JEE Exam  >  JEE Tests  >  JEE Main Practice Test- 14 - JEE MCQ

JEE Main Practice Test- 14 - JEE MCQ


Test Description

75 Questions MCQ Test - JEE Main Practice Test- 14

JEE Main Practice Test- 14 for JEE 2024 is part of JEE preparation. The JEE Main Practice Test- 14 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Practice Test- 14 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Practice Test- 14 below.
Solutions of JEE Main Practice Test- 14 questions in English are available as part of our course for JEE & JEE Main Practice Test- 14 solutions in Hindi for JEE course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt JEE Main Practice Test- 14 | 75 questions in 180 minutes | Mock test for JEE preparation | Free important questions MCQ to study 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- 14 - Question 1

Let f(x) be a one-to-one function such that f(1) = 3, f(3) = 1, f '(1) = – 4 and f '(3) = 2. If g = f –1, then the slope of the tangent line to 1/g at x = 1 is

Detailed Solution for JEE Main Practice Test- 14 - Question 1

JEE Main Practice Test- 14 - Question 2

The value of  is equal to

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

JEE Main Practice Test- 14 - Question 3

If g (x3 + 1) = x6 + x3 + 2, then the value of g(x2 – 1) is

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

g(x3 + 1) = x6 + x3 + 2 = (x3 + 1)2 – x3 + 1
= (x+ 1)– (x+ 1 – 1) + 1 = (x+ 1)– (x+ 1) + 2
Put x3 + 1 = t
So, g(t) = t2 – t + 2
⇒ g(x2 – 1) = (x2 – 1)2 – (x2 – 1) + 2
= x4 – 3x2 + 4. 

JEE Main Practice Test- 14 - Question 4

Suppose that f (0) = 0 and f ' (0) = 2, and let g (x) = f (- x + f (f (x))). The value of g ' (0) is equal to

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

g (x) = f (- x + f (f (x))) ;
f (0) = 0; f ' (0) = 2
g ' (x) = f ' (- x + f (f ( x )))· [- 1 + f ' (f (x))· f ' (x )]
g ' (0) = f ' (f (0))· [- 1 + f '(0) · f '(0)]
= f ' (0) [- 1 + (2)(2)]
= (2) (3) = 6 

JEE Main Practice Test- 14 - Question 5

The value of the definite integral,

Detailed Solution for JEE Main Practice Test- 14 - Question 5

JEE Main Practice Test- 14 - Question 6

A line L is perpendicular to the curve  at its point P and passes through (10, –1). The coordinates of the point P are

Detailed Solution for JEE Main Practice Test- 14 - Question 6




only (D) satisfies (1) and (2) both.

JEE Main Practice Test- 14 - Question 7


then the sum of the square of reciprocal of all the values of x where f(x) is non-differentiable, is equal to

Detailed Solution for JEE Main Practice Test- 14 - Question 7



Clearly f(x) is non differentiable at x = 1/9, 1
∴ sum of squares of reciprocals
= 92 + 1 = 82 Ans.

JEE Main Practice Test- 14 - Question 8


h(x) = {x},k(x) = 5log(x + 3) then in [0, 1], Lagranges Mean Value Theorem is NOT applicable to
[Note : where [x] and {x} denote the greatest integer and fractional part function of x respectively]

Detailed Solution for JEE Main Practice Test- 14 - Question 8

f is not differentiable at x = 1/2
g is not continuous in [0, 1] at x = 0 & 1 h is not continuous in [0, 1] at x = 1
k (x) = (x + 3)ln5 = (x + 3)p where 2 < p < 3

JEE Main Practice Test- 14 - Question 9

If the function f (x) = ax e–bx has a local maximum at the point (2, 10), then

Detailed Solution for JEE Main Practice Test- 14 - Question 9

f (2) = 10, hence 2ae–2b = 10
⇒ ae–2b = 5 ....(1)
f ' (x) = a [e–bx – bx e–bx] = 0
f ' (2) = 0
a(e–2b – 2be–2b) = 0
ae–2b (1 – 2b) = 0
⇒ b = 1/2 or a = 0 (rejected)
from (1) if b = 1/2; a = 5e
∴ a = 5e and b = 1/2 Ans.]

