datavorous/entrance-exam-dataset huggingface dataset:
In majority of the angiosperms, pollen is released in a two-celled stage. The two cells are

Options:
A the gamete and generative cell
B the vegetative cell and tube nucleus
C two male gametes
D the vegetative cell and generative cell

The correct answer is:
the vegetative cell and generative cell

Solution:
A pollen grain on maturation undergoes one unequal mitosis to form a smaller generative cell and a larger vegetative cell. In 60% angiosperms, pollen grain is liberated at this two-celled stage. Thus, a mature pollen grain consists of 2 cells. In the remaining species, the generative cell divides mitotically to give rise to the two male gametes before pollen grains are shed (3-celled stage).

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In man, recessive characters like haemophilia are mainly transmitted through

Options:
A Y-chromosome
B Chromosome number 12
C X-chromosome
D Chromosome number 11

The correct answer is:
X-chromosome

Solution:
Haemophilia is a sex linked recessive disease, which shows its transmission from unaffected carrier female to some of the male progeny has been widely studied. In this disease, a single protein that is a part of the cascade of proteins involved in the clotting of blood is affected.

Topic: Biology, Principles of Inheritance and Variation, NEET
In man the axial skeleton is made up of

Options:
A 80 bones
B 100 bones
C 103 bones
D 106 bones

The correct answer is:
80 bones

Solution:
It forms the main axis. It consists of the skull – 22 bones, vertebral column – 26 vertebrae, sternum – 1, and ribs – 24, Hyoid - 1, and ear ossicles - 6.

Topic: Biology, Locomotion and Movement, NEET
In many situations, to find the electric field using the gauss law, we can use the simplified equation Eâ=qenclosedÎµ0Aâ, where Îµ0 is the permittivity of free space, A is the area of Gaussian surface and qenclosed is the net charge enclosed by the Gaussian surface. Among the following situations, this equation can be used

Options:
A only when the Gaussian surface is an equipotential surface
B only when E→ is constant on the surface
C only when the Gaussian surface is an equipotential surface and E→ is constant of the surface
D for any choice of Gaussian surface

The correct answer is:
only when the Gaussian surface is an equipotential surface and E→ is constant of the surface

Solution:
Magnitude of electric field is constant $\&$ the surface is equipotential

Topic: Physics, Electrostatics, JEE Main
In materials like aluminium and copper, the correct order of magnitude of various elastic modulii is :

Options:
A Young's modulii < shear modulii < bulk modulii
B Bulk modulii < shear modulii < Young's modulii
C Shear modulii < Young's modulii < bulk modulii
D Bulk modulii < Young's modulii < shear modulii

The correct answer is:
Shear modulii < Young's modulii < bulk modulii

Solution:
For aluminium and copper magnitude of Shear modulii is less than Young's modulii and bulk modulii is maximum.

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2014 (09 Apr Online)
In \( \mathrm{H} \)-atom, if ' \( \mathrm{x} \) ' is the radius of the first Bohr orbit, de Broglie wavelength of an electron in the \( 3^{\mathrm{rd}} \) orbit is:

Options:
A \( 3 \pi \mathrm{x} \)
B \( 6 \pi \mathrm{x} \)
C \( \frac{9 x}{2} \)
D \( \frac{x}{2} \)

The correct answer is:
\( 6 \pi \mathrm{x} \)

Solution:
According to the Bohr model, the first radius for hydrogen atom:

r1=0.529A°=x

The third radius of the Bohr model:r3=0.529 n2 A° = 0.52932A°=9x

Hence, the de Broglie wavelength:

λ=2πr3n=2×π×(9x)3=6πx

Topic: Chemistry, Structure of Atom, JEE Main
In \( \mathrm{HVZ} \) reaction substitution takes place at:

Options:
A carbon atom of \( -\mathrm{COOH} \) group
B \( \alpha \)-carbon atom
C Both \( \mathrm{A} \) and \( \mathrm{B} \)
D None of the above

The correct answer is:
\( \alpha \)-carbon atom

Solution:
The Hell Volhard Zelinsky halogenation reaction halogenates carboxylic acids at the α-carbon atom. Carboxylic acids react with chlorine or bromine in presence of small amount of red phosphorous to give alpha halo carboxylic acids.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
In Mayer's relation, CP-CV=R
'R' stands for

Options:
A translational kinetic energy of 1 mol gas
B rotational kinetic energy of 1 mol gas
C vibrational kinetic energy of 1 mol gas
D work done to increase the temperature of 1 mol gas by one degree

The correct answer is:
work done to increase the temperature of 1 mol gas by one degree

Solution:
PV = RT (For 1 mol) …(i)
PV+ΔV=RT+1 …(ii)
From equation (i) and (ii),
PΔV=R
i.e., W = R (work done).

