datavorous/entrance-exam-dataset huggingface dataset:
Which of the following esters cannot undergo Claisen self condensation?

Options:
A $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOC}_{2} \mathrm{H}_{5}$
B $\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{COOC}_{2} \mathrm{H}_{5}$
C $\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{CH}_{2} \mathrm{COOC}_{2} \mathrm{H}_{5}$
D $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOC}_{2} \mathrm{H}_{5}$

The correct answer is:
$\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{COOC}_{2} \mathrm{H}_{5}$

Solution:
Claisen condensation is given by esters having two $\alpha$ -hydrogen atoms.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Which of the following excretory organs is found in crustraceans?

Options:
A Flame cells
B Nephridia
C Green glands
D Malpighian tubules

The correct answer is:
Green glands

Solution:
Protonephridia or flame cells are the excretory structures in Platyhelminthes (Flatworms, e.g., Planaria), rotifers, some annelids and the cephalochordate – Amphioxus. Protonephridia are primarily concerned with ionic and fluid volume regulation, i.e., osmoregulation. Nephridia are the tubular excretory structures of earthworms and other annelids. Nephridia help to remove nitrogenous wastes and maintain a fluid and ionic balance. Malpighian tubules are the excretory structures of most of the insects including cockroaches. Malpighian tubules help in the removal of nitrogenous wastes and osmoregulation. Antennal glands or green glands perform the excretory function in crustaceans like prawns.

Topic: Biology, Excretory Products and their Elimination, NEET
Which of the following exhibit only+ 3 oxidation state?

Options:
A $\mathrm{U}$
B Th
C $\mathrm{Ac}$
D $\mathrm{Pa}$

The correct answer is:
$\mathrm{Ac}$

Solution:
$\mathrm{Ac}(89)=[\mathrm{Rn}]\left[6 d^{1}\right]\left[7 s^{2}\right]$

Topic: Chemistry, Structure of Atom, JEE Main
Which of the following exhibits only +3 oxidation state?

Options:
A $U$
B $\mathrm{Th}$
C Ac
D $\mathrm{Pa}$

The correct answer is:
Ac

Solution:
${ }_{89} \mathrm{Ac}=[\mathrm{Rn}] 6 d^1, 7 s^2$
Oxidation state $=+3$
${ }_{90} \mathrm{Th}=[\mathrm{Rn}] 6 d^2, 7 s^2$
Oxidation states $=+3,+4$
${ }_{91} \mathrm{~Pa}=[\mathrm{Rn}] 6 d^3, 7 s^2$
Oxidation states $=+3,+4,+5$
${ }_{92} \mathrm{U}=[\mathrm{Rn}] 6 d^4, 7 s^2$
Oxidation states $=+3,+4,+5,+6$

Topic: Chemistry, d and f Block Elements, NEET, NEET 2012 (Mains)
Which of the following exhibits tautomerism?

Options:
A CH32NH
B CH33CNO
C R3CNO2
D RCH2NO2

The correct answer is:
RCH2NO2

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
Which of the following exhibits the weakest intermolecular forces

Options:
A $\mathrm{He}$
B $\mathrm{HCl}$
C $\mathrm{NH}_3$
D $\mathrm{H}_2 \mathrm{O}$

The correct answer is:
$\mathrm{He}$

Solution:
Generally zero group elements are linked by the Vander. Waal's force. Hence these show weakest intermolecular forces.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Which of the following exists as covalent crystals in the solid state ?

Options:
A Sulphur
B Phosphorus
C Iodine
D Silicon

The correct answer is:
Silicon

Solution:
Silicon is covalent or network solid.
Each Si atom is in sp3 hybridisation.
Sulphur (S8) – molecular solid.
Phosphorus (P4) – molecular solid.
Iodine (I2) – molecular solid.

Topic: Chemistry, Solid State, JEE Main, JEE Main 2013 (07 Apr)
Which of the following experiments suggests that the simplest living organisms could not have originated spontaneously from the non-living matter?

Options:
A Larvae could appear in decaying organic matter
B Microbes could appear in stored meat
C Microbes appeared, from unsterilized organic matter
D Meat was not spoiled, when heated and kept sealed in a vessel.

The correct answer is:
Meat was not spoiled, when heated and kept sealed in a vessel.

