JEE Exam  >  JEE Tests  >  Mock Tests for JEE Main and Advanced 2024  >  JEE Main Practice Test- 2 - JEE MCQ

JEE Main Practice Test- 2 - JEE MCQ


Test Description

75 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Main Practice Test- 2

JEE Main Practice Test- 2 for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2024 preparation. The JEE Main Practice Test- 2 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Practice Test- 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Practice Test- 2 below.
Solutions of JEE Main Practice Test- 2 questions in English are available as part of our Mock Tests for JEE Main and Advanced 2024 for JEE & JEE Main Practice Test- 2 solutions in Hindi for Mock Tests for JEE Main and Advanced 2024 course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt JEE Main Practice Test- 2 | 75 questions in 180 minutes | Mock test for JEE preparation | Free important questions MCQ to study Mock Tests for JEE Main and Advanced 2024 for JEE Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you? Download the App
JEE Main Practice Test- 2 - Question 1

The volumes of containers A and B, connected by a tube and a closed valve are V and 4 V, respectively. Both the containers A and B have the same ideal gas at pressures (temperatures) 5.0 ×105 Pa(300 K) and 1.0 ×105Pa (400 K), respectively. The valve is opened to allow the pressure to equalise, but the temperature of each container is kept constant at its initial value. Find the common pressure in the containers.

Detailed Solution for JEE Main Practice Test- 2 - Question 1

∴ p1 > p2 so, the equalise pressure of the p is reduced from p1

Given, T1 = 300

T2 = 400

V2 = 4V

p1 = 5 x 105

JEE Main Practice Test- 2 - Question 2

A block of mass 10 kg is suspended through two light spring balances as  shown in figure

Detailed Solution for JEE Main Practice Test- 2 - Question 2

The FED of the spring balances and the block are as shown in figure.

T1=10g T1=T2 ⇒ T2=10g where, T1 and Tare readings of spring balances as shown in figure.

JEE Main Practice Test- 2 - Question 3

If E = 100 sin (100t) volt and are the instantaneous values of voltage and current, then the R.M.S values of voltage and current are respectively.

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

The instantanous valur of voltae is 

E= 100sin (100 t) V ...(1)

 (compare it with

E = Esin (ωt) V

we get,

E= 100 V, ω = 100 rads-1

the r.m.s value of voltage is 

The instantaneous value of current is 

compare it with, I=I0sin(ω t + Φ)

we get

I0 = 100 mA, ω = 100 rads-1

The r.m.s value of current is

JEE Main Practice Test- 2 - Question 4

Two spherical bodies of mass M and 5 M and radii R and 2 R, respectively are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is

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

Initial separation between both centers = 12R
Final separation (when collision occurs) = 3R
Thus both the centers need to travel 9R combinely in such a way that their COM does not move as no external force is acting over the 2 mass system.
Thus let say if smaller mass travel x distance, bigger would eventually travel rest 9R - x
Thus we we write the equation of displacement of the COM, and taking direction of displacement of bigger mass as positive, we get
M (-x) + 5M (9R - x) / (M + 5M) = 0
Thus we get, -Mx + 45MR - 5Mx = 0
We get 6x = 45R
Thus x = 7.5R

JEE Main Practice Test- 2 - Question 5

Mark the correct option.

Detailed Solution for JEE Main Practice Test- 2 - Question 5

A charge moving along a circle is equivalent to a current carrying coil, but with respect to magnetic field on the axis of circle. It is equivalent only for the average value of the magnetic field and not for instantaneous values. While if two charge particles are moving symmetrically along a circle at diametrically opposite points then the average as well as instantaneous magnetic field on its axis is same as due to a current carrying coil.

JEE Main Practice Test- 2 - Question 6

The length of a given cylindrical wire is increased by 100%. Due to the consequent decrease in diameter the change in the resistance of the wire will be

Detailed Solution for JEE Main Practice Test- 2 - Question 6

Initial volume = final volume

JEE Main Practice Test- 2 - Question 7

Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum. They use different lengths of the pendulum and /or record time for different number of oscillations. The observations are shown in the table.

