datavorous/entrance-exam-dataset huggingface dataset:
Match the items given in Column I with those in Column II and select the correct option given below:

Column - I	Column - II

a.	Fibrinogen	i.	Osmotic balance

b.	Globulin	Ii.	Blood clotting

c.	Albumin	iii.	Defense mechanism

Options:
A a - i, b - iii, c - ii
B a - i, b -ii, c - iii
C a - iii, b - ii, c -i
D a - ii, b - iii, c - i

The correct answer is:
a - ii, b - iii, c - i

Solution:
Albumins, globulins and fibrinogen are the important proteins present in the plasma. Fibrinogen is an important blood clotting factor. Globulin is primarily involved in the defence mechanism of the body. Gamma globulins are antibodies that function in the immune responses of the body. Albumins maintain the osmotic balance needed to draw water from surrounding tissue fluid into the capillaries. This action is required to maintain blood volume and pressure.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2018
Match the items given in Column I with those in Column II and select the correct option given below:

Column - I	Column - II

a.	Tidal volume	i.	2500 - 3000 mL

b.	Inspiratory Reserve volume	ii.	1100 - 1200 mL

c.	Expiratory Reserve volume	iii,	500 - 550 mL

d.	Residual volume	iv.	1000 - 1100 mL

Options:
A a - i, b - ii, c - ii, d - iii
B a - iii, b -i, c - iv, d - ii
C a - iii, b - ii, c - i, d - iv
D a - iv, b - iii, c - ii, d - i

The correct answer is:
a - iii, b -i, c - iv, d - ii

Solution:
Tidal Volume (TV) is the volume of air inspired or expired during a normal breath. It is about 500-550 mL.- a-iii
Inspiratory Reserve Volume (IRV) is the extra amount of air that can be inspired directly after a normal inspiration. It is about 2500-3000 mL.- b-i
Expiratory Reserve volume (ERV) is the extra amount of air that can be expired forcibly after a normal expiration. It is about 1000-1100 mL.- c-iv
Residual Volume (RV) is the volume of air which remains still in the lung after the most forceful expiration. It is about 1100- 1200 mL.- d-ii
Therefore, option (2) is correct a - iii, b -i, c - iv, d - ii

Topic: Biology, Breathing and Exchange of Gases, NEET, NEET 2018
Match the items given in List A with those of the items of List B
The correct match is
$\begin{array}{llll}A & B & C & D\end{array}$

Options:
A I IV V III
B V I II IV
C IV II V I
D III I IV V

The correct answer is:
IV II V I

Solution:
$\begin{array}{llll}\text { A } & \text { B } & \text { C } & \text { D } \\ \text { IV } & \text { II } & \text { V } & \text { I }\end{array}$
(A) $|x|+|x-2|$ is continuous for all real values of $x$.
(B) cosech $x$ is defined for $x \in R-\{0\}$.
So, it is continuous, $\forall x \in R-\{0\}$.
(C) $x-[x]=\{x\}$ is discontinuous at all integers.
(D) $\sqrt{2-x}$ has RHL does not exist at $x=2$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Match the items in Column I with its main use listed in Column II:

Column I	Column II

A	Silica gel	i.	Transistor

B	Silicon	ii.	Ion-exchanger

C	Silicone	iii.	Drying agent

D	Silicate	iv.	Sealant

Options:
A A - iii, B - i, C - iv, D -ii
B A - iv, B - i, C - ii, D - iii
C A - ii, B - i, C - iv, D - iii
D A - ii, B - iv, C - i, D - iii

The correct answer is:
A - iii, B - i, C - iv, D -ii

Solution:
Silica gel is used as a drying agent. Silicon is used in Transistors. Silicone is a sealant while silicates including Zeolites are ion exchangers.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2016 (09 Apr Online)
Match the items in Column I with its main use listed in Column II:

