datavorous/entrance-exam-dataset huggingface dataset:
The earth (mass = 6Ã1024 kg) revolves around the sun with angular velocity 2Ã10-7 rad s-1 in a circular orbit of radius 1.5Ã1011 m. The force exerted by the sun on the earth (in newton) is

Options:
A zero
B 18×1025
C 27×1039
D 36×1021

The correct answer is:
36×1021

Solution:
Gravitational force exerted by the sun on the earth will provide the required centripetal force.
F=mRÏ2
=6Ã1024Ã1.5Ã10112Ã10-72
=36Ã1021 N

Topic: Physics, Gravitation, NEET
The Earthâs atmosphere consists primarily of oxygen 21% and nitrogen 78%. The rms speed of oxygen molecules O2 in the atmosphere at a certain location is 535 m/s. The rms speed of the nitrogen molecules at this location will be [Given 1 amu=1.66Ã10-27 kg, molecular mass of O2=32, molecular mass of N2=28]

Options:
A 572 m/s
B 437 m/s
C 835 m/s
D 715 m/s

The correct answer is:
572 m/s

Solution:
The rms speed of N2,vrms, N2=3RTMN2

For O2,vrms, O2=3RTMO2
Thus,

vrms, N2vrms, O2=MO2MN2=3228

vrms, N2=572 m/s

Topic: Physics, Thermodynamics, NEET
The earth's magnetic field always has a vertical component except at the

Options:
A magnetic equator
B magnetic poles
C geographic north pole
D latitude 45o

The correct answer is:
magnetic equator

Solution:
At magnetic equator angle of dip becomes zero so total magnetic field at equator becomes horizontal component and verticle component becomes zero.

Topic: Physics, Magnetic Properties of Matter, NEET
The earth's magnetic field at a certain point is 0.70 gauss. This field is to be annulled by the magnetic field at the centre of a circular conducting loop 5.0 cm in radius. The required current is about

Options:
A 0.66 A

B 5.6 A

C 0.28 A

D 2.8 A

The correct answer is:
5.6 A

Solution:
The magnetic field at the centre of closed loop is
B⁡ = μ 0 I⁡ 2 r⁡ = 4 π × 1 0 - 7 × I⁡ 2 × 5 × 1 0 - 2
or 0 . 7 0 × 1 0 - 4 = 4 π × 1 0 - 7 × I⁡ 1 0 × 1 0 - 2
∴ I⁡ = 5 . 6 A

Topic: Physics, Magnetic Effects of Current, NEET
The earthworm in the ecosystem is called

Options:
A Producers
B Detritivores
C Carnivores
D Herbivores

The correct answer is:
Detritivores

Solution:
Small invertebrate animals called detritivores or detritophages feed on detritus, e.g., earthworms, termites they bring about its fragmentation. Breakdown of detritus into smaller particles by detritivores called fragmentation.

Topic: Biology, Ecosystem, NEET
The ease of adsorption of the hydrated alkali metal ions on an ion-exchange resins follows the order

Options:
A $\mathrm{Li}^{+} < \mathrm{K}^{+} < \mathrm{Na}^{+} < \mathrm{Rb}^{+}$
B $\mathrm{Rb}^{+} < \mathrm{K}^{+} < \mathrm{Na}^{+} < \mathrm{Li}^{+}$
C $\mathrm{K}^{+} < \mathrm{Na}^{+} < \mathrm{Rb}^{+} < \mathrm{Li}^{+}$
D $\mathrm{Na}^{+} < \mathrm{Li}^{+} < \mathrm{K}^{+} < \mathrm{Rb}^{+}$

The correct answer is:
$\mathrm{Rb}^{+} < \mathrm{K}^{+} < \mathrm{Na}^{+} < \mathrm{Li}^{+}$

Solution:
Ease of adsorption of the hydrated alkali metal ions on an ion-exchange resins decreases as the size of alkali metal ions increases.
Since, the order of size of alkali metal ions
$\mathrm{Li}^{+} < \mathrm{Na}^{+} < \mathrm{K}^{+} < \mathrm{Rb}^{+}$
Thus, the ease of adsorption follows the order
$\mathrm{Rb}^{+} < \mathrm{K}^{+} < \mathrm{Na}^{+} < \mathrm{Li}^{+}$