Solution:
Spontaneous generation or anomalous generation explains the ordinary formation of living organisms without descent from similar organisms. Typically, the idea was that certain forms such as fleas could arise from inanimate matter such as dust, or that maggots could arise from dead flesh. According to Redi’s experiment, maggots formed only from the egg of flies which swarmed into the open jars and laid their eggs in the meat. The meat remaining sterile in the closed jar disproved spontaneous generation of maggots.

Topic: Biology, Evolution, NEET
Which of the following explain the poling process?

Options:
A Reduction of metallic oxide impurities to metal by Al
B Reduction of metallic oxide impurities to metal by gaseous hydrocarbon
C Electrolytic reduction of metallic oxide to metal
D Removal of volatile oxides from the molten metal

The correct answer is:
Reduction of metallic oxide impurities to metal by gaseous hydrocarbon

Solution:
In poling process molten impure metal is stirred with green wood. Green wood releases hydrocarbon gases which reduced the impurities of metal oxides to metal.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Which of the following explanation is best for not placing hydrogen with alkali metals or halogen

Options:
A The ionization energy of hydrogen is high for group of alkali metals or halogen
B Hydrogen can form compounds
C Hydrogen is a much lighter element than the alkali metals or halogens
D Hydrogen atom does not contain any neutron

The correct answer is:
The ionization energy of hydrogen is high for group of alkali metals or halogen

Solution:
The ionization energy of hydrogen is too high for group of alkali metals but too low for halogen group.

Topic: Chemistry, Hydrogen, JEE Main
Which of the following expression at constant pressure represents Charle's law

Options:
A $V \propto \frac{1}{T}$
B $V \propto \frac{1}{T^2}$
C $V \propto T$
D $V \propto d$

The correct answer is:
$V \propto \frac{1}{T}$

Solution:
$V \propto \frac{1}{T}$ charles law state that volume of an ideal gas is direclty proportional to absolute temperature

Topic: Chemistry, States of Matter, JEE Main
Which of the following expression is correct for packing fraction of NaCl if a- ions from one face and one diagonal are removed?

Options:
A 133πr-3+163πr+38(r++r-)3
B 133πr-3+43πr+38(r++r-)3
C 163πr-3+133πr+38(r++r-)3
D 43πr-3+133πr+38(r++r-)3

The correct answer is:
133πr-3+163πr+38(r++r-)3

Solution:
Effective number of remaining Cl-=4-2Ã18+12=134
Effective number of Na+=4
Packing fraction =Occupied volumeTotal volume
=134Ã43Ïr-3+4Ã43Ïr+38(r++r-)3
133Ïr-3+163Ïr+38(r++r-)3

Topic: Chemistry, Solid State, JEE Main
Which of the following expressions correctly represents the equivalent conductance at infinite dilution of $\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$. Given that $\Lambda_{\mathrm{Al}^{3+}}^{\circ}$ and $\Lambda_{\mathrm{SO}_4^{2-}}^{\circ}$ are the equivalent conductances at infinite dilution of the respective ions?

Options:
A $2 \Lambda_{\mathrm{Al}^{3+}}^{\circ}+3 \mathrm{\Lambda}_{\mathrm{SO}_4^{2-}}^{\circ}$
B $\Lambda_{\mathrm{Al}^{3+}}^{\circ}+\Lambda_{\mathrm{SO}_4^{2-}}^{\circ}$
C $\left(\Lambda_{\mathrm{Al}^{\circ+}}^{\circ}+3 \Lambda_{\mathrm{SO}_4^{2-}}^{\circ}\right) \times 6$
D $\frac{1}{3} \Lambda_{\mathrm{Al}^{3+}}^{\circ}+\frac{1}{2} \Lambda_{\mathrm{SO}_4^{2-}}^{\circ}$

The correct answer is:
$\Lambda_{\mathrm{Al}^{3+}}^{\circ}+\Lambda_{\mathrm{SO}_4^{2-}}^{\circ}$

Solution:
$\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3 \rightleftharpoons 2 \mathrm{Al}^{3+}+3 \mathrm{SO}_4^{2-}$
Since equivalent conductances are given only for ions, the equivalent conductance at infinite dilution,
$$
\Lambda_{\mathrm{eq}}^{\infty}=\Lambda_{\mathrm{Al}^{3+}}^{\circ}+\Lambda_{\mathrm{SO}_4^{2-}}^{\circ}
$$

Topic: Chemistry, Electrochemistry, NEET, NEET 2010 (Mains)
Which of the following expressions correctly represents the equivalent conductance at infinite dilution of $\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$. Given that $\Lambda_{\mathrm{Al}^{3+}}^{\circ}$ and $\Lambda_{\mathrm{SO}_4^{2-}}^{\circ}$ are the equivalent conductances at infinite dilution of the respective ions?

