datavorous/entrance-exam-dataset huggingface dataset:
The three boxes in this diagram represent the three major biosynthetic pathways in aerobic respiration. Arrows represent net reactants or products
Arrows numbered 4,8 and 12 can all be

Options:
A $\mathrm{NADH}$
B ATP
C $\mathrm{H}_2 \mathrm{O}$
D $\mathrm{FAD}^{+}$orFADH ${ }_2$

The correct answer is:
ATP

Solution:
Pathway $A$ is glycolysis, pathway $B$ is the krebs' cycle and pathway $C$ is oxidative phosphorylation
Arrow 1 - ADP or $\mathrm{NAD}^{+}$
Arrow 2 - Pyrunate
Arrow 3 - NADH
Arrow 4 - ATP
Arrow 5 - ADP, $\mathrm{NAD}^{+}$or $\mathrm{FAD}$
Arrow 6 and 7 - $\mathrm{FADH}_2$ and $\mathrm{NADH}$ (either one can be 6 or 7 )
Arrow 8 - ATP or $\mathrm{CO}_2$
Arrow 9 and $10-\mathrm{O}_2$ and ADP (either one can be 9 or 10 )
Arrow 11 and $12-\mathrm{H}_2 \mathrm{O}$ and ATP (either one can be 11 or 12)

Topic: Biology, Respiration in Plants, NEET, NEET 2013 (All India)
The three distances AB,AC and A A ′ in the given BCC lattice are :

Options:
A a , 2 a , 3 a 2
B a , 3 a 2 , 2 a
C 3 a 2 , 2 a , a
D a , a 2 , 3 a 2

The correct answer is:
a , 2 a , 3 a 2

Solution:
Edge length =AB=AD=BC=CD=a
â´ AC = AB 2 + BC 2 = 2 a

Body diagonal is AG.

Consider a triangle âACG

AG= AC 2 + CG 2
= 2a 2 + a 2 â
= 3 a
Now, AA â² + A â² G = 2 AA â² = 3 a
â´ AA â² = 3 a 2

Topic: Chemistry, Solid State, NEET
The three phases of the uterine cycle in order are

Options:
A Menstrual → Luteal Ovulatory
B Follicular → Luteal → Ovulatory
C Menstrual → Proliferative → Ovulatory
D Ovulatory → Follicular → Luteal

The correct answer is:
Menstrual → Proliferative → Ovulatory

Solution:
Menstrual cycle is the reproductive cycle in females. The name is derived from the word 'mesum' meaning one month. Thus this cycle is repeated every month. This cycle includes changes in the ovary and the uterus simultaneously.The beginning of the cycle is the bleeding phase also known as Menstrual phase. This continues for 3-5 days. During this phase there is shedding of endometrium and unfertilisedovum.This is followed by follicular phase in ovary which shows formation of new follicles and proliferative phase in the uterus that begins repair of endometrium. These phases are then followed by ovulation.Following this is the luteal phase in ovary that shows formation of corpus luteum and secretory phase in the uterus which shows formation of secretory glands as the uterus is preparing itself to receive the fertilised ovum. Fertilisation will mark the cessation of the cycle temporarily however if no fertilisation is seen the same cycle is repeated.Thus the phases are
Menstrual → Proliferative/Follicular → Ovulatory → Secretory/Luteal

Topic: Biology, Human Reproduction, NEET
The three tiny bones present in middle ear are called ear ossicles. Write them in the correct sequence beginning from ear drum.

Options:
A Incus, stapes, malleus
B Malleus, incus, stapes
C Stapes, malleus, incus
D Stapes, incus, malleus

The correct answer is:
Malleus, incus, stapes

Solution:
Malleus, incus, stapes are the ear ossicles.

Topic: Biology, Locomotion and Movement, NEET
The threshold frequency ν0for a metal is 7.0 ×1014s–1. Calculate the kinetic energy of an electron emitted when radiation of frequency ν =1.0 ×1015s–1hits the metal.

