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Part Test - 6 (JEE Advanced)


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Part Test - 6 (JEE Advanced) - Question 1

The wavelengths of Kα x-rays of two metals ‘A’ and ‘B’ are    and   respectively, where ‘R’ is
Rydberg constant. The number of elements lying between ‘A’ and ‘B’ according to their atomic numbers is

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 1

Part Test - 6 (JEE Advanced) - Question 2

A free neutron decays to a proton but a free proton does not decay to a neutron out side nucleus. This is beacuse 

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 2

As proton mass is less than neutron mass proton does not decay to neutron outside nucleus.

Part Test - 6 (JEE Advanced) - Question 3

An α a particle with a kinetic energy of 2.1 eV makes a head on collision with a hydrogen atom moving towards it with a kinetic energy of 8.4 eV. The collision. 

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 3

For completely inelastic collision both come to rest after collision and net energy of
4E + E = 10.5 eV is lost. But electron in ground state of H-atom can accept only an energy of 10.2 eV.
Hence the collision may be inelastic but it can never be perfectly inelastic.

Part Test - 6 (JEE Advanced) - Question 4

A monochromatic radiation of wavelength λ is incident on a sample containing He+. As a result the Helium sample starts radiating. A part of this radiation is allowed to pass through a sample of atomic hydrogen gas in ground state. It is noticed that the hydrogen sample has started emitting electrons whose maximum Kinetic Energy is 37.4 eV. (hc = 12400 eV Å) Then λ is - 

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 4

Part Test - 6 (JEE Advanced) - Question 5

The work function of a certain metal is  When a monochromatic light of wavelength λ < λ0 is
incident such that the plate gains a total power P. If the efficiency of photoelectric emission is η%
and all the emitted photoelectrons are captured by a hollow conducting sphere of radius R already
charged to potential V, then neglecting any interaction between plate and the sphere, expression of
potantial of the sphere at time t is :

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 5

Part Test - 6 (JEE Advanced) - Question 6

The Kinetic energy must an α-particle possess to split a deutron H2 whose binding energy is Eb = 2.2 MeV - 

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 6

Part Test - 6 (JEE Advanced) - Question 7

Assuming that about 20 MeV of energy is released per fusion reaction, 1H2 + 1H30n1 + 2He4, the mass of 1H2 consumed per day in a future fusion reactor of power 1 MW would be approximately 

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 7

Part Test - 6 (JEE Advanced) - Question 8

Nuclei of radioactive element A are produced at rate ' t2 ' at any time t. The element A has decay constant λ. Let N be the number of nuclei of element A at any time t. At time t = t0,  is minimum. Then the number of nuclei of element A at time t = t0 is    

*Multiple options can be correct
Part Test - 6 (JEE Advanced) - Question 9

If nuclear charge is doubled in a hydrogen like atom, which of the following statement(s) are consistent with Bohr’s theory?  

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 9

*Multiple options can be correct
Part Test - 6 (JEE Advanced) - Question 10

Ionization energy of a hydrogen-like ion B is less than that of hydrogen like ion A. Let r, u, E and L represent the radius of the orbit, speed of the electron, energy of the atom and orbital angular momentum of the electron respectively. In ground state  

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 10

*Multiple options can be correct
Part Test - 6 (JEE Advanced) - Question 11

Mark the correct options.

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 11

K x-ray is emitted when electron jumps from outer shell to K shell. This is equivalent to hole jumping from K shell
to outer shell. λK < λL

*Multiple options can be correct
Part Test - 6 (JEE Advanced) - Question 12

An electron makes a transition from n = 2 to n = 1 state in a hydrogen like atom. 

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 12

Part Test - 6 (JEE Advanced) - Question 13

Statement–1 :  undergoes 2α decays, 2β decays (negative β) and 2 γ decays. As a result the
daughter product is  .
Statement–2 : In α decay the mass number decreases by 4 unit and atomic number decreases by 2 unit.
In  β decay (negative β) the mass number remains unchanged and atomic number increases by 1 unit. In
γ decay, mass number and atomic number remains unchanged.

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 13

Statement-2 is true by definition and correctly explains the staement-1, namely,  undergoes 2 α decays, 2β
decays (negative β) and 2 γ decays. As a result the daughter product is  .

Part Test - 6 (JEE Advanced) - Question 14

Statement–1 : In the duration electron jumps from first excited state to ground state in a stationary isolated hydrogen atom, angular momentum of the electron about the nucleus  is conserved.