Options:
A $2 \Lambda_{\mathrm{Al}^{3+}}^{\circ}+3 \mathrm{\Lambda}_{\mathrm{SO}_4^{2-}}^{\circ}$
B $\Lambda_{\mathrm{Al}^{3+}}^{\circ}+\Lambda_{\mathrm{SO}_4^{2-}}^{\circ}$
C $\left(\Lambda_{\mathrm{Al}^{\circ+}}^{\circ}+3 \Lambda_{\mathrm{SO}_4^{2-}}^{\circ}\right) \times 6$
D $\frac{1}{3} \Lambda_{\mathrm{Al}^{3+}}^{\circ}+\frac{1}{2} \Lambda_{\mathrm{SO}_4^{2-}}^{\circ}$

The correct answer is:
$\Lambda_{\mathrm{Al}^{3+}}^{\circ}+\Lambda_{\mathrm{SO}_4^{2-}}^{\circ}$

Solution:
$\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3 \rightleftharpoons 2 \mathrm{Al}^{3+}+3 \mathrm{SO}_4^{2-}$
Since equivalent conductances are given only for ions, the equivalent conductance at infinite dilution,
$$
\Lambda_{\mathrm{eq}}^{\infty}=\Lambda_{\mathrm{Al}^{3+}}^{\circ}+\Lambda_{\mathrm{SO}_4^{2-}}^{\circ}
$$

Topic: Chemistry, Electrochemistry, JEE Main
Which of the following expressions corresponds to simple harmonic motion along a straight line, where $x$ is the displacement and $a, b, c$ are positive constants?

Options:
A $a+b x-c x^{2}$
B $b x^{2}$
C $a-b x+c x^{2}$
D $-b x$

The correct answer is:
$-b x$

Solution:
In linear S.H.M., the restoring force acting on particle should always be proportional to the displacement of the particle and directed towards the equilibrium position. i.e., $\mathrm{F} \propto \mathrm{x}$
or $\mathrm{F}=-\mathrm{bx}$ where $\mathrm{b}$ is a positive constant.

Topic: Physics, Oscillations, JEE Main
Which of the following expressions corresponds to simple harmonic motion along a straight line, where $\mathrm{x}$ is the displacement and $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are positive constants?

Options:
A
$a+b x-c x^2$

B
$\mathrm{bx}^2$

C
$a-b x+c x^2$

D
$-b x$

The correct answer is:
$-b x$

Solution:
In linear S.H.M., the restoring force acting on particle should always be proportional to the displacement of the particle and directed towards the equilibrium position. i.e., $\mathrm{F} \propto \mathrm{x}$
or $\mathrm{F}=-\mathrm{bx}$ where $\mathrm{b}$ is a positive constant.

Topic: Physics, Oscillations, JEE Main, JEE Main 2014 (12 Apr Online)
Which of the following expressions is correct in case of a CsCl unit cell (edge length ‘a’)?

Options:
A rCs++rCl-=a2
B rCs++rCl-=32a
C rCs++rCl-=a2
D rCs++rCl-=a

The correct answer is:
rCs++rCl-=32a

Solution:
CsCl has body centered type structure in which Cs+ occupies at corner of a cube and Cl– occupies the centre of the cube.

2rCs++2rCl-=3a (where a is the edge length of the cube)

rCs++rCr=32a

Topic: Chemistry, Solid State, JEE Main, JEE Main 2023 (15 Apr Shift 1)
Which of the following expressions represents the first law of thermodynamics

Options:
A $\Delta E=-q+W$
B $\Delta E=q-W$
C $\Delta E=q+W$
D $\Delta E=-q-W$

The correct answer is:
$\Delta E=q+W$

Solution:
$\Delta E=q+w$.

Topic: Chemistry, Thermodynamics (C), JEE Main
Which of the following factors are favorable for the formation of oxyhemoglobin?