Options:
A 3.533 ×10–19J
B 2.658 ×10–19J
C 1.988 ×10–19J
D 0.421 ×10–19J

The correct answer is:
1.988 ×10–19J

Solution:
Correct Option is : (C)
1.988 ×10–19J

Topic: Chemistry, Structure of Atom, NEET
The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV and the stopping potential for a radiation incident on this surface is 5 V. The incident radiation lies in

Options:
A ultra-violet region
B infra-red region
C visible region
D x-ray region

The correct answer is:
ultra-violet region

Solution:
hv0=6.2 eV, eV0=5 eV
from Einstein's photoelectric equation
hv=hv0+eV0= 6.2+5 = 11.2 eV
â hcÎ»=11.2
â´ Î»=hc11.2=1108.9â«
It belongs to ultra-violet region.

Topic: Physics, Dual Nature of Matter, NEET
The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV and the stopping potential for a radiation incident on this surface is 5 V . The incident radiation lies in

Options:
A ultra-violet region
B infrared region
C visible region
D x- ray region

The correct answer is:
ultra-violet region

Solution:
We know that
E=w0+KEmax

hcÎ»=5 eVâ¡+6.2 eVâ¡=11.2 eV
Î»=1240011â2âââ«
Î»=1107âââ«

This lies in Ultra Violet light Region.

Topic: Physics, Dual Nature of Matter, NEET
The threshold frequency for certain metal is 3.3Ã1014 Hz . If the light of 8.2Ã1014 Hz frequency is incident on the metal, then the cut - off voltage of the photoelectric current will be

Options:
A 3 V

B 4 V

C 2 V

D 2.5 V

The correct answer is:
2 V

Solution:
In the photoelectric effect, energy is conserved
In the photoelectric effect, Einstein's equation is given by
Photon energy = kinetic energy of electron + work function
i.e. hν=eVS+hν0
where, VS is the stopping potential and ν0 is the threshold frequency
∴ VS=heν-ν0 .... (i)
Given, ν0=3.3×1014 Hz, ν=8.2×1014 Hz
h=6.6×10-34 J s , e=1.6×10-19C
Substituting these values in equation (i). we get
VS=6.6×10-348.2×1014-3.3×10141.6×10-19
VS=6.6×4.91.6×10-1=2 V
⇒ VS=2 V

Topic: Physics, Dual Nature of Matter, NEET
The threshold frequency for certain metal is 3.3×1014 Hz. If the light of frequency 8.2×1014 Hz is incident on the metal, the cut-off voltage of the photoelectric current will be

Options:
A 4.9 V

B 3.0 V
C 2.0 V
D 1 V

The correct answer is:
2.0 V

Solution:
From relation
eVs=hv-v0
or Vs= threshold or cut off voltage
=hev-v0
= 6.6×10-341.6×10-19 (8.2-3.3)×1014
= 6.6×4.9×10-11.6 = 2 V

Topic: Physics, Dual Nature of Matter, NEET
The threshold frequency of a photoelectric metal is $v_0$. If light of frequency $4 v_0$ is incident on this metal, then the maximum kinetic energy of emitted electrons will be:

Options:
A $h v_0$
B $2 h v_0$
C $3 h v_0$
D $4 h \mathrm{v}_0$

The correct answer is:
$3 h v_0$

Solution:
By using Einstein's photoelectric equation, we have,
$\begin{aligned}
& \mathrm{KE}=h v-\phi \\
& \mathrm{KE}=h v-h v_0 \\
& =\mathrm{h}\left(4 v_0\right)-h v_0 \\
& =3 h v_0 \\
& \text { where, } \quad h=\text { Planck's constant, } \\
& v \& v_0=\text { frequencies, and } \\
& \phi=\text { work function. } \\
&
\end{aligned}$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2022 (Phase 2)
The throat swab of a patient with a sore throat, hoarseness of voice, and difficulty in breathing was taken. The organism was observed under the microscope and was identified as Corynebacterium diphtheriae. The following is an image of the slide.

What is the nature of the organisms?