Statement–2 : As the electron jumps from first excited state to ground state, in a hydrogen atom, the electrostatic force on electron is always directed towards the nucleus.

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 14

As electron jumps from n = 2 to n = 1, angular momentum  does not remain conserved. Hence
statement-1 is false.

Part Test - 6 (JEE Advanced) - Question 15

The orbital and spin angular momentum of the atom influence its magnetic structure and these properties are most directly studied by placing the atom in a magnetic field. Also, a magnetic field can affect the wavelengths of the emitted photons.

The angular momentum vector associated with an atomic state can take up only certain specified directions in space. This concept of space quantization was shown by Otto Stern and Walthor Gerlach in their experiment.

In the experiment, silver is vapourized in an electric oven and silver atoms spray into the evacuated apparatus through a small hole in the oven wall. The atoms which are electrically neutral but have a magnetic moment, are formed into a narrow beam as they pass through a slit in a screen. The beam, thus collimated, then passes between the poles of an electromagnet and finally, deposits its silver atoms on a glass plate that serves as a detector. The pole faces of the magnet are shaped to make the magnetic field as nonuniform as possible.

In a non-uniform magnetic field, there is a net force on a magnetic dipole. Its magnitude and direction depends on the orientation of the dipole. Thus the silver atoms in the beam are deflected up or down, depending on the orientation of their magnetic dipole moments with respect to the z–direction.

The potential energy of a magnetic dipole in a magnetic field where is magnetic dipole moment of the atom. From symmetry, the magnetic field at the beam position has no x or y components i.e. 

The net force Fz on the dipole is 

Thus, the net force depends, not on the magnitude of the field itself, but on its spatial derivative or gradient.

The Results

If space quantization did not exist, then could take on any value from + to –, the result would be a spreading out of the beam when the magnet was turned ON. However, the beam was split cleanly into two subbeams, each subbeam corresponding to one of the two permitted  orientations of the magnetic moment of the silver atom, as shown.

In a silver atom, all the spin and orbital magnetic moments of the electrons cancel, except for those of the atom's single valance electron. For this electron the orbital magnetic  moment is zero because orbital angular momentum is zero (because for electrons of s–orbit, L = 0), leaving only the spin magnetic moment. This can take up only two orientations in a magnetic field, corresponding to ms = +1/2 and ms = – 1/2. Hence there are two subbeams – and not some other number. 

Q. A hydrogen atom in ground state passes through a magnetic field that has a gradient of 16mT/m in the vertical direction. If vertical component magnetic moment of the atom is  9.3 × 10–24 J/T, then force on it due to the magnetic moment of the electron is :

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 15

Part Test - 6 (JEE Advanced) - Question 16

The orbital and spin angular momentum of the atom influence its magnetic structure and these properties are most directly studied by placing the atom in a magnetic field. Also, a magnetic field can affect the wavelengths of the emitted photons.

The angular momentum vector associated with an atomic state can take up only certain specified directions in space. This concept of space quantization was shown by Otto Stern and Walthor Gerlach in their experiment.

In the experiment, silver is vapourized in an electric oven and silver atoms spray into the evacuated apparatus through a small hole in the oven wall. The atoms which are electrically neutral but have a magnetic moment, are formed into a narrow beam as they pass through a slit in a screen. The beam, thus collimated, then passes between the poles of an electromagnet and finally, deposits its silver atoms on a glass plate that serves as a detector. The pole faces of the magnet are shaped to make the magnetic field as nonuniform as possible.

In a non-uniform magnetic field, there is a net force on a magnetic dipole. Its magnitude and direction depends on the orientation of the dipole. Thus the silver atoms in the beam are deflected up or down, depending on the orientation of their magnetic dipole moments with respect to the z–direction.

The potential energy of a magnetic dipole in a magnetic field where is magnetic dipole moment of the atom. From symmetry, the magnetic field at the beam position has no x or y components i.e. 

The net force Fz on the dipole is 

Thus, the net force depends, not on the magnitude of the field itself, but on its spatial derivative or gradient.

The Results

If space quantization did not exist, then could take on any value from + to –, the result would be a spreading out of the beam when the magnet was turned ON. However, the beam was split cleanly into two subbeams, each subbeam corresponding to one of the two permitted  orientations of the magnetic moment of the silver atom, as shown.