Least count for length = 0.1 cm

Least count for time = 0.1 s

If EI, EII and EIII are the percentage errors in g, i.e.,

for students I, II and III, respectively, then

Detailed Solution for JEE Main Practice Test- 2 - Question 7

The period of oscillation (T) of a simple pendulum of length ℓ is given by

T = 2π√(ℓ/g)

Therefore, g = 4π2 ℓ/T2 so that the fractional error in g is given by

∆g/g = (∆ℓ/ℓ) + 2(∆T/T)

[The above expression is obtained by taking logarithm of both sides and then differentiating. Note that the sign of the second term on the RHS is changed from negative to positive since we have to consider the maximum possible error].

Here ∆ℓ = 0.1 cm and ∆T = 0.1 s

The percentage error is 100 times the fractional error so that

EI = ∆g/g = [(0.1/64) + 2(0.1/128)]×100 = 5/16 %,

EII = ∆g/g = [(0.1/64) + 2(0.1/64)]×100 = 15/32 % and

EIII = ∆g/g = [(0.1/20) + 2(0.1/36)]×100 = 19/18 %

Thus EI is minimum. 

JEE Main Practice Test- 2 - Question 8

Write down the expression for capacitance of a spherical capacitor whose conductors radii are Rand R2(R2>R1),when inner sphere is grounded.

Detailed Solution for JEE Main Practice Test- 2 - Question 8

JEE Main Practice Test- 2 - Question 9

A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45° with the initial vertical direction is

Detailed Solution for JEE Main Practice Test- 2 - Question 9

Here, position 1 and position 2 can be calculated by using 

work- energy Theorem


JEE Main Practice Test- 2 - Question 10

Acceleration of each block is given as g/5√2. Find the magnitude and direction of force exerted by string on pulley. (μ = 0.4)

Detailed Solution for JEE Main Practice Test- 2 - Question 10

Let coefficient of friction be u. and and 3m block is moving down the incline, then Acceleration  

 

Force exerted by string/on pulley is √2T as shown in figure

∴ F = 6mg/5

JEE Main Practice Test- 2 - Question 11

A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.

Detailed Solution for JEE Main Practice Test- 2 - Question 11

for closed pipes,

so in this case 6 possibilities

JEE Main Practice Test- 2 - Question 12

Three identical spheres, each of mass 1 kg are kept as shown in the figure below, touching each other, with their centres on a straight line. If their centres are marked P, Q, R respectively, the distance of centre of mass of the system from P is

Detailed Solution for JEE Main Practice Test- 2 - Question 12

Radius

r1= 0

r2 = PQ

r3 = PR

the distance of centre of mass of the system from 'P' is given by

JEE Main Practice Test- 2 - Question 13

A resistor R and 2μF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed (log10 2.5 = 0.4)

Detailed Solution for JEE Main Practice Test- 2 - Question 13

Wen Neon bulb is fulled with gas.

No, current frows through it because the resistance is infine.

JEE Main Practice Test- 2 - Question 14

A α−particle passes through a potential difference of 2×106V and then it becomes incident on a silver foil. The charge number of silver is 47. The energy of incident particles will be: (in joule)

Detailed Solution for JEE Main Practice Test- 2 - Question 14

Energy of incident particles ie, alpha particles will be 

JEE Main Practice Test- 2 - Question 15

Direction: Question is based on the following paragraph.

A wire of length L, mass m and carrying a current is suspended from point O as shown. An infinitely long wire carrying the same current I is at a distance L below the lower end of the wire. Given, I = 2A, L= 1m and m = 0.1 kg (ln 2 = 0.693)

 

What is angular acceleration of the wire just after it is released from the position shown?

Detailed Solution for JEE Main Practice Test- 2 - Question 15

 

JEE Main Practice Test- 2 - Question 16

The figure shows three circuits with identical batteries, inductors and resistances. Rank the circuits according to the currents through the battery just after the switch is closed, greatest first

Detailed Solution for JEE Main Practice Test- 2 - Question 16

In circuit ( 1 ), on closing the switch, the current in the inductor is zero due to self induction, ie, i1=0.

In circuit (2), on closing the switch the current in the inductor is zero due to self-induction.
image


Therefore,

i= i′ = E / R1
In circuit (3), on closing the switch, the current in the inductor is again zero due to the same reason.
image


Therefore,

i= i′ = E / (R1 + R2) 

Thus, it is obvious that,

i2>i3>i1(=0)

JEE Main Practice Test- 2 - Question 17

A bullet when fired into a fixed target loses half of its velocity after penetrating 20 cm.How much further it will penetrate before coming to rest?