Options:
A They are spherical organisms without a well-developed nucleus
B They are rod shaped organisms without a well-developed nucleus
C They are spherical organisms with a well-developed nucleus
D They are rod shaped organisms with a well-developed nucleus

The correct answer is:
They are rod shaped organisms without a well-developed nucleus

Solution:
Corynebacterium diphtheria is a bacterium. That means it is a prokaryote without a well-developed nucleus. When you observe these are bacilli, that means they are rod-shaped bacilli.

Topic: Biology, Human Health and Diseases, NEET
The thymus gland secretes the........hormone called........which play a major role in the differentiation of........lymphocytes.

Options:
A Gylcoproteinaceous, thymosin, T
B Steroid , thymosin, T
C Peptide, thymosin, T
D Peptide , thymosin, B

The correct answer is:
Peptide, thymosin, T

Solution:
Thymosin is the peptide hormone secreted by the thymus gland. This hormone stimulates the development of T-cells or T lymphocytes. Thymus is the place where the T-cells gets activated as they are produced in bone marrow cells.

Topic: Biology, Chemical Coordination and Integration, NEET
The Ti plasmid, is often used for making transgenic plants. The plasmid is found in:

Options:
A Azotobacter
B Rhizobium of the roots of leguminous plants
C Agrobacterium
D Yeast as a $2 \mu \mathrm{m}$ plasmid.

The correct answer is:
Agrobacterium

Solution:
$T i$ plasmid is isolated from Agrobacterium tumefaciens and has been modified into a cloning vector which is not pathogenic to plant but still is able to use the mechanisms to deliver genes of interest into plants.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2004
The Ti plasmid, is often used for making transgenic plants. The plasmid is found in

Options:
A Azotobacter.
B Rhizobium.
C Agrobacterium.
D Yeast.

The correct answer is:
Agrobacterium.

Solution:
Ti plasmid (tumor inducing) is isolated from the soil bacterium Agrobacterium tumefaciens is effectively used as vector for gene transfer to plant cells. The part of Ti plasmid transferred into plant cell DNA, is called the TDNA. This TDNA with desired DNA spliced into it, is inserted into the chromosomes of the host plant where it produces copies of itself, by migrating from one chromosomal position to another at random.

Topic: Biology, Biotechnology - Principles and Processes, NEET
The "Ti plasmid" of Agrobacterium tumefaciens stands for

Options:
A Tumor independent plasmid
B Tumor inducing plasmid
C Temperature independent plasmid
D Tumour inhibiting plasmid

The correct answer is:
Tumor inducing plasmid

Solution:
The correct answer is option (2) as Ti plasmid of Agrobacterium tumefaciens is tumor inducing plasmid, containing T-DNA which causes tumor in several dicot plants.
Options (1), (3) and (4) are not correct.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2024
The time by a photoelectron to come out after the photon strikes is approximately

Options:
A 10-1 s
B 10-4 s
C 10-10 s
D 10-16 s

The correct answer is:
10-10 s

Solution:
Emission of photo-electron starts from the surface after incidence of photons in about 10-10 s

Topic: Physics, Dual Nature of Matter, NEET
The time of reverberation of a room $\mathrm{A}$ is one second. What will be the time (in seconds) of reverberation of a room, having all the dimensions double of those of room $\mathrm{A}$ ?

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

The correct answer is:
2

Solution:
Reverberation Time
$$
T=\frac{0 \cdot 61 V}{A S}
$$
Where $V=$ Volume of room in cubic metre
$A=$ Average coefficient of the room
$S=$ Total surface area of room
Then,
$$
\begin{aligned}
& T \propto \frac{V}{S} \\
& \text { or } \frac{T_1}{T_2}=\left(\frac{V_1}{V_2}\right)\left(\frac{S_2}{S_1}\right)=\left(\frac{V}{8 V}\right)\left(\frac{45}{5}\right) \\
& =\frac{1}{2} \\
& \therefore T_2=2 T_1=2 \times 1=2 \mathrm{sec} . \\
& \left(\because T_1=1 \mathrm{sec}\right) \\
&
\end{aligned}
$$

