JEE Main Mock Test - 12 - JEE MCQ

# JEE Main Mock Test - 12 - JEE MCQ

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## 75 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Main Mock Test - 12

JEE Main Mock Test - 12 for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2024 preparation. The JEE Main Mock Test - 12 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Mock Test - 12 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Mock Test - 12 below.
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JEE Main Mock Test - 12 - Question 1

### In an LCR circuit as shown below, both switches are open initially. Later, switch S1 is closed and S2 is kept open. (q is charge on the capacitor and τ = RC is capacitive time constant). Which of the following statements is correct?

Detailed Solution for JEE Main Mock Test - 12 - Question 1

Switch S1 is closed and switch S2 is kept open. Now, capacitor is charging through a resistor R.

Charge on a capacitor at any time t,

q = q0(1 - e-t/τ)

q = CV(1 - e-t/τ) [As q0 = CV]

At t = τ/2, q = CV(1 - e-τ/2τ) = CV(1 - e-1/2)

At t = τ, q = CV(1 - e-τ/τ) = CV(1 - e-1)

At t = 2τ, q = CV(1 - e-2τ/τ) = CV(1 - e-2)

JEE Main Mock Test - 12 - Question 2

### In an ac generator, a coil with N turns, all of the same area A and total resistance R, rotates with frequency ω in a magnetic field B. The maximum value of emf generated in the coil is

Detailed Solution for JEE Main Mock Test - 12 - Question 2

In an ac generator, emf induced, e = NABω sinωt

Hence, amplitude of induced emf or maximum emf = NABω

JEE Main Mock Test - 12 - Question 3

### A rectangular loop has a sliding connector PQ of length ℓ and resistance RΩ and is moving with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I1, I2 and I are

Detailed Solution for JEE Main Mock Test - 12 - Question 3
Emf induced across PQ is ε = Bℓv. The equivalent circuit diagram is as shown in the figure.

Applying Kirchhoff's first law at junction Q, we get I = I1 + I2 ... (i)

Applying Kirchhoff's second law for the closed loop PLMQP, we get

-I1R - IR + ε = 0

I1R + IR = Bv ... (ii)

Again, applying Kirchhoff's second law for the closed loop PONQP, we get

-I2R - IR + ε = 0

I2R + IR = Bℓv ... (iii)

Adding equations (ii) and (iii), we get

2IR + I1R + I2R = 2Bℓv

2IR + R(I1 + I2) = 2Bℓv

2IR + IR = 2Bℓv (Using (i))

3IR = 2Bℓv

Substituting this value of I in equation (ii), we get I1 = Bℓv/3R

Substituting the value of I in equation (iii), we get I2 = Bℓv/3R

JEE Main Mock Test - 12 - Question 4

Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown in the figure. The net electrostatic energy of the configuration is zero if Q is equal to

Detailed Solution for JEE Main Mock Test - 12 - Question 4
Since the hypotenuse of the triangle is √2a, the net electrostatic energy is

For U = 0, we require

Hence, option 2 is the correct answer.

JEE Main Mock Test - 12 - Question 5

A block is kept on a frictionless inclined surface with angle of inclination α. The incline is given an acceleration 'a' to keep the block stationary. Then, 'a' is equal to

Detailed Solution for JEE Main Mock Test - 12 - Question 5

The incline is given an acceleration a.

Acceleration of the block is to the right.

Pseudo acceleration 'a' acts on block to the left.

Equate resolved parts of 'a' and 'g' along the incline.

ma cos α = mg sin α

or, a = g tan α

JEE Main Mock Test - 12 - Question 6

The diameter of a drop of liquid fuel changes with time due to combustion according to the following relationship.

While burning, the drop falls at its terminal velocity under Stokes flow regime. Find the distance that it will travel before complete combustion.

Detailed Solution for JEE Main Mock Test - 12 - Question 6
Given, diameter of the liquid drop,

At complete combustion

D = 0

∴ t = tb

And at t = 0, D = D0

Thus, the initial diameter of the drop is D0 and changes according to given equation and the time for compelete combustion is tb.

