datavorous/entrance-exam-dataset huggingface dataset:
If the head of cockroach is removed, it may live for few days because

Options:
A the cockroach does not have nervous system.
B the head holds a small proportion of a nervous system while the rest is situated along the ventral part of its body.
C the head holds a 1/3rd of a nervous system while the rest is situated along the dorsal part of its body.
D the supra-oesophageal ganglia of the cockroach are situated in ventral part of abdomen.

The correct answer is:
the head holds a small proportion of a nervous system while the rest is situated along the ventral part of its body.

Solution:
NERVOUS SYSTEM:

The nervous system of the cockroach consists of a series of fused, segmentally arranged ganglia joined by paired longitudinal connectives on the ventral side.

Brain: It is present in the head. It is made up of a pair of supra oesophageal which is the present above esophagus. It supplies nerves to antennae and compound eyes. The sub-oesophageal ganglia are present below the esophagus. These ganglions are connected together by paired longitudinal connectives on the ventral side.

Double Ventral Nerve Cord: They arise from sub-oesophageal ganglia and run-up to the posterior end of the body. These nerve cords are connected with the help of 9 pairs of segmented ganglion three in the thorax and 6 in the abdominal segment.

The nervous system of cockroaches is spread throughout the body. The head holds a bit of the nervous system while the rest is situated along the ventral side of the body.

Topic: Biology, Structural Organisation in Animals, NEET, NEET 2020 (Phase 1)
If the heartbeat of a person is about 75 beats per minute, and stroke volume is 70 mL per minute, then the cardiac output will be

Options:
A 5000 mL
B 5250 mL
C 5500 mL
D 5150 mL

The correct answer is:
5250 mL

Solution:
The stroke volume multiplied by the heart rate (no. of beats per min.) gives the cardiac output.
In the given question:
The stroke volume is 70 mL, while the heart rate is 75 beats per minute.
Cardiac output = Stroke volume x Heart rate
= 70 x 75 = 5250 mL

Topic: Biology, Body Fluids and Circulation, NEET
If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is:

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

The correct answer is:
1:2

Solution:
Case 1:

T1=T

Case 2:

T2=2T

We know, speed of transverse wave is related to tension by

v=Tμ, here, μ is mass per unit length of the string.

∴ v1v2=T1T2=T2T=12=1:2

Topic: Physics, Waves and Sound, NEET, NEET 2022 (Phase 1)
If the instantaneous rate of appearance of NO2g is 0.0400 M/s at some moment in time, what is the rate of disappearance of N2O5g in M/s?
N2O5gâ2NO2g+12O2g

Options:
A 0.02
B 0.01
C 0.04
D 0.08

The correct answer is:
0.02

Solution:
N2O5 â2NO2+12O2

rateâ=â d( N 2 O 5 ) dt â=â 1 2 d( N O 2 ) dt =â2 d( O 2 ) dt
Rate of disappear of NO2 = d( N O 2 ) dt =0.04
d( N 2 O 5 ) dt = 1 2 d( N O 2 ) dt
=â 0.04 2 â=â0.02

Topic: Chemistry, Chemical Kinetics, NEET
If the ionic product of NiOH2 is 1.9×10-15 , the molar solubility of NiOH2 in 1.0 M NaOH is

Options:
A 1.9Ã-18M
B 1.9Ã10-13 M
C 1.9Ã10-15M
D 1.9×10-14M

The correct answer is:
1.9×10-15M

Solution:
NaOH â Na + + OH - C M CM
Ni OH 2 â Ni 2 + + 2 OH - x x 2x
â´ Ksp=Ni2+ OH-2
=xx+C2
Ksp=xC2 (neglecting higher power of x )
x=KspC2=1.9Ã10-1512

Topic: Chemistry, Ionic Equilibrium, NEET
If the ionization enthalpy and electron gain enthalpy of an element are 275 and 86 kcal mol-1 respectively, then the electronegativity of the element on the Pauling scale is

