datavorous/entrance-exam-dataset huggingface dataset:
An astronomical refracting telescope is being used by an observer to observe planets in normal adjustment. The focal lengths of the objective and eye piece used in the construction of the telescope are $20 \mathrm{~m}$ and $2 \mathrm{~cm}$ respectively. Consider the following statements about the telescope:
(a) The distance between the objective and eye piece is $20.02 \mathrm{~m}$
(b) The magnification of the telescope is $(-) 1000$
(c) The image of the planet is erect and diminished
(d) The aperture of eye piece is smaller than that of objectie
The correct statements are:

Options:
A (a), (b) and (c)
B (b), (c) and (d)
C (c), (d) and (a)
D (a), (b) and (d)

The correct answer is:
(a), (b) and (d)

Solution:
Given,
$\begin{aligned}
& f_e=2 \mathrm{~cm}, \text { and } \\
& f_o=20 \mathrm{~m}=2000 \mathrm{~cm}
\end{aligned}$
For normal adjustment,
$\begin{aligned}
\text {Magnification power } & =\frac{-f o}{f e}=\frac{-2000}{2} \\
& =-1000 \\
\text {Length of telescope } & =f o+f e \\
& =2000+2 \\
& =2002 \mathrm{~cm} \\
& =20.02 \mathrm{~m}
\end{aligned}$
The image formed is inverted and magnified, and the aperture of objective is larger than eye piece of the telescope.
$\left(f_o~ \& ~f_e\right.$ are focal lengths of objective & eye piece respectively.)

Topic: Physics, Ray Optics, NEET, NEET 2022 (Phase 2)
An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of

Options:
A large focal length and large diameter
B large focal length and small diameter
C small focal length and large diameter
D small focal length and small diameter

The correct answer is:
large focal length and large diameter

Solution:
For astronomical refracting telescope
Angular magnification is more for large focal length
Of objective lens M.P=f0fe
Resolving power =D1.22λ
Resolving power is high for large diameter.
So, objective lens should have large focal length (f.) and large diameter D

Topic: Physics, Ray Optics, NEET, NEET 2018
An astronomical telescope has a large aperture to

Options:
A reduce spherical aberration
B have high resolution
C increase span of observation
D have low dispersion

The correct answer is:
have high resolution

Solution:
Large aperture leads to high resolution of telescope.

Topic: Physics, Ray Optics, NEET
An astronomical telescope has an angular magnification of magnitude 5 for an object. The separation between the objective and the eye-piece is 36 cm. The final image is formed at infinity. The focal length f0 of the objective and fe of the eye-piece are

Options:
A 45 cm and -9 cm
B 50 cm and 10 cm
C 72 cm and 5 cm
D 30 cm and 6 cm

The correct answer is:
30 cm and 6 cm

Solution:
Magnification, m=-5=-f0feâf0=5fe
âL=f0+fe=36
â6fe=36â´fe=6 cm
âf0=5fe=30 cm

Topic: Physics, Ray Optics, NEET
An astronomical telescope has an angular magnification of magnitude 5 for an object. The separation between the objective and the eye-piece is 36 cm. The final image is formed at infinity. The focal length f0 of the objective and fe of the eye-piece are

Options:
A 45 cm and -9 cm

B 50 cm and 10 cm
C 72 cm and 2 cm
D 30 cm and 6 cm

The correct answer is:
30 cm and 6 cm

Solution:
Magnification, m=-5=-f0feâf0=5fe
âL=f0+fe=36
â6fe=36â´fe=6 cm
âf0=5fe=30 cm

Topic: Physics, Ray Optics, NEET
An astronomical telescope has objective and eye-piece lenses of powers 0.5 D and 20 D respectively. The magnifying power of the Telescope is

Options:
A 30
B 10
C 40
D 20

The correct answer is:
40

Solution:
Given , the power of objective lens,
P0 = 0.5 D
The power of eye-piece lens,
Pe= 20 D
The magnifying power of an astronomical telescope
M=f0fe
or M=PeP0 â´P=1f
=200.5=40

Topic: Physics, Ray Optics, NEET
An astronomical telescope has objective and eye-piece lenses of powers 0.5 D and 20 D respectively. The magnifying power of the telescope is

Options:
A 10
B 20
C 30
D 40

The correct answer is:
40

Solution:
Optical power (also referred to as dioptric power, refractive power, focusing power, or convergence power) is the degree to which a lens, mirror, or other optical system converges or diverges light. It is equal to the reciprocal of the focal length of the device, P=1f

Pobjective=0.5 D

fobjective=2

Peye-piece=20 D

feye-piece=0.05

The magnifying power (M) of an astronomical telescope is given by:

M= focal length of objectivefocal length of eye-piece

M=20.05

M=2005

M=40

Topic: Physics, Ray Optics, NEET
An athletic coach told his team that muscle multiplied with speed equal to power. What dimensions does he view for muscle?

