datavorous/entrance-exam-dataset huggingface dataset:
Drugs like LSD, barbiturates, amphetamines, etc., are used as medicines to help patients with mental illness. However, excessive doses and abusive usage are harmful. Enumerate the major adverse effects of such drugs in humans.

Options:

None

Solution:
Drugs like LSD, barbiturates have several harmful effects. They include
(i) Reckless behaviour, vandalism and violence.
(ii) Withdrawal, isolation, depression, fatigue, aggressive behaviour.
(iii) Problems in social adjustment.
(iv) Lack of interest in personal hygiene, fluctuations in weight and appetite.
(v) Withdrawal symptoms can be severeand life threatening.
(vi) Anxiety, shakiness, nausea and sweating, loss of mind control.
(vii) Excessive doses of drugs may lead to coma and death may occur due to respiratory faliure, heart faliure or cerebral haemorrhage.

Topic: Biology, Human Health and Diseases
Due to overexploitation, one of the endangered species of Indian medicinal plants is

Options:
A Garlic
B Nepenthes
C Podophyllum
D Ocimum

The correct answer is:
Podophyllum

Solution:
Podophyllum hexandrum is a Himalayan medicinal plant belonging to the family Berberidaceae. It is one of the endangering species. The overexploitation of this plant takes place for the treatment of many cancers. The plant produces an important compound etoposide, a podophyllotoxin derivative, which is anti-cancerous in nature.

Topic: Biology, Biodiversity and its Conservation, NEET
Duramen is present in

Options:
A inner region of secondary wood
B part of sap wood
C outer region of secondary wood
D region of pericycle

The correct answer is:
inner region of secondary wood

Solution:
After certain years of growth, the xylem elements of the stems of a number of trees develop dark brown colouration, especially in the central or innermost layers. This region comprises dead element with highly lignified walls and is called heart wood or duramen. It is present in inner region of secondary wood.

Topic: Biology, Anatomy of Flowering Plants, JIPMER, JIPMER 2009
During a certain plane period a state out of a total budget of Rs 1400 crores had spent $28 \%$ of the total amount on Agriculture, $35 \%$ on Industry, $12 \%$ on Energy and $8 \%$ on Social Welfare, 105 crores on Education and the balance amount on Transport. What is the amount spent on Transport in crores of rupees?

Options:
A 123
B 145
C 165
D 133

The correct answer is:
133

Solution:
Total Budget $=1400$ crores Total expenditure $=28 \%+35 \%+12 \%+8 \%+105 \mathrm{cr}$.
$=83 \%+105$ crores
Thus, total expenditure $=\left(1400 \times \frac{83}{100}+105\right) \mathrm{cr}$.
$=(1162+105) \mathrm{cr}$
$=1267 \mathrm{cr}$
Now, Balance amount $=(1400-1267) \mathrm{cr} .=133 \mathrm{cr}$. Hence, Amount spent an Transport in crores ofrupees is 1

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 2)
During a cloudy day, a primary and a secondary rainbow may be created, then the:

Options:
A primary rainbow is due to double internal reflection and is formed above the secondary one.
B primary rainbow is due to double internal reflection and is formed below the secondary one.
C secondary rainbow is due to double internal reflection and is formed above the primary one.
D secondary rainbow is due to single internal reflection and is formed above the primary one.

The correct answer is:
secondary rainbow is due to double internal reflection and is formed above the primary one.

Solution:
Secondary rainbow is formed due to double internal reflection and is formed above the primary rainbow. The secondary rainbow is brighter than the primary rainbow.

Topic: Physics, Ray Optics, NEET, NEET 2022 (Phase 2)
During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at 40.0 cm using a standard resistance of 90 Î© ,as shown in the figure. The least count of the scale used in the metre bridge is 1mm. The unknown resistance is

Options:
A 60Â±0.15Â Î©
B 135Â±0.56Â Î©
C 60Â±0.25 Î©
D 135Â±0.23Â Î©

The correct answer is:
60Â±0.25 Î©

Solution:
R=x100-x90
∴R=60 Ω
dRR=100x100-xdx
∴dR=10040600.1×60
=0.25 Ω
Alternatively
dRR=0.140+0.160
∴dR=0.25 Ω

Topic: Physics, Current Electricity, NEET
During anaerobic digestion of organic waste, such as in producing biogas, which one of the following is left undegraded ?

