datavorous/entrance-exam-dataset huggingface dataset:
A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρl . The cork is depressed slightly and then released.What is the time period of the simple harmonic oscillation of the cork. where ρ is the density of cork

Options:
A $T=2\pi \sqrt{\frac{h\rho }{{{\rho }_{1}}g}}$
B $T=\pi \sqrt{\frac{2h\rho }{{{\rho }_{1}}g}}$
C $T=2\pi \sqrt{\frac{2h\rho }{{{\rho }_{1}}g}}$
D $T=\pi \sqrt{\frac{h\rho }{{{\rho }_{1}}g}}$

The correct answer is:
$T=2\pi \sqrt{\frac{h\rho }{{{\rho }_{1}}g}}$

Solution:
Correct Option is : (A)
$T=2\pi \sqrt{\frac{h\rho }{{{\rho }_{1}}g}}$

Topic: Chemistry, Oscillations, NEET
A cylindrical rod of length l and cross-sectional radius r is placed at a distance of 50 r from a infrared point source S of power 1.25 kW as shown in the figure. The lateral surface of the rod is perfectly insulated from the surroundings. If the cross-section A absorbs 80 % of the incident energy and the temperature difference between the ends of the rod is constant, then the rate of heat flow through the rod in steady state is

Options:
A 0.2 J/s
B 0.125 J/s
C 0.1 J/s
D 0.25 J/s

The correct answer is:
0.1 J/s

Solution:
The rate at which energy is absorbed at A,

PA=Psource4π50r2×πr2×0.8=0.1 W

Since the rod is in steady-state, energy absorbed per second at A flows through to B.

So heat flow per sec is 0.1 J/s

Topic: Physics, Thermal Properties of Matter, NEET
A cylindrical vessel contains a liquid of density Ï up to a height h. The liquid is closed by a piston of mass m and area of cross-section A. There is a small hole at the bottom of the vessel. The speed v with which the liquid comes out of the hole is

Options:
A 2gh
B 2gh+mgρA
C 2gh+mgA
D 2gh+mgA

The correct answer is:
2gh+mgρA

Solution:
Applying Bernoulli's theorem at 1 and 2 : difference in pressure energy between 1 and 2 = difference in kinetic energy between 1 and 2
or Ïgh + mgA = 12Ïv2
or v=2gh+2mgÏA= 2gh+mgÏA

Topic: Physics, Mechanical Properties of Fluids, NEET
A cylindrical vessel made of thermally insulating material is divided into two equal parts by an insulating and movable piston. Both parts contain ideal monoatomic gas. The gas in the left part is supplied heat such that the volume of right part becomes one-eight of its initial volume. Work done by the gas in the right part is

Options:
A -92 PV
B 92 PV
C –3PV
D 3PV

The correct answer is:
-92 PV

Solution:
ÎQ=ÎU+Â ÎWÂ
Î³=53
TVÎ³â1=T′V′Î³â1
TV53-1=T′V853-1
T′=4T
ÎU=nCvÎT=n3R2(4T-T)=92nRT
ÎQ=ÎU+ÎW=0
ÎW=âÎU
=-92nRT

Topic: Physics, Thermodynamics, NEET
A cylindrical vessel of 90 cm height is filled upto the brim. It has four holes 1, 2 , 3, 4 which are, respectively, at heights of 20 cm, 30 cm, 45 cm and 80 cm from the horizontal floor PQ. The water falling at the maximum horizontal distance from the vessel comes from

Options:
A hole number 4
B hole number 3
C hole number 2
D hole number 1

The correct answer is:
hole number 3

Solution:
The maximum horizontal distance from the vessel comes from hole number 3
v=2gh â h is height of hole from top
horizontal distance x=vt=2gh 2H-hg
x=2 hH-h

Topic: Physics, Mechanical Properties of Fluids, NEET
A cylindrical vessel of radius R and height H and open at the top is completely filled with water. A small circular hole of radius r is made near the bottom of vessel. The time taken for 25 % of water to flow out is