Options:
A MLT-2
B ML2T-2
C MLT-3
D L

The correct answer is:
MLT-2

Solution:
Accordind to the question
Muscle×Speed=Power
Muscle = Power Speed = MLT - 2

Topic: Physics, Units and Dimensions, NEET
An atom has 26 electrons and its atomic weight is 56. The number of neutrons in the nucleus of the atom will be

Options:
A 26
B 30
C 36
D 56

The correct answer is:
30

Solution:
Atomtextc number is 26 =P=e

atomtextc weight is 56 =P+N
so representation is 26X56
N=A-E=56-26=30

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
An atom has electronic configuration $1 s^2$ $2 p^6 3 s^2 3 p^6 3 d^3 4 s^2$, in which group it will be placed?

Options:
A Fifth
B Fifteenth
C Second
D Third

The correct answer is:
Fifth

Solution:
Since the $3 d$ contains 3 electrons, thus the given elements is Vanadium, which is placed in $5^{\text {th }}$ group and $4^{\text {th }}$ period.
The number of each period represents the number of energy levels that have electrons in them for atoms of each element in that period. The group number in the periodic table represents number of valence electrons of the elements in a certain group.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET, NEET 2002
An atom having atomic mass number 13 has 7 neutrons. What is the atomic number of the atom?

Options:
A 5
B 6
C 7
D 8

The correct answer is:
6

Solution:
Correct Option is : (B)
6

Topic: Chemistry, Structure of Atom, NEET
An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.

Options:
A (i)6 and (ii)[He]2s22p2
B (i)28 and (ii)[Ar]3d10
C (i)29 and (ii)[Ar]3d104s1
D (i)29 and (ii)[Ar]3d104s2

The correct answer is:
(i)29 and (ii)[Ar]3d104s1

Solution:
Correct Option is : (C)
(i)29 and (ii)[Ar]3d104s1

Topic: Chemistry, Structure of Atom, NEET
An atomic orbital has n = 3. What are the possible values of l ?

Options:
A 0,1,2,3
B 1,2,3
C 0,1,2
D 1,2

The correct answer is:
0,1,2

Solution:
Correct Option is : (C)
0,1,2

Topic: Chemistry, Structure of Atom, NEET
An automobile moves on a road with a speed of 54 km h-1 . The radius of its wheels is 0.45âm and the moment of inertia of the wheel about its axis of rotation is 3 kg m2 . If the vehicle is brought to rest in 15 s , the magnitude of average torque transmitted by its brakes to the wheel is:

Options:
A 8.58 kg m2 s-2
B 10.86Â kgÂ m2Â s-2
C 2.86Â kgÂ m2Â s-2
D 6.66 kg m2 s-2

The correct answer is:
6.66 kg m2 s-2

Solution:
ωi=150.45=1003 ωf=0
ωf=ωi+αt
0=1003+-α(15)
α=10045
τ=Iα=3×10045=6.66 N m

Topic: Physics, Rotational Motion, NEET, NEET 2015 (Phase 2)
An azeotropic solution of two liquids has boiling point lower than either of them when it

Options:
A shows negative deviation from Raoultâs law
B shows no deviation from Raoultâs law
C shows positive deviation from Raoultâs law
D is saturated

The correct answer is:
shows positive deviation from Raoultâs law

Solution:
In case of positive deviation from Raoult's law, the observed vapour pressure is greater than the ideal vapour pressure and boiling point of azeotrope becomes lower than either of pure liquid.

Topic: Chemistry, Solutions, NEET
An azo dye is fixed on fabrics by the process applicable in

Options:
A Vat dyes
B Mordant dyes
C Developed dyes
D Substantive dyes

The correct answer is:
Developed dyes

Solution:
Ingrain or developed dyes are those which are synthesized directly on the fabric e.g. azo dyes. Ingrain dyeing is particularly suitable for cotton fabric.

