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# JEE Main Mock- 3

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## 75 Questions MCQ Test JEE Main & Advanced Mock Test Series | JEE Main Mock- 3

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JEE Main Mock- 3 - Question 1

### Three charges e, e and - 2e are placed at the corners of the triangle as shown. The net dipole moment of the system is

Detailed Solution for JEE Main Mock- 3 - Question 1

At A, the x-components of the two dipole moments cancel and y-components add.

∴ Py 2Pcos α

JEE Main Mock- 3 - Question 2

### A mass is attached to one end of a spring of spring constant k. The spring is stretched and then released such that its amplitude of oscillation is A. For a displacement y from the mean position, if the kinetic energy is 44% of its potential energy, then y in terms of A is

Detailed Solution for JEE Main Mock- 3 - Question 2

JEE Main Mock- 3 - Question 3

### At a given instant there are 25% of undecayed radioactive nuclei in a sample. After 20s, the number of undecayed nuclei reduces to 12.5%. The extra time required in which the number of undecayed nuclei will further reduce to 3.125% of the sample is-

Detailed Solution for JEE Main Mock- 3 - Question 3

Half-life of the radioactive sample, t1/2 = time in which the number of undecayed nuclei becomes half = 20 sec (given).
The material further reduces to 3.125% in n half-lives is given by

where Nis % of undecayed nuclie after n half lifes , and A is % of undecayed nuclie initially.
here Nf = 3.125 %
and Ni = 12.5 %
so,
so time taken t = nt1/2 = 2 x 20 = 40 second

JEE Main Mock- 3 - Question 4

The escape velocity for an atmospheric particle 2000 km above the earth's surface is (Radius of the earth = 6.4 x 106 m and g = 10 m s- 2 )

Detailed Solution for JEE Main Mock- 3 - Question 4

Escape velocity ve
At the above the Earth,s surface = and

JEE Main Mock- 3 - Question 5

A circular disc of radius r and thickness r/6 has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and recasted to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is

Detailed Solution for JEE Main Mock- 3 - Question 5

Volume oc circular disc = surface of area x thickness

The mass of the circular disc and sphere are the same

where R is the radius of the sphere,

Moment of interia of a circluar disc

Moment of interia of solid sphere

JEE Main Mock- 3 - Question 6

When an UV light of 1015 Hz and intensity 2 W/m2 is directed at a metal surface, photoelectrons emitted were found to have a maximum kinetic energy of 1.6 eV. If the work functions for different materials are as follows: Potassium 2.2 eV, Sodium 2.3 eV, Lithium 2.5 eV and Calcium 3.2 eV, identify the metal in the given problem.

Detailed Solution for JEE Main Mock- 3 - Question 6

E = hv = 6.63 x 10-34 x 1015
= 6.63 x 10-19 J
= 4.1 eV
( dividing by e = 1.6 x 10-19)
Emax = hv - hv0
⇒ hv0 = 4.1 - 1.6
= 2.5 eV
⇒ Lithium

JEE Main Mock- 3 - Question 7

Two sources of sound S1 and S2 each emitting waves of wavelength A are kept symmetrically on either side of the centre 0 of a circle ABCD such that S1O = S2O = K . When the detector is moved along the circumference of the circle, the number of maxima recorded by the detector in one revolution is

Detailed Solution for JEE Main Mock- 3 - Question 7

For the point A, the path difference is zero and for the point B, it is 2 λ. In between A and B there will be a point E where the path difference is λ.

Hence at the points A, E and B there will be maximum sound due to S1 and S2.
In a circular path, it will be easier to locate points like F, C, G, D and H so that the total number of maxima amounts to 8.