Options:
A 2.8
B 0.0
C 4.0
D 2.6

The correct answer is:
2.8

Solution:
I.E. + E.A = 275 + 86 = 361 kcal mol-1
=361×4.184=1510.42 kJ mol-1
electronegativity =I.E.+E.A.540
So, electronegativity =1510.42540=2.797=2.8

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
If the kinetic energy of a free electron doubles, its de-Broglie wavelength changes by the factor

Options:
A 12
B 2
C 12
D 2

The correct answer is:
12

Solution:
We know
λ=hmv
and K=12mv2=mv22m
mv= 2mK
Thus, λ =h2mK
⇒ λ∝1K
∴ λ2λ1=K1K2=K12K1 ∵K2=2K1
⇒ λ2λ1=12
∴ λ2=λ12

Topic: Physics, Dual Nature of Matter, NEET
If the kinetic energy of a moving particle is E, then the de-Broglie wavelength is

Options:
A λ=h2mE
B λ= 2mEh
C λ=h2mE
D λ=hE2mE

The correct answer is:
λ=h2mE

Solution:
E=12mv2 or &nbspmv=2mE
so Î»=hmv=h2mE

Topic: Physics, Dual Nature of Matter, NEET
If the kinetic energy of a satellite orbiting around the earth is doubled then

Options:
A the satellite will escape into the space.
B the satellite will fall down on the earth
C radius of its orbit will be doubled
D radius of its orbit will become half.

The correct answer is:
the satellite will escape into the space.

Solution:
12mv12=212mv02
v1=2v0
this is the value of escape velocity, hence the satellite will escape out from gravitation of earth

Topic: Physics, Gravitation, NEET
If the lattice parameter for a crystalline structure is $3.6 Å$, then the atomic radius in fcc crystal is

Options:
A $1.81 Å$
B $2.10 Å$
C $2.92 Å$
D $1.27 Å$

The correct answer is:
$1.27 Å$

Solution:
Atomic radius is the distance between two nearest neighbours. For fcc crystal Atomic radius $=\frac{1}{2 \sqrt{2}} \times$ lattice parameter
or $r=\frac{a}{2 \sqrt{2}}$
or $r=\frac{3.6}{2 \sqrt{2}} Å=1.27 Å$

Topic: Chemistry, Solid State, NEET, NEET 2008 (Screening)
If the length of a DNA molecule is 1.1 metres, what will be the approximate number of base pairs?

Options:
A 6.6×109 bp
B 3.3×106 bp
C 6.6×106 bp
D 3.3×109 bp

The correct answer is:
3.3×109 bp

Solution:
The two chains of DNA are coiled in a right-handed helicle fashion. The pitch of the helix is 3.4 nm and has a diameter of 2 nm. In the B-form of DNA, the distance between a bp in a helix is approximately 0.34 nm which means there are about 10 nucleotides present in a complete turn.

The total length of DNA molecule in nm = 1.1×109 nm.

The approximate number of base pairs = =1.1×1090.34=3.3×109

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2022 (Phase 1)
If the length of rod A is 3.25 Â±0.01 cm and that of B is 4.19 Â±0.01 cm, then the rod B is longer than rod A by

Options:
A 0.94Â Â±0.00Â cm
B 0.94Â Â±0.01Â cm
C 0.94 Â±0.02 cm
D 0.94Â Â±0.005Â cm

The correct answer is:
0.94 Â±0.02 cm

Solution:
Given, length of rod A is
LA=3.25Â±0.01
Of B is LB=4.19Â±0.01
Then, the rod B is longer than rod A by a length
âl=LB-LA
âl=4.19Â±0.01-3.25Â±0.01
âl=0.94Â±0.02 cm

Topic: Physics, Units and Dimensions, NEET
If the level of glucose drops in the blood, the body releases __________ to bring it back to normal.

