datavorous/entrance-exam-dataset huggingface dataset:
Embryonal tissue of angiosperms contains:

Options:
A Chloroplasts
B Chromatophores
C Chromoplasts
D Leucoplasts

The correct answer is:
Leucoplasts

Solution:
Chloroplast are organelles required for conducting photosynthesis.
Chromoplast provide colour to the cell.
Leucoplast are colourless plastids useful in storing carbohydrates and proteins. Embryonal tissue contain leucoplast as it helps in its growth and development by using the stored nutrients.

Topic: Biology, Cell - The Unit of Life, NEET
Emergency contraceptives are effective if used within

Options:
A 72 hrs of coitus
B 72 hrs of ovulation
C 72 hrs of menstruation
D 72 hrs of implantation

The correct answer is:
72 hrs of coitus

Solution:
Administration of progestogens or progestogen-estrogen combinations or IUDs within 72 hours of coitus has been found to be very effective as emergency contraceptives as they could be used to avoid possible pregnancy due to rape or casual unprotected intercourse.

Topic: Biology, Reproductive Health, NEET
Emerson's enhancement effect and red drop have been instrumental in the discovery of:

Options:
A Photophosphorylation and non - cyclic electron transport
B Two photosystems operating simultaneously
C Photophosphorylation and cyclic electron transport
D Oxidative phosphorylation

The correct answer is:
Two photosystems operating simultaneously

Solution:
The red drop effect occurs due to decreased functioning of PS-II beyond 680 nm and when both PS-I and PS-II are functioning together there is an enhancement in quantum yield. In PS I the reaction centre chlorophyll a has an absorption peak at 700 nm.

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2016 (Phase 1)
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as- v = 3.29 × 1015(Hz) [ 1/32– 1/n2] Calculate the value of n if the transition is observed at 1285 nm and find the region of the spectrum.

Options:
A n=5 and the spectrum lies in the infrared region.
B n=5 and the spectrum lies in the ultraviolet region.
C n=6 and the spectrum lies in the infrared region.
D n=6 and the spectrum lies in the microwave region.

The correct answer is:
n=5 and the spectrum lies in the infrared region.

Solution:
Correct Option is : (A)
n=5 and the spectrum lies in the infrared region.

Topic: Chemistry, Structure of Atom, NEET
EMP can produce a total of how many ATP under aerobic condition.

Options:
A 6
B 8
C 24
D 38 ATP

The correct answer is:
8

Solution:
Glycolysis is commonly known as the EMP pathway after the name of scientist Embden-Meyerhoff-Parnas. In glycolysis, 8ATP are produced of which 4 are formed from substrate-level phosphorylation and 6 are formed by oxidation of NADH in ETS two molecules of ATP are consumed, thus net gain is 8ATP.

Topic: Biology, Respiration in Plants, NEET
Emulsification of fat is carried out by

Options:
A Pancreatic juice
B HCl
C Bile
D Mucus of intestine

The correct answer is:
Bile

Solution:
The bile salts are produced in liver and are stored in the gallbladder. Bile salts like sodium glycocholate and sodium taurocholate reduce the surface tension of large fat globules and break them into fine-minute fat droplets. This process is known as the emulsification of fat and it happens in the small intestine.

Topic: Biology, Digestion and Absorption, NEET
End product D of the reaction sequence is
CH3Br→KCNA→H2O/H+B→SOCl2C→(C2H5)2CdD

Options:
A CH3CH2-O-C2H5
B C2H5COOCH3
C CH3COC2H5
D CH3COOC2H5

The correct answer is:
CH3COC2H5

Solution:
CH 3 Br → KCN CH 3 CN A → H 3 O + CH 3 COOH B → SOCl 2 Py CH 3 COCl → (C 2 H 5 ) 2 Cd C 2 H 5 COCH 3 2-Butanone

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
End product of glycolysis are

Options:
A 2 ATP, 2 Pyruvic acid
B 2 ATP, 2 FADH2 , 2 Pyruvic acid
C 2 ATP, 2 NADH+H+ , 2 Pyruvic acid
D 6 ATP, 2 NADH+H+ , 2 Pyruvic acid

The correct answer is:
2 ATP, 2 NADH+H+ , 2 Pyruvic acid

Solution:
Glycolysis is the process of breakdown of one molecule of glucose to form 2 molecules of pyruvic acid. During this process oxidation occurs by dehydrogenation, hydrogen thus released is accepted by NAD forming 2 NADH+H+ .

