datavorous/entrance-exam-dataset huggingface dataset:
Match the entries from Column I and Column II and choose the correct order.

Options:
A \(\begin{array}{lll} \mathrm{A} & \mathrm{B} & \mathrm{C} \\ 3 & 2 & 1 \\ \end{array}\)
B \(\begin{array}{lll} \mathrm{A} & \mathrm{B} & \mathrm{C} \\ 1 & 2 & 3 \\ \end{array}\)
C \(\begin{array}{lll} \mathrm{A} & \mathrm{B} & \mathrm{C} \\ 3 & 1 & 2 \end{array}\)
D \(\begin{array}{lll} \mathrm{A} & \mathrm{B} & \mathrm{C} \\ 2 & 1 & 3 \end{array}\)

The correct answer is:
\(\begin{array}{lll} \mathrm{A} & \mathrm{B} & \mathrm{C} \\ 3 & 1 & 2 \end{array}\)

Solution:
The correct matching between Column I and Column II is
(A) Leclanche cell-anode reaction :
\(\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+2 e^{-}\)
Cathode :
\(\mathrm{MnO}_2+\mathrm{NH}_4^{+}+e^{-} \longrightarrow \mathrm{MnO}(\mathrm{OH})+\mathrm{NH}_3\)
(B) Fuel cell : \(2 \mathrm{H}_2+\mathrm{O}_2 \longrightarrow \mathrm{H}_2 \mathrm{O}+\) Energy (gets converted into electrical energy).
(C) Ni-Cd cell : Rechargeable (secondary) cell. Hence, option (c) is correct.

Topic: Chemistry, Electrochemistry, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 2)
Match the entries in List I with those in List II.

Options:
A (a)
B (b)
C (c)
D (d)

The correct answer is:
(a)

Solution:
(a) A-(iii), B-(iv), C-(i), D-(ii)
(a) There are four fundamental interactions or forces. They are gravitation, electromagnetic, the weak interaction and the strong interaction. Out of which weak interaction and electromagnetic interaction arises from electroweak interaction.
(b) Force between two molecules separated by a distance near about the sum of the molecular radii is called electromagnetic interaction.
(c) Nuclear binding force is strong interaction because it is the attractive force that binds the elementary particles called quarks together to form nucleons.
(d) Interaction of bodies of astronomical proportions is called gravitational attraction.

Topic: Physics, Laws of Motion, TS EAMCET, TS EAMCET 2018 (04 May Shift 2)
Match the following.

Options:

None

Solution:
(a) Octopus The appendages in invertebrates that are used for grasping food and for locomotion are tentacles.
(b) Crocodile for locomotion, and swimming limbs are used.
(c) Catta Fins are means of locomotion and are used to generation optimum thrust thus controlling the subsequent motion.
(d) Ctenoplana Locomotory organs formed by strong cilia with fused bases are comb plates.

Topic: Biology, Animal Kingdom
Match the following

(a)	Inhibitor of catalytic activity	(i)	Ricin

(b)	Possess peptide bonds	(ii)	Malonate

(c)	Cell wall material in fungi	(iii)	Chitin

(d)	Secondary metabolite	(iv)	Collagen

Choose the correct option from the following

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

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

Solution:
Chitin: It is the second most abundant organic substance. In chitin the basic unit is a nitrogen containing glucose derivative known as N-acetyl glucosamine. The exoskeletons of arthropods and cell wall of fungi have chitin.

Collagen is the most abundant animal protein and is the chief component of connective tissue. Proteins are the biopolymers made up of monomeric amino acids joined by amide linkage or peptide bond.

Malonate is the competitive inhibitor of the enzyme succinic dehydrogenase.

Abrin and ricin are the toxins produced as secondary metabolites.

Topic: Biology, Biomolecules (B), NEET, NEET 2020 (Phase 1)
Match the following and select the correct option.

