JEE  >  JEE Main 2018 Question Paper with Solutions (15 April - Morning)

# JEE Main 2018 Question Paper with Solutions (15 April - Morning)

Test Description

## 90 Questions MCQ Test JEE Main & Advanced Mock Test Series | JEE Main 2018 Question Paper with Solutions (15 April - Morning)

JEE Main 2018 Question Paper with Solutions (15 April - Morning) for JEE 2022 is part of JEE Main & Advanced Mock Test Series preparation. The JEE Main 2018 Question Paper with Solutions (15 April - Morning) questions and answers have been prepared according to the JEE exam syllabus.The JEE Main 2018 Question Paper with Solutions (15 April - Morning) MCQs are made for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main 2018 Question Paper with Solutions (15 April - Morning) below.
Solutions of JEE Main 2018 Question Paper with Solutions (15 April - Morning) questions in English are available as part of our JEE Main & Advanced Mock Test Series for JEE & JEE Main 2018 Question Paper with Solutions (15 April - Morning) solutions in Hindi for JEE Main & Advanced Mock Test Series course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt JEE Main 2018 Question Paper with Solutions (15 April - Morning) | 90 questions in 180 minutes | Mock test for JEE preparation | Free important questions MCQ to study JEE Main & Advanced Mock Test Series for JEE Exam | Download free PDF with solutions
 1 Crore+ students have signed up on EduRev. Have you?
JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 1

### A body of mass m is moving in a circular orbit of radius R about a planet of mass M. At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius R/2, and the other mass, in a circular orbit of radius 3R/2. The difference between the final and initial total energies is

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 1

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 2

### In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 2

Given,
5 complete rotations of screw = 0.25cm
So 1 rotation of screw = 0.05
Hence, 1 main scale division = 0.05 cm
and 1 circular scale = 0.05/100 division = 5 × 10−4 cm.
Now Reading is 4 main scale and 30 circular scale divisions
So , thickness  = 4 × 0.05 + 30 × 5 × 10−4
= 0.2150 cm.

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 3

### The B-H curve for a ferromagnet is shown in the figure. The ferromagnet is placed inside a long solenoid with 1000 turns/ cm. The current that should be passed in the solenoid to demagnetise the ferromagnet completely is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 3

From given diagram
Coercivity = −100 (A/m)
Hence
Reverse magnetic field required to demagnetize the substance
= µo H
= µo 100
so µo 100 = µo.N.I

I = 1mA

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 4

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are λ1 and λ2, their de Broglie wavelength in the frame of reference attached to their centre of mass is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 4

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 5

Light of wavelength 550 nm falls normally on a slit of width 22.0 x 10-5cm. The angular position of the second minima from the central maximum will be (in radians):

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 5

b sin θ = nλ
b sin θ = 2λ

sin θ = 1/2
θ = π/6

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 6

A uniform rod AB is suspended from a point X, at a variable distance x from A, as shown. To make the rod horizontal, a mass in is suspended from its end A. A set of (m, x) values is recorded. The appropriate variables that give a straight line, when plotted, are:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 6

= y = mx + c

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 7

A Helmholtz coil has a pair of loops, each with N turns and radius R. They are placed coaxially at distance R and the same current I flows through the loops in the same direction. The magnitude of magnetic field at P, midway between the centres A and C, is given by [Refer to figure given below]:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 7

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 8

An ideal capacitor of capacitance 0.2 µF is charged to a potential difference of 10 V. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self-inductance 0.5 mH. The current at a time when the potential difference across the capacitor is 5 V, is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 8

Qo = 0.2 × 10 µC = 2µC

Vo = 10V
Ei of capacitor = 1/2 × 0.2 µf × (10v)2
= 10µJ
= Ef of capacitor = 1/2 × 0.2µf × (5v)2
= 2.5 µj
⇒ Einductor  = 7.5 µJ = 1/2 Li2
⇒ 7.5 × 10−6 = 1/2 × 0.5 × 10−3 × i2
⇒ 30 × 10−3 = i2
⇒ i = √3/10 = 0.17 A

