datavorous/entrance-exam-dataset huggingface dataset:
A process where the diploid cell of an ovule outside the embryo sac developing into another embryo sac without reduction is known as

Options:
A Parthenogenesis
B Apospory
C Double fertilization
D Endosperm formation

The correct answer is:
Apospory

Solution:
A process where the diploid cell of an ovule outside the embryo sac developing into another embryo sac without reduction is known as apospory. When diploid cells of nucellus divide to form diploid embryo sac which contains diploid egg cell, from which embryo develops without fertilization or gametic fusion, then it is called apospory. This mechanism of seed formation without gametic fusion or fertilization is called apomixes. It is commonly present in members of Asteraceae and grasses.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
A progressive wave is represented by y=12sin⁡(5t-4x) cm. On this wave, how far away are the two points having a phase difference of 90°?

Options:
A Ï2 cm
B π 4 cm
C Ï8Â Â cm
D Ï16Â cm

The correct answer is:
π8 cm

Solution:
y=12 sin (5t-4x)
Comparing in y=Asinâ¡wt-kx
We have A=12, w=5 and k=4
wt-kx= phase difference =Ï2
∴| 5t−4x |= π 2
When t=0, 4x=Ï2
â´x=Ï8 cm

Topic: Physics, Waves and Sound, NEET
A projectile is fired from the surface of the Earth of radius R with a velocity ηve, where ve is the escape velocity and η<1. Neglecting the air resistance, the orbital velocity of the projectile is

Options:
A ve1-Î·2
B veη25
C 25veη
D 2η5ve

The correct answer is:
ve1-η2

Solution:
Velocity at the surface=vs=Î·ve
Velocity in orbit=v0
By conversation of mechanical energy,
12mvs2+-GMmR=12mv02-GMmrâ¦1
By Newton's law, GMmr2=mv02râââââ¦2
Solving equations (1) and (2)
v0=ve1-Î·2

Topic: Physics, Gravitation, NEET
A projectile is fired from the surface of the earth with a velocity of 5 m s-1 and angle θ with the horizontal. Another projectile fired from another planet with a velocity of 3 m s-1 at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is: (given =9.8 m s-2)

Options:
A 3.5
B 5.9
C 16.3
D 110.8

The correct answer is:
3.5

Solution:
The Equation of the Trajectory of a Projectile

y=xtanθ-gx22u2cosθ

y ,x are Variable and θ is constant

so gu2=constant

g52=a32
a=9.8×925
a=3.5

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2014
A projectile is fired with velocity v0 at an angle 60o with horizontal. At top of its trajectory it explodes into two fragments of equal mass. If one fragment retraces the path then the speed of the other fragment is

Options:
A 2v0
B 5v02
C v0
D 3v02

The correct answer is:
3v02

Solution:
Let the velocity of the other fragment be v→
P→i=P→f
⇒ 2mv02i^=-mv02i^+mv→
⇒ v→=3v02i^

Topic: Physics, Center of Mass Momentum and Collision, NEET
A projectile is fired with velocity v0 at an angle 60° with horizontal. At top of its trajectory, it explodes into three fragment of equal masses. First fragment retraces the path and the second moves vertically upward with speed 3v02 . The speed of third fragment is

Options:
A 3v02
B 5v02
C v0
D 2v0

The correct answer is:
5v02

Solution:
Let velocity of the third fragment be v→
P→i=P→f
⇒ 3m.v02i^=-mv02i^+m.3v02j^+mv→
⇒ v=4v02i^-3v02j^
⇒ v=5v02

Topic: Physics, Center of Mass Momentum and Collision, NEET
A projectile is fired with velocity v0 at an angle 60° with horizontal. At top of its trajectory it explodes into three fragments of equal masses. First fragment retraces the path, second moves vertically upwards with speed 3v02. Speed of the third fragment is

