sin(π × 107C)t + B0sin (2π × 107C)t
Since there are two EM waves with different frequency, to get maximum kinetic energy, we take the photon with higher frequency.
B1 = B0sin(π × 107C)t  v1 = 107/2 x c 
B2 = B0sin(2π × 107C)t v2 = 107C
where C is speed of light C = 3 × 108 m/s
v2 > v1
so KE of photoelectron will be maximum for photon of higher energy.
v= 107C Hz
hv = φ + KEmax
energy of photon
Eph = hv = 6.6 × 10-34 × 107 × 3 × 109
Eph = 6.6 × 3 × 10-19J

KEmax = Eph - φ
= 12.375 - 4.7 = 7.675 eV  7.72 eV

JEE Main Mock Test - 1 - Question 9

One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27°C. The work done on the gas will be:

Detailed Solution for JEE Main Mock Test - 1 - Question 9

Work done in isothermal process on the gas,

JEE Main Mock Test - 1 - Question 10

The electric field at a point associated with a light wave is E = (100 Vm −1) sin [(3.0 ×1015s−1)t] sin [(6.0×1015s−1)t]. If this light falls on a metal surface having a work function of 2.0 eV, what will be the maximum kinetic energy of the photo electrons ?

Detailed Solution for JEE Main Mock Test - 1 - Question 10

JEE Main Mock Test - 1 - Question 11

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 11

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 12

Both the nucleus and the atom of some element are in their respective first excited states. They get de-excited by emitting photons of wavelengths λN and λA, respectively. The ratio λNis closest to:

Detailed Solution for JEE Main Mock Test - 1 - Question 12

Energy of the emitted photon,

where EA and EN are energies of photons from atom and nucleus, respectively.

EA is order of eV and EN is order of MeV

JEE Main Mock Test - 1 - Question 13

A gravity meter can detect change in acceleration due to gravity (g) of the order of 10-9%. Calculate the smallest change in altitude near the surface of the earth that results in a detectable change in g. Radius of the earth R = 6.4 x 106m.

Detailed Solution for JEE Main Mock Test - 1 - Question 13

JEE Main Mock Test - 1 - Question 14

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 14

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 • Y⇒ 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 15

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 15


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 16

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 16

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 17

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 17

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 18

Consider the following figure:

Which of the following labelled points in the figure given above indicate an unstable state of an object?

Detailed Solution for JEE Main Mock Test - 1 - Question 18

Explanation: 

  • From the given graph we can see that the potential energy is minimum at point B and D
  • Whereas it has maximum potential energy at point A and C respectively
  • Hence point A and C will be an unstable state since they have maximum potential energy compared to B and D.
JEE Main Mock Test - 1 - Question 19

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: 

JEE Main Mock Test - 1 - Question 20

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 20


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

Therefore, 
So, d = R2/4h

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 21

A physical quantity Q is given by

The percentage error in A,B, C, D are 1%, 2%, 4%, 2% respectively. Find the percentage error in Q.


Detailed Solution for JEE Main Mock Test - 1 - Question 21

= 22 or 22 %

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 22

In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10–10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.


Detailed Solution for JEE Main Mock Test - 1 - Question 22

Mass of H atom = m
Mass of Cl atom = 35.5m
Let the centre of mass of the system lie at a distance x from the Cl atom.
Distance of the centre of mass from the H atom = (1.27 – x)

Let us assume that the centre of mass of the given molecule lies at the origin. Therefore, we can have:
[ m(1.27 – x) + 35.5mx ] / (m + 35.5m)  =  0
m(1.27 – x) + 35.5mx =  0
1.27 - x = -35.5x
∴ x = -1.27 / (35.5 - 1)  =  -0.037 Å
Here, the negative sign indicates that the centre of mass lies at the left of the molecule. Hence, the centre of mass of the HCl molecule lies 0.037Å from the Cl atom.

