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JEE Advanced Mock Test - 4 (Paper II) - JEE MCQ


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54 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Advanced Mock Test - 4 (Paper II)

JEE Advanced Mock Test - 4 (Paper II) for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2024 preparation. The JEE Advanced Mock Test - 4 (Paper II) questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Mock Test - 4 (Paper II) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Mock Test - 4 (Paper II) below.
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*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 1

In a Young's double slit experiment,

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 1
If a point source is placed symmetrically from both the slits, then interference will be observed on the screen. if the source is placed unsymmetrically, then both slits will receive same type of input, Thus, interference will be observed in this case as well. If intensity of light is kept same at both the slits, the darker and brighter patterns will be more contrast. But if the two slits are illuminated by two independent sources, then the darker and the brighter pattern will not be observed because the interference will not be in proper sequence to project it on the screen.
*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 2

The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statements is/are correct?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 2

(1) For an isothermal process, temperature is constant.

∴ T1 = T2

(2) For an adiabatic process, temperature decreases.

∴ T3 < T1

(3) In Wisothermal, ΔU = 0; so energy gets converted into work done, whereas in an adiabatic process, Q = 0.

∴ Wisothermal > Wadiabatic

(4) For an isothermal process, change in internal energy is constant, but not in an adiabatic process.

Thus, ΔUisothermal = 0, ΔUadiabatic = −ve

∴ ΔUisothermal > ΔUadiabatic

Hence, the correct options are 1, 3 and 4.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 3

A tank of large base area is filled with water up to a height of 5 m. A hole of 2 cm2 cross section in the bottom allows the water to drain out in continuous stream. For the situation, mark the correct statement(s). [ρwater = 1000 kg/m3, g = 10 m/s2]

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

As the water falls down in the stream, the speed of the stream will increase due to gravity.

So according to equation of continuity (av = a constant), the area of the stream will decrease to keep the product of area and speed constant.

⇒ Option (a) is correct.

Velocity of water just coming out of the hole is v1 =

Velocity of water as it falls down 5m more

v2 = = 14 m/s

∵ v2 =

Now using the equation of continuity, we have

a1v1 = a2v2

⇒ 2 × 7√2 = a2 × 14

⇒ a2 = √2cm2 = 1.414 cm2

⇒ Option (c) is correct.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 4

Among the following, the state function(s) is/are:

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 4
Among the given, state functions are internal energy and molar enthalpy.

Path functions are irreversible expansion work and reversible expansion work.

The value of the state function depends on the initial and final states of the system. It is independent of the path followed.

The value of the path function depends on the path followed.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 5

A cylindrical vessel of 90 cm height is kept filled up to the brim. It has four holes 1, 2, 3 and 4 which are, respectively, at heights of 20 cm, 30 cm, 40 cm and 50 cm from the horizontal floor PQ. The water falling at the maximum horizontal distance from the vessel comes from

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 5
The maximum horizontal distance form the vessel comes from hole numbers 3 and 4. Now, ν = √2gh, where h is the height of the hole from the top.

Horizontal distance,

Substituting the values we can see that whole three and four will have maximum range.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 6

A driver in a stationary car blows a horn which produces monochromatic sound waves of frequency 1000 Hz normally towards a reflecting wall. The wall approaches the car with a speed of 3.3 m s−1.

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

The frequency received by the wall is

The reflected frequency will be same as the frequency received by the wall. Since the wall is moving towards the driver, the frequency heard by the driver will be,

Now the percentage increase,

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 7

A vertical glass capillary tube, open at both ends, contains some water. Which of the following shapes may not be taken by the water in the tube ?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 7
The two free liquid surface must provide a net upward force due to surface tension to balances the weight of the liquid column.

Usually, the water contact angle is smaller than 90o with glass surface and so the surface will be hydrophilic. Shape of water surface on top will be concave. If gravity is taken into account, water will take the shape as shown in option D.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 8

Monochromatic light of wavelength 4000 A∘. is incident on an isolated neutral metal sphere of work function 2 eV. Choose the correct statements(s).