JEE Main Practice Test- 14 - Question 10


then the value of x satisfying the equation f (x, 10) = f (x, 11), is

Detailed Solution for JEE Main Practice Test- 14 - Question 10


JEE Main Practice Test- 14 - Question 11

Detailed Solution for JEE Main Practice Test- 14 - Question 11



JEE Main Practice Test- 14 - Question 12

Number of integral solutions of the equation

[Note : where [x] denotes the greatest integer less than or equal to x and sgn x denotes signum function of x.]

Detailed Solution for JEE Main Practice Test- 14 - Question 12


Hence two integral solution will satisfy above equation.

JEE Main Practice Test- 14 - Question 13

The area bounded by the curve y = x2 + 4x + 5 , the axes of co-ordinates & the minimum ordinate is

Detailed Solution for JEE Main Practice Test- 14 - Question 13

y = x2 + 4x + 5 = (x+2)2 + 1


JEE Main Practice Test- 14 - Question 14

The differential equation of all parabolas having their axis of symmetry coinciding with the axis of x has its order and degree respectively

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

equation (x - a)2 + y2 = (x - b)2 [S = (a, 0) ; D : x = b
y2 = (b2 - a2) + 2x (a - b)
differentiate twice to get

JEE Main Practice Test- 14 - Question 15

Number of roots of the equation x2 –2x–log2 |1 – x | = 3 is

Detailed Solution for JEE Main Practice Test- 14 - Question 15

x2 – 2x – 3 = log2 | 1 – x | 4 points 

JEE Main Practice Test- 14 - Question 16

Let F(x) be the primitive of w.r.t. x. If F(10) = 60 then the value of F(13), is

Detailed Solution for JEE Main Practice Test- 14 - Question 16




given F(10) = 60 = 2 [29 + 1] + C
⇒ C = 0


F(13) = 2 [29 × 2 + 4 × 2]
= 4 × 33 = 132

JEE Main Practice Test- 14 - Question 17


is continuous at x = 0, then
[Note : {x} denotes fractional part of x.]

Detailed Solution for JEE Main Practice Test- 14 - Question 17



∴ k = 0

Note that f (x) is discontinuous at

JEE Main Practice Test- 14 - Question 18


[Note : where C is constant of integration.]

Detailed Solution for JEE Main Practice Test- 14 - Question 18

JEE Main Practice Test- 14 - Question 19

Point 'A' lies on the curve  and has the coordinate  where x > 0. Point B has the coordinates (x, 0). If 'O' is the origin then the maximum area of the triangle AOB is

Detailed Solution for JEE Main Practice Test- 14 - Question 19


JEE Main Practice Test- 14 - Question 20


Which one of the following statement is correct?

Detailed Solution for JEE Main Practice Test- 14 - Question 20

f (x) will be continuous where 3 sin x + a2 – 10a + 30 = 4cos x

or (a – 5)2 + 5 = 4cos x – 3sin x
∴ a = 5 and 4 cos x – 3 sin x = 5

or cos (x + θ) = 1, where tan θ = 3/4

*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 21

(Instruction to attempt numerical value (integer) type question: If your answer is 100 write 100 only. Do not write 100.0)

If   then the value of must be


Detailed Solution for JEE Main Practice Test- 14 - Question 21

*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 22

The sum of all the real roots of the equations
|x − 2|2 + |x − 2| − 2 = 0 is …..


Detailed Solution for JEE Main Practice Test- 14 - Question 22

*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 23

If Pn = cosn x + sinn x, then 2.P6 - 3.P4 + 1 = ..


Detailed Solution for JEE Main Practice Test- 14 - Question 23

*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 24

If x = 2 + t3, y = 3t2 and is a constant then the value of 343n must be


Detailed Solution for JEE Main Practice Test- 14 - Question 24

*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 25

The value of 49A + 5B, where A = 1 - log72 and B = - log5 4 is


Detailed Solution for JEE Main Practice Test- 14 - Question 25

JEE Main Practice Test- 14 - Question 26

A transverse wave is travelling along a horizontal string. The first picture shows the shape of the string at an instant of time. This picture is superimposed on a coordinate system to help you make any necessary measurements. The second picture is a graph of the vertical displacement of one point along the string as a function of time. How far does this wave travel along the string in one second?