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 23

A bullet is fired from a gun at a speed of 5000 m/s. At what height should the gun be aimed above a goal if it has to strike the goal at a distance of 500 m? Take g=10m/s2


Detailed Solution for JEE Main Mock Test - 1 - Question 23

Let t be the time taken to cover the horizontal distance of 500 m from the gun to the target ,then
t = 500 / 5000 = .1 s
During this time,bullet will fall down vertically due to acceleration due to gravity.
Vertical distance covered
h = 1/2gt2 = .005 m
= 5cm.
Hence the gun should be aimed at 5cm above the target

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 24

The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510nT. What is the amplitude of the electric field part of the wave?


Detailed Solution for JEE Main Mock Test - 1 - Question 24

The amplitude of the magnetic field of an electromagnetic wave in a vacuum, B0 = 510nT = 510 x 10-9 T

Speed of light in a vacuum, c = 3 x 108 m/s

The relation between the amplitude of the electric field and the magnetic field is c = E0/B0

E = cBo = 3 x 108 x 510 x 10-9 = 153 N/C

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 25

A radio can tune in to any station in the 7.5 MHz to 12MHz bands. What is the corresponding wavelength band?


Detailed Solution for JEE Main Mock Test - 1 - Question 25

A radio can tune to minimum frequency, v1 = 7.5 MHz = 7.5 x 106 Hz

Maximum frequency, v2 = 12MHz = 12 x 106 Hz

Speed of light, c = 3 x 108 m/s

Corresponding wavelength for v1 can be calculated as: 
Corresponding wavelength for v2 can be calculated as: 

JEE Main Mock Test - 1 - Question 26

Electromagnetic radiations having λ = 310 Å are subjected to a metal sheet having work function = 12.8 eV. What will be the velocity of photoelectrons with maximum Kinetic energy.

Detailed Solution for JEE Main Mock Test - 1 - Question 26



V2 = 9.56 × 1012
V = 3.09 × 106 m/sec.

JEE Main Mock Test - 1 - Question 27

Consider the ground state of Cr atom (X = 24). The number of electrons with the azimuthal quantum numbers, ℓ = 1 and 2 are, respectively [2004]

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

Electronic configuration of Cr atom (z = 24) = 1s2, 2s2 2p6, 3s2 3p6 3d5, 4s1
when ℓ = 1, p - subshell,
Numbers of electrons = 12
when ℓ = 2, d - subshell,
Numbers of electrons = 5

JEE Main Mock Test - 1 - Question 28

Which of the following angle corresponds to sp2 hybridisation?

Detailed Solution for JEE Main Mock Test - 1 - Question 28

sp2 hybridisation gives three sp2 hybrid orbitals which are planar triangular forming an angle of 120° with each other.
The electronic configurations of three elements A, B and C are given below.
Answer the questions from 14 to 17 on the basis of these configurations.
A ls22s22p6
B ls22s22p63s23p3
C ls22s22p63s23p

JEE Main Mock Test - 1 - Question 29

Azide ion [N3-] exhibits an (N−N) bond order of 2 and may be represented by the resonance structures I,II and III given below:

The most stable resonance structure is

Detailed Solution for JEE Main Mock Test - 1 - Question 29

JEE Main Mock Test - 1 - Question 30

C—Cl bond in (vinyl chloride) is stabilised in the same way as in 

Detailed Solution for JEE Main Mock Test - 1 - Question 30


Due to delocalisation of π-electrons, (C—Cl) bond is stable and it does not show SN reactions; Cl directly attached (C=C) bond, i.e. vinyl group.

JEE Main Mock Test - 1 - Question 31

The following equations are balanced atomwise and chargewise.

(i) Cr2O72- + 8H+ + 2H2O2 → 2Cr3+ + 7H2O + 2O2

(ii) Cr2O72- + 8H+ + 5H2O2→ 2Cr3+ + 9H2O + 4O2

(iii) Cr2O72- + 8H+ + 7H2O2→ 2Cr+ + 11H2O + 5O2

The precise equationl equations representing the oxidation of H2O2 is/are

Detailed Solution for JEE Main Mock Test - 1 - Question 31

The correct answer is option A
Cr2O72- converts into Cr3+ in acidic medium I.e. in H+ medium.
First balance the Cr atom on both sides and then Oxygen atom. H+ is in excess due to acidic medium.
Add H+ as +ve charge to balance the charge on both sides.

