datavorous/entrance-exam-dataset huggingface dataset:
A coin is placed on a horizontal platform, which undergoes horizontal SHM about a mean position O. If the coin placed on platform does not slip, coefficient of friction between the coin and the platform is Î¼. The amplitude of oscillation is gradually increased. The coin will begin to slip on the platform for the first time

Options:
A at the mean position
B at the extreme position of oscillations
C for an amplitude of μg/ω2
D for an amplitude of g/μ ω2

The correct answer is:
for an amplitude of μg/ω2

Solution:
Let O be the mean position and x be the distance of coin fromO. The coin will slip if centrifugal force on coin just becomes equal to force of friction ie, mxω2=μmg
The coin will slip if x= maximum = amplitude A

Topic: Physics, Oscillations, NEET
A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency Ï. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time

Options:
A at the highest position of the platform
B at the mean position of the platform
C for an amplitude of g ω 2
D for an amplitude of g 2 ω 2

The correct answer is:
for an amplitude of g ω 2

Solution:
In vertical simple harmonic motion, maximum acceleration a ω 2 and so the maximum force m a ω 2 will be at extreme positions. At highest position, force will be towards mean position and so it will be downwards. At lowest position, force will be towards mean position and so it will be upwards. This is opposite to weight direction of the coin. The coin will leave contact with the platform for the first time when m a ω 2 ≥ m g at the lowest position of the platform.

Topic: Physics, Oscillations, NEET
A coin placed on a rotating table just slips if it is placed at a distance 4r from the centre. If we double the angular velocity of the table, then the coin will just slip when it is away from the centre at a distance equal to

Options:
A 4r
B 2r
C r
D r/4

The correct answer is:
r

Solution:
∴ω=µgr ∴ω∝1r ⇒r∝1ω2
Since ω is double thus,
r′=4r4=r

Topic: Physics, Laws of Motion, NEET
A collection of plants and seeds having diverse alleles of all the genes of a crop is called

Options:
A germplasm
B gene library
C genome
D herbarium

The correct answer is:
germplasm

Solution:
A germplasm is a collection of genetic resources for an organisms. For plants, the germplasm may be stored as a seed collection. It includes diverse alleles of all the genes of an organism.

Topic: Biology, The Living World, NEET, NEET 2011 (Screening)
A colonial algae among the following is:

Options:
A Chlamydomonas
B Ulothrix
C Volvox
D Spirogyra

The correct answer is:
Volvox

Solution:
Volvox is a polyphyletic genus of chlorophyte green algae in the family Volvocaceae. It forms spherical colonies of up to 50,000 cells. They live in a variety of freshwater habitats.

Topic: Biology, Plant Kingdom, NEET
A colorblind man XcY marries a woman who is carrier for hemophilia XXh. Which of the following is TRUE for their progenies?

Options:
A 25% female progenies carry the genes for both haemophilia and colourblindness.
B 25% male progenies carry only the gene for haemophilia.
C 25% female progenies carry only the gene for colourblindness.
D All of these

The correct answer is:
All of these

Solution:
X-lined disorders are inherent disorders due to mutation in the X-chromosome which causes the phenotype to be always expressed in male and in female only in homozygous condition.
Colour blindness and haemophilia are X-linked recessive disorder. In colour blindness, the affected individual has is not able to distinguish colour especially read and green. In haemophilia, the individual has delayed clotting time for blood which can be fatal.
Colour blind father XcY and hemophilic XXh . carrier mother will produce the following combination of progenies-
Probabilities of son - XY/XhY
Probabilities of daughter - XcX/XcXh
25% female progenies carry the genes for both haemophilia and colourblindness.
25% male progenies carry only the gene for haemophilia.
25% female progenies carry only the gene for colourblindness.
25% male progenies are normal without any X-linked disorder.