Column I	Column II

A.	Silica gel	i.	Transistor

B.	Silicon	ii.	Ion-exchanger

C.	Silicone	iii.	Drying agent

D.	Silicate	iv.	Sealant

Options:
A A - iii, B - i, C - iv, D - ii
B A - iv, B - i, C - ii, D - iii
C A - ii, B - i, C - iv, D - iii
D A - ii, B - iv, C - i, D - iii

The correct answer is:
A - iii, B - i, C - iv, D - ii

Solution:
Silica gel is used as a drying agent. Silicon is used in Transistors. Silicone is a sealant while silicates including Zeolites are ion exchangers.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Match the items in Column-I with those in Column II
Select the correct option from the following

Options:
A 1.(iii) 2.(iv) 3.(ii) 4.(i)
B 1.(iii) 2.(ii) 3.(iv) 4.(i)
C 1.(iv) 2.(iii) 3.(ii) 4.(i)
D 1.(iv) 2.(ii) 3.(iii) 4.(i)

The correct answer is:
1.(iv) 2.(iii) 3.(ii) 4.(i)

Solution:
The correct matches are
Podocytes are cells in Bowman's capsule in kidneys. They have filtration slits through which the blood is filtered. Protonephridia help in osmoregulation. Nephridia in annelids help in osmoregulation and excretion. Renal calculi are kidney stones which mainly consist of crystallised oxalates.

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2019 (Odisha)
Match the items of List -I to the items of List -II

List-I	List-II

(A)	The number of ways of not selecting n-r things from n different things	(I)	1+n+C2n+…+Crn

(B)	n-r+1·Cr-1n	(II)	r+1·Cr+1n

(C)	The number of ways of selecting atleast n-r things from n different things	(III)	rCrn

(D)	n-rCr-1n-1+Crn-1	(IV)	2n-1-n-C2n-…-Crn

(V)	Cn-rn

The correct match is:

Options:
A

A	B	C	D

V	III	IV	II

B

A	B	C	D

I	II	IV	III

C

A	B	C	D

V	III	I	II

D

A	B	C	D

I	V	IV	III

The correct answer is:

A	B	C	D

V	III	I	II

Solution:
A The number of ways of not selecting n-r things from n different things

=Crn=Cn-rn

B Cr-1nCrn=Cr-1nCn-rn

⇒Cr-1nCn-rn=n!r-1!×n-r+1!×r!×n-r!n!

⇒Cr-1nCn-rn=rn-r+1

⇒n-r+1Cr-1n=rCn-rn

⇒n-r+1Cr-1n=rCrn

C Number of ways of selecting atleast n-r tnings from n different things

=C0n+C1n+C2n+....+Cn-rn+Cn-r+1n+...+Cnn-C0n+C1n+C2n+....+Cn-r-1n

=Cn-rn+Cn-r+1n+Cn-r+2n+......+Cn-1n+Cnn

=C0n+C1n+.......+Cr-2nCr-2n+Cr-1n+Crn

∵Crn=Cn-rn

D

n-rCr-1n-1+Crn-1

=n-rCrn

=n-rn!r!×n-r!

=n!r!×n-r-1!

=r+1n!r+1!×n-r-1!

=r+1Cr+1n

Topic: Mathematics, Permutation Combination, JEE Main
Match the items of List-I with those of the items of List-II

List-I	List-II

A	Range of sec-11+cos2x,
[·] denote greatest integer function	I	odd function

B	Domain of fx
where fx+1x=x2+1x2	II	0,12

C	fx+y=fx+fy ; f1=5	III	sec-15,sec-14

D	sin-1x-cos-1x+sin-11-x=0⇒x∈	IV	R

V

sec-11,sec-12

Options:
A ABCDVIVIII
B ABCDIIIIVIII
C ABCDVIIIIIIV
D ABCDIIIIIIIV

The correct answer is:
ABCDVIVIII

Solution:
A)

Range of cos2x ∈0,1

⇒1 +cos2x ∈1,2

⇒1+ cos2x ∈1,2

⇒sec-11 + cos2x ∈sec-11,sec-12

B)

Given that

fx+1x=x2+1x2

⇒fx+1x=x+1x2-2

⇒fx = x2 - 2

i.e., it's a quadratic polynomial so domain is all real numbers.