JEE Main Mock Test - 1 - Question 32

In the general electronic configuration -

(n - 2)f1-14 (n - 1)d0-1 ns2, if value of n = 7 the configuration will be -

Detailed Solution for JEE Main Mock Test - 1 - Question 32

The correct answer is Option B.

General electronic configuration is given:
(n − 2)f1-14(n − 1)d01ns2 , where(n=7)
5f1-146d01ns2
The seventh period (n=7) is similar to the sixth period with the successive filling up of the 7s, 5f, 6d and 7p orbitals and includes most of the man-made radioactive elements. This period will end at the element with atomic number 118 which would belong to the noble gas family. Filling up of the 5f orbitals after actinium (Z=89) gives the 5f-inner transition series known as the Actinide Series.
 

JEE Main Mock Test - 1 - Question 33

Which of the following reactions is an example of a redox reaction?

Detailed Solution for JEE Main Mock Test - 1 - Question 33

A redox reaction involves the increase in the oxidation state of one species (oxidation) and a decrease in the oxidation state of another species (reduction).

Xenon undergoes oxidation, while oxygen undergoes reduction.

JEE Main Mock Test - 1 - Question 34

Which of the following tests cannot be used for identifying amino acids?

Detailed Solution for JEE Main Mock Test - 1 - Question 34

Barfoed's test is a chemical test used for detecting the presence of monosaccharides. It is based on the reduction of Copper (II) acetate to Copper (I) oxide, which forms a brick-red precipitate

Biuret test , Xanthoproteic test , Ninhydrin test are used for identifying Amino Acids

JEE Main Mock Test - 1 - Question 35

Which dicarboxylic acid in presence of a dehydrating agent is least reactive to give an anhydride :

Detailed Solution for JEE Main Mock Test - 1 - Question 35


This reaction is least likely to take place because 7 membered cyclic anhydride is very unstable.

JEE Main Mock Test - 1 - Question 36

The major organic compound formed by the reaction of 1,1,1-trichloroethane with silver powder is:

Detailed Solution for JEE Main Mock Test - 1 - Question 36

JEE Main Mock Test - 1 - Question 37

Biochemical Oxygen Demand (BOD) value can be a measure of water pollution caused by the organic matter. Which of the following statements is correct?

Detailed Solution for JEE Main Mock Test - 1 - Question 37

Clean water has BOD value less than 5 ppm. Polluted water has BOD value higher than 10 ppm. The option (A) represents correct statement.

JEE Main Mock Test - 1 - Question 38

The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is

Detailed Solution for JEE Main Mock Test - 1 - Question 38

During the electrolysis of aqueous solution of s-block elements, H2 gas is obtained at cathode. Alkaline earth metals are strong reducing agents. So, their oxides and halides cannot be reduced by any other elements.

JEE Main Mock Test - 1 - Question 39

The actinoids exhibit more oxidation states in general than the lanthanoids. This is because:

Detailed Solution for JEE Main Mock Test - 1 - Question 39

In actinoids, the outermost 5f orbitals are away from the nucleus. This makes them similar in energies to 6d and 7s orbitals. Thus, electrons can easily transition between these orbitals, and once they reach the valence orbital they can be ionized, This leads to the variability in oxidation states. In the case of lanthanoids, the outermost 4f orbitals are too close to the nucleus.

JEE Main Mock Test - 1 - Question 40

The mass of carbon anode consumed (giving the only carbon dioxide) in the production of 270 kg of aluminium metal from bauxite by the Hall process is

[atomic mass Al = 27]

Detailed Solution for JEE Main Mock Test - 1 - Question 40

In hall and Heroult process 2Al2O3 → 4Al + 3O2
4C + 3O2 → 2CO2 + 2CO↑
Only for removal of CO2 following equation is possible

∵ For 108 kg of Al, 36 kg of C is required in above reaction
∴ For 270 kg of Al require amount of C = 36/108 x 270 = 90 Kg

JEE Main Mock Test - 1 - Question 41

Which of the following polymers can be made by condensation polymerisation reaction?