JEE Main Mock- 3 - Question 8

A boat capable of a speed v in still water wants to cross a river of width d. The speed of the water current increases linearly from zero at either bank to a maximum of u at the middle of the river. When the boat is rowed at right angles to the bank, its downstream drift is

Detailed Solution for JEE Main Mock- 3 - Question 8

Let us choose a coordinate system with its origin as the starting point of the boat, the + x-axis points downstream and + y-axis points at right angles to the bank of the river.

x-motion of the boat is due to the water current velocity vcurrent while the y-motion is caused solely by the velocity v of the boat.
The above two motions are independent of each other and can be treated separately.
Assuming that the boat starts at time t = 0, the y coordinate after a time t is y = v t ... (i) The speed of water current is a function of y and is given by

Substituting for y from equation (i)
By symmetry, its value at the middle of the river is D/2.
The time required to reach the middle of the river is
Separating the variable in equation (iii) and intergrating

JEE Main Mock- 3 - Question 9

Figure shows the plot of a potential energy function of a conservative system U versus x. Which of the following statements is correct ?

Detailed Solution for JEE Main Mock- 3 - Question 9

Force F versus x is drawn in figure. At points E1 points E1,E2,E3 and E4
F=0
F has the greatest magnitude nearer to point E1.
If U is minimum i.e.,  = positive, equilibrium is stable.
If U is maximum i.e.,
negative, equilibrium is stable.
If U is constant i.e., equilibrium is neutral.
If U is constant i.e.,equilibrium is neutral.

JEE Main Mock- 3 - Question 10

A sphere with some cavity has outer radius R. It rolls down an inclined plane without slipping and attains a speed v at the bottom. When this sphere slides down without rolling on the frictionless inclined plane of same height its speed at the bottom is 5v/4. The radius of gyration of the sphere is

Detailed Solution for JEE Main Mock- 3 - Question 10

Let k be the radius of gyration and M the mass of the sphere. The kinetic energy should be the same in both the cases

JEE Main Mock- 3 - Question 11

The antenna current of an AM transmitter is 8 A when only carrier wave is sent but the current increases to 8.88 A when the carrier wave is sinusoidally modulated. The percentage of modulation is

Detailed Solution for JEE Main Mock- 3 - Question 11

IT = 8.88 A, IC = 8 A; ma = ?

JEE Main Mock- 3 - Question 12

A square metal wire loop of side 20 cm and resistance 1 Ω is moved with a constant velocity Vo in a uniform magnetic field of induction B = 4 Wb/m2. The magnetic field lines are perpendicular to the plane of the loop and directed inwards. The loop is connected to a network of resistors each of value 2 Ω . The resistance of load wires AB and CD are negligible. To get a current of 2 mA in the loop, the speed of motion of the loop is

Detailed Solution for JEE Main Mock- 3 - Question 12

The network of resistors in a balanced Wheatstone's network. Hence the resistance EF is ineffective. The equivalent resistance R' of the network is

The resistance of the square loop is 1 Ω
∴ effective resistance of the circuit R = 2 + 1 = 3 Ω
The emf induced in the loop
∴ speed of the loop v0

i = 2 mA = 2 x 10-3 A ; I = 0.2 m B = 4 Wb/m2
Speed of the loop

JEE Main Mock- 3 - Question 13

Four identical hollow cylindrical columns of steel, support a big structure of mass 60,000 kg. The inner and outer radii of each column are 40 cm and 50 cm respectively. When the load distribution is uniform, the compressional strain on each column is (Young’s modulus of steel is 2 x 1011 Pa)

Detailed Solution for JEE Main Mock- 3 - Question 13

Total mass supported by the columns = 60,000 kg
Total weight supporte d = 60,000 x 10 N
Compressional force one achcolumn ,
Cross-sectional area of each colum n,

a = 0.283 m2
Young modulus,
∴ compressional strain

= 0.265 x 10-5
= 2.65 x 10-6

JEE Main Mock- 3 - Question 14

A steel ball of mass 60 g falls from a height 0.8 m on the horizontal surface of a massive slab. The coefficient of restitution between the ball and the slab is e = 0.4; the total momentum imparted to the slab by the ball after numerous bounces is

Detailed Solution for JEE Main Mock- 3 - Question 14

By Newton’s law of collision, the velocity v′ after the first impact is v′ = ev (upward direction).