Topic: Chemistry, Thermodynamics (C), JEE Main
In Melamine the total number of N-atoms having sp2 hybridisation are?

Options:

The correct answer is:
3

Solution:
In Melamine 3-N-atoms are sp2 hybridised.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
In metallurgy the term "gangue" is used for

Options:
A Contamination of undesired earthy materials.
B Contamination of metals, other than desired metal
C Minerals which are naturally occuring in pure form
D Magnetic impurities in an ore.

The correct answer is:
Contamination of undesired earthy materials.

Solution:
Gangue is generally an unwanted earthy material or impurity that is present in an ore. This can be found in ore deposits where the mineral is present. During the process of extraction, these impurities are mixed up with ore in the form of stone, sand, rock, etc.

The gangue must be separated from the mineral with the help of ore dressing or mineral dressing. Concentration of ore is one such process used to remove the impurities, which may depend on the physical and chemical properties. Example: Gravity separation and electromagnetic separation are some of the common methods which are used for the concentration of ore.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2022 (29 Jul Shift 1)
In meter bridge experiment for measuring unknown resistance 'S', the null point is obtained at a distance 30 cm from the left side as shown at point D. If R is 5.6 kΩ, then the value of unknown resistance 'S' will be _____ Ω.

Options:

The correct answer is:
2400

Solution:
Using balanced wheatstone condition in meter bridge

RS=7030

Given here, R=5.6 kΩ.

The, S30=5.6×10370

S=37×5.6×103=2400 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (27 Jul Shift 1)
In meter bridge experiment, null point was obtained at a distance ' $l$ ' from left end. The values of resistances in the left and right gap are doubled and then interchanged. The new position of the null point is

Options:
A $(100-l)$
B $(100-2 l)$
C $(100-3 l)$
D $\left(100-\frac{l}{2}\right)$

The correct answer is:
$(100-l)$

Solution:
$\begin{aligned}
& \text { For a Metre bridge, } \\
& \frac{\mathrm{X}}{\mathrm{R}}=\frac{l_{\mathrm{X}}}{l_{\mathrm{R}}}
\end{aligned}$
After doubling and interchanging the left and right gap resistances, the ratio of the resistances do not change.
This means the null point does not change.
$\therefore \quad$ The new position will remain at $(100-l)$.

Topic: Physics, Experimental Physics, MHT CET, MHT CET 2023 (10 May Shift 2)
In meter bridge or Wheatstone bridge for measurement of resistance, the known and the unknown resistances are interchanged. The error so removed is

Options:
A End correction
B Index error
C Due to temperature effect
D Random error

The correct answer is:
End correction

Solution:
In meter bridge experiment, it is assumed that the resistance of the $L$ shaped plate is negligible, but actually it is not so. The error created due to this is called, end error. To remove this the resistance box and the unknown resistance must be interchanged and then the mean reading must be taken.

Topic: Physics, Current Electricity, JEE Main
In methyl alcohol solution, bromine reacts with ethylene to yield $\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{OCH}_3$ in addition to 1,2-dibromoethane because

Options:
A The ion formed initially may react with $\mathrm{Br}^{-}$or $\mathrm{CH}_3 \mathrm{OH}$
B The methyl alcohol solvates the bromine
C The reaction follows Markownikoff's rule
D This is a free-radical mechanism

The correct answer is:
The ion formed initially may react with $\mathrm{Br}^{-}$or $\mathrm{CH}_3 \mathrm{OH}$

Solution:
$\mathrm{CH}_2=\mathrm{CH}_2+\mathrm{Br}_2 \xrightarrow{\mathrm{CH}_3 \mathrm{OH}}$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
In mildly alkaline medium, thiosulphate ion is oxidized by \( \mathrm{MnO}_{4}^{-} \)to " \( \mathrm{A} " . \) The oxidation state of sulphur in " \( \mathrm{A} " \) is