C)

fx+y = fx + fy

⇒ fx = kx

f1 = 5 = 5k ⇒ k =1

⇒fx = x

f-x = -fx

i.e., odd function.

D)

Given that

sin-1x-cos-1x+sin-11-x=0

put x = 0 & 12

sin-10 - cos-10 + sin-11 = 0- π2 + π2 = 0

&

sin-112 - cos-112 + sin-112 = π6 - π3 + π6 = 0

i.e., x=0,12.

Topic: Mathematics, Inverse Trigonometric Functions, TS EAMCET, TS EAMCET 2020 (09 Sep Shift 1)
Match the list-I with list-II

List-I
(Class of Drug)	List-II
(Example)

(a)	Antacid	(i)	Novestrol

(b)	Artificial sweetener	(ii)	Cimetidine

(c)	Antifertility	(iii)	Valium

(d)	Tranquilizers	(iv)	Alitame

Options:
A a−ii, b−iv, c−i, d−iii
B a−iv, b−i, c−ii, d−iii
C a−iv, b−iii, c−i, d−ii
D a−ii, b−iv, c−iii, d−i

The correct answer is:
a−ii, b−iv, c−i, d−iii

Solution:
(a) Antacid : Cimetidine
(b) Artifical Sweetener : Alitame
(c) Antifertility : Novestrol
(d) Tranquilizers : Valium

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Match the List I with List II

List I	List II

A	Intrinsic Semiconductor	I	Fermi-level near valence band

B	n-type semiconductor	II	Fermi-level at middle of valence and conduction band

C	p-type semiconductor	III	Fermi-level near conduction band

D	Metals	IV	Fermi-level inside conduction band

Choose the correct answer from the options given below:

Options:
A (A) → I, (B) → II, (C) → III, (D) → IV
B (A) → II, (B) → I, (C) → III, (D) → IV
C (A) → II, (B) → III, (C) → I, (D) → IV
D (A) → III, (B) → I, (C) → II, (D) → IV

The correct answer is:
(A) → II, (B) → III, (C) → I, (D) → IV

Solution:
(A) The Fermi-level in an intrinsic semiconductor is nearly midway between the conductive and valence band.So, A→II.

(B) In the case of n-type semiconductor, the Fermi level is present just below the conduction band. So, B→III.

(C) Fermi level in a p-type semiconductor lies close to the top of the valence band. So, C→I

(D) In metals and semimetals the Fermi level lies inside at least one band i.e., inside conduction band. So, D→IV.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Match the List-I (Phenomenon associated with electromagnetic radiation) with List-II (Part of electromagnetic spectrum) and select the correct code from the choices given below this lists:

Options:
A
(I)-(A), (II)-(B), (III)-(B),,(IV)-(C)

B
(I)-(A), (II)-(B), (III)-(C), (IV)-(C)

C
(I)-(D), (II)-(C), (III)-(A), (IV)-(B)

D
(I)-(B), (II)-(A), (III)-(D), (IV)-(A)

The correct answer is:
(I)-(B), (II)-(A), (III)-(D), (IV)-(A)

Solution:
Wavelength emitted by atomic hydrogen in interstellar space - Part of short radio wave of electromagnetic spectrum. Doublet of sodium - visible radiation.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2014 (11 Apr Online)
Match the List-I with List-II :

Cations	Group reaction

P→Pb2+,Cu2+	H2 S gas in presence of dilute HCl

Q→Al3+,Fe3+	NH42CO3 in presence of NH4OH

R→Co2+,Ni2+	NH4OH in presence of NH4CI

S→Ba2+,Ca2+	H2 S in presence of NH4OH

Options:
A P→i,Q→iii,R→ii,S→iv
B P→iv,Q→ii,R→iii,S→i
C P→iii,Q→i,R→iv,S→ii
D P→i,Q→iii,R→iv,S→ii