Column I	Column-II

[A]	Vegetative cell	1.	Sporopollenin

[B]	Generative cell	2.	Spindle-shaped cell

[C]	Exine	3.	Large sized and has abundant food reserve

[D]	Intine	4.	Cellulose and pectin

Options:
A A:1,Â B:2,Â C:3,Â D:4Â
B A: 3, B:4, C:2, D:1
C A:3,Â B:2,Â C:1,Â D:4Â
D A:2,Â B:1,Â C:4,Â D:3

The correct answer is:
A:3, B:2, C:1, D:4

Solution:
Pollen grain consists of two cells, Vegetative cell is large and rich in reserve food material. Generative is small spindle shaped and its nucleus divides to form two male gametes. Pollen grain is surrounded by two wall layer, outer exine which is made up of sporopollenin and inner intine which is made up of pectocellulose.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Match the following based on valence bond theory.
\begin{array}{llll}
\hline & Hybridisation & Geometry & \multicolumn{1}{c}{\begin{array}{l}
Complex \\
structure
\end{array}} \\
\hline (A) & s p^3 & (i) \begin{array}{l}
Square \\
planar
\end{array} & (p) \left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-} \\
\hline (B) & d^2 s p^3 & (ii) Tetrahedral & (q) \left.[\mathrm{ZnCl}]_4\right]^{2-} \\
\hline (C) & d s p^2 & (iii) Octahedral & (r) \left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_4\right]^{2+} \\
\hline & & & (s) \left[\mathrm{Ag}(\mathrm{CN})_2\right]^{-} \\
\hline
\end{array}

Options:
A (A-ii-q), (B-iii-p), (C-i-r)
B (A-ii-q), (B-iii-r), (C-i-s)
C (A-i-q), (B-iii-p), (C-ii-r)
D (A-ii-r), (B-iii-s), (C-i-q)

The correct answer is:
(A-ii-q), (B-iii-p), (C-i-r)

Solution:
According to VBT,
Hence, correct match is (A-ii-q), (B-iii-p), (C-i-r).

Topic: Chemistry, Coordination Compounds, AP EAMCET, AP EAMCET 2021 (25 Aug Shift 2)
Match the following
\begin{array}{ll}
\begin{array}{l}
List- I \\
(Hybridisation)
\end{array} & \begin{array}{l}
List- II \\
(Shape)
\end{array} \\
A. \mathrm{dsp}^2 & I. Square planar \\
B. \mathrm{sp}^3 & II. Tetrahedral \\
C. \mathrm{d}^2 \mathrm{sp}^3 & III. Octahedral \\
D. \mathrm{Sp}^3 \mathrm{~d} & IV. Trigonal bipyramidal
\end{array}

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

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

Solution:
$\begin{aligned} & \mathrm{sp}^3-\text { tetrahedral } \\ & \mathrm{sp}^3 \mathrm{~d} \text { - trigonal bipyramidal } \\ & \mathrm{dsp}^3-\mathrm{square} \text { planar } \\ & \mathrm{sp}^3 \mathrm{~d}^2 / \mathrm{d}^2 \mathrm{sp}^3-\text { octahedral }\end{aligned}$

Topic: Chemistry, Chemical Bonding and Molecular Structure, AP EAMCET, AP EAMCET 2023 (18 May Shift 2)
Match the following
\begin{array}{ll}
\multicolumn{1}{l}{\begin{array}{l}
List-I \\
(Process)
\end{array}} & \begin{array}{l}
List-II \\
(Catalyst)
\end{array} \\
(A) Ostwald's process & (I) \mathrm{NO} \\
(B) Lead Chamber process & (II) \mathrm{Fe} \\
(C) Deacon's process & (III) \mathrm{Rh} \\
(D) Haber's process & (IV) \mathrm{CuCl}_2
\end{array}
The correct answer is

