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


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90 Questions MCQ Test JEE Main & Advanced Mock Test Series | JEE Main 2018 Question Paper with Solutions (16th April-Morning)

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JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 1

The relative uncertainty in the period of a satellite orbiting around the earth is 10–2. If the relative uncertainty in the radius of the orbit is negligible the relative uncertainty in the mass of the earth is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 1

From kepler's Law 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 2

At some instant a radioactive sample S1 having an activity 5 μCi has twice the number of nuclei as another sample S2 which has an activity of 10 μCi. The half lives of S1 and S2 are :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 2

Given : N1 = 2N2

Hence  5years and 20 year 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 3

Two moles of helium are mixed with an moles of hydrogen. If Cp/Cv = 3/2 for the mixture then the value of n is

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 3


⇒ 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 4

Unpolarized light of intensity I is incident on a system of two polarizers, A followed by B. The intensity of emergent light is I/2. If a third polarizer C is placed between A and B the intensity of emergent light is reduced to I/3. The angle between the polarizers A and C is θ, then

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 4

A and B have same alignment of transmission axis. 
Lets assume c is introduced at θ angle 


⇒ 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 5

The de-Broglie wav elength (λB) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state (λG) by :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 5

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 6

In the given circuit the current through zener diode is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 6

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 7

The end correction of a resonance column is 1cm. If the shortest length resonating with the tuning fork is 10cm, the next resonating length should be :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 7

Given : e = 1 cm 
For first resonance 

For second resonance 

⇒ ℓ2 = 3 × 11 - 1 = 32 cm 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 8

Two sitar strings A and B playing the note 'Dha' are slightly out of tune and produce beats of frequency 5Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease by 3Hz. If the frequency of A is 425 Hz. the original frequency of B is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 8

Frequency of B is either 420Hz or 430Hz As tension in B is increased its frequency will increase. If frequency is 430Hz, beat frequency will increase If frequency is 420 Hz beat frequency will decrease, hence correct answer is 420Hz 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 9

A power transmission line feeds input power at 2300V to a step down transformer with its primary windings having 4000 turns giving the output power at 230V. If the current in the primary of the transformer is 5A and its efficiency is 90% the output current would be :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 9

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 10

A body of mass m starts moving from rest along x-axis so that its velocity varies as u = a√s where a is a constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first t seconds after the start of the motion is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 10


JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 11

Suppose that the angular velocity of rotation of earth is increased. Then as a consequence:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 11

g' = g – ω2Rcos2φ
Where φ is latitude there will be no change in gravity at poles as φ = 90° 
At all other points as ω increases g' will decrease. 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 12

Both the nucleus and the atom of some element are in their respective first excited states. They get deexcited by emitting photons of wavelengths λNrespectively. The ratio λN is closest to : 

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 12


where Ea and EN are energies of photons from atom and nucleus respectively. EN is of the order of MeV and Ea in few eV. 
So

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 13

A plane electromagnetic wave of wavelength λ has an intensity I. It is propagating along the positive Y-direction. The allowed expressions for the electric and magnetic fields are given by :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 13

If E0 is magnitude of electric field then 1/2ε0E2 x C = I

direction of  

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 14

A charge q is spread uniformly over an insulated loop of radius r. If it is rotated with an angular velocity ω with respect to normal axis then magnetic moment of the loop is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 14


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

A heating element has a resistance of 100Ω at room temperature. When it is connected to a supply of 220V a steady current of 2A passes in it and temperature is 500°C more than room temperature. What is the temperature coefficient of resistance of the heating element?

