datavorous/entrance-exam-dataset huggingface dataset:
A high density of a protected animal in a national park can result in

Options:
A Mutualism
B Intraspecific competition
C Emigration
D Predation

The correct answer is:
Intraspecific competition

Solution:
Competition between organisms of the same species is called intraspecific competition. If the number of organisms gets increased then the competition for food, shelter, and other resources will also get increased.
Mutualism is an obligatory and mutually beneficial relationship occurring between two interacting organisms. Predation is the act of capturing, killing, and eating. It is a negative interaction between two organisms. Migration is individuals from a population for settling into a new area.

Topic: Biology, Organisms and Populations, NEET
A high density of tiger population in an area can result in

Options:
A Migration
B Interspecific competition
C Intraspecific competition
D Protocooperation

The correct answer is:
Intraspecific competition

Solution:
A high density of tiger population in an area can result in intraspecific competition.

Topic: Biology, Organisms and Populations, NEET
A high value of BOD (Biochemical Oxygen Demand) indicates that the

Options:
A water is highly polluted
B water is less polluted
C consumption of organic matter in the water is higher by the microbes
D water is pure

The correct answer is:
water is highly polluted

Solution:
Biological oxygen demand is the measure of how much oxygen is required by aerobic microbes to break down organic material in aquatic system. If the BOD value is high, then the water is considered to be highly polluted otherwise water body is healthy and conducive for the organisms in it.

Topic: Biology, Microbes in Human Welfare, NEET
A historic convention on biological diversity held in Rio de Janerio in 1992 is known as

Options:
A The earth summit
B Montreal protocol
C Geneva convention
D Janerio convention

The correct answer is:
The earth summit

Solution:
The United Nations conference in environment and development is also known as the Rio Summit and Earth Summit. This was a major United Nations conference held in Rio de Janerio from June 3 to June 14, 1992. 172 governments participated, with 108 sending their heads of state or government.

Topic: Biology, Biodiversity and its Conservation, NEET
A hole is made at the bottom of the tank filled with water (density = 1000 kg/m3). If the excess pressure at the bottom of the tank is three atmospheres (1 atmosphere =105 N/m2), then the velocity of efflux is

Options:
A 400 m/s
B 200 m/s
C 600 m/s
D 500 m/s

The correct answer is:
600 m/s

Solution:
The excess pressure at the bottom of the tank, p=Ïgh=3 atm =3Ã105 N/m2
So, the height of the liquid in container
h=3Ã10510Ã103=30010
Now, velocity of efflux = 2gh= 2Ã10Ã30010
= 600 m/s

Topic: Physics, Mechanical Properties of Solids, NEET
A hollow charged metal sphere has a radius r. If the potential difference between its surface and a point at a distance 3r from the centre is V , then the electric field intensity at distance 3r from the centre is -

Options:
A V3r
B V4r
C V6r
D V2r

The correct answer is:
V6r

Solution:
VA-VB=kqr-kq3r=V
kqrÃ 23=V
EB=kq3r2=32Vr3r2=3Vr2Ã9Ãr2=V6r

Topic: Physics, Electrostatics, NEET
A hollow charged metal sphere has radius r. If the potential difference between its surface and a point at distance 3r from the centre is V, then the electric field intensity at distance 3r from the centre is

Options:
A V6r
B V4r
C V3r
D V2r

The correct answer is:
V6r

Solution:
VA-VB=kQr-kQ3r=V=2kQ3r
E =kQ3r2=kQ9r2âE=V6r

Topic: Physics, Electrostatics, NEET
A hollow cylinder has a charge q coulomb within it. If ϕ is the electric flux in units of Vm associated with the curved surface B , the flux linked with the plane surface A in units of Vm will be

