datavorous/entrance-exam-dataset huggingface dataset:
The mature seeds of plants such as gram and peas, possess no endosperm, because

Options:
A These plants are not angiosperms
B There is no double fertilization in them
C Endosperm is not formed in them
D Endosperm gets used up by the developing embryo during seed development.

The correct answer is:
Endosperm gets used up by the developing embryo during seed development.

Solution:
The mature seeds of plants such as gram and peas, possess no endosperm, because endosperm gets used up by the developing embryo during seed development.

Topic: Biology, Morphology of Flowering Plants, NEET
The maximum amount of electrolytes and water $(70-80 \%)$ from the glomerular filtrate is reabsorbed in which part of the nephron?

Options:
A Ascending limb of loop of Henle
B Distal convoluted tubule
C Proximal convoluted tubule
D Descending limb of loop of Henle

The correct answer is:
Proximal convoluted tubule

Solution:
From the Bowman's capsule, the glomerular filtrate enters the proximal convoluted tubule (PCT). PCT is surrounded by a network of peritubular capillaries and is the seat of reabsorption. About $75 \%$ of glomerular filtrate is normally reabsorbed in PCT before reaching the loop of Henle. The reabsorbed materials include glucose, amino acids, vitamins, hormones, sodium, potassium, chlorides, phosphates, bicarbonates, most of water and some urea etc.

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2012 (Screening)
The maximum covalent character is shown by

Options:
A AlCl3
B MgCl2
C CsCl
D LaCl3

The correct answer is:
AlCl3

Solution:
Among the given choice Al is least electropositive therefore, the bond between Al and Cl will be least ionic or most covalent or the difference in electronegativeity of two atom is less than 1.8.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young's modulus, respectively, are $8 \times 10^8 \mathrm{~N} \mathrm{~m}^{-2}$ and $2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$, is:

Options:
A 0.4 mm
B 40 mm
C 8 mm
D 4 mm

The correct answer is:
4 mm

Solution:
In the case for maximum elongation,
Stress $=$ Elastic limit
$\begin{aligned}
\delta_{\max }=\frac{\sigma_{\text {elastic }} \times L}{\text { Young's modulus }}=\frac{8 \times 10^8 \times 1}{2 \times 10^{11}} & =4 \times 10^{-3} \\
& =4 \mathrm{~mm}
\end{aligned}$
i.e. maximum elongation is 4 mm

Topic: Physics, Mechanical Properties of Solids, NEET, NEET 2024
The maximum energy in thermal radiation from a source occurs at the wavelength 4000ââ« . The effective temperature of the source is (Wien's constant, b=2.93Ã 10 â3 m K )

Options:
A 7325 K
B 800 K
C 10 4  K
D 10 6  K

The correct answer is:
7325 K

Solution:
According to Wien's displacement law
λ m = b T ⇒T= b λ m = 2.93× 10 −3 4000× 10 −10 =7325 K

Topic: Physics, Thermal Properties of Matter, NEET
The maximum growth rate occurs in:

Options:
A stationary phase
B senescent phase
C lag phase
D exponential phase.

The correct answer is:
exponential phase.

Solution:
In microbial growth curve, lag initial phase is characterized by cellular activity but not growth. exponential or log phase is the time when the cells are dividing by binary fission and doubling in numbers after each generation time. In stationary phase, the growth curve begins to decline as the available nutrients become depleted and waste products start to accumulate. The terminal irreversible stage of ageing is called senescence. This is the last phase of life span, that ultimately leads to death.
Related Theory
Aging is a progressive decline with time whereas senescence occurs throughout the lifespan, including during embryogenesis.

