JEE Advanced Mock Test - 3 (Paper I) - JEE MCQ

# JEE Advanced Mock Test - 3 (Paper I) - JEE MCQ

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## 54 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Advanced Mock Test - 3 (Paper I)

JEE Advanced Mock Test - 3 (Paper I) for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2024 preparation. The JEE Advanced Mock Test - 3 (Paper I) questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Mock Test - 3 (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 - 3 (Paper I) below.
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JEE Advanced Mock Test - 3 (Paper I) - Question 1

### Match the following:

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 1

RMS value of alternating current =

x = x0sinωt cosωt

x = x0 / 22sinωt cosωt

x = x0 / 2sin2ωt

RMS value,

x = x0sinωt + x0cosωt

RMS value,

JEE Advanced Mock Test - 3 (Paper I) - Question 2

### A body is thrown with a velocity of 10 m s−1 at an angle of 45° to the horizontal. The radius of curvature of its trajectory in t = 1/√2 s after the body began to move is,

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 2

For projectile, we can write,

ux = 10cos45° = vx (ax = 0)

From the equation of motion at time t = 1√2 sec in y-direction,

Thus, the particle is at the highest point.

Velocity of the particle at the given time is

Let the radius of curvature be R.

Then, from centripetal acceleration,

or

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

### The figure below shows a conducting rod of negligible resistance that can slide on smooth U-shaped rail made of wire of resistance 2 Ω/m. The position of the conducting rod at t = 0 is shown. A time-dependent magnetic field B = 4t T is switched on at t = 0.The current in the loop at t = 0 due to induced emf is

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 3

Induced current,

=

= 0.21 A

Using Lenz's law, the upper end of the rod is negative which makes the current flow clockwise.

JEE Advanced Mock Test - 3 (Paper I) - Question 4

A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half-submerged in a liquid of density at equilibrium position. When the cylinder is given a small downward push and released, it starts oscillating vertically with a small amplitude. If the force constant of the spring is k, the frequency of oscillation of the cylinder is

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 4

At equilibrium, let the extension of string be x. Then

Spring Force + up thrust = weight of the block

When the block is further displaced by “y” in downward direction and then released , resultant force acting in upward direction is

Hence,

Comparing with a = ω2y

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 5

Assume that the nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in the figure. Use this plot to choose the correct answer(s) from the choices given below.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 5

Let us examine the four alternatives separately.

(1) for 1 < A < 50, on fusion mass number of the resulting nucleus will be less than 100.

(2) For 51 < A < 100, on fusion mass number the resulting nucleus is between 100 and 200. B/A increases, energy will be released.

(3) On fission for 100 < A < 200, the mass number for fission nuclei will be between 50 to 100. B/A decreases, no energy will be released.

(4) On fission for 200 < A < 260, the mass number for fission nuclei will be between 100 to 130, B/A will increases, energy will be released.

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 6

A ray of light is incident on a reflecting surface. The ray moves in horizontal direction and is reflected vertically after striking the surface. If the surface is denoted by , what are the coordinates of the point(s) where the ray is incident?

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

Consider the diagram for reflection at general point A shown below.

Here, we have,

2i = 90°

i = 45°

So, dy/dx at A should be tan 45°. Therefore,

Therefore,

Therefore, the point is . Another point, just opposite on the surface will also act same and its coordinates are .

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 7

The potential energy function of a particle moving in one dimension is U = k , where a and k are constants. Then,

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 7

Force corresponding to potential energy is

For equilibrium, F = 0. Therefore,

For stable equilibrium,

Potential energy at x = -a is

Hence this is the required solution.

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 8

A spherical metal shell A of radius RA and a solid metal sphere B of radius RB (< RA) are kept far apart and each is given charge '+Q'. Now they are connected by a thin metal wire. Then

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 8

The potential for both the objects is same.

So,

Or

As RB < />A, we get QA > QB

The above equation represents that 2 is the correct option.