Options:
A q 2 ε 0
B Ï 3
C q ε 0 - Ï
D 1 2 q Îµ 0 - Ï

The correct answer is:
1 2 q ε 0 - ϕ

Solution:
Let Ï A , Ï B and Ï C are the electric flux linked with A,âBâand C .
According to gauss theorem, Ï total = âq Îµ 0
Ï A + Ï B + Ï C = q Îµ 0
Since Ï A = Ï C ,
â´ 2 Ï A + Ï B = q Îµ 0
or 2 Ï A = q Îµ 0 - Ï B
or 2 Ï A = q Îµ 0 - Ï
(Given Ï B = Ï )
â´ Ï A = 1 2 q Îµ 0 - Ï

Topic: Physics, Electrostatics, NEET
A hollow cylinder has a charge $q$ coulomb within it. If $\phi$ is the electric flux in units of voltmeter associated with the curved surface B, the flux linked with the plane surface $\mathrm{A}$ in units of voltmeter will be

Options:
A $\frac{q}{2 \varepsilon_0}$
B $\frac{\phi}{3}$
C $\frac{q}{\varepsilon_0}-\phi$
D $\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right)$

The correct answer is:
$\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right)$

Solution:
Step 1: Total flux through cylinder using Gauss Law
By Gauss Law, total flux through the cylinder is given by:
\(\phi_{\text {total }}=\frac{\mathrm{q}}{\varepsilon_0} \ldots (1)\)
Step 2: Flux through plane surface \(A\)
As surfaces \(\mathrm{A} \& \mathrm{C}\) is symmetrical so \(\phi_{\mathrm{A}}=\phi_{\mathrm{C}}\)
\(\phi_{\text {total }}=\phi_{\mathrm{A}}+\phi_{\mathrm{B}}+\phi_{\mathrm{C}}=\frac{\mathrm{q}}{\varepsilon_{\mathrm{o}}} \quad [\text {Using Equation (1)] }\)
Given Flux through curved surface \(\phi_B=\phi\)
So,
\(\begin{aligned}
& 2 \phi_A+\phi=\frac{q}{\varepsilon_0} \\
& \Rightarrow \phi_A=\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right)
\end{aligned}\)

Topic: Physics, Electrostatics, NEET, NEET 2007
A hollow cylinder with inner radius R, outer radius 2R and mass M is rolling with speed of it's axis v. Its kinetic energy is

Options:
A 56Mv2
B 52Mv2
C 12Mv2
D 1316Mv2

The correct answer is:
1316Mv2

Solution:
KE=12Mv2+12IÏ2
KE=12Mv2+12I v2R2 ...(1)
Consider solid cylinder of radius 2R and mass M', from which a cylinder of radius R of mass M"=M'4 is taken out from a common axis.
Mass of hollow cylinder M = M'-M''
M=M'-M'4=3M'4 â M=3M'4 â M'=4M3 & M"=M3
I2R=IR+I â I=I2R-IR â I=12M'2R2-12M"R2
I=12Ã4M3Ã4R2-12ÃM3R2 â I=52MR2 ... (2)
KE=1316Mv2

Topic: Physics, Rotational Motion, NEET
A hollow hemisphere of mass 4m is placed as shown in figure. Two point masses m each are fixed to it at points P and Q on diametrically opposite points. The position of center of mass of the system is

Options:
A 0,-R2
B 0,-R3
C 0,-2R3
D 0,-R4

The correct answer is:
0,-R3

Solution:
Centre of mass for cut portion from a body
Xcm=m1 x1-m2 x2m1+m2
Xcm=mR-mR6m=0
Ycm=4mÃ-R26m=- R/3

Topic: Physics, Center of Mass Momentum and Collision, NEET
A hollow insulated conducting sphere is given a positive charge of 10 Î¼C. What will be the electric field at the centre of the sphere? The radius of the sphere is 2 m.