Options:
A 2Hg 3-1
B R2r2 2Hg 1+23
C R2r2 2Hg 1-32
D r2R2 2Hg 3-1

The correct answer is:
R2r2 2Hg 1-32

Solution:
T=Aa 2Hg=R2r2 2Hg
t1=R2r2 2g×3H4
∴ Δt=T-t1=R2r2 2Hg 1-32

Topic: Physics, Mechanical Properties of Fluids, NEET
A cylindrical wire of radius R has current density varying with distance r from its axis as Jr=J01-r2R2. The total current through the wire is

Options:
A πJ0R22
B 2πJ0R23
C 4πJ0R23
D none of these

The correct answer is:
πJ0R22

Solution:
I=â«0RJ 2Ïrdr
I=ÏJ0R22

Topic: Physics, Current Electricity, NEET
A daniel cell is balanced at 125 cm length of a potentiometer wire. When the cell is circuited by a resistance of 2 Î©, the balance length is obtained at 100 cm. The internal resistance of the Daniel cell is

Options:
A 43Ω
B 1.5Ω
C 1.25Ω
D 0.5Ω

The correct answer is:
0.5Ω

Solution:
r=Rlll2-1
=2125100-1
=254-1
=2Ã14
=0.5 Î©

Topic: Physics, Current Electricity, NEET
A decay chain of the nucleus U92238 involves eight α -decays and six β -decays. The final nucleus at the end of the process will be

Options:
A Z=84;A=206
B Z=84;A=224
C Z=82;A=206
D Z=82;A=200

The correct answer is:
Z=82;A=206

Solution:
(i) When α -particle is emitted from a radioactive nucleus then atomic number decreases by 2 and atomic mass decreases by 4 .
(ii) When β -particle emitted from a radioactive nucleus then atomic number is increased by 1 and atomic mass unaffected.
Here, atomic number Z=92
After emitting 8 α -particles and 6β -particles,
Z=92-2×8+6=82
and atomic mass, A=238
∴A=238-4×8=206

Topic: Physics, Nuclear Physics, NEET
A decrease in blood pressure/volume will not cause the release of

Options:
A Renin
B Atrial Natriuretic Factor
C Aldosterone
D ADH

The correct answer is:
Atrial Natriuretic Factor

Solution:
A decrease in blood pressure/volume stimulates the hypothalamus to release ADH (Anti-Diuretic Hormone) as well as JGA (Juxtaglomerular Apparatus) cells to release renin. Renin by the renin-angiotensin mechanism activates the adrenal cortex to release aldosterone. Atrial Natriuretic Factor (ANF) is produced by the atria of a heart during increased blood pressure/volume. It can cause vasodilation and thereby, decrease blood pressure.

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2017
A decrease in intrapulmonary pressure to less than atmospheric pressure causes

Options:
A inspiration
B expiration
C emphysema
D infarction

The correct answer is:
inspiration

Solution:
Intrapleural pressure is the pressure of the air within the pleural cavity, between the visceral and parietal pleurae.
Air flows into the lungs largely due to a difference in pressure; atmospheric pressure is greater than intra-alveolar pressure, and intra-alveolar pressure is greater than intrapleural pressure. Air flows out of the lungs during expiration based on the same principle; pressure within the lungs becomes greater than the atmospheric pressure.

