datavorous/entrance-exam-dataset huggingface dataset:
Consider the following reactions:
(i) $\mathrm{H}_{(a q)}^{+}+\mathrm{OH}_{(a q)}^{-} \rightarrow \mathrm{H}_2 \mathrm{O}_{(\mathrm{i})}, \Delta \mathrm{H}$ $=-\mathrm{X}_1 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(ii) $\mathrm{H}_{2(a q)}+\frac{1}{2} \mathrm{O}_{2(a q)} \rightarrow \mathrm{H}_2 \mathrm{O}_{(\mathrm{I})} \Delta \mathrm{H}$ $=-\mathrm{X}_2 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(iii) $\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2(g)} \rightarrow \mathrm{CO}_{(g)}+\mathrm{H}_2 \mathrm{O}, \Delta \mathrm{H}$ $=-\mathrm{X}_3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(iv) $\mathrm{C}_2 \mathrm{H}_{2(g)}+\frac{5}{2} \mathrm{O}_{2(g)} \rightarrow 2 \mathrm{CO}_{2(g)}+\mathrm{H}_2 \mathrm{O}_{(i) \text {, }}$ $\Delta \mathrm{H}=+\mathrm{X}_4 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Enthalpy of formation of $\mathrm{H}_2 \mathrm{O}_{(l)}$ is:

Options:
A $+\mathrm{X}_3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
B $-\mathrm{X}_4 \mathrm{~kJ} \mathrm{~mol}^{-1}$
C $+\mathrm{X}_1 \mathrm{~kJ} \mathrm{~mol}^{-1}$
D $-\mathrm{X}_2 \mathrm{~kJ}^{\mathrm{mol}}{ }^{-1}$

The correct answer is:
$-\mathrm{X}_2 \mathrm{~kJ}^{\mathrm{mol}}{ }^{-1}$

Solution:
The given reaction represents the formation of $\mathrm{H}_2 \mathrm{O}$, and the $\mathrm{X}_2$ represents the enthalpy of formation of $\mathrm{H}_2 \mathrm{O}$. The enthalpy of formation is the heat evolved or absorbed when one mole of substance is formed from its constituent atoms.
Related Theory
A negative enthalpy change represents an exothermic change where energy is released from the reaction, a positive enthalpy change represents an endothermic reaction where energy is taken in from the surroundings.

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2007
Consider the following reactions
i H + aq + OH − aq → H 2 O l ,ΔH=− X 1 kJmol −1
ii H 2 g +1/2 O 2 g → H 2 O l ,ΔH=− X 2 kJmol −1
iii CO 2 g + H 2 g →CO g + H 2 O l ,ΔH=− X 3 kJmol −1
iv C 2 H 2 g +5/2 O 2 g →2 CO 2 g + H 2 O l ,ΔH=+ X 4 kJmol −1
Enthalpy of formation of H 2 O l is

Options:
A +X3kJmol-1
B -X4kJmol-1
C +X1kJmol-1
D -X2kJmol-1

The correct answer is:
-X2kJmol-1

Solution:
The amount of heat absorbed or released when 1 mole of a substance is directly obtained from its constituent element is called the heat of formation of enthalpy of formation.
Equation (i) represents neutralisation reaction, (iii) represents hydrogenation reaction and (iv) represents combustion reaction.

Topic: Chemistry, Thermodynamics (C), NEET
Consider the following reactions in which all the reactants and the products are in gaseous state.
2PQ⇌P2+Q2;K1=2.5×105
PQ+1/2R2⇌PQR;K2=5×10-3
The value of K3 for the equilibrium
1/2P2+1/2Q2+1/2R2⇌PQR is