From Stokes law, the velocity of drop,

Also

JEE Main Mock Test - 12 - Question 7

The half-life period of a radioactive element X is the same as the mean life time of another radioactive element Y. Initially, they have the same number of atoms. Then,

Detailed Solution for JEE Main Mock Test - 12 - Question 7
T1/2 (half life of X) = τY (mean life of Y)

⇒ λx = λY(0.693)

Y will decay faster than X.

JEE Main Mock Test - 12 - Question 8

An object is 1 metre in front of the curved surface of a plano-convex lens whose flat surface is silvered. A real image is formed 120 cm in front of the lens. What is the focal length of the lens?

Detailed Solution for JEE Main Mock Test - 12 - Question 8

Here,

u = 100 cm

v = 120 cm

Here,

FL - focal length of the lens

FM - focal length of the mirror

JEE Main Mock Test - 12 - Question 9

Consider a two particle system with particles having masses m1 and m2. The first particle is pushed towards the centre of mass through a distance d. By what distance should the second particle be moved, so as to keep the centre of mass at the same position?

Detailed Solution for JEE Main Mock Test - 12 - Question 9
Let m2 be moved by x, so as to keep the centre of mass at the same position.

∴ m1d + m2 (- x) = 0

or, m1d = m2x

JEE Main Mock Test - 12 - Question 10

An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V1 and contains ideal gas at pressure P1 and temperature T1. The other chamber has volume V2 and contains ideal gas at pressure P2 and temperature T2. If the partition is removed without doing any work on the gas, then the final equilibrium temperature of the gas in the container will be

Detailed Solution for JEE Main Mock Test - 12 - Question 10

As this is a simple mixing of gas, PV = nRT for adiabatic as well as isothermal changes.

The total number of molecules is conserved.

Final state = (n1 + n2)RT

JEE Main Mock Test - 12 - Question 11

In a hydrogen atom, the electron moves around the nucleus in a circular orbit of radius 5 × 10−11 m. Its time period is 1.5 × 10−16s. The current associated with the electron motion is (charge of electron is 1.6 × 10−19 C)

Detailed Solution for JEE Main Mock Test - 12 - Question 11

Time taken by charge to cross a point on the orbit = 1.5 × 10−16 s

JEE Main Mock Test - 12 - Question 12

A cubical vessel open from top of side L is filled with a liquid of density ρ then the torque of hydrostatic force on a side wall about an axis passing through one of bottom edges is-

Detailed Solution for JEE Main Mock Test - 12 - Question 12
Force by hydrostatic pressure

and centre of pressure is at height L/3.

JEE Main Mock Test - 12 - Question 13

A certain ideal gas undergoes a polytropic process PVn = constant such that the molar specific heat during the process is negative. If the ratio of the specific heats of the gas be γ, then the range of values of n will be

Detailed Solution for JEE Main Mock Test - 12 - Question 13
The molar specific heat capacity in a polytropic process PVn = constant is given by

From this equation, we see that C will be negative when n < γ="" n="" />< γ="" />

n > 1, simultaneously, i.e., 1 < n="" />< γ.="" since="" γ="" for="" all="" ideal="" gases="" is="" greater="" than="" 1,="" if="" n="" /> γ or n < 1,="" then="" />v will be positive.

JEE Main Mock Test - 12 - Question 14

A ball is projected from the bottom of an inclined plane of inclination 30o, with a velocity of 30 m s−1, at an angle of 30o with the inclined plane. If g = 10 ms−2, then the range of the ball on given inclined plane is

Detailed Solution for JEE Main Mock Test - 12 - Question 14

Range on incline plane,

JEE Main Mock Test - 12 - Question 15

A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be

Detailed Solution for JEE Main Mock Test - 12 - Question 15
Frequency of LC oscillation

JEE Main Mock Test - 12 - Question 16

If the radius of the earth were to shrink by one percent and its mass remains the same, the acceleration due to gravity on the earth's surface would

Detailed Solution for JEE Main Mock Test - 12 - Question 16

g will increase if R decreases.