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

Resistance after temperature increases by 500°C = 220/2 = 110Ω
110 = 100 (1 + α500) 

a = 2 × 10-4 °C-1

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

A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity ω the maximum e.m.f. induced in the coil will be :

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

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 17

A ray of light is incident at an angle of 60° on one face of a prism of angle 30°. The emergent ray of light makes an angle of 30° with incident ray. The angle made by the emergent ray with second face of prism will be :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 17

δ = I + e - A
30 = 60 + e – 30° 
⇒ e = 0  
So angle with face = 90° 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 18

A galvanometer with its coil resistance 25Ω requires a current of 1mA for its full deflection. In order to construct an ammeter to read up to a current of 2A the approximate value of the shunt resistance should be :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 18

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 19

An oscillator of mass M is at rest in the equilibrium position in a potential V = 1/2 k(x – X)2. A particle of mass m comes from right  with speed u and collides completely inelastically with M and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after 13 collisions is : (M = 10, m = 5, u = 1, k = 1)

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 19

In first collision mu momentum will be imparted to system. In second collision when momentum of (M + m) is in opposite direction mu momentum of particle will make its momentum zero. on 13th collision

⇒ 
⇒ 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 20

One mole of an ideal monatomic gas is taken along the path ABCA as shown in the PV diagram. The maximum temperature attained by the gas along the path BC is given by :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 20

Equation of line BC 







JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 21

In a circuit for finding the resistance of a galvanometer by half deflection method a 6V battery and a high resistance of 11kΩ are used. The figure of merit of the galvanometer produces a deflection of θ = 9 divisions when current flows in the circuit. The value of the shunt resistance that can cause the deflection of θ/2 is

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 21







S = 110Ω

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 22

In the following circuit the switch S is closed at t = 0. The charge on the capacitor C1 as a function of time will be given by 

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 22

q = CeqE [1 – e–t/RCeq)]
Both capacitor will have same charge as they are connected in series. 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 23

Let  The magnitude of a coplanar vector  is given by :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 23


Solving equation (i) and (ii) we get 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 24

A particle executes simple harmonic motion and is located at x = a, b and c at times t0, 2t0 and 3t0 respectively. The frequency of the oscillation is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 24

a = Asinωt0
b = Asin2ωt0
c = Asin3ωt0
a + c = A[sinωt0 + sin3ωt0] = 2Asin2ωt0cosωt0


JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 25

A thin circular disk is in the xy plane as shown in the figure. The ratio of its moment of inertia about z and z' axes will be :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 25

We know that, the moment of inertia of the circular disk about the center is

Using parallel axis theorem,

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 26

Two identical conducting spheres A and B carry equal charge. They are separated by a distance much larger than their diameters and the force between them is F. A third identical conducting sphere C is uncharged. Sphere C is first touched to A then to B and then removed. As a result the force between A and B would be equal to :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 26


 when A and C are touched charge on both will be q/2
Then when B and C are touched 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 27

Two particles of the same mass m are moving in circular orbits because of force given by  The first particle is at distance r =  1 and  the second at r = 4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to 

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 27




JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 28

The percentage errors in quantities P, Q, R and S are 0.5%, 1%, 3% and 1.5% respectively in the measurement of a physical quantity  The maximum percentage error in the value of A will be :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 28


= 3 × 0.5 + 2 × 1 + 1/2 × 3 + 1.5 
= 1.5 + 2 + 1.5 + 1.5 
ΔA/A = 6.5%

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 29

A carrier wave of peak voltage 14V is used for transmitting a message signal given to achieve a modulation index of 80% will be :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 29

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 30

A small soap bubble of radius 4cm is trapped inside another bubble of radius 6cm without any contact. Let P2 be the pressure inside the inner bubble and P0 the pressure outside the outer bubble. Radius of another bubble with pressure difference P2 – P0 between its inside and outside would be :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 30




r = 2.4cm 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 31

For standardizing NaOH solution, which of the following is used as a primary standard?

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 31

Oxalic acid is used as a primary standard for NaOH standardizing.

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 32

Products A and B formed in the following reactions are respectively :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 32

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 33

When XO2 is fused with an alkali metal hydroxide in presence of an oxidizing agent such as KNO3 ; a dark green product is formed which disproportioates in acidic solution to afford a dark purple solution. X is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 33

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 34

The major product B formed in the following reaction sequence is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 34

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 35

In a complexometric titration of metal ion with ligand M (Metal ion) + L (Ligand) → C (Complex) end point is estimated spectrophotometrically (through light absorption). If 'M' and 'C' do not absorb light and only 'L' absorbs, then the titration plot between absorbed light (A) versus volume of ligand 'L' (V) would look like :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 35

Initially ligand consumed by metal due to formation of complex. So absorbed light (A) remain constant, after complex formation is completed, extra volume of ligand solution increases ligand concentration and also increases absorbed light.