JEE Main Mock- 3 - Question 15

In nuclear reaction, energy released per fission is 200 MeV. When uranium 235 is used as nuclear fuel in a reactor having a power level of 1 MW, the amount of fuel needed in 30 days will be

Detailed Solution for JEE Main Mock- 3 - Question 15

Energy produced by the reactor in 1 day = 106 x 86400 J
Energy released per fission = 200 x 106 x 1.6 x 10-19 J
No. of fissions required (i.e.,) no. of 235U atoms fissioned in a month

Mass of 235U having the requisite no. of atoms

JEE Main Mock- 3 - Question 16

The temperature at which the speed of sound in oxygen will be same as the speed of sound in nitrogen at 25 °C-

Detailed Solution for JEE Main Mock- 3 - Question 16

Since the speeds of sound equal

∴ t = 340.6 - 273 = 67.6 °C

JEE Main Mock- 3 - Question 17

An alternating voltage having frequency of 50 cycles/sec and maximum voltage 220 V is supplied to a circuit containing a pure inductance of 0.02 H and a pure resistance of 10 Ω in series. The value of maximum current in circuit is-

Detailed Solution for JEE Main Mock- 3 - Question 17

Impendance of L.R series circut

Value of current

JEE Main Mock- 3 - Question 18

A beam of light traveling in water strikes a glass plate which is also immersed in water. When the angle of incidence is 50 ° the reflected beam is found to be plane polarised. The refractive index of water is 4/3. The refractive index of the glass plate is (Given tan 50 ° = 1.198)

Detailed Solution for JEE Main Mock- 3 - Question 18

by brewster's law,

This is the refractive index of glass w.r.to water.

∴ refractive index of glass

JEE Main Mock- 3 - Question 19

A reversible heat engine converts one fourth of heat input into work. When the temperature of the sink is reduced by 200 K, its efficiency is doubled. The temperature of the source is

Detailed Solution for JEE Main Mock- 3 - Question 19

⇒ T= 800 K

JEE Main Mock- 3 - Question 20

A man pulls a loaded sledge of mass 60 kg along a horizontal surface at constant velocity as shown in figure. The coefficient of kinetic friction between the sledge and the surface is 0.15. The tension in the rope during pulling when it makes an angle of cp = 40 ° with the horizontal is (cos 40 0 = 0.766, sin 40 ° = 0.643 and g = 10 ms'2)

Detailed Solution for JEE Main Mock- 3 - Question 20

Free body diagram

T cos φ = fμKR...(1)
T sin φ + R = mg...(2)

*Answer can only contain numeric values
JEE Main Mock- 3 - Question 21

Gravitational acceleration on the surface of planet is where  is the gravitational acceleration on the surface of earth. The average mass density of the planet is 2/3 times that of the earth. If escape speed on the surface of the earth is taken to be 11kms–1, escape speed on the surface of the planet in kms–1, will be:

Detailed Solution for JEE Main Mock- 3 - Question 21

let the gravitational acceleration on the surface be g' and desity be ρ'

*Answer can only contain numeric values
JEE Main Mock- 3 - Question 22

A point object ‘O’ is placed in a medium of refractive index μ1 = 1.4. S1 and S2 are two concentric spherical surfaces of radii 1m and 2m. To the right of ‘O’ contains a medium of refractive index μ2 = 1.5 between the interfaces S1 and S2. Find the object distance of O form S1 (in meter) so that an image of ‘O’ as seen by observer from air, coincides with O.

Detailed Solution for JEE Main Mock- 3 - Question 22

R = d tan C

*Answer can only contain numeric values
JEE Main Mock- 3 - Question 23

A radioactive substance A decays in to B. It is known that only A is present at t = 0. Find the number of half lives at which the probability of getting B in the mixture is 15 times that of finding A, if we pick out randomly from the sample.