Options:
A 2.5×103
B 5×103
C 5×10-3
D 1.0×10-5

The correct answer is:
1.0×10-5

Solution:
Given , 2PQ⇌P2+Q2;
K1=[P2][Q2][PQ]2=2.5×105
PQ+1/2R2⇌PQR;
K2=[PQR][PQ][R2]1/2
=5×10-3
For the required equilibrium
1/2P2+1/2Q2+1/2R2⇌PQR;
K3=[PQR][P2]1/2[Q2]1/2[R2]1/2
K3=K2×1K1
=5×10-3×12.5×105=5×10-30.5×103
=1×10-5

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Consider the following reactions,

The compound (P) is

Options:
A
B
C
D

The correct answer is:

Solution:
Here compound (P) is and the reactions are as follows

Topic: Chemistry, Amines, NEET
Consider the following recuction processes
Zn2++2e-→Zn(s);E°=-0.76V
Ca2++2e-→Ca(s);E°=-2.87V
Mg2++2e-→Mg(s);E°=-2.36V
Ni2++2e-→Ni(s);E°=-0.25V
The reduction power of the metals increases in the order

Options:
A Ca < Zn < Mg < Ni
B Ni < Zn < Mg < Ca
C Zn > Mg < Ni < Ca
D Ca < Mg < Zn < Ni

The correct answer is:
Ni < Zn < Mg < Ca

Solution:
Reducing power â Tendency to undergo oxidation
∝ 1 reduction potential values
so, reducing power decreases as Ca > Mg > Zn > Ni

Topic: Chemistry, Electrochemistry, NEET
Consider the following relations for emf of an electrochemical cell

(i) EMF of cell = (oxidation potential of anode) - (Reduction potential of cathode)

(ii) EMF of cell = (oxidation potential of anode) + (Reduction potential of cathode)

(iii) EMF of cell = (reduction potential of anode) + (oxidation potential of cathode)

(iv) EMF of cell = (oxidation potential of anode) - (oxidation potential of cathode)

Which of these expressions are correct?

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

The correct answer is:
(ii) and (iv)

Solution:
EMF of a cell = reduction potential of cathode - reduction potential of anode
= reduction potential of cathode + oxidation potential of anode
= oxidation potential of anode - oxidation potential of cathode.

Topic: Chemistry, Electrochemistry, NEET
Consider the following sequence of reaction and identify the final product (Z).
CH 3 −CH= CH 2 → HBr P → Aq.NaOH Q → Oxidation K 2 Cr 2 O 7 / H + R

Options:
A CH3CH2CH3
B CH3CH2CHO
C CH3CH2COOH
D CH3COCH3

The correct answer is:
CH3COCH3

Solution:
This conversion occurs as follows to give acetone

Topic: Chemistry, Hydrocarbons, NEET
Consider the following sequence of reaction and identify the final product (Z).

CH 3 −CH =  CH 2 → HBr X → Aq. NaOH Y → NaOH +  I 2 Z

Options:
A CH3CHICH3
B CH3CH2CH2I
C CHI3
D CH3COCH3

The correct answer is:
CHI3

Solution:
This conversion occurs as follows to give iodoform
CH 3 −CH =  CH 2 → HBr CH 3 − CH | Br − CH 3     X → Aq.NaOH CH 3 − CH | OH − CH 3     Y → NaOH +  I 2 CHI 3 Z +  CH 3 COONa

Topic: Chemistry, Hydrocarbons, NEET
Consider the following sequence of reaction and identify the final product (Z).
CH 3 COOH → Ca (OH) 2 X → Δ Y → NH 2 OH Z

Options:
A Ethanol oxime
B Methanal oxime
C Propanone oxime
D 2-nitrosopropane

The correct answer is:
Propanone oxime

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, NEET
Consider the following sequence of reaction. Indentify the final product (Y)
CH3CH2CH3→Cl2/hv(X)→aq.OH-(Y)

Options:
A propan-1-ol
B propan-2-ol
C mixture of both propan-1-ol and propan-2-ol
D ethanol