I. high pO2
II. low pCO2
III. lesser H+ concentration
IV. lower temperature

Options:
A I, II and III
B I, III and IV
C I and IV
D I, II, III and IV

The correct answer is:
I, II, III and IV

Solution:
Transport of oxygen takes place in two ways. 3% is transported through the plasma and the remaining 97% is through the formation of oxyhemoglobin. This formation takes place in the blood vessels near alveoli and near tissues the oxyhemoglobin breaks to release oxygen into the tissues i.e. dissociation of HbO2 takes place. The factors which favor its formation include high pO2 as the amount of oxygen will be higher. The high partial pressure of oxygen means low partial pressure of carbon dioxide. Near the alveoli the metabolism is too low and hence hydrogen ions are less produced and also the temperature is less as no heat is evolved. These factors favor the binding of oxygen with hemoglobin.

Topic: Biology, Breathing and Exchange of Gases, NEET
Which of the following factors has not aided in the high species diversity of tropics?

Options:
A Less seasonal variation promoting niche specialisation
B Availability of more solar energy
C More pest pressure
D Less migration

The correct answer is:
Less migration

Solution:
There are four main reasons for more species diversity in tropics, long evolutionary time for speciation, less seasonal variation and predictable climate led to niche specialization, more input of solar energy led to more productivity thus more diversity and warm humid condition resulted in more pest pressure that led to variation and speciation.

Topic: Biology, Biodiversity and its Conservation, NEET
Which of the following factors is of no significance for roasting sulphide ores to the oxides and not subjecting the sulphide ores to carbon reduction directly?

Options:
A
Metal sulphides are thermodynamically more stable than $\mathrm{CS}_2$

B
$\mathrm{CO}_2$ is thermodynamically more stable than $\mathrm{CS}_2$

C
Metal sulphides are less stable than the corresponding oxides

D
$\mathrm{CO}_2$ is more volatile than $\mathrm{CS}_2$

The correct answer is:
Metal sulphides are thermodynamically more stable than $\mathrm{CS}_2$

Solution:
No Solution Available

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2008
Which of the following factors is responsible for the formation of concentrated urine?

Options:
A Low levels of antidiuretic hormone.
B Maintaining hyperosmolarity towards inner medullary interstitium in the kidneys.
C Secretion of erythropoietin by Juxtaglomerular complex.
D Hydrostatic pressure during glomerular filtration.

The correct answer is:
Maintaining hyperosmolarity towards inner medullary interstitium in the kidneys.

Solution:
A hyperosmotic renal medullary interstitium is critically important in concentrating the urine and provides the driving force for the reabsorption of water from the collecting duct.

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2019 (All India)
Which of the following factors may be regarded as the main cause of lanthanide contraction

Options:
A Poor shielding of one of $4 f$ electron by another in the subshell
B Effective shielding of one of $4 f$ electrons by another in the subshell
C Poorer shielding of $5 d$ electrons by $4 \mathrm{f}$ electrons
D Greater shielding of $5 d$ electron by $4 f$ electrons

The correct answer is:
Poor shielding of one of $4 f$ electron by another in the subshell

Solution:
The correct option is A Poor shielding of 4f- electrons as compared to other electrons. This phenomenon is associated with the intervention of the 4f orbitals, which must be filled before the 5d series of elements begin. The filling of 4f before 5d orbital results in a regular decrease in atomic radii called lanthanoid contraction. This is because of poor shielding of one of the 4f - electrons by another in the sub - shell.

Topic: Chemistry, d and f Block Elements, JEE Main
Which of the following factors may be regarded as the main cause of lanthanide contraction?

Options:
A
Poor shielding of one of 4f electron by another in the subshell

B
Poor shielding of one of 4f electron by another in the subshell

C
Poorer shielding of 5d electrons by 4f electrons

D
Greater shielding of 5d electrons by 4f electrons

The correct answer is:
Poor shielding of one of 4f electron by another in the subshell

Solution:
No Solution Available

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2005
Which of the following factors tends to increase the stability of metal ion complexes

Options:
A Higher ionic radius of the metal ion
B Higher charge/size ratio of the metal ion
C Lower ionisation potential of the metal ion
D Lower basicity of the ligand

The correct answer is:
Higher charge/size ratio of the metal ion

Solution:
Greater the charge on central metal ion greater is the stability of complex

Topic: Chemistry, Coordination Compounds, JEE Main
Which of the following factors will not cause change in gene frequency?

