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JEE Advanced Mock Test - 6 (Paper I) - JEE MCQ


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54 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Advanced Mock Test - 6 (Paper I)

JEE Advanced Mock Test - 6 (Paper I) for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2024 preparation. The JEE Advanced Mock Test - 6 (Paper I) questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Mock Test - 6 (Paper I) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Mock Test - 6 (Paper I) below.
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JEE Advanced Mock Test - 6 (Paper I) - Question 1

Let there be a spherical symmetric charge distribution with charge density varying as ρ(r) = up to r = R and ρ(r) = 0 for r > R, where 'r' is the distance from origin. The electric field at a distance r (< r)="" from="" the="" origin="" is="" given="" />

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 1
Let dq be the small charge element on the spherical surface of radius r (< r)="" having="" thickness="" />

Hence, electric field can calculated using Gauss’s law.

JEE Advanced Mock Test - 6 (Paper I) - Question 2

A block of mass √3/10 kg is placed on a rough horizontal surface as shown in the figure.

A force of 1 N is applied at one end of the block, and the block remains stationary. The normal force exerted by the surface on block acts through (g = 10 m/s2)

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 2
For equilibrium, resultant force = 0 and resultant torque = 0

For translational equilibrium,

F cos 60o = f

F sin 60o + N = Mg

For rotational equilibrium,

Torque about point A = 0

Mg x 10 = N x

JEE Advanced Mock Test - 6 (Paper I) - Question 3

Three resistances each of 4 Ω are connected to form a triangle. The resistance between any two terminals is

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 3
The two resistance are connected in series and the resultant is connected in parallel with the third resistance.

JEE Advanced Mock Test - 6 (Paper I) - Question 4

A ray of light is incident on the interface between water and glass at an angle i and refracted parallel at the water surface, then value of μg will be

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 4

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 5

A particle of mass m moves on the x-axis as follows:

It starts from rest at t = 0 from the point x = 0, and comes to rest at t = 1 at the point x = 1. No other information is available about its motion at intermediate times (0 < t < 1). If denotes the instantaneous acceleration of the particle, then

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 5
Since the body is at rest at x = 0 and x = 1, α cannot be positive for all time in the interval 0 ≤ t ≤ 1.

Therefore, first the particle is accelerated and then retarded.

Now, total time t = 1 s (given)

Total displacement s = 1 m (given)

s = Area under v-t graph

∴ Height or vmax = 2s/t = 2 m/s is also fixed.

If height and base are fixed, area is also fixed.

In case 2: Acceleration = Retardation = 4 m/s2 (slope of the graph)

In case 1: Acceleration > 4 m/s2, while retardation < 4="" />2

In case 3: Acceleration < 4="" />2 and Retardation > 4 m/s2

Hence, |α| ≥ 4 at some point(s) in its path.

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 6

A simple pendulum of length L and mass (bob) M is oscillating in a plane about a vertical line between angular limits -ϕ and +ϕ. For an angular displacement θ (|θ| < ϕ), the tension in the string and the velocity of the bob are T and v, respectively. The relation(s) that hold(s) good under the above conditions is/are:

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 6
Motion of pendulum is part of a circular motion. In circular motion, resolving of forces occurs in two perpendicular directions; first along radius (towards centre) and second along tangent.

Along radius, Centripetal force,

Along tangent, Resultant force = maT, where aT is the tangential acceleration.

Mg sinθ = MaT

aT = g sinθ

∴ The correct options are (2) and (3).

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 7

A uniform bar of length 6a and mass 8m lies on a smooth horizontal table. Two point masses m and 2m moving in the same horizontal plane with speed 2 v and v, respectively, strike the bar [as shown in the figure] and stick to the bar after collision. Denoting angular velocity (about the centre of mass), total energy and centre of mass velocity respectively by ω, E and vc, we have after collision:

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 7
Using conservation of linear momentum. we have

Initial momentum = Final momentum

pi = mi(2v) - 2m(v) = ⇒ 0 pf = 0 hence velocity of the centre of mass of the combined system after collision is zero.

