Part Test 6 - NEET 2020


180 Questions MCQ Test NEET Mock Test Series & Past Year Papers | Part Test 6 - NEET 2020


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This mock test of Part Test 6 - NEET 2020 for NEET helps you for every NEET entrance exam. This contains 180 Multiple Choice Questions for NEET Part Test 6 - NEET 2020 (mcq) to study with solutions a complete question bank. The solved questions answers in this Part Test 6 - NEET 2020 quiz give you a good mix of easy questions and tough questions. NEET students definitely take this Part Test 6 - NEET 2020 exercise for a better result in the exam. You can find other Part Test 6 - NEET 2020 extra questions, long questions & short questions for NEET on EduRev as well by searching above.
QUESTION: 1

A radioactive nuclide A decays to nuclide B which further decays to C. Their disintegration constant are λ1 and λ2 respectively. At t = 0 only nuclei A are present. Number of nuclei A at t = 0  is N0

Solution:

Rate of decay of A keeps on decreasing continuously because concentration of A decreases with time ⇒ A is false Initial rate of production of B is λ1N° and rate of decay is zero. With time, as the number of B atoms increase, the rate of its production decreases and its rate of decay increases. Thus the number of nuclei of B will first increase and then decrease B is the correct choice
The initial activity of B is zero whereas initial activity of A is λ1N0 ⇒ C is false.
As time t → ∞ : NA = 0, NB = 0 and NC = N0 ⇒ D is false 

QUESTION: 2

An enclosure filled with helium is heated to a temperature of 400 K. Helium atom emerges out of the enclosure. The mean debroglie wavelength of the helium atoms is :

Solution:


⇒  ⇒ 
Substituting values :  

QUESTION: 3

When a metallic surface is illuminated with monochromatic light of wavelength λ, the stopping potential is 5 V0. When the same surface is illuminated with light of wavelength 3λ, the stopping potential is V0. Then the work function of the metallic surface is :

Solution:


QUESTION: 4

No. of identical photons incident on a perfectly black body of mass m kept at rest on smooth horizontal surface. Then the acceleration of the body if n number of photons incident per second is (Assume wavelength of photon to be λ) :

Solution:

Change in momentum due to photon = h/λ
F = rate of change of momentum 
 ⇒ 

QUESTION: 5

If we assume that penetrating power of any radiation/particle is inversely proportional to wavelength of the particle then :            

Solution:


∴ For higher m and q ;
λ will be smaller.
For an 'α' particle; both 'm' and 'q' are higher hence lesser is the wavelength. 
As, (penetrating power) ∝ Energy From above; penetrating power of an α-particle is more than that of a proton. 

QUESTION: 6

Two radioactive sources A and B initially contain equal number of radioactive atoms. Source A has a half-life of 1 hour and source B has a half-life of 2 hours. At the end of 2 hours, the ratio of activity of source A to that of B is : 

Solution:

QUESTION: 7

The radionuclide 238U decays by emitting an alpha particle.    

238U → 234Th + 4He
The atomic masses of the three isotopes are. 
238U      238.05079 amu
238U      234.04363 amu
4He       4.00260 amu

What is the maximum kinetic energy of the emitted alpha particle? Express your answer in Joule. 

Solution:

Mass defect  = (238.05079 - 234.04363 - 4.00260) u = 4.56 x 10-3 u
= 4.56 x 10-3 x 1.66 x 10-27 = 7.57 x 10-30 kg
mc2 = 7.57 x 10-30 x 9 x 1016 = 6.8 x 10-13 . 

QUESTION: 8

Choose the correct statement.        

Solution:

The correct statement in nuclear force is not a central force. 

QUESTION: 9

A beam of electrons striking a copper target produces X-rays.Its spectrum is as shown. Keeping the voltage same if the copper target is replaced with a different metal, the cut-off wavelength and characteristic lines of the new spectrum will change in comparision with old as :    

Solution:

Cut- off wavelength of emitted X-rays depends on maximum kinetic energy of incident electrons on the target and is independent of nature of target. The characteristic lines depend on nature of material of target.

