datavorous/entrance-exam-dataset huggingface dataset:
Trichoderma harzianum is used as a biocontrol agent against various plant pathogens. To which of the following class of fungi does it belong?

Options:
A Ascomycetes
B Zygomycetes
C Deuteromycetes
D Basidiomycetes

The correct answer is:
Deuteromycetes

Solution:
The Deuteromycetes reproduce only by asexual spores known as conidia. The mycelium is septate and branched. Some members are saprophytes or parasites while a large number of them are decomposers of litter and help in mineral cycling. Some examples are Alternaria, Colletotrichum, and Trichoderma.

Topic: Biology, Biological Classification, NEET
Triploblastic, unsegmented, acoelomate exhibit bilateral symmetry reproducing both asexually and sexually, with some parasitic forms. This description is characteristic of the phylum

Options:
A Cnidaria
B Platyhelminthes
C Annelida
D Ctenophora

The correct answer is:
Platyhelminthes

Solution:
Triploblastic means having three germ layers: ectoderm, mesoderm, and endoderm. This is seen for the first time in the phylum Platyhelminthes. These are flatworms that exhibit bilateral symmetry. These do not show any body cavity hence are acoelomates. Example: Tapeworms, liver flukes etc. These reproduce by asexual and sexual manner. Thus only this phylum is triploblastic but are acoelomates.

Topic: Biology, Animal Kingdom, NEET
Triploid tissue will be observed in -

Options:
A nucellus
B endosperm
C tapetum
D endothecium

The correct answer is:
endosperm

Solution:
Double fertilization is a unique characteristic observed in the Angiosperm group of plants. In this, two fertilizations are observed - 1st between the female and male gametic cells forming zygote and 2nd is between the second male gamete and polar nuclei forming the primary endosperm nucleus. The ploidy of PEN is 3n (male gamete [1n] + polar nuclei [2n] ).

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Trisomy is represented by

Options:
A 2n-1
B 2n-2
C 2n+2
D 2n+1

The correct answer is:
2n+1

Solution:
The chromosomal disorder occurs when there is an addition or deletion of chromosome in the individual. Trisomy is represented as 2n+ 1 as it has one extra chromosome than the normal diploid set of just 2n.

Topic: Biology, Principles of Inheritance and Variation, NEET
Triticale, the first man-made cereal crop, has been obtained by crossing wheat with:

Options:
A barlcy
B rye
C pearl millet
D sugarcane

The correct answer is:
rye

Solution:
Triticale is the first man-made cereal crop. It has been obtained by crossing wheat (Triticum sp.) with rye (Secale cereale).

Topic: Biology, Strategies for Enhancement in Food Production, NEET, NEET 2006
Tritium, a radioactive isotope of hydrogen, emits which of the following particles?

Options:
A Betaβ-
B Alpha α
C Gamma (γ)
D Neutron n

The correct answer is:
Betaβ-

Solution:
Tritium decays to 23He+by β- decay.

H13→He+23+e0-1+ν(Antineutrino)

Topic: Chemistry, Hydrogen, NEET, NEET 2021
t-RNA is also known as:

Options:
A Adapter molecule
B Messenger RNA
C Soluble RNA
D Both (A) and (C)

The correct answer is:
Both (A) and (C)

Solution:
tRNA is the smallest RNA with a sedimentation coefficient of 3.8â4S; thus, it is last to be centrifuged. Hence, it is known as soluble RNA. Crick named it as adapter molecule.

Topic: Biology, Molecular Basis of Inheritance, NEET
Tropic movement is due to

Options:
A Cell elongation
B Cell division
C Both Cell elongation and Cell division
D Cell thickening

The correct answer is:
Cell elongation

Solution:
Tropic movements are caused by cell elongation and do not affect the whole plant. Different tropic movements are phototropism, geotropism, chemotropism, aerotropism (e.g., positively aerotropic pneumatophores), thermotropism (towards warmth, e.g., peduncles of Tulip, Anemone), rheotropism, galvanotropism and traumatotropism.

Topic: Biology, Plant - Growth and Development, NEET
True-breeding red-eyed Drosophila flies with plain thoraxes were crossed with pink-eyed flies with striped thoraxes.