Detailed Solution for JEE Main Practice Test- 2 - Question 17

 

JEE Main Practice Test- 2 - Question 18

Directions: Question are Assertion - Reaction type each of these contains two statements: Statement I (Assertion), Statement II (Reason) Each of these questions also has four alternative choices, only one of which is correct. You have to select the correct choices from the codes a, b, c and d given below:

Statement I : The isothermal curves intersect each other at a certain point

Statement II: The isothermal changes takes place slowly, so the isothermal curves have very little slope.

Detailed Solution for JEE Main Practice Test- 2 - Question 18

To carry out isothermal process, a perfect gas is compressed or allowed to expand very slowly. Isothermal curves never intersect each other as they have very little slope.

JEE Main Practice Test- 2 - Question 19

A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be

Detailed Solution for JEE Main Practice Test- 2 - Question 19

∴ increase in capiacitance

JEE Main Practice Test- 2 - Question 20

A ball is dropped from a bridge at a height of 176.4 m over a river. After 2s, a second ball is thrown straight downwards. What should be the initial velocity of the second ball so that both hit the water simultaneously?  (Take g=10m/s2)

Detailed Solution for JEE Main Practice Test- 2 - Question 20

for second ball

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 21

Directions: The answer to this question is a single-digit integer, ranging from 0 to 9. Enter the correct digit in the box given below. 

Q. A 40 cm diameter pipe branches into two pipes of diameters 10 cm and 20 cm each. If the average velocities of water that flows through 10 cm and 20 cm pipes are 6 m/s and 2 m/s, respectively, then the speed of the water that flows through the 40 cm pipe (in m/s) is 

(Answer up to nearest integer)


Detailed Solution for JEE Main Practice Test- 2 - Question 21

Let x be the velocity of water flowing through 40 cm diameter branch
V(40): Volume of water flowing through 40cm diameter
V(20): Volume of water flowing through 20cm diameter
V(10): Volume of water flowing through 10cm diameter
V(40) = V(20) + V(10) - - - (1)
Volume of water flowing = (Cross-sectional area) * (Velocity of water)
(1) → x*π (0.20)2 = 2*π (0.10)2 + 6*π (0.05)2
x*(0.04) = 2*(0.01) + 6*(0.0025)
x = (0.035) / (0.04) = 35/40 = 0.875
Nearest integral value is 1

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 22

Directions: The answer to this question is a single-digit integer, ranging from 0 to 9. Enter the correct digit in the box given below. 

Q. Two point objects A and B are 40 cm apart. A convex lens L of focal length 15 cm should be placed in between them such that the images due to the two point objects coincide. If the least distance between L and A is x, then the value of x – 1 is


Detailed Solution for JEE Main Practice Test- 2 - Question 22

Let VA be the image distance due to the object A. X is the distance between the object A and the lens.

image due to the second object:

As one of the images should be virtual in order to satisfy the given condition.


the lesser value is 10. Hence, the value of x-1 = 9

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 23

Directions: The answer to this question is a single-digit integer, ranging from 0 to 9. Enter the correct digit in the box given below. 

Q. An air-filled parallel plate capacitor having circular plates has a capacitance of 10 pF. When the radii of the plates are increased two times, the distance between them is halved and if a medium of dielectric constant k is introduced, the capacitance increases 16 times. The value of k is


Detailed Solution for JEE Main Practice Test- 2 - Question 23

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 24

Directions: The answer to this question is a single-digit integer, ranging from 0 to 9. Enter the correct digit in the box given below. 

Q. Two parallel identical plates carry equal and opposite charges having a uniform charge of 88.9 C. Positive plate is fixed on the ceiling of a box and the negative plate has to be suspended. If the area of the plates is 6.35 sq. m and 'm' is the mass of the negative plate, then the value of [m] in kg, where [ ] stands for maximum integer value, is


Detailed Solution for JEE Main Practice Test- 2 - Question 24

force of attraction between the plates = weight of the negative plate for it to be suspended

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 25

Directions: The answer to the following question is a single digit integer ranging from 0 to 9. Enter the correct digit in the box given below.

A radioactive sample S1 having an activity of 5 μ Ci and half life of 20 years has twice the number of nuclei as another sample S2, which has an activity of 10 μCi. The half lives (in years) of S2 is


Detailed Solution for JEE Main Practice Test- 2 - Question 25

JEE Main Practice Test- 2 - Question 26

Directions: Questions are based on the following paragraph.