Topic: Chemistry, Chemistry in Everyday Life, NEET
An eagle flies at constant velocity horizontally across the sky, carrying a mouse and releases the mouse while in flight. From the eagle's perspective, the mouse falls vertically with speed v1. From an observer on the ground's perspective, the mouse falls at an angle with speed v2. What is the speed of the eagle with respect to the observer on the ground?

Options:
A v1+v2
B v1-v2
C v12-v22
D v22-v12

The correct answer is:
v22-v12

Solution:
Let the speed of the eagle is v with respect to ground. Diagram of velocities, according to the questions, where, v1 is speed of mouse with respect to the eagle and v2 is speed of mouse with respect to the ground.
Thus, we have
v2=v2+v12+2vv1cos⁡90o
∴v2=v2+v12
v22=v2+v12
∴v=v22-v12

Topic: Physics, Motion In Two Dimensions, NEET
An earthen pitcher loses 1 g of water per minute due to evaporation. If the water equivalent of the pitcher is 0.5 kg and the pitcher contains 9.5âkg of water, then calculate the time required for the water in pitcher to cool to 28Â°C from its original temperature of 30Â°C. Neglect the effect of radiation. Latent heat of vaporization of water in this range of temperature is 580ââcal/g and specific heat of water is 1 cal/g/Â°C.

Options:
A 30.5  min
B 41.2  min
C 38.6  min
D 34.5  min

The correct answer is:
34.5  min

Solution:
Heat lost by (Water + Pitcher)
Q1=m+MÃsÃâT
=9.5+0.5Ã103Ã30-28
=20Ã103 cal
Heat gained for the water to evaporate:
(Let t be time in min)
Q2=mÃL
=dmdtÃtL
=1ÃtÃ580
=580t cal
So,
Q1=Q2
580t=20Ã103
t=20000580=34.5 min

Topic: Physics, Thermal Properties of Matter, NEET
An earth-orbiting satellite has a solar energy collecting panel with total area 5 m2. If solar radiations are perpendicular and completely absorbed, the average force associated with the radiation pressure is
(Solar constant=1.4 kWm-2)

Options:
A 2.33Ã10-3Â N
B 2.33Ã10-4Â N
C 2.33Ã10-5 N
D 2.33Ã10-6Â N

The correct answer is:
2.33Ã10-5 N

Solution:
We know that, Power=IÃarea=(1.4Ã103 )Ã5
Force F=Powerc=1.4Ã103 Ã53Ã108
=2.33Ã10-5 N

Topic: Physics, Dual Nature of Matter, NEET
An earthquake generates both transverse S and longitudinal P sound waves in the earth. The speed of S waves is about 6 km s-1 and that of P waves is about 9 km s-1. A seismograph records P and S waves from an earthquake. The first P wave arrives 5 minute before the first S wave. The epicenter of the earthquake is located at a distance of about -

Options:
A 54 km
B 540 km
C 5400 km
D 72 km

The correct answer is:
5400 km

Solution:
Let is= speed of transverse wave
and ip= speed of longitudinal wave
dis-dip=t
d6-d9=5(60)
d = 5400 km

Topic: Physics, Waves and Sound, NEET
An ecosystem which can be easily damaged but can recover after some time if damaging effect stops will be having:

Options:
A low stability and high resilience
B high stability and low resilience
C low stability and low resilience
D high stability and high resilience.

The correct answer is:
low stability and high resilience

Solution:
Stability is the power of a system to be in their state against unfavourable factor the capability of regaining its original shape or position after being deformed. Hence, it has low stability and high resilience.

Topic: Biology, Ecosystem, NEET, NEET 2004
An ecosystem which can be easily damaged but can recover after some time if damaging effect stops will be having

Options:
A Low stability and high resilience
B High stability and low resilience
C Low stability and low resilience
D High stability and high resilience

The correct answer is:
Low stability and high resilience

Solution:
An ecosystem having low stability can be easily damaged. An ecosystem having high resilience will take less time to recover.

Topic: Biology, Ecosystem, NEET
An ecotype is a type of species in which environmentally induced variations are

Options:
A Temporary
B Genetically fixed
C Genetically not related
D None of the above

The correct answer is:
Genetically fixed

Solution:
A local population adapted genetically to its particular environment is called ecotype. There may be several ecotypes of the same organism which show variation amongst them.

Topic: Biology, Ecosystem, NEET
An electric bulb rated for 500 W at 100 V is used in a circuit having a 200 V supply. The resistance R that must be put in series with the bulb, so that the bulb is drawn 500 W is

Options:
A 18 Î©
B 20 Î©
C 40 Î©
D 700 Ω

The correct answer is:
20 Î©

Solution:
P=VI, I=PV
or I=500 W100 V=5 A
Now, 5R=100
or R=20 Î©.

