Description

This mock test of JEE Main 2021 Mock- 2 for JEE helps you for every JEE entrance exam.
This contains 75 Multiple Choice Questions for JEE JEE Main 2021 Mock- 2 (mcq) to study with solutions a complete question bank.
The solved questions answers in this JEE Main 2021 Mock- 2 quiz give you a good mix of easy questions and tough questions. JEE
students definitely take this JEE Main 2021 Mock- 2 exercise for a better result in the exam. You can find other JEE Main 2021 Mock- 2 extra questions,
long questions & short questions for JEE on EduRev as well by searching above.

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

Solution:

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 :

Solution:

Angular momentum is given by

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 v_{1}, v_{2} such that the stone hits the box.

Solution:

Position vector of box and stone are

When stone hits the block,

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

Solution:

V_{I} = -9cm/s towards right.

So away from the mirror.

QUESTION: 5

A uniform electric field E is present horizontally along the paper throughout the region and uniform magnetic field B_{0} 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 t_{0}. Find the speed of projection if particle moves with constant velocity after t_{0} . (Given qE = mg)

Solution:

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

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

Solution:

From F.B.D of P

N + T sin θ = mg ... (1)

T cos θ = f_{max} = 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)

QUESTION: 7

A monochromatic light source of wavelength λ is placed at S. Three slits S_{1}, S_{2} and S_{3} are equidistant from the source S and the point P on the screen. S_{1}P-S_{2}P=λ/6 and S_{1}P-S_{3}P = 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

Solution:

Take base SS_{3}P

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

Solution:

For energy in radius

Where n is integer and k is constant energy,

now

Now this implies that for R_{max} , K=1

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.

Solution:

Current gain of the circuit is

Input impedance of the circuit is

Load AC resistance is

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 E_{1} and E_{2} are the electric fields in the copper and iron wires respectively, then which of the following is correct?

Solution:

(where symbols have standard meaning)

Since current is same (series)

QUESTION: 11

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

Solution:

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 :-

Solution:

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

QUESTION: 13

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

Solution:

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

QUESTION: 14

The displacement of a particle is represented by the equation y= 40cos^{3 }ωt. The motion is

Solution:

QUESTION: 15

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

Solution:

In the circuit equivalent, Resistance across the battery is

Thus current through the battery is

Thus current 1.5A (from the figure) will be divided in 10Ω & 5Ω in inverse ratio thus

Thus current in branch AB is

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.)

Solution:

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 56^{o}C. Then the difference of temperature across wall X is

Solution:

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

Hence temperature difference across the walls X is (θ_{1 }-θ_{ }) = 21^{0} C

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

Solution:

Energy of one drop

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

Energy of 10^{6} drops,

Potential on the combined drop

= (potential of each initial drop) × (n^{2/3})

V′ = V (10^{6})^{2/3} = 10^{4} V

Energy of combined drop,

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.

Solution:

when particle Za emits 5 alpha then

Using conservation of linear momentum

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

Solution:

Total bandwidth = 150 kHz

f_{a}(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

=(150×10^{3})/(10×10^{3})=15

*Answer can only contain numeric values

QUESTION: 21

When a certain metallic surface is illuminated with monochromatic light of wavelength λ, the stopping potential for photoelectric current is 3V_{0} and when the same surface is illuminated with light of wavelength 2λ, the stopping potential is V_{0}. The threshold wavelength of this surface for photoelectric effect is Kλ. Calculate the value of K.

Solution:

*Answer can only contain numeric values

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 ω_{2. }Find the ratio (Given d << D)

Solution:

Fringe with proportional to wave length

*Answer can only contain numeric values

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 K_{a} 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).

Solution:

*Answer can only contain numeric values

QUESTION: 24

What is the minimum height (in 10^{2} m) of a brick column of uniform cross section for which column breaks due to its own weight?

[P_{atmospheric} = 100 kPa, ρ = 1.8 × 10^{3}kg/m^{3}. Breaking stress σ = 3.7 M Pa]

Solution:

*Answer can only contain numeric values

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.

Solution:

Frequency of direct sound heared by observer

Wave length of reflected sound wave

No. of beats heared by person

QUESTION: 26

Which of the following can act as reducing agent

Solution:

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.

QUESTION: 27

If P,Q,R and S are elements of 3^{rd} 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.

Solution:

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.

QUESTION: 28

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

Solution:

For meta form of an acid to exist it must be capable of giving one H_{2}O molecule and also contain at least one ‘H’ after donation of water.

QUESTION: 29

Bakelite is a

Solution:

Structure of bakelite

QUESTION: 30

If mechanism of reaction is

Where k is rate constant then, what is

Solution:

QUESTION: 31

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

Solution:

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

QUESTION: 32

Solution:

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

Step 2:

LiAlH_{4} reduces —CN to —CH_{2}—NH_{2}

Step 3:

creates carbocation which then expands and get stabilized.

Hence, option (A) is correct.

QUESTION: 33

k_{sp} for AgCl is-

Given: T = 25 °C

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

Solution:

QUESTION: 34

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

Solution:

Doping of NaCl with 10^{–2} mol% SrCl_{2} means that 100 mol of NaCl are doped with 10^{–2} mol of SrCl_{2}.

100 mol of NaCl doped with = 10^{–2} mol SrCl_{2}

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 × 10^{23} mol^{–1}

= 6.02 × 10^{19} mol^{–1} of NaCl

Hence, option B is correct.

QUESTION: 35

Product A is;

Solution:

Mechanism:

Carbocation is formed.

(ii)

Hence option (C) is correct answer.

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 Δ_{vap} H = 30.775 KJ/mol. Which of the following option is correct regarding percent of association of acetic acid in benzene.