Red-eye and plain thorax X Pink-eye and striped thorax

The F1 flies were then test crossed against the double recessive.

The following F2 generation resulted from the cross:

80	16	12	92

Red-eye	Red-eye	Pink-eye	Pink-eye

Plain thorax	Striped thorax	Plain thorax	Striped thorax

What percentage number of recombinants resulted from the testcross?

Options:
A 12
B 14
C 16
D 28

The correct answer is:
14

Solution:
A test cross is a process to examine and identify whether the organism is heterozygous or not. In it, the organisms are crossed with the recessive homozygous as it will ensure the expression of the dominant character in heterozygous form for F1 generation enabling us to know the zygosity. A test cross is a type of back cross.
The total number of offsprings: 80+92+16+12 = 200
Recombinants are those which have a combination other than parents.
F1 is a hybrid having phenotype - Red eye and the plain thorax.
Recessive parent - Pink eye striped thorax.
Total recombinants - Red eye stripe thorax and Pink eye plain thorax: 16 + 12 = 28
Percentage of recombinants in the F1 of testcross = Total number of recombinantsTotal number of progeny x 100 = 28200 x 100 = 14%

Topic: Biology, Principles of Inheritance and Variation, NEET
Trypsin, insulin and morphine are, respectively,

Options:
A a protein, enzyme and alkaloid
B an alkaloid, enzyme and protein
C an enzyme, hormone and alkaloid
D a hormone, alkaloid and protein

The correct answer is:
an enzyme, hormone and alkaloid

Solution:
Trypsin is a digestive enzyme produced by acinar cells of the pancreas, and it hydrolyzes the proteoses and peptones in peptides.
Insulin is a peptide hormone produced by the beta cells of Islets of Langerhans and controls the blood sugar level.
Morphine, an alkaloid, is an opiate narcotic and is derived from opium. Morphine is a most valuable analgesic but it causes addiction. Morphine has sleep and dream-inducing properties.

Topic: Chemistry, Biomolecules, NEET
Trypsinogen is activated into trypsin by an enzyme ______X______ secreted by _________Y________. Choose the option which correctly describes ‘X' and ‘Y'.

Options:
A

X	Y

(A)	Trypsinase	Small intestine

B

X	Y

(B)	Intestinal Lipase	Small intestine

C

X	Y

(C)	Enterokinse	Intestinal mucosa

D

X	Y

(D)	Steapsin	Small Intestine

The correct answer is:

X	Y

(C)	Enterokinse	Intestinal mucosa

Solution:
The secretion of the brush border cells of the mucosa, along with the secretions of the goblet cells constitute the intestinal juice or succus entericus.
Intestinal mucosa also secretes enterokinase enzyme which convert inactive trypsinogen into active trypsin.

Topic: Biology, Digestion and Absorption, NEET
Trypsinogen is converted into active trypsin by the action of:

Options:
A Cholecystokinin
B Enterocrinin
C Enterokinase
D Secretin

The correct answer is:
Enterokinase

Solution:
Trypsin and chymotrypsin are proteolytic endopeptidases. They are secreted in inactive forms as trypsinogen and chymotrypsinogen. The enterokinase secreted by the intestine converts inactive trypsinogen to trypsin.

Topic: Biology, Digestion and Absorption, NEET
T.S. of a plant material shows the following anatomical features.

I. Vascular bundles are conjoint, scattered and surrounded by sclerenchymatous bundle sheaths.
II. Phloem parenchyma is absent.

What will you identify it as?

Options:
A Dicot stem
B Monocot stem
C Dicot root
D Monocot root

The correct answer is:
Monocot stem

Solution:
In a monocot stem, vascular bundles are conjoint, collateral and closed. These vascular bundles are scattered in ground tissue, and each vascular bundls is surrounded by bundle sheath.

Topic: Biology, Anatomy of Flowering Plants, NEET
Tubectomy is method of sterilization in which:

Options:
A small part of the Fallopian tube is removed or tied up
B ovaries are removed surgically
C small part of vas deferens is removed or tied up
D uterus is removed surgically

The correct answer is:
small part of the Fallopian tube is removed or tied up

Solution:
Tubectomy is a sterilization method of birth control in which a small part of the Fallopian tube is removed or tied up. This ensures the sperm and egg will not fuse with each other.