Topic: Physics, Waves and Sound, NEET, NEET 2006
The time period of a geo-stationary satellite is $24 \mathrm{~h}$, at a height $6 R_E-R_E$ is the radius of earth) from surface of earth. The time period of another satellite whose height is $2.5 \mathrm{R}_{\mathrm{E}}$ from surface will be

Options:
A $6 \sqrt{2} \mathrm{~h}$
B $12 \sqrt{2} \mathrm{~h}$
C $\frac{24}{2.5} \mathrm{~h}$
D $\frac{12}{2.5} \mathrm{~h}$

The correct answer is:
$6 \sqrt{2} \mathrm{~h}$

Solution:
From Kepler's third law, the time period of revolution of satellite around earth is
$$
\mathrm{T}^2 \propto \mathrm{r}^3 \text { or } \mathrm{T} \propto \mathrm{r}^{3 / 2}
$$
where, $\mathrm{r}$ is the radius of satellite's orbit.
Here, $\mathrm{r}_1=6 \mathrm{R}_{\mathrm{E}}+\mathrm{R}_{\mathrm{E}}, \mathrm{T}_1=24 \mathrm{~h}$
$\mathrm{r}_2=2.5 \mathrm{R}_{\mathrm{E}}+\mathrm{R}_{\mathrm{E}}, \mathrm{T}_2=$ ?
where $\mathrm{R}_{\mathrm{E}}=$ radius of earth
So, from Eq. (i), we get
$$
\begin{aligned}
\frac{\mathrm{T}_1}{\mathrm{~T}_2} & =\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^{3 / 2} \\
\frac{24}{\mathrm{~T}_2} & =\left(\frac{6 \mathrm{R}_{\mathrm{E}}+\mathrm{R}_{\mathrm{E}}}{2.5 \mathrm{R}_{\mathrm{E}}+\mathrm{R}_{\mathrm{E}}}\right)^{3 / 2}=\left(\frac{7}{3.5}\right)^{3 / 2} \\
\Rightarrow \quad \mathrm{T}_2 & =\frac{24}{(2)^{3 / 2}}=\frac{24}{2 \sqrt{2}}=\frac{12}{\sqrt{2}}=6 \sqrt{2} \mathrm{~h}
\end{aligned}
$$

Topic: Physics, Gravitation, NEET, NEET 2019 (Odisha)
The time period of a particle executing S.H.M. is 8 s. At t=0 it is at the mean position. The ratio of the distance covered by the particle in the 1st second to the 2nd second is

Options:
A 12+1
B 2
C 12
D 2+1

The correct answer is:
2+1

Solution:
T=8 s
Ï=2ÏT=Ï4
Equation of S.H.M. y=Asinâ¡Ï4t
y1=Asinâ¡Ï4=A2
y2=Asinâ¡Ï2=A
Distance travelled in 2nd second =y2-y1
=A-A2
So the ration would be =A-A2A2
=q-11=2-1 2+12+1=12+1

Topic: Physics, Oscillations, NEET
The time period of a satellite of earth is 5 h. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become

Options:
A 40 h
B 20 h
C 10 h
D 80 h

The correct answer is:
40 h

Solution:
Using Kepler's IIIrd law
T 2 ∝ r 3
T ′ 2 T 2 = ( r ′ ) 3 r 3
⇒ T '2 T 2 = ( 4R ) 3 ( R ) 3
T '2 =64  T 2
⇒ T '2 =8T⇒T'=8×5=40 h

Topic: Physics, Gravitation, NEET
The time period of a satellite of earth is 5 h. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become

Options:
A 10 h
B 18 h
C 40 h
D 20 h

The correct answer is:
40 h

Solution:
According to Kepler's law
T2∝r3
or 52∝r3 …(i)
and T′2∝4r3 …ii
From Eqs.(i) and (ii), we have
25T′2=r364r3
T=1600=40 h