Topic: Chemistry, Surface Chemistry, NEET, NEET 2012 (Screening)
The ease of hydrolysis in the compounds

is of the order

Options:
A I > II > III > IV
B IV > III > II > I
C I > II > IV > III
D II > I > IV > III

The correct answer is:
I > II > III > IV

Solution:
The ease of hydrolysis is as follows
CH3COCl(I)>CH3CO-O-COCH3(II)>CH3COOC2H5(III)>CH3CONH2(IV)

Topic: Chemistry, General Organic Chemistry, NEET
The edible part of coconut fruit is:

Options:
A perisperm
B endosperm
C endocarp
D mesocarp

The correct answer is:
endosperm

Solution:
The cellular and nuclear endosperm is the edible part of the drupe fruit of coconut.

Topic: Biology, Morphology of Flowering Plants, NEET
The effect of global warming includes,

(i) Increased capacity of air to accommodate moisture

(ii) Altered wind movement pattern

(iii) Increased frequency of extreme events like floods, drought, etc.

(iv) Altered range of diseases, vectors, pathogens, etc.

Options:
A (i) only
B (ii) and (iv)
C (i) and (ii)
D (i), (ii), (iii) and (iv)

The correct answer is:
(i), (ii), (iii) and (iv)

Solution:
Global warming is the most leading deleterious changes in the environment that results in odd climatic changes (El Nino effect). The overall increase in the earth's temperature will alter the wind movement pattern, thus increased frequency of extreme events like flood , drought, etc. this will also effect the distribution of organisms with altered range of diseases, vectors, pathogens, etc.

Topic: Biology, Environmental Issues, NEET
The effect of rotation of the Earth on the value of acceleration due to gravity is

Options:
A Maximum at both poles
B Minimum at both poles
C Maximum at equator and minimum at the poles
D Minimum at the equator and maximum at the poles

The correct answer is:
Maximum at equator and minimum at the poles

Solution:
As, g′=g-ω2 Rcos2⁡λ
Now, for poles λ=90° and for equation λ=0°
(i) For λ=90
gpole′=g-ω2 Rcos2⁡90°=g
g′=g
There is no effect on g at the poles.
(ii) For λ=0
gequator=g-ω2 Rcos2⁡0°
= g-ω2 R
The effect of the rotation of Earth on the value of g at the equator is maximum.

Topic: Physics, Gravitation, NEET
The effect of uncertainty principle is significant only for motion of microscopic particles and is negligible for the macroscopic particles. Justify the statement with the help of a suitable example.( mass = 10-6kg ). Calculate dx.dv.

Options:
A 0.58 * 10-28m2s-1
B 0.78 * 10-28m2s-1
C 0.48 * 10-28m2s-1
D 0.22 * 10-28m2s-1

The correct answer is:
0.58 * 10-28m2s-1

Solution:
Correct Option is : (A)
0.58 * 10-28m2s-1

Topic: Chemistry, Structure of Atom, NEET
The effective capacitance between A and B of an infinite chain of capacitors joined as shown in the figure -

Options:
A 3-1 Î¼F
B 5-12 Î¼F
C 5- 1 μF
D 1Â Î¼F

The correct answer is:
5-12 Î¼F

Solution:
Ceq=C(C+Ceq)C+C+Ceq
= C2 + CCeq= 2CCeq + Ceq2
Ceq2+CCeq-C2 = 0 âCeq = -CÂ±C2+4C22
= 5â12C

Topic: Physics, Electrostatics, NEET
The effective capacitance of two capacitors of capacitances C1 and C2 C2>C1 connected in parallel is 256 times the effective capacitance when they are connected in series. The ratio C2C1 is

Options:
A 32
B 43
C 53
D 256

The correct answer is:
32

Solution:
Cparallel=256Cseries
⇒C1+C2=256C1C2C1+C2
⇒6C1+C22=25C1C2
⇒6C12+6C22+12C1C2=25C1C2
⇒6C12C22+6+12C1C2=25C1C2
⇒6C12C22-13C1C2+6=0
This is quadratic equation
By solving we get C2C1=32

Topic: Physics, Electrostatics, NEET
The effective capacitances of two capacitors are 3 $\mu \mathrm{F}$ and $16 \mu \mathrm{F}$, when they are connected in series and parallel respectively. The capacitance of two capacitors are:

Options:
A $10 \mu \mathrm{F}, 6 \mu \mathrm{F}$
B $8 \mu \mathrm{F}, 8 \mu \mathrm{F}$
C $12 \mu \mathrm{F}, 4 \mu \mathrm{F}$
D $1.2 \mu \mathrm{F}, 1.8 \mu \mathrm{F}$

The correct answer is:
$12 \mu \mathrm{F}, 4 \mu \mathrm{F}$

Solution:
given: \(C_S=3 M f\)
\(\begin{aligned}
& C_P=16 M f \\
& \text {as } C_S=\left[\left(C_1 C_2\right) /\left(C_1+C_2\right)\right]=3 \\
& C_P=C_1+C_2=16(2) \\
& \therefore \text {from (1), }\left[\left(C_1 C_2\right) /(16)\right]=3 \\
& C_1 C_2=48 \\
& \text {from (2), } C_2=16-C_1 \\
& C_2=16-\left(48 / C_2\right) \\
& \therefore C_2+\left(48 / C_2\right)=16 \\
& \therefore C_2^2+48-16 C_2=0 \\
& \therefore C_2=12 \mu F, C_2=4 \mu F
\end{aligned}\)
i.e. capacitance are \(4 \mu \mathrm{F} ~\&~ 12 \mu \mathrm{F}\)

Topic: Physics, Capacitance, NEET, NEET 2022 (Phase 2)
The effective resistance between P and Q for the following network is

Options:
A 112 Î©
B 21Â Î©
C 12 Î©
D 121 Î©

The correct answer is:
12 Î©

Solution:
In the given circuit diagram, 3 Î© and 3 Î© are in series
3+3=6 Î©
6 Î© and 6 Î© are in parallel
6Ã66+6=3 Î©
This 3 Î© is in series with 4 Î© and 5 Î©
4+3+5=12 Î©

Topic: Physics, Current Electricity, NEET
The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is 0.25 Ω. What will be the effective resistance if they are connected in series?

Options:
A 0.25 Ω
B 0.5 Ω
C 4 Ω
D 1 Ω

The correct answer is:
4 Ω

Solution:
Let the resistance of each wire be R

Effective resistance in parallel combination,

RP=R4

⇒R=4RP

⇒R=4×0.25=1 Ω

Effective resistance is series combination,

RS=4R

⇒Rs=4×1=4 Ω

Topic: Physics, Current Electricity, NEET, NEET 2021
The efficiency of a Carnot engine depends upon

Options:
A the temperature of the sink only
B the temperatures of the source and sink
C the volume of the cylinder of the engine
D the temperature of the source only

The correct answer is:
the temperatures of the source and sink

Solution:
First, recall the formula of efficiency of Carnot engine in terms of source temperature and sink temperature,η=1-T2T1, here the temperature of the source is always higher than the temperature of the sink but when we change source and sink temperate then the efficiency will change,T1= the temperature of the source and, T2= the temperature of the sink.

Topic: Physics, Thermodynamics, NEET, NEET 2020 (Phase 2)
The efficiency of a Carnot engine working between 800 K and 500 K is -

Options:
A 0.4
B 0.625
C 0.375
D 0.5

The correct answer is:
0.375

Solution:
Efficiency, Î·=1-T2T1=1-500800
=38=0.375

Topic: Physics, Thermodynamics, NEET
The efficiency of a fuel cell is given by:

Options:
A $\frac{\Delta \mathrm{G}}{\Delta \mathrm{S}}$
B $\frac{\Delta \mathrm{G}}{\Delta \mathrm{H}}$
C $\frac{\Delta \mathrm{S}}{\Delta \mathrm{G}}$
D $\frac{\Delta \mathrm{H}}{\Delta \mathrm{G}}$

The correct answer is:
$\frac{\Delta \mathrm{G}}{\Delta \mathrm{H}}$

Solution:
$\varepsilon_{\mathrm{thermo}}=\frac{\Delta \mathrm{G}_{\mathrm{rxn}}}{\Delta \mathrm{H}_{\mathrm{rxn}}}=\frac{\text { useful } \mathrm{E}}{\text { total } \mathrm{E}}$.
Related Theory
Fuel cells are more efficient than thermal plants. The efficiency of fuel cells to produce electricity is $70 \%$. Hydrogen-oxygen cell, methyl-oxygen-alcohol cell and propaneoxygen cell are some of the examples of fuel cells.