From Gauss law, the electric field inside a spherical shell is zero, so option 1 is correct.

Now

Thus

Thus, option 3 is also correct.

Electric fields on the surface of shell and sphere are

EA =

and

EB =

Thus,

Or

EA < />B

So, option 4 is also correct.

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 9

The spring is compressed by a distance a and released. The block again comes to rest when the spring is elongated by a distance b. During this process:

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 9
Work done by the spring on block = loss in spring potential energy

=

This is also work done against friction = μ mg (a + b)

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 10

In a photoelectric effect experiment, the maximum kinetic energy of the ejected photoelectrons is measured for various wavelengths of the incident light. Diagram shows a graph of this maximum kinetic energy Kmax as a function of the wavelength λ of the light falling on the surface of the metal. Which of the following statement/s is/are correct?

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 10

Cut-off wavelength, λ0 = 250nm

Threshold frequency,

Work function of the metal,

Maximum kinetic energy of photoelectrons corresponding to light of wavelength 100 nm is

= 7.432 eV ≈ 7.4 eV

Photoelectric effect takes place only for light of wavelength less than 250 nm, whereas

λred ≈ 700 nm

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 11

Two identical objects A and B are at temperatures TA and TB respectively. Both objects are placed in a room with perfectly absorbing walls maintained at a temperature T (TA > T > TB). The objects A and B attain the temperature T eventually. Select the correct statements from the following–

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 11
Heat transfer from a body with a high temperature to a body with a lower temperature, when bodies are not in direct physical contact with each other or when they are separated in space, is called heat radiation.

Every object emits and absorbs the radiations simultaneously, if energy emitted is more than energy absorbed, temperature falls and vice versa.

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 12

Three points are located at the vertices of an equilateral triangle whose side equals a. They all start moving simultaneously with velocity v constant in modulus, with the first point heading continually for the second, the second for the third, and the third for the first. How soon will the points converge?

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 12
Since, the magnitude of velocity remains constant. So, instantaneous tangential acceleration is zero for each particle. So, only centripetal force is present on each particle. The vector sum of forces acting on the particles is zero. So, at every instant the momentum of system remains constant. If the mass of each particle are same. Then the momentum of the system is zero at every instant. So, the centre of mass (centroid) of system remains in rest.

Hence, collision takes place at the centre of mass of system (centroid).

Component of velocity towards O is V cos30

The time of converging = AO / Vcos30

Where

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 13

A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 kg and 0.72 kg. Taking g = 10 m s−2, find the work done (in Joule) by string on the block of mass 0.36 kg during the first second after the system is released from rest.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 13

T−0.36 g = 0.36 a = 0.36g / 3

∴ T=0.48 g

Now,WT = TS cos 0o (on 0. 36 kg mass)

=

= 8 J

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 14

Two plane mirrors shown in figure are making an angle of 60° to each other. A light ray falls on one of the mirrors as shown in figure. The light ray is incident parallel to angular bisector of mirrors. How many reflections does the light ray undergo?

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 14

The light ray XA first reflects at A and after reflection ray goes along AB as shown in figure and strikes the 2nd mirror at B. From geometry, it can be easily found that AB is the normal to 2nd mirror at B. So, it means light ray will retrace its path after reflection at B and the light ray will suffer total 3 reflection first one at A, then at B and the again at A.

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 15

A point object is placed in front of a thin biconvex lens, of focal length 20 cm. When placed in air, the refractive index of material of lens is 1.5 The further surface of the lens is silvered and is having radius of curvature of 25 cm. The position of final image of object is at 25/x cm from lens. Determine the value of x?