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 9

Take the mean distance of the moon and the sun from the earth to be 0.4 x 106 km and 150 x 106 km respectively. Their masses are 8 x 1022 kg and 2 x 1030 kg respectively. The radius of the earth is 6400 km. Let ΔF1 be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and ΔF2 be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to ΔF1/ΔF2 is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 9

=  (375)3 (4 × 10−8)
= (0.37 × 103)3(4 × 10−8)
= 1.64

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 10

An automobile, travelling at 40 km/h, can be stopped at a distance of 40m by applying brakes. If the same automobile is travelling at 80 km/h, the minimum stopping distance, in metres, is (assume no skidding):

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 10

u = 40 Km/hr
μ = 100/9 m/s
V2 − u2 = 2a × 40
a = −1.54 m  m/s2
Now.

S = 160m

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 11

A charge Q is placed at a distance a/2 above the centre of the square surface of edge a as shown in the figure

The electric flux through the square surface is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 11

Through whole cube = Q/∈o
Through one face = Q/6∈o

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 12

A force of 40 N acts on a point B at the end of an L-shaped object, as shown in the figure. The angle θ that will produce maximum moment of the force about point A is given by

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 12

τA = f cosθ(4) + Fsinθ(2)
τA = 2F(sinθ + 2cosθ
A/dθ = 2F[cosθ − 2sinθ] = 0
Tan θ = 1/2

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 13

The equivalent capacitance between A and B in the circuit given below, is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 13

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 14

A monochromatic beam of light has a frequency v = 3/2π × 1012 Hz and is propagating along the direction  It is polarized along the k̂ direction. The acceptable form for the magnetic field is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 14

Wave propagation vector should be along

So, b is the only option which satisfies the above condition.

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 15

A particle is oscillating on the X-axis with an amplitude 2 cm about the point x0 = 10 cm, with a frequency ω. A concave mirror of focal length 5 cm is placed at the origin (see figure).

Identify the correct statements.
(A)The image executes periodic motion.
(B)The image executes non-periodic motion.
(C)The turning points of the image are asymmetric w.r.t. the image of the point at x =10 cm.
(D)The distance between the turning points of the oscillation of the image is 100/21 Cm.

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 15

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 16

A tuning fork vibrates with frequency 256 Hz and gives one beat per second with the third normal mode of vibration of an open pipe. What is the length of the pipe? (Speed of sound in air is 340 ms-1

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 16

Third normal mode of frequency in open pipe,
f = 3vs/2l
Where, Vs = 340m/s
Get L = 2m
Or
L = 200 cm

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 17

A thin uniform tube is bent into a circle of radius r in the vertical plane. Equal volumes of two immiscible liquids, whose densities are ρ1 and ρ21> ρ2), fill half the circle. The angle θ between the radius vector passing through the common interface and the vertical is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 17

h1ρ1 − 1 = h2ρ2
(R Cos θ + R sinθ)ρ2 = (R cosθ − R sinθ)ρ1
1 + ρ2) Sin θ = (ρ1 − ρ2 )Cosθ

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 18

In a common emitter configuration with suitable bias, it is given that RL is the load resistance and RBE is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by:
β is current gain, IB, IC and IE are respectively base, collector and emitter currents.

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 18

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 19

A given object takes n times more time to slide down a 45° rough inclined plane as it takes to slide down a perfectly smooth 45° incline. The coefficient of kinetic friction between the object and the incline Is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 19

So

Sin θ = n2 sinθ − µn2 cosθ
1 = n2 − µn2
1 − n2 = −µn2
n2 − 1 = −µn2
µ = 1 − 1/n2

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 20

One mole of an ideal monatomic 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 2018 Question Paper with Solutions (15 April - Morning) - Question 20

= 300R ln 2

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 21

A Carnot's engine works as a refrigerator between 250 K and 300 K. It receives 500 cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 21

Given,
Q2 = 500 cal.
ηCarnot = 1 - T2/T1
= 1/6
and (COP)HP = 1/ηCarnot = 6
(Cop)HP − (Cop)Ref = 1
So, (cop)Ref = 5 = Q2/w
so w = 100 cal
Or w = 420 J

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 22

In the given circuit all resistances are of value R ohm each. The equivalent resistance between A and B is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 22