QUESTION: 10

Both the frequency and the intensity of a beam of light falling on the surface of photoelectric material are increased by a factor of two. This will :            

Solution:

The maximum K.E. of ejected photoelectrons is (K.E)max = hv – φ0
If the frequency of photon is doubled, maximum kinetic energy of photon electrons becomes 

Photo current 
If intensity and frequency both are doubled. the photocurrent remains same. 

QUESTION: 11

The photon radiated from hydrogen corresponding to 2nd line of Lyman series is absorbed by a hydrogen like atom ‘X’ in 2nd excited state. As a result the hydrogen like atom ‘X’ makes a transition to nth orbit. Then,

Solution:

Energy of nth state in Hydrogen is same as energy of 3nth state in Li++.
∴ 3 → 1 transition in H would give same energy as the 3 x 3 → 1 x 3 transition in Li++

QUESTION: 12

The angular momentum of an electron in first orbit of Li++ ion is :    

Solution:

Angular momentum  (∵ n = 1)

QUESTION: 13

An electron of mass ' m ', when accelerated through a potential V has de-Broglie wavelength λ. The de-Broglie wavelength associated with a proton of mass M accelerated through the same potential difference will be:

Solution:

 ⇒  
K = qV is same for both proton and electron. 

QUESTION: 14

Two hydrogen atoms are in excited state with electrons residing in n = 2. First one is moving towards left and emits a photon of energy E1 towards right. Second one is moving towards left with same speed and emits a photon of energy E2 towards left. Taking recoil of nucleus into account during emission process

Solution:

In the first case K.E. of H-atom increases due to recoil whereas in the second case K.E. decreases due to recoil.
∴ E2 > E1 

QUESTION: 15

In a hydrogen atom following the Bohr’s postulates the product of linear momentum and angular momentum is proportional to (n)x where ‘n’ is the orbit number. Then ‘x’ is :

Solution:

Linear momentum ⇒
angular momentum ⇒  mr α n
∴ product of linear momentum and angular momentum α n0 

QUESTION: 16

One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is :

Solution:

For 2nd line of Balmer seires in hydrogen specturm

For Li2+ : 
which is satisfied by transition n = 12 to n = 6

QUESTION: 17

The ratio of wave length of a photon of energy E and deBroglie wavelength of an electron of mass 'm' having the kinetic energy also E is : (Speed of light = c)         

Solution:

For an electron with kinetic energy E : 
 and for a photo  : 
∴ 

QUESTION: 18

An orbital electron in the ground state of hydrogen has an angular momentum L1 and an orbital electron in the first orbit in the ground state of lithium (double ionised positively charged) has an angular momentum L2 .Then :

Solution:

Doubly ionised positively lithium ion is a hydrogen like atom :
 
which depends only on the value of n. Also for both n = 1.

QUESTION: 19

If in the first orbit of a hydrogen atom the total energy of the electron is -21.76 x 10-19J, then its electric potential energy will be :

Solution:

Potential energy = 2 Total energy = 2 x – 21.76 x 10-19 = – 43.52 x 10-19

QUESTION: 20

In an x - ray tube, if the accelerating potential difference is changed, then:      

Solution:

As the accelerating potential difference is changed only the minimum wavelength changes.
It has no effect on wavelengths of characteristic x-rays (whether they are produced or not) 

QUESTION: 21

All electrons ejected from a surface by incident light of wavelength 200 nm can be stopped before travelling 1 m in the direction of uniform electric field of 4 N/C. The work function of the surface is:

Solution:

The electron ejected with maximum speed vmax are stopped by electric field E = 4N/C after travelling a distance d = 1m 
∴ 
The energy of incident photon  = 1240/200 = 6.2 eV
From equation of photo electric effect

∴ 

QUESTION: 22

The energy ratio of two Kα  photons obtained in x-ray from two metal targets of atomic numbers Z1 and Z2 is: 

Solution:

The energy of Kα x-ray photons is directly propotional to (Z-1)2 . The energy ratio of two Kα photons obtained in x-ray from two metal targets of atomic numbers