Topic: Biology, Plant - Growth and Development, NEET, NEET 2004
The maximum height attained by a projectile when thrown at an angle Î¸ with the horizontal is found to be half the horizontal range. Then Î¸ is equal to

Options:
A tan - 1 2
B π 6
C π 4
D tan - 1 1 2

The correct answer is:
tan - 1 2

Solution:
Maximum height, Hâ¡ 0 = u 2 sin 2 Î¸ 2 g
Range, Râ¡ = u 2 sin 2 Î¸ g = 2u 2 sin Î¸ cos Î¸ g
Given, H 0 = R 2
â´ u 2 sin 2 Î¸ 2 g = 2 u 2 sin Î¸ cos Î¸ 2 g
â sin Î¸ = 2 cos Î¸
â tan Î¸ = 2
â´ Î¸ = tan - 1 2

Topic: Physics, Motion In Two Dimensions, NEET
The maximum intensity in Young's double-slit experiment is I0. Distance between the slits is d=5λ, where λ is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance D=10d?

Options:
A I02
B 34I0
C I04
D I0

The correct answer is:
I02

Solution:
Îx at P=dxD=d22D=5Î»22Ã10Ãd
Îx=5Î»22Ã10Ã5Î»=Î»4
ÎÏ=2ÏÎ»ÃÎx=Ï2
I0=4I
â I=I04
Given, I0=10, d=5Î» and D=10d.
Path difference =Îx at P=xdD

At x=d2
Îx=d22D
Îx=5Î»22Ã10d
Îx=5Î»22Ã10Ã5Î»=Î»4
ÎÏ=2ÏÎ»Îx
ÎÏ=Ï2
Inet=I+I+2Icosâ¡Ï2=2I=2I04=I02 as Imax=I0=4I â I=I04

Topic: Physics, Wave Optics, NEET
The maximum intensity in Young's double-slit experiment is I0 . Distance between the slits is d=5Î» , where Î» is the wavelength of monochromatic light used in the experiment. The intensity of the light in front of one of the slits on a screen at a distance D=10d is

Options:
A I02
B 34I0
C I0
D I04

The correct answer is:
I02

Solution:
Path difference at point P is
Îx=dxD=d22D
Îx=(5Î»)22Ã10d
Îx=(5Î»)22Ã10Ã5Î»=Î»4
Therefore, phase difference is
ÎÏ=2ÏÎ»ÃÎx=Ï2
If I is the intensity due to each slit, then the maximum intensity I0 is,
I0=4I
â I=I04
Therefore, intensity at point P is
Inet=I+I+2IIcosâ¡Ï2
Inet=2I=I02

Topic: Physics, Wave Optics, NEET
The maximum kinetic energy of a photoelectron is 3 eV. Its stopping potential is

Options:
A 3 V
B 1 V
C 6 V
D 2 V

The correct answer is:
3 V

Solution:
Here,
Kmax=3 eV
Maximum kinetic energy is given by:
Kmax=eV03 eV=eÃV03 V=V0
â´ Stopping potential V0=3 V

Topic: Physics, Dual Nature of Matter, NEET
The maximum kinetic energy of electrons emitted in the photoelectric effect is linearly dependent on the _______________ of the incident radiation.

Options:
A amplitude
B period
C wavelength
D frequency

The correct answer is:
frequency

Solution:
According to law of photoelectric effect, 12mv2=hν−hν0 i.e., the maximum kinetic energy of electrons emitted in the photoelectric effect is linearly dependent on the frequency of incident radiation.

Topic: Physics, Dual Nature of Matter, NEET
The maximum kinetic energy of emitted photoelectrons depends on

Options:
A Intensity of incident radiation
B Frequency of incdent radiation
C Speed of incident radiation
D Number of photons in the incident radiation

The correct answer is:
Frequency of incdent radiation

Solution:
KE= 1 2 m v max 2 =h( ν− ν o )
K. E max =hÎ½âh Î½ o
â´ KEmax depends on frequency of incident radiation

Topic: Physics, Dual Nature of Matter, NEET
The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in volt is

Options:
A 2
B 4
C 6
D 8

The correct answer is:
4

Solution:
Stopping potential is the negative potential applied to stop the electrons having maximum kinetic energy. Therefore stopping potential will be 4 V