Options:
A 3v02
B 5v02
C v0
D 2v0

The correct answer is:
5v02

Solution:
From conservation of momentum at topmost point
Let velocity of third fragment be vâ
Pâi=Pâf
â 3m.v02i^=-mv02i^+m.3v02j^+mvâ
â vâ=4v02i^-3v02j^
âv=5v02

Topic: Physics, Center of Mass Momentum and Collision, NEET
A projectile is given an initial velocity of i^+2j^ m s-1. The Cartesian equation of its path is g=10 m s-1 (Here, i^ is the unit vector along horizontal and j^ is unit vector vertically upwards)

Options:
A y=2x-5x2
B y=x-5x2
C 4y=2x-5x2
D y=2x-25x2

The correct answer is:
y=2x-5x2

Solution:
tanθ=usinθucosθ=21
The desired equation is :
y=xtanθ-gx22u2cos2θ
y=x2-10x22(22+12)2152
y=2x-5x2.

Topic: Physics, Motion In Two Dimensions, NEET
A projectile is given an initial velocity of i^+2j^ m s-1, where i is along the ground and j is along vertically upward direction. If g=10 m s-2, the equation of the trajectory is

Options:
A y=x-5x2

B y=2x-5x2

C 4y=2x-5x2

D 4y=2x-25x2

The correct answer is:
y=2x-5x2

Solution:
Initial velocity = u â
u â = u x i â & u y j â = i â + 2 j â
Where ux = velocity along ground
uy = velocity along vertical direction
at time t, let (x, y) be the coordinates of the projectile
x=uxt so t=xux
But ux=1 and uy=2
t=x1=x
y=uyt-12gt2
y=2x-12Ã10x2
y=2x-5x2 is the equation of the trajectory.

Topic: Physics, Motion In Two Dimensions, NEET
A projectile is given an initial velocity of i^+2j^. What is the Cartesian equation of its path? g = 10 m s-2 (Here,i^ is the unit vector along horizontal and j^ is unit vector vertically upwards)

Options:
A y=2x-5x2
B y=x-5x2
C 4y=2x-5x2
D y=2x-25x2

The correct answer is:
y=2x-5x2

Solution:
tan⁡θ=u sinθucos⁡θ=21

The desired equation is:
y=xtanθ-gx22u2cos2⁡θ
Or y=x2-10x2222+122152
Or y=2x-5x2

Topic: Physics, Motion In Two Dimensions, NEET
A projectile is projected from the ground by making an angle of 60Â° with the horizontal. After 1 s projectile makes an angle of 30Â° with the horizontal. The maximum height attained by the projectile is (Take g=10 m s-2)

Options:
A 45 2 m
B 45 4 m
C 43 2 m
D 43 4 m

The correct answer is:
45 4 m

Solution:
tan 30°=v0sin 60°-gtv0cos 60°⇒13=v032-g×tV02
⇒ v 0 2 = 3v 0 2 - gt 3
⇒ v 0 = gt 3 = g 3 .
∴Hmax=v02sin2θ2g=300×3/420=454m

Topic: Physics, Motion In Two Dimensions, NEET
A projectile is projected with velocity kve in vertically upward direction from the ground into space (ve is the escape velocity and k<1). If air resistance is considered to be negligible then, the maximum height, from the centre of the earth, to which it can go will be (R is the radius of the earth)

Options:
A Rk2+1
B Rk2-1
C R1-k2
D Rk+1

The correct answer is:
R1-k2

Solution:
Kinetic energy = potential energy
12mkve2=mgh1+hR
12mk22gR=mgh1+hRh=Rk21-k2
Height of projectile from the earth surface = h
Height from the centre = r=R+h=R+Rk21-k2
On solving, r=R1-k2

Topic: Physics, Gravitation, NEET
A projectile is thrown with a velocity of 18 m/s at an angle of 60o with horizontal. The interval between the moments when speed is 15 m/s is g=10 m/s2

