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# JEE Main Mock- 4

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## 75 Questions MCQ Test JEE Main & Advanced Mock Test Series | JEE Main Mock- 4

JEE Main Mock- 4 for JEE 2022 is part of JEE Main & Advanced Mock Test Series preparation. The JEE Main Mock- 4 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Mock- 4 MCQs are made for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Mock- 4 below.
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JEE Main Mock- 4 - Question 1

### A planet of core density 3ρ and outer curst of density ρ has small tunnel in core. A small Particle of mass m is released from end A then time required to reach end B :

Detailed Solution for JEE Main Mock- 4 - Question 1

At some distance from centre inside core

Now time for A to B =

JEE Main Mock- 4 - Question 2

### A small circular wire loop of radius a is located at the centre of a much larger circular wire loop of radius b as shown above (b> >a). Both loops are coaxial and coplanar. The larger loop carries a time (t) varying current I = I0 cos ωt where I0 and ω are constants. The large loop induces in the small loop an emf that is approximately equal to which of the following.

Detailed Solution for JEE Main Mock- 4 - Question 2

Magnetic flux passing through inner loop due to current in outer loop-

So, emf developed,

JEE Main Mock- 4 - Question 3

### A spring mass system is placed on a frictionless horizontal surface as shown in the figure. The spring is expanded by 1/10m and the blocks are given velocities as shown, then maximum extension of spring is :

Detailed Solution for JEE Main Mock- 4 - Question 3

As there is no loss of energy
Initial mechanical energy = Final mechanical energy
For initial Mechanical energy

Where  and vrelative =5 m/s

= 25 + 25 = 50
For final Mechanical energy
When there is maximum extension relative velocity between block = 0

JEE Main Mock- 4 - Question 4

An electromagnetic wave of frequency f = 7.3MHz passes from a vacuum into a dielectric medium with permittivityε = 9. Then,

Detailed Solution for JEE Main Mock- 4 - Question 4

As permittivity is greater than the permittivity of free space so the refractive index of medium is

Wavelength of the electromagnetic wave in medium will be
Velocity of wave will be.

JEE Main Mock- 4 - Question 5

A particle is projected from point A towards a building of height h as shown at an angle of 60 ° with horizontal. It strikes the roof of building at B at an angle of 30 ° with the horizontal. The speed of projection is

Detailed Solution for JEE Main Mock- 4 - Question 5

Let u be the speed of projection and v be the speed at B

Applying law of conservation of energy,

Using equation (i) in equation (ii),
we have

JEE Main Mock- 4 - Question 6

Three identical bulbs each of resistance 2Ω are connected as shown. The maximum power that can be consumed by individual bulb is 32W, then the maximum power consumed by the combination is :

Detailed Solution for JEE Main Mock- 4 - Question 6

Resistance of B1, B2 and B3 are same which is = 2Ω
As P ∝i2 (when ‘R’ constant)
So, maximum power consumed by bulb B3
Which is = 32 W
⇒i2R = 32
⇒i2 . (2) = 32
⇒i = 4 Ampere
So, current passing through B1 + B2 = i/2 = 2
Total power consumed in circuit
P = (i/2)2 (2) + (i/2)2 (2) + i2(2)
= (2)2 (2) + (2)2 (2) + (4)2 (2)
= 48W

JEE Main Mock- 4 - Question 7

A carrier wave has power of 1675 kW. If the side band power of a modulated wave subjected to 60%. Then find the amplitude modulation level

Detailed Solution for JEE Main Mock- 4 - Question 7

As you know that, sideband power is,  Where Pis power of carrier wave, Ps sideband power and ‘m’ is modulation rate,

JEE Main Mock- 4 - Question 8

Two identical conducting spheres each having radius r are placed at large distance. lnitially charge on one sphere is q, while charge on another sphere is zero when they are connected by conducting wire as shown in figure then find total heat produced when switch S is closed :

Detailed Solution for JEE Main Mock- 4 - Question 8

When sphere connected by wire, charge will flow from one sphere to another until potential of both sphere become same.
Or,
As both sphere is identical in dimension and material, charge will divide equally

Now, Initial potential energy of system
(Self energy of spherical conductor )
Final potential energy of system

Heat generation in conduction = Ui - uf

JEE Main Mock- 4 - Question 9

Four wire A, B, C and D each of length l = 10 cm and each of area of cross section is 0.1 m2 are connected in the given circuit. Then, the position of null point is

Given that resistivity

Detailed Solution for JEE Main Mock- 4 - Question 9

When S1 and S2 both open and jockey has no deflection then let current in upper circuit is ” i”.