Options:
A Zero
B 5 μC m-2
C 20 Î¼C m-2
D 8 Î¼C m-2

The correct answer is:
Zero

Solution:
Charge resides on the outer surface of a conducting hollow sphere of radius R. We consider a spherical surface of radius r<R.
By Gauss theorem
∫sE→.ds→=1ε0× charge enclosed
E×4πr2=1ε0×0⇒E=0
i.e. electric field inside a hollow sphere is zero.

Topic: Physics, Electrostatics, NEET
A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the center

Options:
A increases as r increases for r<R and for r>R.
B zero as r increases for r<R , decreases as r increases for r>R.
C zero as r increases for r<R , increases as r increases for r>R.
D decreases as r increases for r<R and for r>R.

The correct answer is:
zero as r increases for r<R , decreases as r increases for r>R.

Solution:
Inside the sphere, electric field is zero.
Outside the sphere, electric field is E=14πε0Qr2
∴ It decreases with r, outside the sphere.

Topic: Physics, Electrostatics, NEET, NEET 2019 (All India)
A hollow vertical cylinder of radius R and height h has a smooth internal surface. A small particle is placed in contact with the inner side of the upper rim at a point P. It is given a horizontal speed v0 tangential to the rim. It leaves the lower rim at point Q, vertically below P. The number of revolutions made by the particle will be

Options:
A h2πR
B v02gh
C 2πRh
D v02πR2hg

The correct answer is:
v02πR2hg

Solution:
Time to reach the bottom,
∵t=2hg
∴2πRv0×n=2hg
⇒ n=v02πR2hg

Topic: Physics, Laws of Motion, NEET
A Hominid fossil discovered in Java in 1891, now extinct, having cranial capacity of about 900cc was:

Options:
A Homo erectus
B Neanderthal man
C Homo sapiens
D Australopithecus

The correct answer is:
Homo erectus

Solution:
Human evolution form	Cranial capacity

Australopithecus	500-600 cc

Homo habilis	650-800 cc

Homo erectus	900 cc

Homo neanderthalensis (Neanderthal man)	1400 cc

Cro-Magnon man	1600 cc

Homo sapiens	1350 cc

Topic: Biology, Evolution, NEET, NEET 2020 (Phase 2)
A homogeneous solid cylindrical roller of radius R and mass M is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is:

Options:
A F3 mR
B 3F2 m R
C 2F3 m R
D F2 m R

The correct answer is:
2F3 m R

Solution:
i
frR=Iα=mR22α …(ii)
For pure rolling
a=αR …(iii)
From i, (ii) and (iii)
F-mR α2=mαR
F=32mR  α
α=2F3mR

Topic: Physics, Rotational Motion, NEET
A homonym is

Options:
A Two or more names for the same taxon
B Species name repeats the generic name
C Identical name of two different species
D Name given to a taxon in local language

The correct answer is:
Identical name of two different species

Solution:
The identical name of two different taxa is called homonym. There is a rule in the International code of zoological nomenclature is that first name to be published in the senior homonym that should be used and the other names are junior homonyms that must be replaced with new names.

Topic: Biology, The Living World, NEET
A homopolymer has only one type of building block called monomer repeated ‘n’ number of times. A heteropolymer has more than one type of monomer. Proteins are heteropolymers usually made of

Options:
A 20 types of monomers
B 40 types of monomers
C 30 types of monomers
D only one type of monomer

The correct answer is:
20 types of monomers

Solution:
Proteins are heteropolymers usually made of 20 types of monomers.

Topic: Biology, Biomolecules (B), NEET
A homozygous tall pea plant with yellow pods is crossed with a homozygous dwarf pea plant with the green pods. The offsprings produced are selfed. In the F2 generation, the ratio of plants having parental combinations to new combinations is

Options:
A 10Â :6
B 6 :10
C 9Â :7
D 7Â :9

The correct answer is:
6 :10

Solution:
Genotype of homozygous tall plant with yellow pod is TTgg while genotype of dwarf plant with green pod is ttGG. All off springs they produce are hybrids TtGg (Tall with green pods). On selfing, they produce four types of plants: tall with green pod : tall with yellow pod : dwarf with green pod : dwarf with yellow pod in the ratio 9:3:3:1. The parental combinations tall with yellow pods and dwarf with green pods account to 6 out of 16 while recombinants tall with green pods and dwarf with yellow pods account to 10 out of 16.