Options:
A 125s
B 45s
C 12 s
D 25s

The correct answer is:
125s

Solution:
Vx=18Ãcos 60o=9 m/s
â´Vy=152-92=12 m/s
â´Ît=2Ã12g=125s

Topic: Physics, Motion In Two Dimensions, NEET
A projectile is thrown with a velocity of 20 m/s , at an angle of 60Â° with the horizontal. After how much time the velocity vector will make an angle of 45Â° with the horizontal

Options:
A 3Â s
B 13Â s
C 3+1Â s
D 3-1 s

The correct answer is:
3-1 s

Solution:
As the horizontal component is the same for a projectile
Let projectile velocity is v when particle makes angle 45Â° with horizontal,
hence
v=ucosâ¡Î¸secâ¡Ï=20cosâ¡60Â°secâ¡45Â°
v=20.12.2=102 m/s
t=vy-uyg=103-1010=3-1 s

Topic: Physics, Motion In Two Dimensions, NEET
A projectile is thrown with velocity u=20 m/sÂ± 5% at an angle 60 o . If the projectile comes back on the ground at the same level then which of following cannot be a possible answer for range.
Consider g=10â m/s 2

Options:
A 34.6Â m
B 37.5Â m
C 32.0Â m
D 39.0 m

The correct answer is:
39.0 m

Solution:
Range of projectile R=u2sin2Î¸g
for âÎ¸= 60 o , R= 20 2 sin 120 o g =20 3 =34.64âm
for % errorâRR=2âuu
ÎR= 2Ã5 100 Ã 20 3 =2 3 = 3.464âm
So, range R=34.64Â±3.46âm
Rmin=31.1 m and R max 38.1âm

Topic: Physics, Motion In Two Dimensions, NEET
A projectile of mass m is fired from the surface of the earth at an angle α = 6 0 ∘ from the vertical. The initial speed v0 is equal to GMeRe. The maximum height that projectile can rise from surface of earth is Rex. Find the value of x ?

G = Universal Gravitational constant
Me = Mass of earth
Re = Radius of earth

Options:
A 2
B

4

C

6

D

8

The correct answer is:
2

Solution:
By conservation of angular momentum about C at position P and F
mv 0 3 2 R e = mv r max
â v = 3 R e v 0 2 r max ...(i)
By conservation of mechanical energy between P and F
1 2 mv 0 2 - GM e m R e = 1 2 mv 2 - GM e m r max ...(ii)
Put (i) in (ii)
1 2 mv 0 2 - GM e m R e = 1 2 m 3 R 2 v 0 2 4 r max 2 - GM e m r max
v 0 = GM e R
â 1 2 m GM e R e - GM e m R e = 3 8 mR e 2 r max 2 GM e R - GM e m r max
â 1 2 R e - 1 R e = 3 R 8 r max 2 - 1 r max
â - 1 2 R e = 3 R e - 8 r max 8 r max 2
â 4 r max 2 - 8 R e r max + 3 R e 2 = 0
Solving the quadratic in rmax
we get
r max = 8 R e Â± 1 6 R e 8 = 8 R e Â± 4 R e 8
â r max = 8 R e + 4 R e 8 = 1 2 R e 8 = 3 R e 2
â´ h max = r max - R e = 3 R e 2 - R e = R e 2
â´ x = 2

Topic: Physics, Gravitation, NEET
A prokaryotic autotrophic nitrogen fixing symbiont is found in

Options:
A Cycas
B Cicer
C Pisum
D Alnus

The correct answer is:
Cycas

Solution:
The coralloid root of Cycas is symbiotically associated with nitrogen fixing blue-green algae, Anabaena cycadae and Nostoc punctiforme. These blue green algae (cyanobacteria) are prokaryotic, photosynthetic and autotrophic.

Topic: Biology, Mineral Nutrition, NEET, NEET 2011 (Screening)
A. Proteins are polymers of ___a___.
B.Lipids are not strictly ___b___.
C. Polysaccharides are long chains of ___c___.
D. Nucleic acid is a polymer of ___d___.

Fill in the blanks.