Detailed Solution for JEE Main Mock Test - 1 - Question 41

Condensation polymers are formed by repeated condensation reactions between two different bi-functional or tri-functional monomeric units. In these polymerisation reactions, the elimination of small molecules such as water, alcohol, hydrogen chloride, etc. takes place. Dacron, Nylon-6, 6 and Bakelite are produced by condensation polymerisation. On the other hand, Polyethylene is formed by breakage of the double bond and there is no elimination of molecules.

JEE Main Mock Test - 1 - Question 42

The artificial sweetener containing chlorine that has the appearance and taste as that of sugar and is stable at cooking temperature is:

Detailed Solution for JEE Main Mock Test - 1 - Question 42

The artificial sweetener containing chlorine that has the appearance and taste as that of sugar and is stable at cooking temperature is sucralose. Sucralose is derived from sugar through a multi-step patented manufacturing process that selectively substitutes three atoms of chlorine for three hydroxyl groups on the sugar molecule. This change produces a sweetener that has no calories, yet is 600 times sweeter than sucrose. Sucralose tastes like sugar. Its chemical formula is C12H19Cl3O8.

JEE Main Mock Test - 1 - Question 43

The inert gas atom in which the total number of d-electrons is equal to the difference in numbers of total p- and s-electrons, is:

Detailed Solution for JEE Main Mock Test - 1 - Question 43

He, Ne, and Ar do not have any d electrons, while the difference of their s and p electrons is not zero.
The electronic configuration of krypton (atomic number is 36) is: 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2, 4p6
Total s-electrons are 8 and total p-electrons are 18. Thus, the difference is 10 and total d-electrons are 10 which are equal.

JEE Main Mock Test - 1 - Question 44

Which of the following is a Lewis acid?

Detailed Solution for JEE Main Mock Test - 1 - Question 44

In Trimethyl borane there are only 6 electrons in the valence shell of boron. Also it has one empty p-orbital to accomodate a lone pair of electrons.
Hence, it is a Lewis acid.

JEE Main Mock Test - 1 - Question 45

Which of the following species is not paramagnetic?

Detailed Solution for JEE Main Mock Test - 1 - Question 45

NO ⇒ One unpaired electron is present in π* molecular orbital.
CO ⇒ No unpaired electron is present.
O2 ⇒ Two unpaired electrons are present in π* molecular orbitals.
B2 ⇒ Two unpaired electrons are present in π bonding molecular orbitals.

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 46

How many chiral centers are in the following compound?


Detailed Solution for JEE Main Mock Test - 1 - Question 46

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 47

29.2 % (w/w) HCl stock solution has density of 1.25 g mL1. The molecular weight of HCl is 36.5 g mol-1. The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is


Detailed Solution for JEE Main Mock Test - 1 - Question 47

Let mass of the stock solution = 100g

Mass of HCl in 100 g of 29.2 % (w/w) HCl stock solution = 29.2 g

Volume of the stock solution = 100g/1.25 g mL= 80 g

Number of moles of HCl in stock solution = 29.2/36.5 = 0.8

Molarity of the stock solution = (0.82/80)×1000 = 10 M

Using,

M1V1 = M2V2

10× V1 =0.4×200

or

V1 = 8mL

Hence, the volume required is 8 mL.

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 48

The elevation of boiling point of 0.10 m aqueous CrCl3.xNH3 solution is two times that of 0.05 m aqueous CaCl2 solution. The value of x is ______.

[Assume 100% ionisation of the complex and CaCl2, co-ordination number of Cr as 6, and that all NHmolecules are present inside the co-ordination sphere] (Answer up to the nearest integer)


Detailed Solution for JEE Main Mock Test - 1 - Question 48

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 49

The volume, in mL, of 0.02 M K2Cr2O7 solution required to react with 0.288 g of ferrous oxalate in acidic medium is _______.