Topic: Physics, Dual Nature of Matter, NEET
The maximum kinetic energy of the emitted photoelectrons in photoelectric effects is independent of:

Options:
A Frequency of incident radiation
B Wavelength of incident radiation
C Work function of material
D Intensity of incident radiation

The correct answer is:
Intensity of incident radiation

Solution:
By photoelectric equation
$\begin{aligned} & \mathrm{KE}_{\max }=h f-\phi_0 \\ & \mathrm{KE}_{\max }=\frac{h c}{\lambda}-\frac{h c}{\lambda_0}\end{aligned}$
$\mathrm{KE}_{\max }$ is independent of intensity of incident radiation.

Topic: Physics, Dual Nature of Matter, NEET, NEET 2023 (Manipur)
The maximum number of 90° angles between bond pair–bond pair of electrons is observed in

Options:
A sp3d2hybridisation
B sp3 dhybridisation
C dsp3 hybridisation
D dsp2 hybridisation

The correct answer is:
sp3d2hybridisation

Solution:
The maximum number of 90° angles between bond pair–bond pair of electrons is observed in sp3d2 hybridisation. This molecule is made up of 6 equally spaced sp3d2 hybrid orbitals arranged at 90° angles.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The maximum number of carbon atoms arranged linearly in the molecule, CH3-C≡C-CH=CH2 is

Options:
A 5
B 4
C 2
D 3

The correct answer is:
4

Solution:
The way in which the molecule is represented in question isn't how the actual shape is. The actual shape of the molecule is given below.

A compound is said to be linear when all bond angles between Carbon atoms is 180o. The actual shape of the molecule is draw above. We can clearly see that the 4th Carbon from left has sp2 hybridization and hence will have a bond angle of 120o. Thus, the 5th Carbon is not linear with 4th carbon. Hence, only 4 Carbons are arranged linearly.

Topic: Chemistry, General Organic Chemistry, NEET
The maximum number of carbon atoms arranged linearly in the molecule, CH3-C≡C-CH=CH2 is

Options:
A 2
B 3
C 4
D 5

The correct answer is:
4

Solution:
Two carbons of triple bond and the two carbons directly attached to triply bonded carbons are linearly arranged. Hence, the total carbon atoms arranged linearly is 4.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The maximum number of electrons that can exist in the orbital for which n=3,l=1 and m=-1 is:

Options:
A 10
B 14
C 2
D 6

The correct answer is:
2

Solution:
n=3,1=1⇒3p

As the value of m is also fixed, it means we are dealing with one particular orbital of p i.e. p x , p y or pz. Total 2 electron can fit in any orbital.

Topic: Chemistry, Structure of Atom, NEET
The maximum number of electrons that can have principal quantum number, n=3 and spin quantum number, ms=-12, is

Options:
A 3
B 5
C 7
D 9

The correct answer is:
9

Solution:
When n = 3 , l = 0, 1, 2 i.e., there are 3s, 3p and 3d-orbital’s. If all these orbitals are completely occupied as
Total 18 electrons, 9 electrons with s=+12 and 9 with s=-12.

Topic: Chemistry, Structure of Atom, NEET
The maximum number of electrons with n = 3 and l = 3 is

Options:
A 11
B 6
C 10
D 0

The correct answer is:
0

Solution:
As 3f-orbital does not exist as value of l cannot be equal to n, hence, no electron can be accommodated.

Topic: Chemistry, Structure of Atom, NEET
The maximum number of individuals of a population which can be supported with optimum resources for their survival is

Options:
A Controlling capacity
B Carrying capacity
C Holding capacity
D Both a and b

The correct answer is:
Carrying capacity

Solution:
Carrying Capacity (K) is the maximum number of individuals of a population which can be supported with optimum resources for their survival carrying is called carrying capacity of the environment. Growth of a population depends upon its biotic potential, death rate and birth rate. The carrying capacity of earth for human population is considered to be about 50 billions.