Options:
A lipids
B lignin
C hemi-cellulose
D cellulose

The correct answer is:
lignin

Solution:
During bio-gas formation, digestion of cellulose is slow (ratc-limiting) and most of the lignin Is not decomposed. Lignin is a complex polymeric molecule, made up of phenyl propanoid units. This complexity has thus far proven as resistant to microbial degradation. After cellulose lignin is the most abundant plant polymer. It forms 20 to $30 \%$ of the wood of the tree. Cellulose is a large chained polymer of glucose molecules which are linked with each other by glycosidic bonds. Hemicellulose are branched polymers of glucose, xylose, galactose, mannose and arabinose.

Topic: Biology, Respiration in Plants, NEET, NEET 2003
During biological nitrogen fixation, nitrogen is converted to

Options:
A ammonia
B nitrate
C nitrite
D amino acid

The correct answer is:
ammonia

Solution:
Very few living organisms can utilise the nitrogen in the form N2, available abundantly in the air. Only certain prokaryotic species are capable of fixing nitrogen. Reduction of nitrogen to ammonia by living organisms is called biological nitrogen fixation. The enzyme, nitrogenase which is capable of nitrogen reduction, is present exclusively in prokaryotes. Such microbes are called N2- fixers.

Topic: Biology, Mineral Nutrition, NEET
During spermatogenesis, second meiotic division occurs during

Options:
A Spermatogonia A → Spermatogonia B
B Primary spermatocyte → Secondary spermatocyte
C Spermatogonia B → Primary spermatocyte
D Secondary spermatocyte → Spermatid

The correct answer is:
Secondary spermatocyte → Spermatid

Solution:
During spermatogenesis, some spermatogonia (termed as primary spermatocytes) periodically undergo meiosis. A primary spermatocyte completes meiosis-I (reduction division) and gives rise to 2 equal-sized cells called secondary spermatocytes (n = 23). Both secondary spermatocytes undergo meiosis-II (equational division) to produce 2 equal spermatids (n) each.

Topic: Biology, Human Reproduction, NEET
During spermatogenesis, secondary spermatocytes are result of

Options:
A Mitosis
B Meiosis-I
C Meiosis -II
D Growth

The correct answer is:
Meiosis-I

Solution:
Having completed the first meiotic division, these cells give rise to daughter cells called secondary spermatocytes, which divide to complete meiosis to form round spermatids.
The round spermatids do not divide but undergo a complex metamorphosis, called spermiogenesis, to become spermatozoa that are released into the lumen of the seminiferous tubule by a process called spermiation.

Topic: Biology, Human Reproduction, NEET
During the treatment of sewage water, the chemical process is carried out in which step of water treatment?

Options:
A Tertiary
B Secondary
C Primary
D Quaternary

The correct answer is:
Tertiary

Solution:
Treatment of sewage water is carried out in three stages: primary treatment is a physical process in which filtration and sedimentation are carried out to remove floating debris and girt. Secondary treatment is a biological treatment in which aerobic heterotrophic microbes are used, which consumes the major part of organic matter present in the effluent. âTertiary treatment: Physiochemical process in which chlorine gas , zirconium, ozone gas, perchlorate salts, UV rays or reverse osmosis etc. are used to remove DDT, pesticides, pathogens and remove turbidity present in wastewater.

Topic: Biology, Microbes in Human Welfare, NEET
During which stage of sewage treatment microbes are not used?

Options:
A Primary treatment
B Secondary treatment
C Tertiary treatment
D Both a and c

The correct answer is:
Both a and c

Solution:
Secondary treatment/biological treatment is when the primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and the air is pumped into it. This allows the vigorous growth of useful aerobic microbes into flocs.

Topic: Biology, Microbes in Human Welfare, NEET
During which stage of sewage treatment microbes are used?

Options:
A Primary treatment
B Secondary treatment
C Tertiary treatment
D All of these

The correct answer is:
Secondary treatment

Solution:
Primary treatment is a physical process that involves only filtration and sedimentation.
Secondary treatment or Biological treatment: The primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and the air is pumped into it. This allows the vigorous growth of useful aerobic microbes into flocs (masses of bacteria associated with fungal filaments to form mesh-like structures). While growing, these microbes consume the major part of the organic matter in the effluent. This significantly reduces the BOD (biochemical oxygen demand) of the effluent.