In a silver atom, all the spin and orbital magnetic moments of the electrons cancel, except for those of the atom's single valance electron. For this electron the orbital magnetic  moment is zero because orbital angular momentum is zero (because for electrons of s–orbit, L = 0), leaving only the spin magnetic moment. This can take up only two orientations in a magnetic field, corresponding to ms = +1/2 and ms = – 1/2. Hence there are two subbeams – and not some other number. 

Q. A hydrogen atom in ground state passes through a magnetic field that has a gradient of 16mT/m in the vertical direction. If vertical component magnetic moment of the atom is  9.3 × 10–24 J/T, then force on it due to the magnetic moment of the electron is :

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 16

Part Test - 6 (JEE Advanced) - Question 17

The orbital and spin angular momentum of the atom influence its magnetic structure and these properties are most directly studied by placing the atom in a magnetic field. Also, a magnetic field can affect the wavelengths of the emitted photons.

The angular momentum vector associated with an atomic state can take up only certain specified directions in space. This concept of space quantization was shown by Otto Stern and Walthor Gerlach in their experiment.

In the experiment, silver is vapourized in an electric oven and silver atoms spray into the evacuated apparatus through a small hole in the oven wall. The atoms which are electrically neutral but have a magnetic moment, are formed into a narrow beam as they pass through a slit in a screen. The beam, thus collimated, then passes between the poles of an electromagnet and finally, deposits its silver atoms on a glass plate that serves as a detector. The pole faces of the magnet are shaped to make the magnetic field as nonuniform as possible.

In a non-uniform magnetic field, there is a net force on a magnetic dipole. Its magnitude and direction depends on the orientation of the dipole. Thus the silver atoms in the beam are deflected up or down, depending on the orientation of their magnetic dipole moments with respect to the z–direction.

The potential energy of a magnetic dipole in a magnetic field where is magnetic dipole moment of the atom. From symmetry, the magnetic field at the beam position has no x or y components i.e. 

The net force Fz on the dipole is 

Thus, the net force depends, not on the magnitude of the field itself, but on its spatial derivative or gradient.

The Results

If space quantization did not exist, then could take on any value from + to –, the result would be a spreading out of the beam when the magnet was turned ON. However, the beam was split cleanly into two subbeams, each subbeam corresponding to one of the two permitted  orientations of the magnetic moment of the silver atom, as shown.

In a silver atom, all the spin and orbital magnetic moments of the electrons cancel, except for those of the atom's single valance electron. For this electron the orbital magnetic  moment is zero because orbital angular momentum is zero (because for electrons of s–orbit, L = 0), leaving only the spin magnetic moment. This can take up only two orientations in a magnetic field, corresponding to ms = +1/2 and ms = – 1/2. Hence there are two subbeams – and not some other number. 

Q. A hydrogen atom in ground state passes through a magnetic field that has a gradient of 16mT/m in the vertical direction. If vertical component magnetic moment of the atom is  9.3 × 10–24 J/T, then force on it due to the magnetic moment of the electron is :

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 17

Part Test - 6 (JEE Advanced) - Question 18

In column-I, consider each process just before and just after it occurs. Initial system is isolated from all other bodies. Consider all product particles (even those having rest mass zero) in the system. Match the system in column-I with the result they produce in column-II.

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 18

(A)    In the given spontaneous radioactive decay, the number of protons remain constant and all conservation principles are obeyed.

(B)    In fusion reaction of two hydrogen nuclei, a proton is decreased as positron shall be emitted in the reaction. All the three conservation principles are obeyed.

(C)    In the given fission reaction the number of protons remain constant and all conservation principles are obeyed.

(D)    In beta negative decay, a neutron is converted into a proton and the electron is ejected out.    

*Answer can only contain numeric values
Part Test - 6 (JEE Advanced) - Question 19

A radioactive sample has 12.0 × 1018 active nuclei at a certain instant. Number of nuclei still in the same active state after two half-lives is n ×1018. Find n. 