Topic: Biology, Organisms and Populations, NEET
The maximum number of molecules is present in:

Options:
A $15 \mathrm{~L}$ of $\mathrm{H}_2$ gas at STP
B $5 \mathrm{~L}$ of $\mathrm{N}_2$ gas at STP
C $0.5 \mathrm{~g}$ of $\mathrm{H}_2$ gas
D $10 \mathrm{~g}$ of $\mathrm{O}_2$ gas

The correct answer is:
$15 \mathrm{~L}$ of $\mathrm{H}_2$ gas at STP

Solution:
At STP $22.4 \mathrm{~L} \mathrm{H}_2=6.023 \times 10^{23}$ molecules
$$
\begin{aligned}
15 \mathrm{~L} \mathrm{H}_2 & =\frac{6.023 \times 10^{23} \times 15}{22.4} \\
& =4.033 \times 10^{23} \\
5 \mathrm{~L} \mathrm{~N}_2 & =\frac{6.023 \times 10^{23} \times 5}{22.4} \\
& =1.344 \times 10^{23} \\
0.5 \mathrm{~g} \mathrm{H}_2 & =\frac{6.023 \times 10^{23} \times 0.5}{2} \\
& =1.505 \times 10^{23} \\
10 \mathrm{~g} \mathrm{of}_2 & =\frac{6.023 \times 10^{23} \times 10}{32} \\
& =1.882 \times 10^{23} .
\end{aligned}
$$

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2004
The maximum number of moles of Ba3PO42 that can be formed if 2 mole BaCl2 is mixed with 1 mole Na3PO4 is

Options:
A 0.7
B 0.25
C 0.33
D 0.50

The correct answer is:
0.50

Solution:
The balanced chemical reaction is

3BaCl2+2Na3PO4→Ba3PO42+6NaCl

According to the reaction,

3 mole BaCl2 reacts with 2 mole Na3PO42

Thus, 2 mole BaCl2 will require 23×2=43=1.33 mole Na3PO4
Since sodium phosphate is taken in less quantity then the required quantity, here, Na3PO4 is the limiting reactant. The stoichiometric calculations are done based on the limiting reagent.

According to the reaction.
2 mole Na3PO4 gives 1 mole Ba3PO42

So, 1 mole Na3PO4 will give 0.5 mole Ba3PO42.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double - slit experiment is

Options:
A infinite
B five
C three
D zero

The correct answer is:
three

Solution:
For maxima, dsinÎ¸=nÎ»
â2Î»sinÎ¸=nÎ»
â2sinÎ¸=n
â´sinÎ¸=n2
âµsinÎ¸<1
ân2<1
ân<2
n=0,1 maximum number of possible maxima = 3
n=0,Â±1

Topic: Physics, Wave Optics, NEET
The maximum number of reducing hydrogens are contained in which of the following molecule/s?

Options:
A H3PO3
B H3PO2
C H4P2O7
D H3PO4

The correct answer is:
H3PO2

Solution:
Maximum number of P – H bonds are is present in H3PO2 .

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
The maximum oxidation number Phosphorus can show is

Options:
A +5
B +8
C +6
D +3

The correct answer is:
+5

Solution:
Correct Option is : 1
+5

Topic: Chemistry, Redox Reactions, NEET
The maximum percentage of available volume that can be filled in a face centered cubic system by an atom is

Options:
A 74%
B 68%
C 34%
D 26%

The correct answer is:
74%

Solution:
Packing Fraction for cubic close packing is N atoms × V atom V unitcell = 6. 4 3 πr 3 3 (3) 1 2 2 (2r) 2 (2/3) 1 2 .4r =0.74⇒74%

Topic: Chemistry, Solid State, NEET
The maximum possible number of hydrogen bonds a water molecule can form is

Options:
A 2
B 4
C 3
D 1

The correct answer is:
4

Solution:
A water molecule can form at the most four H-bonds :

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The maximum power is dissipated for an ac in a/an

Options:
A Inductive circuit
B Capacitive circuit
C Resistive circuit
D LC circuit

The correct answer is:
Resistive circuit

Solution:
$P=V_{\text {ms }} I_{\mathrm{ms}} \cos \phi$
In purely resistive circuit voltage and current are in same phase therefore phase difference is zero.
So, $\cos \phi=1$
$\therefore$ Power dissipated is maximum.