Options:
A insulin from beta cells of pancreas
B calcitonin from thyroid gland
C glucagon from alpha cells of pancreas
D adrenaline from adrenal medulla

The correct answer is:
glucagon from alpha cells of pancreas

Solution:
When the blood glucose level goes down, glucagon is secreted by the alpha cells of pancreas. Glucagon signals the liver to break down glycogen to glucose and release it into the blood stream. Thus bringing the level of glucose back to normal.

Topic: Biology, Chemical Coordination and Integration, NEET
If the light is made unavailable then the biosynthetic phase of photosynthesis

Options:
A will immediately stop.
B will stop after sometime.
C will stop after few hours.
D will have no effect and continue.

The correct answer is:
will stop after sometime.

Solution:
Effect of Light on Photosynthesis
- Effect of light is categorized under light quality, light intensity and the duration of exposure to light.
- At low light intensities, there is a linear relationship between incident light and CO2 fixation rates.
- At higher light intensities, the rate gradually does not show further increase as other factors become limiting.
- Light saturation occurs at 10 % of the full sunlight. Thus, light is rarely a limiting factor in nature
(except for plants in shade or in dense forests).
- Incident light beyond a point causes the breakdown of chlorophyll and a decrease in photosynthesis.

Topic: Biology, Photosynthesis in Higher Plants, NEET
If the magnetic dipole moment of an atom of diamagnetic material, paramagnetic material and ferromagnetic material is denoted by Î¼d, Î¼p and Î¼f respectively, then

Options:
A Î¼d=0 and Î¼pâ 0
B Î¼dÂ â 0Â andÂ Î¼p=0
C Î¼p=0Â andÂ Î¼fâ 0
D Î¼dâ Â andÂ Î¼fâ 0

The correct answer is:
Î¼d=0 and Î¼pâ 0

Solution:
The magnetic dipole moment of a diamagnetic material is zero as each of its pair of electrons has opposite spins, i.e., Î¼d=0. Paramagnetic substances have dipole moment > 0, i.e., Î¼p â 0, because of the excess of electrons in its molecules spinning in the same direction.
Ferromagnetic substances are very strong magnets and they also have a permanent magnetic moment, i.e., Î¼fâ 0.

Topic: Physics, Magnetic Properties of Matter, NEET
If the magnetic dipole of an atom of diamagnetic material, paramagnetic material are denoted by $\mu_d, \mu_p$ and $\mu_f$ respectively, then:

Options:
A $\mu_d=0$ and $\mu_p \neq 0$
B $\mu_d \neq 0$ and $\mu_p=0$
C $\mu_p=0$ and $\mu_f \neq 0$
D $\mu_d \neq 0$ and $\mu_f \neq 0$

The correct answer is:
$\mu_d=0$ and $\mu_p \neq 0$

Solution:
$\mu_{\mathrm{p}} \neq \mu_f \neq 0$ and $\mu_a=0$

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2005
If the magnetic field is parallel to a surface, then the magnetic flux through the surface is

Options:
A zero
B small but not zero
C infinite
D large but not infinite

The correct answer is:
zero

Solution:
ϕ=B→.A→
The area vector A→ is perpendicular to the surface. So, when the surface is parallel to the magnetic field, the angle between B→ and A→ is 90°.
Hence ϕ=0.

Topic: Physics, Magnetic Effects of Current, NEET
If the magnetic moment of a dioxygen species is 1.73 B.M, it may be

Options:
A O2 or O2-
B O2,O2- or O2-
C O2,O2+
D O2- or O2+

The correct answer is:
O2- or O2+

Solution:
Magnetic moment =1.73 BM
μ=n(n+2)
1.73=n(n+2)
On solving n = 1
Unpaired electron (n) =1
O2+=σ1s2,σ*1s2,σ2s2,σ*2s2,σ2p22,π2px2=π2py2,π*2px1=π*2py
O2-=σ1s2,σ*1s2,σ2s2,σ*2s2,σ2p22,π2px2=π2py2,π*2px2=π*2py1
Hence O2- and O2+ have one unpaired electron.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
If the magnitude of energy of the electron in nth Bohr orbit is En and Jn is its angular momentum, then

