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# JEE Main Mock- 2

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## 75 Questions MCQ Test JEE Main & Advanced Mock Test Series | JEE Main Mock- 2

JEE Main Mock- 2 for JEE 2022 is part of JEE Main & Advanced Mock Test Series preparation. The JEE Main Mock- 2 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Mock- 2 MCQs are made for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Mock- 2 below.
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JEE Main Mock- 2 - Question 1

### A point object moves on a circular path such that distance covered by it is given by function  Meter (t in second). The ratio of the magnitude of acceleration at t = 2 sec. And t = 5 sec. is 1: 2 then radius of the circle is

Detailed Solution for JEE Main Mock- 2 - Question 1

JEE Main Mock- 2 - Question 2

### At time t = 0, a 2 kg particle has position vector m relative to the origin. Its velocity is given by  The torque acting on the particle about the origin at t = 2s, is :

Detailed Solution for JEE Main Mock- 2 - Question 2

Angular momentum is given by

JEE Main Mock- 2 - Question 3

### A box is floating in the river of speed 5m/s. The position of the block is shown in the figure at t=0. A stone is thrown from point O at time t = 0 with a velocity  Find the value of v1, v2 such that the stone hits the box.

Detailed Solution for JEE Main Mock- 2 - Question 3

Position vector of box and stone are

When stone hits the block,

JEE Main Mock- 2 - Question 4

A luminous point object is moving along the principal axis of a concave mirror of focal length 12 cm towards it. When its distance from the mirror is 20 cm its velocity is 4 cm/s . The velocity of the image in cm/s at that instant

Detailed Solution for JEE Main Mock- 2 - Question 4

VI = -9cm/s towards right.
So away from the mirror.

JEE Main Mock- 2 - Question 5

A uniform electric field E is present horizontally along the paper throughout the region and uniform magnetic field B0 is present horizontally (perpendicular to plane of paper in inward direction) right to the line AB. A charge particle having charge q and mass m is projected vertically upward and it crosses the line AB after time t0. Find the speed of projection if particle moves with constant velocity after t0 . (Given qE = mg)

Detailed Solution for JEE Main Mock- 2 - Question 5

Let velocity of particle at ‘P’ is ‘v’ making an angle ‘θ’ with x-axis
(P is the point where particle is crossing AB line)
⇒As after reaching point ‘P’, velocity of particle remain constant, so net force acting on particle will be zero at ‘P’

Now, horizontal velocity at point ‘P’ is due to Electric force qE Only

JEE Main Mock- 2 - Question 6

Two blocks P and Q are connected by a light inextensible string passing over a smooth pulley fixed as shown in the Figure. The coefficient of friction of blocks P and Q to the table is μ = 0.3; the mass of the block Q is 20 kg. The mass of the block P for the block just to slip is

Detailed Solution for JEE Main Mock- 2 - Question 6

From F.B.D of P
N + T sin θ = mg ... (1)
T cos θ = fmax = μN ... (2)
From (1), N = mg − T sin θ ... (3)
From (2) and (3),
T cos θ = μ (mg − T sin θ)
T (cos θ + μ sin θ) = μ mg ... (4)
From F.B.D. of Q
T = 20 g ... (5)

JEE Main Mock- 2 - Question 7

A monochromatic light source of wavelength λ is placed at S. Three slits S1, S2 and S3 are equidistant from the source S and the point P on the screen. S1P-S2P=λ/6 and S1P-S3P = 2λ/3. If I be the intensity at P when only one slit is open, the intensity at P when all the three slits are open is

Detailed Solution for JEE Main Mock- 2 - Question 7

Take base SS3P

JEE Main Mock- 2 - Question 8

On a hypothetical planet satellite can only revolve in quantized energy level i.e. magnitude of energy of a satellite is integer multiple of a fixed energy. If two successive orbit have radius R and 3R/2 what could be maximum radius of satellite

Detailed Solution for JEE Main Mock- 2 - Question 8

Where n is integer and k is constant energy,
now

Now this implies that for Rmax , K=1

JEE Main Mock- 2 - Question 9

In a transistor amplifier when the signal changes by 2 V, the base current changes by 150 μA and collector current by 15 mA. If collector load Find the voltage gain of amplifier.

Detailed Solution for JEE Main Mock- 2 - Question 9

Current gain of the circuit is

Input impedance of the circuit is

JEE Main Mock- 2 - Question 10

Copper and iron wires of same length and diameter are in series and connected across a battery. The resistivity of copper is about one-sixth of the iron. If E1 and E2 are the electric fields in the copper and iron wires respectively, then which of the following is correct?