Topic: Chemistry, Electrochemistry, NEET, NEET 2007
The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is

Options:
A 6.25%
B 20%
C 26.8%
D 12.5%

The correct answer is:
26.8%

Solution:
efficiency of ideal heat engine, Î·=1â T 2 T 1
T 2 : temperature of sink
T 1 : temperature of source
Efficiency in Percentage, Î·=T1-T2T1Ã100
T1=373, T2=273
Î·=100373Ã100=26.8%

Topic: Physics, Thermodynamics, NEET, NEET 2018
The efficiency of the Carnot engine is 50% and the temperature of sink is 500 K. If the temperature of the source is kept constant and its efficiency raise to 60%, then the required temperature of the sink will be

Options:
A 100 K

B 600 K
C 400 K
D 500 K

The correct answer is:
400 K

Solution:
Efficiency (Î·) of a Carnot engine is given by Î· = 1 - T 2 T 1 , where T1 is the temperature of the source and T2 is the temperature of the sink.
Here, T2 = 500 K.
â´ 0.5 = 1 - 500 T 1 â T 1 = 1000 K
Now, Î· Â´ = 0.6 = 1 - T 2 Â´ 1000 T 2 Â´ is the new sink temperature
â T 2 Â´ = 400 K .

Topic: Physics, Thermodynamics, NEET
The efficiency of the Carnot engine is $50 \%$ and the temperature of the sink is $500 \mathrm{~K}$. If the temperature of the source is kept constant and its efficiency raised
to $60 \%$, then the required temperature of the sink will be:

Options:
A $100 \mathrm{~K}$
B $600 \mathrm{~K}$
C $400 \mathrm{~K}$
D $500 \mathrm{~K}$

The correct answer is:
$400 \mathrm{~K}$

Solution:
$\begin{aligned}
& \% \eta=\left(1-\frac{T_2}{T_1}\right) \times 100 \\
& \therefore \text { For } 50 \%, \frac{50}{100}=1-\frac{50}{11} \\
& \Rightarrow T_1=1000 \mathrm{~K} \\
& \text { aim for } 60 \%, \frac{60}{100}=1-\frac{T_2}{1000} \\
& \Rightarrow T_2=400 \mathrm{~K}
\end{aligned}$

Topic: Physics, Thermodynamics, NEET, NEET 2002
The efficiency of the cell to take up rDNA is increased by

Options:
A Monovalent cation
B Divalent cation
C Monovalent anion
D Divalent anion

The correct answer is:
Divalent cation

Solution:
Divalent cation like calcium increases the permeability of bacterial cell wall for DNA so that the DNA readily enters the cell.
In order to force bacteria to take up the plasmid, the bacterial cells must first be made âcompetentâ to take up DNA. This is done by treating them with a specific concentration of a divalent cation, such as calcium, which increases the efficiency with which DNA enters the bacterium through pores in its cell wall. Recombinant DNA can then be forced into such cells by incubating the cells with recombinant DNA on ice, followed by placing them briefly at 42 0C (heat shock) and then putting them back on ice. This enables the bacteria to take up the recombinant DNA.

Topic: Biology, Biotechnology - Principles and Processes, NEET
The egrets always forage close to where the cattle are grazing. The interaction between them can be called

Options:
A Brood Parasitism
B Parasitism
C Mutualism
D Commensalism

The correct answer is:
Commensalism

Solution:
This is the interaction in which one species benefits and the other is neither harmed nor benefited. The egrets always forage close to where the cattle are grazing because the cattle, as they move, stir up and flush out insects from the vegetation that otherwise might be difficult for the egrets to find and catch. The egrets get benefitted while the cattle are neither benefitted nor harmed.

Topic: Biology, Organisms and Populations, NEET
The eighth cranial nerve is

Options:
A Optic - Sensory nerve
B Auditory - Motor nerve
C Optic - Motor nerve
D Auditory - Sensory nerve

The correct answer is:
Auditory - Sensory nerve

Solution:
Cranial Nerve	Nature of nerve	Function

Optic	Sensory	Transmits visual signals from the retina of the eye to the brain.