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 8

KEmax = hυ − W

= hc/λ - W

= (12400/4000 - 2)eV

= 1.1 eV

Emitted electron can have kinetic energy from 0 to 1.1 eV.

After sometime the metal plate gets '+'charge and the electrong cant overcome the potential so photoelectric effect stops

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 9

A long current carrying straight wire having current I1 = 104 ampere is placed at the centre of an another current carrying loop having current I2 = 104 ampere. Straight wire is perpendicular to the plane of the loop as shown in the figure. The torque (in N m) acting on the loop about its center is 10x. Find the value of x. (Radius R = 1 m)


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 9

Net torque acting,

=

= 2 × 10−7 × 104 × 104 × (1) = 20 N m

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 10

During an adiabatic process, the pressure of a gas is found to be proportional to cube of its absolute temperature, Cv / Cp is


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 10
Given in the question that during an adiabatic process, the pressure P of a gas is found to be proportional to cube of its absolute temperature T,

∴ P ∝ T3

⇒ P = kT3

⇒PT-3 = k ...(1)

where kk is the constant of proportionality.

Also, during adiabatic process, pressure and temperature are related as

P1−γTγ = constant

where γ = CP / CV

Comparing equations (1) and (2), we get

Hence, Cv/Cp = 1/γ = 2/3 = 00.67

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 11

A part of circuit in a steady state along with the currents flowing in the branches, the values of resistances, etc. is shown in the figure. Calculate the energy (in mJ) stored in the capacitor C (4μF).

(Round off up to 2 decimal places)


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 11
Using Kirchhoff's first law at junction a and b, we have found the current in other wires of the circuit on which currents were not shown.

Now, to calculate the energy stored in the capacitor, we will have to first find the potential difference Vab across it.

∴ Va - 3 x 5 - 3 x 1 + 3 x 2 = Vb

∴ Va - Vb = Vab = 12 V

= ½(4 x 10-6)(12)2 J = 0.29 mJ

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 12

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 rod moves towards right at 10 cm/s. The current (in A) in the loop at t = 3 s due to induced emf is

(Round off up to 2 decimal places)


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 12

Induced emf, e = Blv = 12 × 0.30 × 0.10 = 0.36 V

Resistance of U-shaped rail = 2 W, as its length is 100 cm

Induced current, I = e/R = 0.36/2 = 0.18 A

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 13

A sample of air is initially at 533 K and 700 N/m2, and occupies a volume of 0.028 m3. The air is expanded at constant pressure to three times. A polytropic process with n = 1.5 is then carried out followed by an isothermal process which completes the cycle.

The efficiency of the cycle (in %) is:


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 13

= -58.4 J

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 14

A sonometer wire resonates with a given turning fork forming standing waves with five antinodes between two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by M, the wire resonates with the same tuning fork forming three antinodes for the same positions of bridges. The value of M (in kg) is


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 14

As per the given condition,

Or M = 25 kg

JEE Advanced Mock Test - 4 (Paper II) - Question 15

Answer the following by appropriately matching the lists based on the information given:

Two wires, each carrying a steady current I are shown in four configurations in List I. Some of the resulting effects are described in List II.

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 15

When P is situated midway between the wires, magnetic fields due to both the wires will be in opposite direction and cancel other.

(A) → (q), (r)

When point P is situated at the midpoint of line joining the centres of circular wires, which have the same radii, magnetic fields at point P will be in same direction.

(B) → (p)

In the given case, magnetic fields due to both wires will be in opposite direction and cancel out.

(C) → (q), (r)

In the above case, magnetic fields due to wires will be in opposite direction and wires will repel each other.

(D) → (q), (p)

JEE Advanced Mock Test - 4 (Paper II) - Question 16

Answer the following by appropriately matching the lists based on the information given:

List I contains a list of processes involving expansion of an ideal gas and List II describes the thermodynamic change during this process.