Options:
A En∝Jn3
B En∝1Jn–3
C En∝Jn2
D En∝1Jn2

The correct answer is:
En∝1Jn2

Solution:
Enâ1n2,Jnân
⇒En∝1Jn2

Topic: Physics, Atomic Physics, NEET
If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is:

Options:
A 0o
B 90°
C 45°
D 180°

The correct answer is:
90°

Solution:
A→+B→=A→-B→

A→2+B→2+2A→B→ cosθ=A→2+B→2-2A→B→ cosθ

4A→B→ cosθ=0
cosθ=0
θ=90°

Topic: Physics, Mathematics in Physics, NEET, NEET 2016 (Phase 1)
If the majority of offsprings exhibit a dominant character, then the parents must be:

Options:
A Both hybrids
B One dominant and one recessive
C Both recessive
D Both dominant

The correct answer is:
Both hybrids

Solution:
Both parents are hybrid or heterozygous for a given gene, as it represents a 3:1 ratio.

Topic: Biology, Principles of Inheritance and Variation, NEET
If the male plant is tetraploid and female plant is diploid, what will be the ploidy level of endosperm after fertilization?

Options:
A 3n
B 4n
C 5n
D 6n

The correct answer is:
4n

Solution:
Endosperm is formed by union of female central cell with male gamete. Diploid female plant will form haploid female gametophyte and large central cell with two polar nuclei giving 2n stage. Tetraploid male plant will produce 2n male gamete. When zn male gamete fuse with two polar nuclei of central cell of female it will form endosperm with ploidy level 4n.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
If the map distance between genes P and Q is 4 units, between P and R is 11 units, and between Q and R is 7 units, the order of genes on the linkage map can be traced as follows:

Options:
A
B
C
D

The correct answer is:

Solution:
Genes P, Q and R are linked, thus present on the same chromosome. P gene and R gene are farthest from each other having a distance of 11 units where as P gene and Q gene are closest with just 4 units and Q is 7 units away from R. Hence, Q gene is present between P and R and the map is as follows -

Topic: Biology, Principles of Inheritance and Variation, NEET
If the margin of sepals overlaps one another but not in any particular direction than the aestivation is called:

Options:
A Vexillary
B Imbricate
C Twisted
D Valvate

The correct answer is:
Imbricate

Solution:
Arrangement of sepals and petals in bud condition with respect to each other is called aestivation. If the margin of sepals overlaps one another but not in any particular direction than the aestivation is called imbricate aestivation.

Topic: Biology, Morphology of Flowering Plants, NEET
If the mass defect of 5 B 11 is 0.081 u, its average binding energy (in MeV) is

Options:
A 8.60  MeV
B 6.85  MeV
C 6.60  MeV
D 5.86  MeV

The correct answer is:
6.85  MeV

Solution:
Given, Î m for 5 B 1 1 = 0.0821 u
Number of nucleons = 11
âBinding energy =( 931×Δm )MeV
= 9 3 1 Ã 0.081
= 75.411 MeV
Average binding energy
= Binding energy Number of nucleons
= 75.411 11
= 6.85 MeV

Topic: Physics, Nuclear Physics, NEET
If the mass defect of 5 B 11 is 0.081 u, its average binding energy (in MeV) is

Options:
A 8.60  MeV
B 6.85ââMeV
C 6.60  MeV
D 5.86  MeV

The correct answer is:
6.85  MeV

Solution:
Given, Î m for 5 B 1 1 = 0.0821 u
Number of nucleons = 11
âBinding energy =( 931ÃÎm )MeV
= 9 3 1 Ã 0.081
= 75.411 MeV
Average binding energy
= Binding energy Number of nucleons
= 75.411 11
= 6.85 MeV

Topic: Physics, Nuclear Physics, NEET
If the mass of A=10  kg, coefficient of static friction =0.22, coefficient of kinetic friction =0.2, then minimum mass of B to start motion is