Auditory	Sensory	Mediates sensation of sound, rotation, and gravity (essential for balance and movement). More specifically, the vestibular branch carries impulses for equilibrium and the cochlear branch carries impulses for hearing.

Topic: Biology, Neural Control and Coordination, NEET
The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal(in eV)

Options:
A 7.12 eV
B 3.12 eV
C 5.69 eV
D 4.48 eV

The correct answer is:
4.48 eV

Solution:
Correct Option is : (D)
4.48 eV

Topic: Chemistry, Structure of Atom, NEET
The electric and magnetic field of an electromagnetic wave are:

Options:
A in opposite phase and perpendicular to each other
B in opposite phase and parallel to each other
C in phase and perpendicular to each other
D in phase and parallel to each other

The correct answer is:
in phase and perpendicular to each other

Solution:
In EM waves $\vec{E}$ and $\vec{B}$ are in same phase and perpendicular to each other.

Topic: Physics, Electromagnetic Waves, NEET, NEET 2007
The electric and the magnetic field, associated with an electromagnetic wave, propagating along the $+z$-axis, can be represented by

Options:
A $\left[\mathbf{E}=E_0 \hat{\mathbf{k}}, \mathbf{B}=B_0 \hat{\mathbf{i}}\right]$
B $\left[\mathbf{E}=E_0 \hat{\mathbf{j}}, \mathbf{B}=B_0 \hat{\mathbf{j}}\right]$
C $\left[\mathbf{E}=E_0 \hat{\mathbf{j}}, \mathbf{B}=B_0 \hat{\mathbf{k}}\right]$
D $\left[\mathbf{E}=E_0 \hat{\mathbf{i}}, \mathbf{B}=B_0 \hat{\mathbf{j}}\right]$

The correct answer is:
$\left[\mathbf{E}=E_0 \hat{\mathbf{i}}, \mathbf{B}=B_0 \hat{\mathbf{j}}\right]$

Solution:
$\vec{\mu}=\overrightarrow{\mathbf{E}} \times \overrightarrow{\mathbf{B}}$
$$
=E_0 \hat{\mathbf{i}}+B_0 \hat{\mathbf{j}}=E_0 B_0 \hat{\mathbf{k}}
$$
$\overrightarrow{\mathbf{E}} \times \overrightarrow{\mathbf{B}}$ points in the direction of wave propagation.

Topic: Physics, Electromagnetic Waves, NEET, NEET 2011 (Screening)
The electric and the magnetic field, associated with an EM wave propagating along the +z-axis, can be represented by

Options:
A E⁡→=E⁡0i⁡^B⁡→=B⁡0j⁡^
B E⁡→=E⁡0k⁡^B⁡→=B⁡0i⁡^
C E⁡→=E⁡0j⁡^B⁡→=B⁡0i⁡^
D E⁡→=E⁡0j⁡^B⁡→=B⁡0k⁡^

The correct answer is:
E⁡→=E⁡0i⁡^B⁡→=B⁡0j⁡^

Solution:
The electromagnetic wave is propagating along the +z axis.
Since the electric and magnetic fields are perpendicular to each other and also perpendicular to the direction of propagation of wave.
Also, E⁡ → × B⁡ → gives the direction of wave propagation.
E⁡→=E⁡0i⁡^,B=B⁡0j⁡^∵i⁡^×j⁡^=k^

Topic: Physics, Electromagnetic Waves, NEET
The electric field and electric potential at a point are E and V respectively, then -

Options:
A If E=0, V must be zero
B If V=0, E must be zero
C If E≠0, V may be zero
D If V≠0, E cannot be zero

The correct answer is:
If E≠0, V may be zero

Solution:
A suppose a point where E=0 between two positive charges but net potential is the scalar sum of potential of two charges.
B suppose a point where V=0 between one positive and one negative charge but net electric field is in the direction from positive to negative charge.
C Suppose a point between one positive and one negative charge, where E≠0 but net potential may be zero depending on location of point between them.
D Suppose a point where V≠0 between two positive charges but we cannot say that E cannot be zero because at neutral point we must say that E=0