The correct answer is:
propan-2-ol

Solution:
The rate of abstraction of hydrogen atom from an alkane follows the order 3°>2°>1° . Therefore, (X) is

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Consider the following sequence of reaction.
Na+ NH 3 g → X → N 2 O Y → Δ Z Gasprue
Identify (Z) gas

Options:
A N2
B NH3
C O2
D None of these

The correct answer is:
N2

Solution:
Here gas Z is nitrogen
Na+ NH 3 g → Δ NaNH 2 X → N 2 O NaN 3 Y → Δ N 2 ↑ Z
Hint : NaN3 (sodium azide) on heating at 300Â°C easily gives nitrogen gas and sodium residue.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
Consider the following sequence of reaction
Ph-NO2→Sn/HClX→NaNO2/HClP→CuBr/HBrQ
The final product (Q) is

Options:
A chlorobenzene
B bromobenzene
C benzyl bromide
D benzyl chloride

The correct answer is:
bromobenzene

Solution:
Ph-NO2→Sn/HClPh-NH2→NaNO2/HCl
Ph-N=N-Cl→CuBr/HBrPh-Br

Topic: Chemistry, Amines, NEET
Consider the following sequence of reactions:
C 6 H 5 CH 3 → Cl 2 p− ClC 6 H 4 CH 3 → NBS (X) → Mg (Y) → (i) ethylene oxide (Z)
Final product (Z) is

Options:
A p-ClC6H4CH2CH2OH
B p-ClC6H4CH2CH2CH2OH
C p-(HOCH2CH2)-C6H4CH2Br
D p-(HOCH2CH2)-C6H4CH3

The correct answer is:
p-ClC6H4CH2CH2CH2OH

Solution:
N-bromosuccinimide (NBS) is used for brominating alkene at allylic position (H atom attached to a carbon atom next to a double bond). Grignard formation occurs preferntially with the more reactive benzylic halide
C 6 H 5 CH 3 → Cl 2 p− ClC 6 H 4 CH 3 → NBS p− ClC 6 H 4 CH 2 Br → Mg p− ClC 6 H 4 CH 2 MgBr
→ (i) ethylene oxide p− ClC 6 H 4 CH 2 CH 2 CH 2 OH

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Consider the following sets of quantum number

n	l	m	s

(i)	3	0	0	+1/2

(ii)	2	2	1	+1/2

(iii)	4	3	-2	-1/2

(iv)	1	0	-1	-1/2

(v)	3	2	3	+1/2

Which of the following sets of quantum number is not possible?

Options:
A (i), (ii), (iii) and (iv)
B (ii), (iv) and (v)
C (i) and (iii)
D (ii), (iii) and (iv)

The correct answer is:
(ii), (iv) and (v)

Solution:
(i) represents an electron in 3s orbital.
(ii) is not possible as value of l varies from 0, 1, .... (n – 1).
(iii) represents an electron in 4f orbital.
(iv) is not possible as value of m varies from -l…+l.
(v) is not possible as value of m varies from -l…+l, it can never be greater than l.

Topic: Chemistry, Structure of Atom, NEET
Consider the following sets of quantum numbers:
\(\begin{array}{|c|c|c|c|c|} \hline & n & l & m & s \\ \hline \text {(i) } & 3 & 0 & 0 & +1 / 2 \\ \text {(ii) } & 2 & 2 & 1 & +1 / 2 \\ \text {(iii) } & 4 & 2 & -2 & -1 / 2 \\ \text {(iv) } & 1 & 0 & -3 & -1 / 2 \\ \text {(v) } & 3 & 2 & 3 & +1 / 2 \\ \hline \end{array}\)
Which of the following sets of quantum number is not possible?