Topic: Biology, Breathing and Exchange of Gases, NEET
A deep rectangular pond of surface area $\mathrm{A}$, containing water (density $=\rho$, specific heat capacity $=s$ ), is located in a region where the outside air temperature is a steady value at the $-26^{\circ} \mathrm{C}$. The thickness of the frozen ice layer in this pond, at a certain instant is $\mathrm{x}$. Taking the thermal conductivity of ice as $\mathrm{K}$, and its specific latent heat of fusion as L, the rate of increase of the thickness of ice layer, at this instant would be given by

Options:
A $26 \mathrm{~K} / \rho r(\mathrm{~L}-4 \mathrm{~s})$
B $26 \mathrm{~K} /\left(\rho x^2-L\right)$
C $26 K /(\rho x L)$
D $26 K / \rho r(L+4 s)$

The correct answer is:
$26 K /(\rho x L)$

Solution:
Key Idea If area of cross-section of a surface is not uniform or if the steady state condition is not reached, the heat flow equation can be applied to a thin layer of material perpendicular to direction of heat flow.
The rate of heat flow by conduction for growth of ice is given by,
$$
\frac{d \theta}{d t}=\frac{\left.K A\left(\theta_0-\theta_0\right)\right]}{x}
$$
where, $d \theta=\rho A d x L, \theta_0=0$ and $\theta_1=-\theta$
Given, $\theta_0=0^{\circ} \mathrm{C}, \theta_1=-26^{\circ} \mathrm{C}$
The rate of increase of thickness can be calculated from Eq.
$$
\begin{aligned}
\frac{\mathrm{d} \theta}{\mathrm{dt}} & =\frac{\mathrm{KA}\left(\theta_0-\theta_1\right)}{\mathrm{x}} \\
\Rightarrow \quad \frac{\rho \mathrm{AdxL}}{\mathrm{dt}} & =\frac{\mathrm{KA}\left(\theta_0-\theta_1\right)}{\mathrm{x}} \\
\Rightarrow \quad \frac{\mathrm{dx}}{\mathrm{dt}} & =\frac{\mathrm{KA}\left(\theta_0-\theta_1\right)}{\rho \mathrm{AxL}} \\
& =\frac{\mathrm{K}[0-(-26)]}{\rho \times \mathrm{L}}=\frac{26 \mathrm{~K}}{\rho \times \mathrm{L}}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2019 (Odisha)
A deficiency of which vitamin can lead to the disease known as rickets?

Options:
A Vitamin A
B Vitamin C
C Vitamin D
D Vitamin K

The correct answer is:
Vitamin D

Solution:
Rickets is a condition characterized by weakened and deformed bones, often caused by a deficiency of vitamin D, which is important for calcium absorption and bone health.

Topic: Chemistry, Biomolecules, NEET
A dehydration reaction links two glucose molecules to produce maltose. If the formula for glucose is C6H12O6 then what is the formula for maltose?

Options:
A C12H24O12
B C12H22O11
C C12H24O11
D C12H20O10

The correct answer is:
C12H22O11

Solution:
Maltose is a common disaccharide that is formed by the dehydration reaction of two glucose molecules. The glycosidic bond present in maltose is α-1,4. During the formation, from two glucose molecules, one molecule of water is removed.

Thus, the formula of maltose becomes C12H22O11

Topic: Biology, Biomolecules (B), NEET, NEET 2022 (Phase 1)
A deuteron is bombard on ${ }_8 \mathrm{O}^{16}$ nuclei than $\alpha$-particle is emitted then the product nuclei is:

Options:
A ${ }_7 \mathrm{~N}^{13}$
B ${ }_5 \mathrm{~B}^{10}$
C ${ }_4 \mathrm{Be}^9$
D ${ }_7 \mathrm{~N}^{14}$

The correct answer is:
${ }_7 \mathrm{~N}^{14}$

Solution:
${ }_8 \mathrm{O}^{16}+{ }_1 \mathrm{H}^1 \longrightarrow{ }_2 \mathrm{X}^4+{ }_2 \mathrm{He}^4$
Use conservation of charge and mass
The conservation of mass number gives
$$
16+2=A+4 \Rightarrow A=14
$$
The conservation of atomic number gives.
$$
8+1=2+2 \Rightarrow 2=7
$$
Therefore, ${ }_2 \mathrm{X}^4$ is ${ }_7 \mathrm{~N}^{14}$

Topic: Physics, Nuclear Physics, NEET, NEET 2002
A device is inserted in the uterus for contraceptive effect. Which of these effects won't be caused by it?