Topic: Biology, Principles of Inheritance and Variation, NEET
A hoop rolls on a horizontal ground without slipping with a linear speed V. Speed of a particle P on the circumference of the hoop at angle θ is

Options:
A 2 v sin Î¸ 2
B v sin θ
C 2 v cos θ 2
D v cos θ

The correct answer is:
2 v sin θ 2

Solution:
This can assumed as a pure rotation about point of contact say O with angular velocity
Ï = v R , R is the radius of hoop.
speed of P will be :
v p = op Ï = 2 R sin Î¸ 2 Ï
or v p = 2 R Ï sin Î¸ 2
or v p =2vsin( θ 2 )

Topic: Physics, Rotational Motion, NEET
A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity 4 m s-1.
The ball strikes the water surface after 4 s. The height of bridge above water surface is
(Take g=10 m s-2):

Options:
A 68 m
B 56 m
C 60 m
D 64 m

The correct answer is:
64 m

Solution:
The formula to calculate the vertical height travelled by the ball is given by

H=ut-12gt2 ...1

Substitute the values of the known parameters into equation (1) to calculate the required height.

H=4 ms-1×4 s-12×10 ms-2×4 s2=-64 m

Hence, the required height of the bridge above the surface of the river is 64 m.

Topic: Physics, Motion In One Dimension, NEET, NEET 2023 (All India)
A horizontal force 10 N is applied to a block $A$ as shown in figure. The mass of blocks $A$ and $B$ are 2 kg and 3 kg respectively. The blocks slide over a frictionless surface. The force exerted by block $A$ on block $B$ is :

Options:
A 4N
B 6N
C 10N
D Zero

The correct answer is:
6N

Solution:
$\begin{aligned} & F=\left(M_1+M_2\right) a \\ & a=\frac{10}{2+3}=2 \mathrm{~ms}^{-2} \\ & F^{\prime}=M_2(2)=3 \times 2 \mathrm{~N}=6 \mathrm{~N}\end{aligned}$

Topic: Physics, Laws of Motion, NEET, NEET 2024
A horizontal overhead power line is at a height of 4 m from the ground and carries a current of 100 A from east to west. The magnetic field directly below it on the ground is

Options:
A 2.5×10-7 T northwards
B 2.5×10-7 T southwards
C 5×10-6 T northwards
D 5Ã10-6 T southwards

The correct answer is:
5Ã10-6 T southwards

Solution:
By Ampere's theorem, BâÂ·2Ïd=Î¼0i
Bâ=Î¼0i2Ïd=4ÏÃ10-7Ã100 A2ÏÃ4 m
=5Ã10-6 T southwards

Topic: Physics, Magnetic Effects of Current, NEET
A horizontal overhead power line is at a height of 4 m from the ground and carries a current of 100 A from east to west. The magnetic field directly below it on the ground is Î¼0=4ÏÃ10-7 H m-1

Options:
A 2.5Ã10-7 T northward
B 5.0Ã10-7 T southward
C 5.0×10-7 T northward
D 5.0Ã10-6 T southward

The correct answer is:
5.0Ã10-6 T southward

Solution:
By Ampere's theorem, BâÂ·2Ïdâ¡=Î¼0iâ¡
Bâ=Î¼0iâ¡2Ïdâ¡=4ÏÃ10-7Ã100A2ÏÃ4m=50Ã10-7 T
âB=5Ã10-6 T southwards.