Options:
A a- macromolecules, b- amino acids, c- sugars, d- nucleotides
B a- amino acids, b- macromolecules, c- sugars, d- nucleotides
C a- nucleotides, b- sugars, c- nucleotides, d- macromolecules
D a- sugars, b- nucleotides, c- amino acids, d-macromolecules

The correct answer is:
a- amino acids, b- macromolecules, c- sugars, d- nucleotides

Solution:
The unit of protein is amino acid. They bind with peptide bond with the help of ribosome. Lipids are micromolecules but not present after filtration. Polysaccharides are polymers of monosaccharides.

Topic: Chemistry, Biomolecules, NEET
A proton, a deuteron and an α-particle are projected perpendicular to the direction of a uniform magnetic field with same kinetic energies. The ratio of the radii of the circular paths described by them is

Options:
A 1 :2 :1
B 1Â :2Â :2
C 2Â :2Â :1
D 2Â :1Â :1

The correct answer is:
1 :2 :1

Solution:
According to the question, magnetic field is uniform and the kinetic energy for each particle is same.
We know that,
r=mvqB=2mK.E.qB
râ m q
rp:rd:rÎ±= 1 1 : 2 1 : 4 2
=1 :2 :1

Topic: Physics, Magnetic Effects of Current, NEET
A proton, a deuteron and an α particle with the same kinetic energies enter in a region of uniform transverse magnetic field. The ratio of the radii of their circular paths is

Options:
A 1:2:1
B 1:2:2
C 2:1:1
D 2:2:1

The correct answer is:
1:2:1

Solution:
Radius of circular path of proton,
rp=2mpKeB
Radius of circular path of deuteron,
rd=2(2mp)KeB
Radius of circular path of α particle,
rα=2(4mp)K(2e)B
Therefore
rp:rd:rα=1:2:1

Topic: Physics, Magnetic Effects of Current, NEET
A proton and a deuteron with the same initial kinetic energy enter a magnetic field in a direction perpendicular to the direction of the field. The ratio of the radii of the circular trajectories described by them is

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

The correct answer is:
1:2

Solution:
For a moving charge in a perpendicular magnetic field,
mv2r=Bqv
â r= mv Bq = p Bq = 2mK qB
rprd=mpmd= 12=12

Topic: Physics, Magnetic Effects of Current, NEET
A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. if the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be:

Options:
A 0.5 MeV
B 1..5 MeV
C 1 MeV
D 4 MeV

The correct answer is:
1 MeV

Solution:
R=mVqB=2mkEqB
Since R is same so KE ∝q2m
So KE of α particle will be 224=1 MeV

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2015 (Phase 2)
A proton and an electron are released from an infinite distance apart and they get attracted towards each other. Which of the following statement about their kinetic energy is true?

Options:
A Kinetic energy of electron is more than that of proton
B Kinetic energy of electron is less than that of proton
C Kinetic energy of electron = kinetic energy of proton
D None of the above is true as it depends on the distance between the particles

The correct answer is:
Kinetic energy of electron is more than that of proton

Solution:
Proton and electron move under the influence of their mutual electrostatic attraction. Therefore momentum is conserved i.e. constant, hence K=p22m. As electron has less mass, it possesses greater kinetic energy.

Topic: Physics, Electrostatics, NEET
A proton and an Î±-particle, accelerated through the same potential difference, enter in a uniform magnetic field normally. If the radius of the proton orbit is 10 cm, then the radius of Î± -particle is

Options:
A 10 cm
B 102 cm
C 20 cm
D 5 2 cm

The correct answer is:
102 cm

Solution:
R=mvqB
RαRp=mαvαqαmpvpqp=2vαvp
Beacuse
qα=2qp
mα=4mp
as they are acclerated by same potential
Equation of α=qαV=2qp V
12mα×Vα2=212mPVp2
VαVP=12
Hence RαRP=2×12=2
Rα=2RP =102 cm

Topic: Physics, Atomic Physics, NEET
A proton and an $\alpha$-particle are accelerated from rest to the same energy. The de-Broglie wavelengths $\lambda_p$ and $\lambda_\alpha$ are in the ratio