Topic: Physics, Alternating Current, NEET, NEET 2023 (Manipur)
The maximum prescribed concentration of cadmium in drinking water in ppm is

Options:
A 0.005
B 3
C 2
D 5

The correct answer is:
0.005

Solution:
The US EPA has established a Maximum Contaminant Level (MCL) of 0.005 milligrams per litre (mg/L) for cadmium in drinking water.

Topic: Chemistry, Environmental Chemistry, NEET
The maximum refractive index of a prism which permits the passage of light through it, when the refracting angle of the prism is 90°,is

Options:
A 3
B 2
C 32
D 32

The correct answer is:
2

Solution:
For passing the ray from prism,
μ<cosecA2
μ<cosec90°2
μ<2
μmax⁡=2

Topic: Physics, Ray Optics, NEET
The maximum temperature that can be achieved in blast furnace is:

Options:
A upto 1200 K
B upto 2200 K
C upto 1900 K
D upto 5000 K

The correct answer is:
upto 2200 K

Solution:
Hot air is blown from the bottom of the furnace and coke is burnt to give temperature up to about 2200K in the lower portion itself. It is the highest temperature in the blast furnace.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET, NEET 2021
The maximum transverse velocity and maximum transverse acceleration of a harmonic wave in a one-dimensional string are 1 ms-1 and 1 ms-2 respectively. The phase velocity of the wave is 1 ms-1 . The waveform is

Options:
A sin⁡x-t
B sin⁡x-2t
C sin⁡x2-t
D sin⁡x-t2

The correct answer is:
sin⁡x-t

Solution:
âµ Wave equation, y=A0sinâ¡kx-Ït ....(i)
where, k= angular wave number =2ÏÎ»
A0= amplitude
âµ vmax=A0Ï
amax=Ï2A0
âµgiven that vmax=1ms, a=1ms2
â´Ï=1
k=ÏV=1
â´ From Equation (i),
y=sinâ¡x-t

Topic: Physics, Waves and Sound, NEET
The maximum usable energy per molecule of glucose metabolised will be generated during

Options:
A	Aerobic	respiration	by	germinating seeds
B	Production of methanol by entericbacteria
C	Fermentation into ethanol by yeast
D	Glycolysis in the skeletal muscle of a sprinter performing a hundred metredash

The correct answer is:
Aerobic	respiration	by	germinating seeds

Solution:
The energy yield during aerobic respiration (36 ATP) is 18 times of that during anaerobic respiration (2 ATP).

Topic: Biology, Respiration in Plants, NEET
The maximum velocity (in m s-1) with which a car driver must traverse a flat curve of the radius of 150 m and the coefficient of friction 0.6, to avoid skidding is

Options:
A 60
B 30
C 15
D 25

The correct answer is:
30

Solution:
Using the relation
mv2r=Î¼R, R=mg
mv2r=Î¼mgâ¹v2=Î¼rg
v2=0.6Ã150Ã10
v=30 m s-1

Topic: Physics, Laws of Motion, NEET
The maximum velocity of a particle, executing simple harmonic motion with an amplitude 7 mm is 4.4 m s-1. The period of oscillation is

Options:
A 100 s
B 0.01 s
C 10 s
D 0.1 s

The correct answer is:
0.01 s

Solution:
Maximum velocity Î½m=aÏ=a2ÏT
â´ T=2ÏaÎ½m=2Ã227Ã7Ã10-34Â·4

T=10-2=0.01 s

Topic: Physics, Oscillations, NEET
The maximum velocity of a simple harmonic motion represented by y=3 sinâ¡(100t+Ï6)m is given by