Topic: Biology, Principles of Inheritance and Variation, NEET
A colorless substance ‘A’ (C6H7N) is sparingly soluble in water and gives a water soluble compound ‘B’ on treating with mineral acid. On reacting with CHCl3and alcoholic potash ‘A’ produces an obnoxious smell due to the formation of compound ‘C’. Reaction of ‘A’ with benzenesulfonyl chloride gives compound ‘D’ which is soluble in alkali. With NaNO2and HCl, ‘A’ forms compound ‘E’ which reacts with phenol in alkaline medium to give an orange dye ‘F’. Identify compounds ‘A’ to ‘C’.

Options:
A A = Phenol, B = Anilium chloride , C = Benzene isonitrile
B A = Aniline, B = Anilium chloride , C = Benzene
C A = Aniline, B = Benzene , C = Benzene isonitrile
D A = Aniline, B = Anilium chloride , C = Benzene isonitrile

The correct answer is:
A = Aniline, B = Anilium chloride , C = Benzene isonitrile

Solution:
Correct Option is : (D)
A = Aniline, B = Anilium chloride , C = Benzene isonitrile

Topic: Chemistry, Amines, NEET
A colorless substance ‘A’ (C6H7N) is sparingly soluble in water and gives a water soluble compound ‘B’ on treating with mineral acid. On reacting with CHCl3and alcoholic potash ‘A’ produces an obnoxious smell due to the formation of compound ‘C’. Reaction of ‘A’ with benzenesulfonyl chloride gives compound ‘D’ which is soluble in alkali. With NaNO2and HCl, ‘A’ forms compound ‘E’ which reacts with phenol in alkaline medium to give an orange dye ‘F’. Identify compounds ‘D’ to ‘F’.

Options:
A D = N-Phenyl Benzene Sulphonamide, E = Benzenediazonium chloride and F = Azobenzene
B D = Benzenesulfonamide, E = Benzenediazonium chloride and F = p-Hydroxy azobenzene
C D = N-Phenyl Benzene Sulphonamide, E = Benzenediazonium chloride and F = p-Hydroxy azobenzene
D D = N-Phenyl Benzene Sulphonamide, E = Benzene chloride and F = p-Hydroxy azobenzene

The correct answer is:
D = N-Phenyl Benzene Sulphonamide, E = Benzenediazonium chloride and F = p-Hydroxy azobenzene

Solution:
Correct Option is : (C)
D = N-Phenyl Benzene Sulphonamide, E = Benzenediazonium chloride and F = p-Hydroxy azobenzene

Topic: Chemistry, Amines, NEET
A colour-blind girl is rare because she will be born only when

Options:
A Her mother and maternal grand father were colour blind
B Her father and maternal grand father were colour blind
C Her mother is colour blind and father has normal vision
D Parents have normal vision but grand parents were colour blind

The correct answer is:
Her father and maternal grand father were colour blind

Solution:
The genotype of the mother was to be either XC XC or XCX and that of father XCY so that the daughter becomes colour blind.

Topic: Biology, Principles of Inheritance and Variation, NEET
A colour-blind man has a colour-blind sister but a normal brother. Which of these is correct about their parents?

Options:
A Father is colour blind and mother is a carrier
B Father is normal and mother is colour blind
C Father and mother both are colour blind
D Father and mother both are normal

The correct answer is:
Father is colour blind and mother is a carrier

Solution:
Colour blind man has one X chromosome with the gene for colour blindness which he receives from his mother. His colourblind sister has two X chromosomes each with the gene for colourblindness. She has received one of these from her father and one from her mother. Thus from this information, we can conclude that their father is colourblind and their mother has at least one X chromosome for colourblindness. Also, they have a normal brother. That means their mother has one X chromosome without colour-blindness genes. Hence mother is a carrier.

Topic: Biology, Principles of Inheritance and Variation, NEET
A colourblind man marries a normal vision woman whose father was colourblind. They have a first child as a son. What are the chances that a child would be colourblind?

Options:
A 25%
B 50%
C 100%
D 0%

The correct answer is:
50%

Solution:
Clour blindness is X linked genetic recessive disorder in which a carrier female passes its X- linked recessive gene to 50% of its male progeny and 50% male progeny are normal.