So, RA = 1Ω
RB = 3Ω
RC = 6Ω
RD = 1Ω

When jockey is touched, and there is no deflection then potential difference between ‘A’ and the point should be same in both path-

So, in lower circuit as S2 is open
No current will flow in this circuit
⇒Potential difference between A and P
Will be = 2V

⇒i1 . 3= i2 . 6 and i1 + i2 = 1

Potential difference between A and P

For this point ‘P’ should be half way in wire C.
Similarly when jockey touched on wire ‘B’, Null point will be obtained at middle point of wire B.

JEE Main Mock- 4 - Question 10

The radius of curvature of spherical surface is 10 cm. The spherical surface separates two media of refractive indices μ2 = 1.3 and μ3 = 1.5 as shown in Figure. The medium of refractive index 1.3 extends upto 78 cm from the spherical surface. A luminous point object O is at the distance of 144 cm from the spherical surface in the medium of refractive index μ1 = 1.1. The image formed by the spherical surface is at

Detailed Solution for JEE Main Mock- 4 - Question 10

Distance of object from the spherical surface

For the refraction at the spherical Surface

JEE Main Mock- 4 - Question 11

A transformer has an efficiency of 80%. It is connected to a power input of 4 kW and 100 V. if the secondary voltage is 240 V, then the secondary current is

Detailed Solution for JEE Main Mock- 4 - Question 11

JEE Main Mock- 4 - Question 12

Magnetic moments of two identical magnets are M and 2M respectively. Both are combined in such a way that their similar poles are same side. The time period in this case is ‘T1’ .If polarity of one of the magnets is reversed its period becomes ‘T2’ then find out ratio of their time periods T1/T2.

Detailed Solution for JEE Main Mock- 4 - Question 12

Case 1:

Msystem = M + 2M = 3M
Isystem = I + I = 2I
Case 2:

Msystem = 2M – M = M
Isystem = 2I

JEE Main Mock- 4 - Question 13

The maximum current in a galvanometer can be 10 mA. It’s resistance is 10Ω. To convert it into an ammeter of 1 Amp. A resistor should be connected in

Detailed Solution for JEE Main Mock- 4 - Question 13

JEE Main Mock- 4 - Question 14

A soap bubble (surface tension = T) is charged to a maximum surface density of charge = σ, when it is just going to burst. Its radius R is given by:

Detailed Solution for JEE Main Mock- 4 - Question 14

Pressure due to surface tension
Pressure due to electrostatic force
Just before the bubble burst
Pressure due to surface tension = Pressure due to electrostatic force

JEE Main Mock- 4 - Question 15

One mole of an ideal gas  heated by law p = αV where P is pressure of gas, V is volume, α is a constant.
The heat capacity of gas in the process is-

Detailed Solution for JEE Main Mock- 4 - Question 15

JEE Main Mock- 4 - Question 16

A sonometer wire resonates with a given tuning fork forming stationary waves with 5 antinodes, between 2 bridges, when a mass of 9 kg is suspended from the wire. When the mass is replaced by another mass m, the wire with the same tuning fork forms three antinodes, for the same position of the bridges. The value of m is

Detailed Solution for JEE Main Mock- 4 - Question 16

where μ is the mass per unit length of wire and T = weight of mass(Mg) and p is number of antinodes.
Since wire and the tuning fork are the same in both cases, so

JEE Main Mock- 4 - Question 17

In a damped oscillator the amplitude of vibrations of mass m = 150 grams falls by 1/e times of its initial value in time t0 due to viscous forces. The time t0 and the percentage loss in mechanical energy during the above time interval t0 respectively are (Let damping constant be 50 grams/s)

Detailed Solution for JEE Main Mock- 4 - Question 17

JEE Main Mock- 4 - Question 18

A thermodynamic process of one mole ideal monoatomic gas is shown in figure. The efficiency of cyclic process ABCA will be :

Detailed Solution for JEE Main Mock- 4 - Question 18

Heat absorbed

JEE Main Mock- 4 - Question 19

An infinite current carrying wire, carrying current I is bent in V shape, lying in x-y plane as shown in figure. Intensity of magnetic field at point P will be (take OP = 2a)

Detailed Solution for JEE Main Mock- 4 - Question 19

B at P due to 1 wire

due to both wire = 2( is in same direction)

JEE Main Mock- 4 - Question 20

A silicon diode is connected in series with resistance and battery. What should be the value of battery if the reading of ammeter 3.62 A.