Options:
A 300 m s-1
B 3Ï6 m s-1
C 100 m s-1
D Ï6 m s-1

The correct answer is:
300 m s-1

Solution:
The given equation is written as,
y=3sinâ¡100 t+Ï6 ...(i)
The general equation of simple harmonic motion is written as
y=asinâ¡(Ït+â) ...(ii)
Equating Eqs. (i) and (ii), we get
a=3, Ï=100
Maximum velocity, v=aÏâ =3Ã100=300 m s-1

Topic: Physics, Oscillations, NEET
The maximum velocity of electrons emitted from a metal surface is v when the frequency of light falling on it is f. The maximum velocity when the frequency becomes 4f is

Options:
A 2v
B >2v
C <2v
D between 2v and 4v

The correct answer is:
>2v

Solution:
According to Einstein's equation for P.E.E.
For 1st case,
12mv2=hf-W, where f=frequency of incident light and W=work function
For 2nd case,
12mv12=h4f-W
12mv12=4hf-4W+3W
12mv12=4hf-W+3W
12mv12=412mv2+3W
12mv12=12m2v2+3W
This equation tells us on comparison that,
v1>2v

Topic: Physics, Dual Nature of Matter, NEET
The maximum voltage in DC circuit is 282 V. The effective voltage in AC circuit will be

Options:
A 200 V

B 300 V

C 400 V

D 564 V

The correct answer is:
200 V

Solution:
Maximum voltage is AC circuit
V0=282 V
V=V02=2822
V=2821.41=28200141
V=200 V

Topic: Physics, Alternating Current, NEET
The maximum volume of air a person can breathe in after a forced expiration is known as

Options:
A expiratory capacity
B vital capacity
C inspiratory capacity
D total lung capacity

The correct answer is:
vital capacity

Solution:
Vital Capacity (VC) is the maximum volume of air a person can breathe in after a forced expiration. This includes ERV, TV and IRV or the maximum volume of air a person can breathe out after a forced inspiration.

Topic: Biology, Breathing and Exchange of Gases, NEET, NEET 2019 (Odisha)
The maximum volume of air a person can breathe in after a forceful expiration is called

Options:
A Vital Capacity
B Total Lung Capacity
C Inspiratory Capacity
D Functional Residual Capacity

The correct answer is:
Vital Capacity

Solution:
Vital capacity VC : The maximum volume of air a person can breathe in after a forceful expiration or the maximum volume of air a person can breathe out after a forceful inspiration is called vital capacity.
VC = ERV+TV+IRV or EC+IRV or, IC+ERV
Inspiratory capacity: It is the total volume of air that a person can inspire after a normal expiration.

Topic: Biology, Breathing and Exchange of Gases, NEET
The maximum volume of air a person can breathe in post a forced expiration is called ___.

Options:
A Vital theory
B Vital capacity
C Vital volume
D None of the above

The correct answer is:
Vital capacity

Solution:
The maximum volume of air a person can breathe in post a forced expiration is called vital capacity

Topic: Biology, Breathing and Exchange of Gases, NEET
The maximum wavelength of radiation at 2000 K is 4 Î¼m. What will be the maximum wavelength emitted at 2400 K?

Options:
A 3.3 μm
B 0.66 μm
C 1 m
D 1 μm

The correct answer is:
3.3 μm

Solution:
λm2λm1=T1T2 (Wien's displacement law)
λm24=20002400, λm2=4×2024=3.3 μm

Topic: Physics, Thermal Properties of Matter, NEET
The mean free path for a gas, with molecular diameter d and number density n can be expressed as:

Options:
A 12nπd2
B 12n2πd2
C 12n2π2d2
D 12nπd

The correct answer is:
12nπd2

Solution:
Mean free path for a gas sample

λm=12πd2n

where d is diameter of a gas molecule and n is molecular density

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2020 (Phase 1)
The mean free path l for a gas molecule depends upon diameter, d of the molecule as:

Options:
A l∝1d2
B l∝d
C l∝d2
D l∝1d

The correct answer is:
l∝1d2

Solution:
The mean free path is an average distance over which a moving particle substantially changes its direction or energy, typically as a result of one or more successive collisions with other particles.