Topic: Biology, Principles of Inheritance and Variation, NEET
A colour-blind man marries a woman who is carrier for haemophilia. Which of the following is true for their progenies?

Options:
A 25% of the progenies carry the genes for both haemophilia and colour-blindness
B 25% of the progenies carry only the gene for haemophilia
C 25% of the progenies carry only the gene for colour-blindness
D All of these

The correct answer is:
All of these

Solution:
Genotype of Male is XcY and Genotype of Female is XhX .
The genotypes of their children will be

Xc	Y

Xh	XcXh	XhY

X	XcX	XY

Their children XcX (daughter carrier for colourblindness),
XcXh (daughter carrier for colourblindness and haemophilia), XY (normal son), XhY (haemophiliac son)
Thus, 25% of their children will carry the genes for both haemophilia and colour-blindness XcXh, 25% of their children will carry the gene only for haemophilia XhY and 25% of the progenies carry only the gene for colour-blindness XcX

Topic: Biology, Principles of Inheritance and Variation, NEET
A colourblind man marries a woman with normal sight who has no history of colour blindness in her family. What is the probability of their sons being colourblind?

Options:
A 1
B Nil
C 0.25
D 0.5

The correct answer is:
Nil

Solution:
Colour Blidness : It is a sex-linked recessive disorder due to defect in either red or green cone of eye resulting in failure to discriminate between red and green colour. This defect is due to mutation in certain genes present in the X chromosome. It occurs in about 8 per cent of males and only about 0.4 per cent of females. This is because the genes that lead to red-green colour blindness are on the X chromosome.

When a colourblind man marries a woman with normal sight who has no history of colour blindness in her family. In their progeny, all the daughters would be a colourblind carrier, and all sons would be normal.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2015 (Phase 2)
A colourblind man marries a woman with normal sight who has no history of colour blindness in her family. What is the probability of their sons being colourblind?

Options:
A 1
B Nil
C 0.25
D 0.5

The correct answer is:
Nil

Solution:
When a colourblind man marries a woman with normal sight who has no history of colour blindness in her family. In their progeny, all the daughters would be a colourblind carrier, and all sons would be normal.

Topic: Biology, Principles of Inheritance and Variation, NEET
A colourblind man marries with a daughter of colourblind father, the generation will be:

Options:
A There will be no daughter colourblind
B All sons will be colourblind
C All daughter will be colourblind
D Half sons will be colourblind

The correct answer is:
Half sons will be colourblind

Solution:
A colourblind man marries with a daughter of the colourblind father. The generation will be half sons colourblind. Daughter of a colourblind person is a carrier of this disease, i.e., one X-chromosome have recessive genes. Therefore, half of the sons will be colourblind.

Topic: Biology, Principles of Inheritance and Variation, NEET
A colourless fuming liquid A can be prepared by passing SO2 over phosphorous pentachloride. The liquid can readily be hydrolysed to give sulphurous acid. The compound A is

Options:
A SOCl2
B SO2Cl2
C SCl2
D SCl4

The correct answer is:
SOCl2

Solution:
PCl5+SO2→SOCl2+POCl3
Both SCl4 and SOCl2 , on hydrolysis may give H2SO3 . But SCl4 is a yellow solid liquiefies at room temperature.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
A colourless liquid A(b.p.184°C) is sparingly soluble in warm water to which it gives feebly alkaline. On treating with NaNO2 and dil HCl in the cold solution, it yields a solution which reacts with an alkaline solution of β- naphthol to give an orange yellow precipitate. Compound A is–

Options:
A C6H5N2Cl
B C6H5NHNH2
C n-C4H9NH2
D C6H5NH2

The correct answer is:
C6H5NH2

Solution:
Aniline undergoes diazotization reaction in the presence of NaNO2 and HCl , forms benzene diazonium salt. Then followed by diazo coupling reaction with beta nephathol.