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 16

When P is situated midway between the wires, magnetic fields due to both the wires will be in opposite direction and cancel other.

(A) → (q), (r)

When point P is situated at the midpoint of line joining the centres of circular wires, which have the same radii, magnetic fields at point P will be in same direction.

(B) → (p)

In the given case, magnetic fields due to both wires will be in opposite direction and cancel out.

(C) → (q), (r)

In the above case, magnetic fields due to wires will be in opposite direction and wires will repel each other.

(D) → (q), (p)

JEE Advanced Mock Test - 4 (Paper II) - Question 17

If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 17

On a system of particles if,

No other conclusions can be drawn.

JEE Advanced Mock Test - 4 (Paper II) - Question 18

When a potential difference of 103V is applied between A and B , a charge of 0.75 mC is stored in the system of capacitors. The value of C is (μF)

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 18

q = CeqV

0.75C + 1.5 = C + 1

0.5 = 0.25 C

⇒ C = 2

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 19

A 100 mL mixture of CO and CO2 is passed through a tube containing red hot charcoal. The volume now becomes 160 mL. The volumes are measured under the same conditions of temperature and pressure. Amongst the following, select the correct statement (s).

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 19

CO2 (g) + C(s) ⟶ 2 CO (g)

Vol of CO2 = x mL

Vol of CO = (100 - x) mL

Final volume = (100 - x + 2x) mL

= 100 + x = 160 ∴ x = 60 mL

Volume of CO = 40 mL

Mole-fraction of CO = 40 / 100 = 0.4

Mol % CO2 = 60%

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 20

The structure of AZT (azidothymine) is given below. It is used to treat AIDS patients. It fights an AIDS infection but does not cure it.

Which of the following statements are correct about AZT?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 20
For (​It shows acidic property), H atom attached to N is acidic due to the presence of the two electron-withdrawing (C = O) groups. Its ion is resonance stabilised.

It gives white precipitates with Anlyd. ZnCl2 + conc.HCl (Lucas reagent). 1o ROH does not give Lucas test at room temperature.

(Azide ion (N3) cannot replace Br by SN2 reaction) as strong (C-N) bond is not broken by Br​.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 21

The compound which gives corresponding metal on strong heating is/are

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 21

AsH3 unstable at high temperature [Ni(CO)4] decomposes at high temperature in Mond's process.

HgO decomposes at high temperature to form Hg with evolution of O2

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 22

Amphoteric nature of aluminum is employed in which of the following process for extraction of Aluminum?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 22
Baeyer's process and hall's process are used to remove Fe2O3 impurity from Bauxite.
*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 23

Which of the following Bronsted bases is/are stronger than the phenoxide ion?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 23

H2O, NH3 and C2H5OH are weaker acids than phenol and hence, their conjugate bases are stronger than the phenoxide ion.

The order of acidic strength is

NH3 < C2H5OH < H2O < C6H5OH < CH3COOH

Therefore, the order of basic strength is

NH2- > C2H5O- > OH- > C6H5O- > CH3COO-

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 24

The decreasing order of hyperconjugative effect is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 24
Hyperconjugation effect depends on the number of alpha-H-atoms.

Both options (a) and (b) are arranged in the order of decreasing number of alpha-hydrogen atoms and hence, in the decreasing order of the hyperconjugating effect.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 25

Which of the following orders is/are correct?

I. SbH3 > NH3 > AsH3 > PH3 - Boiling point

II. NH3 > PH3 > AsH3 > SbH3 - Thermal stability

III. NH3 > PH3 > AsH3 > SbH3 - Basic character

IV. NH3 > PH3 > AsH3 > SbH3 - Bond angle

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 25
All the orders regarding the hydrides of group 15 are correct.

The boiling point increases with the increase in the molecular mass, but the boiling point of NH3 is unexpectedly high because of H-bonding.

Thermal stability decreases down the group due to an increase in the bond length.

The basic character decreases down the group due to the decrease in the electronegativity of the central atom.