Due to short circuit current will flow along
B-C-D-E-F-A
So
R eq = R + R
R eq = 2R

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 23

A solution containing active cobalt 60/27 Co having activity of 0.8 µCi and decay constant λ is injected in an animal's body. If 1 cm3 of blood is drawn from the animal's body after 10 hrs of injection, the activity found was 300 decays per minute. What is the volume of blood that is flowing in the body? (1 Ci = 3.7 x 1010 decays per second and at t = 10 hrs e –λt = 0.84)

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 23

(dn/dt) = −λNo
0.8 µ Ci = −λNo
N1 = No(0.84)
v → N1
v cm3 → N1/v

v = 3.7/5 × 0.84 × 0.84 × 104 cm3
= 0.5 × 104 cm3 = 5 × 103cm3 = 5 lit

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 24

In a meter bridge, as shown in the figure, it is given that resistance Y =12.5 Ω and that the balance is obtained at a distance 39.5 cm from end A (by Jockey J). After interchanging the resistances X and Y, a new balance point is found at a distance I2 from end A. What are the values of X and l2

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 24

For balanced wheat stone bridge
x(100 − l1) = y × l1
So x(100 − 39.5) = 12.5(39.5)
x = 8.16 Ω
If x and y inter changed
y(100 − l2) = xl2
12.5(100 − l2) = 8.16 l2
get l2 = 60.5 cm

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 25

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 2018 Question Paper with Solutions (15 April - Morning) - Question 25

E1 = ionization energy of ionized He
E2 = 2. 2E2
E1 = 13.6 ev
E2 = 13.6/2.2 = 6.18 ev
Total = E1 + E2 = 13.6 ev + 6.18 ev
= 20 ev

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 26

The number of amplitude modulated broadcast stations that can be accomodated in a 300 kHz band width for the highest modulating frequency 15 kHz will be:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 26

The station will require a band width of 30 khz
So
No. of stations = 300/30
= 10

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 27

The velocity-time graphs of a car and a scooter are shown in the figure. (i) The difference between the distance travelled by the car and the scooter in 15 s and (ii) the time at which the car will catch up with the scooter are, respectively.

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 27

s1 = 1/2 × 3 × (15)2 = 1.5 × 225
= 225 + 112.5
= 337.5 m
s2 = v2t
s2 = 15 × 30 = 450 m ⟹ s2 − s1 = 112.5 m
For catching up ⟹ s1 = s2
30t = 1/2 × 3 × t2
20 = t

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 28

A body of mass M and charge q is connected to a spring of spring constant k. It is oscillating along x direction about its equilibrium position, taken to be at x = 0, with an amplitude A. An electric field E is applied along the x-direction. Which of the following statements is correct?

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 28

Energy at the extreme
= 1/2 kA2 = T. E
After switching on electric field
New mean position ⟹ kxo = qE
xo = qE/k
So entreme position also shifts by qE/k

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 29

A planoconvex lens becomes an optical system of 28 cm focal length when its plane surface is silvered and illuminated from left to right as shown in Fig-A.
If the same lens is instead silvered on the curved surface and illuminated from other side as in Fig. B, it acts like an optical system of focal length 10 cm. The refractive index of the material of lens is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 29

When plane surface is silvered
Focal length f1 =
and f1 = 28 cm
When curved surface is silvered
Focal length f2 = R/2μ ... (ii)

µ = 1.55

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 30

The relative error in the determination of the surface area of a sphere is α. Then the relative error in the determination of its volume is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 30

Area of sphere (A) = 4πR2
taking log
ln A = In(4π) + 2In(R)
differentiating both sides

now, similarly

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 31

The main reduction product of the following compound with NaBH4 in methanol is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 31

Sodium borohydride will reduce ketone to alcohol. It will not reduce amide group and

C=C double bond.
Hence, option (B) is the correct answer.

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 32

A white sodium salt dissolves readily in water to give a solution which is neutral to litmus. When silver nitrate solution is added to the aforementioned solution, a white precipitate is obtained which does not dissolve in dil. nitric acid. The anion is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 32

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 33

Which of the following statements about colloids is False?