Topic: Biology, Microbes in Human Welfare, NEET
$\mathrm{E}_{\mathrm{Fe} 2+/ \mathrm{Fe}}^{\mathrm{o}}=-0.441 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Fe}^{3+} \mathrm{Fe}^{2+}}^{\mathrm{O}}=0.771$
$\mathrm{V}$, the standard EMF of the reaction $\mathrm{Fe}$ $+2 \mathrm{Fe}^{3+} \rightarrow 3 \mathrm{Fe}^{2+}$ will be:

Options:
A $0.111 \mathrm{~V}$
B $0.330 \mathrm{~V}$
C $1.653 \mathrm{~V}$
D $1.212 \mathrm{~V}$

The correct answer is:
$1.212 \mathrm{~V}$

Solution:
For the cell reaction
$$
\begin{aligned}
& \mathrm{Fe}+2 \mathrm{Fe}^{3+} \rightarrow 3 \mathrm{Fe}^{2+} \\
& \text { Anode reaction is } \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \\
& \text {Cathode reaction in } 2 \mathrm{Fe}^{3+}+2 \mathrm{e}^{-} \rightarrow \\
& 2 \mathrm{Fe}^{2+} \\
& \mathrm{E}_{\mathrm{Cell}}^{\circ}=\mathrm{E}_{\text {Cathode }}^{\circ}-\mathrm{E}_{\text {Anode }}^{\circ} \\
& =0.771-(-0.441) \\
& \mathrm{E}_{\mathrm{Cell}}^{\circ}=1.212 \mathrm{~V} \\
&
\end{aligned}
$$
Related Theory
If the cell potential is negative, the reaction is reversed. In this case, the electrode of the galvanic cell should be written in a reversed order.
Caution
The positive potential indicates a spontaneous reaction

Topic: Chemistry, Electrochemistry, NEET, NEET 2006
Each of $\mathrm{A}$ and $\mathrm{B}$ tosses two coins. What is the probability that they get equal number of heads?

Options:
A $\frac{3}{16}$
B $\frac{5}{16}$
C $\frac{4}{16}$
D $\frac{6}{16}$

The correct answer is:
$\frac{5}{16}$

Solution:
If both get one head then it is $\frac{1}{4} \times \frac{1}{4}$
and if both get two heads then it is $\frac{1}{2} \times \frac{1}{2}$
$\Rightarrow$ Prob (getting same number of heads) $=\frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{2}$
$=\frac{1}{16}+\frac{1}{4}=\frac{5}{16}$

Topic: Mathematics, Probability, NDA, NDA 2007 (Phase 2)
Each of the following statements concerning testosterone is true, except

Options:
A It is synthesized by Leydig cells.
B It is required for spermatogenesis.
C It is required for secondary development in males.
D It is excluded from the adluminal compartment of the seminiferous epithelium.

The correct answer is:
It is excluded from the adluminal compartment of the seminiferous epithelium.

Solution:
Testosterone is the primary male sex steroid. It is an androgenic steroid synthesized by interstitial Leydig cells in the testes. It is required for spermatogenesis, functional differentiation of the excurrent ducts of the testes and the development of male secondary sexual characteristics. Androgen-binding protein synthesized by Sertoli cells assures that a high concentration of testosterone is present in the adluminal compartment of the seminiferous epithelium.

Topic: Biology, Human Reproduction, NEET
Each of the following statements concerning the seminiferous epithelium is correct except.

Options:
A Before puberty, it contains spermatids
B It consists of Sertoli cells and spermatogenic cells
C It has a robust basement membrane
D Its basal compartment contains spermatogonia

The correct answer is:
Before puberty, it contains spermatids

Solution:
It consists of a population of Sertoli cells and a spermatogenic cell line. Sertoli cells span the thickness of the seminiferous epithelium. Sertoli cell characteristics including the distribution of their intercellular junctions, divide the seminiferous epithelium into a basal compartment containing only spermatogonia and an adluminal compartment containing primary and secondary spermatocytes, spermatids, and spermatozoa. The differentiation of spermatogonia into spermatids does not begin until puberty. Thus, the seminiferous epithelium does not contain spermatids prior to puberty like all epithelial layers, the basal surface of the seminiferous epithelium is bounded by the basement membrane.