Topic: Biology, Respiration in Plants, NEET
End product of the following sequence of reactions is:

Options:
A
B
C
D

The correct answer is:

Solution:
The first step is an allylic bromination of cycloalkene.(free radical substitution)
NBS is a Br2 source: Br - Br â h Î½ 2 Br â

Topic: Chemistry, Hydrocarbons, NEET
End product S of the reaction sequence is
CH3-CH2Br→KCNP→H2O/H+Q→SOCl2R→(C2H5)2CdS

Options:
A CH3CH2-CH2-O-C2H5
B C2H5COOCH2-CH3
C CH3-CH2-COC2H5
D CH3COOC2H5

The correct answer is:
CH3-CH2-COC2H5

Solution:
Here the final product S is pentan-3-one
CH 3 − CH 2 Br → KCN CH 3 − CH 2 CN P → H 3 O + CH 3 − CH 2 COOH Q → SOCl 2 CH 3 − CH 2 COCl R
→(C2H5)2PdC2H5COCH2CH3Pentan-3-oneS

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Endemic plants are those which are:

Options:
A Cosmopolitan in distribution
B Restricted to certain area
C Found in arctic region
D Gregarious in habit

The correct answer is:
Restricted to certain area

Solution:
Species restricted to small area are called endemic. It is estimated that 18% of plants in India are endemic and found nowhere in the else in the world.

Topic: Biology, Biodiversity and its Conservation, NEET
Endocarp is hard and stony in the drupe of

Options:
A Mangifera indica and Cocos nucifera
B Mangifera indica and Pheonix dactylifera
C Pheonix dactylifera and Cocos nucifera
D Pheonix dactylifera and Cucumis sativa

The correct answer is:
Mangifera indica and Cocos nucifera

Solution:
In mango (Mangifera indica) and coconut (Cocos nucifera), the fruit is known as a drupe. They develop from monocarpellary superior ovaries and are one seeded. In mango the pericarp is well differentiated into an outer thin epicarp, a middle fleshy edible mesocarp and an inner stony hard endocarp. In coconut which is also a drupe, the mesocarp is fibrous and endocarp is stony hard. The fruit of date (Pheonix dactylifera) is berry with a papery endosperm. The fruit of cucumber (Cucumis sativa) is pepo

Topic: Biology, Morphology of Flowering Plants, NEET
'Endomitosis' refers to

Options:
A Division of nucleus without chromosomal division
B Division of chromosome without nuclear division
C Division of cytoplasm
D None of these

The correct answer is:
Division of chromosome without nuclear division

Solution:
'Endomitosis' refers to the replication of DNA without any nuclear division. In this process, the division of chromosomes will be seen without nuclear division and resulting in an increased number of chromosomes in the cell.

Topic: Biology, Cell Cycle and Cell Division, NEET
Endonucleases cut at :

Options:
A Hydrogen bond
B Glycosidic bond
C Sugar phosphate backbone
D Ori site

The correct answer is:
Sugar phosphate backbone

Solution:
Restriction endonucleases will bind to the DNA and cut each of the two strands of the double helix at specific points in their sugar -phosphate backbones.

Topic: Biology, Biotechnology - Principles and Processes, NEET
Endosperm, a product of double fertilization in angiosperms is absent in the seeds of

Options:
A Coconut
B Orchids
C Maize
D Castor

The correct answer is:
Orchids

Solution:
Endosperm, a product of double fertilization in angiosperms, is absent in the seeds of orchids.