Detailed Solution for Part Test - 6 (JEE Advanced) - Question 19

In one half-life the number of active nuclei reduces to half the original number. Thus, in two half-lives the number is
reduced to (1/2)(1/2) of the original number. The number of remaining active nuclei is, therefore,
12 × 1018 × (1/2)(1/2)= 3 × 1018 = 3

*Answer can only contain numeric values
Part Test - 6 (JEE Advanced) - Question 20


Detailed Solution for Part Test - 6 (JEE Advanced) - Question 20

*Answer can only contain numeric values
Part Test - 6 (JEE Advanced) - Question 21

In the figure shown electromagnetic radiations of wavelength 200nm are incident on the metallic plate A. The photo electrons are accelerated by a potential difference 10V. These  electrons strike another metal plate B from which electromagnetic radiations are emitted. The minimum wavelength of the emitted photons is 100nm. The work function of the metal ‘A’ is  x eV then find x + 2.2 use hc = 12400 eVÅ, use Rch = 13.6 eV.


Detailed Solution for Part Test - 6 (JEE Advanced) - Question 21

*Answer can only contain numeric values
Part Test - 6 (JEE Advanced) - Question 22

 An X-ray tube is working at potential of 20 kV. The potential difference is decreased to 10 kV. It is found that
the difference of the wavelength of Kα X-ray and the most energetic continuous X-ray becomes 4 times the
difference before the change of voltage. Find the atomic number of the target element. Take b = 1 and  =0.54.


Detailed Solution for Part Test - 6 (JEE Advanced) - Question 22

Part Test - 6 (JEE Advanced) - Question 23

Which of the following statement is correct with respect to the metal carbonyls of Ist  transition series?

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 23

Order of C–O bond strength; [Mn(CO)6]+ > [Cr(CO)6] > [V(CO)6]- > [Ti(CO)6]2- and [Ni(CO)4] > [Co(CO)4]- > [Fe(CO)4]2-
(A) True statement.
(B) As + ve charge on the central metal atom increases, the less readily the metal can donate electron density into
the π* orbitals of CO ligand to weaken the C–O bond.
(C) In the carbonylate anions, the metal has a greater electron density to be dispersed, with the result that M–Cπ
bonding is enhanced and the C–O bond is diminished in strength.

Part Test - 6 (JEE Advanced) - Question 24

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 24

Part Test - 6 (JEE Advanced) - Question 25

 Concentrated sulphuric acid is added followed by heating of each of the following test tubes labelled (i) to (v). 

Which of the following statement is incorrect about these observations ? 

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 25

Part Test - 6 (JEE Advanced) - Question 26

The following flow diagram represents the extraction of magnesium from sea water.

Which of the following options describes the correct reactants, products and reaction conditions ?

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 26

Part Test - 6 (JEE Advanced) - Question 27

When CS2 layer containing both Br2 and I2 (2 : 1) is shaken with excess of Cl2 water, the violet colour due to I2 disappears and a pale yellow colour appears in the solution. The disappearance of violet colour and appearance of pale yellow colour is due to the formation of :

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 27

Part Test - 6 (JEE Advanced) - Question 28

Which of the following has both face-mer and optical isomers ?

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 28

Part Test - 6 (JEE Advanced) - Question 29

An inorganic chloride (X) on heating with a sodium hydroxide solution liberates a colourless and non-inflammable gas having characterisitic odour. The salt (X) gives a red colouration with 4-nitrobenzene-diazonium chloride in the presence of sodium hydroxide solution. X is :

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 29

Part Test - 6 (JEE Advanced) - Question 30

A metal nitrate (X) gives a white precipitate with ammonia solution but the precipitate gets dissolved on adding ammonium salts. Lead dioxide and concentrated nitric acid on boiling with a dilute solution of metal nitrate (X) produces a violet-red (or purple) colour solution. Small amount of metal nitrate (X) gives a amethyst-red bead with borax in oxidising flame when cold. The cation of metal nitrate is :

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 30

*Multiple options can be correct
Part Test - 6 (JEE Advanced) - Question 31

Which of the following molecules or ions may act as a bidentate ligand ?

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 31

CO32- and NO3-are flexidentate ligands having 1 or 2 donor oxygen atoms (Source : Shriver Atkins)
C2O42-  – 2 donor oxygen atoms.
CH3C = N  – 1 donor nitrogen atom.

*Multiple options can be correct
Part Test - 6 (JEE Advanced) - Question 32

Which of the following statements is incorrect ?

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 32

(B) The statement is correct ; non-chiral because there is mirror plane through the metal bisecting the dien
ligand.

*Multiple options can be correct
Part Test - 6 (JEE Advanced) - Question 33

Which of the following statements is(are) false ?