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

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

Solution:
Ostwald's process :-
$$
4 \mathrm{NH}_3(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g})
\stackrel{\text { Pt/Rh gauge catalyst }}{500 \mathrm{~K}, 9 \text { bar }}$$
$4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$
Lead Chamber process:
$$
\begin{aligned}
& \mathrm{NO}+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{NO}_2 \\
& \mathrm{NO}_2+\mathrm{SO}_2 \rightarrow \mathrm{NO}+\mathrm{SO}_3 ; \mathrm{SO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_4
\end{aligned}
$$
Deacon's process :
$$
4 \mathrm{HCl}+\mathrm{O}_2 \underset{723 \mathrm{~K}}{\stackrel{\mathrm{CuCl}_2}{\longrightarrow}} 2 \mathrm{Cl}_2+2 \mathrm{H}_2 \mathrm{O}
$$
Haber's process :
$\mathrm{N}_2+3 \mathrm{H}_2 \underset{\mathrm{Fe}}{\stackrel{250 \mathrm{~atm}}{\rightleftharpoons}} 2 \mathrm{NH}_3$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), AP EAMCET, AP EAMCET 2023 (18 May Shift 2)
Match the following
\begin{array}{lll}
& \begin{array}{l}
List I \\
(Complex)
\end{array} & \begin{array}{l}
List II \\
(Color)
\end{array} \\
A. \left[\mathrm{Ni}(\mathrm{en})_3\right]^{2+} & I. & Green \\
B. \left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_4(\mathrm{en})\right]^{2+} & II. Blue \\
C. \left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} & III. Pale blue \\
D. \left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6(\mathrm{en})_2\right]^{2+} & IV. Violet
\end{array}

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

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

Solution:
The splitting energy order is (due to strong-field ligand 'en') :-
$$
\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} < \left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_4(\mathrm{en})\right]^{2+} < \left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_2(\mathrm{en})_2\right]^{2+} < $$
$\left[\mathrm{Ni}(\mathrm{en})_3\right]^{2+}$
Thus, the colour shifts from large wave length to short or small frequency to high frequency which is from green to pale blue to blue to violet.

Topic: Chemistry, d and f Block Elements, AP EAMCET, AP EAMCET 2023 (19 May Shift 1)
Match the following entities of column I with their respective orders of column II and choose the correct combination form the option.
$\begin{array}{ll|ll}
\hline & \text{Column I} & & \text{Column II} \\
\hline \text{A.} & \text{Wheat} & \text{1.} & \text{Primate} \\
\text{B.} & \text{Mango} & \text{2.} & \text{Diptera} \\
\text{C.} & \text{Housefly} & \text{3.} & \text{Sapindales} \\
\text{D.} & \text{Man} & \text{4.} & \text{Poales} \\
\hline
\end{array}$

Options:
A $\begin{array}{llll}A & B & C & D \\ 4 & 3 & 2 & 1\end{array}$
B $\begin{array}{llll}A & B & C & D \\ 1 & 2 & 4 & 3\end{array}$
C $\begin{array}{llll}A & B & C & D \\ 3 & 4 & 2 & 1\end{array}$
D $\begin{array}{llll}A & B & C & D \\ 2 & 4 & 1 & 3\end{array}$

The correct answer is:
$\begin{array}{llll}A & B & C & D \\ 4 & 3 & 2 & 1\end{array}$

Solution:
$\begin{array}{l|l|l|l|l}
\hline \text{Common Name} & \text{Scientific Name} & \text{Family} & \text{Order} & \text{Class} \\
\hline \text{Wheat} & \text{Triticum aestivum} & \text{Poaceae} & \text{Poales} & \text{Monocotyledons} \\
\text{Mango} & \text{Mangifera indica} & \text{Anacardiaceae} & \text{Sapindales} & \text{Dicotyledons} \\
\text{Housefly} & \text{Musca domestica} & \text{Muscidal} & \text{Diptera} & \text{Insecta} \\
\text{Man} & \text{Homo} sapiens & \text{Hominidal} & \text{Primata} & \text{Mammalia} \\
\hline
\end{array}$