(Molar mass of Fe = 56 g mol-1)
(Answer up to the nearest integer)


Detailed Solution for JEE Main Mock Test - 1 - Question 49

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 50

3-Methyl pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is:
(Answer up to the nearest integer)


Detailed Solution for JEE Main Mock Test - 1 - Question 50

The product 2-bromo-3-methyl pentane is asymmetrical with 2 chiral carbons.
Number of optical isomers = 2n = 22 = 4

JEE Main Mock Test - 1 - Question 51

The domain of the function  is (where [x] denotes greatest integer function)

Detailed Solution for JEE Main Mock Test - 1 - Question 51


JEE Main Mock Test - 1 - Question 52

Domain of definiti on of the function  

Detailed Solution for JEE Main Mock Test - 1 - Question 52


JEE Main Mock Test - 1 - Question 53

A real valued function f(x) satisfies the functional equation f(x – y) = f(x) f(y) – f(a – x) f(a + y) where a is a given constant and f(0) = 1, f(2a – x) is equal to

Detailed Solution for JEE Main Mock Test - 1 - Question 53

JEE Main Mock Test - 1 - Question 54

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 - 1 - Question 54

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 - 1 - Question 55

If z = x- iy and = p+ iq, then  is equal to [2004]

Detailed Solution for JEE Main Mock Test - 1 - Question 55

= p+ iq ⇒ z = p3 + (iq)3 + 3 p (iq)( p+ iq)

⇒ x - iy = p3 - 3pq2 + i(3p2q-q3)

∴ x = p3 - 3pq2 ⇒ 

y = q3 - 3p2q ⇒ 

 

JEE Main Mock Test - 1 - Question 56

If z ≠ 1 and is real, then the point represented by the complex number z lies : [2012]

Detailed Solution for JEE Main Mock Test - 1 - Question 56

Now   is real ⇒  Im  = 0

⇒  

⇒​  Im [(x2 – y2 + 2ixy) (x – 1) – iy)] = 0

⇒  2xy (x – 1) – y (x2 – y2) = 0

⇒ y(x2 + y2 – 2x) = 0 ⇒  y = 0;  x2 + y2 – 2x = 0

∴ z lies either on real axis or on a circle through origin.

 

JEE Main Mock Test - 1 - Question 57

Calculate the least whole number, which when subtracted from both the terms of the ratio 5 : 6 gives a ratio less than 17 : 22.

Detailed Solution for JEE Main Mock Test - 1 - Question 57

Given:

Initial ratio = 5 ∶ 6

Final ratio should be less than 17 ∶ 22

Calculation:

Let the least whole number that is needed to be subtracted be a.

According to the question,

(5 - a)/(6 - a) < 17/22

⇒ 5 × 22 - 22a < 17 × 6 - 17a 

⇒ 110 - 22a < 102 - 17a 

⇒ 110 - 102 < - 17a + 22a 

⇒ 8 < 5a 

⇒ 8/5 = 1.6 < a 

∴ The least whole number must be 2.

JEE Main Mock Test - 1 - Question 58

Given, 6x + 2(6 - x) > 2x - 2 < 5x/2 - 3x/4, then x can take which of the following values?

Detailed Solution for JEE Main Mock Test - 1 - Question 58

⇒ 6x + 2(6 - x) > 2x - 2

⇒ 6x + 12 - 2x > 2x - 2

⇒ 2x > - 14

⇒ x > - 7     ----(1)

⇒ 2x - 2 < 5x/2 - 3x/4

⇒ 2x - 2 < 7x/4

⇒ 8x - 7x < 8

⇒ x < 8     ----(2)

From (1) and (2),

- 7 < x < 8

∴ x = 5 satisfies the given conditions from the above options.

JEE Main Mock Test - 1 - Question 59

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that atleast 3 men are there on the committee. In how many ways can it be done ?

Detailed Solution for JEE Main Mock Test - 1 - Question 59

Given: 

(7 men + 6 women) 5 persons are to be chosen for a committee.

Formula used: nCr = n!/(n - r)! r!