Topic: Physics, Gravitation, NEET
The time period of a satellite of Earth is 5 hours. If the separation between the Earth and the satellite is increased to 4 times the previous value, the new time period will become -

Options:
A 40 hours
B 20 hours
C 10 hours
D 80 hours

The correct answer is:
40 hours

Solution:
T2Â âÂ r3Â ;Â T′2Â âÂ 4r3 or T2/T′2=64
or
T′/T=8
or T′=8Â ÃÂ 5h=40Â h

Topic: Physics, Gravitation, NEET
The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become -

Options:
A 40 hours
B 20Â hours
C 10Â hours
D 80Â hours

The correct answer is:
40 hours

Solution:
T2∝ r3; T′2∝ 4r3 or T′2T2=64 or T′T=8
or T′=8 × 5 h=40 h

Topic: Physics, Gravitation, NEET
The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become

Options:
A 40 hr
B 20 hr
C 10 hr
D 8 hr

The correct answer is:
40 hr

Solution:
T2âr3; Tâ²2â4r3
or Tâ²2T2=64
or Tâ²T=8
or Tâ²= 8 Ã 5h = 40 h

Topic: Physics, Gravitation, NEET
The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become

Options:
A 10 hrs
B 80 hrs
C 40 hrs
D 20 hrs

The correct answer is:
40 hrs

Solution:
According to Kepler's law T2âr3
â´ T1T22=r1r23=143=164 or T1T2=18
or T2=8T1=8Ã5=40 hrs

Topic: Physics, Gravitation, NEET
The time period of mass suspended from a spring is $T$. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be:

Options:
A $\frac{T}{4}$
B $T$
C $\frac{T}{2}$
D $2 T$

The correct answer is:
$\frac{T}{2}$

Solution:
Let $K=$ force constant of spring and $K^{\prime}=$ force constant of each part then
$$
\begin{gathered}
\frac{1}{K}=\frac{4}{K^{\prime}} \\
\Rightarrow \quad K^{\prime}=4 K \\
\therefore \text { Time period }=2 \pi \sqrt{\frac{m}{4 K}} \\
=\frac{1}{2} \times 2 \pi \sqrt{\frac{m}{4 K}}=\frac{T}{2}
\end{gathered}
$$

Topic: Physics, Oscillations, NEET, NEET 2003
The time period of oscillation of a magnet is 2 sec. When it is magnetised so that its pole strength is 4 times, its period will be

Options:
A 1 sec
B 2Â sec
C 4Â sec
D 8Â sec

The correct answer is:
1 sec

Solution:
Time period of oscillation of a magnet is,
T=2ÏIMB
T2T1=M1M2
T22=M4M
T2=1 sec

Topic: Physics, Magnetic Properties of Matter, NEET
The time period of oscillation of a simple pendulum is given by T=2πlg . The length of the pendulum is measured as l=10±0.01 cm and the time period as T=0.5±0.02 s . The percentage error in the value of g is

Options:
A 5%
B 8%
C 7%
D None of these

The correct answer is:
8%

Solution:
ÎggÃ100=ÎllÃ100+2ÎTTÃ100
Îl=0.01 cm
ÎT=0.02s
l=10 cm
T=0.5 s
Îl l Ã100= 0.01 10 Ã100=0.1
ÎT T Ã100= 0.02 0.5 Ã100=4
Δgg×100=0.1+2×4≈8%

Topic: Physics, Units and Dimensions, NEET
The time period of oscillations of a block attached to a spring is t1. When the spring is replaced by another spring, the time period of the block is t2. If both the springs are connected in series and the block is made to oscillate using the combination, then the time period of the block is

Options:
A T=t1+t2
B T2=t12+t22
C T-1=t1-1+t2-1
D T-2=t1-2+t2-2

The correct answer is:
T2=t12+t22

Solution:
When springs are in series, k=k1k2k1+k2
For first spring, t1=2πmk1
For Second spring, t2=2πmk2
∴ t12+t22=4π2mk1+4π2mk2=4π2mk1+k2k1k2
or t12+t22=2πmk1+k2k1k22
or t12+t22=T2 .