Vcm = 0

Using conservation of the angular momentum. we have

Initial angular momentum = Final angular momentum

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 8

A solid sphere of uniform density and radius 4 units is located with its centre at the origin O of coordinates. Two spheres of equal radii 1 unit, with their centres respectively at A (-2, 0, 0) and B (2, 0, 0), are taken out of the solid leaving behind spherical cavities as shown in the figure. Then,

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 8
Due to the solid sphere of a uniform density, gravitational field is zero at the centre of the sphere. The cavities A and B can be treated as negative masses and they are situated on opposite sides of the centre O. Hence, gravitational forces exerted by the cavity masses on a mass at O are opposite in direction. Hence, resultant force on mass at O is zero. Thus, gravitational force due to this object at the origin O is zero.

Consider a circle y2 + z2 = 36; the centre of the circle is (0,0,0) and the radius of the circle is 6 units. The circle lies in y-z plane. For the point outside the sphere or situated at the sphere, the mass of the sphere can be assumed to be situated at the centre. All the points of the circle y2 + z2 = 36 are equidistant from the centre; hence, gravitational potential is same at all the points in the given circle.

Consider the circle y2 + z2 = 4; its centre lies at (0, 0, 0) and the radius is 2 units. It lies in y-z plane perpendicular to x-axis. All the points of the circle y2 + z2 = 4 are equidistant from the centre; hence, gravitational potential is same at all the points in the given circle.

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 9

Let [ε0] denote the dimensional formula of the permittivity of the vacuum, and [μ0] that of the permeability of the vacuum. If M = mass, L = Length, T = time and I = electric current. Choose the correct option.

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 9

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 10

Velocity & acceleration vector of a charged particle moving in a magnetic field at some instant are Select the correct alternatives -

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 10

x = −1.5

B is perpendicular to plane of a & velocity

∴ B is along Z direction

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 11

Two inductors having self inductances L1, L2 and mutual inductance M are arranged in parallel combination as shown in the diagram. The sense of the helix is same for both the inductors.

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 11
The effective circuit is shown in the diagram.

Applying Kirchhoff's loop law equations, we get

Solving these we get,

Solving these we get,

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 12

In the V−T graph shown in figure

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 12
Volume of the gas is increasing, therefore work done will be positive.

Temperature of the gas is decreasing, therefore internal energy of the gas is decreasing.

From the given information we cannot conclude whether Q is positive or negative. Because W is positive and ΔU is negative.

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper I) - Question 13

To find the distance d over which a signal can be seen clearly in foggy conditions, a railway engineer uses dimensions and assumes that the distance depends on the mass density p of the fog, intensity (power/area) S of the light from the signal and its frequency f. The engineer finds that d is proportional to S1/n. The value of n is


Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 13

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper I) - Question 14

Three charges Q, q and q are placed at the vertices of a right-angled isosceles triangle such that the potential energy of the system is zero. The value of Q in terms of q is given by the expression, Then, the value of m is


Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 14

Potential energy of the system

As PE of the given system = 0

Hence the value of m = 2

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper I) - Question 15

A convex lens, when inserted between an object and a screen, which are at a fixed distance of 243 cm from each other, forms images of magnification 8. Then, the difference between the focal length of the lens and the object distance (in cm) is


Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 15

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper I) - Question 16

A particle is suspended by a light vertical inelastic string of length ll from a fixed support. At its equilibrium position, it is projected horizontally with a speed . What is the ratio of the tension in the string in its horizontal position to that in string when the particle is vertically above the point of support ?


Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 16
For vertical circular motion, student should always remember that

At horizontal position,

…….(i)

At highest point

From above equations, we get

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper I) - Question 17

Two particles A and B are located at points (0, −10√3) and (0, 0) in xy plane. They start moving simultaneously at time t = 0 with constant velocities m s−1 and m s−1 respectively. Time when they are closest to each other is found to be K/2 seconds. Find K. All distances are given in meter.


Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 17
As observed by B, motion of A is along AM and BM is the shortest distance between them. Relative displacement of A w.r.t to B is AM = AB cos 30o

= 1.5 s

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper I) - Question 18

The equation of a standing wave propagating along a string fixed at both ends is given by y = (4 cm) sin(0.314 cm−1) cos(3.14 s−1)t. The linear mass density of the string is 10 g cm−1. The average power transmitted through the string is n × 10−4 W. Find n.


Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 18
Net energy transmitted through a stationary wave is zero as stationary waves are basically just 2 equal waves with different phase difference moving in opposite directions
JEE Advanced Mock Test - 6 (Paper I) - Question 19

Identify (X) and (Z) in the above sequence of reaction.

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 19

JEE Advanced Mock Test - 6 (Paper I) - Question 20

Directions: The following question is based on the paragraph given below.

P and Q are isomers of dicarboxylic acid C4H4O4. Both decolorise Br2/H2O. On heating, P forms the cyclic anhydride. Upon treatment with dilute alkaline KMnO4, P as well as Q could produce one or more than one from S, T and U.

The compounds formed from P and Q respectively are:

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 20
Since both the dioic acids decolourise Br2/H2O, these are unsaturated dioic acids and display geometrical isomerism.

And also, 'P' forms a cyclic anhydride; therefore, it is a cis isomer.

'P' is maleic acid and 'Q' is fumaric acid.

On treatment with KMnO4 solution, both 'P' and 'Q' form tartaric acid which exists as three stereoisomers: S, T and U.

Maleic acid 'P' yields the meso isomer 'S', which is optically inactive.

Fumaric acid 'Q' yields the optically active isomer (T, U).

Hence, option (2) is correct.

JEE Advanced Mock Test - 6 (Paper I) - Question 21

1 mole of an ideal gas A(Cv, m = 3R) and 2 moles of an ideal gas B are (Cv,m = 3/2 R) taken in a container and expanded reversible and adiabatically from 1 liter to 4 liters starting from initial temperature of 320 K. ΔE or ΔU For the process is :

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 21

for adiabatic process ΔU = ΔW

JEE Advanced Mock Test - 6 (Paper I) - Question 22

During the process of electrolytic refining of copper, some metals present as impurity settle as 'anode mud'. These are

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 22
During electrolysis, noble metals (inert metals) like Ag, Au and Pt are not affected and separate as anode mud from the impure anode
*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 23

Consider the following equilibrium.

8 g of SO3 is kept in a container at 527oC. The equilibrium pressure and density are 1.6 atm and 1.6 gL-1 , respectively.

(R = 0.08 L. atm/mol. K)

With reference to the given data, which of the following facts about the reaction is/are true?

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 23

Using PM = dRT

Hence, (3) is true.

Or, x = 0.05 and nT = 0.125

Hence, option (2) is correct.

Statement (4) is incorrect because at equilibrium, Q = Keq and ΔG = 0

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 24

The cation precipitated by ammonium chloride and aqueous ammonia is

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 24
Bi+3 + 3NH4OH → Bi(OH)3 (white ppt) + 3N

Pb+2 + 2NH4OH → Pb(OH)2 (white ppt) + 2N

Fe3+ + 3NH4OH → Fe(OH)3 (reddish brown ppt) + 3N

Due to common ion effect of N , the OH- ion concentration is not enough to precipitate Mg+2 as Mg(OH)2 because of its high solubility product.

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 25

Which of the following statements is incorrect?

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 25

(1) Replacement of -OH by a halogen in an alcohol is nucleophilic substitution reaction. It is the protonted alcohol which acts as a substrate.

(2) Alcohols are acidic enough to react with active metals to liberate hydrogen gas. They are basic enough to accept a proton from strong acids.

(3) Secondary alcohols, on oxidation, give a ketone containing the same number of carbon atoms. Further oxidation will give carboxylic acid containing lesser number of carbon atoms.