Topic: Physics, Magnetic Effects of Current, NEET
A horizontal pipeline carries water in streamline flow. At a point along the pipe, where the cross-sectional area is 10 cm2, the water velocity is 1 m s-1 and the pressure is 2000 Pa, The pressure of water at another point where the cross-sectional area is 5 cm2, is. (Density of water = 103 kg m3)

Options:
A 500
B 400
C 300
D 600

The correct answer is:
500

Solution:
From the continuity equation
v2=(A1A2)v1=(105)(I)=2 m s-1

Applying Bernoulliâs theorem at 1 and 2
P2+12pv22=p1+12pv12
â´ p 2 = p 1 + 1 2 Ï vâ¡ 1 2 - vâ¡ 2 2 = 2 0 0 0 + 1 2 Ã 1 0 3 1 - 4
â´Pâ¡2=500 Pa

Topic: Physics, Mechanical Properties of Fluids, NEET
A horizontal pipeline carries water in streamline flow. At a point where the cross-sectional area is 10 cm2 the water velocity is 1 m s-1 and pressure is 2000 Pa. The pressure of water at another point where the cross-sectional area is 5 cm2, is

Options:
A 200 Pa

B 400 Pa
C 500 Pa
D 800 Pa

The correct answer is:
500 Pa

Solution:
Since the cross-sectional area is halved, therefore, velocity is doubled.
Now, p1=2000 Pa, v1=1 m s-1
p2=?, v2=2 m s-1
Again p2+12×1000×2×2=2000+12×1000×1×1
or p2=2000+5001-4=500 Pa

Topic: Physics, Mechanical Properties of Fluids, NEET
A horizontal platform with an object placed on it is executing SHM in the vertical direction. The amplitude of oscillation is 2.5 cm. What must be the least period of these oscillations so that the object is not detached? (Given : g=10 m s-2 )

Options:
A π s
B π5 s
C π10 s
D π15 s

The correct answer is:
π10 s

Solution:
At the extreme location, the forces on the object are mg and mωâ2 A in opposite directions
For an object to always be in contact will platform
m ω 2 A≤mg
or ω= g A or T 2π = A g
or T=2πAg=2π2.51000=π10 s

Topic: Physics, Oscillations, NEET
A horizontal platform with an object placed on it is executing SHM in the vertical direction. The amplitude of oscillation is 2.5 cm. What must be the least period of these oscillations so that the object is not detached?

Options:
A Ï s
B Ï5 s
C Ï10 s
D Ï15 s

The correct answer is:
Ï10 s

Solution:
mω2a=mg, ω=ga, T=2πag
T=2π2.51000=π10 s

Topic: Physics, Oscillations, NEET
A horizontal ray of light is incident on the right-angled prism with prism angle $6^{\circ}$. If the refractive index of the material of the prism is 1.5 , then the angle of emergence will be

Options:
A $4^{\circ}$
B $6^{\circ}$
C $9^{\circ}$
D $10^{\circ}$

The correct answer is:
$9^{\circ}$

Solution:
Deviation produced by a thin prism
$\delta=(\mu-1) A=\left(\frac{3}{2}-1\right) 6^{\circ}=3^{\circ}$
$\begin{aligned} \text { Angle of emergence }=e & =6^{\circ}+3^{\circ} \\ & =9^{\circ}\end{aligned}$

Topic: Physics, Ray Optics, NEET, NEET 2023 (Manipur)
A hormone responsible for the normal sleep-wake cycle is

Options:
A Epinephrine
B Gastrin
C Melatonin
D Insulin

The correct answer is:
Melatonin

Solution:
Option (c) is the answer.

Topic: Biology, Chemical Coordination and Integration, NEET
A hormone with the seat of activity in liver, converting glucose to glycogen in produced by

Options:
A thymus.
B pancreas.
C pituitary.
D parathyroid.

The correct answer is:
pancreas.