Options:
A Increasing phagocytosis of sperms
B Suppressing motility of sperms
C Preventing implantation of the blastocyst
D Preventing ovulation

The correct answer is:
Preventing ovulation

Solution:
Intrauterine devices are contraceptives inserted by doctors or expert nurses in the uterus through the vagina. Copper releasing IUDs such as the Copper T act by increasing phagocytosis of sperms within the uterus and the Copper ions released suppress sperm motility and the fertilizing capacity of sperms. Ovulation is controlled by FSH and LH hormones. Copper T doesn't prevent ovulation.

Topic: Biology, Reproductive Health, NEET
A device that converts energy of combustion of fuels like hydrogen and methane, directly into electrical energy is known as:

Options:
A Fuel cell
B Electrolytic cell
C Dynamo
D Ni-Cd cell

The correct answer is:
Fuel cell

Solution:
A fuel cell is a device that converts the energy of combustion of fuels like hydrogen and methane, directly into electrical energy
Electrolytic cell converts electrical energy into chemical energy.
Dynamo is an electrical generator that produces direct current with the use of a commutator.

Topic: Chemistry, Electrochemistry, NEET, NEET 2015 (Phase 1)
A device that converts energy of combustion of fuels like hydrogen and methane, directly into electrical energy is known as

Options:
A Electrolytic cell
B Dynamo
C Ni-Cd cell
D Fuel Cell

The correct answer is:
Fuel Cell

Solution:
A device that converts energy of combustion of fuels, directly into electrical energy is known as fuel cell.
Example H2+O2 fuel cell.

Topic: Chemistry, Electrochemistry, NEET
A diabetic person carries a packet of glucose with him always, because

Options:
A glucose increases the blood sugar level slowly
B glucose reduces the blood sugar level
C glucose increases the blood sugar level almost instantaneously
D glucose reduces the blood sugar level slowly

The correct answer is:
glucose increases the blood sugar level almost instantaneously

Solution:
The sugar level in the blood of a diabetic person may suddenly fall. Immediate intake of glucose increases the blood sugar level almost instantaneously.

Topic: Chemistry, Biomolecules, NEET
A diagnostic trait for identification of Fabaceous flower is

Options:
A tetradynamous androecium.
B inferior ovary.
C perianth.
D vexillary aestivation in corolla.

The correct answer is:
vexillary aestivation in corolla.

Solution:
Fabaceae, i.e. Leguminosae is the family of plant having legumes, peas group in them. This group is the third largest group and economically important family of flowering plants. In this group the following floral characteristics are observed.
The flower is bisexual and zygomorphic with superior ovary and diadelphous anther. The sepals are fused and show imbricate or valvate aestivation. Corolla with 5 petals show vexillary aestivation. Aestivation is the arrangement of the petals/sepals around the flower. In this pattern, the anterior petal is innermost and posterior petal is outermost and largest. There is large petal which has two lobes fused. There are two wings and one small keel or carina. This aestivation is also called descending imbricate.

Topic: Biology, Morphology of Flowering Plants, NEET
A diagram of areolar connective tissue is given below.

Identify the option with correct labeling.

Options:
A P = Fibroblast, R = Collagen fibres
B R = Collagen fibres, S = Macrophages
C Q = Fibroblast, S = Mast cell
D P = Macrophage, S = Fibroblast.

The correct answer is:
Q = Fibroblast, S = Mast cell

Solution:
Areolar tissue possesses fibroblast cells, mast cells and macrophages. It acts as a supporting framework for epithelia. It is located beneath the skin as a panniculus carnosus.