Topic: Biology, The Living World, JIPMER, JIPMER 2013
Match the following fundamental forces of nature with their relative strength.
The correct match is

Options:
A (a)
B (b)
C (c)
D (d)

The correct answer is:
(a)

Solution:
The four fundamental forces of nature are as follows,
(i) Strong nuclear force which is between nuclear particles like, quarks. Its strength is very high and of the order of 1.
But it's range is very short.
(ii) Weak nuclear force which is also between nuclear particles, however its strength is low and of the order of $10^{-13}$. This force is usually found in radioactive nuclei. It is also a short range force.
(iii) Electromagnetic force which is the force between the electrically charged particles. Its strength is of the order of $10^{-2}$. But its range is very high, upto infinity.
(iv) Gravitational force which is the force of attraction between bodies due to gravity. Its strength is very low and of the order of $10^{-39}$. But it's range is very high upto infinity.
So, the correct sequence of match is $\mathrm{A} \rightarrow$ (ii), $\mathrm{B} \rightarrow$ (iv), $\mathrm{C} \rightarrow$ (i) and $\mathrm{D} \rightarrow$ (iii).

Topic: Physics, Laws of Motion, TS EAMCET, TS EAMCET 2019 (03 May Shift 1)
Match the following lists.
\begin{array}{|c|c|c|c|}
\hline & List I & & List II \\
\hline A & \begin{array}{l}\text { Zeroth law of } \\
\text { thermodynamics }\end{array} & 1 & \begin{array}{l}\text { Direction of flow of } \\
\text { heat }\end{array} \\
\hline B & \begin{array}{l}\text { First law of } \\
\text { thermodynamics }\end{array} & \| & Work done is zero \\
\hline\underline{\mathrm{C}} & Free expansion of a gas & III & Thermal equilibrium \\
\hline D & \begin{array}{l}\text { Second law of } \\
\text { Thermodynamics }\end{array} & IV & \begin{array}{l}\text { Law of conservation } \\
\text { of energy }\end{array} \\
\hline
\end{array}
The correct answer is
$\begin{array}{lccccccccc} & \text { A } & \text { B } & \text { C } & \text { D } \end{array}$

Options:
A $\begin{array} \text { II } & \text { IV } & \text { III } & \text { I }\end{array}$
B $\begin{array}& \text { III } & \text { IV } & \text { II } & \text { I }\end{array}$
C $\begin{array}& \text { III } & \text { I } & \text { II } & \text { IV }\end{array}$
D $\begin{array}& \text { I } & \text { III } & \text { IV } & \text { II }\end{array}$

The correct answer is:
$\begin{array}& \text { III } & \text { IV } & \text { II } & \text { I }\end{array}$

Solution:
(i) Zeroth law of thermodynamics states about
thermal equilibrium of different states in
contact.
(ii) First law is energy conservation law.
(iii) In free expansion of gases, work done is
always zero.
(iv) Second law of thermodynamics discusses
about flow of heat in thermal bodies.

Topic: Physics, Thermodynamics, AP EAMCET, AP EAMCET 2018 (22 Apr Shift 1)
Match the following table.
\begin{array}{|c|c|c|c|}
\hline & List-I & & List-II \\
\hline A. & \begin{array}{l}\text { Michelson-Morley } \\
\text { experiment }\end{array} & I. & \begin{array}{l}\text { The existence of } \\
\text { anti-matter }\end{array} \\
\hline B. & \begin{array}{l}\text { Stern-Gerlach } \\
\text { experiment }\end{array} & II. & \begin{array}{l}\text { The existence of } \\
\text { de-Broglie matter waves }\end{array} \\
\hline C. & \begin{array}{l}\text { Davisson-Germer } \\
\text { experiment }\end{array} & III. & Electrons have spins \\
\hline D. & \begin{array}{l}\text { Anderson discovery } \\
\text { of positron }\end{array} & \mathrm{IV}. & \begin{array}{l}\text { The non-existence of } \\
\text { ether }\end{array} \\
\hline
\end{array}
$\begin{array}{llll}A & B & C & D\end{array}$