The correct answer is:
P→i,Q→iii,R→iv,S→ii

Solution:
Group reagents for precipitating ions:

Group	Cations	Group reagent

Zero	NH4+	None

I	Pb2+	Dilute HCl

II	Pb2+, Cu2+, As3+	H2 S gas in presence of dil. HCl

III	Al3+, Fe3+	NH4OH in presence of NH4Cl

IV	Co2+, Ni2+, Mn2+, Zn2+	H2S in presence of NH4OH

V	Ba2+, Sr2+, Ca2+	(NH4)2CO3 in presence of NH4OH

VI	Mg2+	None

So, the correct answer is P→i, Q→iii, R→iv, S→ii

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Match the List-I with List-II.

List I	List II

A	AC generator	I	Presence of both L and C

B	Transformer	II	Electromagnetic Induction

C	Resonance phenomenon to occur	III	Quality factor

D	Sharpness of resonance	IV	Mutual Inductance

Choose the correct answer from the options given below:

Options:
A A-IV, B-II, C-I, D-III
B A-II, B-I, C-III, D-IV
C A-II, B-IV, C-I, D-III
D A-IV, B-III, C-I, D-II

The correct answer is:
A-II, B-IV, C-I, D-III

Solution:
(A) An AC generator works on the electromagnetic induction principle. So, A→II

(B) The transformer works on the principle of mutual induction. So, B→IV

(C) Resonance occurs when the value of inductive and capacitive reactance have equal magnitude but a phase difference of 180°. So, C→I

(D) The sharpness of resonance is defined using the quality factor which explains how fast energy decay in an oscillating system. D→III

Topic: Physics, Alternating Current, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Match the List-I with List-II:

List I	List II

A	Microwaves	I	Radioactive decay of the nucleus

B	Gamma rays	II	Rapid acceleration and deceleration of electron in aerials

C	Radio waves	III	Inner shell electrons

D	X-rays	IV	Klystron valve

Choose the correct answer from the options given below:

Options:
A A-I, B-II, C-III, D-IV
B A-IV, B-I, C-II, D-III
C A-I, B-III, C-IV, D-II
D A-IV, B-III, C-II, D-I

The correct answer is:
A-IV, B-I, C-II, D-III

Solution:
(A) A Klystron valve or the magnetron valve are generally used for the production of electromagnetic waves and that electromagnetic wave is Microwave. So, A→IV.

(B) Gamma rays are produced during the disintegration of radioactive atomic nuclei and the decay of certain subatomic particles.So, B→I

(C) Radio waves can be produced by rapid acceleration and deceleration of electrons in aerials (conductors). These can be detected by receivers of aerials. So, C→II

(D) X-rays are produced when a metal target or any other element is bombarded with high energy particles such as electrons, photons or ions. When high energy electrons is incident on a bound electron in an atom, then bound electron emits from the inner shell of that atom. So, D→III

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Match the metal ions given in Column 'I' with the spin magnetic moments of the ions given in Column 'II'.

Column I	Column II

a. Co3+	i. 8 B.M.

b. Cr3+	ii. 35 B.M.

c. Fe3+	iii. 3 B.M.

d. Ni2+	iv. 24 B.M.

v. 15B.M

Options:
A abciviiiÂ Â Â Â diii
B abciiiviÂ Â Â Â dii
C abciiiiii div
D abcivvii di

The correct answer is:
abcivvii di

Solution:
Magnetic moment μ=nn+2 B.M.
(i) Co3+→1s22s22p63s23p64s03d6
n=4
μ= 44+2=24 B.M.
(ii) Cr+3→1s22s22p63s23p64s03d3
n=3
μ= 33+2= 15 B.M.
(iii) Fe3+→1s22s22p63s23p64s03d5
n=5
μ= 55+2= 35 B.M.
(iv) Ni+2→1s22s22p63s23p64s03d8
n=2
μ= 22+2= 8 B.M.