Detailed Solution for JEE Main Mock- 2 - Question 10

(where symbols have standard meaning)

Since current is same (series)

JEE Main Mock- 2 - Question 11

E, m, p and G denote energy, mass, angular momentum and gravitational constant. Then has the dimensions of

Detailed Solution for JEE Main Mock- 2 - Question 11

JEE Main Mock- 2 - Question 12

A circular conducting loop of radius R carries a current I. Another straight infinite conductor carrying current I passes through the diameter of this loop as shown in the figure. The magnitude of force exerted by the straight conductor on the loop is :-

Detailed Solution for JEE Main Mock- 2 - Question 12

Take an arc of angle which make angle ‘θ’ with vertical. (dθ very small → point ‘P’)

⇒ Magnetic field due to infinite wire at this point

(where r = R sinθ i.e. perpendicular distance of ‘P’ from line)

⇒ force dF on this Point (Arc of length dl = Rdθ) is

Point (small arc) symmetric to point ‘P’ will also have same force dF

Force balancing on the arc having angle 2θ

Total force = 2 dF sinθ

Force on whole circular loop

JEE Main Mock- 2 - Question 13

Find the intensity of electromagnetic wave if the electric field in electromagnetic wave is

Detailed Solution for JEE Main Mock- 2 - Question 13

Electric field of electromagnetic wave Intensity of Electromagnetic wave in terms of maximum electric field is

JEE Main Mock- 2 - Question 14

The displacement of a particle is represented by the equation y= 40cosωt. The motion is

Detailed Solution for JEE Main Mock- 2 - Question 14

JEE Main Mock- 2 - Question 15

In the circuit shown in figure find the current in branch BD of the circuit :

Detailed Solution for JEE Main Mock- 2 - Question 15

JEE Main Mock- 2 - Question 16

A particle undergoes from position O(0, 0, 0) to A (a, 2a, 0) via path in x-y plane under the action of a force which varies with particle’s (x, y, z) coordinate as  Work done by the force  is: (all symbols have their usual meaning and they are in SI unit.)

Detailed Solution for JEE Main Mock- 2 - Question 16

JEE Main Mock- 2 - Question 17

A wall made up of two layer X and Y. The thickness of the two layers is the same, but materials are different. The thermal conductivity of X is thrice than that of Y. In thermal equilibrium, the temperature difference between the two ends is 56oC. Then the difference of temperature across wall X is

Detailed Solution for JEE Main Mock- 2 - Question 17

Suppose the thickness of each wall is ‘t’. Then

Hence temperature difference across the walls X is (θ1  ) = 210 C

JEE Main Mock- 2 - Question 18

One million small identical drops of water, all charged to the same potential, are combined to form a single large drop. If E is the sum of the electrostatic energy of each small drop, the combined energy of the large drop is

Detailed Solution for JEE Main Mock- 2 - Question 18

Energy of one drop

where q = CV is the charge of one drop, C is the capacitance.
Energy of 106 drops,

Potential on the combined drop
= (potential of each initial drop) × (n2/3)
V′ = V (106)2/3 = 104 V
Energy of combined drop,

JEE Main Mock- 2 - Question 19

An unknown particle  originally at rest emits 5 alpha particles with speed 11385 km/h. Find the recoil speed of the unknown daughter nucleus.

Detailed Solution for JEE Main Mock- 2 - Question 19

when particle Za emits 5 alpha then

Using conservation of linear momentum

JEE Main Mock- 2 - Question 20

The number of AM broadcast stations that can be accommodated at a 150 kHz band width, if the highest frequency modulating carrier is 5 kHz is

Detailed Solution for JEE Main Mock- 2 - Question 20

Total bandwidth = 150 kHz
fa(max) = 5 kHz
Any station being modulated by a 5 kHz will produce an upper-side frequency 5 kHz above its carrier and a lower-side frequency 5 kHz below its carrier.
Thus  one station needs a bandwidth of 10 kHz.
Number of stations accommodated = Total BW /BW per station
=(150×103)/(10×103)=15

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

When a certain metallic surface is illuminated with monochromatic light of wavelength λ, the stopping potential for photoelectric current is 3V0 and when the same surface is illuminated with light of wavelength 2λ, the stopping potential is V0. The threshold wavelength of this surface for photoelectric effect is Kλ. Calculate the value of K.