When ammonium vanadate is heated with oxalic add solution, a compound Z is formed. A sample of Z was titrated with KMnO4 solution in hot acidic solution. The resulting liquid was reduced with SO2, the excess SOboiled off, and the liquid again titrated with KMnO4. The ratio of the volumes of KMnO4 used in the two titrations was 5 : 1. KMnO4 oxidises all oxidation state of vanadium to Vanadium (+V) and SO2 reduces vanadium (+V) to vanadium (+IV). Read the above experiment and answer the following questions. If vanadium exists as , reduced species by SO2 would be

Detailed Solution for JEE Main Practice Test- 2 - Question 26

 is reduced to +4 oxidation state which is 

JEE Main Practice Test- 2 - Question 27

In a cubic dosed packed structure of mixed oxides, the lattice is made up of oxide ions, one eighth of tetrahedral/voids are occupied by divalent ions (A2+), while one half of the octahedral voids are occupied by trivalent ions(B3+)What is the formula of the oxide ?

Detailed Solution for JEE Main Practice Test- 2 - Question 27

Let number of oxides = x

Number of octahedral void = x

Number of tetrahedral void = 2x

Number of 

 

 Hence, formula of oxide is AB2O4.

JEE Main Practice Test- 2 - Question 28

Which of the following sequence of reaction is the best means to furnish the conversion RCH2OH→RCH2NH2

Detailed Solution for JEE Main Practice Test- 2 - Question 28

JEE Main Practice Test- 2 - Question 29

The standard heat of combustion of carbon(s), sulphur (s) and carbon disulphide (l) are -393.3, -293.72 and - 1108.76 kJ/mol respectively. The standard heat of formation of carbon disulphide(l) is

Detailed Solution for JEE Main Practice Test- 2 - Question 29

on putting various enthalpy of formation in equation III 

(reactants) - 1108.76 = [-393.3 + 2(-293.72)] - 

=-1108.76 = -393.3 -2 x 293.72 - 

JEE Main Practice Test- 2 - Question 30

The volume of 0.1 M oxalic acid that can be completely oxidised by 20 mL of 0.025 M KMnO4 solution is

Detailed Solution for JEE Main Practice Test- 2 - Question 30

JEE Main Practice Test- 2 - Question 31

α-D(+)-glucose and β-D-(+)-glucose are

Detailed Solution for JEE Main Practice Test- 2 - Question 31

Anomers are diastereo isomers of cyclic forms of sugars or similar molecules differing in the configuration at the anomeric carbon (C-1 atom of aldose or the  C-2 atom of a 2-ketose).The cyclic forms of carbohydrates can exist in two forms , α – and β- based on the position of the substituent at the anomeric centre.

JEE Main Practice Test- 2 - Question 32

Match List I with List II and select the correct answer using the codes given below the lists:

Detailed Solution for JEE Main Practice Test- 2 - Question 32

Cyanide process is used for the extraction of Au, floatation process uses pine oil as a foaming agent, electrolytic reduction is used in the extraction of Al and zone refining process produces ultra pure Ge.

JEE Main Practice Test- 2 - Question 33

Which one of the following is the structure of polyacrylonitrile?

Detailed Solution for JEE Main Practice Test- 2 - Question 33

Polyacrylonitrile has molecular formula (C3H3N)n.

Hence its structure is

JEE Main Practice Test- 2 - Question 34

In the hydrocarbon 

The state of hybrization of carbons 1,3 and 5 are in the following sequence :

Detailed Solution for JEE Main Practice Test- 2 - Question 34

Among the four (I) and (II) have chirality . so, here, optical isomers are obtained.

JEE Main Practice Test- 2 - Question 35

If for  then the value of keq for the reaction  will be

Detailed Solution for JEE Main Practice Test- 2 - Question 35

for reaction

JEE Main Practice Test- 2 - Question 36

For the elementary reaction M → N, the rate of disappearance of M increases by factor of 8 upon doubling the concentration of M. The order of the reaction with respect to M is

Detailed Solution for JEE Main Practice Test- 2 - Question 36

On doubling cone of [M], Rate becomes 8 times

JEE Main Practice Test- 2 - Question 37

Calgon used as a water softener is

Detailed Solution for JEE Main Practice Test- 2 - Question 37

JEE Main Practice Test- 2 - Question 38

In which of the following case, increase in concentration of ion cause increase in Ecell?

Detailed Solution for JEE Main Practice Test- 2 - Question 38

JEE Main Practice Test- 2 - Question 39

Directions: Questions  are based on the following paragraph.