Detailed Solution for JEE Main Practice Test- 14 - Question 26

From the graphs λ = 9cm
T = 3 sec

JEE Main Practice Test- 14 - Question 27

A cyclic process of an enclosed gas of constant mass is represented by volume (V) against absolute temperature (T) as shown. If P represents pressure, the graph representing the same process can be

Detailed Solution for JEE Main Practice Test- 14 - Question 27

Co mb ina ti on of is ob ori c, is oc hor ic & isothermal.

JEE Main Practice Test- 14 - Question 28

A closed organ pipe is vibrating in its second overtone. The length of the pipe is 10cm and maximum amplitude of vibration of particles of the air in the pipe is 2mm. Then the amplitude of S.H.M. of the particles at 9cm from the open end is:

Detailed Solution for JEE Main Practice Test- 14 - Question 28

4L/5 = λ ⇒ λ = 8cm
hus 2 cm corresponds to Δϕ = z/2
1 cm corresponds to Δϕ = z/4

JEE Main Practice Test- 14 - Question 29

A sound source S and observers O1, O2 are placed as shown. S is always at rest and O1, O2 start moving with velocity v0 at t = 0. At any later instant, let f1 and f2 represent apparent frequencies of sound received by O1 and O2, respectively. The ratio f1/f2 is

Detailed Solution for JEE Main Practice Test- 14 - Question 29


JEE Main Practice Test- 14 - Question 30

Equal masses of three liquids A, B and C have temperatures 10oC, 25oC and 40oC respectively. If A and B are mixed, the mixture has a temperature of 15oC. If B and C are mixed, the mixture has a temperature of 30oC,. If A and C are mixed the mixture will have a temperature of

Detailed Solution for JEE Main Practice Test- 14 - Question 30

msA (15 – 10) = msB (25 – 15) sA = 2sB
msB (30 – 25) = msC (40 – 30)
sB = 2sC ⇒ sA = 4sC
msA (T – 10) = msC (40 – T)
⇒ 4(T – 10) = 40 – T
T = 16°C

JEE Main Practice Test- 14 - Question 31

A steel rod is 4.000 cm in diameter at 30ºC. A brass ring has an interior diameter of 3.992 cm at 30ºC. In order that the ring just slides onto the steel rod, the common temperature of the two should be nearly (αsteel = 11 × 10-6/ºC and αbrass = 19 × 10-6/ºC)

Detailed Solution for JEE Main Practice Test- 14 - Question 31

For ring just slides on to the steel rod the diameter of rod and ring should be equal to each other and suppose due to Δθ increment in temperature the diameter of both are equal then
4 (1+  αs Δθ) = 3.992 (1 + αBrass Δθ) 4 + 4 × 11 × 10-6 × Δθ
= 3.992 + 3.992 × 20 × 10–6 × Δθ
4 + 44 × 10–6 Δθ = 3.992 + 79.84 × 10–6
× Δθ
0.008 = 35.84 × 10–6 Δθ


so if temperature increased by 223°C then ring will start to slide and this temperature will equal to
θ = 30° + Δθ = 30 + 253 = 283°C
θ = 283°C ≈ 280°C

JEE Main Practice Test- 14 - Question 32

A hot liquid is kept in a big room. The logarithm of the numerical value of the temperature difference between the liquid and the room is plotted against time. The plot will be very nearly.

Detailed Solution for JEE Main Practice Test- 14 - Question 32

From N.Law of collision ℓn (T - T0) = -kt + ℓn (Ti - T0) (y = -mx + x) equation of straight line.

JEE Main Practice Test- 14 - Question 33

A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C.The temperature of the surroundings is 20°C.

Detailed Solution for JEE Main Practice Test- 14 - Question 33

By average form of Newton’s Law of cooling:


Solving (i) and (ii) we get t = 9 minute We should apply actual result.
By Newton’s law of cooling :



From (i) and (ii) we get 21/5 = 22/t
⇒ 1/5 = 2/t
⇒ t = 10 min.