Topic: Biology, Reproductive Health, NEET, NEET 2014
Tubectomy is method of sterilization in which:

Options:
A small part of the fallopian tube is removed or tied up.
B ovaries are removed surgically.
C small part of vas deferens is removed or tied up.
D uterus is removed surgically.

The correct answer is:
small part of the fallopian tube is removed or tied up.

Solution:
Tubectomy is a sterlization method of birth control in which small part of the fallopian tube is removed or tied up.

Topic: Biology, Reproductive Health, NEET
Tubulin protein is synthesized in which phase of cell cycle?

Options:
A Prophare
B Metaphase
C Interphase
D Telophase

The correct answer is:
Interphase

Solution:
Interphase is the portion of the cell cycle that is not accompanied by observable changes under the microscope and includes the G1, S and G2 phases. During interphase, the cell grows (G1), replicates its DNA (S) and prepares for mitosis (G2).

Topic: Biology, Cell Cycle and Cell Division, NEET
Turgor pressure becomes equal to the wall pressure when

Options:
A Water leaves the cell
B No exchange of water takes place
C Water enters the cell
D Solute goes from the cell into water

The correct answer is:
No exchange of water takes place

Solution:
When turgor pressure is equal to the wall pressure the establishment of equilibrium (isotonic solution) takes place between the solution, and hence osmosis doesn't take place.

Topic: Biology, Plant - Growth and Development, NEET
Turner syndrome and Klinefelter syndrome are all related to

Options:
A both 'X' chromosome and 'Y' chromosome
B 'Y' chromosomes
C 'X' chromosomes
D neither 'X' chromosome nor 'Y' chromosome

The correct answer is:
'X' chromosomes

Solution:
45, X is known as Turner syndrome, is a condition where human females have only one X chromosome and 47, XXY is known as Klinefelter syndrome, is a condition where human males have an extra X chromosome.

Topic: Biology, Principles of Inheritance and Variation, NEET
Turner syndrome is an example of

Options:
A Structural chromosomal aberration - Due to non-disjunction
B Numerical chromosomal aberration - Due to disjunction
C Aneuploidy - Due to non-disjunction
D Both (B) and (C)

The correct answer is:
Aneuploidy - Due to non-disjunction

Solution:
In aneuploidy, any change in the number of chromosomes in an organism would be different than the multiple of a basic set of chromosomes. It commonly arises due to non-disjunction (absence of separation of two homologous chromosomes during cell division) of the two chromosomes of homologous pair so that one gamete comes to have an extra chromosome (n+1) while the other because deficient in one chromosome (n-1).

Topic: Biology, Principles of Inheritance and Variation, NEET
Turner's syndrome in humans is caused by:

Options:
A Autosomal aneuploidy
B Sex chromosome aneuploidy
C Polyploidy
D Point mutation

The correct answer is:
Sex chromosome aneuploidy

Solution:
Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome(s), called aneuploidy. For example, Down’s syndrome results in the gain of extra copy of chromosome 21. Similarly, Turner’s syndrome results due to loss of an X chromosome in human females.

Topic: Biology, Principles of Inheritance and Variation, NEET
T.W. Engelmann split the light into its components using a

Options:
A green glass
B convex lens
C prism
D concave lens

The correct answer is:
prism

Solution:
Using a prism he split light into its spectral component and illuminated a green alga Cladophora placed in suspension of aerobic bacteria.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

Options:
A 660 V
B 1320 V
C 1980 V
D 1520 V

The correct answer is:
1980 V

Solution:
The radius of the bigger drop will be 43πR3=27×43πr3⇒R=3r, the potential of the small drop, V=Kqr. The potential of the big drop Q=nq, V'=KQR⇒V'=KnqR

As we know,V' =n2/3V=(27)2/3×220=1980 V

Topic: Physics, Electrostatics, NEET, NEET 2021
Twisted aestivation and axile placentation are found in:

Options:
A China rose
B Dianthus
C Sunflower
D Mustard

The correct answer is:
China rose

Solution:
China rose belongs to the Malvaceae family that has a superior ovary. The arrangement of sepals and petals is called aestivation. In twisted aestivation, the first petals overlap the next petals and so on. Along with the monadelphous condition is found in China rose in which filaments are fused to form a single bundle. Axile placentation is found in china rose in which there is a septum present and ovules are present at the center.