Detailed Solution for JEE Main Mock- 3 - Question 23

*Answer can only contain numeric values
JEE Main Mock- 3 - Question 24

A single electron orbits round a stationary nucleus of charge Ze, where Z is a constant and - e is the electron charge. It is observed that it requires 47.2 eV to excite the electron from second Bohr orbit to third orbit. Then to ionise the atom from ground state, required wavelength of electromagnetic(EM) radiations is found to be 4nÅ. Find the value of n. [Take energy of electron in ground state of hydrogen atom is - 13.6 eV.]

Detailed Solution for JEE Main Mock- 3 - Question 24

*Answer can only contain numeric values
JEE Main Mock- 3 - Question 25

An inductor L = 50 mH carrying an initial current l0 = 2.5 amp is connected across a non linear resistor. The voltage across resistor is related to current as V = 10 l2. After how much time (in ms) current through inductor becomes 1.25 amp.

Detailed Solution for JEE Main Mock- 3 - Question 25

JEE Main Mock- 3 - Question 26

The intermediate formed in the following reaction is-

Detailed Solution for JEE Main Mock- 3 - Question 26

The reaction is an example of cine subsitution or elimination addition

JEE Main Mock- 3 - Question 27

The compound present in brorax bead is

Detailed Solution for JEE Main Mock- 3 - Question 27

This test is meant for detecting the coloured cations. When borax is heated on a platinum loop it swells up into a white porous mass first which on subsequent heating melts into a shining transparent glass bead (B2O3).

When this glass bead is heated with coloured compounds, a characteristic colour of metaborate is formed.

However, in a reducing flame, a different colour is obtained.

JEE Main Mock- 3 - Question 28

The correct decreasing order of acidic nature is

Detailed Solution for JEE Main Mock- 3 - Question 28

Acidic character depends on stability of conjugate base. More stable conjugate base, more will be the acidic character

Electronegativity of Cl > Br > I So stability order of conjugate bases CIO- > BrO- > IO- Hence acidic character HCIO > HBrO > HIO

JEE Main Mock- 3 - Question 29

In  the formal charge on each oxygen atom and P - 0 bond order respectively are

Detailed Solution for JEE Main Mock- 3 - Question 29

on exist ;n 4 resonating structures. Bond order of

3 negative charges of the ion is beings h are d by 4 oxygen atoms,
∴ formal charge on an O atom = -3/4

JEE Main Mock- 3 - Question 30

Which one of the following compounds is not aromatic?

Detailed Solution for JEE Main Mock- 3 - Question 30

Aromaticity of a compound can be decided by Huckel's rule.
Huckel's rules are as follows:
1. Compound must be cyclic
2. The cyclic compound must have resonance and complete delocalization of electrons i.e., there must be conjugation.
3. The total number of pi electrons must fit the formula = 4n + 2, where n is a non - negative integer.
The important summary of this concept is as follows: (Remember this summary)

JEE Main Mock- 3 - Question 31

The actinoids exhibit more number of oxidation states in general than the lanthanoids because

Detailed Solution for JEE Main Mock- 3 - Question 31

shielding effect of 5f < shielding effect of 4f
so Zeff (z effect) in actinoidis < Zeff (z effect) in lanthanoid
so extraction of electrons in actinoid is easy than lanthanoid which has 4f and better shielding.

JEE Main Mock- 3 - Question 32

The compound when kept in pentane converts to a constitutional isomer. Identify that
isomer.