JEE Main Mock Test - 12 - Question 17

A black body, at 200 K, is found to have maximum energy at a wavelength of 14 μm. When its temperature is raised to 1000 K, the wavelength at which maximum energy is emitted is

Detailed Solution for JEE Main Mock Test - 12 - Question 17

JEE Main Mock Test - 12 - Question 18

The time constant of charging of the capacitor shown in the diagram is

Detailed Solution for JEE Main Mock Test - 12 - Question 18
The given circuit is equivalent to two simple circuits connected in parallel with the battery as shown below.

τc = 2RC

JEE Main Mock Test - 12 - Question 19

Spherical wavefronts shown in figure, strike a plane mirror. Reflected wavefronts will be as shown in

Detailed Solution for JEE Main Mock Test - 12 - Question 19
According to Huygens principle, ray of light will be normal to the wavefront.

JEE Main Mock Test - 12 - Question 20

A simple pendulum has a time period T1 when on the earth's surface and T2 when taken to a height 2R above the earth's surface where R is the radius of the earth. The value of

Detailed Solution for JEE Main Mock Test - 12 - Question 20
The periodic time of a simple pendulum is given by,

When taken to height 2R,

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 21

The magnetic field due to a current-carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 μT. Its value at the centre of the loop is CμT. What is the value of C?

Detailed Solution for JEE Main Mock Test - 12 - Question 21
Magnetic field at any point on the axis of coil,

At the centre of coil,

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 22

In a potentiometer experiment, the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of 2Ω, the balancing length becomes 120 cm. The internal resistance of the cell, in ohm, is

Detailed Solution for JEE Main Mock Test - 12 - Question 22
The internal resistance of a cell is given by

Hence:

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 23

The height, in terms of R (radius of Earth), at which the acceleration due to gravity becomes g/9 (where g = acceleration due to gravity on the surface of Earth), is nR. What is the value of n?

Detailed Solution for JEE Main Mock Test - 12 - Question 23
The acceleration due to gravity at a height h from the ground is given as g/9.

At the surface of Earth, the acceleration due to gravity is

At r = R + h:

Acceleration due to gravity

∴ r = 3R

The height above the ground = h = r - R = 2R

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 24

A particle executes S.H.M. given by x = 0.24cos(400t − 0.5) in SI units. Find amplitude in cmcm.

Detailed Solution for JEE Main Mock Test - 12 - Question 24
Here, x = 0.24 cos (400 t - 0.5) ...(i)

The standard equation for S.H.M. is

x = r cos(2 π v t − ϕ)…(ii)⁡

Comparing the equations (i) and (ii),we have

r = 0.24 m

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 25

The apparent depth of needle lying at the bottom of the tank which is filled with water to a height of 15.5 cm is measured by a microscope to be 8.5 cm. If water is replaced by a liquid of refractive index 1.94 to the same height as earlier, then the displacement of the microscope needed to establish the focus on the needle again is n μmn μm. The value of n is

Detailed Solution for JEE Main Mock Test - 12 - Question 25

Actual depth of the needle in water,

h1 = 15.5 cm

Apparent depth, H = 15.5/1.94 = 7.99 cm

Here, H is less than h2. Thus to focus the needle again, the microscope should be moved up by the distance 8.5 − 7.99 = 0.51 cm

JEE Main Mock Test - 12 - Question 26

The radiations from a naturally occurring radioactive substance, as seen after deflection by a magnetic field in one direction, are

Detailed Solution for JEE Main Mock Test - 12 - Question 26
Deflection in the magnetic field confirms that the particles are charged. It does not confirm the charge on the particles.

Hence, option (4) is correct.

α-rays consist of positively charged particles (He++) and β-rays consist of negatively charged particles . Since they are oppositely charged, so they get deflected in opposite directions. γ-rays are neutral (carry no charge). So, they remain undeflected.

JEE Main Mock Test - 12 - Question 27

Both the molecules possess dipole moment in which of the following pairs?

Detailed Solution for JEE Main Mock Test - 12 - Question 27
H2O and SO2 both possess dipole moment due to bent structure.