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 36

The major product of the following reaction is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 36

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 37

Among the following, the incorrect statement is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 37

In amylose1,4 α-glycosidic linkage is present.

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 38

In the extraction of copper from its sulphide ore, metal is finally obtained by the oxidation of cuprous sulphide with :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 38

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 39

Among the oxides of nitrogen :
N2O3, N2O4 and N2O5 ; the molecule(s) having nitrogen-nitrogen bond is/are :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 39

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 40

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

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 40

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 41

The most polar compound among the following is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 41

the bond dipole vector of C–F bond is not subtractive.

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 42

In Wilkinson's catalyst, the hybridization of central metal ion and its shape are respectively :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 42

Wilkinson catalyst
[RhCl(PPh3)3]

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 43

At 320 K, a gas A2 is 20 % dissociated to A(g). The standard free energy change at 320 K and 1 atm in J mol–1 is approximately : (R = 8.314 JK–1 mol–1 ; ln 2 = 0.693 ; ln 3 = 1.098)

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 43



ΔGº = –2.303 × 8.314 × 320 log10 0.2 = 4281 J/mole

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 44

Which of the following complexes will show geometrical isomerism?

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 44

*Multiple options can be correct
JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 45

Which of the following statements is false ?

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 45

When temperature is increased, black body emit high energy radiation, from higher wavelength to lower wavelength.
Rydberg constant has unit length–1 (i.e. cm–1)

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 46

When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount of p-aminophenol produced is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 46


JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 47

For which of the following processes, ΔS is negative ?

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 47

N2 (g, 1 atm) → N2 (g, 5 atm)

for isothermal process


ΔS < 0

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 48

An unknown chlorohydrocarbon has 3.55 % of chlorine. If each molecule of the hydrocarbon has one chlorine atom only ; chlorine atoms present in 1 g of chlorohydrocarbon are :
(Atomic wt. of Cl = 35.5 u ; Avogadro constant = 6.023 × 1023 mol–1)

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 48

CxHyCl
% Cl = 3.55


JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 49

The incorrect statement is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 49

Due to common ion effect, sufficient S2– concentration is not produced and no ppt formed of NiS.

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 50

The mass of a non-volatile, non-electrolyte solute (molar mass = 50 g mol–1) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75 %, is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 50



JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 51

The incorrect geometry is represented by :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 51

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 52

Assuming ideal gas behaviour, the ratio of density of ammonia to that of hydrogen chlroide at same temperature and pressure is : (Atomic wt. of Cl 35.5 u)

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 52

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 53

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

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 53

→ Phenelzine contains hydrazine
→ Chloroxylenol contains phenol
→ Uracil is the pyrimidine base
→ Ranitidine contains furan ring

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 54

The gas phase reaction 2NO2(g) → N2O4(g) is an exothermic reaction. The decomposition of N2O4, in equilibrium mixture of NO2(g) and N2O4(g), can be increased by :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 54

2NO2(g) → N2O4(g) ΔH = (–)
By addition of an inert gas at constant pressure, volume increases, so reaction moving in backward direction and decomposition of N2O4 increases.

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 55

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

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 55

Physical adsorption is multilayer adsorption.

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 56

A group 13 element 'X' reacts with chlorine gas to produce a compound XCl3. XCl3 is electron deficient and easily reacts with NH3 to form Cl3X → NH3 adduct; however, XCl3 does not dimerize. X is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 56

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 57

The major product of the following reaction is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 57


Inversion takes place at the carbon containing bromine atom.

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 58

If 50 % of a reaction occurs in 100 second and 75 % of the reaction occurs in 200 second, the order of this reaction is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 58


First order reaction as half life is constant.