Topic: Physics, Oscillations, NEET
The time period of vibrations of a dip needle in the vertical plane in the magnetic meridian is 3 s. When the same magnetic needle is made to vibrate in the horizontal plane, the time period is 32 s. Then the angle of dip will be

Options:
A 90o
B 60o
C 45o
D 30o

The correct answer is:
60o

Solution:
Time period of a dip needle vibration due to magnetic field is given by
T=2πIMB
In magnetic meridian 3=2πIMB .......(1)
In horizontal plane, 32=2πIMBH .......(2)
332=BHB=Bcos⁡θB
cos⁡θ=12
Therefore angle of dip is θ=60o

Topic: Physics, Magnetic Properties of Matter, NEET
The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 1010s–1. Calculate k at 318K and Ea

Options:
A Ea= 76.750 kJ mol–1and k = 3.6528 × 10–2s–1
B Ea= 86.985 kJ mol–1and k = 0.9965 × 10–2s–1
C Ea= 76.750 kJ mol–1and k = 0.9965 × 10–2s–1
D Ea= 86.985 kJ mol–1and k = 3.6582 × 10–2s–1

The correct answer is:
Ea= 76.750 kJ mol–1and k = 0.9965 × 10–2s–1

Solution:
Correct Option is : (C)
Ea= 76.750 kJ mol–1and k = 0.9965 × 10–2s–1

Topic: Chemistry, Chemical Kinetics, NEET
The time required for a 50 Hz alternating current to increase from 0 to 70.7% of its peak value is

Options:
A 2.5 ms
B 10 ms
C 20 ms
D 14.14 ms

The correct answer is:
2.5 ms

Solution:
We know that i=i0sin⁡ωt
i=i0sin⁡100πt
0.707i=i0sin⁡100πt
12=sin⁡100πt
π4=100πt
t=1400
t=2.5 ms

Topic: Physics, Alternating Current, NEET
The time required to coat aluminium metal on the surface of a square plate of length 20 cm about 5 mm thickness on both sides by using molten AlCl3 solution and 10 A current is nearly [Specific gravity of Al=1.8 g  mL −1 ]

Options:
A 107.2 hr
B 214.4 hr
C 53.6 hr
D 71.5 hr

The correct answer is:
214.4 hr

Solution:
m=20×20×0.5×2×1.8=720 gm ;
m=996500×10×t
t=214.44 hr

Topic: Chemistry, Electrochemistry, NEET
The tissue is a group of cells having

A different origin

B same origin

C performs the same function

D absent in plants

Options:
A A and C true
B B and C true
C D true
D A true

The correct answer is:
B and C true

Solution:
All cells are same in origin or mother cell are same and doing similar work and form tissue. The group of similar cells form tissue.

Topic: Biology, Structural Organisation in Animals, NEET
The tissue lining the inner surface of a hollow organ like a fallopian tube is:

Options:
A Columnar epithelium
B Cuboidal epithelium
C Ciliated epithelium
D Brush bordered epithelium

The correct answer is:
Ciliated epithelium

Solution:
The inner surface of hollow visceral organs like bronchioles, fallopian tube and ventricles of brain are lined by ciliated columnar epithelium. The function of cilia is to move particles or mucus in a specific direction over the epithelium.

Topic: Biology, Structural Organisation in Animals, NEET
The tissue with the power of division and regeneration, throughout life, is _________.

Options:
A Epithelial tissue
B Areolar tissue
C Muscular tissue
D Myelinated nerve fibers

The correct answer is:
Epithelial tissue

Solution:
Epithelial tissue covers the outside of the body and lines organ cavities. The tissue always rests upon underlying connective tissue and are closely packed with each other so there is very little intercellular space. Due to absence/less of intercellular spaces, blood vessels, lymph vessels and capillaries are unable to pierce this tissue so blood circulation is absent in epithelium. Hence cells depend on their nutrients on underlying connective tissue. The cells of this tissue can divide throughout life with high regeneration capacity as they process the stem cells.