JEE Main Mock Test - 12 - Question 28

Which of the following solutions will have pH close to 1.0?

Detailed Solution for JEE Main Mock Test - 12 - Question 28
In option (4),

pH = 1

JEE Main Mock Test - 12 - Question 29

Out of H2S2O3, H2S4O6, H2SO5 and H2S2O8, peroxy acids are

Detailed Solution for JEE Main Mock Test - 12 - Question 29
A peroxy acid (sometimes called peracid) is an acid which contains an acidic -OOH group. H2SO5 and H2S2O8 are peroxy acids as shown in their structures.

JEE Main Mock Test - 12 - Question 30

Geometrical shapes of the complexes formed by the reaction of Ni2+ with Cl-, CN- and H2O, respectively, are

Detailed Solution for JEE Main Mock Test - 12 - Question 30
Ni2+ + 4Cl- → [NiCl4]2- - tetrahedral due to sp3 hybridisation.

Ni2+ + 4CN- → [Ni(CN)4]2- - dsp2 hybridisation, it is square planar.

Ni2+ + 6H2O → [Ni(H2O)6]2+ - sp3d2 hybridisation, it is octahedral.

JEE Main Mock Test - 12 - Question 31

Propan-1-ol can be prepared from propene by

Detailed Solution for JEE Main Mock Test - 12 - Question 31
Propan-1-ol can be prepared from propene by anti-Markownikoff addition as shown below

(The above reaction follows Markownikoff's rule and 2-propanol will be obtained instead of 1-propanol.)

JEE Main Mock Test - 12 - Question 32

How many structures of F are possible?

Detailed Solution for JEE Main Mock Test - 12 - Question 32

Thus, we find 3 possible structures of F.

JEE Main Mock Test - 12 - Question 33

The order of acidity of the following compounds is

Detailed Solution for JEE Main Mock Test - 12 - Question 33
Presence of OH groups, ortho to COOH, increases the acidity of the compound. The main reason being the stabilisation of carboxylate ion through hydrogen bonding with OH group. Presence of OH group at p-position reduces acidity. Thus, the correct order should be IV > II > I > III.

This is the correct order.

JEE Main Mock Test - 12 - Question 34

In the following reaction, predict the compounds X and Y.

Detailed Solution for JEE Main Mock Test - 12 - Question 34
The complete reaction is as follows

JEE Main Mock Test - 12 - Question 35

Which of the following does not exist as a zwitter ion?

Detailed Solution for JEE Main Mock Test - 12 - Question 35
The lone pair of electrons on the -NH2 group is donated towards the benzene ring due to resonance effect. As a result, acidic character of -COOH group and basic character of -NH2 group decrease. Therefore, the weakly acidic -COOH group cannot transfer an H+ ion to the weakly basic -NH2 group. Thus, o- and p-aminobenzoic acids do not exist as zwitter ions.
JEE Main Mock Test - 12 - Question 36

XeF6 on partial hydrolysis with water, produces a compound X. The same compound X is formed when XeF6 reacts with silica. The compound X is:

Detailed Solution for JEE Main Mock Test - 12 - Question 36
During the hydrolysis the oxidation of xenon not changes in xenon hexafluoride.

JEE Main Mock Test - 12 - Question 37

How many EDTA (ethylenediaminetetraacetate) unit(s) are required to make an octahedral complex with a Ca2+ ion?

Detailed Solution for JEE Main Mock Test - 12 - Question 37
As EDTA is hexadentate ligand. Hence, only one EDT (ethylenediamine tetraacetate) molecule is required to make an octahedral complex with Ca2+.