QUESTION: 23

Because of the space charge in a diode valve,

Solution:
QUESTION: 24

A point source of light is used in a photoelectric effect.If the source is removed farther from the emitting metal,the stopping potential

Solution:
QUESTION: 25

The X-ray beam coming from an X-ray tube

Solution:
QUESTION: 26

Cutoff wavelength of X-rays coming from a Coolidge tube depends on the

Solution:
QUESTION: 27

The mass number of a nucleus is

Solution:
QUESTION: 28

The radii of nuclei of two atoms are in ratio 3/2 . Assuming them to be Hydrogen like atom, the ratio of their orbital radius for K shell will be (assume no. of proton = No. of neutron for each atom)     

Solution:
QUESTION: 29

A semiconducting device is connected in a series circuit with a battery and a resistance. A current is found to pass through the circuit. If the polarity of the battery is reversed, the current drops to almost zero. The device may be

Solution:
QUESTION: 30

A radioactive nucleus ' X ' decays to a stable nucleus ' Y '. Then the graph of rate of formation of ' Y' against time ' t ' will be : 

Solution:
QUESTION: 31

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

Solution:


 ⇒ z1 = 26
 ⇒ z2 = 31
Therefore, 4 elements lie between A and B.

QUESTION: 32

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

Solution:

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

QUESTION: 33

An α 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.

Solution:

 
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. 

QUESTION: 34

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 -

Solution:

Maximum energy of radiation incident on H-sample = KEmax of electron + 13.6 eV = 51eV this energy corresponds to the transition n = 4 → n = 1 in Helium
For electrons of the He to get excited to n = 4 

QUESTION: 35

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 : 

Solution:

Potential of the sphere at any time

QUESTION: 36

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

Solution:

Loss in K.E. will be used in splitting of Deutron.
Let us calculate the minimum K.E. required for the loss of 2.2 MeV. For this we must assume perfectly inelastic collision. 
 = 
maximum loss of K.E. will be in perfectly inelastic collision. By energy conservation. 
⇒ 

⇒ 
∴  

QUESTION: 37

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

Solution:

no of moles 1H2 consumed

∴ m = 0.1 g

QUESTION: 38

A hydrogen atom is in the 4th excited state, then:

Solution:

The hydrogen atom is in n = 5 state. 

Max. no of possible photons = 4
To emit photon in ultra violet region, it must jump to n = 1, because only Lyman series lies in u.v. region. Once it jumps to n = 1 photon, it reaches to its ground state and no more photons can be emitted. So only one photon in u. v. range can be emitted.
If H atom emits a photon and then another photon of Balmer series, option D will be correct.

QUESTION: 39

In the photoelectric experiment, if we use a monochromatic light, the I - V curve is as shown. If work function of the metal is 2eV, estimate the power of light used. (Assume efficiency of photo emission = 10-3%, i.e. number of photoelectrons emitted are 10-3% of number of photons incident on metal.) 

Solution:

The energy of incident photons is given by
hv = eVs + φ0 = 2 + 5 = 7 eV (Vs is stopping potential and φ0 is work function)
Saturation current (η is photo emission efficiency) 
∴ P = 7W.

QUESTION: 40

If first excitation potential of a hydrogen like atom is V volt, then the ionization energy of this atom will be :

Solution:

First excitaion energy  
∴ 
∴  

QUESTION: 41

The element which has kα x-rays line of wavelength 1.8 Å is 

Solution:

 ⇒   

QUESTION: 42

When an impurity is doped into an intrinsic semiconductor,the cunductivity of the semiconductor

Solution:
QUESTION: 43

A radioactive nucleus can decay by either emitting an α particle or by emitting a β particle. Probability of α decay is 75% while that of β decay is 25%. The decayconstant of α decay is λ1 and that of β decay is 

Solution:

QUESTION: 44

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 is minimum.  Then the number of nuclei of element A at time t = t0 is 

Solution:



⇒ 

QUESTION: 45

A cobalt (atomic no. = 27) target is bombarded with electrons, and the wavelengths of its characteristic x-ray spectrum are measured. A second weak characteristic spectrum is also found, due to an impurity in the target. The wavelengths of the Kα lines are 225.0 pm (cobalt) and 100.0 pm (impurity). Atomic number of the impurity is (take b = 1)

Solution:

Using Mosely's law for both cobalt and impurity
 ⇒  
⇒  and  
⇒ 

QUESTION: 46

Match column-I with column-II and select the correct answer using the codess given below :

Solution:

(a) Tin-cassiterite, SnO2
(c) Iron-siderite, FeCO3
Therefore, (B) option is correct.