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 34

A sample of NaCIO3 is converted by heat to NaCl with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as AgCl. The mass of AgCI (in g) obtained will be: (Given: Molar mass of AgCl= 143.5 g mol -1

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 34

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 35

In which of the following reactions, an increase in the volume of the container will favour the formation of -products?

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 35

Volume increases P decreases reaction proceed in that direction where number of gaseous moles increases.

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 36

The reagent(s) required for the following conversion are:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 36

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 37

An ideal gas undergoes a cyclic process as shown in Figure.

ΔUBC = −5 Kj mol−1, qAB = 2Kj mol−1
WAB = −5Kj mol−1, WCA = 3Kj mol−1
Heat absorbed by the system during process CA is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 37

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 38

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

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 38

graphite → sp2 → % S → 33 % , % p = 67 %
diamond → Sp3 → % S → 25 %, % p = 75 %

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 39

The correct combination is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 39

(1) [Ni (Cl)4]2 → d8 (Ni)2+
Cl is weak

Field ligand → So due to unpraised is paramagnetic
(2)
Ni(CO)4 → Tetrachedral

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 40

The major product of the following reaction is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 40

The reaction undergoes acylation first followed by substitution Intramolecular.

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 41

When an electric current is passed through acidified water, 112 mL of hydrogen gas at N.T.P. was collected at the cathode in 965 seconds. The current passed, in ampere, is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 41

at NTP 1 mol  = 22. 4 l
112 ml  H2 ⇒
⇒ 1/200 mol of H2
H2O → H2 + 1/2 O2
2H + 2e→ H2
1 mol of H2 required 2 mole e̅
1/200 mol of H2 require 2/200 = 1/100 mol of e̅
1/100 mol of e̅ = 1/100 × 6.022 × 1023e̅ × 1.6 × 10−19

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 42

Which of the following is a Lewis acid?

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 42

Lewis acid → which has vacant orbital,
So B(CH3)3

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 43

The correct match between List-I and List-II is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 43

O – P ⇒ difference in Boiling point  ⇒ Steam distillation
Coloured impurity → Chromatography

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 44

Xenon hexafluoride on partial hydrolysis produces compounds 'X' and 'Y'. Compounds 'X' and 'Y' and the oxidation state of Xe are respectively:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 44

Xe F6 + H2O → XeO F4 + 2HF
XeF6 + 2H2O → XeO2F2 + 4HF

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 45

N2O5 decomposes to NO2 and O2 and follows first order kinetics. After 50 minutes, the pressure inside the vessel increases from 50 mmHg to 87.5 mmHg. The pressure of the gaseous mixture after 100 minute at constant temperature will be:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 45

2N2O5 → 4NO2 + O2
p − 2x - 4x - x
pt = p − 2x + 4x + x
pt = p + 3x

at  t = 0,   pt = p = 50 mmHg
at  t = 50 mm,   pt = 87.5 mmHg
p + 3x = 87.5
p = 87.5 − 3x
50 = 87.5 − 3x
12.5 = x

p − 2x = 50 − 2(12.5) = 25

Since K will remain same

50 = 50 × 4 − 8y
50 = 200 − 8y
8y = 150
y =  18.75
pt = p + 3y
= 50 + 3 (18.73) = 106.25 mmHg

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 46

For Na+, mg2+, F- and O2-; the correct order of increasing ionic radii is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 46

When negative charge increase, the radius of ion increases.

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 47

Which of the following will not exist in zwitter ionic form at pH =7?

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 47

The N atom of amide is not basic

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 48

The IUPAC name of the following compound is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 48

Basic Nomenclature.