JEE Main Mock Test - 12 - Question 38

Electrometallurgical process is used to extract

Detailed Solution for JEE Main Mock Test - 12 - Question 38
Because Na is very reactive and can not be extracted by means of the reduction by C, CO etc. So, it is extracted by electrolysis of molten NaCl Solution.
JEE Main Mock Test - 12 - Question 39

For one mole of a Van der Waals gas, when b = 0 and T = 300 K, the PV vs. 1/V plot is shown below. The value of the Van der Waals constant a (atm litre2 mol-2) is:

Detailed Solution for JEE Main Mock Test - 12 - Question 39

Van der Waals equation for 1 mole of real gas is

Given that b = 0

Following y = mx + c for the curve PV vs 1/v

Slope = −a

∴ a = 1.5

JEE Main Mock Test - 12 - Question 40

Cr2+ and Mn3+ both have d4 configuration. Thus

Detailed Solution for JEE Main Mock Test - 12 - Question 40

Extra stability is gained when Mn3+ is reduced to Mn2+ and is thus an oxidizing agent.

JEE Main Mock Test - 12 - Question 41

Three elements X, Y, and Z have atomic numbers 19, 37, and 55 respectively. Then the correct statements(s) is/are

Detailed Solution for JEE Main Mock Test - 12 - Question 41
The elements X (19), Y (37) and Z (55) are the elements of same group (IA). As atomic number, increases, the ionization potential decrease down the group. Since, the position of Y is in between X and Z. Thus, the ionization potential of Y will also be in between X and Z.
JEE Main Mock Test - 12 - Question 42

An electron in an atom jumps in such a way that its kinetic energy changes from x to x/9. The change in its potential energy (magnitude) will be-

Detailed Solution for JEE Main Mock Test - 12 - Question 42
Change in kinetic energy is

For Bohr model of atom

The kinetic energy in a state is equal to half of the potential energy in magnitude.

∴ Potential energy

= 2 Kinetic energy

JEE Main Mock Test - 12 - Question 43

The amount (in grams) of sucrose (mol wt. = 342 g) that should be dissolved in 100 g water in order to produce a solution with a 105.0 oC difference between the freezing point and boiling point is

(Given that Kf = 1.86 K kg mol−1 and Kb = 0.51 K kg mol−1 for water)

Detailed Solution for JEE Main Mock Test - 12 - Question 43
Boiling point (Tb) = 100 + ΔTb = 100 + kbm

Freezing point (Tf) = 0 − ΔTf = −kfm

Tb − Tf = (100 + kbm) − (−kfm)

105 = 100 + 0.51 m + 1.86 m

∴ Weight of sucrose to be dissolved in 100 g water

JEE Main Mock Test - 12 - Question 44

The rate of decomposition for methyl nitrite and ethyl nitrite can be given in terms of rate constant k1 and k2 respectively. The energy of activation for the two reactions are 152.30 kJ mol−1 and 157.7 kJ mol−1 as well as frequency factors are 1013 and 1014 respectively for the decomposition of methyl and ethyl nitrite. Calculate the temperature at which rate constant will be the same for the two reactions.

Detailed Solution for JEE Main Mock Test - 12 - Question 44

For methyl nitrite k1 = 1013 e[−152300/(8.314×T)]

For ethyl nitrite k2 = 1014 e[−157700/(8.314×T)]

If k1 = k2 then

1013 e[−152300/(8.314×T)] = 1014 e[−157700/(8.314×T)]

T = 282 K

JEE Main Mock Test - 12 - Question 45

Choose from the indicated protons, the one that is most acidic.

Detailed Solution for JEE Main Mock Test - 12 - Question 45

Deprotonation of hydrogen labelled 44 produces a conjugate base which has more stable resonating structure and a resonance structure in which negative charge is present on the electronegative oxygen atom. Therefore, hydrogen labelled 44 is most acidic.

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 46

The relative lowering of vapour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molar mass of the substance in grams will be

Detailed Solution for JEE Main Mock Test - 12 - Question 46

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 47

Degree of dissociation of 0.1 N CH3COOH is 10-x. The value of x is(Given: Ka = 1 × 10-5)

Detailed Solution for JEE Main Mock Test - 12 - Question 47

Therefore, x = 2

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 48

The pH of a 0.01 M solution of a weak acid having degree of dissociation 12.5% is

(Nearest integer)

Detailed Solution for JEE Main Mock Test - 12 - Question 48
[H+] = Molarity × Degree of dissociation

We know,

pH = -log [H+]

= -log (1.25 × 10-3)

= -log 1.25 + 3

= 3 - 0.0970

= 2.9030

≈ 3

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 49

The number of chiral carbon centres in penicillin is _________.