When ammonium vanadate is heated with oxalic add solution, a compound Z is formed. A sample of Z was titrated with KMnO4 solution in hot acidic solution. The resulting liquid was reduced with SO2, the excess SOboiled off, and the liquid again titrated with KMnO4. The ratio of the volumes of KMnOused in the two titrations was 5 : 1. KMnO4 oxidises all oxidation state of vanadium to Vanadium (+V) and SO2 reduces vanadium (+V) to vanadium (+IV). Read the above experiment and answer the following questions.

Q. What is the oxidation state of vanadium in the compound Z?

Detailed Solution for JEE Main Practice Test- 2 - Question 39

vanadate ion   is reduce to Vx (species Z) by  

ion in acidic medium. Vis oxidised by 

∴  to which in turn isoxidised to 

volumes of   used in Eqs (1) , (2) are in ratio , of 

  ∴ 

JEE Main Practice Test- 2 - Question 40

Direction: Question is assertion reason type. These question contains two statements Statement I (Assertion), Statement II (Reason). These question also has four alternative choices, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given below:

Statement I : Detection of chlorine in 2, 4, 6 -trinitrochlorobenzene can be done directly by addition of aq. AgNO3 solution.

Statement II: C-Cl bond is weakened by electron withdrawing - NO2 group

Detailed Solution for JEE Main Practice Test- 2 - Question 40

2,4, 6 - trinitrocholorobenzene.The presence of electron withdrawing group (like -NO2) makes the nucleophilic aromatic substitution easier, as it decreases the strength of C-Cl bond. Thus,gives ppt. of AgCl with aq. AgNo3

JEE Main Practice Test- 2 - Question 41

Silver (atomic weight = 108 g mol-1 ) has a density of 10.5 g cm-3. The number of silver atoms on a surface of area 10-12 mcan be expressed in scientific notation as y * 10x . The value of x is

Detailed Solution for JEE Main Practice Test- 2 - Question 41

since the number of atom of silver 

JEE Main Practice Test- 2 - Question 42

Names of some compounds are given. Which one is not correct in IUPAC system?

Detailed Solution for JEE Main Practice Test- 2 - Question 42

Correct IUPAC name is 4-ethyl-3-methyl heptane.

JEE Main Practice Test- 2 - Question 43

The structure of isobutyl group in an organic compound is :

Detailed Solution for JEE Main Practice Test- 2 - Question 43

Structure of 1 butene-3yne we know that double bond contains one sigma and one pi-bond while a triple bond contains one sigma and two pi-bonds. So, total number of sigma and pi-bonds.

= (5+1+1) σ + (0+1+2) π i.e 7σ, 3π

1-butene-3yne contains 7 sigma and 3 Pi-bonds

JEE Main Practice Test- 2 - Question 44

consider the follwing reaction,

the value of x, y and z in the reaction respectively are

Detailed Solution for JEE Main Practice Test- 2 - Question 44

Given

after balancing the equation, we get

∴ The value of x, y and z in the given equation are 2,5 and 16 respectively

JEE Main Practice Test- 2 - Question 45

The correct statement about the following disaccharide is

Detailed Solution for JEE Main Practice Test- 2 - Question 45

In the given structures A and B ring A is 6 membered ring containg one oxygen with α – glycosidic linkage, hence it is pyranose where as ring B is 5- membered ring containing one oxygen is furanos with β – glycosidic linkage.

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 46

Directions: The answer to the following question is a single digit integer ranging from 0 to 9. Enter the correct digit in the box given below.

Q. X x 10-2 moles of NaCl need to be added to precipitate out PbCl2 from 1 L solution of Pb(NO3)2 having concentration 0.01M. If Ksp of PbCl2 at 298K is 1.6 x 10-5, then the value of X will be


Detailed Solution for JEE Main Practice Test- 2 - Question 46

Hence [NaCl] required 4 x 10-2

Nimber of moles of NaCl to be added = 4 x 10-2

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 47

Directions: The answer to the following question is a single digit integer ranging from 0 to 9. Enter the correct digit in the box given below.

Q. For a diatomic molecule having bond distance , the dipole moment is 1.2 D. What is the integral value obtained as a product of the fraction of charge separated and 16?


Detailed Solution for JEE Main Practice Test- 2 - Question 47

fraction of electric charge  on ach atom 

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 48

Directions: The answer to the following question is a single digit integer ranging from 0 to 9. Enter the correct digit in the box given below.