JEE Main Practice Test- 14 - Question 34

A sinusoidal wave (longitudinal or transverse) is propagating through a medium in the direction of -ve x-axis. The parameters of the waves are A, ω and k. The particle at x = λ/4 executes the motion y(t) = A sinωt. Possible equation of the wave is

Detailed Solution for JEE Main Practice Test- 14 - Question 34

Let equation of wave as it is moving along - ve x-axis is y = A sin(kx + ωt +α) But, y(λ/4, t) = A sinωt Comparing then kx + α = 0 ⇒ a =-π/2

JEE Main Practice Test- 14 - Question 35

PV versus T graph of equal masses of H2, He and CO2 is shown in figure.

Choose the correct alternative

Detailed Solution for JEE Main Practice Test- 14 - Question 35

PV/T = tanθ = nR
∴ slope α no. of moles

JEE Main Practice Test- 14 - Question 36

A monoatomic gas is taken from A to C as shown in the figure. The temperature of gas at B is 27°C, then the change in internal energy of the gas is

Detailed Solution for JEE Main Practice Test- 14 - Question 36

JEE Main Practice Test- 14 - Question 37

A rope hangs from a rigid support. A pulse is set by jiggling the bottom end. We want to design a rope in which velocity v of pulse is independent of z, the distance of the pulse from fixed end of the rope. If the rope is very long the desired function for mass per unit length µ(z) in terms of µ0 (mass per unit length of the rope at the top (z = 0), g, v and z is :

Detailed Solution for JEE Main Practice Test- 14 - Question 37

∑Fz = 0 (T + dT) + µgdz - T = 0
dT = -µgdz .......... (i)
also T = µv2
dT = dµv2 + 2vdv dµ
As v is independent of z
dv = 0
dT = v2dµ .......... (ii)
from equation (1) and (2) we get

JEE Main Practice Test- 14 - Question 38

In the Pressure versus Volume graph shown, in the process of going from a to b 60 J of heat is added, and in the process of going from b to d 20 J of heat is added. In the process of going from a to c to d, what is the total heat added?

Detailed Solution for JEE Main Practice Test- 14 - Question 38

Qabd – Qacd
= (Wabd – Wacd) + (DUabd – DUacd)
= Wabd – Wacd + 0
(internal energy change is same for two paths)
= area of abdca = 15 J
Qabd = Qab + Qbd = 60 + 20 = 80 J Qacd = Qabd –15= 65 J

Alternative :
From the First law of Thermodynamics, one has
ΔUa→c→d = Qa→c→d + Wa→c→d = (60 J + 20 J) + [-(8Pa) (3m3)] ⇒ 56 J.
Since energy is a state variable, ΔUa→c→d = Qa→c→d + Wa→c→d ⇒ 56 J
= Q + [-(3Pa) (3m3)] ⇒ Qa→c→d = 65 J

JEE Main Practice Test- 14 - Question 39

A glass tube of 1.0 meter length is filled with water. The water can be drained out slowly at the bottom of the tube. If a vibrating tuning fork of frequency 500Hz is brought at the upper end of the tube and the velocity of sound is 330m/s then the total number of resonances obtained will be

Detailed Solution for JEE Main Practice Test- 14 - Question 39

JEE Main Practice Test- 14 - Question 40

A body of mass 25 kg is dragged on a rough horizontal floor for one hour with a speed of 2km/h.The coefficient of friction between the body and the surface in contact is 0.5 and half the heat produced is absorbed by the body. If specific heat of body is 0.1 cal g–1 (0C)–1, then the rise in temperature of body is

Detailed Solution for JEE Main Practice Test- 14 - Question 40

Friction force = 0.5 × 25 × 10 = 125 N distance moved = 2 × 103
∴ work done against friction = 250 × 103 J
∴ Heat given to the body = 125 × 103 J

JEE Main Practice Test- 14 - Question 41

There is a rectangular metal plate in which two cavities in shape of rectangle and circle are made, as shown in figure, with dimensions. P & Q are centres of these cavities. On heating the plate which of the following quantities increases.