Detailed Solution for JEE Main Mock Test - 12 - Question 49

Star marked atoms are chiral centers.

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 50

In borax number of B−O−B bonds are XX and number of boron atoms are Y. Then X+YX+Y is

Detailed Solution for JEE Main Mock Test - 12 - Question 50
Borax is a powdery white substance, also known as sodium borate, sodium tetraborate, or disodium tetraborate. It's widely used as a household cleaner and a booster for laundry detergent. It's a combination of boron, sodium, and oxygen. Boric acid is made from the same chemical compound as borax and even looks like it

X = 5

Y = 4

X + Y = 9

JEE Main Mock Test - 12 - Question 51

Two finite sets have m and n elements. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. The values of m and n, respectively, are

Detailed Solution for JEE Main Mock Test - 12 - Question 51
Total number of subsets of a set of m elements = 2m

Total number of subsets of a set of n elements = 2n

According to the question,

2m = 2n + 56

2m - 2n = 56

2n(2m - n - 1) = 56

2n(2m - n - 1) = 8 × 7

2n(2m - n - 1) = 23(23 - 1)

n = 3, m - n = 3

m = 6, n = 3

JEE Main Mock Test - 12 - Question 52

For complex numbers z1 = x1 + iy1 and z2 = x2 + iy2, (where i = √-1), we write z1 ∩ z2 for x1 ≤ x2 and y1 ≤ y2, then for all complex numbers z with 1 ∩ z, we have

Detailed Solution for JEE Main Mock Test - 12 - Question 52
As given for z1= x1 + iy1 and z2 = x2 +iy2, the symbol z1 ∩ z2 is used if x1 ≤ x2. Therefore for 1 ∩ z,we have 1 ≤ x and 0 ≤ y as 1 = 1 + 0.i and z = x + iy.

JEE Main Mock Test - 12 - Question 53

In a plane, there are 37 straight lines, out of which 13 pass through the point A and 11 pass through the point B. Besides, no three lines pass through the same point, no line passes through both points A and B, and no two lines are parallel. The number of intersection points the lines have is equal to

Detailed Solution for JEE Main Mock Test - 12 - Question 53

The number of points of intersection of 37 lines is 37C2.

But 13 straight lines out of the given 37 straight lines pass through the same point A.

Therefore instead of getting 13C2 points, we get only one point A. Similarly 11 straight lines out of the given 37 straight lines intersect at point B.

Therefore instead of getting 11C2 points, we get only one point B.

So, the number of intersection points = 37C2 - 13C2 - 11C2 + 2 = 535 (2 is added because of points A and B)

JEE Main Mock Test - 12 - Question 54

In a sequence of (4n + 1) terms, the first (2n + 1) terms are in AP, whose common difference is 2, and the last (2n + 1) terms are in GP, whose common ratio is 0.5. If the middle terms of the AP and GP are equal, then the middle term of the sequence is

Detailed Solution for JEE Main Mock Test - 12 - Question 54

(4n + 1) term are

Middle term of AP = (n + 1)th term

According to the question

∴ Middle term of the sequence

JEE Main Mock Test - 12 - Question 55

then the global maximum and local minimum values of f(x) for x ∈ [-2, 2], respectively, are

Detailed Solution for JEE Main Mock Test - 12 - Question 55

Since the given function is continuous but not differentiable at x = 0,

⇒ x = 0 is the point of local minima and f(0) = 1.

Also, there is no other critical point.

Now, f(-2) = 11, f(2) = 4 + cos 2

On comparing values of function at x = -2, 0, 2:

Global maximum occurs at x = -2 ⇒ f(-2) = 11

Hence, global maximum and local minimum values are 11 and 1, respectively.