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 59

The major product of the following reaction is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 59

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 60

Which of the following compounds will most readily be dehydrated to give alkene under acidic condition?

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 60

 will most readily be dehydrated to give conjugated alkene.

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 61

If  then dy/dx is equal to : 

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 61

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 62

Let N denote the set of all natural numbers. Define two binary relations on N as R1 = {(x, y) ∈ N × N : 2x + y = 10} and R2 = {(x, y) Î N × N : x + 2y = 10}. Then

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 62

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 63

The coefficient of x2 in the expansion of the product (2– x2)· ((1 + 2x + 3x2)6 + (1 – 4x2)6) is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 63

coefficient of x2 = 2 coefficient of x2 in ((1 + 2x + 3x2)6 + (1 – 4x2)6) – constant term 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 64

If the area of the region bounded by the curves, y = x2, y = 1/x and the lines y = 0 and x = t (t > 1) is 1 sq. unit, then t is equal to :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 64



JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 65

If the length of the latus rectum of an ellipse is 4 units and the distance between a focus and its nearest vertex on the major axis is 3/2 units, then its eccentricity is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 65

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 66

The number of numbers between 2,000 and 5,000 that can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is not allowed) and are multiple of 3 is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 66

There are 4 places to be filled with the given digits. The thousands place can have only 2, 3 and 4 since the number has to be greater than 2000. For the remaining 3 places, we have pick out digits such that the resultant number is divisible by 3. The divisibility criteria for 3 states that sum of digits of the number should be divisible by 3. Case 1: if we pick 2 for thousand place. The remaining digits we can pick such that sum of digits at all places is a multiple of 3 are: 0,1 and 3 as 2 + 1 + 0 + 3 = 6 is divisible by 3. 0,3 and 4 as 2 + 3 + 0 + 4 = 9 is divisible by 3. In both the above combination, the remaining three digits can be arranged in 3! ways. Total number=2×3!=12 Case 2: If we pick 3 for thousands place. The remaining digits we can pick such that sum of digits at all places is a multiple of 3 are: 0,1and 2 as 3 + 1 + 0 + 2 = 6 is divisible by 3. 0,2 and 4 as 3 + 2 + 0 + 4 = 9 is divisible by 3. In both the above combinations, the remaining three digits can be arranged in 3! ways. Total number=2×3!=12
Case 3: if we pick 4 for thousand place. The remaining digits we can pick such that sum of digits at all places is a multiple of 3 are: 0,2 and 3 as 4 + 2 + 0 + 3 = 9 is divisible by 3. In the above combination, the remaining three digits can be arranged in 3!  ways. Total number = 3! = 6.
Total number of numbers between 2000 and 5000 divisible by 3 are 12 + 12 + 6 = 30

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 67

Two different families A and B are blessed with equal number of children. There are 3 tickests to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to the children of the family B is 1/12, then the number of children in each family  is : 

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 67

Let n number of children are there in each family 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 68

 equals : 

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 68


JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 69

Let p, q and r be real numbers (p ≠ q, r ≠ 0), such that the roots of the equation  are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 69

(2x + p + q) r = (x + p) (x + q) 
x2 + (p + q – 2r) x + pq – pr – qr = 0 
p + q = 2r    ……..(i) 
α2 + β2 = (α + β)2 - 2αβ
= 0 – 2 [pq – pr – qr]  = – 2pq + 2r (p + q)  = – 2pq + (p + q)2 = p2 + q2 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 70

Let 1/x1 , 1/x2 ,..., 1/xn (xi ≠ 0 for i = 1, 2, &., n) be in A.P. such that x1 = 4 and x21 = 20. If n is the least positive integer for which xn > 50, then  is equal to : 

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 70


JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 71

The differential equation representing the family of ellipses having foci either on the x-axis or on the y-axis, centre at the origin and passing through the point (0, 3) is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 71

Equation of ellipse 

 since it passes through (0, 3)
⇒b2=9
∴ Eqution of ellipse becomes
⇒ 
......(2)