Solution:

Given to (boiling point of benzene) = 80.10 °C = 353.1K molality (m) = 0.5

Δ_{vap} H = 30.775 KJ/mol

MW_{1} (benzene) = 78

QUESTION: 37

Ore of Y would be

Solution:

Cinnabar (HgS) is Ore of Hg.

Siderite is FeCO_{3}

Malachite is Cu_{2}CO_{3}(OH)_{2}

Hornsilver is AgCl

Hence, option (B) is correct.

QUESTION: 38

Product is:

Solution:

step 1:

Step 2: Ring expansion to make more stable carbocation.

Step 3:

QUESTION: 39

Solution:

(i) NBS reagent is used for allylic substitution.

(ii) NBS is used for Bromination.

QUESTION: 40

[X_{3}B ← NH_{3}], in which of the following boric halide, tendency to accept electrons from nitrogen of ammonia will be least (X = halogens)

Solution:

In BF_{3}, due to F back bonding to B, least Lewis acid character is observed. So out of above four, BF_{3} will have least tendency to accept electron from NH_{3}.

Hence, option (D) is correct.

QUESTION: 41

E_{1}, E_{2} and E_{3} are activation energies then, which of the following is correct.

Solution:

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

Here double bonds are in conjugation, so it will be difficult to hydrogenate then. E_{2} will also be high but not so high as E_{1}.

Here, no conjugation or aromaticity present in reaction so process will be easy. E_{3} will be lowest.

E_{1} > E_{2} > E_{3}.

Hence, option D is correct.

QUESTION: 42

The ionic molar conductivities of ions are x, y and z S cm^{2} mol^{–1}, respectively then value of of (NaOOC – COOK) is

Solution:

According KOOC to Kohlrausch law:

QUESTION: 43

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

Solution:

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,

QUESTION: 44

X + HNO_{3}→ Y + NO_{2} + H_{2}O + S

Y + ammonium molybdate → yellow ppt.

Identity which of the following is (X):

Solution:

As_{2}S_{5} + HNO_{3}→ H_{3}AsO_{4} + NO_{2} + H_{2}O + S

H_{3}ASO_{4} + (NH_{4})_{2}MoO_{4}→(NH_{4})_{2}4sO4 12MoO_{3}

Yellow ppt.

Hence option (A) correct

QUESTION: 45

Oleum in water is treated with 0.5*l* of 2.75 M Ca(OH)_{2} solution. The resulting solution required 15.7 gm of H_{3}PO_{3 }(Assume strong acid) solution for complete neutralization. Calculate the amount of free SO_{3} in 100 gms of oleum

Solution:

*Answer can only contain numeric values

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, K_{sp} (AgCl) = 10^{–10}] are

Solution:

*Answer can only contain numeric values

QUESTION: 47

Half-life of a reaction is 20 sec. If t_{2} is second half-life of reaction assuming it to be zero order and t_{3} is third half-life assuming it to be 2^{nd} order reaction, then t_{3}/t_{2} = ?

Solution:

*Answer can only contain numeric values

QUESTION: 48

K_{a} 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)

Solution:

*Answer can only contain numeric values

QUESTION: 49

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

Solution:

“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 SN_{2} substitution.

*Answer can only contain numeric values

QUESTION: 50

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

Solution:

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

Solution:

I. Ellipse.

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

Eccentricity = e_{1} = 2.

Focus is at

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

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

Solution:

Let denote the event that the i^{th} person answers the question correctly.

P(at least one of them answers correctly)

QUESTION: 53

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

Solution:

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

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

Solution:

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

QUESTION: 55

Sum to n terms of the series

Solution:

QUESTION: 56

The value of the limit is

Solution:

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

Solution:

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

QUESTION: 58

If 0<x<1, then is equal to

Solution:

Consider the triangle with sides 1, x and

so,

QUESTION: 59

The function defined by

Solution:

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.

QUESTION: 60

Let a_{1}, a_{2}, a_{3}, .... be terms of an A.P

Solution:

⇒ (2a_{1}−d) (p − q) = 0

As p is not equal to q

QUESTION: 61

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

Solution:

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 true__, __r is false__ and __P can be true or false__

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

Solution:

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

QUESTION: 63

Solution:

QUESTION: 64

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

Solution:

Let the centre be at

Thus, the equation of the ellipse becomes

QUESTION: 65

If then range of f(x) is -

Solution:

Minimum value of And maximum value of

so

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

Solution:

Equation of ellipse is

QUESTION: 67

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

Solution:

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

Solution:

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

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

Solution:

Let x_{1}, x_{2}, ……….., x_{n} be the n terms.

QUESTION: 70

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

Solution:

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 x^{2} – y^{2} = x^{2}y^{2}

*Answer can only contain numeric values

QUESTION: 71

Given that

Solution:

*Answer can only contain numeric values

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

Solution:

*Answer can only contain numeric values

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.

Solution:

*Answer can only contain numeric values

QUESTION: 74

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

Solution:

*Answer can only contain numeric values

QUESTION: 75

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

Solution:

### NTA 101 JEE MAIN MOCK TESTS

Doc | 756 Pages

### Detailed Syllabus of JEE (Main)-2021

Doc | 24 Pages

### Official Notification (NTA)- JEE Main 2021

Doc | 13 Pages

- JEE Main 2021 Mock- 2
Test | 75 questions | 180 min

- JEE Main 2021 Mock- 7
Test | 75 questions | 180 min

- JEE Main 2021 Mock- 8
Test | 75 questions | 180 min

- JEE Main 2021 Mock- 4
Test | 75 questions | 180 min

- JEE Main 2021 Mock- 10
Test | 75 questions | 180 min