Options:
A Mutation
B Gene flow
C Large population
D Genetic recombination

The correct answer is:
Large population

Solution:
The conditions necessary for gene frequencies to remain constant are: 1. Mating must be completely random. 2. Mutations must not occur. 3. Migrations of individual organisms into and out of the population must not occur. 4. The population must be very large. 5. All genes must have an equal chance of being passed to the next generation.
5 factors are known to affect Hardy-Weinberg equilibrium:
1. Gene migration or gene flow 2. Genetic drift 3. Mutation 4. Genetic recombination 5. Natural selection

Topic: Biology, Evolution, NEET
Which of the following facts about the complex $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$ s wrong ?

Options:
A
The complex is paramagnetic

B
The complex is an outer orbital complex

C
The complex gives white precipitate with silver nitrate solution

D
The complex involves $d^2 s p^3$ hybridization and is octahedral in shape.

The correct answer is:
The complex is an outer orbital complex

Solution:
$\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$ involves $\mathrm{d}^2 \mathrm{sp}^3$ hybridization and it is an inner orbital complex.

Topic: Chemistry, Coordination Compounds, JEE Main
Which of the following facts about the complex $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$ s wrong ?

Options:
A
The complex is paramagnetic

B
The complex is an outer orbital complex

C
The complex gives white precipitate with silver nitrate solution

D
The complex involves $d^2 s p^3$ hybridization and is octahedral in shape.

The correct answer is:
The complex is an outer orbital complex

Solution:
$\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$ involves $\mathrm{d}^2 \mathrm{sp}^3$ hybridization and it is an inner orbital complex.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2011
Which of the following families are included under order Polymoniales?

Options:
A Fabaceae
B Convolvulaceae
C Liliaceae
D More than one option is correct

The correct answer is:
Convolvulaceae

Solution:
Plant families such as Convolvulaceae, Solanaceae are included in the order Polymoniales mainly based on the floral characters.

Topic: Biology, The Living World, NEET
Which of the following feature is common to bacteria, mosses and fungus?

Options:
A Mode of nutrition
B Presence of cell wall
C Autotrophic
D Body organization

The correct answer is:
Presence of cell wall

Solution:
Cell wall is common to bacteria, mosses and fungus. In bacteria cell wall is made up of peptidoglycan. In mosses cell wall is made up of mainly cellulose and in fungus cell wall is made up of chitin.

Topic: Biology, Biological Classification, NEET
Which of the following features is common between birds and mammals?

Options:
A Pigmented skin and bony endoskeleton
B Pulmonary respiration and four-chambered heart
C Viviparity and 12 pairs of cranial nerves
D Homeotherms and four-chambered heart

The correct answer is:
Homeotherms and four-chambered heart

Solution:
Birds and mammals are homeotherms. Birds are warm-blooded, which means that they can maintain their own body temperature irrespective of the surrounding. This feature is evolved in them due to the development of a 4-chambered heart.

Topic: Biology, Animal Kingdom, NEET
Which of the following fluorides of Xenon is impossible?

Options:
A $\mathrm{XeF}_6$
B $\mathrm{XeF}_4$
C $\mathrm{XeF}_3$
D $\mathrm{XeF}_2$

The correct answer is:
$\mathrm{XeF}_3$

Solution:
Xenon is an inert gas. Its electronic configuration is $[\mathrm{Kr}]_4 \mathrm{~d}_{10} 5 \mathrm{~s}_25 \mathrm{p}_6$.
All orbitals that are filled have paired electrons.
When one, two or three electrons are promoted from 5 p (filled) to $5 \mathrm{~d}$ (empty) orbitals, two, four and six half-filled orbitals are formed.
Xenon can combine with even number of $\mathrm{F}$ atoms to form $\mathrm{XeF}_2, \mathrm{XeF}_4$ or $\mathrm{XeF}_6$. It cannot combine with odd number of $\mathrm{F}$ atoms.
Thus, formation of $\mathrm{XeF}_3$ and $\mathrm{XeF}_5$ is not possible.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Which of the following food items can be used as natural food preservatives?