(4) Primary alcohols, on oxidation, give an aldehyde containing the same number of carbon atoms. Further oxidation will give carboxylic acid containing the same number of carbon atoms.

Hence, only option (1) is incorrect.

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 26

With reference to the scheme given, which of the given statements about T, U, V and W is/are correct?

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 26
T is soluble in hot aqueous NaOH.

Molecular formula of W is C10H18O4.

V gives effervescence on treatment with aqueous NaHCO3.

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 27

Which statements are correct?

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 27
In irreversible adiabatic process

In reversible adiabatic process as q = 0

No change in entropy.

Reaction is spontaneous i.e. ΔGo is −ve (highly)

If Q < />

Reaction will process to equilibrium with a high value of K and almost complete and at equilibrium Q = K and ΔG = 0

Reactants almost changed to products.

Highly negative ΔGo value reaction at equilibrium reactants almost change to products and K>>1 (i.e. irreversible process)

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 28

Which of the following statements are correct about phenol-formaldehyde resin ?

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 28
Novolac or resol is linear polymer and it is thermoplastic material.

Bakelite is cross linked polymer and P/F < 1.="" the="" structure="" has="" been="" mentioned="" />

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 29

In which of the following arrangements, the order is not according to the property indicated against it?

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 29
Among the isoelectronic species, more the positive charge on the cation smaller is the size and more the negative charge on the anion larger is the size.

Al3+ < Mg2+ < Na+ < F

The electron gain enthalpy decreases down the group, but the second element of the group has the highest negative value of electron gain enthalpy. Hence, I < Br < F < Cl is the correct order.

The metallic radius increases down the group. Hence, the order

Li < Na < K < Rb

The order should be B < C < O < N, as N has stable electronic configuration of 1s2, 2s2, 2p3.

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 30

Work done (in kJ) by the gas in the following cyclic process is:

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 30
w = πab

w = 11000 J

w = 11 kJ

Net work done will be positive.

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper I) - Question 31

On complete combustion of a 9 L mixture of ethane and propane, 21 L of CO2 at STP is produced. The molar ratio of ethane and propane in the mixture is


Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 31
Avogadro's law states that under same conditions of temperature and pressure, equal moles of gases occupy equal volumes. Hence, in chemical reactions involving gaseous reactants and products, the volume ratio is equal to the molar ratio.

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper I) - Question 32

How many open chain structures are possible for N-Methly butanamine (including it) that are referred to as metamers?


Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 32
Molecular formula of N-methyl butanamine is C5H13N. Primary, secondary and tertiary amines are different functional groups and metamerism cannot coexist with functional isomerism. Hence, we shall consider only those isomers of N-methyl butanamine which are secondary (2°) amines.

Every molecule of group A is a metamer of every molecule in group B.

Hence, there are six such isomers.

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper I) - Question 33

100 g of is reduced to 3.125 g in 25 days. Half life of bismuth-210 (in days) is


Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 33
Radioactive reactions follow first order kinetics.

Second method:

In a first order kinetics,

At: Mass or concentration of the reactant at time t

Ao: Initial mass or concentration of the reactant

n is number of half lives elapsed.

Hence, number of half lives elapsed = 5

t1/2 = 25 days/5 = 5 days

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper I) - Question 34

In Goldschmidt aluminothermic process, the thermite mixture contains _____ parts of Fe2O3 and 1 part aluminium by mass.


Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 34
Aluminothermic process is an exothermic chemical reaction using AlAl as a reducing agent. In the reaction, the reactants are, Fe2O3 + 2Al → 2Fe + Al2O3

So, 3 parts of Fe2O3 and one part Al by mass.

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper I) - Question 35

Prostaglandin E1 is a compound produced by the body to regulate a variety of processes including blood clotting, fever, pain and inflammation.

How many stereogenic centres are present in the given compound?


Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 35

Asymmetric stereogenic carbons have 4 different groups attached to it. Carbons numbered 1, 4, 5 and 6 are such Carbons.