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 49

The minimum volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution (Ksp of PbCl2 = 3.2 x 10-8; atomic mass of Pb= 207 u) is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 49

Ksp of PbCl2 is 3.2 × 10-8
PbCl2 is  3.2 × 10−8
Pbcl2(s) ⇌ Pb2+(aq) + 2Cl(aq)
t = 0           1              0  0.
At equilibrium  1 – S  S 2S.
Ksp = [S] [2S]2
3.2 × 10−8 = 4s3
S3 = 0.8 × 10−8
S3 = 8 × 10−9
S = 2 × 10−3
Solubility = W/V
∴ Solubility of PbCl2 in gL−1 = 2 × 10−3 × 278
= 556 × 10−3gL−1
0.556 gL−1
0.556/0.1 = 1/x
x = 0.1/0.556
= 0.18L

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 50

Ejection of the photoelectron from metal in the photoelectric effect experiment can be stopped by applying 0.5 V when the radiation of 250 nm is used. The work function of the metal is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 50

= 4.375 eV

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 51

Which of the following arrangements shows the schematic alignment of magnetic moments of antiferromagnetic substance?

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 51

Basic knowledge of Antiferromagnetic

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 52

In hydrogen azide (above) the bond orders of bonds (I) and (II) are:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 52

Hydrogen azide: HN3

Both works correct

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 53

The copolymer formed by addition polymerization of styrene and acrylonitrile in the presence of peroxide is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 53

(C6H8)n − (C3H3N)m

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 54

Which of the following will most readily give the dehydrohalogenation product?

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 54

Most probable is C6H5 Br

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 55

In the molecular orbital diagram for the molecular ion, N2+, the number of electrons in the σ2p molecular orbital is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 55

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 56

Which of the following is the correct structure of Adenosine?

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 56

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 57

Identify the pair in which the geometry of the species is T-shape and square-pyramidal, respectively:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 57

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 58

The decreasing order of bond angles in BF3, NH3, PF3 and I3- is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 58

I3- :180

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 59

For which of the following reactions, ΔH is equal to ΔU?

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 59

ΔH = ΔU + ΔngRT.
[If Δng = 0 then ΔH =ΔU]

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 60

The increasing order of nitration of the following compounds is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 60

Nitration is electrophilic aromatic substitution reaction. Methoxy and amino groups are strongly activating groups.  Methyl group is weakly activating group.
Since among methyl and methoxy group, methoxy group is more reactive than methyl group, (c) is more reactive than (d).
Even-though amino group is strongly activating group, it gets protonated in presence of acid to form anilinium ion which is strongly deactivating. Hence, (a) is less reactive than (c) and (d).
Note:
The activating groups increases the electron density on benzene ring and increases the rate of electrophilic aromatic substitution reaction. The deactivating groups decreases the electron density on benzene ring and decreases the rate of electrophilic aromatic substitution reaction.

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 61

The set of all α ∈ R, for which w =is a purely imaginary number , for all z ∈ C Satifying |z| = 1 and Re z ≠ 1, is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 61

As ω is purely imaginary

If Re(z)≠1
then, α = 0

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 62

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 63

An aeroplane flying at a constant speed, parallel to the horizontal ground, √3 km above it, is observed at an elevation of 60° from a point on the ground. If, after five seconds, its elevation from the same point, is 30°, then the speed (in km/hr) of the aeroplane, is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 63

M 5 sec = 2km
Speed = 2/5 km/sec

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 64

The value of the integral

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 64

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 65

A variable plane passes through a fixed point (3, 2, 1) and meets x, y and z axes at A, B and C respectively. A plane is drawn parallel to yz-plane through A, a second plane is drawn parallel zx-plane through B and a third plane is drawn parallel to xy-plane through C. Then the locus of the point of intersection of these three planes, is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 65

E. q. , of variable plane
a(x − 3) + b(y − 2) + c(2 − 1) = 0

Plane paroled to y2 plane passing through !

Intersection of the x three

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 66

An angle between the plane, x + y + z = 5 and the line of intersection of the planes, 3x + 4y + z – 1 = 0 and 5x + 8y + 2z + 14 = 0, is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 66

Perpendicular vector to plane
Î +̂J  + k̂
Parallel vector to line
= I(8 − 8) − Ĵ (6 − 5) + k̂ (24 − 20)
= −Ĵ + 4k̂

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 67

If n is the degree of the polynomial,  m is the coefficient of xn in it, then the ordered pair (n, m)is equal to ∶

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 67

Rationalize and get

Degree = 12
Coff = 2 [ 8C0 +  8Cr + −  8C8]54
8C0 +  8C2 +  8C4 + −  8C4 = 24−1
Coff = 2 (28−1)  54
Coff = 2 2754
= 2 × 23 × 24 × 54
= 16 (10)4​
= (20)4