The bond angle decreases down the group as the size of the central atom increases and the electronegativity decreases. As a result the electron density around the central atom decreases which decreases the electronic repulsions and hence the bond angle.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 26

In a quantitative determination of iron present in an ore, an analyst converted 0.42 g of the ore into its ferrous form. This required 42.0 ml of 0.1 N solution of KMnO4 (in the acidic medium) for titration. Choose the correct statement(s).

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 26

(1) Number of G. Eq. of Fe = Number of G. Eq. of KMnO4

Number of G. Eq. of Fe = Number of G. Eq. of KMnO4 =

Hence, option (1) is correct.

(2) = 56%

Hence, option (2) is incorrect.

(3) Number of moles of KMnO4 = No. of moles of KMnO4 / 5

=

Hence, option (3) is incorrect.

(4)

So, one mole of KMnO4 will oxidise 5 moles of Fe+2 ions.

Hence, option (4) is correct.

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 27

50 mL of 0.1 M solution of a salt reacted with 25 mL of 0.1 M solution of sodium sulphite. The half reaction for the oxidation of sulphite ion is:

If the oxidation number of metal in the salt was 3, then what would be the new oxidation number of the metal?


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 27

Meq of sodium sulphite = Meq of salt ⇒ 25 x 0.1 x 2 = 50 x 0.1 x n

n = 1 (Where n represents valence factor for metal involving the number of electrons gained)

Thus, M+3 + e- → M+2

Hence, new oxidation number of metal 'M' = 2

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 28

When two oppositely charged ions approach each other, the ion smaller in size attracts outermost electrons of the other ion and repels its nuclear charge. The electron cloud of anion no longer remains symmetrical, but is elongated towards the cation. Due to that, sharing of electrons occurs between the two ions to some extent and the bond shows some covalent character. The value of dipole moment can be used for determining the amount of ionic character in a bond. Thus, percentage ionic character

The dipole moment of LiH is 1.964 × 10-29 C.m. and the interatomic distance between Li and H in this molecule is 1.596 . What is the % ionic character in LiH?


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 28

% ionic character =

x 100

76.8%

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 29

Assume that you are using an open-end manometer filled with mineral oil rather than mercury. What is the gas pressure in the bulb (in millimetre of Hg) if the level of mineral oil in the arm connected to the bulb is 237 mm higher than the level in the arm connected to the atmosphere and the atmospheric pressure is 746 mm Hg? (The density of mercury is 13.6 g/ml and that of mineral oil is 0.822 g/ml.)

(Round off up to 1 decimal place)


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 29
The gas pressure in the bulb equals the difference between the outside pressure and the manometer reading. The manometer indicates that the pressure of the gas in the bulb is less than the atmospheric pressure because the liquid level is higher on the side connected to the sample.

Mercury is more dense than mineral oil by a factor of

A given pressure will hold a column of mercury only 1/16.5 times the height of a column of mineral oil.

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 30

SO2Cl2 and Cl2 are introduced into a 3 L vessel. Partial pressure of SO2Cl2 and Cl2 at equilibrium are 1 atm and 2 atm respectively. The value of Kp for the following reaction is 10.

The total pressure in atm at equilibrium would be :


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 30

∴ x = 5 atm

peq = Σp = 1 + 5 + 2 = 8 atm

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 31

'a' moles of NaNa reacts with sufficient water in an open vessel at 300 K. The work done by the liberated gas H2 is equivalent to 4988.4 J. What is aa?


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 31

W = −Pext(ΔV) = −Δ(PV)

= −Δ(nRT)

= −RT (Δn)

For above reactionΔn = a/2

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 32

3 A current was passed through an aqueous solution of an unknown salt of Pd for 1 hour. 2.977 g of Pdn+ was deposited at the cathode. Find nn. (Atomic weight of Pd=106.4)


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 32

Pdn+ + ne → Pd

For

n = 4

JEE Advanced Mock Test - 4 (Paper II) - Question 33

Complex compounds are molecular compounds which retain their identities even when dissolved in water. They do not give all the simple ions in solution but instead furnish complex ions. The complex compounds are often called co-ordination compounds because certain groups called ligands are attached to the central metal ion by co-ordinate or dative bonds. Co-ordination compounds exhibit isomerism, both structural and stereoisomerism. The structure, magnetic property, colour and electrical properties of complexes are explained by various theories.