Options:
A $2: 1$
B $1: 1$
C $\sqrt{2}: 1$
D $4: 1$

The correct answer is:
$2: 1$

Solution:
The de-Broglie wavelength associated with a charged particle is given by
$$
\lambda=\frac{\mathrm{h}}{\mathrm{p}}
$$
where, $\mathrm{h}=$ planck's constant and $\mathrm{p}=$ momentum $=\sqrt{2 \mathrm{mKE}}$ (here, KE is the kinetic energy of the charged particle)
$$
\Rightarrow \quad \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKE}}}
$$
For proton and $\alpha$-particle, the wavelengths are respectively given as,
$$
\begin{aligned}
\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{KE}_{\mathrm{p}}}} \text { and } \lambda_\alpha=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_\alpha \mathrm{KE}_\alpha}} \\
\therefore \quad \quad \frac{\lambda_{\mathrm{p}}}{\lambda_\alpha}=\frac{\sqrt{2 \mathrm{~m}_\alpha \mathrm{KE}_\alpha}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{KE}_{\mathrm{P}}}}
\end{aligned}
$$
Here, $\mathrm{KE}_\alpha=\mathrm{KE}_{\mathrm{p}}$ and $\mathrm{m}_\alpha=4 \mathrm{~m}_{\mathrm{p}}$
Substituting these above mentioned relations in Eq. (i), we get
$$
\Rightarrow \quad \frac{\lambda_{\mathrm{P}}}{\lambda_\alpha}=\sqrt{\frac{4 \mathrm{~m}_{\mathrm{P}}}{\mathrm{m}_{\mathrm{p}}}}=\frac{2}{1} \quad \text { or } \quad \lambda_{\mathrm{P}}: \lambda_\alpha=2: 1
$$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2019 (Odisha)
A proton and an α -particle are accelerated through the same potential difference V. The ratio of their de-Broglie wavelength is

Options:
A 2
B 22
C 3
D 23

The correct answer is:
22

Solution:
λ=h2mqV
As V is same.
λ∝1mq
λpλα=mαqαmpqp=4mp×2qpmpqp
=22

Topic: Physics, Dual Nature of Matter, NEET
A proton and an α-particle are accelerated, using the same potential difference. How are the deBroglie wavelengths λpand λarelated to each other?

Options:
A λp= λa
B λp= 2λa
C λp= 4λa
D λp= √8λa

The correct answer is:
λp= √8λa

Solution:
Correct Option is : (D)
λp= √8λa

Topic: Chemistry, Dual Nature of Matter, NEET
A proton and an α-particle are projected normally with the same speed into a magnetic field. What will be the ratio of radii of the trajectories of the proton and α-particle?

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

The correct answer is:
1 : 2

Solution:
In perpendicular magnetic field
Magnetic force = centripetal force
qvB=mv2r âr=mvqB
For proton q1=e, m1=m
For Î±-particle q2=2e, m2=4m
â´ r1r2=m1vq1B Ãq2Bm2v
r1r2=me Ã2e4m
r1r2=12

Topic: Physics, Magnetic Effects of Current, NEET
A proton and an α -particles are accelerated through the same potential difference. The ratio of de Broglie wavelength of proton to that of α -particle will be

Options:
A 2:1
B 1:1
C 1:2
D 22:1

The correct answer is:
22:1

Solution:
λ= h 2×m×K.E.
λp=h2mpeV
λα=h2mα2eV
=h2(4mp)2eV
λpλα=8=22

Topic: Chemistry, Structure of Atom, NEET
A proton beam passes without deviation through a region of space where there are uniform transverse mutually perpendicular electric and magnetic field E and B. Then the beam strikes a grounded target. The force imparted by the beam on the target if the beam current is equal to I is