Topic: Chemistry, Structure of Atom, JEE Main
Match the metals (column I) with the coordination compound(s)/enzyme(s) (column II) :

Options:
A (A)-(iii); (B)-(iv); (C)-(i); (D)-(ii)
B (A)-(i); (B)-(ii); (C)-(iii); (D)-(iv)
C (A)-(ii); (B)-(i); (C)-(iv); (D)-(iii)
D (A)-(iv); (B)-(iii); (C)-(i); (D)-(ii)

The correct answer is:
(A)-(iii); (B)-(iv); (C)-(i); (D)-(ii)

Solution:
Wilkinson catalyst: $\left[\mathrm{Rh}(\mathrm{PPh})_{3} \mathrm{Cl}\right]$

Chlorophyll : $\mathrm{C}_{55} \mathrm{H}_{72} \mathrm{O}_{5} \mathrm{~N}_{4} \mathrm{Mg}$

Vitamin $\mathrm{B}_{12}$ contains $\mathrm{Co}$

Carbonic anhydrase contains a Zn ion.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Match the mode of dark reactions with their correct example.

a	Calvin Cycle	i	Kalanchoe

b	C4 Plants	ii	Zea mays

c	CAM plants	iii	Triticum

Options:
A aÂ âÂ (ii),Â bÂ âÂ (iii)Â ,Â cÂ âÂ (i)
B aÂ âÂ (i)Â ,Â bÂ âÂ (ii)Â ,Â cÂ âÂ (iii)
C a â (iii) , b â (ii) , c â (i)
D aÂ âÂ (i)Â ,Â bÂ âÂ (ii)Â ,Â cÂ âÂ (ii)

The correct answer is:
a â (iii) , b â (ii) , c â (i)

Solution:
The carbon dioxide assimilation during photosynthesis is of two main types. In those plants in which the first product of carbon dioxide fixation is 3-phosphoglyceric acid (a C3 acid)is known as the C3 pathway or Calvin cycle, and those in which the first product is oxaloacetic acid (a C4 acid) is the C4 pathway. CAM (Crassulacean Acid Metabolism) pathway is a method of carbon fixation evolved by some plants in dry circumstances where stomata open during the night to take in carbon dioxide to prevent water loss. Kalanchoe belongs to family Crassulaceae and shows the CAM pathway of carbon fixation. Triticum (wheat plant) uses the standard method of Carbon fixation that is the C3 pathway. Maize (Zea mays) shows the C4 pathway as it grows in dry tropical climates.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Match the ores (column A) with the metals (column B) :

Options:
A (I) - (a); (II) - (b); (III) - (c); (IV) - (d)
B (I) - (c); (II) - (d); (III) - (b); (IV) - (a)
C (I) - (c); (II) - (d); (III) - (a); (IV) - (b)
D (I) - (b); (II) - (c); (III) - (d); (IV) - (a)

The correct answer is:
(I) - (c); (II) - (d); (III) - (b); (IV) - (a)

Solution:
Siderite $=\mathrm{FeCO}_{3}$

Calamine $=\mathrm{ZnCO}_{3}$

Malachite $=\mathrm{CuCO}_{3} \cdot \mathrm{Cu}(\mathrm{OH})_{2}$

Kaolinite $=\mathrm{Al}_{2} \mathrm{Si}_{2} \mathrm{O}_{5}(\mathrm{OH})_{4}$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Match the organic compounds in column - I with the Lassaigne's test result in column - II appropriately:

Column – I	Column - II

A.	Aniline	i.	Red colour with FeCl3

B.	Benzene sulfonic acid	ii.	Violet color with sodium nitroprusside

C.	Thiourea	iii.	Blue color with and acidic solution of FeSO4

Options:
A A - ii ; B - iii ; C - i
B A - iii ; B - ii ; C - i
C A - ii ; B - i ; C - iii
D A - iii ; B - i ; C - ii