Topic: Physics, Current Electricity, NEET
An electric current i enters and leaves a uniform circular wire of radius a through diametrically opposite points. A charged particle q moving along the axis of the circular wire passes through its centre at speed v. The magnetic force acting on the particle when it passes through the centre has a magnitude

Options:
A qvμ0i2a
B qvμ0i2πa
C qvμ0ia
D zero

The correct answer is:
zero

Solution:
Magnetic field due to each half is equal in magnitude but opposite in direction so net field at the centre of the circular wire is zero. Therefore magnetic force acting on the particle when it passes through the centre is zero.

Topic: Physics, Magnetic Effects of Current, NEET
An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii of the wires are in the ratio of 4/3 and 2/3, then the ratio of the currents passing through the wire will be

Options:
A 3
B 8/9
C 2
D 1/3

The correct answer is:
1/3

Solution:
V = I 1 R 1 = I 1 ρ ℓ 1 π r 1 2
V = I 2 R 2 = I 2 ρ ℓ 1 π r 2 2
⇒ I 1 R 1 = I 2 R 2
⇒ I 1 I 2 = R 2 R 1 = ρ ℓ 2 π r 2 2 × π r 1 2 ρ ℓ 1 = ℓ 2 ℓ 1 r 1 r 2 2 = 3 4 × 2 3 2 = 1 3

Topic: Physics, Current Electricity, NEET
An electric dipole is fixed at the origin of coordinates. Its moment is directed in the positive x-direction. A positive charge is moved from the point (r, 0) to the point (âr, 0) by an external agent. In this process, the work done by the agent is -

Options:
A positive and inversely proportional to r
B positive and inversely proportional to r2
C negative and inversely proportional to r
D negative and inversely proportional to r2

The correct answer is:
negative and inversely proportional to r2

Solution:
The potential due to the dipole is positive at
A(r, 0) and negative at B(âr, 0). Also this potential is inversely proportional to r2. Work is done by agent is also inversely proportional to r2
Wext.AâB= q (VB â VA) = âve

Topic: Physics, Electrostatics, NEET
An electric dipole is kept on the axis of a uniformly charged ring at distance R2 from the centre of the ring. The direction of the dipole moment is along the axis. The dipole moment is P, the charge of the ring is Q and the radius of the ring is R. The force on the dipole is nearly

Options:
A ​ 4kPQ 3 3 R 2
B 4kPQ 3 3 R 3
C 2kPQ 3 3 R 3
D Zero

The correct answer is:
Zero

Solution:
At x= R 2 , the electric field due to the ring is maximum.

The force on a dipole in a non-uniform field is

F=pdEdx

But at the point of maxima, dEdx=0

This means F=0

Topic: Physics, Electrostatics, NEET
An electric dipole is placed as shown in the figure.

The electric potential (in 102 V) at point P due to the dipole is (ε0=permittivity of free space and 14πε0=K)

Options:
A 83qK
B 38qK
C 58qK
D 85qK

The correct answer is:
38qK

Solution:
The net electric potential at P can be calculated as follows-

Vp=Kq5 cm-3 cm×1 m102 cm-Kq5 cm+3 cm×1 m102 cm=Kq×102×12-18=3Kq8×102 V

Topic: Physics, Electrostatics, NEET, NEET 2023 (All India)
An electric dipole is placed at an angle of 30Â° with an electric field intensity 2Ã105 N C-1. It experiences a torque equal to 4âNâm . The charge on the dipole, if the dipole length is 2ââcm , is

Options:
A 8 mC
B 2 mC
C 5 mC
D 7 Î¼C

The correct answer is:
2 mC

Solution:
Torque on the dipole in uniform electric field is given as τ=PEsin⁡θ
⇒τ=qlEsin⁡θ
⇒q=τlEsin⁡θ=42×10-2×0.5×2×105=2×10-3 C

Hence, charge on dipole is 2 mC.