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 68

If β is one of the angles between the normals to the ellipse, x2 + 3y2 = 9 at the points (3 cosθ, √3 sinθ)and(−3 sin θ√3 cosθ) ′; θϵ (0, π/2) ; then 2 cot β /sin 2θ is equal to:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 68

2x dx + 6y dy = 0
2x dx = −cy dy

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 69

In a triangle ABC, coordinates of A are (1, 2) and the equations of the medians through B and C are respectively, x + y = 5 and x = 4. Then area of ∆ABC (in sq. units) is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 69

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 70

The mean of a set of 30 observations is 75. If each observation is multiplied by a non-zero number λ and then each of them is decreased by 25, their mean remains the same. Then λ is equal to :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 70

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 71

A circle passes through the points (2, 3) and (4, 5). If its centre lies on the line, y — 4x + 3 = 0, then its radius is equal to :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 71

Let A(2, 3) & B(4, 5) be the points through which circle is passing & radius lies on the line y-4x + 3 = 0.
Let centre be (n, k), this point must satisfy the line y – 4x - + 3 = 0. Hence
k − 4n + 3 = 0
k = 4n − 3.
So, Centre coordinates be 0(n, 4n -3)
Now, OA = OB (both one radius)
A(2, 3) B(4, 5)
⇒ OA = OB
⇒ OA2 = OB2
(n − 2)2 + (4n − 3 − 3)2 = (n − 4)2 + (4n − 3 − 5)2
n2 + 4 − 4n + 16n2 + 36 − 48n = n2 + 16 − 8n + 16n2 + 64 − 64n
40 − 52n = 80 − 72n
72n − 52n = 80 − 40
20n = 40
n = 2
Hence k = 4n − 3
k = 8 − 3
k = 5
Centre coordinates O(2, 5)& !(2, 3)
radius = OA = (2 − 2)2 + (5 − 3)2 = 2

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 72

Let S be the set of all real values of k for which the system of linear equations
x+ y + z = 2
2x + y - z = 3
3x + 2y + kz = 4
has a unique solution.
Then S is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 72

Therefore,set S=equal to R-{0}

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 73

If tanA and tanB are the roots of the quadratic equation, 3x2 —10x — 25 = 0, then the value of 3 sin2(A +B) —10 sin(A +B).cos(A+B) —25 cos2(A + B)

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 73

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 74

If x2 +  y2 + sin y = 4, then the value of d2y/dx2 at the point ( -2,0) is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 74

x2 + y2 + sin y = 4
Diff above eqn with respect to x.

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 75

If x1 , x2 … . , xn and 1/h1, 1/h......1/hare two A. P. s such that x3 = h2 = 8 and x8 = h7 = 20 , then x5 . h10 equals:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 75

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 76

If are unit vectors such that  then  is equal to:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 76

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 77

If the tangents drawn to the hyperbola 4y2 = x2 + 1 intersect the co-ordinate axes at the distinct points A  and B, then the locus of the mid-point of AB is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 77

4y2 = x2 + 1
Point 4yy1 = xx1 + 1 with 4y12 = x12 + 1
x axis
y axis
Mid point h =   k =

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 78

The area (in sq. units) of the region {x ϵ R ∶ x ≥ 0, y ≥ x − 2 and y ≥ √x} is ∶

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 78

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 79

Let A be a matrix such that A ∙  isa scalar matrix and |3A| = 108. Then A2 equals:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 79

a = 2c + 3d
c = 0
2a + 3b = 0
|3A| = 108
|A| = 12
A2
c = 0
d = 2
a = 6
b = -4
A2

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 80

If f  = 2x + 1, (x ϵ R − {1, −2}), then ∫ f(x)dx is equal to:
(where C is a constant of integration)

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 80

= 2x + 1
Let t =
tx + 2t = x − 4
2t + 4 = x(1 − t)
x =

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 81

If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm2) of this cone is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 81