Topic: Biology, Structural Organisation in Animals, NEET
A diagram showing axon terminal and synapse is given. Identify correctly at least two of $A-D$

Options:
A A-Receptor C-Synaptic vesicles
B B-Synaptic connection $D-K^{+}$
C A-Neurotransmitter B-Synaptic cleft
D C-Neurotransmitter $\mathrm{D}-\mathrm{Ca}^{2+}$

The correct answer is:
A-Receptor C-Synaptic vesicles

Solution:
A-receptor, B-synaptic cleft, C-synaptic vesicles and $D-\mathrm{Ca}^{2+}$.

Topic: Biology, Neural Control and Coordination, NEET, NEET 2013 (All India)
A diamagnetic material in a magnetic field moves:

Options:
A from stronger to the weaker parts of the field
B from weaker to the stronger parts of the field
C perpendicular to the field
D in none of the above directions

The correct answer is:
from stronger to the weaker parts of the field

Solution:
A diamagnetic material in a magnetic field moves from stronger to the weaker part of the field.

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2003
A diatomic gas belonging to group 15 combines with a halogen to form a trihalide which is fairly stable and inert. The trihalide is

Options:
A NCl3
B PCl3
C BiF3
D NF3

The correct answer is:
NF3

Solution:
It is Nitrogen gas and is diatomic in nature. Its halide NF3 is fairly stable where as NCl3 is unstable

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
A diatomic gas is used in carnot heat engine having efficiency 80%. Find the ratio of initial volume to final volume of gas during adiabatic expansion.

Options:
A 1532
B 1352
C 1552
D 1525

The correct answer is:
1552

Solution:
η=1-T2T1

0.8=1-T2T1

T2T1=0.2

T2V2γ-1=T1V1γ-1

V2V1γ-1=T1T2=5

V1V2=1552

Topic: Physics, Thermodynamics, NEET
A diatomic ideal gas undergoes a thermodynamic change according to the P-V diagram shown in the figure. The total heat given to the gas is nearly (use ln2=0.7 )

Options:
A 2.5P0V0
B 1.4P0V0
C 39P0V0
D 1.1P0V0

The correct answer is:
39P0V0

Solution:
QAB=ÎUAB+WAB, where WAB=0
ÎUAB=f2nRÎT
âÎUAB=f2ÎPV
âÎUAB=52ÎPV
âÎUAB=2.5ÎPV
According to the first law of thermodynamics QAB=ÎUAB+WAB
Where WAB=0
therefore QAB=ÎUAB
âQAB=2.5P0V0
Process BC
QBC=ÎUBC+WBC
â QBC=0+2P0V0ln2
QBC=1.4P0V0
Qnet=QAB+QBC=3.9P0V0

Topic: Physics, Thermodynamics, NEET
A diatomic molecule has a dipole moment of 1.2 D. If its bond length is equal to 10-10 m then the fraction of an electronic charge on each atom will be

Options:
A 45 %
B 55 %
C 75 %
D 25 %

The correct answer is:
25 %

Solution:
µ=q×d
1.2×10-18=q×10-8cm
q=1.2×10-1810-8=1.2×10-10
% charge =qe×100=1.2×10-104.8×10-10×100=25%

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
A diatomic molecule is formed by two atoms which may be treated as mass points m1 and m2 joined by a massless rod of length r. Then the moment of inertia of molecule about an axis passing through centre of mass and perpendicular to the rod is

Options:
A Zero
B m1+m2r2
C m1m2m1+m2r2
D m1+m2m1m2r2

The correct answer is:
m1m2m1+m2r2

Solution:
With respect to centre of mass,
Effective mass =m1m2m1+m2
∴I=m1m2m1+m2r2

Topic: Physics, Rotational Motion, NEET
A dicotyledonous plant bears flowers but never produces fruits and seeds. The most probable cause for the above situation is

Options:
A Plant is dioecious and bears only pistillate flowers.
B Plant is dioecious and bears both pistillate and staminate flowers
C Plant is monoecious
D A plant is dioecious and bears only staminate flowers