Detailed Solution for JEE Main Mock- 4 - Question 20

Rearranging the circuit,

As we know the potential drop across silicon diode in forward biased is 0.7V. Current flowing through the circuit is

*Answer can only contain numeric values
JEE Main Mock- 4 - Question 21

The relation between Internal energy U, pressure P and volume V of a gas in an adiabatic process is U = a + bPV where a = b = 3. The greatest integer of the ratio of specific heats [γ] is

Detailed Solution for JEE Main Mock- 4 - Question 21

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JEE Main Mock- 4 - Question 22

Number of teeth in front and rear sprocket (tooth wheel) of a bicycle is 40 and 20 respectively. If angular speed of front sprocket is 4 radian per second then find the angular speed of the rear sprocket Assume that spacing between the teeth is same for both sprocket.

Detailed Solution for JEE Main Mock- 4 - Question 22

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JEE Main Mock- 4 - Question 23

Consider the lens shown in fig with radii of curvature of the lens equal to 10 cm and 20 cm. Refractive index of the material of the lens is 1.5 and x axis is the principal axis of the lens. Find the magnification produced by lens for the object placed at a distance of 20 cm from the lens.

Detailed Solution for JEE Main Mock- 4 - Question 23

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JEE Main Mock- 4 - Question 24

Interference pattern is formed on a screen by Young's double slits S1 and S2 Illuminated by monochromatic light of wavelength λλ. Points P and P’ are on the screen closest to and on either side of central maxima, where intensity is one fourth that at the central maxima. If . Find x where x is the coefficient of d.

Detailed Solution for JEE Main Mock- 4 - Question 24

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JEE Main Mock- 4 - Question 25

When the radiation emitted by Li++ in transition from n = 4 to n = 3 is made to fall on a metal surface, the emitted photo electrons moves in a circular track of radius  cm when moved in a transverse magnetic field  Find work function of the metal (in eV). (Given that Rydberg's constant

Detailed Solution for JEE Main Mock- 4 - Question 25

JEE Main Mock- 4 - Question 26

Consider the following equilibrium
N2O4 (g) ⇔ 2NO2(g)
Then select the correct graph: (graph represent relation between concertation of NO2 and N2O4 at equilibrium)

Detailed Solution for JEE Main Mock- 4 - Question 26

Keq =
[NO2]2 = keq [N2O4]
It resembles graph of y2 = 4 ax
So option (b) is correct

JEE Main Mock- 4 - Question 27

If two different non-axial d-orbitals having XZ nodal plane form π bond by overlapping each other, then internuclear axis will be;

Detailed Solution for JEE Main Mock- 4 - Question 27

Two different non-axial d-orbitals will lie in planes perpendicular to each other, such d-orbital will not formed
Hence, (d) option is correct.

JEE Main Mock- 4 - Question 28

The given osazone can be formed by

Detailed Solution for JEE Main Mock- 4 - Question 28

Formation of osazone involves only C1 and C2 carbon atoms. The rest of the carbon atoms remains uneffected.
Glucose and fructose differ only in the configurations of first and second carbon atoms remaining positions are similar. So they form same osazone.

JEE Main Mock- 4 - Question 29

The cyclotrimetaphosphoric acid is

Detailed Solution for JEE Main Mock- 4 - Question 29

Hence, option (C ) is correct.

JEE Main Mock- 4 - Question 30

In a face centered cubic lattice of edge length 'a' number of next to next nearest neighbor to corner atom and distance of corner atom to next to next nearest atom is respectively-

Detailed Solution for JEE Main Mock- 4 - Question 30

Nearest neighbor
Next nearest neighbor = a
Next to next nearest neighbor

No of such neighbors = 24

JEE Main Mock- 4 - Question 31

CrO42(-) (yellow) changes to Cr2O72(-) (orange) in pH= x and Cr2O72(-) (orange) changes to CrO42(-) (yellow) in ph = y. Then x and y can be:

Detailed Solution for JEE Main Mock- 4 - Question 31

Hence, option (a) is correct.