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 15
We know that a thin silvered lens is equivalent to a combination of two lenses and a mirror. This equation system is working as a mirror whose

focal length is given by

where, fL = 20 cm and fM = −25 / 2cm

Let v is the image distance from lens, then

⇒ v = −12.5 cm

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 16

A nail is located at a certain distance vertically below the point of suspension of a simple pendulum. The pendulum bob is given velocity m/s horizontally. Calculate the distance (in m) of the nail from point of suspension, such that the bob will just perform revolution with the nail as centre. Assume, length of pendulum be 5 m.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 16

Minimum velocity at the lowest point for complete revolution,

v =

5 x 10(5 - x) = 100 x 2

5 - x = 4

x = 1 m

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 17

Illuminate the surface of a certain metal alternately with wave length λ1 = 0.35 μm and λ2 = 0.54 μm. It is found that the corresponding maximum velocity is of photo electron having a ratio n = 2. Find the work function of that metal (in eV).

(Round off up to 2 decimal places)

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 17

Using equation of two wave lengths,

- W

As, v1 = 2v2

3W = 4

3W =

= 5.64 eV

W =

= 1.88 eV

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 18

A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the figure. The stick applies a force of 2 N on the ring and rolls t without slipping with an acceleration of 0.3 m/s2. The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is (P/10). The value of P is

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 18

Let the friction between surface and sphere be fs.

Let the friction between stick and sphere be fa.

Let N be the normal reaction between sphere and stick.

Now, N - fs = ma

Also N = 2N

fs = 1.4N

Writing torque about the centre, we get

(fs - fa)r = Iα

a = αr and I = mr2

On solving, fa = 0.8 = μN

μ = 0.4 = 4/10

So, P = 4

JEE Advanced Mock Test - 3 (Paper I) - Question 19

Directions: The following question has four choices, out of which ONLY ONE is correct.

Consider the cell:

Cd(s) | Cd2+ (1.0 M) || Cu2+ (1.0, M) | Cu(s)

If we want to make a cell with a more positive voltage using the same substances, we should

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 19

Cd(s) + Cu+2(aq) → Cd+2(aq) + Cu(s)

According to Nernst equation,

According to the equation, decrease in the concentration of Cd+2 or increase in the concentration of Cu+2 would increase the voltage.

Hence, option (d) is correct.

JEE Advanced Mock Test - 3 (Paper I) - Question 20

Given below are two cleavage reactions:

(i) (CH3)3COCH3 → CH3I + (CH3)3COH

(ii) (CH3)3COCH3 → CH3OH + (CH3)3CI

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 20

The low polarity of solvent ether favours the SN2 mechanism.

The O-CH3 bond breaks and methyl iodide and tertiary butyl alcohol are formed as products.

The presence of a solvent with high polarity such as water favours the SN1 mechanism.

The (CH3)3-O bond breaks and methyl alcohol is formed.

The mechanism is

JEE Advanced Mock Test - 3 (Paper I) - Question 21

In which of the following molecules / ions resonance structures are equivalent:

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 21

Only format ion has equivalent resonance structures. Charge is delocalised on the similar type of elements in the equivalent resonating structures. Resonating structures are not equivalent in other compounds. Equivalent resonance is more dominant than normal resonance. Equivalent resonance stabilised the species more as compared to the normal resonance.

JEE Advanced Mock Test - 3 (Paper I) - Question 22

What is the ratio of closed packed atoms to tetrahedral holes in a cubic close packing?

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 22

Every constituting atom has two tetrahedral voids.

In CCP lattice,

The number of atoms = 8 × 1/8 + 6 × 12 = 4

Hence, the number of tetrahedral voids = 4 × 2 = 8

Thus, the ratio = 4 : 8 =1 : 2

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 23

What will be the effect of addition of a catalyst at constant temperature?

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 23

, Kf and Kb are affected equally and Keq is not affected by the presence of a catalyst.

ΔHreaction remains constant. There is no effect of catalyst since,

Both will change up to the same extent.

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 24

Which of the following statements are correct about the reaction given below?

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 24

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 25

Which of the following is/are correct below the polymer (E) ?