The correct answer is:
A plant is dioecious and bears only staminate flowers

Solution:
Dioecious plants are those which have separate male and female individuals. The male plant bears only staminate flowers. It produces pollen but lacks ovary thus never produces fruit.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
A difference of 2.3 eV separates two energy levels in an atom. The frequency of radiation emitted when the electron makes a transition is

Options:
A 6.7 × 1014 Hz

B 4.6 × 1014 Hz

C 5.6 Ã 1014 Hz

D 7.7 × 1014 Hz

The correct answer is:
5.6 Ã 1014 Hz

Solution:
Given, the difference in energy level E = 2.3 eV
= 2.3 Ã 1.6 Ã 10-19 J
âPlanck's constant h = 6.63 Ã 10-34 J s
Let v be the frequency, then E = hf
or f=Eh=2.3Ã1.6Ã10-196.63Ã10-34
= 5.6 Ã 1014 Hz

Topic: Physics, Atomic Physics, NEET
A dikaryon is formed when

Options:
A Meiosis is arrested
B The two haploid cells do not fuse immediately
C Cytoplasm does not fuse
D None of the above

The correct answer is:
The two haploid cells do not fuse immediately

Solution:
When two haploid cells do not fuse immediately, dikaryon is formed.

Topic: Biology, Biological Classification, NEET
A dikaryotic stage is observed in

Options:
A Phycomycetes
B Zygomycetes
C Deuteromycetes
D Basidiomycetes

The correct answer is:
Basidiomycetes

Solution:
In Ascomycetes and Basidiomycetes, plasmogamy is not immediately followed by karyogamy as a result a dikaryotic cell is produced. In ascomycetes, the dikaryotic cell divides to form small dikaryotic hyphae known as ascogenous hyphae, which produces ascospores, in Basidiomycetes the dikaryotic cell divides to form a well developed dikaryotic mycelium, which produces basidium at a later stage.

Topic: Biology, Biological Classification, NEET
A dilute aqueous solution of Na2SO4 is electrolysed using platinum electrodes. The product at the anode and cathode are

Options:
A O2,H2
B S2O82-,Na
C O2,Na
D S2O82-,H2

The correct answer is:
O2,H2

Solution:
During the electrolysis of aq. Na2SO4,H2 is evolved at the cathode as ER.P. of sodium (-2.71 V) is much lower than that of water (-0.83 V) while O2 is evolved at the anode as E0R.P. potential of SO42- ion (-0.2 V) is much more than that of water (-1.23 V).

Topic: Chemistry, s Block Elements, NEET
A dilute aqueous solution of N a 2 S O 4 is electrolyzed using platinum electrodes. The products at the anode and cathode are :

Options:
A

S 2 O 8 2â ,ââ H 2

B

S 2 O 8 2− ,  Na

C

O 2 ,  Na

D

O 2 ,ââ H 2

The correct answer is:

O 2 ,ââ H 2

Solution:
H 2 O is more readily reduced at cathode than N a + . It is also more readily oxidized at anode than S O 4 2â . Hence, the electrode reactions are

2 H 2 O+2 e â â H 2 â+2O H â ââ[ atââcathode ] âââââââââââââââ H 2 O â 1 2 O 2 â+2 H + +2 e â ââ[ atââanode ]

Topic: Chemistry, Electrochemistry, NEET
A dilute solution of H2SO4with platinum electrodes: What is the product formed at the cathode during the electrolysis of a dilute solution of H2SO4with platinum electrodes?