Options:

The correct answer is:
6

Solution:
MnO4-+S2O32-→MnO2+SO42-

Oxidation state of 'S' in SO42-

=+6

Topic: Chemistry, Redox Reactions, JEE Main
In mildly alkaline medium, thiosulphate ion is oxidized by MnO4- to "A". The oxidation state of sulphur in "A" is_

Options:

The correct answer is:
6

Solution:
MnO4-+S2O32-→MnO2+SO42-

Oxidation state of 'S' in SO42-

=+6

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2021 (26 Feb Shift 2)
In Millikan's oil drop experiment an oil drop carrying a charge \( Q \) is held stationary by a potential difference \( 2400 \mathrm{~V} \) between the plates. To keep a drop of half the radius stationary the potential difference had to be made \( 600 \mathrm{~V} \). What is the charge on the second drop?

Options:
A \( \frac{Q}{4} \)
B \( \frac{Q}{2} \)
C \( Q \)
D \( \frac{3 Q}{2} \)

The correct answer is:
\( \frac{Q}{2} \)

Solution:
Let radius of drop is r. According to Millikan, for balance of drop
QE=mg, where Q is the charge on the drop, E is electric field, m is the mass of the drop and g is the acceleration due to gravity.

or QVd=43πr3ρg, where E=Vd,

where, V is potential difference and ρ is charge density of drop

∴ Q1Q2=r1r23×V2V1

∴ QQ2=rr23×6002400=2

∴ Q2=Q2

Topic: Physics, Electrostatics, JEE Main
In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius 2.0×10-5 m and density 1.2×103 kg m-3? Take viscosity of liquid =1.8×10-5 N s m-2. (Neglect buoyancy due to air).

Options:
A 5.8×10-10 N
B 3.9×10-10 N
C 1.8×10-10 N
D 3.8×10-11 N

The correct answer is:
3.9×10-10 N

Solution:
Use the concept the viscous force is the same as the weight of the body,

⇒F=mg=ρ×43πr3×g=3.9×10-10 N.

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2021 (27 Aug Shift 1)
In mitochondria, protons accumulate in the

Options:
A outer membrane
B inner membrane
C intermembrane space
D matrix

The correct answer is:
intermembrane space

Solution:
During the passage of two electrons from one an electron carrier (NADH) to another on ETC, enough energy is liberated that pumps out three pairs of protons $\left(\mathrm{H}^{+}\right)$from matrix (inner chamber) to outer chamber (inter membrane space) of mitochondria. The accumulation of protons in inter membrane space results their higher concentration than in the matrix creating a proton gradient $(\Delta \mathrm{pH})$ and electric potential $(\Delta \psi)$ across the membrane.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2011 (Mains)
In mixture A and B component show -ve deviation as

Options:
A
$\Delta V_{\text {mix }}>0$

B
$\Delta \mathrm{H}_{\text {mix }} < 0$
C
A - B interaction is weaker than A - A and B - B interaction

D
A - B interaction is stronger than A - A and B - B interaction

The correct answer is:
$\Delta \mathrm{H}_{\text {mix }} < 0$

Solution:
$\left[\Delta \mathrm{H}_{\text {mix }} < 0\right]$

Topic: Chemistry, Solutions, JEE Main
In mixture A and B component show -ve deviation as

Options:
A
$\Delta V_{\text {mix }}>0$

B
$\Delta \mathrm{H}_{\text {mix }} < 0$
C
A - B interaction is weaker than A - A and B - B interaction

D
A - B interaction is stronger than A - A and B - B interaction

The correct answer is:
$\Delta \mathrm{H}_{\text {mix }} < 0$

Solution:
$\left[\Delta \mathrm{H}_{\text {mix }} < 0\right]$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2002
In Mond's process for refining of nickel, the following reactions occur. The value of 'x' is ______.
Ni + xCO → 330 K - 350 K Ni CO x → 450 K - 407 K Ni s  +  xCO g

Options:

The correct answer is:
4

Solution:
Ni + 4CO → 330 K - 350 K Ni CO 4 → 450 K - 407 K Ni (s)  +  4CO g

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
In most of the flowering plants, seeds does not contain any remnant of