JEE Main Mock- 4 - Question 32

Identify position most favorable for aromatic substitution (EAS)

Detailed Solution for JEE Main Mock- 4 - Question 32

is activating group so it would activate ortho and para position for E.A.S but only para position is
empty so EAS will take place there.
Hence, option (b) is correct

JEE Main Mock- 4 - Question 33

Polymer of prop – 2 – ene nitrile is-

Detailed Solution for JEE Main Mock- 4 - Question 33

Prop- 2 – ene nitrile

JEE Main Mock- 4 - Question 34

Molar mass of a substance, 1g of which when dissolved in 100g of water gave a solution whose boiling point is 100.1oC at a pressure of 1 atm (kb of water = 0.52 k kg mol–1) is-

Detailed Solution for JEE Main Mock- 4 - Question 34

The given values
WB = 1g kb = 0.5 2 k kg mol–1
WA = 100
Tb = 100. 1–100 = 0. 1
Using equation

JEE Main Mock- 4 - Question 35

Arrange the following in increasing order of their Pka values

Detailed Solution for JEE Main Mock- 4 - Question 35

Because acidic character depends on stability of conjugate ion.

3 resonating structure

2 resonating structure

Acidic character of x > y > z
So ka → x > y > z
Hence pka x < y < z

JEE Main Mock- 4 - Question 36

Reduction potential of following electrode Pt, H2 (4atm) | H2SO4 (0.01M) is-

Detailed Solution for JEE Main Mock- 4 - Question 36

H2 (4 atm) →2H(+) (0.01M) + 2e(-)

Hence option (b) is correct

JEE Main Mock- 4 - Question 37

If molecular weight of As2S3 is M. Then in the following reaction
As2S3 + 7NaClO3 + 12NaOH → 2Na3AsO4 + 7NaClO + 3Na2SO4 + 6H2O.
the equivalent weight of As2S3 is

Detailed Solution for JEE Main Mock- 4 - Question 37

Both As and S go under oxidation
As2S3→ AsO43(–) + 4e(–) (oxidation of As)
As2S3→ 3SO42(–) + 24e(–) (oxidation of S)
Total electrons = 24 × 4 = 28 = n-factor

JEE Main Mock- 4 - Question 38

Which will give only one product (excluding stereoisomer) when undergoing E2 reaction ?

Detailed Solution for JEE Main Mock- 4 - Question 38

Which are identical product
Hence option (d) is correct

JEE Main Mock- 4 - Question 39

Assume that a particular amino acid has isoelectric point as 6.0. In a solution at PH 1.0 which of the following will predominate

Detailed Solution for JEE Main Mock- 4 - Question 39

In acidic medium z witter ion

Hence option (a) is correct

JEE Main Mock- 4 - Question 40

C – 2 and C – 3 configuration of

is -

Detailed Solution for JEE Main Mock- 4 - Question 40

According to CIP rule
Clockwise is R configuration and Anti-clockwise is S configuration

Hence option (c) is correct

JEE Main Mock- 4 - Question 41

End product in the reaction is,

Detailed Solution for JEE Main Mock- 4 - Question 41

Step 1.
Acid base reaction

Step 2.

Hence (a) option is correct.

JEE Main Mock- 4 - Question 42

Which of the following does not occur in Bessemer’s converter

Detailed Solution for JEE Main Mock- 4 - Question 42

n Bessemer’s converter
2CuFeS2 + O2→Cu2S + 2 FeS + SO2 does not take place
Hence, option (c) is correct

JEE Main Mock- 4 - Question 43

For reaction CO(g) + NO2(g)  CO2(g) +NO(g) Energy profile diagram is given below.

What is the 'activation energy' of the reaction?

Detailed Solution for JEE Main Mock- 4 - Question 43

Activation energy is energy from reactant to highest energy barrier
So activation energy is ‘x’ hence A option is correct

JEE Main Mock- 4 - Question 44

Which of the following is buffer-

Detailed Solution for JEE Main Mock- 4 - Question 44

Buffer is conjugate acid and salt of conjugate base pair.
Hence option (C) is correct answer.