Options:
A (i), (ii), (iii) and (iv)
B (ii), (iv), and (v)
C (i), and (iii)
D (ii), (iii) and (iv)

The correct answer is:
(ii), (iv), and (v)

Solution:
(ii) is not possible for any value of $n$ because $l$ varies from 0 to $(n-1)$ thus for $n=2$, can be only $0,1,2$.
(iv) is not possible because for $l=0$, $m=0$.
(v) is not possible because for $l=2$, $m$ varies from -2 to +2 .
Related Theory
The value of the spin quantum number is independent of principal, azimuthal and magnetic quantum number. It is never possible that two electrons in the same orbital will have the value of all four quantum numbers the same.

Topic: Chemistry, Structure of Atom, NEET, NEET 2007
Consider the following species,
1. O 2 2− ,  2. CO + ,3. O 2 +
Among these , sigma bond alone is present in

Options:
A 1 alone
B 2 alone
C 3 alone
D 1 and 2

The correct answer is:
1 alone

Solution:
O 2 2− ;  bond order = 1
CO + ;bond order =3
O 2 + ;bond order = 1.5
Thus, only O22- have Ï bonds alone.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Consider the following species: CN+,CN-,NO and CN
Which one of these will have the highest bond order?

Options:
A CN+
B CN-
C NO
D CN

The correct answer is:
CN-

Solution:
Using M.O theory
NO:(σ1s)2,(σ*1s)2,(σ2s)2,(σ*2s)2,(σ2pz)2,(π2px)2=(π2py)2,(π*2px)1=(π*2py)0
BO=10-52=2.5
CN-:(σ1s)2,(σ*1s)2,(σ2s)2,(σ*2s)2,(π2px)2=(π2py)2,(σ2pz)2
BO=10-42=3
CN:(σ1s)2,(σ*1s)2,(σ2s)2,(σ*2s)2,(π2px)2=(π2py)2,(σ2pz)1
BO=9-42=2.5
CN+:(σ1s)2,(σ*1s)2,(σ2s)2,(σ*2s)2,(π2px)2=(π2py)2
BO=8-42=2

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Consider the following species,
CN+, CN-1, NO and CN
Which one of these will have the highest bond order?

Options:
A CN+
B CN-
C NO
D CN

The correct answer is:
CN-

Solution:
Ion/Species	Total electron	Bond order

NO	15	2.5

CN-	14	3

CN+	12	2

CN	13	2.5

Number of e− in NO=7+8=15
NO=σ1s2<σ *1s2<σ2s2<σ *2s2<σ2Pz2<π2Py2=π2Px2<π*2Py1=π *2Px0
Bond order =(number of electron in bonding molecular orbital − number of electron in anti bonding molecular orbital)2
BO=6−12=2.5 .
Number of e− in CN=6+7=13
NO=σ1s2<σ *1s2<σ2s2<σ *2s2<σ2Pz2<π2Py2=π2Px2<π*2Py1=π *2Px0.
Bond order = 5 2 =2.5
Number of e− in CN+=6+7−1=12
NO=σ1s2<σ *1s2<σ2s2<σ *2s2<σ2Pz2<π2Py2=π2Px2<π*2Py1=π *2Px0
Bond order = 4−0 2 =2
CN−= number of e− in CN−=6+7+1=14
CN-=σ1s2<σ *1s2<σ2s2<σ *2s2<σ2Pz2<π2Py2=π2Px2<π*2Py1=π *2Px0
Bond order=6−02=3 .
CN− as highest bond order.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2018
Consider the following statement for the given reaction
1. The reaction is SN1
2. The reaction intermediate is carbocation
3. The major product will be 2, 3-dimethyl-3-pentanol
4. The major product has one stereogenic centre

Options:
A 1, 2 and 3 are correct
B 2, 3 and 4 are correct
C 1, 2, 3 and 4 are correct
D 1, 3 and 4 are correct

The correct answer is:
1, 2, 3 and 4 are correct

Solution:
Alkyl halide whose β -carbon is 3° and 4° gives SN1 reaction.
It means all statements regarding this reaction are correct.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Consider the following statement :