Detailed Solution for JEE Main Mock- 3 - Question 32

JEE Main Mock- 3 - Question 33

The vapour pressure of water at 20 °C is 17.5 torr. What will be the no. of moles of water present in one litre of air at 20 °C and 40% relative humidity.(R =0.082 L atm/mol-k)

Detailed Solution for JEE Main Mock- 3 - Question 33

Relative humidity (RH) =
∴ Partial pressure of H2O = RH × Vapour pressure of H2O

JEE Main Mock- 3 - Question 34

0.400 g of impure CaSO4 (Molar mass = 136) solution when treated with excess of barium chloride solution, gave 0.617 g of anhydrous BaSO4 (Molar mass = 233). The percentage of CaSO4 present in the sample is

Detailed Solution for JEE Main Mock- 3 - Question 34

JEE Main Mock- 3 - Question 35

At temperature 300 K for reaction  C2H4(g) + 3O2(g)→ 2CO2(g) + 2H2O(l); ΔH = − 336.2 kcal. The approximate value of ΔU  at 300 K for the same reaction (R = 2 cal degree− 1 mol− 1) will be-

Detailed Solution for JEE Main Mock- 3 - Question 35

C2H4(g) + 3 O2(g)→2 CO2(g) + 2 H2O(l)
At 300 K ΔH = − 336.2 kcal
Δng = 2 − (1 + 3) = − 2
T = 300 K, ΔH = − 336.2 kcal,
R = 2 × 10−3 kcal degree−1 mol−1
ΔH = ΔU+ ΔngRT
− 336.2 = ΔU + (− 2) × 300 × 2 × 10−3
ΔU = − 336.2 + 1.2 = − 335.0 kcal

JEE Main Mock- 3 - Question 36

The cations present in 'A'

Detailed Solution for JEE Main Mock- 3 - Question 36

JEE Main Mock- 3 - Question 37

20 g helium is compressed isothermally and reversibly at 27 °C from a pressure of 2 atmosphere to 20 atmosphere. The change in the value of heat energy change during the process is (R = 2 cal kelvin-1 mol-1)

Detailed Solution for JEE Main Mock- 3 - Question 37

JEE Main Mock- 3 - Question 38

Which of the following type of compound is used to increase the hardness of the silicone polymer.

Detailed Solution for JEE Main Mock- 3 - Question 38

Hydrolysis followed by condensation polymerization of RSiCl3 produces 3-D cross-linked silicones which are hard.

JEE Main Mock- 3 - Question 39

Consider the reaction 2A + B →products, when the concentration of A alone was doubled, the half-life of the reaction did not change. When the concentration of B alone was doubled, the rate was not altered. The unit of rate constant for this reaction is

Detailed Solution for JEE Main Mock- 3 - Question 39

When concentration of B doubled rate does not changes
So,

Now when Concentration of A doubled half life does not change
it means reaction is first order so

order of reaction =1
so unit of rate constant = s -1

JEE Main Mock- 3 - Question 40

The products A and B of the above reaction are respectively

Detailed Solution for JEE Main Mock- 3 - Question 40

JEE Main Mock- 3 - Question 41

0.066 g of a metal was deposited when a current of 2 ampere passed for 100 seconds through an aqueous solution of its metal ion. The equivalent mass of the metal is-

Detailed Solution for JEE Main Mock- 3 - Question 41

Total charge passed in Coulombs = It = 2 X100 = 200 Coulombs
200 Coulombs of charge deposits 0.066 g of metal.
Equivalent mass of the metal is that mass of metal deposited by 1 Farad or 96500 Coulombs.
Equivalent mass of metal

JEE Main Mock- 3 - Question 42

A mixture of gaseous nitrogen and gaseous hydrogen attains equilibrium with gaseous Ammonia according to the reaction

It is found that at equilibrium when temperature is 673 K , atmospheric pressure is 100, moles of gaseous nitrogen and gaseous hydrogen are in ratio 1:3 and mole percent of gaseous ammonia is 24 . The equilibrium constant Kp for the reaction at the above condition is-

Detailed Solution for JEE Main Mock- 3 - Question 42

Partial pressure of
=24.0 atm
Pressure of N2 + H2 = 100 – 24 = 76 atm
Partial pressure of
Partial pressure of N2

JEE Main Mock- 3 - Question 43

Detailed Solution for JEE Main Mock- 3 - Question 43

JEE Main Mock- 3 - Question 44

Study the following reactions and identify the reactant 'R'.