Detailed Solution for JEE Main Mock- 2 - Question 21

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

A monochromatic beam of electrons accelerated by a potential difference V falls normally on the plane containing two narrow slits separated by a distance d. The interference pattern is observed on a screen parallel to the plane of the slits and at a distance of D from the slits. Fringe width is found to be w1w1. When electron beam is accelerated by the potential difference 4V the fringe width becomes ω2. Find the ratio  (Given d << D)

Detailed Solution for JEE Main Mock- 2 - Question 22

Fringe with proportional to wave length

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

A target element A is bombarded with electrons and the wavelengths of the characteristic spectrum and measured. A second characteristic spectrum is also obtained, because of an impurity in the target. The wavelength of the Ka lines are 196 pm (element A) and 169 pm (impurity). If the atomic number of impurity is z = (10 x – 1). Find the value of x. (atomic number of element A is 27).

Detailed Solution for JEE Main Mock- 2 - Question 23

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JEE Main Mock- 2 - Question 24

What is the minimum height (in 102 m) of a brick column of uniform cross section for which column breaks due to its own weight?
[Patmospheric = 100 kPa, ρ = 1.8 × 103kg/m3. Breaking stress σ = 3.7 M Pa]

Detailed Solution for JEE Main Mock- 2 - Question 24

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

A sound source emits frequency of 175 Hz when moving towards a rigid wall with speed 5 m/s and observer is moving away from wall with same speed 5 m/s. Both source and observer moves on a straight line which is perpendicular to the wall. The number of beats per second heard by the observer will be [Speed of sound = 355 m/s] Source is in between observer and wall.

Detailed Solution for JEE Main Mock- 2 - Question 25

Frequency of direct sound heared by observer
Wave length of reflected sound wave

No. of beats heared by person

JEE Main Mock- 2 - Question 26

Which of the following can act as reducing agent

Detailed Solution for JEE Main Mock- 2 - Question 26

Stability of metal carbonyl is deduced from EAN rule.
EAN rules states – those complex’s in which effective atomic number is numerically equal to atomic number of noble gas element found in same period in which metal is situated, are most stable.
EAN (effective atomic number) =

A. [Co(CO)4]
EAN = 27 – (–1) + 2 × 4
= 36 (inert gas atomic number)
Since, it already has EAN = 36 so it won’t give up or take in any electron.
B. EAN = 25 – (0) + 2 × 6 = 37
It has EAN = 37 so to attain inert gas electronic configuration it will donate one electron

EAN = 25 – (+1) + 2 × 6 = 36
It will act as reducing agent.
C. [Mn(CO)5]
EAN = 25 – (0) + 2 × 5 = 35
To gain inert gas electronic configuration, it will accept one electron.

It will act as oxidizing agent.
D. [Cr(CO)6]
EAN = 24 – (0) + 2 × 6 = 36
It already has inert gas configuration so it won’t exchange electrons.
Hence option (B) is correct.

JEE Main Mock- 2 - Question 27

If P,Q,R and S are elements of 3rd period of p–block in modern periodic table and among these one element is metal and rest are non-metal and their order of electronegativity is also given as
P < Q < R < S .Then in which of the following release of H+ is relatively easier.

Detailed Solution for JEE Main Mock- 2 - Question 27

In any X – O – H, we take into account electronegative difference between X and O, and O and H. If electronegative difference between X and O is greater than O and H, X – O bond will be more polar and will break easily giving OH. If electronegative difference between O and H is larger than X and O, O – H bond will be more polar and easier to break.
Electronegativity of S is largest, so electronegative difference between S and O will be least in S – O – H. It will be easier to break O – H bond, giving

Hence option (B) is correct.

JEE Main Mock- 2 - Question 28

In which of the following ‘meta form’ of ‘-ic acid’ is not possible.

Detailed Solution for JEE Main Mock- 2 - Question 28

For meta form of an acid to exist it must be capable of giving one H2O molecule and also contain at least one ‘H’ after donation of water.

JEE Main Mock- 2 - Question 29

Bakelite is a

Detailed Solution for JEE Main Mock- 2 - Question 29

Structure of bakelite

JEE Main Mock- 2 - Question 30

If mechanism of reaction is

Where k is rate constant then, what is

Detailed Solution for JEE Main Mock- 2 - Question 30

JEE Main Mock- 2 - Question 31

which of the following species undergo non–redox thermal decomposition reaction on heating

Detailed Solution for JEE Main Mock- 2 - Question 31

Non–redox decomposition implies no change in oxidation number
On taking a look at these options we observe only in option (D) there has been no change in oxidation number.
Hence, option (D) is correct

JEE Main Mock- 2 - Question 32

Detailed Solution for JEE Main Mock- 2 - Question 32

Step 1:: HCN attacks the carbonyl and forms hydrocyanin compound.