Options:
A outer integument
B inner integument
C nucellus
D micropyle

The correct answer is:
nucellus

Solution:
During formation of seed, the outer integument gets converted to the outer, thick, hard seed coat called testa. The inner integument gets converted into an inner, thin, membranous seed coat called tegmen. The micropyle is retained in the seed. Water enters the seed from the micropyle and initiates germination. In most of the flowering plants, during the embryo formation, the nucellus is complete used up and there is no remnant of nucellus in the seed. In some plants such as castor and black pepper, small amount of nucellus may remain. It is called perisperm.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In Mycorrhiza, the fungal hyphae

Options:
A Remain restricted to the root surface
B Remain restricted up to the root cortex
C Pass into the root interior
D Enter the root apex and reach all parts

The correct answer is:
Remain restricted up to the root cortex

Solution:
The mycorrhizal association is an association between a fungus and the root of a higher plant, e.g., pine, birch, Eucalyptus, Ficus etc. The actively growing roots of higher plants get infected by fungi. The fungus hyphae get their food from cortical cells of the root.

Topic: Biology, Biological Classification, NEET
In neutral atom, which particles are equivalent?

Options:
A $e^{-}, e^{+}$
B $p^{+}, e^{+}$
C $e^{-}, p^{+}$
D $p^{+}, n^o$

The correct answer is:
$e^{-}, p^{+}$

Solution:
Correct option is 3. $e^{-}, p^{+}$
Electrons and protons are same in neutral atom.

Topic: Chemistry, Structure of Atom, JEE Main
In neutral or alkaline solution, MnO4- oxidises thiosulphate to

Options:
A S2O72-
B S2O82-
C SO32-
D SO42-

The correct answer is:
SO42-

Solution:
KMnO4 is a strong oxidising agent in acidic,alkaline or neutral medium as well

Thiosulphate is oxidised almost quantitatively to sulphate:

8MnO4+-3S2O3+2-H2O⟶8MnO2+6SO4+2-2OH-

The equivalent weight of KMnO4 in the above reaction =1585

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (27 Jul Shift 2)
In neutral or faintly alkaline medium, KMnO4 being a powerful oxidant can oxidise, thiosulphate almost quantitatively, to sulphate. In this reaction overall change in oxidation state of manganese will be

Options:
A 5
B 1
C 0
D 3

The correct answer is:
3

Solution:
KMnO4 acts as an oxidising agent. In neutral or faintly alkaline medium alkaline KMnO4 is reduced to MnO2.

8MnO4-+3S2O32-+H2O→8MnO2+6SO42-+2OH-

The change in oxidation state of manganese in this reaction is from +7 to +4. So, In this reaction overall change in oxidation state of manganese will be 3.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (29 Jul Shift 1)
In neutral or faintly alkaline solution, 8 moles of permanganate anion quantitatively oxidize thiosulphate anions to produce X moles of a sulphur containing product. The magnitude of X is

Options:

The correct answer is:
6

Solution:
8MnO4-+3S2O32-+H2O→8MnO2+6SO42-+2OH⊖

Topic: Chemistry, d and f Block Elements, JEE Main
In normal adjustment, for a refracting telescope, the distance between objective and eye piece is 30 cm. The focal length of the objective, when the angular magnification of the telescope is 2, will be:

Options:
A 20 cm
B 30 cm
C 10 cm
D 15 cm

The correct answer is:
20 cm

Solution:
Let f0 be the focal length of objective and fe is the focal length of eyepiece.

Given here, f0+fe=30.

Angular magnification of telescope is given by m=f0fe.

So, 2=f0fe⇒f0=2fe

Then, we have f0+f02=30⇒3f02=30

Or, f0=20 cm.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (28 Jul Shift 1)
In npn transistor circuit, the collector current is \( 10 \mathrm{~mA} \). If \( 95 \% \) of the electrons emitted reach the collector, what is the base current ?