Topic: Chemistry, Amines, NEET
A colourless liquid ‘A’ contains H and O elements only. It decomposes slowly on exposure to light. It is stabilised by mixing urea to store in the presence of light. (i) Suggest possible structure of A. (ii) Write chemical equation for its decomposition reaction in light.

Options:
A (i) H2O2; (ii) 2H2O2→ 2H2O + O2
B (i) H2O; (ii) 2H2O → 2H2+ O2
C (i) H2O2; (ii) 2H2O2→ 2H2+ 2O2
D (i) H2O; (ii) 2H2O → 2H2+ 2O2

The correct answer is:
(i) H2O2; (ii) 2H2O2→ 2H2O + O2

Solution:
Correct Option is : (A)
(i) H2O2; (ii) 2H2O2→ 2H2O + O2

Topic: Chemistry, Hydrogen, NEET
A colourless organic compound gives brisk effervescences with a mixture of sodium nitrite and dil. HCl. It could be

Options:
A Oxalic acid
B Acetic acid
C Urea
D Glucose

The correct answer is:
Urea

Solution:
NaNO2+HClâHNO2+NaCl
H 2 NCONH 2 Urea + HNO 2 → CO 2 + NH 3 + H 2 O+ N 2
CO2 gas evolves with brisk effervescence.

Topic: Chemistry, Carboxylic Acid Derivatives, NEET
A column of water within xylem vessels of tall trees does not break under its weight because of:

Options:
A Tensile strength of water
B Lignification of xylem vessels
C Positive root pressure
D Dissolved sugars in water

The correct answer is:
Tensile strength of water

Solution:
The column of water within the Xylem vessel of tall trees does not break under its weight due to the high tensile strength of water. Tensile strength is the ability to resist pulling forces.

Topic: Biology, Transport in Plants, NEET, NEET 2015 (Phase 2)
A combination of logic gates has the truth table below.

Which of the following combinations has this truth table

Options:
A
B
C
D None of these

The correct answer is:

Solution:

Topic: Physics, Semiconductors, NEET
A comet of mass 108 kg travels around the Sun in an elliptical orbit. When it is closest to the Sun it is 2.5×1011 m away and its speed is 2×104 m s-1 . Find the change in kinetic energy when it is farthest from the Sun and is 5×1010 m away from the Sun

Options:
A 38×108 J
B 48×108 J
C 58×108 J
D 56×108 J

The correct answer is:
48×108 J

Solution:
From the conservation of angular momentum,
v1r1=v2r2
⇒ v2=2×104×2.5×10115×1010=105 m s-1
Change in kinetic energy is,
ΔKE=12×1081052-4×108=48×108 J

Topic: Physics, Gravitation, NEET
A comet revolves around the sun in an elliptical orbit. When it is closest to the sun at distance d, then its corresponding kinetic energy is k0. If it is farthest from the sun at distance 3d then the corresponding kinetic energy will be

Options:
A k09
B 8k09
C k04
D 4k09

The correct answer is:
k09

Solution:
v1r1=v2r2= constant
∴ v0d=3dv1
⇒ v1=v03
∴ KE=12mv029
⇒ KE=k09

Topic: Physics, Gravitation, NEET
A common biocontrol agent for the control of plant diseases is

Options:
A Baculovirus
B Bacillus thuringiensis
C Glomus
D Trichoderma

The correct answer is:
Trichoderma

Solution:
Trichoderma is a genus of fungi that is present in all soils, where they are the most prevalent culturable fungi. Many species in this genus can be characterized as opportunistic avirulent plant symbionts. Several strains of Trichoderma have been developed as biocontrol agents against fungal diseases of plants. The various mechanisms include antibiosis, parasitism, inducing host-plant resistance and competition. Most biocontrol agents are from the species $T$. harzianum, $T$. viride and $T$. hamatum. The biocontrol agent generally grows in its natural habitat on the root surface and so affects root disease in particular but can also be effective against foliar diseases.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2010 (Screening)
A common biocontrol agent for the control of plant diseases is:

Options:
A Baculovirus
B Bacillus thuringiensis
C Glomus
D Trichoderma

The correct answer is:
Trichoderma

Solution:
A common biocontrol agent for control of plant diseases is Trichoderma. It is a tree-living fungus that exert biocontrol over serval plant pathogens for the control of plant diseases. It is the natural method of plant and pathogen control.