QUESTION: 47

Which is not a correct statement ? 

Solution:

Sulphide ore is roasted in presence of excess of air or O2 below its melting point to convert into the oxide and to remove the impurities of S, P, Sb etc., as their volatile oxides. In some cases roasting of certain sulphide ores provide directly the metals. 

QUESTION: 48

Which metal is extracted using a hydrometallurgical process involving complexation ?

Solution:

4Ag + 8CN- + 2H2O → 4[Ag(CN)2]- (soluble complex) + 4OH-
2[Ag(CN)2]- + Zn → 2Ag + [Zn(CN)4]2- Hydrometallurgy involves the dissolution of ores in suitable chemical reagent followed by precipitation of the metal from their solution by more electropositive metal.

QUESTION: 49

Which of  the following acids is monobasic? 

Solution:

It has one replaceable hydrogen.

QUESTION: 50

ClO2 is the anhydride of :

Solution:

As it gives mixture of two acids, HCIO2 and HCIO3, it is called as mixed anhydride of HCIO2 and HCIO3.
2ClO2+ H2O → HClO+ HClO3 

QUESTION: 51

.... is obtained when ammonium dichromate is heated.

Solution:

QUESTION: 52

In the Ostwald’s process, nitric acid is prepared by the catalytic oxidation of :

Solution:

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) ; NO + 1/2O2 → NO2
3NO2 + H2O → 2HNO3 + NO.

QUESTION: 53

The thermal stability of hydrides of oxygen family is in order :

Solution:

Down the group (H — E) bond dissociation enthalpy decreases as (H — E) bond length increases because of increase in size of cation. Thus thermal stability of hydrides also decreases. 

QUESTION: 54

Which of the following gaseous molecules is monoatomic ?    

Solution:

Noble gases exist as monoatomic molecules.

QUESTION: 55

The transition elements have a general electronic configuration :

Solution:

General electronic configuration of transition elements is [Noble gas] 

QUESTION: 56

The stable oxidation state of Cr is :

Solution:

Cr3+ is most stable because in aqueous solution it has higher CFSE on account of half filled t32g energy level of 3d orbitals in octahedral spliting.

QUESTION: 57

Which of the following ions gives colourless aqueous solution? 

Solution:

Cu+ is diamagnetic with 3d10 configuration; so no d-d transition is possible and thus colourless.

QUESTION: 58

The developer used in photography is an alkaline solution of :

Solution:

QUESTION: 59

At 300ºC, FeCl3 :

Solution:

QUESTION: 60

What is the charge on the complex [Cr(C2O4)2(H2O)2] formed by Cr(III) ?

Solution:

Chromium has + 3 charge. H2O is neutral ligand while C2O42- is negative ligand. Charge on the complex is algebric sum of the charges on metal ion and ligands present in the coordination sphere.
So, the charge on the complex will be 3 + 2(-2) = - 1

QUESTION: 61

The correct IUPAC name for the compound [Co(NH3)4CI(ONO)]Cl is : 

Solution:

Cation is named first followed by anion. In coordination sphere the ligands are named alphabetically followed by name of metal ion and then its oxidation state in Romman numeral. In [Co(NH3)4Cl(ONO)]+CI-, cobalt is in +3 oxidation state and the complex is cationic. ONO ligand attach to metal ion through O atom ; so — O — is used while naming as nitrito before the name of metal ion. So IUPAC name given in option (D) is correct.

QUESTION: 62

EAN of a metal carbonyl M(CO)x is 36. If atomic number of metal M is 26, what is the value of x ?