JEE Main Mock- 4 - Question 45

Nitrogen dioxide cannot be obtained from

Detailed Solution for JEE Main Mock- 4 - Question 45

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JEE Main Mock- 4 - Question 46

Find the value of ‘x’ in the tremolite asbestos Ca2Mgx (Si4O11)2 (OH)2

Detailed Solution for JEE Main Mock- 4 - Question 46

The silicate is a amphibole with general formula

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JEE Main Mock- 4 - Question 47

Ka for butyric acid is 2 x 10-5. What will be pH of 0.2 M aqueous solution of sodium butyrate?

Detailed Solution for JEE Main Mock- 4 - Question 47

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JEE Main Mock- 4 - Question 48

1.575 gm of oxalic acid (COOH)2. xH2O are dissolved in water and the volume made upto 250 ml. on titration 16.68 ml of this solution requires 25mL of N/15 NaOH solution for complete neutralization. Calculate x.

Detailed Solution for JEE Main Mock- 4 - Question 48

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JEE Main Mock- 4 - Question 49

Ethereal solution of how many of the following pairs of compounds can be separated by aqueous NaHCO3 solution but not by aqueous NaOH solution?

Detailed Solution for JEE Main Mock- 4 - Question 49

I and III

*Answer can only contain numeric values
JEE Main Mock- 4 - Question 50

The sum of coordination number and oxidation state of central metal ion in complex formed, when excess of KCN is added to aqueous solution of copper sulphate is

Detailed Solution for JEE Main Mock- 4 - Question 50

K3 [Cu(CN)4]

JEE Main Mock- 4 - Question 51

There are two balls in an urn whose colours are not known (each ball can be either white or black). A white ball is put into the urn. A ball is drawn from the urn. The probability that it is white is

Detailed Solution for JEE Main Mock- 4 - Question 51

Let E(0 < i < 2) denote the event that urn contains 'i' white and '(2 – i)' black balls.
Let A denote the event that a white ball is drawn from the urn.
We have P(Ei) = 1/3 for i = 0, 1, 2
P (A|E1) = 1/3, P(A|E2) = 2/3, P(A|E3) = 1.
By the total probability rule,
P(A)= P(E1)P(A|E1) + P(E2)P(A|E2) + P(E3)P(A|E3)

JEE Main Mock- 4 - Question 52

The plane passing through the point (−2, −2, 2) and containing the line joining the points
(1, 1, 1) and (1, −1, 2) makes intercepts on the coordinates axes then sum of the lengths of intercepts is

Detailed Solution for JEE Main Mock- 4 - Question 52

JEE Main Mock- 4 - Question 53

If a, b, c are in GP and  are in AP, then a, b, c are the lengths of the sides of a triangle which is

Detailed Solution for JEE Main Mock- 4 - Question 53

a, b, c are in GP b2 = ac
are in AP.
2(log 2b – log 3c) = log a – log 2b + log 3c – log a
log 2b = log 3c 2b = 3c
∴ b2 = ac & 2b = 3c
b = 2a/3 and c = 4a/9
Since (a+b) = 5a/3 > c, (b+c) = 10a/9 > a and
(c + a) = 13a/9 > b, therefore, a, b, c are the sides
of a triangle. Also, as ‘a’ is the greatest side, let us
find angle A of ΔABC

Hence ΔABC is an obtuse angled triangle.

JEE Main Mock- 4 - Question 54

Sum of the series
1 + 3 + 6 + 10 + 15 + ……………………….n terms is

Detailed Solution for JEE Main Mock- 4 - Question 54

JEE Main Mock- 4 - Question 55

The solution of the equation (2x + y + 1) dx + (4x + 2y – 1) dy = 0 is

Detailed Solution for JEE Main Mock- 4 - Question 55

Put 2x + y = X ⇒Therefore, the given equation is reduced to

= x + constant
2(2x + y) + log (2x + y – 1) = 3x + constant
x + 2y + log (2x + y – 1) = C

JEE Main Mock- 4 - Question 56

The area bounded by curves y = f(x), the x-axis and the ordinates x = 1 and x = b is (b - 1) sin (3b + 4). Then f(x) is-