Options:
A Vinegar, ginger, apple and banana
B Garlic, lemon, sugar and vinegar
C Garlic, apple, salt and tamarind
D Ginger, garlic, banana and tamarind

The correct answer is:
Garlic, lemon, sugar and vinegar

Solution:
- Garlic, lemon, sugar, and vinegar are renowned for their multifaceted roles in food preservation.
- Garlic's allicin content, lemon's citric acid, sugar's ability to reduce water activity, and vinegar's acetic acid create a formidable arsenal against microbial growth.
- Their combined effect extends beyond mere preservation, enhancing flavors and textures while safequarding aqainst spoilage.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
Which of the following forms a colloidal solution in water

Options:
A $\mathrm{NaCl}$
B Glucose
C Starch
D Barium nitrate

The correct answer is:
Starch

Solution:
A solution in which a material is evenly suspended in liquid is called Colloidal solution. Salt and Barium Nitrate forms a true solution in water. All colloids show Tyndall effect which is not exhibited by Glucose.
But starch in water follow all the colloidal properties and starch in water is a colloid.

Topic: Chemistry, Surface Chemistry, JEE Main
Which of the following forms stable $+4$ oxidation state?

Options:
A
$\operatorname{La}(Z=57)$

B
$\operatorname{Eu}(Z=63)$

C
$\mathrm{Ce}(Z=58)$

D
$\operatorname{Gd}(Z=64)$

The correct answer is:
$\mathrm{Ce}(Z=58)$

Solution:
Only $\operatorname{Ce}(Z=58)$ shows stable (+4) oxidation state.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2012 (26 May Online)
Which of the following free radicals is most stable

Options:
A Primary
B Methyl
C Secondary
D Tertiary

The correct answer is:
Tertiary

Solution:
Greater the no. of alkyl groups attached to the carbon atom carrying the odd electrons, greater is the delocalization of odd electron and hence more stable is the free radical.

Topic: Chemistry, General Organic Chemistry, JEE Main
Which of the following free radicals is most stable

Options:
A Primary
B Methyl
C Secondary
D Tertiary

The correct answer is:
Tertiary

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
Which of the following frequencies does not belong to FM broadcast.

Options:
A 106 MHz
B 64 MHz
C 99 MHz
D 89 MHz

The correct answer is:
64 MHz

Solution:
Sound waves of different frequencies can be used in different purposes. FM broadcasting frequently makes use of frequency modulation. Additionally, it is utilised in telemetry, radar, seismic prospecting, EEG-based seizure monitoring of neonates, two-way radio systems, sound synthesis, magnetic tape recording, and some video transmission systems.

The range for FM broadcasting lies between 88 MHz to 108 MHz.

Hence, 64 MHz does not belong to the FM broadcast.

Topic: Physics, Communication System, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Which of the following frequencies will be suitable for beyond the horizon communication using sky waves?

Options:
A \( 10 \mathrm{kHz} \)
B \( 10 \mathrm{MHz} \)
C \( 1 \mathrm{GHz} \)
D \( 1000 \mathrm{GHz} \)

The correct answer is:
\( 10 \mathrm{MHz} \)

Solution:
10 MHz frequencies will be suitable for beyond the horizon communication using sky waves.

Topic: Physics, Communication System, JEE Main
Which of the following function is even function

Options:
A $f(x)=\frac{a^x+1}{a^x-1}$
B $f(x)=x\left(\frac{a^x-1}{a^x+1}\right)$
C $f(x)=\frac{a^x-a^{-x}}{a^x+a^{-x}}$
D $f(x)=\sin x$

The correct answer is:
$f(x)=x\left(\frac{a^x-1}{a^x+1}\right)$

Solution:
In $(1)$,
$f(-x)=\frac{a^{-x}+1}{a^{-x}-1}=\frac{1+a^x}{1-a^x}=-\frac{a^x+1}{a^x-1}=-f(x)$
So, it is an odd function.
In $(2)$
$f(-x)=(-x) \frac{a^{-x}-1}{a^{-x}+1}=-x \frac{1-a^x}{1+a^x}=x \frac{a^x-1}{a^x+1}=f(x)$
So, it is an even function.
In (3),
$f(-x)=\frac{a^{-x}-a^x}{a^{-x}+a^x}=-f(x)$
So, it is an odd function.
In (4), $f(-x)=\sin (-x)=-\sin x=-f(x)$
So, it is an odd function.

Topic: Mathematics, Functions, JEE Main