Detailed Solution for JEE Main Practice Test- 14 - Question 41

All dimension will increase

JEE Main Practice Test- 14 - Question 42

A block of wood is floating in water at 0°C. The temperature of only water is slowly raised from 0°C to 10°C. With the rise in temperature of the water the volume of block above water level will be

Detailed Solution for JEE Main Practice Test- 14 - Question 42

To keep Buoyent force constant volume of submerged part must increase.

JEE Main Practice Test- 14 - Question 43

A rectangular block of lead has dimensions 4 cm × 3 cm × 20 cm. A temperature difference of 100°C can be applied to any pair of opposite faces that we choose.

Detailed Solution for JEE Main Practice Test- 14 - Question 43


JEE Main Practice Test- 14 - Question 44

Two identical solid spheres have the same constant temperature. One of the spheres is cut into two identical pieces. These two hemispheres are then separated. The intact sphere radiates an energy Q during a given time interval at temperature T0. During the same interval, the two hemisphere radiate at total energy Q' at temperature T0. Emissivity of all the surfaces is same. The ratio Q'/Q has value :

Detailed Solution for JEE Main Practice Test- 14 - Question 44

u = σeAT4 and σ1e and T are constant

JEE Main Practice Test- 14 - Question 45

A car blowing a horn of frequency 350 Hz is moving normally towards a wall with a speed of 5 m/s. The beat frequency heard by a person standing between the car and the wall is (speed of sound in air = 350 m/s)

Detailed Solution for JEE Main Practice Test- 14 - Question 45

Frequency observed by man is same as “observed” by wall and it reflects the same and as man and wall are relatively at rest, hence man observers same frequency of reflected sound. Hence no beat frequency

*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 46

(Instruction to attempt numerical value (integer) type question: If your answer is 100 write 100 only. Do not write 100.0)

When a plane mirror is placed horizontally on ground at a distance of 60 m from the base of a tower, then the top of the tower and its image in the mirror subtend an angle of 90° at the corner of the mirror nearer to the base of tower. The height of the tower (in m) is:


Detailed Solution for JEE Main Practice Test- 14 - Question 46

*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 47

A bar magnet with magnetic moment 2.5 x 103 JT-1 is rotating in horizontal plane in the space containing magnetic induction B = 4 x 10-5T. The work done in rotating the magnet slowly from a direction parallel to the field to a direction 45from the field, is (in joule, correct to two decimal places)


Detailed Solution for JEE Main Practice Test- 14 - Question 47

The work done (in rotating) by external agent
 = change in potential energy

*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 48

In figure, find the velocity of m1 in ms-1 when m2 falls by 9m.

Given m1 = m; m2 = 2m (take g = 10 ms-2)


Detailed Solution for JEE Main Practice Test- 14 - Question 48

*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 49

The figure shows three particles A, B and C on the x-axis. They are given velocities v1 = 3 m/s, v2 = 2 m/s and v3 = 5 m/s, respectively, in the direction shown. The position of centre of mass of A, B and C at time t = 1 s will be x = k (m), where k is:


Detailed Solution for JEE Main Practice Test- 14 - Question 49

*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 50

Four bricks, each of length l, are put on the top of one another as in the figure in such a way that a part of each extends beyond the one beneath. For the largest equilibrium extensions, the TOTAL overhanging length on the edge of the bottom brick is n l, where n is:


Detailed Solution for JEE Main Practice Test- 14 - Question 50

JEE Main Practice Test- 14 - Question 51

During the process of electrolytic refining of Copper which of the following may obtained as a anode mud?

Detailed Solution for JEE Main Practice Test- 14 - Question 51

Ag, Au (due to less electropositive)

JEE Main Practice Test- 14 - Question 52

How many moles of KOH are required per mole of R–NH2 in the balanced reaction?

Detailed Solution for JEE Main Practice Test- 14 - Question 52


3 mole of KOH are required in carbylamine test.