JEE Main Mock Test - 12 - Question 56

The area of the triangle formed by the tangent and the normal to the parabola y2 = 4ax, both drawn at the same end of the latus rectum and the axis of the parabola, is

Detailed Solution for JEE Main Mock Test - 12 - Question 56

An end of the latus rectum = (a, 2a)

The equation of the tangent at (a, 2a) is: y(2a) = 2a(x + a), i.e. y = x + a

The normal at (a, 2a) is: y + x = 2a + a, i.e. y + x = 3a

Solving y = 0 and y = x + a, we get

x = -a, y = 0

Solving y = 0 and y + x = 3a, we get

x = 3a, y = 0

The area of the triangle with vertices (a, 2a), (-a, 0), (3a, 0) is: (1/2)(4a)(2a) = 4a2

JEE Main Mock Test - 12 - Question 57

is equal to

Detailed Solution for JEE Main Mock Test - 12 - Question 57

JEE Main Mock Test - 12 - Question 58

The lines are coplanar, if the value of λ is

Detailed Solution for JEE Main Mock Test - 12 - Question 58

2(8 + 3λ) + (4 + λ2) + 2(3 - 2λ) = 0

λ2 + 2λ + 26 = 0

D = b2 - 4ac

D = 22 - 4 × 26 = -100

which does not give any real value of λ.

JEE Main Mock Test - 12 - Question 59

The mean of n items is If these n items are successively increased by 2, 22, 23, …, 2n, then the new mean is

Detailed Solution for JEE Main Mock Test - 12 - Question 59

New mean

JEE Main Mock Test - 12 - Question 60

If are three non-coplanar vectors and are vectors defined by the relations then the value of

Detailed Solution for JEE Main Mock Test - 12 - Question 60

∴ Given expression is 1 + 1 + 1 = 3

JEE Main Mock Test - 12 - Question 61

The fourth term of equal to 200, then the value of x satisfying this is

Detailed Solution for JEE Main Mock Test - 12 - Question 61
Since, fourth term of

Taking logarithm on both side

JEE Main Mock Test - 12 - Question 62

The value of

Detailed Solution for JEE Main Mock Test - 12 - Question 62

JEE Main Mock Test - 12 - Question 63

If ln(x + y) = 2xy, then y'(0) is equal to

Detailed Solution for JEE Main Mock Test - 12 - Question 63
Given equation is ln(x + y) = 2xy …(1)

ln x + y = 2xy …1

For x = 0

ln(0 + y) = 2.0.y = 0

⇒ lny = 0

⇒ y = 1

Now, differentiating (1), we get,

At point (0,1), we get,

JEE Main Mock Test - 12 - Question 64

The value of

Detailed Solution for JEE Main Mock Test - 12 - Question 64
Given limit can be written as,

Using L'Hospital' rule,

JEE Main Mock Test - 12 - Question 65

Two vertices of a triangle are (3,−2) and (−2, 3) and its orthocentre is (−6, 1). The coordinates of its third vertex are-

Detailed Solution for JEE Main Mock Test - 12 - Question 65
Let the third vertex be A(α,β)

Using the diagram, OA⊥BC

⇒ Slope of OA × Slope BC = −1

Solving Equations(i)i and (ii)ii, we get

α = −1, β = 6

∴ The third vertex is (−1, 6)

JEE Main Mock Test - 12 - Question 66

The equation of a circle C1 is x2 + y2 − 4x − 2y − 11 = 0. Another circle C2 of radius 1 unit rolls on the outer surface of the circle C1. Then the equation of the locus of the centre of C2 is

Detailed Solution for JEE Main Mock Test - 12 - Question 66
The centre and radius of a circle x2 + y2 + 2gx + 2fy + c = 0 are (−g, −f) and

Hence, the centre of x2 + y2 − 4x −2y − 11 = 0 is A(2, 1) and the radius

If P(α, β) be the centre of C2 of radius r2 = 1

We know that, if two circles with centres A and P and radii r1 and r2 touches each other externally, then distance between their centres AP is equal to the sum of their radii i.e. AP = r1 + r2

The locus is obtained by replacing (α, β) by (x, y)