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 72

The sum of the intercepts on the coordinate axes of the plane passing trhough the point (–2, –2, 2) and containing  the line joining the points (1, –1, 2) and (1, 1, 1), is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 72

Equation of plane passing through three given points is

⇒ - (x + 2) + 3 (y + 2) + 6 (z - 2) = 0
⇒ x – 3y – 6z + 8 = 0
∴ sum of intercepts 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 73

Let A = and B = A20. Then the sum of the elements of the first column of B is : 

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 73



Sum of the elements of first column = 231 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 74

Let A, B and C be three events, which are pair-wise independent and denotes the complement of an event E. If P(A∩B∩C) = 0 and P(C) > 0, then  is equal to :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 74


JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 75

If p → (~p ν ~q) is false, then the truth v alues of p and q are respectively :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 75

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 76

If the function f defined as f(x) =  is continuous at x = 0, then the ordered pair (k, f(0)) is equal  to :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 76



Clearly k = 3 and f(0) = 1 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 77

If the angle between the lines,  and is cos–1  then p is equal to : 

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 77


Angle between both lines is




JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 78

The locus of the point of intersection of the lines, √2 x – y + 4 √2 k = 0 and √2 kx + ky – 4 √2 = 0 (k is any non-zero real parameter), is

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 78


JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 79

A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizonatal road. If it takes 18 min. for the angle of depression of the car to change from 30° to 45° ; then after this, the time taken (in min.) by the car to reach the foot of the tower, is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 79

Let length of tower = h 

⇒ AC' = AB = h
and AC = AB cot 30º = √3 h 
⇒ CC' = (√3 –1) h 
Time taken by the car form C to C' = 18 min 
⇒ time taken by the car to reach the foot of the tower = 
= 9 (√3 + 1) min 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 80

If an angle A of a ΔABC satisfies  5 cosA + 3 = 0, then the roots of the quadratic equaiton, 9x2 + 27x + 20 = 0 are :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 80

5cosA + 3 = 0 ⇒ cosA = - 3/5 clearly A ∈ (90º, 180º)
Now roots of equation  9x2 + 27x + 20 = 0 are -5/3 and -4/3
⇒ Roots secA and tanA 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 81

If a circle C, whose radius is 3, touches externally the circle, x2 + y2 + 2x – 4y – 4 = 0 at the point (2, 2), then the length of the intercept cut by this circle C, on the x-axis is equal to :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 81

 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 82

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

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 82

Let P (2t, t2
equation normal at P to x2 = 4y be 

it passes through (–3,0) 

Point P is (–2,1) 
equation of tangent to x2 = 4y at (–2,1) 
x(–2) = 2 (y + 1) 
x + y + 1 = 0 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 83

If f(x) =  dt then :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 83

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 84

The number of v alues of k for which the system of linear equations,
(k + 2)x + 10y = k
kx + (k + 3) y = k – 1 has no soution, is :

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 84

For no solution 

(k + 2) (k + 3) = 10 k 
k2 – 5k + 6 = 0 ⇒ k = 2,3 
k ≠ 2  for k = 2 both lines identical 
so k = 3 only  
so number of values of k is 1 

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 85

If   (C is a constant of integration), then the ordered pair (K, A) is equal to

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 85


JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 86

The least positive integer n for which  = 1, is 

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 86


ωn = 1 
least positive integer value of n is 3.

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 87

The sum of the first 20 terms of the series  ..... is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 87


JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 88

Let  and a vector be such that  Then equals : 

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 88


JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 89

The mean and the standard deviation (s.d.) of five observations are 9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 89

Here mean = 

Now, standard deviation = 0
∴ all the five terms are same i.e., 9.
Now for changed observation

= 2

JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 90

Let M and m be respectively the absolute maximum and the absolute minimum values of the function, f (x) = 2x3 - 9x2 + 12x + 5 in the interval [0, 3]. Then M - m is equal to:

Detailed Solution for JEE Main 2018 Question Paper with Solutions (16th April-Morning) - Question 90


For maxima or minima put f'(x) = 0



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