Step 2:
LiAlH4 reduces —CN to —CH2—NH2

Step 3:
creates carbocation which then expands and get stabilized.

Hence, option (A) is correct.

JEE Main Mock- 2 - Question 33

ksp for AgCl is-
Given: T = 25 °C

are 17.7, 13.2 and 23.0 cal/mol. (Antilog 0.21 =1.6)

Detailed Solution for JEE Main Mock- 2 - Question 33

JEE Main Mock- 2 - Question 34

If NaCl is doped with 10–2 mol% of SrCl2, which of the following option shows concentration of cation vacancies?

Detailed Solution for JEE Main Mock- 2 - Question 34

Doping of NaCl with 10–2 mol% SrCl2 means that 100 mol of NaCl are doped with 10–2 mol of SrCl2.
100 mol of NaCl doped with = 10–2 mol SrCl2
1 mol of NaCl doped with

Each  ion introduces one cation vacancy,
therefore, cation vacancy: 10–4 mol/mol of NaCl
= 10–4 × 6.023 × 1023 mol–1
= 6.02 × 1019 mol–1 of NaCl
Hence, option B is correct.

JEE Main Mock- 2 - Question 35

Product A is;

Detailed Solution for JEE Main Mock- 2 - Question 35

Mechanism:

Carbocation is formed.

(ii)

Hence option (C) is correct answer.

JEE Main Mock- 2 - Question 36

0.5 molal solution acetic acid (M.W. = 60) in benzene (M.W. = 78) boils at 80.80 °C. The normal boiling point of benzene is 80.10 °C and Δvap H = 30.775 KJ/mol. Which of the following option is correct regarding percent of association of acetic acid in benzene.

Detailed Solution for JEE Main Mock- 2 - Question 36

Given to (boiling point of benzene) = 80.10 °C = 353.1K molality (m) = 0.5
Δvap H = 30.775 KJ/mol
MW1 (benzene) = 78

JEE Main Mock- 2 - Question 37

Ore of Y would be

Detailed Solution for JEE Main Mock- 2 - Question 37

Cinnabar (HgS) is Ore of Hg.
Siderite is FeCO3
Malachite is Cu2CO3(OH)2
Hornsilver is AgCl
Hence, option (B) is correct.

JEE Main Mock- 2 - Question 38

Product is:

Detailed Solution for JEE Main Mock- 2 - Question 38

step 1:

Step 2: Ring expansion to make more stable carbocation.

Step 3:

JEE Main Mock- 2 - Question 39

Detailed Solution for JEE Main Mock- 2 - Question 39

(i) NBS reagent is used for allylic substitution.
(ii) NBS is used for Bromination.

JEE Main Mock- 2 - Question 40

[X3B ← NH3], in which of the following boric halide, tendency to accept electrons from nitrogen of ammonia will be least (X = halogens)

Detailed Solution for JEE Main Mock- 2 - Question 40

In BF3, due to F back bonding to B, least Lewis acid character is observed. So out of above four, BF3 will have least tendency to accept electron from NH3.
Hence, option (D) is correct.

JEE Main Mock- 2 - Question 41

E1, E2 and E3 are activation energies then, which of the following is correct.

Detailed Solution for JEE Main Mock- 2 - Question 41

Here, aromaticity is being ruptured hence activation energy is highest.

Here double bonds are in conjugation, so it will be difficult to hydrogenate then. E2 will also be high but not so high as E1.

Here, no conjugation or aromaticity present in reaction so process will be easy. E3 will be lowest.
E1 > E2 > E3.
Hence, option D is correct.

JEE Main Mock- 2 - Question 42

The ionic molar conductivities of  ions are x, y and z S cm2 mol–1, respectively then value of  of (NaOOC – COOK) is

Detailed Solution for JEE Main Mock- 2 - Question 42

According KOOC to Kohlrausch law:

JEE Main Mock- 2 - Question 43

For a gas obeying the van der waals equation at critical temperature, which of the following is true.