Detailed Solution for JEE Main Mock- 4 - Question 56

1. Area bounded by curve y = f(x), x =1 and x = b is

Now differentiating both sides with respect to b we get
fB. = sin (3b + 4) + 3(b – 1) cos (3b + 4)

JEE Main Mock- 4 - Question 57

If sin 5x + sin 3 x + sin x = 0, then the value of x other than zero, lying between 0 < x < π/2 is

Detailed Solution for JEE Main Mock- 4 - Question 57

1. sin 5x + sin x + sin 3x = 0
2 sin 3x . cos2x + sin3x = 0
sin 3x (2 cos 2x + 1) = 0, 0 ≤ x ≤π/2

From both x = π / 3 (other than 0).

JEE Main Mock- 4 - Question 58

For n > 2 the product   where is equal to

Detailed Solution for JEE Main Mock- 4 - Question 58

Let  Hence the given series S is
S =

JEE Main Mock- 4 - Question 59

The value of K, for which the equation (K–2)x2 + 8x + K + 4 = 0 has both the roots real distinct and negative is:

Detailed Solution for JEE Main Mock- 4 - Question 59

(K–2)x2 + 8x + K + 4 = 0

Let
If the equation f(x) = 0, has real distinct and negative roots then:
D > 0 & f(0) > 0 &
(i) D > 0

64 – 4 (K2 + 2K – 8) > 0
– K2 – 2K + 24 < 0 K2 + 2K – 24 < 0
(K + 6)(K – 4) < 0
–6 < K < 4
(ii)

⇒ K < –4 or K > 2
(iii)

K – 2 > 0
K > 2
∴D > 0 & f(0) > 0 &

–6 < K < 4 & (K < –4 or K > 2) & K > 2
–6 < K < 4 & K < –4 & K > 2) or (K > 2 & K > 2
– 6 < K < 4 &
2 < K < 4
K∈(2, 4)
Among the given options
3 lies in the given range,
Hence K = 3

JEE Main Mock- 4 - Question 60

The equation of the common tangent touching the circle (x-3)2 + y2 = 9 and the parabola y2 = 4x above the x-axis, is

Detailed Solution for JEE Main Mock- 4 - Question 60

Equation of any tangent to parabola

If the tangent to parabola is also tangent to the circle (x-3)2 + y2 = 9 then:

9m4 + 6m2 + 1 = 9m4 + 9m2
3m2 – 1 = 0

From the figure it is evident that for the tangent above x-axis   Hence equation of required common tangent is:

JEE Main Mock- 4 - Question 61

Let f(x) be a continuous function such that f(a – x) + f(x) = 0 for all x[0,a]. Then, the value of the integral is equal to

Detailed Solution for JEE Main Mock- 4 - Question 61

Let

JEE Main Mock- 4 - Question 62

The circles which can be drawn to pass through (1,0) & (3,0) and touching the y-axis, intersect at an angle θ. The value of cos θ is equal to

Detailed Solution for JEE Main Mock- 4 - Question 62

Equation of line joining A (1,0) and B (3,0) is y = 0. Equation of family of circles passing through A and B is:
(x – 1)(x – 3) + (y – 0) (y – 0) + λy = 0
x2 + y2 – 4x + λy + 3 = 0
If above circle touches y-axis then x = 0 is a tangent. Substituting x = 0:
y2 + λy + 3 = 0
Discriminant of above quadratic must be zero.
λ2 – 12 = 0

Hence the circles are:

Thus, the coordinates of C1 and C2 arerespectively. Also the radii of each of the circles is r1 = r2 = 2
We know that the angle of intersection of two circles of radii r1 and r2 is given by
where d is the distance between their centers.