Topic: Biology, Microbes in Human Welfare, NEET
A common characteristic feature of plant sieve tube cells and most of mammalian erythrocytes is

Options:
A Absence of mitochondria
B Presence of cell wall
C Presence of haemoglobin
D Absence of nucleus

The correct answer is:
Absence of nucleus

Solution:
A common characteristic feature of plant sieve tube cells and most of mammalian erythrocytes is absence of nucleus.

Topic: Biology, Cell - The Unit of Life, NEET
A common collagenous connective tissue layer which hold the number of muscle bundles of skeletal muscle is called:

Options:
A Fascia
B Fascicles
C Endomysium
D Perimysium

The correct answer is:
Fascia

Solution:
Each organised skeletal muscle in our body is made of a number of muscle bundles or fascicles held together by a common collagenous connective tissue layer called fascia. Endomysium is connective tissue covering around each muscle fibre. Each fasciculus is surrounded by connective tissue cover called perimysium.

Topic: Biology, Locomotion and Movement, NEET
A common emitter amplifier circuit, built using an NPN transistor, is shown in the figure. Its dc current gain is 250, RC=1 kÎ© and VCC=10 V . The minimum base current for VCE to reach saturation is

Options:
A 10Â Î¼A
B 40 Î¼A
C 7Â Î¼A
D 100Â Î¼A

The correct answer is:
40 Î¼A

Solution:
Given, Î²=250
RC=1000 Î©
VCC=10 V

At saturation, VCE=0
âie=10âV1000âÎ©=10âmA
Current gain factor, Î²=ieiB
âiB=ieÎ²=10âmA250
=40 Î¼A

Topic: Physics, Semiconductors, NEET
A common emitter amplifier has a voltage gain of 50, an input impedance of 100 Î© and an output impedance of 200 Î©. The power gain of the amplifier is

Options:
A 500
B 1000
C 1250
D 100

The correct answer is:
1250

Solution:
AC power gain = Change in output powerChange in input power
= âVcÃâicâViÃâib=âVcâViÃâicâib=AVÃÎ²AC
where AV is voltage gain and Î²AC is AC current gain.
Also,
AV=Î²ACÃresistance gain =RoRi
Given, AV=50, Ro=200 Î©, Ri=100 Î©
Hence, 50 = Î²ACÃ200100
â´ Î²AC=25
Now, AC power gain = AVÃÎ²AC = 50 Ã 25 = 1250

Topic: Physics, Semiconductors, NEET
A common emitter amplifier has a voltage gain of 50, an input impedance of 100 Î© and an output impedance of 200 Î©. The power gain of the amplifier is

Options:
A 500
B 1000
C 1250
D 100

The correct answer is:
1250

Solution:
AC power gain = output powerChange in input power
= âVcÃâicâViÃâib=âVcâViÃâicâib=AVÃÎ²AC
where AV is voltage gain and Î²AC is AC current gain.
Also,
AV=Î²ACÃresistance gain =RoRi
Given, AV=50, Ro=200 Î©, Ri=100 Î©
Hence, 50 = Î²ACÃ200100
â´ Î²AC=25
Now, AC power gain = AVÃÎ²AC = 50 Ã 25 = 1250

Topic: Physics, Semiconductors, NEET
A common emitter amplifier has a voltage gain of 50 , an input impedance of $100 \Omega$ and an output impedance of $200 \Omega$. The power gain the amplifier is

Options:
A 500
B 1000
C 1250
D 50

The correct answer is:
1250

Solution:
Voltage gain $=\beta \times$ impedance gain
$$
\Rightarrow \quad 50=\beta \times \frac{200}{100} \Rightarrow \beta=25
$$
Also, power gain $=\beta^2 \times$ impedance gain
$$
=(25)^2 \times \frac{200}{100}=1250
$$