Options:
A \( 10.53 \mathrm{~mA} \)
B \( 9.53 \mathrm{~mA} \)
C \( 5.6 \mathrm{~mA} \)
D \( 0.53 \mathrm{~mA} \)

The correct answer is:
\( 0.53 \mathrm{~mA} \)

Solution:
IC=10 mA

iC=iE×95100

iE=100iC95=10.53 mA

iB=iE-iC=10.53-10=0.53 mA

Topic: Physics, Semiconductors, JEE Main
In nuclear reactor, the energy transformation occurs:

Options:
A Electrical energy \( \rightarrow \) Nuclear energy
B Nuclear energy \( \rightarrow \) Electrical energy
C Heat energy \( \rightarrow \) Light energy
D Sound energy \( \rightarrow \) Nuclear energy

The correct answer is:
Nuclear energy \( \rightarrow \) Electrical energy

Solution:
A nuclear reactor is a device in which a controlled nuclear chain reaction can be initiated and sustained to produce energy. They can produce steady flow of neutrons generated by the fission of heavy nuclei. These are used at nuclear power plants for the generation of electricity. Hence, in nuclear reactor, nuclear energy is converted into electrical energy.

Topic: Physics, Nuclear Physics, JEE Main
In nuclear reactors, what are the controlling rods made of?

Options:
A Cadmium
B Graphite
C Stainless steel
D Plutonium

The correct answer is:
Cadmium

Solution:
In nuclear reactors, cadmium rods are used as control rods. They are used to absorb the extra nucleons and to absorb neutrons, making them unavailable to cause fission. The other options are not valid.

Topic: Physics, Nuclear Physics, JEE Main
In nucleophilic aliphatic substitution, the nucleophiles are generally

Options:
A Acids
B Bases
C Salts
D Neutral molecules

The correct answer is:
Bases

Solution:
$\mathrm{R}-\mathrm{X}+\underset{\substack{\text { Nucleophile } \\ \text { or base }}}{\mathrm{OH}^{-}} \rightarrow \mathrm{R}-\mathrm{OH}+\mathrm{X}^{-}$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
In nucleophilic substitution reaction, order of halogens as incoming (attacking) nucleophile is:
$\mathrm{I}^{-}>\mathrm{Br}^{-}>\mathrm{Cl}^{-}$
The order of halogens as departing nucleophile should be :

Options:
A
$\mathrm{Br}^{-}>\mathrm{I}^{-}>\mathrm{Cl}^{-}$

B
$\mathrm{I}^{-}>\mathrm{Br}^{-}>\mathrm{Cl}^{-}$

C
$\mathrm{Cl}^{-}>\mathrm{Br}^{-}>\mathrm{I}^{-}$

D
$\mathrm{Cl}^{-}>\mathrm{I}^{-}>\mathrm{Br}^{-}$

The correct answer is:
$\mathrm{I}^{-}>\mathrm{Br}^{-}>\mathrm{Cl}^{-}$

Solution:
Since the leaving group breaks away as a base, it is easier to displace weaker bases as compared to stronger bases. Thus less basic the substituent, the more easily it is displaced.
Since the basic strength of the given groups is in order.
$\mathrm{I}^{-} < \mathrm{Br}^{-} < \mathrm{Cl}^{-}$
Thus the order of halogen leaving groups is
$\mathrm{I}^{-}>\mathrm{Br}^{-}>\mathrm{Cl}^{-}$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2013 (25 Apr Online)
In nucleoside, the base is attached to which position of sugar molecule?

Options:
A C -1
B C -2
C C -3
D C -5

The correct answer is:
C -1

Solution:

Topic: Chemistry, Biomolecules, JEE Main
In older times, people used to think that the Earth was flat. Imagine that the Earth is indeed not a sphere of radius R, but an infinite plate of thickness H. What value of H is needed to allow the same gravitational acceleration to be experienced as on the surface of the actual Earth? (Assume that the Earth's density is uniform and equal in the two models).

Options:
A 2R3
B 4R3
C 8R3
D R3

The correct answer is:
2R3

Solution:
Applying Gauss law
∮E→.ds→=4πGmin
Where E → is gravitational field at ds→
Computing over colsed surface given
E2s=4πG(ρsh)
E= 4πG ρh 2 ... (1)
On the surface of earth with radius R, E= G( ρ 4 3 π R 3 ) R 2
=Gρ 4 3 πR ... (2)
Equating (1) and (2)
4π Gρh 2 =Gρ 4 3 πR
h= 2R 3

Topic: Physics, Gravitation, JEE Main
In oogamy, fertilization involves:

Options:
A a small non-motile female gamete and a large motile male gamete
B a large non-motile female gamete and a small motile male gamete
C a large non-motile female gamete and a small non motile male gamete
D a large motile female gamete and a small non-motile male gamete.