Topic: Biology, Human Reproduction, NEET
Each of the two strings of length 51.6 cm and 49.1 cm are equally tensioned by 20 N force. Mass per unit length of both the strings is the same and equal to 1 g/m. When both the strings vibrate simultaneously in fundamental mode the number of beats is--

Options:
A 7
B 8
C 3
D 5

The correct answer is:
7

Solution:
l 1 = 0.516 m , l 2 = 0. 491 m, T = 20 N .
Mass per unit length, Î¼ = 0.001 kg/m
Frequency , vâ¡ = 1 2 lâ¡ Tâ¡ Î¼
vâ¡ 1 = 1 2 Ã 0 Â· 51 6 2 0 0 Â· 0 0 1
vâ¡ 2 = 1 2 Ã 0 Â· 4 9 1 2 0 0 Â· 0 0 1
â´ Number of beats = Ï 1 - Ï 2 = 7

Topic: Physics, Waves and Sound, NEET
Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to 1 g m-1. When both the strings vibrate simultaneously in fundamental mode, the number of beats is

Options:
A 7
B 8
C 3
D 5

The correct answer is:
7

Solution:
l1=0.516 m,l2=0. 491 m, T=20 N
Mass per unit length, Î¼=0.001 kg m-1
Frequency,Ï=12lâ¡Tâ¡Î¼
Ï1=12Ã0Â·516200Â·001
Ï2=12Ã0Â·491200Â·001
â´ Number of beats =Ï1-Ï2=7

Topic: Physics, Waves and Sound, NEET
Each of thefollowing terms has some anatomical significance. What do these terms mean? Explain with the help of line diagrams.
(a) Plasmodesmata
(b) Middle Lamella
(c) Secondary wall

Options:

None

Solution:
The given terms and their functions include:

Topic: Biology, Anatomy of Flowering Plants
Each plant group of plants has some phylogenetic significance in relation to evolution Cycas, one of the few living members of gymnosperms is called as the 'relic of past'. Can you establish a phylogenetic relationship of $C y$ cas with any other group of plants that justifies the above statement?

Options:

None

Solution:
Cycas is an evergreen plant which resembles palm. It has an unbranched stem and large compound leaves. It exhibits phylogenetic relationship with pteridophyte. Its evolutionary characters include the following:
(i) Growth is redundant.
(ii) Shedding of seed while the embryo is still immature.
(iii) Minimal secondary growth and manoxylic wood.
(iv) Megasporophylls are leaf like.
(v) Sperms are flagellate even when pollen tube is present.
(vi) Leaf bases are persistent.
(vii) Ptysix is circinate.
(viii) Arrangement of microsporangia in well defined archegonia.

Topic: Biology, Plant Kingdom
Ectophloic siphonostele is found in:

Options:
A Osmunda and Equisetum
B Marsilea and Botrychium
C Adiantum and Cucurbitaceae
D Dicksonia and Maiden hair fern.

The correct answer is:
Osmunda and Equisetum

Solution:
In Ectophloic siphonostele, the central pith is surrounded successively by xylem, phloem, pericycle, and endodermis. It is found in Osmunda and Equisetum

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2005
Ectophloic siphonostele is found in

Options:
A Adiantum and Cucurbitaceae
B Osmunda and Equisetum
C Marsilea and Botrychium
D Dicksonia and maiden hair fern

The correct answer is:
Osmunda and Equisetum

Solution:
In the ectophloic siphonostele the xylem surrounds pith and this xylem is surrounded by phloem, pericycle and endoderm is respectively. e.g., Osmunda and Equisetum.

Topic: Biology, Plant Kingdom, JIPMER, JIPMER 2007
Ectophloic siphonostele is found in

Options:
A Osmunda and Equisetum
B Marsilea and Botrychium
C Adiantum and Cucurbitaceae
D Dicksonia and Maidenhair fern.

The correct answer is:
Osmunda and Equisetum

Solution:
Stele is a column containing vascular tissues which is surrounded by pericycle and separated from ground tissue by endodermis.
Siphonostele is medullated protostele or protostele with a central non-vascular pith. Leaf gaps are absent. Siphonostele is of two types :
In Ectophloic siphonostele, central pith is surrounded successively by xylem, phloem, pericycle and endodermis. In amphiphloic siphonostele there is a central pith and xylem is surrounded on either side by phloem, pericycle and endodermis. It is found in Osmunda and Equisetum.