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 25

Product of polymerisation of phenol and formaldehyde depends upon the ratio of moles of phenol to formaldehyde (P/F ratio).

P/F > 1for relative amount of phenol is more than amount of formaldehyde.

P/F < 1for="" relative="" amount="" of="" phenol="" is="" less="" than="" amount="" of="" />

P/F = 1for relative amount of phenol is equal to amount of formaldehyde.

Novolac is used in the manufacture of adhesives. Novolac is thermoplastic polymer.

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 26

Which of the following are possible products (in significant amount)-

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 26

Alkene are not significant products because BrΘ is a poor base. Carbocation rearrangements are common in organic chemistry and are defined as the movement of a carbocation from an unstable state to a more stable state through the use of various structural reorganizational shifts within the molecule.

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 27

Which of the following statements is/are correct regarding defects in solids?

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 27
Frenkel defect is shown by ionic solids. The smaller ion (usually cation) is dislocated from its normal site to an interstitial site. Thus, it is a dislocation effect and is usually favoured by large size difference between cation and anion.

F-centre is formed when anionic vacancies are occupied by electrons.

Schottky defect is a vacancy defect, where equal number of cations and anions are missing. The presence of Schottky defect decreases the density of the crystals.

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 28

Which of the following factors is/are responsible for increase in the rate of surface-catalysed reactions?

1. A catalyst provides proper orientation for the reactant molecules to react.

2. Heat of adsorption of reactants on a catalyst helps the reactant molecules to overcome the activation energy.

3. The adsorption of reactants on the catalyst surface decreases the activation energy of the reaction.

4. Adsorption increases the local concentration of reactant molecules on the surface of the catalyst.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 28
The following two factors are responsible for increase in the rate of surface-catalysed reactions

1. Adsorption is an exothermic process and the heat of adsorption of the reactants on the catalyst helps the reactant molecules to acquire the activation energy for the reaction.

2. Adsorption increases the local concentration of reactant molecules on the surface of the catalyst.

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 29

What happens when a mixture of NaCl and K2Cr2O7 is gently warmed with conc. H2SO4?

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 29

During this process, chromyl chloride is formed which is deep red coloured fuming liquid.

4NaCl + K2Cr2O7 + H2SO4 + 4NaHSO4 + 2KHSO4 + 3H2O

The vapour of CrO2Cl2 when passed into NaOH solution gives a yellow solution of Na2CrO4.

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 30

Which of the following statements is/are correct?

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 30

24Cr = 1s2, 2s2 2p6, 3s2 3p6 3d5, 4s1 = [Ar] 3d5, 4s1

For the magnetic quantum number (m), negative values are possible.

For a given value of 'l', the magnetic quantum can have values -l to +l.

E.g.

For s subshell, l = 0. Hence, m = 0

For p subshell, l = 1. Hence, m = -1, 0, +1

47Ag = 1s2, 2s2 2p6, 3s2 3p6 3d10, 4s2 4p6 4d10, 5s1

Hence, 23 electrons have a spin of one type and 24 of the opposite type.

The oxidation state of N in HN3 is -3.

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 31

Acetylene undergoes linear polymerisation when passed through a solution of cuprous chloride in ammonium chloride. How many acetylene molecules unite to form a molecule of the polymer?

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 31
Acetylene undergoes linear polymerisation reaction when passed through a solution of cuprous chloride in ammonium chloride to give vinylacetylene and divinylacetylene.

This is correct as 3 molecules are involved in polymerisation.

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 32

1.00 L sample of a mixture of methane gas and oxygen gas measured at 25°C and 740 torr was allowed to react at constant pressure in a calorimeter. The calorimeter together with its contents had a heat capacity of 1000 cal/K. The complete combustion of methane to CO2 and water caused a rise in temperature of 0.42 K. Heat of the following reaction is ΔH = - 210.8 kcal.