Topic: Physics, Electrostatics, NEET
The electric field associated with a light wave is E=E0sinâ¡1.57Ã107x-ct where x is in metre and t is in second. If this light is used to produce photoelectric emission from the surface of a metal of work function 1.9 eV, then the stopping potential will be

Options:
A 1.2 V
B 1.5Â V
C 1.75Â V
D 1.9Â V

The correct answer is:
1.2 V

Solution:
From E=E0sin⁡1.57×107x-ct frequency of incident wave is,
v=1.57×107c2π
For stopping potential,
eVs=hv-φ
Vs=6.626×10-341.57×1073×1082π1.6×10-19-1.9
Vs=1.2 V

Topic: Physics, Dual Nature of Matter, NEET
The electric field associated with an electro magnetic wave in vacuum is given by $\mathbf{E}=\mathbf{i}$ $40 \cos \left(k z-6 \times 10^8 t\right)$, where $E$, $z$ and $t$ are in volt/m, metre and second respectively. The value of wave vector $k$ is

Options:
A $2 \mathrm{~m}^{-1}$
B $0.5 \mathrm{~m}^{-1}$
C $6 \mathrm{~m}^{-1}$
D $3 \mathrm{~m}^{-1}$

The correct answer is:
$2 \mathrm{~m}^{-1}$

Solution:
Electromagnetic wave equation
$E=E_0 \cos (k w-\omega t)$
Speed of electromagnetic wave $v=\frac{\omega}{k}$
Given, equation
$\mathbf{E}= 40 \cos \left(k z-6 \times 10^8 t\right) \mathbf{i}$
Comparing Eqs. (i) and (ii), we get
$\omega=6 \times 10^8$
and $E_0=40 \mathbf{i}$
Here, wave factor $k=\frac{\omega}{v}=\frac{6 \times 10^8}{3 \times 10^8}=2 \mathrm{~m}^{-1}$

Topic: Physics, Electromagnetic Waves, NEET, NEET 2012 (Screening)
The electric field at a point on the equatorial plane at a distance r from the centre of a dipole having dipole moment p→ is given by (r>> separation of two charges forming the dipole, ϵ0- permittivity of free space)

Options:
A E→=P→4πϵ0r3
B E→=2P→4πϵ0r3
C E→=-P→4πϵ0r2
D E→=-P→4πϵ0r3

The correct answer is:
E→=-P→4πϵ0r3

Solution:
Use the concept of potential gradient, then the relation, Er=-dVdr⇒Eθ=-1r×dVdθ & V=KPr2cosθ, for general point,
⇒Eθ=-1r×dKPr2cosθdθ ⇒E=-KPr31+3cos2θ, for the equatorial position, θ=90°

⇒E→=-kP→r3=-P→4πϵ0r3.

Topic: Physics, Electrostatics, NEET, NEET 2020 (Phase 2)
The electric field components in Fig. are Ex = a${{x}^{1/2}}$ , Ey = Ez = 0, in which a = 800 N/C ${{m}^{1/2}}$. Calculate the charge within the cube. Assume that a = 0.1 m.

Options:
A 9.27 × ${{10}^{-12}}$C
B 9.27 × ${{10}^{-11}}$C
C 9.27 × ${{10}^{-10}}$C
D 9.27 × ${{10}^{-13}}$C

The correct answer is:
9.27 × ${{10}^{-12}}$C

Solution:
Correct Option is : (A)
9.27 × ${{10}^{-12}}$C

Topic: Chemistry, Electrostatics, NEET
The electric field components in Fig. are Ex = a${{x}^{1/2}}$ , Ey = Ez = 0, in which a = 800 N/C ${{m}^{1/2}}$. Calculate the flux through the cube. Assume that a = 0.1 m.