Options:
A $\begin{array}{llll}\mathbb{N} & \| & \|\end{array}$
B II $\quad$ III $\quad$ IV $\quad$ I
C I II II IV III
D $\begin{array}{llll}\text { IV } & \text { III } & \text { II } & \text { I }\end{array}$

The correct answer is:
$\begin{array}{llll}\text { IV } & \text { III } & \text { II } & \text { I }\end{array}$

Solution:
$\mathbf{A} \rightarrow$ (iv) Michelson Morley experiment was
an attempt to detect the existance of ether. The result of his experiment suggested the non-existance of the ether.
B $\rightarrow$ (iii) The Stern-Gerlach experiment demonstrated that the spatial orientation of angular momentum is quantised. Due to spin motion of electrons, particles have non-zero magnetic moment and hence they are deflected to the magnetic field gradient from a straight path.
C $\rightarrow$ (ii) Davisson - Germer experiment
demonstrates the wave nature of electrons i.e., the existence of de-Broglie matter waves.
$\mathrm{D} \rightarrow$ (i) By studying the tracks of cosmic ray particles in a cloud chamber, Anderson discovered a positively charged particles with a mass seemingly equal to that of an electron, named positron, which suggests the existence of anti- matter.

Topic: Physics, Dual Nature of Matter, TS EAMCET, TS EAMCET 2019 (06 May Shift 1)
Match the following (Take the relative strength of the strongest fundamental forces in nature as one)
The correct match is

Options:
A A-f, B-i, C-e, D-h
B A-f, B-h, C-e, D-h
C A-f, B-h, C-e, D-i
D A-f, B-e, C-h, D-i

The correct answer is:
A-f, B-h, C-e, D-i

Solution:
$$
\text { So, A-f, B-h, C-e, D-i }
$$

Topic: Physics, Gravitation, AP EAMCET, AP EAMCET 2014
Match the following (Take the relative strength of the strongest fundamental forces in nature as one)
The correct match is

Options:
A A-f, B-i, C-e, D-h
B A-f, B-h, C-e, D-h
C A-f, B-h, C-e, D-i
D A-f, B-e, C-h, D-i

The correct answer is:
A-f, B-h, C-e, D-i

Solution:
$$
\text { So, A-f, B-h, C-e, D-i }
$$

Topic: Physics, Gravitation, TS EAMCET, TS EAMCET 2014
Match the following.
(Take the relative strength of the strongest fundamental forces in nature as one)

Options:
A A-f, B-i, C-e, D-h
B A-f, B-h, C-e, D-g
C $\mathrm{A}-\mathrm{f}, \mathrm{B}-\mathrm{h}, \mathrm{C}-\mathrm{e}, \mathrm{D}-\mathrm{i}$
D A-f, B-e, C-h, D-i

The correct answer is:
$\mathrm{A}-\mathrm{f}, \mathrm{B}-\mathrm{h}, \mathrm{C}-\mathrm{e}, \mathrm{D}-\mathrm{i}$

Solution:
Strong nuclear force is the strongest force- relative strength (1); Gravitational force is the weakest forcerelative strength $\left(10^{-39}\right)$; Electromagnetic force - relative strength $\left(10^{-2}\right)$ is stronger than weak nuclear force relative strength $\left(10^{-13}\right)$.

Topic: Physics, Units and Dimensions, TS EAMCET, TS EAMCET 2023 (14 May Shift 2)
Match the following.
The correct answer is
\begin{array}{llllllllll}
& A & B & C & D &
\end{array}

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

The correct answer is:
III V I II

Solution:
(A) Resonance-Ben zene shows the resonance.
(B) Inductive effect 1-chloropropane shows the highest inductive effect.
(C) Electromeric effect Ethene shows the electromeric effect.
(D) Hyper conjugation Eth ane shows the hyper conjugation in these species.