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (I) Mole percentage of methane in the original mixture is

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 32

Heat generated = (0.42 K × 1000 cal/K) = 420 cal

Number of moles of Methane gas in the mixture = 420 cal x

= 1.99 10-3 mol

Ntotal =

= 0.0398 mol

Mole percentage of methane =

= 5

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 33

The coordination number of AI in the crystalline state of AICI3 is

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 33
AICI3 has a 6-coordinate layer lattice with AI3+ occupying cubic close packed sites. Thus, the coordination number of AI in AICI3 is 6.
*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 34

The total number of alkenes possible by dehydrobromination of 3–bromo–3–cyclopentylhexane using alcoholic KOH is –

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 34

The substrate has three different types of C-H. Therefore, first, three structural isomers of alkenes are expected as

The last two alkenes II and III are also capable of showing geometrical isomerism which are

Hence, two geometrical isomers for each of them will be counted giving a total of five alkenes.

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 35

Atomicity of white or yellow phosphorous is 4 and it is represented as P4 molecule. Calculate the value of expression (x).(y) / (z) regarding this molecule.

Where, x : Total number of vertex angles in P4 molecule.

y : Total number of lone pairs of electrons in P4 molecule.

z : Total number of P - P bonds in P4 molecule.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 35
The molecular formula of white phosphorous is P4. In this structure four phosphorous atoms lie at the corners of the tetrahedron. Each of phosphorous atom is linked to other three phosphorous atoms by covalent bonds. The P-P bond length is 2.21 A and P-P-P angle is 60o.

​Number of P-P bonds in P4 molecule is 6.

x = 12

y = 4

z = 6

The value of

= 8

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 36

The periodic table consists of 18 groups. An isotope of copper, on bombardment with a protons, undergoes a nuclear reaction, yielding an element, X as shown below. To which group, element X belongs in the periodic table ?

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 36

Mass number : 63 + 1 = 1 × 6 + 4 + 1 × 2 + A

A = 64 − 12 = 52

Atomic number : 29 + 1 = 6 × 0 + 2 + 2 × 1 + Z

Z = 30 − 4 = 26

Hence, X is in group '8' in the periodic table.

JEE Advanced Mock Test - 3 (Paper I) - Question 37

A continuous, even periodic function f with period 8 is such that f(0) = 0, f(1) = −2, f(2) = 1, f(3) = 2, f(4) = 3, then the value of tan−1tan{f(−5) + f(20) + cos−1(f(−10)) + f(17)} is equal to

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 37

f(x) is given to be an even, periodic function with period equal to 8.

⇒ f(x + 8) = f(x)

1. f(−5) = f(3) = 2

2. f(20) = f(12) = f(4) = 3

3. f(−10) = f(−2) = f(2) = 1

4. f(17) = f(9) = f(1) = −2

f(−5) + f(20) + cos−1(f(−10)) + f(17) = 2 + 3 + cos−1(1)−2 = 3

tan−1(tan(3)) = tan−1(tan(3 − π)) = 3 − π

(using tan−1(tanx) = x − π if x ∈ (π/2, π))

JEE Advanced Mock Test - 3 (Paper I) - Question 38

The following integral is equal to

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 38

Let

⇒ cosecx + cotx = eu & cosec x − cotx = e−u

(eu − e−u)du = −2 cosecxcotxdx

=

JEE Advanced Mock Test - 3 (Paper I) - Question 39

Directions: The following question has FOUR options, out of which ONLY ONE is correct.

The value of is

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 39

JEE Advanced Mock Test - 3 (Paper I) - Question 40

Let the eccentricity of the hyperbola be the reciprocal of that of the ellipse x2 + 4y2 = 4. Also, the hyperbola passes through a focus of the ellipse. Then, the equation of the hyperbola is

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 40

a2 = 4, b2 = 1, e2 = 1 -

Eccentricity of hyperbola =

Focus of ellipse = … (i)

passes through focus ( ).