Q. On complete combustion of a 9 L mixture of ethane and propane, 21 L of CO2 at STP is produced. The molar ratio of ethane and propane in the mixture is


Detailed Solution for JEE Main Practice Test- 2 - Question 48

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 49

Directions: The answer to the following question is a single digit integer ranging from 0 to 9. Enter the correct digit in the box given below.

Q. How many open chain structures are possible for N-Methyl butanamine (including it) that are referred to as metamers.


Detailed Solution for JEE Main Practice Test- 2 - Question 49

Molecular formula of N-methyl butanamine is C5H13N. Primary, secondary and tertiary amines are different functional groups and metamerism cannot coexist with functional isomerism. Hence, we shall consider only those isomers of N-methyl butanamine which are secondary (2o) amines.

Every molecule of group A is a meramer of every molecule in group B

 

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 50

Directions: The answer to the following question is a single digit integer ranging from 0 to 9. Enter the correct digit in the box given below.

Q. 100gm of  is reduced to 3.125 g in 25 days. Half life of bismuth -210 (in days) is


Detailed Solution for JEE Main Practice Test- 2 - Question 50

JEE Main Practice Test- 2 - Question 51

A bag contains a white and b black balls. Two players A and B alternately draw a ball from the bag replacing the ball each time after the draw till one of them draws a white ball and wins the game. A begins the game. If the probability of A winning the game is three times that of B, then the ratio a : b is

Detailed Solution for JEE Main Practice Test- 2 - Question 51

w → drawing white ball at any draw and B that  for a black ball

then

Also P(B wins the game)

JEE Main Practice Test- 2 - Question 52

If Z is the set of integers. Then, the relation R= {(a, b) :1+ ab > 0} on Z is

Detailed Solution for JEE Main Practice Test- 2 - Question 52

so it is reflexive

so, it is symmetric 

So, it is not transitive. 

JEE Main Practice Test- 2 - Question 53

Direction: Question  is Assertion-Reason type question. These question contains two statements: Statement I (Assertion) and Statement II (Reason). These question also has four alternative choices, only one of which is the correct answer. You have to select the correct choice in the cedes (a), (b), (c) and (d) in the given below: 

Statement I: 

Staement II: |x| is non- differential at x = 0

Detailed Solution for JEE Main Practice Test- 2 - Question 53

Given

 found, since condition on x is not given

 also |x| is nondifferential at x = 0

JEE Main Practice Test- 2 - Question 54

For any two real numbers θ and ϕ, we define θRϕ , if and only if sec2 θ - tan2ϕ =1. The relation R is

Detailed Solution for JEE Main Practice Test- 2 - Question 54

JEE Main Practice Test- 2 - Question 55

Direction: Question is Assertion-Reason type question. These question contains two statements: Statement I (Assertion) and Statement II (Reason). These question also has four alternative choices, only one of which is the correct answer. You have to select the correct choice in the cedes (a), (b), (c) and (d) in the given below: 

Detailed Solution for JEE Main Practice Test- 2 - Question 55

 

Now statement II is valid only where  

∴ Statement I is true and statement II is false

JEE Main Practice Test- 2 - Question 56

The value of  

Detailed Solution for JEE Main Practice Test- 2 - Question 56

JEE Main Practice Test- 2 - Question 57

the value of 

Detailed Solution for JEE Main Practice Test- 2 - Question 57

JEE Main Practice Test- 2 - Question 58

if n≥ 2 is an integer  and I is the identity matrix of order 3. then

 

Detailed Solution for JEE Main Practice Test- 2 - Question 58

Given

similarly

 

JEE Main Practice Test- 2 - Question 59

 ax is equal to 

Detailed Solution for JEE Main Practice Test- 2 - Question 59

Put

put

JEE Main Practice Test- 2 - Question 60

The number of real solutions of the equation  2x/2+(√2+1)x=(5+2√2)x/2 is

Detailed Solution for JEE Main Practice Test- 2 - Question 60

Thus no. of real solution is 1.