Arrange the following compounds in order of their molar conductance :

(i) K[Co(NO2)4(NH3)2]

(ii) [Cr(ONO)3(NH3)3]

(iii) [Cr(NO2)(NH3)5]3[Co(NO2)6]2

(iv) Mg[Cr(NO2)5(NH3)]

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 33

Molar conductance will depend on ionic mobility and no of ions produced per unit species on dissociation

(i) K[Co(NO2)4(NH3)2] - 2 Ions / molecule

(ii) [Cr(ONO)3(NH3)3] - 1 unit / molecule

(iii) [Cr(NO2)(NH3)5]3[CO(NO2)6]2 - 5 Ions / molecule

(iv) Mg[Cr(NO2)5(NH3)] - 2 Ions / molecule

out of (i) and (iv) K+ and Mg2+ , Mg2+ due to high solvation will have lower Ionic mobility & lower molar conductance. Molar conductance will have order (ii) < (iv) < (i) < (iii)

JEE Advanced Mock Test - 4 (Paper II) - Question 34

A can be

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 34
Step I is the chromyl chloride test, which is used to detect chloride ions in the qualitative analysis. If any alkali chloride salt is heated with acidified potassium dichromate, it produces red colour fumes of chromyl chloride. It confirms the presence of chloride ions in that salt.

However, for salts such as chlorides of antimony, lead and tin, chromyl chloride test is not applicable. This is because the chlorides of these metals are covalent and they do not generate Cl ions. The chromyl chloride test is applicable only for compounds having chloride ionic bonds. So, the chromyl chloride test is carried out mainly to detect chloride ions in ionic compounds.

When this gas is passed through a sodium hydroxide solution, a yellow solution is obtained due to the presence of sodium chromate.

JEE Advanced Mock Test - 4 (Paper II) - Question 35

Arrange the following by appropriately matching the lists based on the information given in the paragraph.

List - 1 includes organic name reactions.

List - 2 includes appropriate reagents for the reactions shown in List - 1.

Which of the following options has the correct combination considering List - 1 and List - 2?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 35

Rosenmund Reaction:

Finkelstein Reaction:

JEE Advanced Mock Test - 4 (Paper II) - Question 36

Arrange the following by appropriately matching the lists based on the information given in the paragraph.

List - 1 includes various tests to distinguish the compounds.

List - 2 includes the compounds that can be distinguished with the help of tests given in List - 1.

Which of the following options has the correct combination considering List - 1 and List - 2?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 36

Lucas Test:

Tollen Test:

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 37

Directions: The following question has four choices, out of which ONE or MORE is/are correct.

Find the equation(s) of common tangent(s) to y = x2 and y = - x2 + 4x - 4.

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 37

The equation of tangent to y = x2,

y = mx - m2/4

Putting in y = - x2 + 4x - 4,

mx - m2/4 = -x2 + 4x - 4

⇒ x2 + x(m - 4) + 4 - m2/4 = 0

D = 0

Now, (m - 4)2 - (16 - m2) = 0

⇒ 2m (m - 4) = 0

⇒ m = 0, 4

y = 0 and y = 4(x - 1) are the required tangents.

Correct answers: y = 4(x - 1), y = 0

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 38

Directions: The following question has four choices, out of which ONE or MORE is/are correct.

If f(x) = min {1, x2, x3}, then

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 38

f(x) = min {1, x2, x3} which could be presented graphically as given.

⇒ f(x) is continuous for all x and not differentiable at x = 1 due to sharp edge.