Detailed Solution for JEE Main Mock- 2 - Question 43

At critical point only one phase exits. Graph appears as given below

There is a stationary inflection point in PV Diagram (at a given critical temperature)
So,

JEE Main Mock- 2 - Question 44

X + HNO3→ Y + NO2 + H2O + S
Y + ammonium molybdate → yellow ppt.
Identity which of the following is (X):

Detailed Solution for JEE Main Mock- 2 - Question 44

As2S5 + HNO3→ H3AsO4 + NO2 + H2O + S
H3ASO4 + (NH4)2MoO4→(NH4)24sO4 12MoO3
Yellow ppt.
Hence option (A) correct

JEE Main Mock- 2 - Question 45

Oleum in water is treated with 0.5l of 2.75 M Ca(OH)2 solution. The resulting solution required 15.7 gm of H3PO(Assume strong acid) solution for complete neutralization. Calculate the amount of free SO3 in 100 gms of oleum

Detailed Solution for JEE Main Mock- 2 - Question 45

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

The minimum number of moles of solid KCl added to one liter contents of a standard silver …… silver ion electrode to convert it to standard silver …. silver chloride, electrode. [E° Ag+/Ag = 0.8V, Ksp (AgCl) = 10–10] are

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

Half-life of a reaction is 20 sec. If t2 is second half-life of reaction assuming it to be zero order and t3 is third half-life assuming it to be 2nd order reaction, then t3/t2 = ?

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

Ka for weak acid HA is 2×10–5 at 298K what is the pH value of 0.2 M aqueous solution of its salt with a strong base KOH? (log2=0.3)

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

How many of the following reactions are eliminations [majorly]?

Detailed Solution for JEE Main Mock- 2 - Question 49

“a” gives substitution; because molecule dominantly exists in diequatoria conformation; in which is no availability of suitable hydrogen for elimination
“b” gives elimination; because diaxial existace of    -hydrogen encourages elimination.

c, d, e, f, g → conceptual
(b) CN → strong nucleophile [than a base]; with 2° halides dominantly gives SN2 substitution.

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

How many of the following are reducing carbohydrates Glucose, Fructose, Sucrose, Lactose, Maltose, Galactose, Cellobiose, Cellulose, Starch, Arabinose, Ribose

JEE Main Mock- 2 - Question 51

The foci of a hyperbola coincide with the foci of the ellipse .The equation of the hyperbola if its eccentricity is 2, is

Detailed Solution for JEE Main Mock- 2 - Question 51

I. Ellipse.

Focus S is at S (ae,0) or S(4,0) …(i) II. Hyperbola

Eccentricity = e1 = 2.

Focus is at

By assumption (i) & (ii) represent the same position.

JEE Main Mock- 2 - Question 52

Four persons are asked the same question by an interviewer. If each has, independently, a probability 1/6 of answering correctly, then the probability that at least one of them answers correctly is

Detailed Solution for JEE Main Mock- 2 - Question 52

Let  denote the event that the ith person answers the question correctly.
P(at least one of them answers correctly)

JEE Main Mock- 2 - Question 53

Let  for all  is a root of the equation f(x) = 0 then

Detailed Solution for JEE Main Mock- 2 - Question 53

Above is a system of equation in f(x) and f(1/x).

JEE Main Mock- 2 - Question 54

A variable chord passing through the fixed point P on the axis of the parabola  cuts the parabola at the points A & B. The co-ordinates of the point P such that constant is

Detailed Solution for JEE Main Mock- 2 - Question 54

Let the point P be (h, 0). Hence equation of the chord passing through P(h, 0) in parametric form is given by

JEE Main Mock- 2 - Question 55

Sum to n terms of the series

Detailed Solution for JEE Main Mock- 2 - Question 55

JEE Main Mock- 2 - Question 56

The value of the limit  is

Detailed Solution for JEE Main Mock- 2 - Question 56

JEE Main Mock- 2 - Question 57

If z and ω are two non-zero complex numbers such that |zω| = 1 and arg(z) – arg(ω) = π/2, the is equal to

Detailed Solution for JEE Main Mock- 2 - Question 57

so, modulus of this complex number is 1 and argument is -π/2. so complex no. is  "-i"

JEE Main Mock- 2 - Question 58

If 0<x<1, then  is equal to

Detailed Solution for JEE Main Mock- 2 - Question 58

Consider the triangle with sides 1, x and

so,

JEE Main Mock- 2 - Question 59

The function  defined by

Detailed Solution for JEE Main Mock- 2 - Question 59

From graph we can see that, for two value of ‘x’, f(x) has same value so f(x) is not injective.
Also, range of f(x) is [0, 1)
So it is not surjective also.