Topic: Biology, Morphology of Flowering Plants, NEET
Endosperm, a product of double fertilization in angiosperms is absent in the seeds of:

Options:
A Wheat
B Orchids
C Maize
D Castor

The correct answer is:
Orchids

Solution:
Endosperm is formed by triple fusion in angiosperm and it is a nutritive tissue which provided nutrition to the developing embryo. In most of the dicot seeds endosperm is completely consumed during the development of the embryo thus seeds are non endosprmic, exception is castor which is endospermic.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Endosperm is consumed by developing embryo in the seed of

Options:
A coconut
B castor
C pea
D maize

The correct answer is:
pea

Solution:
Endosperm, if present acts as the food storage tissue of a seed. The seeds with endosperm are called endospermic or albuminous (eg, cocos nucifera, Ricinus communis, Zea mays and other cereals, while those without endosperm are called non-endospermic or ex-albuminous seeds (eg, Brassica campestris, Arachis hypogea, Pisum sativum).
In castor (Ricinus communis) the endosperm is whitish soft mass rich with oily reserve food material.
The endosperm in maize is yellowish in colour, the outermost layer of endosperm is called aleurone layer, which is rich in protein.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2008 (Screening)
Endosperm is consumed by developing embryo in the seed of

Options:
A Pea
B Maize
C Coconut
D Castor

The correct answer is:
Pea

Solution:
In some seeds, during the process of the development of the embryo, the food stored up in the endosperm is continuously drawn up by the developing embryo and thus completely exhausted. Such seeds are known as exalbuminous or non-endospermic. The common examples are gram, pea, bean, tamarind, orchid, etc.

Topic: Biology, Morphology of Flowering Plants, NEET
Endosperm is not seen in the seeds of ___________ family.

Options:
A Orchidaceae
B Cactaceae
C Ranunculaceae
D Malvaceae

The correct answer is:
Orchidaceae

Solution:
The endosperm is the tissue produced inside the seeds of most of the flowering plants following fertilization. It surrounds the embryo and provides nutrition in the form of starch, though it can also contain oils and proteins. The endosperm is not there in the mature seeds of the members of the family Orchidaceae, Podostemonaceae, and Trapaceae families.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Endothecium layer of anther lobes is present

Options:
A Outside the epidermis
B Just inside the epidermis
C In the innermost layer
D In the middle region

The correct answer is:
Just inside the epidermis

Solution:
Structure of microsporangium: In a transversesection, a typical microsporangium appears nearcircular in outline. It is generally surrounded by four wall layers the epidermis, endothecium, middle layers and the tapetum. The outer three wall layers perform the function of protection and help in dehiscence of anther to release the pollen. The innermost wall layer is the tapetum. It nourishes the developing pollen grains. Cells of the tapetum possess dense cytoplasm and generally have more than one nucleus.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Energy E of a hydrogen atom with principal quantum number n is given by E = -13.6n2 eV. The energy of a photon ejected when the electron jumps from n=3 state to n=2 state of hydrogen is approximately

Options:
A 1.5 eV
B 0.85 eV
C 3.4 eV
D 1.9 eV

The correct answer is:
1.9 eV

Solution:
Given, En = -13.6n2 eV
The energy of photon ejected when an electron jumps from n=3 state to n=2 state is given by
âE=E3-E2
â´ E3 = - 13.632 eV =- 13.69 eV
E2 = - 13.622 eV =- 13.64 eV
So, âE=E3-E2 = -13.69-- 13.64
=1.9 eV (approximately)

Topic: Physics, Atomic Physics, NEET
Energy $E$ of a hydrogen atom with principal quantum number $n$ is given by $E=\frac{-13.6}{n^2} \mathrm{eV}$. The energy of a photon ejected when the electron jumps for $n=3$ state $n=2$ state of hydrogen is approximately

Options:
A $1.5 \mathrm{eV}$
B $0.85 \mathrm{eV}$
C $3.4 \mathrm{eV}$
D $1.9 \mathrm{eV}$