The correct answer is:
A - iii ; B - ii ; C - i

Solution:
Aniline (C6H5NH2) → it contains (N) element
Benzene sulfonic acid (C6H5SO3H) → it contains (S)element
Thiourea (SC(NH2)2) → it contains (S & N) element
Test for nitrogen - The carbon and nitrogen present in the organic compound on fusion with sodium metal gives sodium cyanide (NaCN) soluble in water. This is converted into sodium ferro cyanide by the addition of sufficient quantity of ferrous sulphate. Ferric ions generated during the process react with Ferro cyanide to form Prussian blue precipitate of ferric ferrocyanide.
Na+C+N→NaCN
6 NaCN + FeSO 4 → Na 4 Fe CN 6 + Na 2 SO 4 Sodium ferrocyanide
3 Na 4 Fe CN 6 + 4 Fe 3 + → Fe 4 Fe CN 6 3 + 1 2 Na + Ferric ferrocyanide
Test of sulphur - If sulphur is present in the organic compound, sodium fusion will convert it into sodium sulphide. Sulphide ions are readily identified using sodium nitroprusside.
2 Na+S→Na2S
Na 2 S + Na 2 Fe CN 5 NO → Na 4 Fe CN 5 NOS Sodium Violet nitroprusside colour
Test for both nitrogen and sulphur - If both nitrogen and sulphur are present in an organic compound, sodium fusion will convert it into sodium thiocyanate which then react with Fe3+ to form blood colour complex FeSCN2+
Na+C+N+S→NaSCN
Fe3++3SCN- →FeSCN3 (Blood red)
Note: If the Lassaigne's extract containing excess of sodium metal, sodium cyanide and sulphides are formed instead of sodium thiocyanate.
NaSCN+2Na→NaCN+Na2S
Here in this type of cases, both sulphur and nitrogen are to be identified in separate tests.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2015 (11 Apr Online)
Match the parts of the brain given in column I with the functions performed by them given in column II and select the correct option from the codes given below.

No.	Part	No.	Function

I	The part forming the major part of the brain	i	Maintenance of balance and equilibrium

II	The part situated at the base of the thalamus	ii	Responsible for motor, sensory and intersensory associations

III	The part with a median lobe called the vermis	iii	Has centers for control of activities like respiration, gastric secretions, etc.

IV	The part which connects with the spinal cord	iv	Control of body temperature and the urge for eating and drinking

Options:
A I-ii, II-iii. III-iv, IV-ii
B I-ii, II-iv, III-i, IV-iii
C I-iii, II-I, III-ii, IV-iv
D I-iv, II-ii, III-iii, IV-i

The correct answer is:
I-ii, II-iv, III-i, IV-iii

Solution:
The part which forms the major part of the brain is the cerebrum. It has areas responsible for motor, sensory and intersensory associations. The part situated at the base of thalamus is the hypothalamus. It controls body temperature and the urge for eating and drinking. The part with a median lobe called vermis is the cerebellum. It is responsible for maintaining the balance and equilibrium of the body.The part of brain which connects with the spinal cord is the medulla oblongata. It has areas which control activities like respiration, gastric secretions and cardiovascular reflexes.

Topic: Biology, Neural Control and Coordination, NEET
Match the polymers in column-A with their main uses in column-B and choose the correct answer:

Column - A	Column - B

A.	Polystyrene	i.	Paints and lacquers

B.	Glyptal	ii.	Rain coats

C.	Polyvinyl chloride	iii.	Manufacture of toys

D.	Bakelite	iv.	Computer discs

Options:
A A - iii , B - i , C - ii , D - iv
B A - ii , B - i , C - iii , D - iv
C A - ii , B - iv , C - iii , D - i
D A - iii , B - iv , C - ii , D - i

The correct answer is:
A - iii , B - i , C - ii , D - iv

Solution:
Polystyrene is a synthetic aromatic hydrocarbon polymer.Polystyrene is clear, hard, and rather brittle.
Glyptal is a polyester obtained by condensation polymerization of glycerol and phthalic anhydride.
Bakelite is electrical nonconductivity and heat-resistant.
Polyvinyl Chloride is a high strength thermoplastic material.