Detailed Solution for JEE Main Mock- 3 - Question 44

JEE Main Mock- 3 - Question 45

A new carbon – carbon bond is formed in:

Detailed Solution for JEE Main Mock- 3 - Question 45

New carbon-carbon bond is formed is Friedal Craft reaction and Reimer-Teimann reaction
Friedal Craft’s reaction

Riemer Teimann Reaction

*Answer can only contain numeric values
JEE Main Mock- 3 - Question 46

How many of the following are reducing sugers?

Detailed Solution for JEE Main Mock- 3 - Question 46

I, IV, VII, VIII

*Answer can only contain numeric values
JEE Main Mock- 3 - Question 47

The co-ordination number of central ion of the complex obtained in the sodium nitroprusside test of sulphide ion is

Detailed Solution for JEE Main Mock- 3 - Question 47

Na4[Fe(CN)4NOS]

*Answer can only contain numeric values
JEE Main Mock- 3 - Question 48

Here X is ( No of moles of CH4)

*Answer can only contain numeric values
JEE Main Mock- 3 - Question 49

A buffer solution is formed by mixing 100 mL 0.01 M CH3COOH with 200 mL 0.02 M CH3COONa. If this buffer solution is made to 1.0 L by adding 700 mL of water. pH will change by a factor of

Detailed Solution for JEE Main Mock- 3 - Question 49

No change in pH

*Answer can only contain numeric values
JEE Main Mock- 3 - Question 50

CH3 (CHOH)2 — COOH in this compound number of optical isomer is

Detailed Solution for JEE Main Mock- 3 - Question 50

In this compound there are 4 optical isomers; this is calculated by multiplying the number of chiral centres by two.

JEE Main Mock- 3 - Question 51

The coefficient of x5 in (1+2x+3x2+……….up to infinite term)-3/2 is

Detailed Solution for JEE Main Mock- 3 - Question 51

(1+2x+3x2+………)–3/2
=((1–x)–2)–3/2
= (1 − x)3 = 1 − 3x + 3x2 − x3
∴ coefficient of x5 = 0.

JEE Main Mock- 3 - Question 52

Sum to the infinity of the series

Detailed Solution for JEE Main Mock- 3 - Question 52

We have

JEE Main Mock- 3 - Question 53

Detailed Solution for JEE Main Mock- 3 - Question 53

Limit is of the form 1, so we have

= (ln a + ln b + ln c) 2/3
= 2/3(ln abc)
∴ Desired limit =

JEE Main Mock- 3 - Question 54

If the progressions 3, 10, 17, ....... and 63, 65, 67, ....... are such that their nth terms are equal, then n is equal to

Detailed Solution for JEE Main Mock- 3 - Question 54

nth term of 1st series =nth term of 2nd series
⇒ 3 + (n − 1) 7 = 63 + (n − 1) 2
⇒ (n − 1)5 = 60
⇒ n -1=12
⇒ n = 13

JEE Main Mock- 3 - Question 55

If the latus rectum of a hyperbola through one focus, subtends 60 ° angle at the other focus, then its eccentricity is

Detailed Solution for JEE Main Mock- 3 - Question 55

Taking only positive value of e as eccentricity cannot be negative.