Calculation:

Ways in which at least 3 men are selected;

⇒ 3 men + 2 women

⇒ 4 men + 1 woman 

⇒ 5 men + 0 woman 

Number of ways = 7C3 × 6C2 + 7C4 × 6C1 + 7C5 × 6C0

⇒ 7!/(3! × 4!) × 6!/(2! × 4!) + 7!/(4! × 3!) × 6!/(1! × 5!) + 7!/(5! × 2!) × 6!/(6!× 0!)

⇒ 35 × 15 + 35 × 6 + 21 

⇒ 735 + 21 = 756

∴ The required no of ways = 756.

JEE Main Mock Test - 1 - Question 60

The area (in sq. units) of the region {x ∈ R : x ≥ 0, y ≥ 0, y ≥ x - 2 and y ≤ √x) is 

Detailed Solution for JEE Main Mock Test - 1 - Question 60

JEE Main Mock Test - 1 - Question 61

Two parabolas with a common vertex and with axes along x-axis and y-axis intersect each other in the first quadrant. If the length of the latus rectum of each parabola is 3, then the equation of the common tangent to the two parabolas is

Detailed Solution for JEE Main Mock Test - 1 - Question 61

Equations of two parabolas are y2 = 3x and x2 = 3y.
Let equation of tangent to y2= 3x be y = mx + 3/4m
It is also tangent to x2 = 3y.

Hence common tangent is y = - x - 3/4
4(x + y) + 3 = 0

JEE Main Mock Test - 1 - Question 62

The letters of the word OUGHT are written in all possible orders and these words are written out as in a dictionary. Find the rank of the word TOUGH in this dictionary.

Detailed Solution for JEE Main Mock Test - 1 - Question 62

The word TOUGH will appear after all the words that start with G, H and O. Then we look at the second letter of the words starting with T and then third.

Hence, the rank of the word TOUGH will be one more than the sum of all the possibilities just mentioned.

Total number of letters in the word OUGHT is 5 and all the five letters are different, the alphabetical order of these letters is G, H, O, T, U.

Number of words beginning with G=4 !=24, Number of words beginning with H=4 !=24, Number of words beginning with O=4 !=24, Number of words beginning with TG=3 !=6, Number of words beginning with TH=3 !=6, Number of words beginning with T O G=2 !=2, Number of words beginning with TOH=2 !=2

Next come the words beginning with TOU and TOUH is the first word beginning with TOU.

∴ Rank of 'TOUGH' in the dictionary =24 + 24 + 24 + 6 + 6 + 2 + 2 + 1 = 89

JEE Main Mock Test - 1 - Question 63

Let P be a point on the parabola x2 = 4y. If the distance of P from the centre of the circle x2 + y2 + 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P is

Detailed Solution for JEE Main Mock Test - 1 - Question 63

Let P = (2t, t2)
For the distance between point P and centre of circle to be minimum, line drawn from the centre of circle to point P must be normal to the parabola at P.
The equation normal at P to x2 = 4y:
y - t2 = -1/t (x - 2t)
It passes through (-3, 0).
0 - t2 = -1/t (-3 - 2t)
t3 + 2t + 3 = 0
(t + 1)(t2 - t + 3) = 0
 t = -1
Point P is (-2, 1).
The equation of tangent to x2 = 4y at (x', y') is:
xx' = 2(y + y')
So, at P(-2, 1),
x(-2) = 2(y + 1)
x + y + 1 = 0

JEE Main Mock Test - 1 - Question 64

Let the population of rabbits surviving at a time t be governed by the differential equation  If p(0) = 100, then p(t) equals

Detailed Solution for JEE Main Mock Test - 1 - Question 64

JEE Main Mock Test - 1 - Question 65

A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag. Its colour is seen and this ball, along with two additional balls of the same colour, is returned to the bag. If now, a ball is drawn at random from the bag, then the probability that this drawn ball is red is:

Detailed Solution for JEE Main Mock Test - 1 - Question 65

E1: Event that first ball drawn is red
E2: Event that first ball drawn is black
E: Event that second ball drawn is red

JEE Main Mock Test - 1 - Question 66

The length of projection of the line segment joining the points (5, -1, 4) and (4, -1, 3) on the plane x + y + z = 7 is

Detailed Solution for JEE Main Mock Test - 1 - Question 66

JEE Main Mock Test - 1 - Question 67

O is the origin and A is the point (a, b, c). Deduce the equation of the plane through A at right angles to OA.