Options:
A 2 kg
B 2.2 kg
C 4.8 kg
D 3.4 kg

The correct answer is:
2.2 kg

Solution:
Let the minimum mass of B is MB.
Force applied by it F=MB g
Friction force on block A
f=Î¼ MA g
For motion to start
MB g=Î¼ MA g
MB=0.22Ã10
=2.2 kg

Topic: Physics, Laws of Motion, NEET
If the mass of A=10 kg, coefficient of static friction =0.22, coefficient of kinetic friction =0.2, then the minimum mass of B to start motion is

Options:
A 2 kg
B 2.2 kg
C 4.8 kg
D 3.4 kg

The correct answer is:
2.2 kg

Solution:
Let the minimum mass of B is MB.
Force applied by it F=MB g
Friction force on block A
f=Î¼ MA g
For motion to start
MB g=Î¼ MA g
MB=0.22Ã10
=2.2 kg

Topic: Physics, Laws of Motion, NEET
If the mass of neutron =1.7Ã10-27 kg, then the de-Broglie wavelength of neutron of energy 3 eV is (h=6.6Ã10-34 J-s)

Options:
A 1.6×10-16 m
B 1.6×10-11 m
C 1.4×10-10 m
D 1.4×10-11 m

The correct answer is:
1.6×10-11 m

Solution:
Given E=3 eV=3Ã1.6Ã10-19 J
We know that Î»=h2mE
=6.6Ã10-342Ã1.7Ã10-27Ã3Ã1.6Ã10-19=1.65Ã10-11 m

Topic: Physics, Dual Nature of Matter, NEET
If the mass of the block A=10 kg and the coefficient of static and kinetic friction is 0.2, then the mass of block B to start the motion is

Options:
A 2 kg
B 2.2 kg
C 4.8 kg
D 200 g

The correct answer is:
2 kg

Solution:
μs=mBmA⇒0.2=mB10⇒mB=2 kg

Topic: Physics, Laws of Motion, NEET
If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is $\frac{x}{2}$ times its original time period. Then the value of $x$ is:

Options:
A $\sqrt{2}$
B $2 \sqrt{3}$
C 4
D $\sqrt{3}$

The correct answer is:
$\sqrt{2}$

Solution:
$\begin{aligned} & T^{\prime}=2 \pi \sqrt{\frac{\ell^{\prime}}{g}} \text { where } \ell^{\prime}=\frac{\ell}{2} \\ & T=2 \pi \sqrt{\frac{\ell}{g}} \\ & T^{\prime}=\frac{x}{2} T \\ & 2 \pi \sqrt{\frac{\ell}{2 g}}=\frac{x}{2} 2 \pi \sqrt{\frac{\ell}{g}} \\ & \frac{1}{\sqrt{2}}=\frac{x}{2} \Rightarrow x=\sqrt{2}\end{aligned}$

Topic: Physics, Oscillations, NEET, NEET 2024
If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct?

Options:
A Time period of a simple pendulum on the Earth would decrease
B Walking on the ground would become more difficult
C Raindrops will fall faster
D g on the Earth will not change

The correct answer is:
g on the Earth will not change

Solution:
msâ²=ms10and Gâ²=10G
gE=Gâ²MER2=10GMER2=10g
â´ Raindrops will fall faster, Walking on the ground would become more difficult, Time period of a simple pendulum on the earth would decrease.

Topic: Physics, Gravitation, NEET, NEET 2018
If the maximum concentration of PbCl2 in water is 0.01 M at 25°C , its maximum concentration in 0.1 M NaCl will be

Options:
A 2×10-3M
B 1.6×10-2M
C 1×10-4M
D 4×10-4M

The correct answer is:
4×10-4M

Solution:
PbCl2âPb2++2Cl
Ksp=[Pb2+][Cl-]2
=(0.01)(2Ã0.01)2
Ksp=4Ã10-6
PbCl2âPb2+(aq)0S+2Cl-(aq)0.1(2S+0.1)
K sp  = [ Pb 2+ ] [Cl − ] 2
0.1+2S≈0.1
4Ã10-6=SÃ(0.1)2
S=4Ã10-4M
Maximum concentration of PbCl2 in 0.1 M NaCl is 4Ã10-4M .