Topic: Chemistry, General Organic Chemistry, AP EAMCET, AP EAMCET 2018 (23 Apr Shift 2)
Match the gases in column-I with their source in column-II.

Column-I	Column- II

(A)	Nitrous oxide (N2O)	(1)	Secondary pollutant from car exhausts

(B)	Chlorofluorocarbons (CFCs)	(2)	Combustion of fossil fuels, wood, etc.

(C)	Methane (CH4)	(3)	Denitrification

(D)	Ozone (O3)	(4)	Refrigerators, aerosol, sprays

(E)	Carbon dioxide (CO2)	(5)	Cattle dung and toilets

Options:
A A B C D E 3 4 5 1 2
B A B C D E 5 1 3 4 2
C A B C D E 4 5 1 2 3
D A B C D E 1 3 4 5 2

The correct answer is:
A B C D E 3 4 5 1 2

Solution:
Column I	Column II

Nitrous oxide (N2O)	Denitrification

Chlorofluorocarbons CFCs	Refrigerators, aerosol, sprays

Methane (CH4)	Cattle dung and toilets

Ozone (O3)	Secondary pollutant from car exhausts

Carbon dioxide (CO2)	Combustion of fossil fuels, wood, etc.

Topic: Biology, Environmental Issues, NEET
Match the organism with its properties in a food chain:

Organism	Role in the food chain	Nutrition	Trophic level

1. Grass	i. Secondary Consumer	a. Herbivores	w. Third

2. Grasshopper	ii. Tertiary Consumer	b. Primary Carnivores	x. Fourth

3. Rodents	iii. Producer	c. Secondary Carnivores	y. First

4. Cats	iv. Primary Consumer	d. Transducers	z. Second

Options:
A 1-iv-c-w ; 2-ii-d-x ; 3-iii-a-y ; 4-i-b-z
B 1-i-b-x ; 2-iii-c-y ; 3-ii-d-x ; 4-iv-a-w
C 1-iii-d-y ; 2-iv-a-z; 3-i-b-w ; 4-ii-c-x
D 1-ii-a-z ; 2-i-b-w ; 3-iv-c-x ; 4-iii-d-y

The correct answer is:
1-iii-d-y ; 2-iv-a-z; 3-i-b-w ; 4-ii-c-x

Solution:
Grass is at the First Trophic level as it is a producer. It can convert sunlight to chemical energy hence it is a transducer. Grasshoppers (herbivores) eat grass and hence are primary consumer at second trophic level. Rodents eat Grasshoppers and thus are secondary consumers and also primary carnivores at third level. Cats eat Rodents and are Tertiary consumer and secondary carnivores at fourth trophic level.

Topic: Biology, Ecosystem, NEET
Match the proteins with their appropriate properties.

Protein	Property

(1)	Collagen	(i)	Hormone

(2)	Trypsin	(ii)	Enzyme

(3)	Insulin	(iii)	Fight Infectious agent

(4)	Receptor	(iv)	Enable glucose transport into cells

(5)	GLUT-4	(v)	Sensory reception (smell, taste hormone)

(6)	Antibody	(vi)	Intercellular ground substance

Options:
A (1) – (iii), (2) – (ii), (3) – (iv), (4) – (i), (5) – (vi), (6) – (v)
B (1) – (vi), (2) – (ii), (3) – (i), (4) – (v), (5) – (iv), (6) – (iii)
C (1) – (i), (2) – (iv), (3) – (ii), (4) – (v), (5) – (iii), (6) – (vi)
D (1) – (ii), (2) – (iv), (3) – (v), (4) – (iii), (5) – (i), (6) – (vi)

The correct answer is:
(1) – (vi), (2) – (ii), (3) – (i), (4) – (v), (5) – (iv), (6) – (iii)

Solution:
Collagen is the most abundant protein found in the animal body forming the matrix or ground substance of connective tissue that joins almost all parts and tissues of the body. Trypsin is the active form of digestive enzyme secreted by pancreas. Insulin is a peptide hormone helping in reducing blood glucose. Receptor is a molecule that receives the signal or assist in sensing stimuli. GLUT-4 is a transporter molecule found in the cell that helps glucose uptake in the cell. Antibody is a protein molecule produced by B lymphocyte to fight pathogens.