Options:
A Hydrogen gas (H2)
B Oxygen gas (O2)
C Solid sulfur (S)
D Sulfate ions (SO4²⁻)

The correct answer is:
Hydrogen gas (H2)

Solution:
Correct Option is : (A)
Hydrogen gas (H2)

Topic: Chemistry, Redox Reactions, NEET
A dioecious flowering plant prevents both:

Options:
A Autogamy and xenogamy
B Autogamy and geitonogamy
C Geitonogamy and xenogamy
D Cleistogamy and xenogamy

The correct answer is:
Autogamy and geitonogamy

Solution:
When unisexual male and female flowers are present on different plants, the condition is called dioecious, and it prevents both autogamy and geitonogamy. Autogamy or self-fertilization refers to the fusion of two gametes that come from one individual. Xenogamy is the fertilization of a flower by pollen from a flower on a genetically different plant. Geitonogamy is considered if a first flower is pollinated by the pollen of the second flower of the same plant. Cleistogamy is self-fertilization that occurs within a permanently closed flower.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2017
A dip needle lies initially in the magnetic meridian when it shows an angle of dip θ at a place. The dip circle is rotated through an angle x in the horizontal plane and then it shows an angle of dip θ′ . Then tan⁡ θ′tan⁡ θ is

Options:
A 1cos⁡x
B 1sin⁡x
C 1tan⁡x
D cos⁡x

The correct answer is:
1cos⁡x

Solution:
In the magnetic meridian, true dip Î¸ is given by,
tanâ¡Î¸=BVBH
When the dip circle is rotated through an angle x in the horizontal plane, Apparent dip Î¸â² is given by.
tanâ¡Î¸â²=BVBHcosâ¡x
â´ tanâ¡Î¸â²tanâ¡Î¸=1cosâ¡x

Topic: Physics, Magnetic Properties of Matter, NEET
A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of vibration is found to be 2 s. The same needle is then allowed to vibrate in the horizontal plane, and the time period is again found to be 2 s. Then the angle of dip is

Options:
A 0Â°
B 30°
C 45°
D 90Â°

The correct answer is:
45°

Solution:
Time period when dip needle vibrates in the vertical plane perpendicular to the magnetic meridian is,
T1=2ÏIMBV
The time period of vibration when needle is allowed to vibrate in the horizontal plane is,
T2=2ÏIMBH
as T1=T2=2 s
â´BV=BH
Angle of dip is,
tanâ¡Î¸=BvBH=1
Î¸=45Â°

Topic: Physics, Magnetic Properties of Matter, NEET
A diploid living organism develops from zygote by which type of the following repeated cell divisions?

Options:
A Meiosis
B Amitosis
C Mitosis
D Segmentation

The correct answer is:
Mitosis

Solution:
The zygote is formed by the fusion of male and female gametes. Hence a zygote is diploid in nature. The male and female gametes are formed by meiosis in a diploid organism. A diploid living organism develops from zygote by repeated mitotic divisions.

Topic: Biology, Cell Cycle and Cell Division, NEET
A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?

Options:
A 2
B 4
C 8
D 6

The correct answer is:
8

Solution:
If an entity is heterozygous at four loci with four different characters, then while meiotic division, they split to form 8 different gametes.

Topic: Biology, Principles of Inheritance and Variation, NEET
A dipole is placed in an electric field as shown. In which direction will it move?

Options:
A towards the left as its potential energy will increase.
B towards the right as its potential energy will decrease.
C toward, the left as its potential energy will decrease.
D towards the right as its potential energy will increase.

The correct answer is:
towards the right as its potential energy will decrease.

Solution:
F1=qE1,F2=qE2

E1>E2⇒F1>F2

Hence, net force is towards right and its potential energy will decrease.
Correction option is (2)

Topic: Physics, Electrostatics, NEET, NEET 2021
A disc has mass M and radius R. How much tangential force should be applied to the rim of the disc so as to rotate the disc with angular velocity ω in time t?

Options:
A MRω4t
B MRω2t
C MRωt
D MRωt

The correct answer is:
MRω2t

Solution:
Ï=IÎ±
FÃR= M R 2 2 Ã Ï t
F= MR 2 Ã Ï t

Topic: Physics, Rotational Motion, NEET