JEE Main Practice Test- 14 - Question 53

An ionic solid 'AB' (Mw = 120 g/mol) has NaCl type structure. The shortest distance between cation and anion is 500 pm. If there is a schottky defect of 2% then the density of crystal in g / cm3 will be : (NA = 6 ×1023)

Detailed Solution for JEE Main Practice Test- 14 - Question 53

Shortest distance between cation and anion = a/2 = 500 pm
a = 1000 pm

JEE Main Practice Test- 14 - Question 54

Which of the following metal is not commercially extracted by self reduction method from their corresponding ore?

Detailed Solution for JEE Main Practice Test- 14 - Question 54

Sulphide ore of Cu, Pb, Hg are reduced by self reduction.

JEE Main Practice Test- 14 - Question 55

What is the major product of the given reaction sequence?

Detailed Solution for JEE Main Practice Test- 14 - Question 55

JEE Main Practice Test- 14 - Question 56

At 27°C, the vapour pressure of an aqueous solution of urea is equal to the osmotic pressure of 5 × 10–3 M aqueous solution of glucose. If vapour pressure of pure water at 27°C is 114 torr then the mole fraction of urea in its solution is - [Given : R = 0.08 L-atm/mol-K]

Detailed Solution for JEE Main Practice Test- 14 - Question 56

Vapour pressure of aquoeus solution of urea = 5 × 10–3 × 0.08 × 300 (∵π = CRT)
= 0.12 atm
= 0.12 × 760 = 91.2 torr

JEE Main Practice Test- 14 - Question 57

Select incorrect statement regarding silver extraction -

Detailed Solution for JEE Main Practice Test- 14 - Question 57

Ag is 3000 times more soluble in Zn in compression of Pb.

JEE Main Practice Test- 14 - Question 58

What is the colour of the dye formed in the given reaction?

Detailed Solution for JEE Main Practice Test- 14 - Question 58

JEE Main Practice Test- 14 - Question 59

In a solid compound, X-particles are present in ccp position, Y-particle are in all octahedral voids and Z-particles are in alternate tetrahedral voids. If all void- particles along one of the body-diagonal are removed then the formula of compound will be given as -

Detailed Solution for JEE Main Practice Test- 14 - Question 59

Zx = 4
Zy = 4
Zz = 4

JEE Main Practice Test- 14 - Question 60

During the extraction process of Copper, Silica is added to roasted ore, in order to remove.

Detailed Solution for JEE Main Practice Test- 14 - Question 60

FeS + O2 (limited) → FeO + SO2
FeO + SiO2 (Flux) → FeSiO3 (Slag)

JEE Main Practice Test- 14 - Question 61

Find the incorrect combination of reaction and reaction name.

Detailed Solution for JEE Main Practice Test- 14 - Question 61


It is Friedel Craft's acylation not Friedel crafts alkylation.

JEE Main Practice Test- 14 - Question 62

For a 1st order reaction, the correct graph showing variation of half-life (t1/2) with inverse of Kelvintemperature (1/T) will be -

Detailed Solution for JEE Main Practice Test- 14 - Question 62


JEE Main Practice Test- 14 - Question 63

During Froath Floatation process, use of depressant is :

JEE Main Practice Test- 14 - Question 64

Which sugar is composed only of β-D-glucose units ?

Detailed Solution for JEE Main Practice Test- 14 - Question 64

Cellulose is a linear polymer of D-glucose units joined by β-glycosidic linkage.

JEE Main Practice Test- 14 - Question 65

Due to non-stoichiometric defect, a sample of cuprous oxide is found to have a composition Cu1.8O.The mole % of Cu2+ in total copper content of the crystal will be -

Detailed Solution for JEE Main Practice Test- 14 - Question 65


Let total Cu ions = 100 if Cu2+ = x
⇒ Cu+ = (100 – x)

1000 = 1800 – 9x
x = 800/9 = 88.88 %

JEE Main Practice Test- 14 - Question 66

A salt impart red colour to the borax bead in reducing flame, what could be the colour of the bead in oxidisingflame.

Detailed Solution for JEE Main Practice Test- 14 - Question 66

JEE Main Practice Test- 14 - Question 67

Which of the following is a condensation copolymer?