Correct answers: f(x) is continuous everywhere and f(x) is not differentiable at one point.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 39

Directions: The following question has four choices, out of which ONE or MORE is/are correct.

If and , then

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 39

Now, it is obvious from the graph that f(x) has local maxima at x = 1 and local minima at x = 2.

g'(x) = f(x) = 0

i.e. ex = 0, which is not possible.

Therefore, 2 - ex-1 = 0

⇒ ex - 1 = 2

⇒ x - 1 = In 2

⇒ x = 1 + In 2

x - e = 0

⇒ x = e

g(x) has local maxima at x = 1 + In 2

g(x) has local minima at x = e

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 40

Directions: The following question has four choices, out of which ONE or MORE is/are correct.

Let A and B be two distinct points on the parabola y2 = 4x. If the axis of the parabola touches a circle of radius r having AB as its diameter, then the slope of the line joining A and B can be

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 40

Let and

C = (Centre of circle)

Slope of line AB =

r = |(t1 + t2)|

Slope of line AB = ±2/r

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 41

Let n be number of distinct objects, kk be number of like boxes and let P(n,k) be the number of ways of putting all the nn objects into the boxes so that no box is empty then which is/are correct ?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 41

Given that the number of ways to put nn objects into kk like boxes = P(n,k)

Let us assume that we pick one object from nn distinct objects. Now there are two possibilities:

Case I: If this object is put into a box alone then the remaining n−1 objects need to be put in k−1 boxes which can be done in P(n−1,k−1) ways.

Case II: If this particular object is not kept in a box alone then the remaining n−1 objects are first distributed in kk boxes in P(n−1,k),k and then this object is put in any one of the k boxes in k ways.

The total number of ways is k⋅P(n−1, k)

Altogether the total number of ways is P(n, k) = P(n−1, k−1) + k⋅P(n−1, k)

For n = 6 & k = 4, we get P(6,4) = P(5,3) + 4⋅P(5,4)

Similarly, for n = 5 & k = 3 we get P(5,3) = P(4,2) + 3P(4,3)

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 42

Consider the system of equations

3x − y + 4z = 3, x + 2y − 3z = −2, 6x + 5y + λz = −3.Where Then

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 42

Given system of equations are

So, Discriminant △⩾0

Also for given system of linear equations

∴ The system of linear equation has unique solution for exactly 7 integer values of λ.

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 43

If ,where [⋅] represents greatest integer function and {⋅} represents fractional part of x, then which of the following is/are true.

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 43

f(x) is injective and discontinuous ( from the graph)

f(x) is surjective (range = co-domain) and non differentiable ( since discontinuous)

Clearly, f(x) is discontinuous (from graph) and bijective function (injective and surjective)

*Multiple options can be correct
JEE Advanced Mock Test - 4 (Paper II) - Question 44

Values of x for which the sixth term of the expansion of

is 567, are

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 44
Given, function

Put

Next, Put

Now, function will become

E = (y + z)7 and 6th term is given by

Since, u=3 satisfies this equation.

Therefore,

x = 2 ± 1 = 3 or 1

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 45

Let an denote the number of all n-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let bn = the number of such n-digit integers ending with digit 1 and cn = the number of such n-digit integers ending with digit 0.

The value of b6 is


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 45

To find b6, we have to find all 6 digit numbers ending with '1' such that no consecutive digits are '0'.

Some of the examples possible are:

1. 1 0 1 1 1 1

2. 1 0 1 0 1 1

3. 1 1 1 1 1 1

Three case possible:

1. One zero-It can be placed in any of the four places.

So, we get '4' such six digit numbers.

2. Two zeros-We get '3' such six digit numbers possible.

3. No zeros-We get only '1' such six digit number.

Hence, b6 = 4 + 3 + 1 = 8

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 46

The number of distinct real roots of x4 - 4x3 + 12x2 + x - 1 = 0 is


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 46

x4 − 4x3 + 12x2 + x − 1 = 0

Let f(x) = x4 − 4x3 + 12x2 + x − 1

f′(x) = 4x3 − 12x2 + 24x + 1

f′(x) will be negative from −∞ to a and will be positive from a to .