(h − R)2 + r2 = R2
r2 = R2 − (h − R)2
R = 3
​r2 = q − (h − 3)2

h(−3h + 12) = 0
3h = 12, h = 0
h = 4.     r = 2√2.
CSA = πrl
= π × 2√2 × √24
= π 2 × √48 = 8π√3

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 82

If b is the first term of an infinite G.P. whose sum is five, then b lies in the interval:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 82

=5
b = 5 (1 − r)
b ε (0,10)          [−1 < r < 1]

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 83

Consider the following two binary relations on the set A = {a, b, c}: R1 = {(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)} and R2 = {(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)}.Then:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 83

R1 = {(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)}
b, c ϵR1  c, a ∉ R1 R1 is not symmetric (b, c), (c, a)ϵR1(b, a) ∉ R1 , R1 is not transitive
R 2 = {(a, b), (b, a), (c, c), (c, a), (c, a), (a, a), (b, b), (a, c)}
∀(a, b)ϵR2(b, a) × R2
Therefore it is symmetric
(c, a), (a, b)ϵR2(c, b) ∉ R 2
Therefore R 2 is not transitive

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 84

Let S = {λ, µ)ϵ R × R ∶ = f(t) = (|λ|e|t| − µ).  sin (2(2|t|), t ϵ R, is a differentiable function}.  Then S is a subset of ∶

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 84

f(t) = (|λ|et − µ) sin 2t          t > 0
−(|λ|e̅t − µ) sin 2t                t < 0
f ′ (t) = 2 cos 2t (|λ|et − µ) + |λ|et sin 2t t > 0
−2 cos 2t (|λ|et − µ) + |λ|e̅t sin 2t     t > 0
f ′(t → ot) = 2(|λ| − µ)
f ′(t → o̅ t) = −2(|λ| − µ)
for differentiability LHD = RHD
2(|λ| − µ) = −2(|λ| − µ)
|λ| = µ
⟹ λ ϵ R µ ϵ R t
(λ, µ)  C    R × [0, ∞)

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 85

A box 'A' contains 2 white, 3 red and 2 black balls. Another box 'B' contains 4 white, 2 red and 3 black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are  drawn from box 'B' is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 85

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 86

If (p ∧ ~q) ∧ (p ∧ r) → ~ p ∨ q is false, then the truth values of p, q and r are, respectively:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 86

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 87

If λ ϵ R is such that the sum of the cubes of the roots of the equation, x2 + (2 - λ)x + (10 - λ) = 0 is minimum, then the magnitude of the difference of the roots of this equation is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 87

α3 + β3 = (α + β)(α2 + β2 − αβ)
= −(2 − λ)((λ + β)2 − 3αβ)
= (λ − 2)((λ − 2)2 + 3(λ − 10))
= (λ − 2)(b2 − 4λ + 4 + 3λ − 30)

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 88

If f(x) =

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 88

= cos x − tan x [x2 − 2x2]
= x2 tan x − x2 cos x
f ′(x) = 2x tan x + x2 sec2 x − 2x cos x + x2 sin x
= 2 tan x + x sec2 x − 2 cos x + x sin x
= 0 + 0 − 2 + 0 = −2

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 89

Two parabolas with a common vertex and with axes along x-axis and y-axis, respectively, 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 2018 Question Paper with Solutions (15 April - Morning) - Question 89

4y = −4x − 3
4(x + y) + 3 = 0

JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 90

n-digit numbers are formed using only three digits 2, 5 and 7. The smallest value of n for which 900 such distinct numbers can be formed, is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (15 April - Morning) - Question 90

For each place we have 3 choices
(i)for n − digits 3 × 3 … n times = 3n > 900
n = 7

## JEE Main & Advanced Mock Test Series

2 videos|324 docs|160 tests
 Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code
Information about JEE Main 2018 Question Paper with Solutions (15 April - Morning) Page
In this test you can find the Exam questions for JEE Main 2018 Question Paper with Solutions (15 April - Morning) solved & explained in the simplest way possible. Besides giving Questions and answers for JEE Main 2018 Question Paper with Solutions (15 April - Morning), EduRev gives you an ample number of Online tests for practice

## JEE Main & Advanced Mock Test Series

2 videos|324 docs|160 tests