Topic: Physics, Electrostatics, NEET, NEET 2016 (Phase 2)
An electric dipole is placed at an angle of 30Â° with an electric field intensity 2Ã105 N C-1. It experiences a torque equal to 4âNâm . The charge on the dipole, if the dipole length is 2ââcm , is

Options:
A 8 mC
B 2 mC
C 5 mC
D 7 Î¼C

The correct answer is:
2 mC

Solution:
Ï=PEsinâ¡Î¸
âÏ=qlEsinâ¡Î¸
âq=ÏlEsinâ¡Î¸
=42Ã10-2Ã0.5Ã2Ã105=2 mC

Topic: Physics, Electrostatics, NEET
An electric dipole is placed at an angle of 30° with an electric field of intensity 2×105 N C-1. It experiences a torque equal to 4 N m. Calculate the magnitude of charge on the dipole, if the dipole length is 2 cm.

Options:
A 2 mC
B 8 mC
C 6 mC
D 4 mC

The correct answer is:
2 mC

Solution:
Torque experienced by the electric dipole is given by,

τ→=p→×E→⇒4=pEsinθ⇒4=ql×2×105×sin30°⇒q=4105×2×10-2=2×10-3 C=2 mC

Topic: Physics, Electrostatics, NEET, NEET 2023 (All India)
An electric dipole is placed at the origin and is directed along the x-axis. At a point P, far away from the dipole, the electric field is parallel to the y-axis. OP makes an angle θ with the x-axis, then -

Options:
A tanÂ Î¸=Â 3
B tan Î¸ =2
C Î¸Â =Â 45Â°
D None of these

The correct answer is:
tan Î¸ =2

Solution:
Direction of electric field from axis of dipole is given by =Î¸ + tanâ1 tanÎ¸2
Here given that E is along y-axis.
Î¸ + tanâ1tanÎ¸2=Ï2
tanÎ¸2=cotÎ¸
tanÎ¸ = 2 cot Î¸âtan2Î¸ = 2
â´tanÎ¸ = 2

Topic: Physics, Electrostatics, NEET
An electric dipole of dipole moment $p$ is aligned parallel to a uniform electric field $E$. The energy required to rotate the dipole by $90^{\circ}$ is:

Options:
A $p^2 E$
B $p E$
C infinity
D $p E^3$

The correct answer is:
$p E$

Solution:
Potential energy of dipole
$$
\begin{aligned}
& U=-p E \\
& \left(\cos \theta_2-\cos \theta_1\right) \\
& \text { Here } \quad \theta_2=0^{\circ} \\
& =-p E \\
& (\cos 90-\cos \theta) \\
& \theta_2=90^{\circ} \\
& \stackrel{L}{U}=P E \\
&
\end{aligned}
$$

Topic: Physics, Electrostatics, NEET, NEET 2013 (Karnataka)
An electric dipole of dipole moment p→ is placed in a uniform electric field E→ such that p→ is perpendicular to E→ . The work done to turn the dipole through an angle of 180o is

Options:
A zero
B pE
C 2pE
D 2pE

The correct answer is:
zero

Solution:
W=Uf-Ui
W=-pEcos270°--pEcos90°
W=0

Topic: Physics, Electrostatics, NEET
An electric dipole, of length 2 cm, is placed with its axis making an angle 30Â° with a uniform electric field. If it experiences a torque of 103 N m, then potential energy of the dipole is

Options:
A â10Â J
B â20Â J
C â30 J
D âÂ 40Â J

The correct answer is:
â30 J

Solution:
τ = PE sin 30
103 = PE2
PE = 203
Potential energy = –PE cos 30
potential energy = –203 × 32
= – 10 × 3 = –30 J

Topic: Physics, Electrostatics, NEET
An electric dipole of length 2 cm is placed with its axis making an angle of 30Â° to a uniform electric field 105 N C-1. If it experiences a torque of 103 N m, then potential energy of dipole is

Options:
A -10Â J
B -20Â J
C -30 J
D -40Â J

The correct answer is:
-30 J

Solution:
We know that
τ=pEsin⁡θ
⇒τ=pEsin⁡30
⇒103=pE2
⇒pE=203
Potential energy, U=-pEcos⁡30°
∴ Potential energy U=-203×32
=-10×3=-30 J

Topic: Physics, Electrostatics, NEET
An electric dipole of moment $\vec{p}$ is lying along a uniform electric field $\vec{E}$. The work done in rotating the dipole by rotating the dipole by 90 degrees:

Options:
A $p E$
B $\sqrt{2} p E$
C $p E / 2$
D $2 P E$

The correct answer is:
$p E$

Solution:
Work done
$$
\begin{aligned}
W & =\int_0^\theta P E \sin \theta d \theta \\
& =P E(1-\cos \theta) \\
\therefore \quad \text { Since } \theta & =90^{\circ} \\
W & =P E(1-\cos 90) \\
Q & =P E
\end{aligned}
$$