A stereogenic centre is that where interchanging the substituents results in a stereoisomer. Hence, the Carbons of the double bond (2 and 3) will be stereogenic centres.

Thus, the total number of stereogenic centres will be 6

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper I) - Question 36

10% of sites on a catalyst have been absorbed by H2. On heating, H2 gas is evolved from the sites and collected at 0.03 atm and 300 K in a small vessel of 2.46 cm3. Number of sites available is 6.0 x 1015 per cm2 and the surface area is 1000 cm2. Find out the molecules of hydrogen occupied per site. (Given: NA = 6 × 1023 mol−1)


Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 36
Adsorbed moles of

= 3 × 10−6

∴ Number of absorbed molecules of H2 = 3 × 10−6 × 6 × 1023 = 18 × 1017

Total number of surface sites available = 6 × 1015 × 1000 = 6 × 1018 cm2

Number of surface sites that is occupied by adsorption of

Number of H2 molecules occupied per site

JEE Advanced Mock Test - 6 (Paper I) - Question 37

Let f(x)

The range of f(-2x) is

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 37
Let f(x) = ax2 + bx

a + b = 3a + 4b + 4a + 5b + 4

⇒ 3a + 4b + 2 = 0 ......(i)

f(2) = 12a + 16b = 8a + 10b + 8

⇒ 2a + 3b + 1 = 0 ......(ii)

By (i) and (ii), a = -2 and b = 1

Hence, f(x) = -2x2 + x

f(-2x) = -2.(2x)2 - 2x < 0,="" for="" all="" x="" ∈="" />

JEE Advanced Mock Test - 6 (Paper I) - Question 38

f(x), g(x) and h(x) are all continuous and differentiable functions in [a, b], a < c < b and f(a) = g(a) = h(a). Point of intersection of the tangent at x = c with chord joining x = a and x = b is on the left of c in y = f(x), on the right in y = h(x) and tangent at x = c is parallel to the chord in the case of y = g(x).

If f’(x) > g’(x) > h’(x)

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 38
According to paragraph,

Hence, f(b) > g(b) > h(b).

JEE Advanced Mock Test - 6 (Paper I) - Question 39

A point P moves such that the tangents PT1 and PT2 from it to the hyperbola 4x2 − 9y2 = 36 are mutually perpendicular. Then the equation of the locus of P is

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 39
4x2 − 9y2 = 36

The two tangent PT1 and PT2 will be perpendicular if the product of roots is −1.

(The point of intersection of perpendicular tangents lies on director circle)

For the hyperbola the equation of the director circle is x2 + y2 = a2 − b2

∴ The equation of the locus of P(α,β) is x2 + y2 = 5

(This is the director circle of the hyperbola).

JEE Advanced Mock Test - 6 (Paper I) - Question 40

The tangent at any point P(a secθ, b tanθ) of the hyperbola makes an intercept of length p between the point of contact and the transverse axis of the hyperbola. If p1, p2 are the lengths of the perpendiculars drawn from the foci on the normal at P, then

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 40
Let F1(ae, 0) and F2(–ae, 0) be the foci of the hyperbola.

Equations of the tangent and normal at P are

Q(a cosθ, 0) is the point of intersection of the tangent at P and the x-axis.

G(ae2 secθ, 0) is the point where the normal at P meets xx-axis.

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 41

The foci of a hyperbola are the vertices of the ellipse and the directrices of the hyperbola pass through the foci of the ellipse. The radius of the auxiliary circle of the hyperbola is

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 41
Focus of hyperbola = Vertex of ellipse = (10, 0) and (-10, 0)

⇒ ae = 10 ...(1)

Directrix of hyperbola pass through focus of ellipse.

⇒ x = a/e pass through (6, 0)

⇒ a/e = 6 ...(2)

From (1) and (2):

a2 = 60 and b2 = a2(e2 - 1) = 40

Equation of the hyperbola is

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 42

The number of distinct integral factor pairs of 10,500 whose product is 10,500 is

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 42
10,500 = 100 x 105

Number of integral factors = (2 + 1)(3 + 1)(1 + 1)(1 + 1) = 48

Number of distinct pairs = 48/2 = 24

(All the pairs are distinct since 10,500 is not a perfect square.)