Topic: Biology, Morphology of Flowering Plants, NEET
Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time combination is connected to a 220 volt a.c. supply line. The power drawn by the combination in each case respectively will be:

Options:
A 50 watt, 10 watt
B 100 watt, 50 watt
C 200 watt, 150 watt
D 50 watt, 200 watt

The correct answer is:
50 watt, 200 watt

Solution:
\(\begin{aligned} & \mathrm{R}_1=\frac{\mathrm{V}^2}{\mathrm{P}_1}=\frac{(220)^2}{100}=484 \Omega \\ & \mathrm{R}_2=\frac{\mathrm{V}^2}{\mathrm{P}_2}=\frac{(220)^2}{100}=484 \Omega \\ & \mathrm{I}=\frac{220}{484 \times 2}=\frac{5}{22} \\ & \mathrm{P}=\mathrm{I}^2\left(\mathrm{R}_1+\mathrm{R}_2\right)=\frac{25}{22 \times 22} \times(484 \times 2) \\ & =50 \mathrm{~W}\end{aligned}\)

\(\begin{aligned} & P=\frac{V^2}{R_1}+\frac{V^2}{R_2} \\ & =\frac{2 \times(220)^2}{484}=200 \mathrm{~W}\end{aligned}\)

Topic: Physics, Alternating Current, NEET, NEET 2003
Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will:

Options:
A Keep floating at the same distance between them
B Move towards each other
C Move away from each other
D Will become stationary

The correct answer is:
Move towards each other

Solution:
Gravitational force between them pulls towards each other.

Gravity, also called gravitation, in mechanics, the universal force of attraction acting between all matter. On Earth all bodies have a weight, or downward force of gravity, proportional to their mass, which Earth's mass exerts on them. Gravity is measured by the acceleration that it gives to freely falling objects.

Topic: Physics, Gravitation, NEET, NEET 2017
Two atoms X and Y are non-polar and electrically symmetrical

What type of intermolecular force of attraction can be developed between them?

Options:
A Dipole-induced dipole forces
B London forces or dispersion forces
C Dipole-dipole forces
D No forces of any kind

The correct answer is:
London forces or dispersion forces

Solution:
In non-polar atom, a dipole may develop momentarily. If the atom X becomes unsymmetrical resulting in development of instantaneous dipole for a very short time, it distorts the electron density of Y which is close to it and dipole is induced in the atom Y. Temporary dipoles X and Y attract each other. This attraction is called London or dispersion force.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Two bacteria found to be very useful in genetic engineering experiments are

Options:
A Caenorhabditis and Agrobacterium
B Nitrobacter and Azotobacter
C Rhizobium and Agrobacterium
D Nitrosomonas and Klebsiella

The correct answer is:
Caenorhabditis and Agrobacterium

Solution:
Caenorhabditis elegans and Agrobacterium tumefaciens are both used as vectors in genetic engineering. Nitrobacter converts nitrites to nitrates. Azotobacter is free-living nitrogen-fixing bacteria. Nitrosomonas converts ammonia to nitrite. Rhizobium is a symbiotic nitrogen-fixing bacteria living in the root nodules of leguminous plants Diplococcus pneumoniae causes pneumonia.

Topic: Biology, Biotechnology - Principles and Processes, NEET
Two balls are thrown simultaneously in the air. The acceleration of the centre of mass of the two balls when in air

Options:
A is equal to g (acceleration due to gravity)
B depends in the speeds of the two balls
C depends on the masses of the two balls
D depends on the direction of motion of the two balls.

The correct answer is:
is equal to g (acceleration due to gravity)

Solution:
We know that,
Acceleration of centre of mass
acm=m1a1+m2a2m1+m2
Here, a1=a2=g (Because balls are thrown in air i.e. under the gravity)
acm=m1g+m2gm1+m2
acm=gm1+m2m1+m2
acm=g

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two balls of mass m and 4m are connected by a rod of length L. The mass of the rod is small and can be treated as zero. The size of the balls also can be neglected. We also assume the centre of the rod is hinged, but the rod can rotate about its centre in the vertical plane without friction. What is the gravity-induced angular acceleration of the rod when the angle between the rod and the vertical line is θ as shown?