The correct order is
â

Column - A	Column - B

A.	Polystyrene	iii.	Manufacture of toys

B.	Glyptal	i.	Paints and lacquers

C.	Polyvinyl chloride chloride	ii.	Raincoats

D.	Bakelite	iv.	Computer discs

Topic: Chemistry, Polymers, JEE Main, JEE Main 2015 (10 Apr Online)
Match the polymers in column-A with their main uses in column-B and choose the correct answer:

Column - A	Column - B

A.	Polystyrene	i.	Paints and lacquers

B.	Glyptal	ii.	Rain coats

C.	Polyvinyl chloride	iii.	Manufacture of toys

D.	Bakelite	iv.	Computer discs

Options:
A A - iii , B - i , C - ii , D - iv
B A - ii , B - i , C - iii , D - iv
C A - ii , B - iv , C - iii , D - i
D A - iii , B - iv , C - ii , D - i

The correct answer is:
A - iii , B - i , C - ii , D - iv

Solution:
This is the correct order
â

Column - A	Column - B

A.	Polystyrene	i.	Manufacture of toys

B.	Glyptal	ii.	Paints and lacquers

C.	Polyvinyl chloride chloride	iii.	Raincoats

D.	Bakelite	iv.	Computer discs

Topic: Chemistry, Polymers, JEE Main
Match the process of decomposition with the event that occurs in it:

(i)	Fragmentation	A	Degradation of detritus by bacterial and fungal enzymes

(ii)	Leaching	B	Precipitation of water-soluble nutrients as unavailable salt

(iii)	Catabolism	C	Break down of detritus into smaller particles

Options:
A iâA,Â iiâB,Â iiiâC
B iâB,Â iiâC,iiiâA
C iâC, iiâB, iiiâA
D iâC,Â iiâA,Â iiiâB

The correct answer is:
iâC, iiâB, iiiâA

Solution:
Fragmentation, leaching, and catabolism are important steps of decomposition. Fragmentation is the process of breakdown of detritus, consisting of remains of dead plants and animals, into smaller particles by detrivores like an earthworm. The water-soluble inorganic nutrients obtained from the process of decomposition get precipitated as unavailable salts. This process is known as leaching. Degradation of detritus into simpler inorganic substances by bacterial and fungal enzymes is called catabolism.

Topic: Biology, Ecosystem, NEET
Match the processes occurs during cell cycle with the names of the events

(P)	Synapsis aligns homologous chromosomes	(i)	Anaphase-II

(Q)	Synthesis of RNA and protein	(ii)	Zygotene

(R)	Action of enzyme recombinase	(iii)	G-phase

(S)	Centromeres do not separate but chromatids move towards opposite poles	(iv)	Anaphase-I

(v)	Pachytene

Options:
A (P) - (i), (Q) - (ii), (R) - (v), (S) - (iv)
B (P) - (ii), (Q) - (iii), (R) - (iv), (S) - (v)
C (P) - (ii), (Q) - (i), (R) - (iii), (S) - (iv)
D (P) - (ii), (Q) - (iii), (R) - (v), (S) - (iv)

The correct answer is:
(P) - (ii), (Q) - (iii), (R) - (v), (S) - (iv)

Solution:
Zygotene is the second stage of Prophase I where the chromosomes start pairing together. This process is called synapsis and the paired chromosomes are known as homologous chromosomes. Pachytene is the stage where bivalent chromosomes clearly appear as tetrads and is characterized by the appearance of recombination nodules which function as the sites of crossing over between non-sister chromatids of the homologous chromosomes. Crossing over is the exchange of genetic material between two homologous chromosomes and is mediated by the enzyme called recombinase. The stage where sister chromatids remain associated but homologous chromosomes separate is called Anaphase I. During Anaphase II the sister chromatids move towards the opposite poles of the cell due to the simultaneous splitting of the centromere.

Topic: Biology, Cell Cycle and Cell Division, NEET