(I) CH 3 OCH â 2 is more stable than CH 3 CH â 2

(II) Me 3 C â is more stable than CH 3 CH 2 CH â 2

(III) CH 2 = CH - CH â 2 is more stable than CH 3 CH 2 CH 2 â

(IV) CH 2 = CH â is more stable than CH 3 CH 2 â

Of these statement:

Options:
A I and II are correct
B III and IV are correct
C I, II and III are correct
D II, III and IV are correct

The correct answer is:
I, II and III are correct

Solution:
(I)ââ CH 3 â O â¢â¢ â¢â¢ â CH 2 â is more stable than CH 3 â CH 2 â , becausse of +M-effect of CH 3 - O â¢ â¢ â¢ â¢ - .
(II) Me 3 C â H is more stable than CH 3 - CH 2 - CH â 2 ,
because Me 3 C â have more 2-hydrogen atoms than CH 3 â CH 2 â CH 2 â , so, it is more stable, due to greater hyperconjugation.
(III) The positive charge is delocalised in C H 2 =CHâC H 2 â , so more stable than, CH 3 â CH 2 â CH 2 â .
(IV) CH 2 = CH â will be less stable than CH 3 - CH 2 â as positive charge is less stable on double bonded 'C' than on single bonded 'C'.

Topic: Chemistry, General Organic Chemistry, NEET
Consider the following statement w.r.t. transcription and select the right choice

(i) The promoter is said to be located towards the 5â²-end of the coding strand.
(ii) The promoter is an RNA sequence that provides a binding site for RNA polymerase.
(iii) The terminator is located towards the 3â²-end of the coding strand.
(iv) The strand that has polarity 3â²â5â² acts as a template strand.

Options:
A i & iii are incorrect
B i & ii are correct
C Only ii is incorrect
D Only iv is correct

The correct answer is:
Only ii is incorrect

Solution:
-Transcription is under the control of the enzyme RNA POLYMERASE (DNA dependent RNA polymerase)
-The promoter is the place which promotes transcription and promoter is situated upstreams of a structural gene at 5’
-Terminator is situated downstream of the structural gene at the 3’ end of the coding strand.
-One of the two strands of DNA takes part in transcription and which strand take part in transcription is known as master strand/antisense strand/non-coding strand/template strand/negative strand

Topic: Biology, Molecular Basis of Inheritance, NEET
Consider the following statements
1. Cs+ ion is more highly hydrated than other alkali metal ions
2. Among the alkali metals Li, Na, K and Rb, Lithium (Li) has the highest melting point
3. Among the alkali metals, only lithium form a stable nitride by direct combination.
O these statements

Options:
A 1, 2 and 3 are correct
B 1 and 2 are correct
C 1 and 3 are correct
D 2 and 3 are correct

The correct answer is:
2 and 3 are correct

Solution:
Li+>Na+>K+>Rb+>Cs+ (order of hydration)
As size of cation increases, solvation decreases.
Li due to its small size can form nitrides as it can stabilize the electron cloud of nitride ion.

Topic: Chemistry, s Block Elements, NEET
Consider the following statements (A) and (B) and identify the correct answer.

(A) A Zener diode is connected in reverse bias when used as a voltage regulator.
(B) The potential barrier of p-n junction lies between 0.1 V to 0.3 V.

Options:
A (A) and (B) both are correct.
B (A) and (B) both are incorrect.
C (A) is correct and (B) is incorrect.
D (A) is incorrect but (B) is correct.

The correct answer is:
(A) is correct and (B) is incorrect.

Solution:
When Zener diode is forward biased, it behaves like an ordinary diode and when reversing biased, it will be a voltage stabilizer. Also, at room temperature, the voltage across the depletion layer for silicon is about 0.6 V−0.7 V and for germanium is about 0.3 V−0.35 V Hence, statement A is correct but statement B is wrong.