∴ a2 = 3

∴ b2 = a2(e2 - 1) = 1.

Equation of hyperbola:

x2 - 3y2 = 3

Focus of hyperbola = (±ae, o) = (±2, 0)

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 41

Directions: The following question has four choices, out of which ONE or MORE can be correct.

Three lines px + qy + r = 0, qx + ry + p = 0 and rx + py + q = 0 are concurrent, if

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 41

Three lines px + qy + r = 0, qx + ry + p = 0 and rx + py + q = 0 are concurrent.

Then,

R1 → R1 + R2 + R3

There (a) and (c) are correct answers.

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 42

Directions: The following question has four choices, out of which ONE or MORE can be correct.

The function f(x) = (et - 1)(t - 1)(t - 2)3(t - 3)5 dt has a local minimum at x equal to

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 42

f(x) = (et - 1)(t - 1)(t - 2)3(t - 3)5 dt

f(x) = d/dx (et - 1)(t - 1)(t - 2)3(t - 3)5 dt

= x(ex - 1)(x - 1)(x - 2)3(x - 3)5 x 1

For local minimum, f '(x) = 0

⇒ x = 0, 1, 2, 3

Let f '(x) = g(x) = x(ex - 1)(x - 1)(x - 2)3(x - 3)5

Using sign scheme rule,

This shows that f(x) has a local minimum at x = 1 and x = 3 and is maximum at x = 2.

Therefore, (b) and (d) are the correct answers.

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 43

Directions: The following question has four choices, out of which ONE or MORE can be correct.

If f(x) = , then

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 43

For -1 ≤ x ≤ 2, we have

f(x) = 3x2 + 12x - 1

⇒ f(x) = 6x + 12 > 0, ∀ -1 ≤ x ≤ 2

Hence, f(x) is increasing in [-1, 2].

Again, function is an algebraic polynomial, so it is continuous at x ∈ (-1, 2) and (2, 3).

For continuity at x = 2:

f(x) = (3x2 + 12x - 1)

= [3 (2 - h)2 + 12(2 - h) - 1]

= [3 (4 + h2 - 4h) + 24 - 12 h - 1]

= (12 + 3 h2 - 12h + 24 - 12h - 1)

= (3h2 - 24h + 35) = 35

and f(x) = (37 - x)

= [37 - (2 + h)] = 35

Also f(2) = 3.22 + 12.2 - 1 = 12 + 24 - 1 = 35

Therefore, LHL = RHL = f(2)

This implies, function is continuous at x = 2.

So, function is continuous in -1 ≤ x ≤ 3.

Now, Rf'(2) =

and Lf'(2) =

Since Rf ' (2) ≠ Lf ' (2), so f' 2) does not exist.

Again, f(x) is increasing in [-1, 2] and is decreasing in (2, 3), it shows that f(x) has a maximum value at x = 2.

Therefore, options (a), (b), (c) and (d) are all correct.

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 44

Let PQ be a chord of the parabola y2 = 4x. A circle drawn with PQ as diameter passes through the vertex V of the parabola. If area of ΔPVQ = 20 square units then coordinates of P are

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 44

Slope of

Slope of

Equation of VQ

Solving it with y2 = 4x

x(t2x - 16) = 0

x = 0, x = 16/t2

Area of ΔPVQ =

PV2. VQ2 = 1600

t2 + 4 = ± 5t

t = ± 1, ± 4

Hence, P (16 ± 8), (1, ±2)

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 45

If m is a positive integer, then is divisible by (where[.] denotes the greatest integer function)

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 45

Let (√3+1)2m = I + f, where I is some integer and 0 ≤ f < />

We have, √3−1 = 2 / √3+1⋅

Therefore, 0 < √3−1="" />< 1.="" let="" f="(√3" −="" />2m

Now,

=

=2m+1K , where K is some integer.

∴ I + f + F is an integer, say J

⇒ f + F = J − I is an integer.