Detailed Solution for JEE Main Mock Test - 1 - Question 67

Given, D.R.'s of OA are a 0, b - 0, c - 0 or a, b, c.

Equation of any plane through A, i.e. (a, b, c) is: A(x – a) + B(y - b) + C(z - c) = 0

But plane is at right angles to OA which therefore is normal and whose D.R.'s are a, b, c.

So, the plane is, a(x - a) + b(y - b) + c(z - c) = 0

ax + by + cz = a2 + b2 + c2

JEE Main Mock Test - 1 - Question 68

Tangents drawn from the point (-8, 0) to the parabola y2 = 8x touch the parabola at P and Q. If F is the focus of the parabola, then the area of the triangle PFQ (in sq. units) is equal to

Detailed Solution for JEE Main Mock Test - 1 - Question 68


Equation of tangent for parabola y2 = 8x:

JEE Main Mock Test - 1 - Question 69

The sides of a rhombus ABCD are parallel to the lines x - y + 2 = 0 and 7x - y + 3 = 0. If the diagonals of the rhombus intersect at P(1, 2) and the vertex A (different from the origin) is on the y-axis, then the ordinate of A is:

Detailed Solution for JEE Main Mock Test - 1 - Question 69


x - y + 2 = 0
7x - y + 3 = 0

Let two sides be:
x - y + d1 = 0
7x - y + d2 = 0

Slope of AP = m

JEE Main Mock Test - 1 - Question 70

If f(x) is a quadratic expression such that f(1) + f(2) = 0, and -1 is a root of f(x) = 0, then the other root of f(x) = 0 is:

Detailed Solution for JEE Main Mock Test - 1 - Question 70

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 71

If the function ƒ(x) = 2 + x2 – e–x and g(x) = ƒ–1(x), then the value of  equals 


Detailed Solution for JEE Main Mock Test - 1 - Question 71

g(ƒ(x)) = x ⇒ g'(ƒ(x)). ƒ'(x) = 1
ƒ(x) = 1 ⇒ x = 0
g'(1). ƒ'(0) = 1
ƒ'(x) = 2x + e–x
ƒ'(0) = 1
g'(1) = 1

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 72

Let   then the value of a + b is:


Detailed Solution for JEE Main Mock Test - 1 - Question 72

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 73

If function ƒ(x) = x3 + ax2 + bx + c is monotonically increasing ∀ x ∈ R, where a & b are prime numbers less than 10, then number of possible ordered pairs (a,b) is


Detailed Solution for JEE Main Mock Test - 1 - Question 73

a, b ∈ {2, 3, 5, 7}
ƒ '(x) = 3x2 + 2ax + b ≥ 0    ∀ x ∈ R
D ≤ 0
4a2 – 12b ≤ 0
4(a2 – 3b) ≤ 0
a = 2    b = 2,3,5,7
a = 3    b = 3,5,7
a = 5    No solution
a = 7    No solution 
Ordered pairs (a,b) are (2,2), (2,3), (2,5), (2,7), (3, 3), (3,5) & (3,7)

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 74

The value of . Find the value of m.


Detailed Solution for JEE Main Mock Test - 1 - Question 74

*Answer can only contain numeric values
JEE Main Mock Test - 1 - Question 75

Let S be the set of all triangles in the xy-plane, each having one vertex at the origin and the other two vertices lie on coordinate axes with integral coordinates. If each triangle in S has area 50 sq. units, then the number of elements in the set S, is _______.


Detailed Solution for JEE Main Mock Test - 1 - Question 75


But all cases are possible, so total number of positive cases = 9 + 9 + 9 + 9 = 36

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