Topic: Chemistry, Ionic Equilibrium, NEET
If the menstrual cycle of a female is 28 days, what will be the time period of periodic abstinence (avoiding of sex) as a natural method of contraception?

Options:
A 13th to 20th day of menstrual cycle
B 5th to 12th day of menstrual cycle
C 17th to 24th day of menstrual cycle
D 10th to 17th day of menstrual cycle

The correct answer is:
10th to 17th day of menstrual cycle

Solution:
Contraception is avoidance of pregnancy. These include Natural Methods, Barrier methods, Intrauterine devices, Chemical methods, Surgical methods and Implants. Natural methods like periodic abstinence, lactational amenorrhea and coitus interruptus prevent meeting of sperm and ovum. Though chances are less effective these methods are popular among couples. Periodic abstinence means periodic avoidance of sex during fertility period. Female is very fertile during her ovulation days which is calculated as total number of days of Menstrual Cycle - 14 days of Secretory phase. Here the menstrual cycle is of 28 days and hence ovulation day is 28-14=14 . i.e. 14th day is the ovulation day. This means ovum is in the oviduct and hence chances of fertilization if the sperm enters is the maximum. Thus one should avoid coitus during this period. Just to be on the safer side two or three days before and after ovulation is also considered as the period to avoid sexual intercourse.

Topic: Biology, Reproductive Health, NEET
If the molecular mass of an amino acid is 150 daltons, the molecular mass of a tripeptide will be

Options:
A 414

B 486

C 504

D 450

The correct answer is:
414

Solution:
Tripeptide is formed by the condensation of three amino acids with the loss of two water molecules. The molecular mass of two water molecules is 18x2=36. Therefore, tripeptide mass will be (150x3)-36=414.

Topic: Chemistry, Biomolecules, NEET
If the monochromatic source in Young's double slit experiment is replaced by white light, then

Options:
A There will be a central dark fringe surrounded by a few coloured fringes
B There will be a central bright white fringe surrounded by a few coloured fringes
C All bright fringes will be of equal width
D Interference pattern will disappear

The correct answer is:
There will be a central bright white fringe surrounded by a few coloured fringes

Solution:
At central point on screen, path difference is zero for all wavelength. So, central bright fringe is white and other fringes depend on wavelength as $\beta=\frac{\lambda D}{d}$.
Therefore, other fringes will be coloured.

Topic: Physics, Wave Optics, NEET, NEET 2024
If the mother is Rh-ve and the father is Rh+ve, the condition of erythroblastosis foetalis may occur. This condition may be avoided by administrating anti-Rh antibodies to

Options:
A foetus, just after the birth.
B mother, just after the birth.
C mother, just before the birth.
D both foetus and mother, just after the birth.

The correct answer is:
mother, just after the birth.

Solution:
If a Rh-ve pregnant mother is having a Rh+ve foetus, then Rh antigens of the foetus are not exposed to the Rh-ve blood of the mother in the first pregnancy because the two bloods are well separated by the placenta. But during the delivery of the first Rh+ve child, some amount of foetal blood is exposed to the maternal blood. In this condition, the mother starts preparing antibodies against Rh antigens in her blood. These antibodies from the mother (Rh-ve) can leak into the blood of the foetus (Rh+ve) during next Rh+ve pregnancies and destroy the foetal RBCs.

Topic: Biology, Body Fluids and Circulation, NEET
If the nitrogen atom had electronic configuration 1s7 it would have energy lower than that of the normal ground state configuration 1s2 2s22p3 because the electrons would be closer to the nucleus. Yet 1s7 is not observed. It violates

Options:
A Heisenberg's uncertainty principle
B Hund's rule
C Pauli exclusion principle
D Bohr postulate of stationary orbits

The correct answer is:
Pauli exclusion principle

Solution:
Not more than two electrons can be present in same atomic orbital. This is Paulis exclusion principle.