JEE Main Mock- 3 - Question 56

Find the solution of the differential equation

given that y=4 at x=3

Detailed Solution for JEE Main Mock- 3 - Question 56

Integrating both sides
⇒Putting x=3 and y=4, we get

⇒Thus we get the solution as

JEE Main Mock- 3 - Question 57

If the mean of a binomial distribution is 25, then its standard deviation lies in the interval

Detailed Solution for JEE Main Mock- 3 - Question 57

Mean, np = 25 and q < 1

⇒0 ≤ σ< 5

JEE Main Mock- 3 - Question 58

The equation of the normal to the ellipse   at the positive end of the latus rectum is

Detailed Solution for JEE Main Mock- 3 - Question 58

The equation of the normal at (x1, y1) to the given ellipse is

Here x1 = ae and y1 = b2/a
So the equation of the normal at positive end of the latus rectum is

JEE Main Mock- 3 - Question 59

The tangent to the circle x2 + y2 = 5 at (1, − 2) also touches the circle x2 + y2 − 8x + 6y + 20 = 0. Then the point of contact is

Detailed Solution for JEE Main Mock- 3 - Question 59

Tangent at (1, − 2) to x2 + y2 = 5 is
x − 2y − 5 = 0 ... (i).
x2 + y2 − 8x + 6y + 20 = 0 are
C (4, − 3) and radius r = √5
Perpendicular distance from
C (4, − 3) to (i) is radius.
∴ (i) is also a tangent to the second circle.
Let P (h, k) be the foot of the drawn circle from C (4, − 3) on (i)

∴ (h, k) = (3, −1)

JEE Main Mock- 3 - Question 60

The equation x3 – 3x + [a] = 0, will have three real and distinct roots if –
(where [ ] denotes the greatest integer function)

Detailed Solution for JEE Main Mock- 3 - Question 60

f(x) = x3 – 3x + [a]
Let [a] = t (where t will be an integer)
f(x) = x3 – 3x + t ……….(i)
⇒ f ’(x) = 3x2 – 3
⇒ f ‘(x) = 0 has two real and distinct solution which are x = 1 and x = -1
so  f(x) = 0 will have three distinct and real solution when  f (1). f(-1) < 0
……………. (ii)
Now,
f(1) = (1)3 -3(1) + t = t – 2
f(–1) = (–1)3 – 3 (–1) + t = t + 2
From equation (ii)
(t –2)  (t + 2) < 0
⇒ t ∈ (-2, 2)
Now t = [a]
Hence [a] ∈ (-2, 2)
⇒ a ∈ [-1, 2)

JEE Main Mock- 3 - Question 61

If  then a2 + b2 is

Detailed Solution for JEE Main Mock- 3 - Question 61

Taking magnitude and squaring on both sides, we get

JEE Main Mock- 3 - Question 62

If g(x) satisfies the conditions of Rolle’s theorem in [1, 2] and g′(x) = f(x), then is equal to

Detailed Solution for JEE Main Mock- 3 - Question 62

As g(x) satisfies the condition of Rolle’s theorem in [1, 2], g(x) is continuous in the interval and
g(1) = g(2)

= g(2) − g(1) = 0

JEE Main Mock- 3 - Question 63

If sin   then x is equal to

Detailed Solution for JEE Main Mock- 3 - Question 63

JEE Main Mock- 3 - Question 64

The equation of the plane through the line of intersection of planes x + y + z + 3 = 0 and 2x − y + 3z + 1 =0 and parallel to the line is

Detailed Solution for JEE Main Mock- 3 - Question 64

Any plane through the given line is 2x − y + 3z + 1 + λ (x + y + z + 3) = 0 If this plane is parallel to the linethen the normal to the plane is also perpendicular to the above line.
∴(l1l2 + m1 m2 + n1 n2 = 0
⇒ (2 + λ) 1 + (λ −1) 2 + (3 + λ) 3 = 0
⇒ λ= -3/2
and the required plane is
2x − y + 3z + 1 + (-3/2) (x + y + z + 3) = 0
⇒ x − 5y + 3z − 7 = 0
or
x − 5y + 3z = 7

JEE Main Mock- 3 - Question 65

Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated is

Detailed Solution for JEE Main Mock- 3 - Question 65

The total number of words that can be formed is 105 and number of these words in which no letters are repeated is 10P5.
Hence the required number
= 105 − 10P5
= 100000 − 10 × 9 × 8 × 7 × 6
= 69760

JEE Main Mock- 3 - Question 66

Which of the following is logically equivalent to ∼ (∼p ⇒ q) ?