Options:
A 1.55 N ${{m}^{2}}{{C}^{-1}}$
B 1.95 N ${{m}^{2}}{{C}^{-1}}$
C 0.05 N ${{m}^{2}}{{C}^{-1}}$
D 1.05 N ${{m}^{2}}{{C}^{-1}}$

The correct answer is:
1.05 N ${{m}^{2}}{{C}^{-1}}$

Solution:
Correct Option is : (D)
1.05 N ${{m}^{2}}{{C}^{-1}}$

Topic: Chemistry, Electrostatics, NEET
The electric field in a certain region is acting radially outward and is given by E=Ar. A charge contained in a sphere of radius a centred at the origin of the field will be given by

Options:
A 4πϵ0Aa2
B Aϵ0a2
C 4πϵ0Aa3
D ϵ0Aa3

The correct answer is:
4πϵ0Aa3

Solution:
We know
ESphere=14πε0qr2
14πε0qa2=Aa
q=4πε0Aa3

Topic: Physics, Electrostatics, NEET, NEET 2015 (Phase 1)
The electric field in a certain region is acting radially outward and is given by E=Ar. A charge contained in a sphere of radius a centred at the origin of the field will be given by

Options:
A 4πϵ0Aa2
B Aϵ0a2
C 4πϵ0Aa3
D ϵ0Aa3

The correct answer is:
4πϵ0Aa3

Solution:
We know
ESphere=14πε0qr2
14πε0qa2=Aa
q=4πε0Aa3

Topic: Physics, Electrostatics, NEET
The electric field intensity at all points in space is given by Eâ=3i^-j^ volts/metre. A square frame LMNO of side 1 metre is shown in the figure. The point N lies in x-y plane. The angle between the line ON and the x-axis is Î¸ = 6 0 â then the flux of electric field through frame is

Options:
A 0 V m

B 1 V m
C 2 V m
D 4 V m

The correct answer is:
2 V m

Solution:
The direction of the electric field is in x-y plane as shown in figure
The magnitude of the electric field is E=Ex2+Ey2=3+1=2 V m-1
The direction of the electric field is given by Î¸=tan-1EyEx=tan-113=30â
Hence electric field is normal to square frame LMNO as shown in the figure.
â´ Electric flux =EâÂ·Aâ=E A cosÎ¸=2Ã1=2 V m

Topic: Physics, Electrostatics, NEET
The electric field lines coming out of an equipotential surface are

Options:
A perpendicular to the surface
B parallel to the surface
C in all directions
D zero

The correct answer is:
perpendicular to the surface

Solution:
The electric field lines are always perpendicular to an equipotential surface.

Topic: Physics, Electrostatics, NEET
The electric field of an electric dipole at a point on its axis, at a distance d from the centre of the dipole, varies as

Options:
A 1d
B 1d2
C 1d3
D 1d32

The correct answer is:
1d3

Solution:
The electric field of an electric dipole at a point on its axis at a distance d from the centre of the dipole is
E=2p4πε0d3, where p is the dipole moment of the dipole.
Hence, E∝1d3

Topic: Physics, Electrostatics, NEET
The electric field of certain radiation is given by the equation E=200 sinâ¡4ÏÃ1010t+sinâ¡4ÏÃ1015t falls in a metal surface having work function 2.0 eV. The maximum kinetic energy in eV of the photoelectrons is [Planck's constant h=6.63Ã10-34 J s and electron charge e=1.6Ã10-19 C ]

Options:
A 3.3
B 4.3
C 5.3
D 6.3

The correct answer is:
6.3

Solution:
∵ According to einstein's photoelectric equation,
12mvmax2=hv-w0 .....(i)
Here, w0 is work function and v is frequency of the photon.
According to question equation, the electric field representing given as
E=200 sin⁡4π×1010t+sin⁡4π×1015t
maximum possible angular frequency,
ω=4π×1015 rad
∵ω=2πv or, v=4π×10152π=2×1015 Hz
Put the value of h, v and W0 in Equation (i)
KEmax=6.63×10-34×2×1015-2×1.6×10-19 J
Hence KEmax=6.3 eV

Topic: Physics, Dual Nature of Matter, NEET
The electric field portion of an electromagnetic wave is given by (all variables in SI units) E=10-4sin⁡6×105t-0.01x. The frequency f and the speed v of electromagnetic wave are

Options:
A f=30π kHz and v=1.5×107 m s-1
B f=90π kHz and v=6.0×107 m s-1
C f=300π kHz and v=6.0×107 m s-1
D f=600π kHz and v=7.5×107 m s-1

The correct answer is:
f=300π kHz and v=6.0×107 m s-1

Solution:
ω=6×105=2πf
f=62π×105=3π×105 Hz
=300π×103 Hz
=300π kHz
Wave speed =ωk
=6×1050.01=6×107 m/s

Topic: Physics, Electromagnetic Waves, NEET