Options:
A mEIeB
B mIEe
C mEIB
D eImB

The correct answer is:
mEIeB

Solution:
F→=0⇒F→eF→m=0⇒Fe=Fm⇒qE=qVB
V=EB···1
F=dpdt=Nmvt···2
I=Net⇒t=NeI···3
From (1), (2) and (3),
F= mEIeB

Topic: Physics, Magnetic Effects of Current, NEET
A proton is moving perpendicular to a uniform magnetic field of 2.5 T with 2 MeV kinetic energy. The force on proton (in N) is (Mass of proton =1.6×10-27 kg ,Charge of the proton =1.6×10-19C )

Options:
A 8×10-12
B 8×10-11
C 3×10-11
D 3×10-10

The correct answer is:
8×10-12

Solution:
Force on moving charge in magnetic field
F=qvBsin⁡θ
But θ=90
F=qvB
Hence velocity v= 2Em ( E is the kinetic energy of proton)
Putting the values we get
F=8×10-12N

Topic: Physics, Magnetic Effects of Current, NEET
A proton is projected with a uniform velocity v along the axis of a current-carrying solenoid, then

Options:
A The proton will be accelerated along the axis
B The proton path will be circular about the axis
C The proton move along helical path
D The proton will continue to move with velocity v along the axis

The correct answer is:
The proton will continue to move with velocity v along the axis

Solution:
The force on a charged particle in magnitude field
F=qvÃB
The proton is moving along the axis of solenoid. Magnetic field inside the solenoid is parallel to the axis of solenoid.
â´ v â â¥ B â
F=qvBsinâ¡Î¸
Î¸=0
F=0
â´ The proton will continue to move with velocity along the axis.

Topic: Physics, Magnetic Effects of Current, NEET
A proton moving with a velocity of 1.25Ã105 m s-1 collides with a stationary helium atom. The velocity of the proton after the collision is

Options:
A 0.75Ã105 m s-1
B 7.5Ã105 m s-1
C -0.75Ã105 m s-1
D 0 m s-1

The correct answer is:
-0.75Ã105 m s-1

Solution:
From momentum conservation
mpu1=mpv1+mHev2
⇒u1=v1+4v2 ...(i)
For elastic collision e=1
e=v2−v1u1
⇒u1=v2−v1 ...(ii)
From equation (i) and equation (ii), we get
v2=25u1 and v1=−35u1
∴v1=−35×1.25×105 m s-1=−0.75×105 m s−1

Topic: Physics, Center of Mass Momentum and Collision, NEET
A proton of mass m and charge +e is moving in a circular orbit in a magnetic field with energy 1 MeV. What should be the energy of Î±-particle (mass = 4m and charge = + 2e), so that it can revolve in the path of the same radius?

Options:
A 1 MeV
B 4 MeV
C 2 MeV
D 0.5 MeV

The correct answer is:
1 MeV

Solution:
r=2mKqB⇒K∝q2m⇒KpKα=qpqα2×mαmp
⇒1Kα=qp2qp2×4mpmp=1⇒Kα=1 MeV.

Topic: Physics, Atomic Physics, NEET
A proton of mass m collides elastically with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of 90° with respect to each other. The mass of unknown particle is:

Options:
A m3
B m2
C 2m
D m

The correct answer is:
m

Solution:
Apply principle of conservation of momentum along x- direction,
mu=mv1cosâ¡45+Mv2cosâ¡45
mu=12mv1+Mv2 ...(i)
along y- direction,
o=mv1sinâ¡45-Mv2sinâ¡45
o=mv1-Mv212 ...(ii)
Coefficient of Resolution (e) = 1 = v2-v1cosâ¡90ucosâ¡45
âv2u2=1
âu=2v2 ...(iii)
Solving equation (i), (ii) & (iii) we get
M=m

Topic: Physics, Center of Mass Momentum and Collision, NEET
A proton when accelerated through a potential difference of V volt has a wavelength Î» associated with it. An Î± - particle in order to have the same Î» must be accelerated through potential difference of