Detailed Solution for JEE Main Mock- 3 - Question 66

∼(p ⇒ q) ≡ p ∧ ∼q
∴ ∼(∼p ⇒ q) ≡ ∼p ∧ ∼q

JEE Main Mock- 3 - Question 67

If  is equal to

Detailed Solution for JEE Main Mock- 3 - Question 67

JEE Main Mock- 3 - Question 68

If  then f(2x) − f(x) is not divisible by

Detailed Solution for JEE Main Mock- 3 - Question 68

Doing R3→ R3−xR2 and
R2→R2− xR1
we get

So,
f(2x) − f(x)
= a[(a + 2x)2 − (a + x)2]
= a (a + 2x − a − x) (a + 2x + a + x)
= ax (2a + 3x)
So, f(2x) - f(x)=ax(2a+3x)
Thus, f(2x) -f(x) is divisible by a, x and (2a+3x).

JEE Main Mock- 3 - Question 69

Consider the matrix  then B1027 equals

Detailed Solution for JEE Main Mock- 3 - Question 69

B1027 = (B4)256 . B3
= (I)256 . B3
= I . B3
= B3
= B . B2

JEE Main Mock- 3 - Question 70

Two vertical poles AL and BM of heights 20 m and 80 m respectively stand apart on a horizontal plane. If A and B be the feet of the poles and AM and BL intersect at P, then the height of P is equal to

Detailed Solution for JEE Main Mock- 3 - Question 70

⇒Z = 16 m

*Answer can only contain numeric values
JEE Main Mock- 3 - Question 71

ABC is a right angle triangle (right angled at B) inscribed in the parabola y2 = 4x. The minimum length of the intercept cut off by the tangents at A and C to the parabola from y-axis is

Detailed Solution for JEE Main Mock- 3 - Question 71

Let A(t1), B(t2) and C(t3) be the points on the parabola

also length of intercept cut off by tangents from y-axis is |t1 - t3|≥ 4

*Answer can only contain numeric values
JEE Main Mock- 3 - Question 72

From a point on the line y= x + c, c (parameter), tangents are drawn to the hyperbola  such that chords of contact pass through a fixed point (x1, y1). Then x1/y1 s equal to

Detailed Solution for JEE Main Mock- 3 - Question 72

*Answer can only contain numeric values
JEE Main Mock- 3 - Question 73

If the sum of the coefficients in the expansion of  (ℓ2x2−2ℓx+1)51 vanishes then ℓ is equal to

Detailed Solution for JEE Main Mock- 3 - Question 73

*Answer can only contain numeric values
JEE Main Mock- 3 - Question 74

A seven-digit number made up of all distinct 8, 7, 6, 4, 2, and y is divisible by 3. Then possible number of order pair (x, y) is

Detailed Solution for JEE Main Mock- 3 - Question 74

8, 7, 6, 4, 2, and y
Any number is divisible by 3 if sum of digits is divisible by 3 i.e. x + y 27 is divisible by 3
and y can take values from 0, 1, 3, 5, 9
possible pairs (5, 1) (3, 0) (9, 0) (9, 3) & (1, 5), (0, 3) (0, 9) (3, 9).

*Answer can only contain numeric values
JEE Main Mock- 3 - Question 75

If z1, z2, z3 are complex numbers such that |z1|=|z2|=|z3|=1, then |z1−z2|2+|z2−z3|2+|z3−z1|2 cannot exceed

Detailed Solution for JEE Main Mock- 3 - Question 75

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## JEE Main & Advanced Mock Test Series

2 videos|324 docs|160 tests