Since 0 < f="" +="" f="" />< 2,="" therefore,="" f="" +="" f="" />

Now,

⇒ 2m+1 and hence 2m divides N.

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 46

The solutions of x2y12 + xyy1 − 6y2 = 0 are

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 46

Here y1 = dy/dx

Given here is a quadratic equation in y1, solving it we get

⇒ y = Cx2

or 1/2logy = C + logx

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 47

If a variable straight line xcosα + ysinα = p which is a chord of the hyperbola subtend a right angle at the centre of the hyperbola, then it always touches a fixed circle whose

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 47

Equation of the pair of straight lines passing through the origin (centre of the hyperbola) and points of intersection of the variable chord and the hyperbola is

They are at right angles if

As p is the length of the perpendicular from the origin on the line xcosα + ysinα = p, line touches the circle with centre at the origin (0,0) and radius equal to .

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 48

The adjoining Figure gives the road plan of lines connecting two parallel roads AB and A1B1. A man walking on the road AB takes a turn at random to reach the road A1B1. It is known that he reaches the road A1B1 from O by talking a straight line. The chance that he moves on a straight line from the road AB to the A1B1 is

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 48

Number of ways man reaches O from road AB = 5.

Number of ways from O to A1B1 by taking a straight line = 2.

∴ Total ways of reaching to A1B1 from AB such that from O to A1B1 he takes a straight line

= 5 × 2 =10.

Favourable ways i.e. the number of ways, when he moves on a straight line from the road AB to the road A1B1 = 2.

∴ Required probability = 2 / 10 = 0.2.

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 49

A bag contains n white and n black balls (all different). Pairs of balls are drawn one-by-one without replacement until the bag is empty. If the number of ways to draw the balls in which each pair consists of one black and one white ball is 576, then the value of n is equal to__

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 49

Number of ways to draw 1st pair = (nC1× nC1)

Number of ways to draw 2nd pair = (n−1C1× n−1C1)

Number of ways to draw 3rd pair =(n−2C1× n−2C1)

Number of ways to draw last i.e., nth pair = (1C1× 1C1)

Combining all the above results, we get

Number of ways to draw n pairs

⇒ n2(n−1)2...22 × 12 = (24)2

⇒ (n!)2 = (24)2

⇒ n! = 24 = 4!

∴ n = 4

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 50

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

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 50

∴ bn > an ⇒1 − an > an ⇒ 2an < />

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 .

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 51

Let are unit vectors such that . If the area of triangle formed by vectors and is A, then what is the value of 16A2?

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 51

Given

Now vector is along the diagonal of the parallelogram which has adjacent side vectors and . Since is also a unit vector, triangle formed by vectors and is an equilateral triangle.

Then, area of triangle is √3 / 4

Then, area of triangle is √3 / 4

16A2 = 3

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 52

Given that α and γ are the roots of the equation Ax2 - 4x + 1 = 0, and β and δ are the roots of the equation Bx2 - 6x + 1 = 0. Find the value of (B - A)/10, such that α, β, γ, δ are in H.P.

(Answer round off upto 1 decimal place)

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 52

As per the given conditions,

Let d be the common difference.

Adding both of the equations, we get

6 - 4 = 2d

or, d = 1

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 53

If 27iz3 - 18z2 - 12z - 8i = 0, and z = x + iy, the greatest value of is

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 53

Using the inequality ,

= 16

∴ The greatest value of

is 4

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 54

If range of the function f(x) = sin-1 x + 2 tan-1 x + x2 + 4x + 1 is [a, b], then the value of a + b is

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 54
Domain of f(x) is [-1, 1]. Therefore

Therefore f(x) is an increasing function. Hence a is maximum value of f(x). Therefore

And b is the maximum value of f(x). Therefore

Therefore the range of f(x) is .

Therefore,

= 4

Hence, it is the required solution.

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