Topic: Biology, Anatomy of Flowering Plants, AIIMS, AIIMS 2008
Edible part of mushroom is

Options:
A basidiocarp
B secondary mycelium
C primary mycelium
D tertiary mycelium

The correct answer is:
basidiocarp

Solution:
The cap and the stalk together constitute the fruiting body called the basidiocarp orsporocarp. Because of the presence of characteristic gills the members of Agaricaceae family are generally called the gill or gilled fungi. They are generally fleshy and are also known as fleshy fungi. Some of these gilled or
fleshy fungi are edible, ordinarily the edible fleshy (or filled) fungi are called the mushrooms.

Topic: Biology, Biological Classification, JIPMER, JIPMER 2009
Efficiency of the following cells is 84%.
A(s)+B2+(aq)⇌A2+(aq)+B(s) ; ΔH=-285 kJ
Then the standard electrode potential of the cell will be

Options:
A 1.20 V
B 2.40 V
C 1.10 V
D 1.24 V

The correct answer is:
1.24 V

Solution:
Efficiency =ΔG°ΔH°=-nFE°ΔH°
0.84=-2×E°×96500-285×1000
E°=+1.24V .

Topic: Chemistry, Electrochemistry, NEET
EFGH is a rhombus such that the angle EFG is $60^{\circ}$. The
magnitude of vectors $\overline{\mathrm{FH}}$ and $\{\mathrm{m} \overline{\mathrm{EG}}\}$ are equal where $\mathrm{m}$ is
a scalar. What is the value of $\mathrm{m}$ ?

Options:
A 3
B $1.5$
C $\sqrt{2}$
D $\sqrt{3}$

The correct answer is:
$\sqrt{3}$

Solution:
Rhombus EFGH, $\angle \mathrm{EFG}=60^{\circ}$
$\angle \mathrm{EFH}=30^{\circ}=\angle \mathrm{HFG}$
From parallelogram of forces
$\overrightarrow{\mathrm{FE}}+\overrightarrow{\mathrm{FG}}=\overrightarrow{\mathrm{FH}}$
Given $|\overrightarrow{\mathrm{FE}}|+|\overrightarrow{\mathrm{FG}}|=\mathrm{a}($ say $)$
$\therefore \overrightarrow{\mathrm{FH}}=2 \cdot \frac{\sqrt{3}}{2} \overrightarrow{\mathrm{a}}=\sqrt{3} \overrightarrow{\mathrm{a}}$
$\overrightarrow{\mathrm{EG}}=\overrightarrow{\mathrm{EF}}+\overrightarrow{\mathrm{EH}}$
$=a \sin 30^{\circ}+a \sin 30^{\circ}=a \cdot \frac{1}{2}+a \cdot \frac{1}{2}=a$
Thus, $\frac{\overrightarrow{\mathrm{FH}}}{\overline{\mathrm{EG}}}=\frac{\sqrt{3} \mathrm{a}}{\mathrm{a}}=\sqrt{3}$
So, $\overrightarrow{\mathrm{FH}}=\sqrt{3} \overrightarrow{\mathrm{EG}}$
$\Rightarrow \quad \mathrm{m}=\sqrt{3}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2012 (Phase 2)
Electrolyte used in Ni-Cd cell

Options:
A KOH
B H2SO4
C LiOH
D Al2O3

The correct answer is:
KOH

Solution:
Electrolyte used in Ni-Cd cell is KOH (potassium hydroxide). It is a type of secondary cell. It consists of a cadmium anode and metal grid containing NiO2 as cathode and KOH solution is electrolyte. It has longer life than the lead storage cell but more expensive to manufacture.

Topic: Chemistry, Electrochemistry, NEET
Elements of a population are classified according to the presence or absence of each of 3 attributes $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$. What is the number of smallest ultimate classes into which the population is divided?

Options:
A 5
B 6
C 8
D 9

The correct answer is:
8

Solution:
Elements of a population are classified according to the presence or absence of each of 3 attributes $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$. Then, the smallest number of smallest ultimate classes into which the population is divided, is $2^{3}=8$

Topic: Mathematics, Sets and Relations, NDA, NDA 2007 (Phase 2)
Eliminating the arbitrary constants $\mathrm{B}$ and $\mathrm{C}$ in the expresion
$y=\frac{2}{3 C}(C x-1)^{3 / 2}+B$, we get