Topic: Physics, Semiconductors, NEET, NEET 2010 (Screening)
A common emitter amplifier has a voltage gain of 50, an input impedance of $100 \Omega$ and an output impedence of 200 $\Omega$. the power gain of the amplifier is:

Options:
A 1000
B 1250
C 100
D 500

The correct answer is:
1250

Solution:
$\begin{aligned}
\text { Voltage gain } & =(\beta)\left(\frac{\mathrm{R}_0}{\mathrm{R}_i}\right)=50 \\
\Rightarrow \quad \beta & =\frac{50 \times 100}{200}=25 \\
\text { Power gain } & =(125)^2 \times \frac{200}{100} \\
& =(125)^2 \times \frac{200}{100} \\
& =625 \times \frac{200}{100} \\
& =1250
\end{aligned}$

Topic: Physics, Semiconductors, NEET, NEET 2007
A common emitter transistor amplifier has a current gain of 50. If the load resistance is 4 kÎ© and input resistance is 500 Î© .then the voltage gain of the amplifier is

Options:
A 160
B 200
C 400
D none

The correct answer is:
400

Solution:
âÎ²=50
âRL=4 kÎ©,Ri=500 Î©
âVce=IcRL,Vbe=IbRi
âÎ²=IcIb=50
âÎ²=IcIb=50
âAV=VceVbe=IeIbÃRLRi=50Ã4Ã1000500=400

Topic: Physics, Semiconductors, NEET
A common test to find the genotype of a hybrid is by:

Options:
A crossing of one $F_2$ progeny with female parent
B studying the sexual behaviour of $\mathrm{F}_1$ progenies
C crossing of one $F_1$ progeny with male parent
D crossing of one $\mathrm{F}_2$ progeny with male parent.

The correct answer is:
crossing of one $F_1$ progeny with male parent

Solution:
A common test to find the genotype of a hybrid is a test cross. Here, $\mathrm{F}_1$ hybrids are crossed with the recessive parent. Phenotypic ratios of $1: 1$ confirm that the $\mathrm{F}_1$ hybrid is heterozygous.
$$
\text { Related Theory }
$$
Caution
This is a misleading question. Students consider the question as wrong, as they miss the word recessive parent. However, they do not realise that test cross is cross between $F_J$, hybrid progeny with recessive parent, and $F_s$ is only present in option $(\mathrm{C})$.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2007
A common test to find the genotype of a hybrid is by:

Options:
A The crossing of one F2 progeny with female parent
B The crossing of one F1 progeny with dominant male parent
C The crossing of one F1 progeny with recessive male parent
D Crossing of one F2 progeny with male parent.

The correct answer is:
The crossing of one F1 progeny with recessive male parent

Solution:
To find the genotype of a hybrid, test cross is done. Cross of F1 hybrid with a homozygous recessive parent is called test cross. just a recessive character is what you require for a test cross rather than being male or female.

Topic: Biology, Principles of Inheritance and Variation, NEET
A common-emitter transistor amplifier has a current gain of 50. If the load resistance is 4 kÎ© and input resistance is 500 Î© the voltage gain of the amplifier is

Options:
A 160
B 200
C 400
D none

The correct answer is:
400

Solution:
Given
β=50
RL=4 kΩ
Ri=500 Ω
VCE=ICRL
VBE=IBRI
We know that in Common Emitter Circuit-
Current Gain
β=IcIb=50
And voltage gain
AV=VceVbe
AV=IcIb×RLRi
AV=50×4×1000500=400

Topic: Physics, Semiconductors, NEET
A community that starts the process of succession in a habitat is called

Options:
A pioneer community
B seral community
C climax community
D ecotonal community

The correct answer is:
pioneer community

Solution:
Species that grow quickly and are low to ground with maximum sunlight intake are referred to as pioneer species. These groups make up the pioneer community and very important to initiate an ecosystem.
Mosses, lichen, and low-growing grasses are the first to colonize an area.

Topic: Biology, Ecosystem, NEET