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 33

(A) Statement is correct 

(B) Statement is false.
2e- + 2H+ + XeF2 → Xe + 2HF
SRP = +2.64V.
(C) Statement is false as Ne atoms being smaller do not trapp in the cavities formed by water molecules (ice) and
thus does not form clatharte compounds. Xe and Kr being bigger form clathrate compounds.
(D) Statement is false according to the following reaction : 2F2 + 2NaOH → OF2(g) + 2NaF + H2O.

*Multiple options can be correct
Part Test - 6 (JEE Advanced) - Question 34

Identify the true statement(s).

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 34

(A) As oxidation state of Mn increases, the difference in values of electronegativities of Mn and O decrease and thus
acidic character increase. In Mn2O7, the Mn is in highest oxidation state + 7. So, it is most acidic.
(B) It is factual.

Part Test - 6 (JEE Advanced) - Question 35

Statement-1 :  Argon is used in the laboratory for handling substances that are air-sensitive.

Statement-2 : Argon is inert towards chemical reactivity due to the completely filled valence shell electronic configuration, high ionization enthalpy and more positive electron gain enthalpy.

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 35

S1 and S2 both are correct statement and S2 is the correct explanation of S1 .

Part Test - 6 (JEE Advanced) - Question 36

Statement-1 : tris(ethane-1, 2-diamine)nickel(II) nitrate is homoleptic outer orbital complex and paramagnetic with two unpaired electrons.

Statement-2 : Complex tris(ethane-1, 2-diamine)nickel(II) nitrate when treated with a neutral solution of a thiosulphate produces a crystalline, green precipitate.

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 36

Statement-1 : [Ni (en)3] (NO3)2 is paramagnetic with two unpaired electrons and has octahedral geometry with sp3d2
hybridisation and as nd orbital participate in hybridisation it is outer orbital complex. As all ligands are of same type,
the complex is termed as homoleptic.

Part Test - 6 (JEE Advanced) - Question 37

Q. What is (Z) ?

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 37

This reaction is Vohhard method for estimation of manganese carried out in presence of ZnSO4 or suspended ZnO
which catalyses the oxidation.

Part Test - 6 (JEE Advanced) - Question 38

Q. The reaction 

     is an example of :

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 38

Part Test - 6 (JEE Advanced) - Question 39

Q. Which of the following products are formed when potassium bromide reacts with (Y) in  alkaline pH ?

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 39

Part Test - 6 (JEE Advanced) - Question 40

Match the compounds listed in column-I with characteristic(s) / type of reaction(s) listed in column-II. 

Column –I                 Column –II

    (A) XeF2                     (p) White solid
    (B) XeF4                     (q) Acts as oxidising agent.
    (C) XeF6                     (r) Undergoes addition reaction.
    (D) XeO3                    (s) Has lone pair(s) of electrons.
                                     (t) Gives disproportionation reaction with H2O or OH– 

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 40

*Answer can only contain numeric values
Part Test - 6 (JEE Advanced) - Question 41

  

(P) In how many M-O bonds, the bond lengths are equal ?

(Q) What is the oxidation state of the central metal ion ?

(R) What is the magnetic moment of the central metal ion ?

(S) How many peroxide linkages are there in the product formed by the reaction of [X] with H2O2 in acidic medium?

   


Detailed Solution for Part Test - 6 (JEE Advanced) - Question 41

*Answer can only contain numeric values
Part Test - 6 (JEE Advanced) - Question 42

Find out the oxidation state of sulphur in the compounds (A) to (C). Give average value without sign, if the compound(s) formed has more than one sulphur atoms. Does not consider the decimal. For example if the value is 3.5 than neglect  the decimal and consider it as 35

  


Detailed Solution for Part Test - 6 (JEE Advanced) - Question 42

*Answer can only contain numeric values
Part Test - 6 (JEE Advanced) - Question 43

Answer the following questions for the product [X].
    (i)    Number of negative monodenate ligand(s).
    (ii)    Effective atomic number of iron in the complex.
    (iii)    Number of unpaired electrons


Detailed Solution for Part Test - 6 (JEE Advanced) - Question 43

*Answer can only contain numeric values
Part Test - 6 (JEE Advanced) - Question 44

Give the number of characteristic bond(s) found in the various oxy-acids of phosphorus as given below.
    (P)    Number of P-O-P bond(s) in cyclotrimetaphosphoric acid.
    (Q)    Number of P-P bond(s) in hypophosphoric acid.
    (R)    Number of P-H bond(s) in hypophosphorus acid.
    (S)    Number of P-OH bond(s) in pyrophosphoric acid.