Options:
A $x\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=\frac{d^{2} y}{d x^{2}}$
B $2 x\left(\frac{d y}{d x}\right) \frac{d^{2} y}{d x^{2}}=1+\left(\frac{d y}{d x}\right)^{2}$
C $\left(\frac{d y}{d x}\right) \frac{d^{2} y}{d x^{2}}=1$
D $\left(\frac{d y}{d x}\right)^{2}+1=\frac{d^{2} y}{d x^{2}}$

The correct answer is:
$2 x\left(\frac{d y}{d x}\right) \frac{d^{2} y}{d x^{2}}=1+\left(\frac{d y}{d x}\right)^{2}$

Solution:
$\mathrm{y}=\frac{2}{3 \mathrm{C}}(\mathrm{Cx}-1)^{\frac{3}{2}}+\mathrm{B}$
On differentiating both sides w.r.t. $\mathrm{x}$, we get
$\frac{d y}{d x}=\frac{2}{3 C} \cdot \frac{3}{2}(C x-1)^{\frac{1}{2}} \cdot C+0=(C x-1)^{\frac{1}{2}}$
$\frac{d y}{d x}=(C x-1)^{\frac{1}{2}}$
On squaring both sides, we get
$\left(\frac{d y}{d x}\right)^{2}=C x-1$
$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}+1=\mathrm{Cx}$
Now, on differentiating w.r.t. $\mathrm{x}$, we get
$2\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \cdot \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{C}$
From eq. (i)
$\left(\frac{d y}{d x}\right)^{2}+1=2 x\left(\frac{d y}{d x}\right) \frac{d^{2} y}{d x^{2}}$

Topic: Mathematics, Differential Equations, NDA, NDA 2014 (Phase 2)
Ellingham diagram is the plot of $\mathrm{X} v s \mathrm{Y}$. what are $\mathrm{X}$ and $\mathrm{Y}$ ?

Options:
A $\Delta \mathrm{G}^{\circ}, \mathrm{T}$
B $\Delta \mathrm{G}^{\circ}, 1 / \mathrm{T}$
C $\Delta \mathrm{S}^{\circ}, \mathrm{T}$
D $\Delta \mathrm{H}^{\circ}, 1 / \mathrm{T}$

The correct answer is:
$\Delta \mathrm{G}^{\circ}, \mathrm{T}$

Solution:
Ellingham diagram is a graphical representation of the variation of change in standard gibbs free energy with temperature.
Thus, $\mathrm{X}=\Delta \mathrm{G}^{\circ}, \mathrm{Y}=\mathrm{T}$

Topic: Chemistry, Thermodynamics (C), TS EAMCET, TS EAMCET 2023 (13 May Shift 1)
Endemic species are

Options:
A Rare species
B Species localised in a specific region
C Cosmopolitan in distribution
D Critically endangered species

The correct answer is:
Species localised in a specific region

Solution:
The two methods used for biodiversity conservation are in-situ conservation and ex-situ conservation. The development of a particular region and conservation of biodiversity can be at times challenging and non feasible. The numbers of species facing danger of extinction are also increasing. In order to overcome this, the species are conserved in those areas where they are found in large numbers. Species which are found localized in a specific area are known as endemic species. These endemic species are confined to that particular region and not found anywhere else. This high degree of endemism makes it easier to conserve these species in their own habitat by creating biodiversity hotspots.

Topic: Biology, Biodiversity and its Conservation, NEET
Endothecium and tapetum in anther are derived from

Options:
A	Primary sporogenous layer
B Primary parietal layer
C Both primary sporogenous layer and primary parietal layer
D None of the above

The correct answer is:
Primary parietal layer

Solution:
Primary parietal cells divide to form two layers of secondary parietal cells. The outer layer of secondary parietal cells differentiates to form endothecium and the inner layer of secondary parietal cells divide to form two layers, one of which forms the middle layers and the other forms the tapetum.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Energy E, angular momentum L and universal gravitational constant G are chosen as fundamental quantities. The dimensions of universal gravitational constant in the dimensional formula of Planck's constant h is

Options:
A zero
B -1
C 53
D 1

The correct answer is:
zero

Solution:
h = GxLyEz
[M1 L2T-1] = [M-1 L3T-2] x [M1 L2T-1]y [M1 L2T-2]z
Comparing the power, we get

1 = - x + y + z ... (i)
2 = 3x + 2y + 2z ... (ii)
-1 = - 2x - y - 2z ... (iii)

On solving Eq. (i), (ii) and (iii), we get x = 0

Topic: Physics, Units and Dimensions, NEET