Options:
A V volt
B 4V volt
C 2V volt
D V/8 volt

The correct answer is:
V/8 volt

Solution:
Î»P=Î»Î±
or h2mpQPVp=h2mÎ±QÎ±VÎ±
â´ m p Q p V p = m Î± Q Î± V Î±
â´ V Î± = m p m Î± Q p Q Î± V = 1 4 1 2 V = V 8

Topic: Physics, Dual Nature of Matter, NEET
A protoplast is a cell:

Options:
A Without nucleus
B Undergoing division
C Without cell wall
D Without Golgi body

The correct answer is:
Without cell wall

Solution:
Plant cell - cell wall = Protoplast
Protoplast is a biological term proposed by Hanstein in 1880 to refer to the entire cell, excluding the cell wall, but currently has several definitions: a plant, bacterial or fungal cell that had its cell wall completely or partially removed using either mechanical or enzymatic means.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2015 (Phase 2)
A protoplast is a cell:

Options:
A Without nucleus
B Undergoing division
C Without cell wall
D Without Golgi body

The correct answer is:
Without cell wall

Solution:
Plant cell - cell wall = Protoplast
Protoplast is a biological term proposed by Hanstein in 1880 to refer to the entire cell, excluding the cell wall, but currently has several definitions: a plant, bacterial or fungal cell that had its cell wall completely or partially removed using either mechanical or enzymatic means.

Topic: Biology, Cell - The Unit of Life, NEET
A protoxin is:

Options:
A A primitive toxin
B A denatured toxin
C The toxin produced by protozoa
D Inactive toxin

The correct answer is:
Inactive toxin

Solution:
A protoxin is an Inactive toxin.

Topic: Biology, Biotechnology and Its Applications, NEET
A protozoan reproduces by binary fission. What will be the number of protozoans in its population after six generations?

Options:
A 64
B 128
C 24
D 32

The correct answer is:
64

Solution:
In the geometric growth pattern, both the progenies of the divided cell participate in the next division cycle. Exponential growth phase is observed in this pattern. The graph in geometric growth is a typical sigmoid curve whereas in arithmetic growth the graph is a straight line.
Binary fission is a type of asexual mode of reproduction where the cell divides in two to form two new identical daughter cells. It is the most common form of reproduction observed in bacterial groups and also in some unicellular eukaryotes like Amoeba and Paramecium (type of protozoans). If a protozoan cell reproduces by binary fission then after 6 generations, there will 64 cells as each cell produces two new cells.
It can be calculated using the formula 2n where n represents number of generation. 26 = 64 .

Topic: Biology, Reproduction in Organisms, NEET
A p-type extrinsic semiconductor is obtained when Germanium is doped with

Options:
A Arsenic
B Boron
C Antimony
D Phosphorous

The correct answer is:
Boron

Solution:
To make a p-type semiconductor, an intrinsic semiconductor is doped with an intrinsic impurity which has 3 valence electrons in its outer orbit.
For example - Gallium, Indium, Boron, Aluminium.
So, correct answer is option (2).

Topic: Physics, Semiconductors, NEET, NEET 2023 (Manipur)
A public park, in the form of a square, has an area of 100Â±0.2 m2. The side of park is

Options:
A (10±0.01) m
B (10±0.1) m
C (10.0±0.1) m
D (10.0±0.2) m

The correct answer is:
(10±0.01) m

Solution:
A=l2l=A=10 mΔAA=2ΔllΔll=12   ΔAAΔl=12(ΔAA)lΔl=12(0.2100)×10=0.01 m

Topic: Physics, Units and Dimensions, NEET
A pulley fixed with ceiling carries a string with blocks of mass m and 3m attached to its ends. The masses of string and pulley are negligible. When the system is released, its centre of mass moves with acceleration

Options:
A g
B g5
C g4
D zero

The correct answer is:
g4

Solution:
The magnitude of the acceleration of each mass with respect to the pulley
a=3mg-mg3m+m=g2
Acceleration of centre of mass =3m g/2-mg/23m+m=g4

Topic: Physics, Laws of Motion, NEET