Topic: Chemistry, Structure of Atom, NEET
If the nuclear radius of ${ }^{27} \mathrm{Al}$ is $3.6 \mathrm{Fermi}$, the approximate nuclear radius of ${ }^{64} \mathrm{Cu}$ in Fermi is

Options:
A 2.4
B 1.2
C 4.8
D 3.6

The correct answer is:
4.8

Solution:
Nuclear radius $r \propto A^{1 / 3}$, where $A$ is mass number
$\begin{aligned}
& r=r_0 A^{1 / 3} \\
& r=r_0(27)^{1 / 3}=3 r_0 \\
& r_0=\frac{3.6}{3}=1.2 \mathrm{fm}
\end{aligned}$
For ${ }^{64} \mathrm{Cu}$
$\begin{aligned}
r & =r_0 A^{1 / 3} \\
& =1.2 \mathrm{fm}(64)^{1 / 3} \\
& =4.8 \mathrm{fm}
\end{aligned}$

Topic: Physics, Nuclear Physics, NEET, NEET 2012 (Screening)
If the nucleus ${ }_{13}^{27} \mathrm{Al}$ has a nuclear radius of about $3.6 \mathrm{fm}$, then ${ }_{32}^{125} \mathrm{Te}$ would have its radius approximately as:

Options:
A $9.6 \mathrm{fm}$
B $12.0 \mathrm{fm}$
C $4.8 \mathrm{fm}$
D $6.0 \mathrm{fm}$.

The correct answer is:
$6.0 \mathrm{fm}$.

Solution:
Nuclear radii \(R=\left(R_0\right) A^{1 / 3}\)
where, \(A\) is the mass number.
\(\frac{R_{\mathrm{Te}}}{R_{\mathrm{Al}}}=\left(\frac{A_{\mathrm{Te}}}{A_{\mathrm{Al}}}\right)^{1 / 3}=\left(\frac{125}{27}\right)^{1 / 3}=\left(\frac{5}{3}\right)\)
or, \(\quad R_{\mathrm{Te}}=\frac{5}{3} \times R_{\mathrm{Al}}=\frac{5}{3} \times 3.6=6 \mathrm{fm}\).
(Given \(R_{\mathrm{Al}}=3.6 \mathrm{fm}\))

Topic: Physics, Nuclear Physics, NEET, NEET 2007
If the number of bivalents is 10 in prophase - I, what will be the number of the chromosomes during late anaphase - II in each cell?

Options:
A 10 with two chromatids each
B 10 with one chromatid each
C 20 with two chromatids each
D 20 with one chromatid each

The correct answer is:
10 with one chromatid each

Solution:
Cell with 10 bivalents consists of 20 chromosomes, thus during anaphase I disjunction of chromosomes results in reduction of chromosome number to half as a result each cell after meiosis I will have 10 chromosomes. During anaphase II centromere splits and chromatids are separated. Thus at the end of anaphase II each cell consists of 10 chromosome with one chromatid.

Topic: Biology, Cell Cycle and Cell Division, NEET
If the number of chromosome present in the egg of a gymnosperm plant is 10, then the number of chromosomes present in the endosperm of the same plant is

Options:
A 30
B 20
C 10
D 40

The correct answer is:
10

Solution:
Endosperm is a tissue formed before the fertilization occurs and is developed from in the ovule during the development of the megaspore in Gymnosperms. This tissue surrounds the developing embryo and provides nourishment to it, as it is its function. The endosperm in gymnosperm is haploid unlike in angiosperm where it is triploid as the endosperm develops from the haploid female gametophyte. The chromosome number of endosperm in gymnosperm will be same as that of the egg.

Topic: Biology, Plant Kingdom, NEET