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 43

In the Argand plane, the locus of z ≠ 1 such that Arg

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 43

∴ This implies that z describes the segment of a circle through at which chord subtends an angle of 2π/3.

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 44

If 1 + 2i is a root of the equation x4 - 3x3 + 8x2 - 7x + 5 = 0, then the other roots are

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 44
If α = 1 + 2i, β = 1 - 2i are the roots, then the quadratic equation is given by

x2 − (1 + 2i + 1 − 2i)x + (1 + 2i)(1 − 2i) = 0

⇒ x2 - 2x + 5 = 0

⇒ x2 - 2x + 5 divides x4 - 3x3 + 8x2 - 7x + 5 = 0.

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 45

The planes ax + 4y + z = 0, 2y + 3z − 1 = 0 and 3x − bz + 2 = 0 will

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 45
Considering the three equations as a 3×33×3 linear system, we have

Similarly, Dy = −ab + 6a − 3, Dz = −4a + 12

If, D ≠ 0 ⇒ ab ≠ 15 the system will have unique solution.

So, Planes will meet in a point.

If a = 3, Dx = Dy = Dz = 0

For ab = 15 and a ≠ 3 or b ≠ 5 at least one of the determinants Dx, Dy and Dz become non zero while D = 0.

So, Planes will have no common point.

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 46

If the equation cx2 + bx − 2a = 0 has no real roots and then -

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 46

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 47

Let mm and nn are distinct natural numbers, then is -

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 47

if one of n and m is odd and other is even

= 0 ; if n and m both are either even or both odd (as n – m and m + n will be even)

*Multiple options can be correct
JEE Advanced Mock Test - 6 (Paper I) - Question 48

The expression (tan4x + 2tan2x + 1) cos2x when x = π/12 can be equal to:

Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 48

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper I) - Question 49

If x, y and z > 1, 2x4 = y4 + z4, xyz = 8, and logy x, logz y and logx z form a GP, then the value of z must be


Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 49
logy x, logz y and logx z are in GP.

⇒ (log y)3 = (log z)3

⇒ log y = log z

⇒ y = z

When y = z, the first two equations give x4 = z4 and xz2 = 8.

We easily get z3 = 8.

⇒ z = 2

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper I) - Question 50

Directions: The answer to the following question is a single-digit integer, ranging from 0 to 9.

Let p(x) be a real polynomial of least degree, which has a local maxima at x = -4 and a local minimum at x = 9. If p(0) = 0 and p(-6) = 108, then what is the value of p'(-4)?


Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 50

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper I) - Question 51

Directions: The answer to the following question is a single-digit integer, ranging from 0 to 9.

The largest value of non-negative integer 'a' for which


Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 51

∴ Largest value of is 2.

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper I) - Question 52

AB is a chord of the parabola y2 = 8x with vertex at A and BC is drawn perpendicular to AB meeting the axis of the parabola at C. The projection of BC on the axis of the parabola is equal to .....


Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 52
Let us assume that AB makes angle θ with positive x-axis.

Also, let the coordinates of B are (x,y)

∴ Projection of BC on the x-axis

= y2/x = 8

So, the required projection is 8 units.

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper I) - Question 53

In ΔABC , the least value of is equal to _____.


Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 53
We have,

A.M ≥ G.M of three numbers

So, using (i)i we get

*Answer can only contain numeric values
JEE Advanced Mock Test - 6 (Paper I) - Question 54

If and bn = 1 − an, then the smallest natural number n, such that bn > an, is


Detailed Solution for JEE Advanced Mock Test - 6 (Paper I) - Question 54

This is not true for n = 1, 3, 5 and is true for n = 2, 4, 6

∴ The minimum value of n is 2, such that bn > an.

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