Solution:

Let x = number of lone pairs of electrons donated to central metal ion.
EAN = number of electrons in metal ion + number of lone pairs of electrons donated by the ligands. 
Therefore, 26 + 2x = 36 or x = 10/2 = 5

QUESTION: 63

Which one of the following complexes produces three moles of silver chloride when its one mole is treated with excess of silver nitrate ?

Solution:

As 3 mole of CI- are present out side the coordination spehre i.e. in ionisation sphere so according to the following reaction, 3CI-+ 3Ag+ → 3AgCI, one mole of complex will give three moles of AgCI precipitate. 

QUESTION: 64

Which statement is incorrect ?

Solution:

CO is stronger field ligand, so it compels for pairing of electrons. 


So, it is tetrahedral but not paramagnetic as all electrons are paired. 

QUESTION: 65

Which has maximum paramagnetic nature ?

Solution:

[Mn(H2O)6]2+ contains maximum number of unpaired electrons i.e. five. so it has maximum paramagnetic character.

QUESTION: 66

Which one of the following complexes has highest stability ?

Solution:

It is a chelate complex having three five-membered rings. It attains extra stability through chelation. 

QUESTION: 67

[Co(NH3)5NO2]Cl2 and [Co(NH3)5ONO]Cl2 are related to each other as :

Solution:

NO2- is an ambidentate ligand and thus it can linkage to metal ion through O as well as through N. Hence it shows linkage isomerism.

QUESTION: 68

When a mixture of solid NaCl, solid K2Cr2O7 is heated with concentrated H2SO4, deep red vapours are obtained. This is due to the formation of : 

Solution:

QUESTION: 69

Precipitate of PbSO4 is soluble in :

Solution:

PbSO4 + 2CH3COONH4 → (NH4)2SO4 + (CH3COO)2Pb.

QUESTION: 70

Nitrate is confirmed by ring test. The brown colour of the ring is due to formation of :

Solution:

2NO3- + 4H2SO4 + 6Fe2+ → 6Fe3+ + 2NO + 4SO42-+ 4H2O.
Fe2+ + NO + SO42- → [Fe(NO)]2+SO42-

QUESTION: 71

Nessler’s reagent is :

Solution:

It is alkaline solution of K2[Hgl4].

QUESTION: 72

In which of the pair the precipitates are red and black coloured respectively and both are soluble in excess KI solution ? 

Solution:

(red) + 2l- → [Hgl4]2- (soluble colourless complex) ; (black) + l- → [Bil4]- (soluble orange complex). 

QUESTION: 73

The solution of sodium meta aluminate on diluting with water and then boiling with ammonium chloride gives:

Solution:


Hydroxide ion concentration is reduced owing to the formation of ammonia (a weak base) which escapes on boiling.

QUESTION: 74

Precipitate of Zn(OH)2 is soluble in :

Solution:

QUESTION: 75

If crimson flame is given when an inorganic mixture is tested by flame test. It may be due to the presence of 

Solution:

Strontium chloride gives crimson colour flame in Bunsen burner. BaCl2 -apple green, CaCl2 -brick red, KCI - lilac(violet). 

QUESTION: 76

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

Solution:

Order of C–O bond strength; [Mn(CO)6]+>[Cr(CO)6] > [V(CP)6]- > [Ti(CO)6]2- and [Nli(CO)4] > [Co(CO)4]- > [Fe(CO)4]2- 
(A) True statement.
(B) As + ye 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. 

QUESTION: 77

Solution:

QUESTION: 78

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 ?

Solution:

(ii) 2NaBr + 2H2SO4 → Na2SO4 + SO2 + Br2 (reddish brown) + 2H2
(iii) 2Cu(NO3)2 + 2H2SO4 → 2CuSO4 (blue solution) + 4NO2 (reddish brown) + O2 + 2H2
(iv) K4[Fe(CN)6] + 6H2SO4 + 6H2O → 2K2SO4 + FeSO4 + 3(NH4)2SO4 + 6CO  (neutral and colourless) CO bums with blue flame.
(v) KCI + H2SO4 → KHSO4 + HCI (colourless) 

QUESTION: 79

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 ?