Topic: Biology, Structural Organisation in Animals, NEET
The tolerance level of a resistor with the colour code red, blue, orange, gold is

Options:
A ±5%
B ±10%
C ±20%
D ±40%

The correct answer is:
±5%

Solution:
The tolerance level of resistance is given by 4th colour
Gold represent ±  5%tolerance

Topic: Physics, Current Electricity, NEET
The total angular width of central maxima in diffraction at a single slit is

Options:
A λa
B 2λa
C 2aλ
D 2a3λ

The correct answer is:
2λa

Solution:
Width of central bright fringe βcentral=2λDa
Therefore α=βcentralD=2λDaD
α=βcentralD=2λa

Topic: Physics, Wave Optics, NEET
The total area covered by all the biodiversity hotspots is

Options:
A 8.1%
B 2.4%
C Less than 2%
D 30%

The correct answer is:
Less than 2%

Solution:
Biodiversity hotspots are the region with very high levels of species richness and a high degree of endemism (that is, species confined to that region and not found anywhere else). Initially, 25 biodiversity hotspots were identified but subsequently, nine more have been added to the list, bringing the total number of biodiversity hotspots in the world to 34.
These hotspots are also regions of accelerated habitat loss. Three of these hotspots â Western Ghats and Sri Lanka, Indo-Burma, and Himalaya â cover our countryâs exceptionally high biodiversity regions. Although all the biodiversity hotspots put together cover less than 2% of the earthâs land area, the number of species they collectively harbour is extremely high and strict protection of these hotspots could reduce the ongoing mass extinctions by almost 30%.

Topic: Biology, Biodiversity and its Conservation, NEET
The total current supplied to the given circuit by the battery is

Options:
A 9 A

B 6 A

C 2 A

D 4 A

The correct answer is:
6 A

Solution:
The equivalent circuit of the given circuit is as shown
Resistances 6 Î© and 2 Î© are in parallel
â´ Râ²=6Ã26+2=32 Î©
Resistances 32 Î© and 1.5 Î© are in series
â´ Râ²â²=32+1.5=3 Î©
Resistances 3Î© and 3Î© are in parallel
â´ R=3Ã33+3=32
The current, I=VR
= 9 3 / 2 = 6 A

Topic: Physics, Current Electricity, NEET
The total electrostatic energy stored in both the capacitors is

Options:
A 9 Î¼J
B 40.5Â Î¼J
C 13.5Â Î¼J
D 18Â Î¼J

The correct answer is:
9 Î¼J

Solution:
According to circuit diagram, capacitors are connected in series combination.
So, equivalent capacitance Ceq=3Ã63+6
=189=2 Î¼F
We know that,
Total electrostatic energy U=12CeqV2
=12Ã2Ã-32
=9 Î¼J

Topic: Physics, Electrostatics, NEET
The total energy of a satellite is

Options:
A always positive
B always negative
C always zero
D positive or negative depending upon radius of orbit

The correct answer is:
always negative

Solution:
The total energy of the satellite is T.E.=-GMm2r

Topic: Physics, Gravitation, NEET
The total energy of an electron in an atom in an orbit is -3.4 eV. Its kinetic and potential energies are, respectively,

Options:
A -3.4  eV, -3.4  eV
B -3.4  eV, -6.8  eV
C 3.4  eV, -6.8  eV
D 3.4  eV,  3.4  eV

The correct answer is:
3.4  eV, -6.8  eV

Solution:
KE=-TE
⇒KE=3.4 eV
PE=2·TE=-6.8 eV

Topic: Physics, Atomic Physics, NEET, NEET 2019 (All India)
The total energy of an electron in the first excited state of H-atom is -3.4 eV, then its potential energy is in this state is (in eV)

Options:
A 6.8
B 3.4
C -6.8
D -3.4

The correct answer is:
-6.8

Solution:
Total energy of electron E=-3.4 eV
Potential energy U=2E
⇒ U=-6.8 eV

Topic: Physics, Atomic Physics, NEET