Topic: Physics, Semiconductors, NEET, NEET 2021
Consider the following statements A and B and identify the correct answer:
A. For a solar-cell, the I-V characteristics lies in the IV quadrant of the given graph.
B. In a reverse biased pn junction diode, the current measured in $(\mu \mathrm{A})$, is due to majority charge carriers.

Options:
A $A$ is incorrect but $B$ is correct
B Both A and B are correct
C Both A and B are incorrect
D $A$ is correct but B is incorrect

The correct answer is:
$A$ is correct but B is incorrect

Solution:
A: Solar cell characteristics
B: In reverse biased $p n$ junction diode, the current measured in $(\mu \mathrm{A})$, is due to minority charge carrier.

Topic: Physics, Semiconductors, NEET, NEET 2024
Consider the following statements :
A. Annelids are true coelomates
B. Poriferans are pseudocoelomates
C. Aschelminthes are acoelomates
D. Platyhelminthes are pseudocoelomates
Choose the correct answer from the options given below :

Options:
A A only
B C only
C D only
D B only

The correct answer is:
A only

Solution:
The correct answer is option no. (1), because annelids are true coelomate animals. Options (2), (3) and (4) are incorrect because poriferans are acoelomates, aschelminths are pseudocoelomates and platyhelminthes are acoelomates.

Topic: Biology, Animal Kingdom, NEET, NEET 2024
Consider the following statements:
(A) Coenzyme or metal ion that is tightly bound to enzyme protein is called prosthetic group.
(B) A complete catalytic active enzyme with its bound prosthetic group is called apoenzyme.
Select the correct option.

Options:
A Both (A) and (B) are correct.
B (A) is true but (B) is false.
C Both (A) and (B) are false.
D (A) is false but (B) is true

The correct answer is:
(A) is true but (B) is false.

Solution:
Many enzymes show catalytic activity only in association with certain non-protein substances. Those substances are known as cofactors (cofactors may be simple metal ions, e.g. Mg++ or K+ etc.). The enzymes which works only in presence of cofactors are called apoenzymes. An organic non-protein cofactor which is easily separable from the apoenzyme is called coenzyme, and the non-protein organic cofactor that is tightly bound to and non-dissociable from the enzyme is termed as prosthetic group, e.g. biotin of carboxylases, haem of hemoglobin, etc.

Topic: Biology, Biomolecules (B), NEET, NEET 2019 (All India)
Consider the following statements (A - D) and select the option which correctly fills the blanks.

A. A group of species belonging to a trophic level that exploits a common resource base in a similar fashion is known as (i).

B. (ii) reduces decomposition and causes the pilling up of detritus.

C. In oceans, productivity is limited by (iii) and (iv).

D. The largest reservoir of calcium is (v).

Options:
A

(i)	(ii)	(iii)	(iv)	(v)

Guild	Oxygen	Nitrogen	Light	Rocks

B

(i)	(ii)	(iii)	(iv)	(v)

Guild	Anaerobic condition	Light	Nitrogen	Rocks

C

(i)	(ii)	(iii)	(iv)	(v)

Guild	Aerobic respiration	Phosphorus	Dark	Soil

D

(i)	(ii)	(iii)	(iv)	(v)

Guild	Anaerobic respiration	Nitrogen	Light	Soil

The correct answer is:

(i)	(ii)	(iii)	(iv)	(v)

Guild	Anaerobic condition	Light	Nitrogen	Rocks

Solution:
Guild refers to the group of species in a trophic level using the same resource in a similar manner. Decomposition is an oxygen-consuming process. The anaerobic condition leads to decrease in it and increases the amount of dead matter. The productivity of the ocean is limited by light and nitrogen. Calcium which is one of the essential elements in an organism has the largest reservoir in rocks (lithosphere).

Topic: Biology, Ecosystem, NEET
Consider the following statements A â D each with one or two blanks.

(I) Bears go into â¦ (1)â¦.. during winter to â¦..(2)â¦.. cold weather.