The mean free path for a gas molecule depends upon the diameter, d of the molecule as: λ=12πd2n∝1d2,

λ= the mean free path
d= the effective diameter of the molecule
n=the number density of molecules

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2020 (Phase 2)
The mean free path of electrons in a metal is $4 \times 10^{-8} \mathrm{~m}$. The electric field which can give on an average $2 \mathrm{eV}$ energy to an electron in the metal will be in unit of $\mathrm{Vm}^{-1}$

Options:
A $8 \times 10^7$
B $5 \times 10^{-11}$
C $8 \times 10^{-11}$
D $5 \times 10^7$

The correct answer is:
$5 \times 10^7$

Solution:
$$
\begin{aligned}
\text { Energy } & =2 \mathrm{eV} \\
\mathrm{eV}_0 & =2 \mathrm{eV} \\
\Rightarrow \quad \mathrm{V}_0 & =2
\end{aligned}
$$
Now, electric field
$$
\begin{aligned}
E & =\frac{2}{4 \times 10^{-8}} \\
& =0.5 \times 10^8 \\
& =5 \times 10^7 \mathrm{Vm}^{-1}
\end{aligned}
$$

Topic: Physics, Current Electricity, NEET, NEET 2009 (Screening)
The mean free path of molecules of a gas (radius ‘ r ’) is inversely proportional to:

Options:
A r3
B r2
C r
D r

The correct answer is:
r2

Solution:
Here we have to calculate that mean free path of molecule is inversely prooportional to what power of r

we know that,

Mean free path l=12nd2
So: l∝1d2⇒l∝1r2

Hence the mean free path of the molecule is inversely proportional to r2.

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2014
The mean kinetic energy of one mole of gas per degree of freedom (on the basis of kinetic theory of gases) is

Options:
A 12Â kÂ T
B 32Â kÂ T
C 32Â RÂ T
D 12 R T

The correct answer is:
12 R T

Solution:
The mean kinetic energy of one mole of gas n degree of freedom.
E=n2RT
The mean kinetic energy of one mole of gas per degree of freedom.
E′=En=n2RTn
E′=12RT

Topic: Physics, Kinetic Theory of Gases, NEET
The mean lives of a radioactive substances are 1620 yrs and 405 yrs for α - emission and β - emission respectively. Find out the time during which three-fourth of a sample will decay if it is decaying both by α - emission and β - emission simultaneously

Options:
A 449 yrs
B 399 yrs
C 549 yrs
D 579 yrs

The correct answer is:
449 yrs

Solution:
Let at some instant of time t, number of atoms of the radioactive substance are N. It may decay either by Î± - emission on by Î²- emission. So, we can write,
- dâ¡ Nâ¡ dâ¡ t net = - dâ¡ Nâ¡ dâ¡ t Î± + - dâ¡ Nâ¡ dâ¡ t Î²
If the effective decay constant is Î», then
Î» Nâ¡ = Î» Î± Nâ¡ + Î» Î² Nâ¡
â´ Î» â¡ = Î» Î± + Î» Î² = 1 1 6 2 0 + 1 4 0 5
= 1 3 2 4 year - 1
Now, Nâ¡ 0 4 = Nâ¡ 0 eâ¡ - Î» t
â´ - Î» t = ln 1 4 = - 1 . 3 8 6
or 1 3 2 4 t = 1 . 3 8 6
â´ t = 449 yrs

Topic: Physics, Nuclear Physics, NEET