Options:
A 5g6Lsin⁡θ
B g6Lcos⁡θ
C 6g5Lsin⁡θ
D g3Lsin⁡θ

The correct answer is:
6g5Lsin⁡θ

Solution:
Key Point The magnitude of torque depends on three quantities, the force applied, the length of the lever arm connecting the axis to the point of force application and the angle between the force vector and the lever arm.
Moment of inertia of the rod and balls system about the centre is
I=4mL22+mL22=54mL2
Net torque on the connected rod of length L
τ=4mgL2sin⁡θ-mgL2sin⁡θ=3mgL2sin⁡θ
We have,
So, gravity induced angular acceleration of the rod
α=τI=3mgL2sin⁡θ54mL2=6gsin⁡θ5L

Topic: Physics, Rotational Motion, NEET
Two balls of masses m and 2m are attached to the ends of a light rod of length L. The rod rotates with an angular speed ω about an axis passing through the centre of mass of system and perpendicular to the plane. Find the angular momentum of the system about the axis of rotation.

Options:
A 23mωL2
B 13ω2Lm
C 23ω2Lm
D 13Lm

The correct answer is:
23mωL2

Solution:
Location of CM,
XCM=2m×0+m×L2m+m=L3
The velocity of ball, A w.r.t. the centre of mass C is
VA=ωXCM=ωL3
VB=ωL-XCM=2ωL3
Thus, angular momentum of a system about C,
L=2mvAXCM+mvBL-XCM
=2m×ωL3×L3+m×2ωL3×2L3
=2mωL23

Topic: Physics, Rotational Motion, NEET
Two balls of radius r each are placed at a distance R and released. The speed of each ball when they collide

Options:
A v=-2Gm1R+1r
B v=-Gm1R-12r
C v=-2Gm1R+12r
D v=-Gm12R+1r

The correct answer is:
v=-Gm1R-12r

Solution:
From mechanical energy conservation,
(Energy)i = (Energy)f
- Gm 2 R = - Gm 2 2 r + 2 1 2 mv 2
v = - Gm 1 R - 1 2 r

Topic: Physics, Gravitation, NEET
Two balls with charges 5 Î¼ C and 1 0 Î¼ C are at a distance of 1 m from each other. In order to reduce the distance between them to 0.5 m, the amount of work to be performed is

Options:
A 45 J

B 0.45×10-6 J

C 1.2×10-4 J

D 0.45 J

The correct answer is:
0.45 J

Solution:
w ext = Î u = u f - u i
Here, u = q 1 q 2 4 Ï Îµ 0 r
u i = 5 Ã 1 0 - 6 1 0 Ã 1 0 - 6 4 Ï Îµ 0 1
= 9 Ã 1 0 9 Ã 5 Ã 1 0 - 6 Ã 1 0 Ã 1 0 - 6
Similarly,
u f = 9 Ã 1 0 9 Ã 5 Ã 1 0 - 6 Ã 1 0 Ã 1 0 - 6 0.5
â´ W ext = u f - u i = 0.45 J

Topic: Physics, Electrostatics, NEET
Two bar magnets having geometry with magnetic moments $M$ and $2 M$ are first placed in such a way that their similar poles are same side then its time period of oscillation is $T_1$. Now the polarity of one of the magnet is reversed then the time period of oscillation is $T_2$, so:

Options:
A $T_1 < T_2$
B $T_1=T_2$
C $T_1>T_2$
D $T_2=\infty$

The correct answer is:
$T_1 < T_2$

Solution:
We can write
$T=2 \pi \sqrt{\frac{1}{M B}} \Rightarrow T \propto \frac{1}{\sqrt{M}}$
Case I : $M_1=2 M+M$
Case II : $M_2=2 M-M$
$\begin{aligned}
& \therefore \frac{T_1}{T_2}=\sqrt{\frac{M}{3 M}}=\frac{1}{\sqrt{3}} \\
& \Rightarrow T_2=\sqrt{3} T_1
\end{aligned}$

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2002
Two bar magnets of the same mass, same length and breadth but having magnetic moments M and 3M are joined together pole for pole and suspended by a string.
The time period of assembly in a magnetic field of strength 1H is 3 s. If now the polarity of one of the magnets is reversed and the combination is again made to oscillate in the same field, the time of oscillation is