(II) A wasp pollinating a fig flower is an example of â¦â¦.(3)â¦â¦ .

Which one of the following options gives the correct fill up for the respective blanks?

Options:
A (1) stable (2) commensalism, (3) marsh
B (1) aestivation, (2) escape, (3) mutualism
C (3) Aestivation, (2) hibernation, (3) Commensalism
D (1) hibernation, (2) escape, (3) Mutualism

The correct answer is:
(1) hibernation, (2) escape, (3) Mutualism

Solution:
Bears going into hibernation during winters is an example of escape in time to survive in cold winers. A Gall Wasp lays its eggs in hypanthodium inflorescence of fig plant and during the act of laying egg it brings about pollination by transferring pollen grains from one inflorescence to other. The fig offers the wasp some of its developing seeds, as food for the developing wasp larvae.

Topic: Biology, Organisms and Populations, NEET
Consider the following statements.

A. Temperature is the most ecologically relevant environmental factor.
B. Temperature increases progressively from the equator towards the poles and from plains to the mountain tops.
C. Diversity is not an important characteristic of living organisms.
D. Mango trees do not and cannot grow in temperate countries like Canada and Germany.

Of the above statements:

Options:
A B and D are correct
B A and B are correct
C A and C are correct
D A and D are correct

The correct answer is:
A and D are correct

Solution:
One of the most important factors of ecology is the temperature that varies seasonally. Also, physiological functions, morphology and the geographical distribution of organisms is affected by temperature. Temperature decreases progressively from the equator towards the poles and from plains to the mountain tops. Mango requires higher temperature for tis growth and development and lower temperatures through enzymes, if influences basal metabolism, activity and other physiological functions of the organism. Thus, Mango trees do not and cannot grow in temperate countries like Canada and Germany. All living organisms have several characteristics like growth, reproduction, sensitivity, nutrition, movement, respiration, excretion. Diversity is considered an important characteristic of living organisms as it provides information on its origin and inter-relationship with other organisms and environment.

Topic: Biology, Organisms and Populations, NEET
Consider the following statements.

A. The mustard flower is hypogynous

B. Rose is perigynous

C. China rose is hypogynous

Which of the statements given above is/are CORRECT?

Options:
A A and C
B B and C
C A, B & C
D None

The correct answer is:
A, B & C

Solution:
In hypogynous flowers, the stamens and other floral parts are attached to the receptacle below the gynoecium with a superior ovary and in perigynous flowers, there is a floral tube which is formed from the fusion of stamens, petals and sepals, this tube is attached to the receptacle below the gynoecium. Examples of hypogynous flowers are mustard, china rose flower and examples of perigynous are rose flowers, tomato, etc.

Topic: Biology, Morphology of Flowering Plants, NEET
Consider the following statements:

(A) The portion of the spectrum between 500 nm and 800 nm is also referred to as photosynthetically active radiation (PAR).

(B) Magnesium, calcium and chloride ions play prominent roles in the photolysis of water

(C) In cyclic photophosphorylation, oxygen is not released (as there is no photolysis of water) and NADPH is also not produced.

Of these statements given above, choose the correct option from the following.

Options:
A A is true but B and C are false
B A and B are false but C is true
C B is true but A and C are false
D A and B are true but C is false

The correct answer is:
A and B are false but C is true

Solution:
Cyclic photophosphorylation occurs when there is a scarcity of NADP+. When NADP+ is scarce then the electron is not transferred from A (Fes) complex to ferredoxin. A (Fes) complex transfer the electron to Cyt b6f complex and then via plastocyanin to P-700. Generation of ATP instead of NADPH takes place.
Magnesium and calcium do not play any role during photosynthesis. A major role is played by manganese and chloride ions.
The most optimum region for photosynthesis is 600-680 nm.

Topic: Biology, Photosynthesis in Higher Plants, NEET