Since between two lines whenever there is angle θ there is always the angle π–θ, therefore,

JEE Main Mock- 4 - Question 63

a, b, c are positive numbers and abc2 has the greatest value 1/ 64. Then

Detailed Solution for JEE Main Mock- 4 - Question 63

We have

Also for the greatest value of abc2 the numbers have to be equal, i.e a = b = c/2
Also given that maximum value = 1/64
so, a + b + c = 1
i.e. a = b = 1/4, c = 1/2

JEE Main Mock- 4 - Question 64

If A and B are two square matrices such that B = –A–1 BA, then (A+B)2 is equal to

Detailed Solution for JEE Main Mock- 4 - Question 64

B = –A–1 BA
AB = A (–A–1BA)

AB = –BA
AB + BA = 0
Now, (A + B)2 = A2 + B2 + AB + BA
(A + B)2 = A2 + B2, (∵AB + BA = 0)

JEE Main Mock- 4 - Question 65

The set of all values of the parameter a for which the points of minimum of the function y = 1 + a2 x – x3
Satisfy the inequality

Detailed Solution for JEE Main Mock- 4 - Question 65

x2 + 5x + 6 < 0 –3 < x < –2
y = 1 + a2 x – x3

(i) Let a > 0:
Hence  Hence y has minima at  for a > 0.

(ii) Let a < 0:
Hence Hence y has minima at  for a < 0.

From (i) & (ii)

JEE Main Mock- 4 - Question 66

The value of  is

Detailed Solution for JEE Main Mock- 4 - Question 66

Substitute (x – 1) = t. Hence

Hence given limit is:

JEE Main Mock- 4 - Question 67

The function

Detailed Solution for JEE Main Mock- 4 - Question 67

f (x) is an even function.
NOTE: Integration of every even function is odd function and integration of every odd function is even function, provided the integrands are integrable and the limit is 0 to x. Similarly differentiation of every odd function is even function and that of every even function is odd function, provided the they are differentiable. Here  which is an odd function, therefore  is even function.

JEE Main Mock- 4 - Question 68

If for a variable line   the condition a–2 + b–2 = c–2 (c is a constant), is satisfied, then the locus of foot of the perpendicular drawn from origin to this is:

Detailed Solution for JEE Main Mock- 4 - Question 68

a–2 + b–2 = c–2

bx + ay – ab = 0
The foot of perpendicular to the above line from O(0,0) is given by:

Using a–2 + b–2 = c–2

x2 + y2 = c2

JEE Main Mock- 4 - Question 69

The eccentricity of the hyperbola whose latus rectum is half of its transverse axis, is

Detailed Solution for JEE Main Mock- 4 - Question 69

Length of latus rectum of hyperbola = Length of the transverse axis

JEE Main Mock- 4 - Question 70

A tangent having slope ofto the ellipse  intersects the major and minor axes in points A and B respectively. If C is the center of the ellipse then the area of the triangle ABC is:

Detailed Solution for JEE Main Mock- 4 - Question 70

Equation of the tangent to  whose slope

From symmetry of ellipse it is obvious that area of the triangle ABC will be same with respect either tangent.
Let us consider

Hence area Δ of the triangle ABC is:

*Answer can only contain numeric values
JEE Main Mock- 4 - Question 71

The number of solutions of the equation  is/are

Detailed Solution for JEE Main Mock- 4 - Question 71

Out of these two values of x, on;y x=1 satisfies the given equation

*Answer can only contain numeric values
JEE Main Mock- 4 - Question 72

If the circle (x - a)+ y2 = 25 intersects the circle x2 + (y - b)2 = 16 in such a way that common chord is of maximum length, then value of a2 + b2 is

Detailed Solution for JEE Main Mock- 4 - Question 72

Radial axis: - 2ax + a2 + 2by - b2 = 9
which passes through (0, b)
i.e. 0 + a2 + 2b2 – b2 = 9 ⇒ a2 + b2 = 9.

*Answer can only contain numeric values
JEE Main Mock- 4 - Question 73

Area bounded by curves = [cos A + cos B + cos C],   (where [.]) denotes the greatest integer function and A, B, C are angles of a triangle) and curve |x-1|+|y|=2 is

Detailed Solution for JEE Main Mock- 4 - Question 73

*Answer can only contain numeric values
JEE Main Mock- 4 - Question 74

If a tangent of slope 2 of the ellipse  is normal to the circle x2 + y2 + 4x + 1 = 0, then the maximum value of ais

Detailed Solution for JEE Main Mock- 4 - Question 74

*Answer can only contain numeric values
JEE Main Mock- 4 - Question 75

Let  then n is equal to ...

Detailed Solution for JEE Main Mock- 4 - Question 75

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## JEE Main & Advanced Mock Test Series

2 videos|324 docs|160 tests