Solution:



 

QUESTION: 80

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 :

Solution:

QUESTION: 81

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

Solution:

(A) [Co (NH3)3 (NO3)3] shows fac - mer isomerism but does not show optical isomerism,
(B) [Co (ox)3]3- does not show fac - mer isomerism but shows optical isomerism.

QUESTION: 82

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 :

Solution:


 (red colouration) + Cl- H2O

QUESTION: 83

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 amethystred bead with borax in oxidising flame when cold. The cation of metal nitrate is :

Solution:


On adding ammonium salts, the reaction proceeds towards left.
5PbO2 + 2Mn2+ + 4H+→ 2MnO4- (violet-red or purple colouration) + 5Pb2+ + 2H2

QUESTION: 84

Match the name of the processes given in column-I with type of metallurgical methods given in column-II. 

Solution:

(a) Hall-Heroult process is the electrolytic reduction of molten Al2Odissolved in cryolite or fluorspar.
(b) Dow's sea water process involves the isolation of Mg from sea water as MgCl2 and then electrolytic reduction of molten MgCl2 dissolved in CaCl2 and NaCl.
(c) Hoop's process is the electrolytic purification of impure aluminium. The cell has three liquid layers middle one contains fluorides of Na+, Ba2+ and Al3+, lower most layer has impure Al.
(d) Mac-Arthur forest cyanide process is the process used for the extraction of gold and silver. Extraction of gold and silver involves leaching the metal with CN. The metal is later recovered by displacement method.

QUESTION: 85

'Spin only' magnetic moment of Ni in [Ni(dmg)2] is same as that found in : 

Solution:

Ni (dmg)2 complex is square planar and dimagnetic.
Pt (II) – 5d8 configuration. Complex is square planar and therefore, dimagnetic So μ = 0

QUESTION: 86

Which of the following complexes shows ionisation as well as hydrate isomerism ?

Solution:


Both compounds have same formula, but gives different ions in solution, hence exhibit, ionization isomerism. It also shows hydrate isomerism as shown above because hydrate isomers differ by whether or not a water molecule(s) is / are directly bonded to the metal ion or merely present as free water molecule(s) in the crystal lattice. 

QUESTION: 87

Which of the following statements is true for the compounds [X]. [Y] and [Z] 7 

Solution:




Na2Cr2O7 + H2SO4 → 2CrO3 (chromic anhydride) + Na2SO4 + H2O. 

QUESTION: 88

The high reactivity and low melting point of white phosphorous is due to :

Solution:

Each phosphours is sp3 hybridised. But there is much deviation in P–P–P bond angle 60° from normal tetrahedral angle (109.5°). So it has angle strain and thus reactive 


P4 exists as discrete molecule and thus it has low van der Waal's force of attraction. This is the reason that P4 has low melting point.

QUESTION: 89

When Cl2 water in excess is added to a salt solution containing chloroform layer turns pale yellow. Salt contains :

Solution:

Br- + Cl2 →  Br2 + 2Cl- ; Br2 + Cl2 → 2BrCl (pale yellow). 
Therefore, salt contains Br-

QUESTION: 90

A metal nitrate solution reacts with dilute hydrochloric acid to give a white precipitate which is soluble in concentrated potassium chloride. White precipitate on passing of hydrogen sulphide gas is converted into black precipitate. The black precipitate on boiling with hydrogen peroxide (3%) is converted again to a white precipitate which is soluble in ammonium acetate. The cation of the metal nitrate is :

Solution:

Pb2+ + 2C- → PbCl↓ (white) ; PbCl2 + 2Cl- → [PbCl4]2- (colourless soluble complex)
PbCl↓ + H2S → PbS ↓ (black) + 2HCI
PbS + 4H2O2 → PbSO4 ↓ (white) + 4H2
PbSO4 ↓ + 2CH3COONH4 → (NH4)2SO4 + (CH3COO)2Pb 

QUESTION: 91

Mark the incorrect about value of  of Z ,related oto biodiversity

Solution:
QUESTION: 92

Mark the incorrect statement -

Solution:
QUESTION: 93

For which of the following texa relationship between species richness and area is  turns out to be rectangular hyperbola-