The correct answer is:
$1.9 \mathrm{eV}$

Solution:
Energy of photon $=E_2-E_2$
$$
\begin{gathered}
=\frac{13.6}{9}-\left(\frac{-13.6}{9}\right)=\frac{5}{30} \times 13.6 \\
=1.9 \mathrm{eV}
\end{gathered}
$$

Topic: Physics, Atomic Physics, NEET, NEET 2004
Energy is being emitted from the surface of a black body at a temperature of 127 °C, at the rate of 1.0×106 J sec-1 m-2 . The temperature of the black body at which the rate of energy emission is 16.0×106 J sec-1 m-2 will be

Options:
A 254 °C
B 508 °C
C 527 °C
D 727 °C

The correct answer is:
527 °C

Solution:
T1=127o
T1=400 K
From Stefan's Law
dQ1dtdQ2dt=T14T24
11614=T1T2
12=400T2
T2=800K
T2=800-273=527 oC

Topic: Physics, Thermal Properties of Matter, NEET
Energy level of A, B, C of a certain atom correspond to increasing values of energy i.e., EA<EB<EC . If λ1,λ2,λ3 are the wavelength of radiation corresponding to the transition C to B, B to A and C to A respectively. Which of the following statement is correct?

Options:
A λ3=λ1+λ2
B λ3=λ1λ2λ1+λ2
C λ1+λ2λ3=0
D λ32=λ12λ22

The correct answer is:
λ3=λ1λ2λ1+λ2

Solution:
We know, Ε2-E1=hcλ
So ΕC-EB=hcλ1 , EB-EA=hcλ2 , EC-EA=hcλ3
EC-EA=(EC-EB)+(EB-EA)
hcλ3-=hcλ1+hcλ2
So λ3=λ1λ2λ1+λ2

Topic: Chemistry, Structure of Atom, NEET
Energy levels $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ of a certain action corresponding to increasing values of energy i.e. $E_A < E_B < E_C$. If $\lambda_1$, $\lambda_2$ and $\lambda_3$ are wavelength of radiations corresponding to transitions $\mathrm{C}$ to $\mathrm{B}, \mathrm{B}$ to $\mathrm{A}$ and $\mathrm{C}$ to $\mathrm{A}$ respectively, which of the following relations is correct?

Options:
A $\lambda_3=\lambda_1+\lambda_2$
B $\lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}$
C $\lambda_1+\lambda_2+\lambda_3=0$
D $\lambda_3^2=\lambda_1^2+\lambda_2^2$

The correct answer is:
$\lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}$

Solution:
$E_A < E_B < F_C$
$$
\begin{aligned}
E & =\frac{h c}{\lambda} \\
E_{C A} & =E_{C B}+E_{B A} \\
\frac{h c}{\lambda_3} & =\frac{h c}{\lambda_1}+\frac{h c}{\lambda_2} \\
\frac{1}{\lambda_3} & =\frac{1}{\lambda_1}+\frac{1}{\lambda_2} \\
\Rightarrow \quad \lambda_3 & =\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}
\end{aligned}
$$

Topic: Physics, Atomic Physics, NEET, NEET 2005
Energy of an electron is given by E = - 2.178 Ã 10 - 1 8 J Z 2 n 2.Wavelength of light required to excite an electron in an hydrogen atom from level n=1 to n=2 will be :
h=6.62Ã10-34 Js and c=3.0Ã108 ms-1

Options:
A 6.500 Ã 1 0 - 7 m
B 8.500 × 1 0 - 7 m
C 1.214 Ã 1 0 - 7 m
D 2.816 Ã 1 0 - 7 m