Solution:
The hormone which is responsible for conversion of excessive glucose present in the body into glycogen is insulin. Glycogen is stored in the liver. Insulin is produced by the beta cells of Islet of Langerhans of pancreas.
Thymus is responsible for production of thymosins, which help in maturation of T lymphocytes.
The pituitary gland is responsible for production of hormones like GH, TSH, FSH, LH, etc. which have their respective functions.
The parathyroid gland is responsible for production of parathyroid hormone, which regulates calcium metabolism.

Topic: Biology, Chemical Coordination and Integration, NEET
A hot air balloon is a sphere of radius 8 m. The air inside is at a temperature of 60°C. How large a mass can the balloon lift when the outside temperature is 20°C? (Assume air is an ideal gas, R = 8.314 J mole–1K-1, 1 atm. = 1.013×105Pa; the membrane tension is 5 N m–1.)

Options:
A 3034.2 N
B 3044.2 N
C 3054.2 N
D 3064.2 N

The correct answer is:
3044.2 N

Solution:
Correct Option is : (B)
3044.2 N

Topic: Chemistry, Mechanical Properties of Fluids, NEET
(a) How much charge is required for the reduction of 1 mol of Al3+to Al?

Options:
A 289,461 C
B 192,974 C
C 482,435 C
D 97,236 C

The correct answer is:
289,461 C

Solution:
Correct Option is : (A)
289,461 C

Topic: Chemistry, Electrochemistry, NEET
(a) How much electricity (in terms of Faraday) is required to produce 20.0 g of Ca from molten CaCl2?

Options:
A 1 F
B 2 F
C 3 F
D 4 F

The correct answer is:
1 F

Solution:
Correct Option is : (A)
1 F

Topic: Chemistry, Electrochemistry, NEET
A human female with Turner's syndrome

Options:
A has 45 chromosomes with XO
B has one additional X chromosome.
C exhibits male characters
D is able to produce children with normal husband.

The correct answer is:
has 45 chromosomes with XO

Solution:
TURNER’s SYNDROME:

Introduction: It was first described in 1938 by Turner. The disorder develops due to monosomy of the sex chromosome. The disease is seen only in females. It is an allosomal chromosomal abnormality.
Cause: The disorder is due to monosomy. It appears due to fusion of abnormal egg (22+0) and a normal sperm (22+X) or a normal egg (22+X) and abnormal sperm (22+0).

Karyotype: [44 + XO]

Symptoms:
1. Short stature.
2. Webbed neck.
3. Flat and broad chest.
4. Rudimentary ovaries (defective ovaries, no production of ova). Sterility in females.
5. No Barr bodies.
6. Average to normal I.Q.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2014
A human male is heterozygous for autosomal genes 'A' and 'B' present on two different chromosomes, and also has hemophilic gene h. What proportion of his sperms will have abh genes?

Options:
A 116
B 14
C 18
D 132

The correct answer is:
18

Solution:
Haemophilic gene h is present on X chromosome. The probability of his sperms having X chromosome or h gene is 12 . He is heterozygous Aa. The probability of his sperms having a is 12 . Similarly the probability of his sperms having b is 12 . Thus, the probability of his sperms having a, b and h gene is 12×12×12=18

Topic: Biology, Principles of Inheritance and Variation, NEET
A human male produces sperms with the genotype $A B, A b, a B$, and ab pertaining to two diallelic characters in equal proportions. What is the corresponding genotype of his person'?

Options:
A AaBB
B $\mathrm{AABb}$
C AABB
D AaBb.

The correct answer is:
AaBb.

Solution:
As the person produces sperm having both the alleles for both the characters. So the person must be heterozygous for both the genes, hence, the genotype will be $\mathrm{AaBb}$.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2007
A human male who is hemizygous for a gene gives rise to

Options:
A brown eyes
B phenylketonuria
C colour-blindness
D cystic fibrosis

The correct answer is:
colour-blindness

Solution:
Colour blindness is a X-linked recessive disorder. In it, the affected individual has is not able to distinguish colour especially read and green. 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. Hence, male considered hemizygous for the colour blindedness gene.