Detailed Solution for Part Test - 6 (JEE Advanced) - Question 44

Part Test - 6 (JEE Advanced) - Question 45

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 45

Part Test - 6 (JEE Advanced) - Question 46

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 46

Part Test - 6 (JEE Advanced) - Question 47

 Image of the point 'P' with position vector  

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 47

Part Test - 6 (JEE Advanced) - Question 48

 Let  are three noncoplanar vectors. If is perpendicular to then minimum value of x2 + y2 is

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 48

Part Test - 6 (JEE Advanced) - Question 49

The plane x + 2y + 3z = 7 is rotated about the line where it cut yz-plane by an angle θ. In the new position the plane contains the point (– 1, 0, 2). Then θ =

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 49

Part Test - 6 (JEE Advanced) - Question 50

A ten digited  number is formed without repeating any digit. The probability that the difference of the digits at equal distances from the begining and the end is always 1, is

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 50

Part Test - 6 (JEE Advanced) - Question 51

There are five urns of the following compositions : 2 urns (composition A1) with 2 white and 3 black balls each, 2 urns (composition A2)  with 1 white and 4 black balls each, 1 urn (composition A3) with 4 white balls and 1 black ball. A ball is chosen from one of the urns taken at random. It turned out to be white. The probability that the ball drawn is from the urn of the third composition, is

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 51

Part Test - 6 (JEE Advanced) - Question 52

Let A, B, C, D be real matrices (not necessarily square) such that AT = BCD, BT = CDA, CT = DAB and DT = ABC, where AT represents transpose of A. Then for the matrix S = ABCD

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 52

*Multiple options can be correct
Part Test - 6 (JEE Advanced) - Question 53

P is the foot of the perpendicular dropped from origin O on the line of intersection of the planes x – 2y + 3z = 5 and 2x + 3y + z + 4 = 0 then

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 53

*Multiple options can be correct
Part Test - 6 (JEE Advanced) - Question 54

k subsets of a set A (having n elements) are chosen at random with replacement. The probability that their intersection is 

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 54

*Multiple options can be correct
Part Test - 6 (JEE Advanced) - Question 55

p and q are chosen randomly from the set {1, 2, 3, ...., 10} with replacement. The probability that the roots  of the equation x2 + px + q = 0 are

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 55

*Multiple options can be correct
Part Test - 6 (JEE Advanced) - Question 56

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 56

Part Test - 6 (JEE Advanced) - Question 57

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 57

Part Test - 6 (JEE Advanced) - Question 58

Statement-1 : If two balls are drawn without replacement from an urn containing 3 red and 2 blue balls, then the probability of the event that 'second ball drawn is blue' is 1/4.

Statement-2 : If two balls are drawn without replacement from an urn containing 3 red and 2 blue balls, then the probability of the event that 'first ball drawn is blue' is equal to the probability of the event that 'second ball drawn is blue' .

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 58

Part Test - 6 (JEE Advanced) - Question 59

Q. The quadratic form of matrix A =   is 

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 59

Part Test - 6 (JEE Advanced) - Question 60

Q. 

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 60

Part Test - 6 (JEE Advanced) - Question 61

Q. If number of distinct terms in a quadratic form is 10, then number of variables in quadratic form is 

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 61

Part Test - 6 (JEE Advanced) - Question 62

Detailed Solution for Part Test - 6 (JEE Advanced) - Question 62

*Answer can only contain numeric values
Part Test - 6 (JEE Advanced) - Question 63


Detailed Solution for Part Test - 6 (JEE Advanced) - Question 63

*Answer can only contain numeric values
Part Test - 6 (JEE Advanced) - Question 64

Find the volume of the tetrahedron, if lengths of two opposite sides are 2 and 4 respectively and the shortest distance between them is 6 and these sides are inclined at 30º.


Detailed Solution for Part Test - 6 (JEE Advanced) - Question 64

*Answer can only contain numeric values
Part Test - 6 (JEE Advanced) - Question 65


Detailed Solution for Part Test - 6 (JEE Advanced) - Question 65

*Answer can only contain numeric values
Part Test - 6 (JEE Advanced) - Question 66

 


Detailed Solution for Part Test - 6 (JEE Advanced) - Question 66

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