The correct answer is:
1.214 × 1 0 - 7 m

Solution:
E 1 =â2.178Ã 10 â18 Ã 1 1 =â2.178Ã 10 â18 J
= 2.178Ã 10 â18 1.6Ã 10 â19 =13.6 eV
E 2 = â2.178Ã 10 â8 ( 2 ) 2 = â2.178Ã 10 â18 4 =0.5448Ã 10 â18 J
=â3.4 eV
ÎE = E 2 - E 1 = - 3 . 4 - - 1 3 . 6 = 1 0 . 2 eV
Î» in Ã = 12375 10.2 =1215.6 Ã â 1214.4Ã 10 â10 m â 1.214Ã 10 â7 m { â´ 1 Ã=10 â10 m }

Topic: Chemistry, Structure of Atom, NEET
Energy per unit volume in a stretched wire is

Options:
A Half of load × strain
B Load × strain
C Stress × strain
D Half of stress × strain

The correct answer is:
Half of stress × strain

Solution:
U=energy per unit volume=12×stress×strain
=12FA×∆ll=12×Y×∆ll2

Topic: Physics, Mechanical Properties of Solids, NEET
Energy per unit volume represents

Options:
A Pressure
B Force
C Thrust
D Work

The correct answer is:
Pressure

Solution:
[Energy][Volume]=[ML2T-2][L3]=[ML-1T-2]
[pressure]=[MLT-2][L2]=[ML-1T-2]

Topic: Physics, Units and Dimensions, NEET
Energy transfer from one trophic level to other, in a food chain is

Options:
A 1%
B 2%
C 10%
D 20%

The correct answer is:
10%

Solution:
As per 10 percent law, only 10% of the energy fixed at one trophic level is passed to next trophic level. For example, when plants are eaten by animals only 10% of energy of the food is fixed into animal flesh; rest is lost through respiration.

Topic: Biology, Ecosystem, NEET
Entamoeba differs from Amoeba in not having:

Options:
A Nucleus
B Pseudopodia
C Ectoplasm
D Contractile vacuole

The correct answer is:
Contractile vacuole

Solution:
Amoeba is a freshwater organism which feeds on alga and plankton, whereas Entamoeba is an internal parasite which lives inside animals and human. The latter is also the causative agent for amoebic dysentery in humans.
Entamoeba lacks both mitochondria and contractile vacuoles, which are distinguishing features from Amoeba.

Topic: Biology, Biological Classification, NEET
Entamoeba differs from Amoeba is not having

Options:
A Nucleus
B Pseudopodia
C Ectoplasm
D Contractile vacuole

The correct answer is:
Contractile vacuole

Solution:
Absence of contractile vacuole in E.histolytica is a distinguish feature from Amoeba.

Topic: Biology, Biological Classification, NEET
Entamoeba histolytica infection occurs through

Options:
A Contaminated water and food
B Sweat
C Bird droppings
D Mosquito bites

The correct answer is:
Contaminated water and food

Solution:
Infection depends upon the intake of food or water contaminated with faecal matter containing cysts of E. histolytica.

Topic: Biology, Human Health and Diseases, NEET
Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, – 393, 81 and 9.7 kJ mol–1respectively. Find the value of ∆rH for the reaction: N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)

Options:
A –777.7 kJ mol-1
B –869.4 kJ mol-1
C –648.3 kJ mol-1
D –579.2 kJ mol-1

The correct answer is:
–777.7 kJ mol-1

Solution:
Correct Option is : (A)
–777.7 kJ mol-1

Topic: Chemistry, Thermodynamics (C), NEET
Enthalpy change for the reaction,
$$
4 \mathrm{H}(g) \longrightarrow 2 \mathrm{H}_2(g) \text { is }-869.6 \mathrm{~kJ}
$$
The dissociation energy of $\mathrm{H}-\mathrm{H}$ bond is

Options:
A $-869.6 \mathrm{~kJ}$
B $+434.8 \mathrm{~kJ}$
C $+217.4 \mathrm{~kJ}$
D $-434.8 \mathrm{~kJ}$