JEE Main Mock- 2 - Question 60

Let a1, a2, a3, .... be terms of an A.P

Detailed Solution for JEE Main Mock- 2 - Question 60

⇒ (2a1−d) (p − q) = 0
As p is not equal to q

JEE Main Mock- 2 - Question 61

If statement (P →q) → (q →r) is false, then truth values of statements p,q and r respectively can be

Detailed Solution for JEE Main Mock- 2 - Question 61

If (P →q) →(q →r) is false then (p →q) must be true and (q →r) must be false.
Now,
If P →q is true then P & q both can be true or false :
If q →r is false then q must be true and r must be false
So when both condition i.e. P →q true and q → r false satisfy then q is truer is false and P can be true or false

JEE Main Mock- 2 - Question 62

A mirror and a source of light are situated at the origin O and at a point on OX respectively. A ray of light from the source strikes the mirror and is reflected. If the direction ratios of the normal to the plane are proportional to 1, –1, 1 then direction cosines of the reflected ray are

Detailed Solution for JEE Main Mock- 2 - Question 62

Let the source of light be situated at A(a,0,0) where  a≠0

Let OA be the incident ray, OB be the reflected ray and ON be the normal to the mirror at O.

Direction ratio of  are proportional to a, 0, 0 and so its direction cosines are 1, 0, 0.

Direction cosines of ON are

Let l, m, n be the direction cosines of the reflected ray OB Then,

Hence, direction cosines of the reflected ray are

JEE Main Mock- 2 - Question 63

Detailed Solution for JEE Main Mock- 2 - Question 63

JEE Main Mock- 2 - Question 64

Find the differential equation of the family of ellipse such that its centre is on the origin

Detailed Solution for JEE Main Mock- 2 - Question 64

Let the centre be at
Thus, the equation of the ellipse becomes

JEE Main Mock- 2 - Question 65

If  then range of f(x) is -

Detailed Solution for JEE Main Mock- 2 - Question 65

Minimum value of And maximum value of

so

JEE Main Mock- 2 - Question 66

The eccentricity of an ellipse with its centre at the origin is 1/2. If one of the directrix is x = 4, then the equation of the ellipse is

Detailed Solution for JEE Main Mock- 2 - Question 66

Equation of ellipse is

JEE Main Mock- 2 - Question 67

The greatest value of λ ≥ 0 for which both the equations 2x2 + ( λ − 1)x + 8 = 0 and x2 − 8x + λ + 4 = 0 have real roots is

Detailed Solution for JEE Main Mock- 2 - Question 67

JEE Main Mock- 2 - Question 68

The number of pairs of solution of the system of equations x+y=2π/3, cosx+cosy=3/2 where x and y are real is

Detailed Solution for JEE Main Mock- 2 - Question 68

We write the second equation as

Since, the range of cos function is [-1,1], so no values of x and y satisfy the second equation.
x∈φ, y∈φ

JEE Main Mock- 2 - Question 69

The mean of n terms is  If first term is increased by 1, second term by 2, and so on, then new mean is

Detailed Solution for JEE Main Mock- 2 - Question 69

Let x1, x2, ……….., xn be the n terms.

JEE Main Mock- 2 - Question 70

The points of contact of the tangents drawn from the origin to the curve y = sin x lie on the curve

Detailed Solution for JEE Main Mock- 2 - Question 70

1. Let (h, k) be a point of contact of the tangents drawn from the origin to y = sin x. Then, (h,k) lies on y = sin x..
∴k = sin h …(i)

Now,

The equation of the tangent at (h,k) is
y – k = (cos h) (x – h)
This passes through (0,0).

Hence, the locus of  is x2 – y2 = x2y2

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

Given that

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

In ∆ABC Orthocentre is (2, 3) Circum centre is (6, 10) and equation of side  is 2x + y = 17. Then the radius of the Circum circle of ∆ABC is

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

C is the centre of the hyperbola  and ‘A’ is any point on it. The tangents at A to the hyperbola meet the line x – 2y = 0 and x + 2y = 0 at Q and R respectively. The value of CQ CR.

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

Total number of even divisors of ‘1323000’ which are divisible by 105 is 2k –10, then k is

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

The area bounded by y = 2 – |2 – x|,y=3/|x| is (k-3ln3)/2, then k =________

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## JEE Main & Advanced Mock Test Series

2 videos|324 docs|160 tests