Topic: Chemistry, Biomolecules, NEET
Match the thermodynamic properties (List-I) with their relation (List-II)

List-I
List-II

P.
Free energy change ÎG
1.
â RTlog e K

Q.
Entropy changes ÎSÂ°
2.
-nFE

R.
ÎHÂ° enthalpy change of a reaction in standard state
3.
RT2dlnKdTP

S.
Standard free energy change (ÎGÂ°)
4.
-dÎGdTP

Select the correct answer from the given codes

Options:
A P-1, Q-2, R-3, S-4
B P-2, Q-4, R-3, S-1
C P-4, Q-2, R-3, S-1
D P-1, Q-2, R-4, S-3

The correct answer is:
P-2, Q-4, R-3, S-1

Solution:
List-I
List-II

P.
Free energy change ÎG
1.
-nFE

Q.
Entropy changes ÎSÂ°
2.
-dÎGdTP

R.
ÎHÂ° enthalpy change of a reaction in standard state
3.
RT2dlnKdTP

S.
Standard free energy change (ÎGÂ°)
4.
-RTlogeK

Topic: Chemistry, Thermodynamics (C), NEET
Match the trophic levels with their correct species examples in grassland ecosystem.

(a)	Fourth trophic level	(i)	Crow

(b)	Second trophic level	(ii)	Vulture

(c)	First trophic level	(iii)	Rabbit

(d)	Third trophic level	(iv)	Grass

Select the correct option.

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

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

Solution:
a) Primary producers (PP): They are autotrophic organisms that fix up the solar energy and manufacture their own food from inorganic raw material. So, they form the base of the food chain constituting the first trophic level (T1). E.g. Grass

b) Primary consumers (PC) or Herbivores: There are animals that feed on green plants or plant products, so they constitute a second trophic level (T2). E.g. Rabbit

c) Secondary consumers (SC) or Primary carnivores: These are animals that feed on herbivores and form third level (T3). E.g. Crow

d) Tertiary consumers (TC) or Secondary carnivores: These are animal that feeds on secondary consumers constitute the fourth trophic level (T4). E.g. Vulture

Topic: Biology, Ecosystem, NEET, NEET 2020 (Phase 1)
Mean and standard deviation of 100 items are 50 and 4, respectively. Find the sum of all the items and the sum of the squares of the items.

Options:

None

Solution:
We have, $\overline{\mathrm{x}}=\frac{\sum \mathrm{x}}{100}=50$
$$
\Rightarrow \quad \Sigma \mathrm{x}=5000
$$
also $\sigma=\sqrt{\frac{\sum \mathrm{x}^2}{100}-(50)^2}=4$
$$
\Rightarrow \quad 16=\frac{\sum x^2}{100}-2500
$$
$$
\Rightarrow \quad \frac{\Sigma x^2}{100}=2516 \quad \Rightarrow \Sigma x^2=251600
$$

Topic: Mathematics, Statistics
Mean and standard deviation of \( 100 \) items are \( 50 \) and \( 4 \) respectively. The sum of all squares of
the items is

Options:
A \( 256100 \)
B \( 261600 \)
C \( 251600 \)
D \( 266000 \)

The correct answer is:
\( 251600 \)

Solution:
(C)
\[
\begin{array}{l}
\sigma=\sqrt{\frac{\sum x i^{2}}{n}-\left(\bar{x}^{2}\right)} \\
4=\sqrt{\frac{\sum x i^{2}}{100}-(50)^{2}} \\
16=\frac{\sum x i^{2}}{100}-2500 \\
\Rightarrow \sum x i^{2}=251600
\end{array}
\]

Topic: Mathematics, Statistics, KCET, KCET 2019