The correct answer is:
$+434.8 \mathrm{~kJ}$

Solution:
$$
\begin{array}{ll}
4 \mathrm{H}(g) & \rightarrow 2 \mathrm{H}_2(g) ; \Delta H=-869.6 \mathrm{~kJ} \\
2 \mathrm{H}_2(g) & \rightarrow 4 \mathrm{H}(g) ; \Delta H=869.6 \mathrm{~kJ} \\
\mathrm{H}_2(g) & \rightarrow 2 \mathrm{H}(g) ; \Delta H=\frac{869.6}{2}=434.8 \mathrm{~kJ}
\end{array}
$$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2011 (Screening)
Enthalpy diagram for a particular reaction is given in Fig. 6.3. Is it possible to decide spontaneity of a reaction from given diagram.

Options:
A Yes
B No
C Yes, but under specific set of conditions
D The graph is incorrect

The correct answer is:
No

Solution:
Correct Option is : (B)
No

Topic: Chemistry, Thermodynamics (C), NEET
Enthalpy (H) of system is equal to

Options:
A Work (W) done by a system
B Product of pressure (P) and volume (V) of gas
C Internal energy (E)+PV
D Internal energy (E)

The correct answer is:
Internal energy (E)+PV

Solution:
Enthalpy (H) is defined as the sum of internal energy and pressure volume work. H=E+PV .

Topic: Chemistry, Thermodynamics (C), NEET
Enthalpy is an extensive property. In general, if enthalpy of an overall reaction A→B along one route is ∆rH and ∆rH1, ∆rH2, ∆rH3..... represent enthalpies of intermediate reactions leading to product B. What will be the relation between ∆rH for overall reaction and ∆rH1, ∆rH2..... etc. for intermediate reactions.

Options:
A ∆rH = ∆rH1- ∆rH2- ∆rH3....
B ∆rH = ∆rH1+ ∆rH2+ ∆rH3....
C ∆rH = ∆rH1+ ∆rH2- ∆rH3....
D ∆rH = ∆rH1- ∆rH2+ ∆rH3....

The correct answer is:
∆rH = ∆rH1+ ∆rH2+ ∆rH3....

Solution:
Correct Option is : (B)
∆rH = ∆rH1+ ∆rH2+ ∆rH3....

Topic: Chemistry, Thermodynamics (C), NEET
Enthalpy of a reaction at 27oC is 15 kJ mol-1. The reaction will be feasible if entropy is

Options:
A 15 J mol-1 K-1
B -50 J mol-1 K-1
C Greater than 50 J mol-1 K-1
D Less than 50 J mol-1 K-1

The correct answer is:
Greater than 50 J mol-1 K-1

Solution:
Î G < Î H - T Î S for spontaneous reaction
Î G should be - ve
T Î S > Î H
i.e., 3 0 0 Î S > 15000 J
ÎS>50 Jâ mol -1 ââ K -1

Topic: Chemistry, Thermodynamics (C), NEET
Enthalpy of atomization of C 2 H 6 g and C 3 H 8 g are 6 2 0 and 880 kJ mol -1 respectively. The C - C and C - H bond energies are respectively

Options:
A 80 and 60 kJ mol -1
B 80 and 90 kJ mol -1
C 70 and 90 kJ mol -1
D 200 and 80 kJ mol -1

The correct answer is:
80 and 90 kJ mol -1

Solution:
Since, C2H6 contains one C-C and 6 C-H bonds, thus 620 is the sum of BE of one C-C and 6 C-H bonds. Thus,
* C 2 H 6 â C - C + 6 C - H Î H = 620 kJ/mol ....... (i)
Similarly,
* C 3 H 8 â 2 C - C + 8 C - H Î H = 880 kJ/mol .... (ii)
Multiply Eq. (i) by 2 then we get
* 2 C 2 H 6 â 2 C - C + 1 2 C - H Î H = 1240 kJ/mol .... (iii)
equation (iii)-(i)
2 C - C + 8 C - H = 880 kJ/mol ...... (ii) 2 C - C + 1 2 C - H = 1240 kJ/mol ...... (iv)
--------------------------------------------------------------------
4 C - H = 360 kJ/mol â´ C - H = 90.0 kJ/mol
On putting the value of C-H, in equation (i) we get
6 Ã 90.0 + C - C = 6 2 0
C - C = 6 2 0 - 5 4 0
= 80 kJ/mol

Topic: Chemistry, Thermodynamics (C), NEET
Enthalpy of CH4+12O2→CH3OH is negative. If enthalpy of combustion of CH4 and CH3OH are x and y respectively, which relation is correct?