Hence, the locus is x2 + y2 − 4x − 2y − 20 = 0

JEE Main Mock Test - 12 - Question 67

The number of real values of k for which the lines and are intersecting is

Detailed Solution for JEE Main Mock Test - 12 - Question 67
Any point on the first line is (4r+k, 2r+1,r−1), and any point on the second line is (r′+k+1 ,−r′,2r′+1) for some values of r and r'. The lines are intersecting if these two points coincide i.e

4r + k = r′ + k + 1, 2r + 1 = −r′, r − 1 = 2r′ + 1 for some r and r'

⇒ 4r − r′ = 1, 2r + r′ = −1, r − 2r′ = 2

Now, 4r − r′ = 1, 2r + r′ = −1 ⇒ r = 0, r′ = −1 which satisfy r − 2 r′ = 2.

⇒ The given lines are intersecting for all real values of k.

JEE Main Mock Test - 12 - Question 68

Consider set A = {1, 2, 3}. Number of symmetric relations that can be defined on A containing the ordered pair (1, 2) and (2, 1) is

Detailed Solution for JEE Main Mock Test - 12 - Question 68
n(A × A) = 9

Number of relations containing (1, 2), (2, 1), (a, a) = 23

Number of relations containing (1, 2), (2, 1), (1, 3), (3, 1), (a, a) = 23

Number of relations containing (1, 2), (2, 1), (2, 3), (3, 2), (a, a) = 23

Number of relations containing (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2), (a, a) = 23

⇒ Total number of symmetric relations = 32.

JEE Main Mock Test - 12 - Question 69

The negation of the statement p→(q∧r) is

Detailed Solution for JEE Main Mock Test - 12 - Question 69
As we know, ~(p → q) = p ∧ ~q

Hence, ~(p→(q ∧ r)) = p ∧(~(q ∧ r))

⇒~(p → (q ∧ r)) = p∧(~q ∨ ~r)

JEE Main Mock Test - 12 - Question 70

is equal to

Detailed Solution for JEE Main Mock Test - 12 - Question 70

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 71

If the system of equations

kx + y + 2z = 1

3x − y − 2z = 2

−2x − 2y − 4z = 3

has infinitely many solutions, then kk is equal to ______ .

Detailed Solution for JEE Main Mock Test - 12 - Question 71
If the system of equations

P1: kx + y + 2z = 1

P2: 3x − y − 2z = 2

P3: −2x − 2y − 4z = 3

Which are equivalent to equations of planes.

We observe 5P2 − P1 = 3P3

So, 15 − K = −6

⇒ K = 21

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 72

If f(x) = cos|x| − 2ax + b is a function which increases for all x, then the maximum value of 2a + 1 is

Detailed Solution for JEE Main Mock Test - 12 - Question 72
f(x) = cos|x| − 2ax+b, f is differentiable everywhere.

f′(x) = −sin x − 2a,

f′(x) ≥ 0 ∀x

⇒ −sinx − 2a ≥ 0

⇒ sin x ≤ − 2a

∵ f increases for all x

∴ −2a must have least value greater than equal to maximum value of sin x.

⇒ −2a ≥ 1

⇒ 2a + 1 ≤ 0

Hence, the maximum value of 2a + 1 = 0

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 73

There are 5 letters and 5 directed envelopes. The number of ways in which all the letters be put in a wrong envelope is _______.(Nearest Integer)

Detailed Solution for JEE Main Mock Test - 12 - Question 73
Derangement theorem state that, the number of ways, if 'n' particular items need to be arranged or organized in a row in such a manner that none of them acquires its usual or correct place is

So, the required number of ways

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 74

The sum of first 9 terms of the series

Detailed Solution for JEE Main Mock Test - 12 - Question 74

*Answer can only contain numeric values
JEE Main Mock Test - 12 - Question 75

If one of the lines of my2 + (1 - m2)xy - mx2 = 0 is a bisector of the angle between the lines xy = 0, then m(m > 0) is

Detailed Solution for JEE Main Mock Test - 12 - Question 75
Equation of bisectors of lines, xy = 0 are y = ± x

Putting y = ± x in my2 + (1 - m2)xy - mx2 = 0, we get

(1 - m2)x2 = 0

⇒ m = ±1

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