Topic: Physics, Electrostatics, NEET, NEET 2006
An electric dipole of moment $p$ is placed in an electric field of intensity $E$. The dipole acquires a position such that the axis of the dipole makes an angle $\theta$ with the direction of the field. Assuming that the potential energy of the dipole to be zero when $\theta=90^{\circ}$, the torque and the potential energy of the dipole will respectively be

Options:
A $p E \sin \theta,-p E \cos \theta$
B $p E \sin \theta,-2 p E \cos \theta$
C $p E \sin \theta, 2 p E \cos \theta$
D $p E \cos \theta,-p E \sin \theta$

The correct answer is:
$p E \sin \theta,-p E \cos \theta$

Solution:
Here, Torque $\tau=p E \sin \theta$
Potential energy of the dipole
$\begin{aligned}
U & =-\int \tau d \theta \\
& =-\int_{\pi / 2}^0 p E \sin \theta d \theta=p E[\cos \theta-0] \\
& =-p E \cos \theta
\end{aligned}$

Topic: Physics, Electrostatics, NEET, NEET 2012 (Screening)
An electric fan has blades of length 30 cm as measured from the axis of rotation. If the fan is rotating at 1200 rpm, the acceleration of a point on the tip of the blade is about

Options:
A 1600 m s-2
B 4740 m s-2
C 2370 m s-2
D 5055 m s-2

The correct answer is:
4740 m s-2

Solution:
Given, f=1200 rpm,
r=30 cm=30100 m
Acceleration of a point at the tip of the blade
= centripetal acceleration
= ω2r=2πf2r
=2×227×1200602×30100
=4740 m s-2

Topic: Physics, Laws of Motion, NEET
An electric field is applied to a semiconductor. Let the number of charge carries be n and the average drift speed be v. If the temperature is increased, then

Options:
A both n and v will increase
B n will increase and v will decrease
C both n and v will decrease
D n will decrease and v will increase

The correct answer is:
n will increase and v will decrease

Solution:
As the temperature is increased, more electrons are able to move from valance band to conduction band, hence the number of charge carriers increases but with increase in the temperature the frequency of collisions increases so the average drift speed decreases

Topic: Physics, Current Electricity, NEET
An electric field is given by E→=yi^+xj^ N C-1. The work done in moving a 1 C charge from rA→=2i^+2j^ m to rB→=4i^+j^ m is

Options:
A + 4 J

B – 4 J

C + 8 J

D Zero

The correct answer is:
Zero

Solution:
A = (2, 2) and B = (4, 1)
Now WA⟶B=qVB-VA ... (1)
∫ABdV=-∫ABE→·d→r
or VB-VA=-∫2241yi^+xj^·dxi^+dyj^+dzk^
or VB-VA=-∫2241ydx+xdy
=-∫2241dxy=-xy2,24,1=0
∴ WA→B=0 [from Eq. (1)]

Topic: Physics, Electrostatics, NEET
An electric field is uniform, and in the positive x direction for positive x, and uniform with the same magnitude but in the negative x direction for negative x. It is given that E = 200$\hat{i}$ N/C for x > 0 and E = –200 $\hat{i}$ N/C for x < 0. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at x = +10 cm and the other is at x = –10 cm. What is the flux through the side of the cylinder

Options:
A 0.96 N ${{m}^{2}}{{C}^{-1}}$
B 2.21 N ${{m}^{2}}{{C}^{-1}}$
C 0 N ${{m}^{2}}{{C}^{-1}}$
D 1.57 N ${{m}^{2}}{{C}^{-1}}$

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

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

Topic: Chemistry, Electrostatics, NEET
An electric field is uniform, and in the positive x direction for positive x, and uniform with the same magnitude but in the negative x direction for negative x. It is given that E = 200$\hat{i}$ N/C for x > 0 and E = –200 $\hat{i}$ N/C for x < 0. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at x = +10 cm and the other is at x = –10 cm . What is the net outward flux through each flat face?

Options:
A 0.96 N ${{m}^{2}}{{C}^{-1}}$
B 2.21 N ${{m}^{2}}{{C}^{-1}}$
C 1.23 N ${{m}^{2}}{{C}^{-1}}$
D 1.57 N ${{m}^{2}}{{C}^{-1}}$

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

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

Topic: Chemistry, Electrostatics, NEET