Options:
A x > y
B x < y
C x = y
D x≥y

The correct answer is:
x > y

Solution:
Enthalpy of combustion is -ve (exothermic)CH4+2O2→CO2+2H2O;ΔH=-x
CH3OH+32O2→CO2+2H2O;ΔH=-y
CH4+12O2→CH3OH;ΔH=y-x
As y-x=-ve
So x>y .

Topic: Chemistry, Thermodynamics (C), NEET
Enthalpy of combustion of carbon to CO2is –393.5 kJ mol–1. Calculate the heat released upon formation of 35.2 g of CO2from carbon and dioxygen gas.

Options:
A –478.2 kJ
B –417.8 kJ
C –356.4 kJ
D –314.8 kJ

The correct answer is:
–314.8 kJ

Solution:
Correct Option is : (D)
–314.8 kJ

Topic: Chemistry, Thermodynamics (C), NEET
Environmental heterophylly is found in:

Options:
A Cotton
B Larkspur
C Coriander
D Buttercup

The correct answer is:
Buttercup

Solution:
Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called plasticity, e.g. heterophylly in cotton, coriander and larkspur. In such plants, the leaves of the juvenile plant are different in shape from those in mature plants. On the other hand, difference in shapes of leaves produced in air and those produced in water in buttercup also represent the heterophyllous development due to environment.

Topic: Biology, Plant - Growth and Development, NEET
Environmental heterophylly is observed in:

Options:
A Cotton
B Coriander
C Larkspur
D Buttercup

The correct answer is:
Buttercup

Solution:
Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called plasticity, e.g., heterophylly in cotton, coriander and larkspur. In such plants, the leaves of the juvenile plant are different in shape from those in mature plants. On the other hand, difference in shapes of leaves produced in air and those produced in water in buttercup also represent the heterophyllous development due to environment

Topic: Biology, Plant - Growth and Development, NEET
Enzyme DNA dependent RNA polymerase is used in

Options:
A Ribosome synthesis
B mRNA synthesis
C DNA synthesis
D Carbohydrate synthesis.

The correct answer is:
mRNA synthesis

Solution:
Transcription is one of the step involved in the process of protein synthesis. In this process, the DNA is used as a template to construct a RNA molecule called messenger RNA a.k.a. mRNA. This mRNA is then translated to make the protein. Making of RNA from DNA is called as transcription. It needs enzyme RNA polymerase which is dependent on DNA.

Topic: Biology, Molecular Basis of Inheritance, NEET
Enzyme peptidyl transferase can be located in:

Options:
A Larger subunit of ribosome
B mRNA catalyses peptide bond formation
C Central part of transfer RNA
D Smaller subunit of ribosome

The correct answer is:
Larger subunit of ribosome

Solution:
Peptide bond formation is catalysed by enzyme peptidyl transferase which is a ribozyme. High molecular weight rRNA, 23S in prokaryotes and 28S in eukaryotes function as a ribozyme. These rRNAs are located in a larger subunit of the ribosome.

Topic: Biology, Molecular Basis of Inheritance, NEET
Enzyme recombinase has a significant role during the gamete formation. In which phase is it active?

Options:
A prophase-I
B prophase-II
C metaphase-I
D anaphase-II

The correct answer is:
prophase-I

Solution:
The crossing over process occurring in the meiotic division is a enzyme mediated process. The enzyme involved is recombinase participates in the process of crossing over during the pachytene phase of prophase -I.

Topic: Biology, Cell Cycle and Cell Division, NEET