Detailed Solution for JEE Main Practice Test- 14 - Question 67

Bakelite is formed from a condensation reaction of phenol with formaldehyde.

JEE Main Practice Test- 14 - Question 68

The abnormal molecular mass of CH3COOH when dissolved in benzene is found to be 80 g/mol. The percentage of CH3COOH present in dimeric form in solution is -

Detailed Solution for JEE Main Practice Test- 14 - Question 68




α = 0.5
% α = 50%


JEE Main Practice Test- 14 - Question 69

Which of the following pair of cations can be separated by using excess NH3 solution?

Detailed Solution for JEE Main Practice Test- 14 - Question 69

Bi3+ + excess NH3 → Bi(OH)3 ↓ (white)
Al3+ + excess NH3 → Al(OH)3 ↓ (white)
Zn2+ + excess NH3 → Zn(NH3)3 (clear)
Hg2+ + excess NH3 → HgO.HgNH2
Pb2+ + excess NH3 → Pb(OH)2
Cu2+ + excess NH3 → Cu(NH3)42+ (clear)
Cd2+ + excess NH3 → Cd(NH3)42+ (clear)

JEE Main Practice Test- 14 - Question 70

Select the cationic detergent amongst the following.

Detailed Solution for JEE Main Practice Test- 14 - Question 70


Cetyltrimethylammoniumbromideiscationic detergent.

*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 71

(Instruction to attempt numerical value (integer) type question: If your answer is 100 write 100 only. Do not write 100.0)

Number of pπ –pπ double bond in given compound CN–CH = CH–CN is _____ .


Detailed Solution for JEE Main Practice Test- 14 - Question 71

4π between C & N and one π between C – C.

*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 72

A compound H2X with molar weight of 80 g is dissolved in a solvent having density of 0.4 g ml–1. Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is


Detailed Solution for JEE Main Practice Test- 14 - Question 72

*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 73

The number of correct options is
a. P2O5 > ZnO > MgO > Na2O2 (acidic strength)
b. Tl2O3 > Tl2O > Ga2O3 > Al2O3 (basic strength)
c. MnO > P2O5 > CrO3 > Mn2O7 (ionic character)
d. H2O > HF > NH3 (melting point)
e. H2O > HF > NH3 (boiling point)


Detailed Solution for JEE Main Practice Test- 14 - Question 73

a. Correct
Acidic character of oxides increases along the period(→). According to Fajan’s rule, higher the oxidation state of the atom, more is the covalent character, and more is the acidic character. The smaller the cation, the more covalent and more acidic the solution.
b. Wrong
TlO2 > Tl2O3 > Ga2O3. According to Fajan’s rule, small charge, large cation and small anion, the more ionic character and thus the more basic is the compound.
c. Correct
The lesser the charge, the more is the ionic character.
d. H2O > NH3 > HF. It is due to more number of hydrogen bonds.
e. Correct
H2O(373) > HF(293) > NH3(238.5)K

*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 74

A single electron orbits around a stationary nucleus of charge +Ze, where Z is a constant and e is the magnitude of electronic charge, It requires 47.2 eV to excite the electron from the second Bohr orbit to the third Bohr orbit. Find the ionisation energy of above electron system in electronvolt.


Detailed Solution for JEE Main Practice Test- 14 - Question 74

*Answer can only contain numeric values
JEE Main Practice Test- 14 - Question 75

The number of sp3 and sp2 hybridisation of underlined atoms in following molecule is x and y respectively then xyxy is
AICl4, Cl2ONH2CH3BF4
PCl4+ClO2, H3O+, H2CO, B2H6


Detailed Solution for JEE Main Practice Test- 14 - Question 75

Sp3: AICl4, Cl2ONH2BF4– = 4
Sp2: H2CO = 1
x/y = 4/1 = 4

Information about JEE Main Practice Test- 14 Page
In this test you can find the Exam questions for JEE Main Practice Test- 14 solved & explained in the simplest way possible. Besides giving Questions and answers for JEE Main Practice Test- 14, EduRev gives you an ample number of Online tests for practice
Download as PDF
Download the FREE EduRev App
Track your progress, build streaks, highlight & save important lessons and more!