Solution:
QUESTION: 94

Consider the following

(a) In rivet popper hypothesis flight safety is proper functioning of ecosystem

(b) Tilman shows increase productivity increase biodivesity

Solution:
QUESTION: 95

Among the given example which have lowest species diversity

Solution:
QUESTION: 96

Biologist Make the stastical comparison to come with the gross estimate of species on earth after -

Solution:
QUESTION: 97

Which might be criteria to delineate species -

Solution:
QUESTION: 98

Which is correct about tropical environment

Solution:
QUESTION: 99

Consider the following matched-

(1) Birds facing extinction –   12 percent

(2)  Nile perch – alien species invasion

(3)  Passenger pigeon  –  overexploitation

How many are correctly matched- 

Solution:
QUESTION: 100

A stable ecosystem have all the feature except-

Solution:
QUESTION: 101

Consider the following statement-

(a) loss in biodiversity lead to decrease in resistance to environment perturbations

(b) In world summit 190 counteries taken pledge to take appropriate measure for conservation of biodiversity

Solution:
QUESTION: 102

Which of the following is not example of ex-situ conservation

Solution:
QUESTION: 103

Which of the following is present in least number  -

Solution:
QUESTION: 104

It is least  possible in  less develop countries to conserve biodiversity as-

Solution:
QUESTION: 105

Consider the following statement -

(a)  Dodo from Mauritius and Quagga from Australia get extinct

(b) hotspot are under more protection than other biodiversity rich region

Solution:
QUESTION: 106

Which is wrong –

Solution:
QUESTION: 107

Which is not result of loss in biodiversity-

Solution:
QUESTION: 108

Consider the following statement-

(1) Value of Z lies is lower  for entire continent

(2) Value of Z not depend on taxonomic group but depend on region considered

Solution:
QUESTION: 109

Harmful effect of Air pollution depend upon-

Solution:
QUESTION: 110

Which may be most deleterious effect of PM 2.5 present in air pollution-

Solution:
QUESTION: 111

Mark the correct statement-

Solution:
QUESTION: 112

In India which of the following act related to environment come into force first-

Solution:
QUESTION: 113

Which of the following is  suspended impurities-

Solution:
QUESTION: 114

In sewage water low level  DO and high level of BOD means-

Solution:
QUESTION: 115

A sewage water have high level of COD, low level of BOD and later  low level of DO means-

Solution:

COD high and low BOD  means more inorganic waste

QUESTION: 116

High concentration of DDT in food chain is responsible for decline in bird population it involve -

Solution:
QUESTION: 117

Which step occur last in Natural Eutrophication of lake-

Solution:
QUESTION: 118

Mark the incorrect-

Solution:
QUESTION: 119

What is not true about  ‘Ecosan’ –

Solution:
QUESTION: 120

Mark the incorrect-

Solution:
QUESTION: 121

Global warming can be  controlled most  effectively by-

Solution:

only trees are effective reservoir to bring CO2 out from atmosphere

QUESTION: 122

Montreal protocol is signed in –

Solution:
QUESTION: 123

Reforestation involve all except-

Solution:
QUESTION: 124

Which is not effect of warm water as a pollutant-

Solution:
QUESTION: 125

Electrostatic precipitator  is useful against-

Solution:
QUESTION: 126

Accelearated eutrophication is due to waste water released from-

Solution:
QUESTION: 127

In Arcata marsh seeded bacteria and fungi remove heavy metal by-

Solution:
QUESTION: 128

COP held in 2016 on climate change occur at-

Solution:
QUESTION: 129

Primary production and primary productivity is different as

Solution:
QUESTION: 130

Which ecosystem have lowest primary productivity-

Solution:

desert have lowest primary productivity

QUESTION: 131

Productivity of which group is lowest-

Solution:

higher is trophic level lower is productivity

QUESTION: 132

In oceans primary productivity occur  in –

Solution:

deep sea have benthic zone where no primary productivity occur

QUESTION: 133

In tropical forest a plant  show slow growth means

Solution:

primary productivity is high but more respiration loss

QUESTION: 134

Low value of NPP in desert  is probably attributed to-

Solution:

desert show low GPP

QUESTION: 135

Leaching is responsible for precipitation of  ………… as unavailable substance for plant-