The correct answer is:
a large non-motile female gamete and a small motile male gamete

Solution:
In most single-celled flagellates, the female and the male or the + and the gametes, respectively have the same size. This state is called isogamy. Primitive multicellular organisms in contrast display a tendency towards gametes of different size. The female gamete is larger than the male gamete, a state called heterogamy or anisogamy. In oogamy, the female gamete, also called egg or oogonium has no flagellum. Its nucleus is surrounded by a voluminous plasma containing a lot of nutriments. The male gamete is small, mobile, and contains only a small amount of plasma, just enough to provide the energy required for the movements.
Related Theory
The evolution towards oogamy is accompanied by the evolution of gametophytes that again protect the development of the gametes. While the male gametes leave the gametophyte after ripening, the egg cell becomes the embryo after fertilization that remains in the protection of the gametophyte. The progression from isogamy to oogamy occurred several times in the course of evolution. It exists in the plant kingdom, the animal kingdom, and with fungi.

Topic: Biology, Reproduction in Organisms, NEET, NEET 2004
In optical communication system operating at \( 1200 \mathrm{~nm} \), only \( 2 \% \) of the source frequency is available for TV transmission having a bandwidth of \( 5 \mathrm{MHz} \). The number of TV channels that can be transmitted is

Options:
A \( 2 \) million
B \( 10 \) million
C \( 0.1 \) million
D \( 1 \) million

The correct answer is:
\( 1 \) million

Solution:
The frequency optical communication
v=cÎ»Â orÂ v=3Ã1081200Ã10-9
v=25Ã1013Â Hz
But only 2% of the source frequency is available for TV transmission
v′=2.5Ã1014Â Ã2%
Or v′=2.5Ã1014Â Ã2100
Or v′=5Ã1012Â Hz
â´MemberÂ ofÂ channels=v′bondÂ width
Or NoÂ ofÂ channels=5Ã10125Ã106=106=1Â million

Topic: Physics, Communication System, JEE Main
In optical communication system operating at 1200 nm, only 2% of the source frequency is available for TV transmission having a bandwidth of 5 MHz. The number of TV channels that can be transmitted is

Options:
A 2 million
B 10 million
C 0.1 million
D 1 million

The correct answer is:
1 million

Solution:
The frequency optical communication
v=cÎ» or v=3Ã1081200Ã10-9
v=25Ã1013 Hz
But only 2% of the source frequency is available for TV transmission
vâ²=2.5Ã1014 Ã2%
Or vâ²=2.5Ã1014 Ã2100
Or vâ²=5Ã1012 Hz
â´ Number of channels=vâ²bandwidth
Or No of channels=5Ã10125Ã106=106=1 million

Topic: Physics, Communication System, JEE Main
In optical communication system operating at 1200 nm, only 2% of the source frequency is available for TV transmission having a bandwidth of 5 MHz. The number of TV channels that can be transmitted is

Options:
A 2 million
B 10 million
C 0.1 million
D 1 million

The correct answer is:
1 million

Solution:
The frequency optical communication
v=cλ or v=3×1081200×10-9
v=25×1013 Hz
But only 2% of the source frequency is available for TV transmission
v′=2.5×1014 ×2%
Or v′=2.5×1014 ×2100
Or v′=5×1012 Hz
∴Number of channels=v′bond width
Or No. of channels=5×10125×106=106=1 million

Topic: Physics, Communication System, JEE Main
In optical communication systems operating at 1200 nm, only 2% of the source frequency is available for TV transmission having a bandwidth of 5 MHz. The number of TV channels that can be transmitted is 10n. The value of n is

Options:

The correct answer is:
6

Solution:
The frequency of optical communication is given by,
v=cÎ» or v=3Ã1081200Ã10-9
v=25Ã1013 Hz
But only 2% of the source frequency is available for TV transmission.
vâ²=2.5Ã1014 Ã2%
vâ²=2.5Ã1014 Ã2100
vâ²=5Ã1012 Hz
â´ Number of channels=vâ²bandwidth
No of channels=5Ã10125Ã106=106

Topic: Physics, Communication System, JEE Main