JEE Main Practice Test- 2 - Question 61

The function 

Detailed Solution for JEE Main Practice Test- 2 - Question 61

Using Leibnitz's rule

on differentiating w.r.t θ we get

 

JEE Main Practice Test- 2 - Question 62

If m1, m2, m3 and m4 are, respectively the magnitudes of the vectors

then the corect order of m1, m2, m3 and m4 is 

Detailed Solution for JEE Main Practice Test- 2 - Question 62

JEE Main Practice Test- 2 - Question 63

 is equal to 

Detailed Solution for JEE Main Practice Test- 2 - Question 63

JEE Main Practice Test- 2 - Question 64

The area bounded by y = x|sin x| and X - axis between x = 0, x = 2π is

Detailed Solution for JEE Main Practice Test- 2 - Question 64

JEE Main Practice Test- 2 - Question 65

the eccenricity of the hyperbola  is

Detailed Solution for JEE Main Practice Test- 2 - Question 65

JEE Main Practice Test- 2 - Question 66

If θ is the angle between the tangents from (-1, 0) to the circle x2 + y2 - 5x + 4y - 2 = 0, then θ is equal to

Detailed Solution for JEE Main Practice Test- 2 - Question 66

we have, angle between the two tangents from (x1,y1)

JEE Main Practice Test- 2 - Question 67

Let Sk be the sum of an infinite GP series whose first term k is k and common ratio is k/(k + 1) (k> 0). Then, the value of  is equal to

Detailed Solution for JEE Main Practice Test- 2 - Question 67

JEE Main Practice Test- 2 - Question 68

Let z1 ≠ z2 and |z1| = |z2|. If z1 has positive real part and z2 has negative imaginary part. Then, 

Detailed Solution for JEE Main Practice Test- 2 - Question 68

Now,


= a purely imaginary or 0 if (x1/x2 )= (y1/y2)

if (x1/x2 )= (y1/y2) then 

x1 + iy1 = k(x2+iy2)

 is purely imaginary.

JEE Main Practice Test- 2 - Question 69

, Where [] denotes the greater function is equal to

Detailed Solution for JEE Main Practice Test- 2 - Question 69

JEE Main Practice Test- 2 - Question 70

Three vectors   forms

Detailed Solution for JEE Main Practice Test- 2 - Question 70

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 71

Directions: The answer to the following question is a single digit integer ranging from 0 to 9. Enter the correct digit in the box given below.

Q. if , then the number of solutions of 


Detailed Solution for JEE Main Practice Test- 2 - Question 71

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 72

Directions: The answer to the following question is a single digit integer ranging from 0 to 9. Enter the correct digit in the box given below.

Q. The points (1, 3) and (5, 1) are two opposites vertices of a rectangle and the other two vertices lie on the line y – 2x + C = 0. Then the value of C is


Detailed Solution for JEE Main Practice Test- 2 - Question 72

Since the diagonals of a rectangle bisect each other, so the point

lies on y = 2x - C, which gives C = 4

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 73

Directions: The answer to the following question is a single digit integer ranging from 0 to 9. Enter the correct digit in the box given below.

Given that α and γ are the roots of the equation Ax2 – 4x + 1 = 0, and β and δ are the roots of the equation Bx2 – 6x + 1 = 0. Find the value of B - A, such that α, β, γ, δ are in H.P.


Detailed Solution for JEE Main Practice Test- 2 - Question 73

As per the given condition

Let d be the common difference

Adding both of the equations, we get

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 74

Directions: The answer to the following question is a single digit integer ranging from 0 to 9. Enter the correct digit in the box given below.

if  the greatest value of  is


Detailed Solution for JEE Main Practice Test- 2 - Question 74

*Answer can only contain numeric values
JEE Main Practice Test- 2 - Question 75

Directions: The answer to the following question is a single digit integer ranging from 0 to 9. Enter the correct digit in the box given below.

if range of the function f(x) = sin–1 x + 2 tan–1 x + x2 + 4x + 1 is [a, b], then the value of a + b is


Detailed Solution for JEE Main Practice Test- 2 - Question 75

therefore, f(x) is an increasing function. Hence, a is minimum value of f(x). therefore

And, b is maximum value of  f(x) . Therefore 

Therefore, the range of f(x) is . Therefore

Hence, it is required solution.

357 docs|148 tests
Information about JEE Main Practice Test- 2 Page
In this test you can find the Exam questions for JEE Main Practice Test- 2 solved & explained in the simplest way possible. Besides giving Questions and answers for JEE Main Practice Test- 2, EduRev gives you an ample number of Online tests for practice

Up next

Download as PDF

Up next

Download the FREE EduRev App
Track your progress, build streaks, highlight & save important lessons and more!