Options:
A 3 s
B 33 s
C 3/3 s
D 32 s

The correct answer is:
32 s

Solution:
In pole for pole combination,
M1=3M+M = 4M
and if the polarity is reversed M2=3M−M=2M
T2=?
T2T1=M1M2=4 M2 M
T2=2T1=32 s

Topic: Physics, Magnetic Properties of Matter, NEET
Two batteries, one of E.M.F. 18 V and internal resistance 2 Î© and other of E.M.F. 12 V and internal resistance 1 Î©, are connected as shown. The voltmeter V will record a reading of

Options:
A 30 V
B 18 V
C 15 V
D 14 V

The correct answer is:
14 V

Solution:
Since the cells are in parallel, V=E1r1+E2r21r1+1r2=182+12112+11=211.5=14 V

Topic: Physics, Current Electricity, NEET
Two batteries, one of emf 18 volts and internal resistance $2 \Omega$ and the other of emf 12 volts and internal resistance $1 \Omega$, are connected as shown. The voltmeter $V$ will record a reading of:

Options:
A 30 volt
B 18 volt
C 15 volt
D 14 volt

The correct answer is:
14 volt

Solution:
Reading of voltmeter
$$
=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}}=\frac{E_1 r_2+E_2 r_1}{r_1+r_2}
$$
Putting value we get 14 volt.

Topic: Physics, Current Electricity, NEET, NEET 2005
Two billiard balls undergo a head-on collision. Ball 1 is twice as heavy as ball 2. Initially, the ball 1 moves with a speed v towards the ball 2 which is at rest. Immediately after the collision, ball 1 travels at a speed of v3 in the same direction. What type of collision has occurred?

Options:
A Inelastic
B Elastic
C Completely inelastic
D Cannot be determined from the information given

The correct answer is:
Elastic

Solution:
v 1 = m 1 - m 2 u 1 m 1 + m 2 + 2 m 2 u 2 m 1 + m 2
v 2 = m 2 - m 1 u 2 m 1 + m 2 + 2 m 1 u 1 m 1 + m 2
u 1 = v m 1 = 2 m v 1 = v 3 | u 2 = 0 m 2 = m v 2 = ?
From conservation of linear momentum
2 m v = 2 m × v 3 + m × v 2
v 2 = 2 v - 2 v 3 ⇒ v 2 = 4 v 3
e = velocity of seperation after collission velocity of apporach before collission ⇒ e = 4 v 3 - v 3 v
e = 1

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two blocks A and B are connected by a massless string (shown in the figure). A force of 30 N is applied on block B. The distance travelled by centre of mass in 2 s starting from rest is

Options:
A 1 m

B 2 m

C 3 m

D 4 m

The correct answer is:
2 m

Solution:
The acceleration of the centre of mass is
aCM=FmA+mB
=3010+20=1 m s-2
â´ s=12acmt2=12Ã1Ã22=2 m

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two blocks A and B are placed one over the other on a smooth horizontal surface. The maximum horizontal force that can be applied on the lower block A, so that A and B move without separation is 49 N. The coefficient of friction between A and B is ( take g=9.8 m s-2 )

Options:
A 0.2
B 0.3
C 0.5
D 0.8

The correct answer is:
0.5

Solution:
Given, F=49 N, MA=7 kg, MB=3 kg, μ= ?
F=(3+7)a
⇒a=4.9  m s-2
Free body diagram of B
fmax=μ(3g)=3a
μ=3a3g=4.9g=4.99.8=0.5

Topic: Physics, Laws of Motion, NEET
Two blocks ' A ' and ' B ' each of mass ' m ' are placed on a smooth horizontal surface. Two horizontal forces F and 2F are applied on the blocks A and B, respectively, as shown in figure. The block A does not slide on block B . The normal reaction acting between the two blocks is

Options:
A F
B F2
C F3
D 3F

The correct answer is:
3F

Solution:
Acceleration of two mass system is a=F2m leftward
FBD of block A

Ncos⁡60o-F=ma
⇒Ncos⁡60o-F=mF2m
⇒N=3F

Topic: Physics, Laws of Motion, NEET