Topic: Biology, Principles of Inheritance and Variation, NEET
A husband and wife have normal vision, although both of their fathers are red-green colour blind, which is inherited as an x-linked recessive trait.

What is the probability that their first child will be?

i. A normal son

ii. A carrier daughter

iii. A colour-blind son

iv. A colour-blind daughter

Options:
A i=Â 14 , ii=14Â ,iii=14Â ,iv=14
B i=14 ,ii=14 ,iii=14, iv=04
C i=04Â ,ii=12Â ,iii=14Â ,Â iv=14
D i=14, ii=12 , iii=14 ,iv=0

The correct answer is:
i=14 ,ii=14 ,iii=14, iv=04

Solution:
Mother is a carrier (because her father is colour blind) and father is normal. Hence, their genotype is Xc X and XY, respectively. Thus, the probability that their first child will be a normal son is 1/4, a carrier daughter is 1/4, colour-blind son is 1/4 and a colour-blind daughter is 0 (both husband and wife have normal vision, though their fathers were colour blind. The probability of their daughter becoming colour blind is 0%. The chances of a daughter becoming colour blind arise only when the father is also colour blind).

Topic: Biology, Principles of Inheritance and Variation, NEET
A hydraulic automobile lift has input and output pistons with diameters of 10 cm and 30 cm. The lift is used to hold up a car with a weight of 1.44Ã104 N.
What is the force on the input piston?

Options:
A 1.6×103 N
B 1.5×103 N
C 1.4×103 N
D 1.8×103 N

The correct answer is:
1.6×103 N

Solution:
Here, the diameter of the input piston, d = 10 cm = 0.1 m
The diameter of output piston, D = 30 cm = 0.3 m
The weight lifted by the output piston, F=1.44×104 N
Let a and A be the areas of a cross-section of the input and output pistons respectively. If f is the force applied on the input piston, then
FA=fa or FπD24=fπd24
Or f=F×d2D2=1.44×104×0.120.32=1.6×103 N

Topic: Physics, Mechanical Properties of Fluids, NEET
A hydrocarbon ‘A’, (C4H8) on reaction with HCl gives a compound ‘B’, (C4H9Cl), which on reaction with 1 mol of NH3gives compound ‘C’, (C4H11N). On reacting with NaNO2and HCl followed by treatment with water, compound ‘C’ yields an optically active alcohol, ‘D’. Ozonolysis of ‘A’ gives 2 mols of acetaldehyde. Identify compounds ‘A’ and ‘B’.

Options:
A A = 2-Butene and B = 2-Chlorobutane
B A = 2-Butyne and B = 2-Chlorobutane
C A = 2-Butene and B = 1-Chlorobutane
D A = 2-Butyne and B = 1-Chlorobutane

The correct answer is:
A = 2-Butene and B = 2-Chlorobutane

Solution:
Correct Option is : (A)
A = 2-Butene and B = 2-Chlorobutane

Topic: Chemistry, Amines, NEET
A hydrocarbon ‘A’, (C4H8) on reaction with HCl gives a compound ‘B’, (C4H9Cl), which on reaction with 1 mol of NH3gives compound ‘C’, (C4H11N). On reacting with NaNO2and HCl followed by treatment with water, compound ‘C’ yields an optically active alcohol, ‘D’. Ozonolysis of ‘A’ gives 2 mols of acetaldehyde. Identify compounds ‘C’ and ‘D’.

Options:
A C = Butan-2-amine and D = Butan-1-ol
B C = Butan-2-amine and D = Butan-2-ol
C C = Butan-3-amine and D = Butan-2-ol
D C = Butan-3-amine and D = Butan-1-ol

The correct answer is:
C = Butan-2-amine and D = Butan-2-ol

Solution:
Correct Option is : (B)
C = Butan-2-amine and D = Butan-2-ol

Topic: Chemistry, Amines, NEET