Solution:
QUESTION: 136

Humification process is faster in –

Solution:

temperate forest show fastest humification

QUESTION: 137

Decomposition is slowest  in dead –

Solution:

wood have lignin so slow decompose

QUESTION: 138

Mark the incorrect-

Solution:

Humus production require DFC

QUESTION: 139

Standing crop of any trophic level is calculated by-

Solution:
QUESTION: 140

Which is correct regarding ecological pyramid-

Solution:
QUESTION: 141

Most costly ecosystem services is among following –

Solution:
QUESTION: 142

Which process not occur separately not with other three -

Solution:
QUESTION: 143

Which is not seral stage-     

Solution:
QUESTION: 144

In a population of 100 warblers 4 have small beak and rest have long beak ,if long beak allele B  is dominant over small beak allele b than frequency of allele b in population if population is in genetic equlilibrium -

Solution:

hardy Weinberg numerical

QUESTION: 145

Which is not example of adavtive radiation in Marsupial -

Solution:
QUESTION: 146

Which of the following is incorrect about man-made selection –

Solution:

man-made selection not give survival advantage in nature

QUESTION: 147

According to Darwin for origin of species which is correct -

Solution:
QUESTION: 148

Which of the following is  common ancestor of turtle and lizards-

Solution:
QUESTION: 149

Which is incorrect about homo-sapiens-

Solution:
QUESTION: 150

Pouch mammals present  in Australia till now –

Solution:
QUESTION: 151

Conifers are maximum in  –

Solution:
QUESTION: 152

Mark the correct –

Solution:
QUESTION: 153

Which gas is not taken up by miller in his experiment -

Solution:
QUESTION: 154

Which gas not released from molten mass present on early atmosphere-

Solution:
QUESTION: 155

Which of the following is an example of convergent evolution -

Solution:
QUESTION: 156

Adaptive radiation can  become example of convergent evolution when -

Solution:
QUESTION: 157

Which is example of antropogenic selection –

Solution:

Correct Answer :- d

Explanation : a) Natural substrates or pleiotropic effects provide adaptive explanations for the selection of resistance mechanisms prior to anthropogenic pesticide exposure, resulting in the fixation of intrinsic resistance, or co‐optable pre‐resistance adaptations. 

b) Use of antibiotics resulted in antibiotic-resistant super-bugs.

c) The combination of high-yield crop varieties and the widespread use of inorganic fertilizers markedly improved crop production, with clear benefits in terms of food security. This has translated to excessive anthropogenic release of reactive nitrogen and phosphate into the environment. Inorganic fertilizers have pushed the global nitrogen and phosphorus cycles well beyond their estimated safe operating space18, with considerable negative effects on biodiversity, human health and the atmosphere.

QUESTION: 158

Which of the following is wrongly matched-

Solution:
QUESTION: 159

Which of the following groups of  beverages are product of distillation of fermemted broth-

Solution:
QUESTION: 160

Consider the following steps in sewage treatment plant

(1) filtration  and sedimentation

(2) anaerobic digestion

(3) digestion of organic waste aerobically

(4) agitation in aeration tank                                      

Mark the correct order-

Solution:
QUESTION: 161

Water soluble waste removed through-

Solution:
QUESTION: 162

Consider the following statements-

a) LAB grow in milk and convert it into curd

b) digestion of milk protein  and coagulation of milk occur with help of acid

c)  digestion of milk protein  complete in curd making

How many statements are correct-

Solution:
QUESTION: 163

Consider the following statements and mark wrong statement  - 

Solution:
QUESTION: 164

Which is correct about swiss cheese-

a)texture ,flavour and taste of this cheese develop from microbes

b) texture in swiss cheese given by propinobacterium

Mark correct option-

Solution:
QUESTION: 165

Which one is/ are correctly mached-

a) Rouquefort cheese and swiss cheese  -  fungi give taste and bacterium give texture

b) Citric acid and cyclosporin A  -  produced by yeast

c) Bread and statins  -  produced  by yeast

Mark the correct option -