So, f(−∞) = +ve

Then, after decreasing till x = a, it cuts the x axis once and then after increasing till x = , it cuts the x axis once.

Total roots = +2

Hence, the correct answer is 2.

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 47

If are vectors such that , then a possible value of is


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 47
Given,

⇒ |λ| = 1

⇒ λ = 1

Thus,

Now,

= (-14 + 6 + 12)

= 4

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 48

Let an ordered pair A be defined as A(x, y) where x∈ prime number, such that x < 10 and y∈ natural numbers and y≤10. If the probability that the ordered pair A satisfies the relation x2 − 3y2 = 1 is P then 60P equals.


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 48

Given, x∈ prime numbers and x<10 and="" y∈="" natural="" numbers="" and="" y="" ≤="">

⇒ x∈{2, 3, 5, 7} & y∈{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Also given that A(x, y)

Total no. of order pair n(A) = 4 × 10 = 40

Let, E be the set of all order pair (x, y) which satisfy x2 − 3y2 = 1

⇒ 60P = 3

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 49

If x5 − x3 + x = a, where x > 0, then the maximum value of 2a − x6 is equal to


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 49

Given, x5 − x3 + x = a, where x > 0

Here, x5,−x3, x is a G.P. whose first term is x5 and common ratio −1 / x2

pply AM ≥ GM on x & 1/x

Hence, the maximum value of 2a − x6 is equal to 1.

*Answer can only contain numeric values
JEE Advanced Mock Test - 4 (Paper II) - Question 50

Let an d where and I3 be the identity matrix of order 3. If the determinant of the matrix is αω2 , then the value of α is equal to _________.


Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 50

Let

Multiplying with P on both sides

Now

Det(A − I) = (w2 + w + w) +7(−w) + w3 = −6w

Det((A − I))2 = 36w2

⇒ α = 36

JEE Advanced Mock Test - 4 (Paper II) - Question 51

Match the statements given in Column I with the values given in Column II.

Which of the following is the only correct option?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 51

A.

is opposite to the side of the maximum length

B.

⇒ f(x) = x

C.

= ㅠ

D.

JEE Advanced Mock Test - 4 (Paper II) - Question 52

Match the statements given in Column I with the values given in Column II.

Which of the following is the only correct option?

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 52

A.

is opposite to the side of the maximum length

B.

⇒ f(x) = x

C.

= ㅠ

D.

JEE Advanced Mock Test - 4 (Paper II) - Question 53

Let ωω be a cube root of unity with ω ≠ 1. A fair die is thrown three times. If r1, r2 and r3 are the numbers obtained on the die, then the probability that ωr1 + ωr2 + ωr3 = 0 is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 53

Total number of events = 6 × 6 × 6 = 216.

Given, ωr1 + ωr2 + ωr3 = 0

If one of r1​,r2​,r3​ ​ takes value from the set {3, 6} other takes values from the set {1, 4} and the third takes value from the set {2, 5}

The total number of these ways

=(​2C1 × ​2C1 × ​2C1​) × 3!.

So, favourable number of elementary events

=(​2C1 × ​2C1 × ​2C1​) × 3! = 48

Hence, required probability will be = 48 / 216 ​ = 2/9

JEE Advanced Mock Test - 4 (Paper II) - Question 54

Let a = cos−1(cos 20), b = cos−1(cos 30) and c = sin−1sin(a + b) then maximum value of sin(2(a + b + c)x) + cos2 ((a + b + c) x) is

Detailed Solution for JEE Advanced Mock Test - 4 (Paper II) - Question 54

From given, a = 20 − 6π, b = 10π−30, c = sin−1sin(4π−10) = 10 − 3π

So a + b + c = π

f(x) = sin(2(a + b + c)x) + cos2((a + b + c)x)

= sin2πx + cos2πx

Now,

⇒ maximum value of f(x)

=

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