datavorous/entrance-exam-dataset huggingface dataset:
Reagent used in the qualitative analysis of IVth group is

Options:
A $\mathrm{HCl}$
B $\mathrm{H}_2 \mathrm{~S}$ (alkaline)
C $\left(\mathrm{NH}_4\right)_2 \mathrm{~S}$
D None of these

The correct answer is:
$\mathrm{H}_2 \mathrm{~S}$ (alkaline)

Solution:
In presence of \(\mathrm{NH}_4 \mathrm{OH}\) dissociation of \(\mathrm{H}_2 \mathrm{S}\) is remarkably high so increases the solubility product of \(I V^{\text {th }}\) group sulphides.
\(\begin{aligned}
& \mathrm{H}_2 \mathrm{~S} \rightarrow 2 \mathrm{H}^{+}+\mathrm{S}^{2-} \\
& \mathrm{NH}_4 \mathrm{OH} \rightarrow \mathrm{NH}_4^{+}+\mathrm{OH}^{-} \\
& \mathrm{OH}^{-}+\mathrm{H}^{+} \rightarrow \mathrm{H}_2 \mathrm{O}
\end{aligned}\)

Topic: Chemistry, Practical Chemistry, JEE Main
Reagent used to convert allyl alcohol to acrolein is

Options:
A
$\mathrm{MnO}_2$

B
$\mathrm{H}_2 \mathrm{O}_2$

C
$\mathrm{OsO}_4$

D
$\mathrm{KMnO}_4$

The correct answer is:
$\mathrm{MnO}_2$

Solution:
$\mathrm{MnO}_2$ being a mild oxidising agent stops the oxidation of $-\mathrm{CH}_2 \mathrm{OH}$ group at aldehyde stage.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2012 (19 May Online)
Reagent(s) which can be used to bring about the following transformation is(are)

Options:
A LiAlH4 in C2H52O
B BH3 in THF
C NaBH4 in C2H5OH
D all of these

The correct answer is:
NaBH4 in C2H5OH

Solution:
LiAlH4/C2H52O : Reduces to esters, carboxylic acid, epoxides, aldehydes and ketones.
BH3 in T.H.F. : Reduces to -COOH and aldehydes into alcohols but do not reduce to
ester and epoxides.
NaBH4 in C2H5OH : Reduces only aldehydes and ketones into alcohols but not to others.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Real part of $e^{e^{i \theta}}$ is

Options:
A $e^{\cos \theta}[\cos (\sin \theta)]$
B $e^{\cos \theta}[\cos (\cos \theta)]$
C $e^{\sin \theta}[\sin (\cos \theta)]$
D $e^{\sin \theta}[\sin (\sin \theta)]$

The correct answer is:
$e^{\cos \theta}[\cos (\sin \theta)]$

Solution:
$e^{e^{i \theta}}=e^{\cos \theta+i \sin \theta}=e^{\cos \theta}\left[e^{i \sin \theta}\right]=e^{\cos \theta}[\cos (\sin \theta)+i \sin (\sin \theta)]$
$\therefore$ Real part of $e^{e^{i \theta}}$ is $e^{\cos \theta}[\cos (\sin \theta)]$

Topic: Mathematics, Complex Number, JEE Main
Rearrangement of an oxime to an amide in the presence of a strong acid is called

Options:
A Curtius rearrangement
B Frie's rearrangement
C Beckmann's rearrangement
D Aldol condensation

The correct answer is:
Beckmann's rearrangement

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Reason of weightlessness in a satellite is

Options:
A zero gravity.
B no atmosphere.
C zero reaction force by satellite surface
D None of the above.

The correct answer is:
zero reaction force by satellite surface

Solution:
Look the free body diagram here, Fc=mv2R, where v is orbital velocity.

It means,

F-N=Fc

⇒GMmR2-N=mv2R.

The formula of orbital velocity,

v=GMR, it means,

⇒GMmR2-N=mGMR2R

⇒N=0, so zero reaction force by satellite surface.

Topic: Physics, Gravitation, JEE Main
Recognise the figure and find out the correct match.

Options:
A a-Antheridial branch, b-Archegonial branch, c- Sphagnum gametophyte.
B a-Antheridial branch, b-Archegonial branch, c- Sphagnum sporophyte
C a-Archegonial branch, b-Antheridial branch, c- Sphagnum sporophyte
D a-Archegonial branch, b-Antheridial branch, c- Sphagnum gametophyte

The correct answer is:
a-Antheridial branch, b-Archegonial branch, c- Sphagnum gametophyte.

Solution:
Sphagnum is a type of moss which provides peat that is used as fuel and packing material.
The given image of the Sphagnum gametophyte shows the antheridial branch which appear near the apex of the main shoot and are longer. Antheridial branch produces the male gametes - antherozoids.
The female gametes (egg cell) are produced at the archegonium. The female sex organ forms at the archegonial branch that is short and ovoid-like shape.

Topic: Biology, Plant Kingdom, NEET
Red light differs from blue light as they have :

Options:
A Different frequencies and different wavelengths
B Different frequencies and same wavelengths
C Same frequencies and same wavelengths
D Same frequencies and different wavelengths

The correct answer is:
Different frequencies and different wavelengths

Solution:
Red light and blue light have different wavelength and different frequency.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Red precipitate is obtained when ethanol solution of dimethylglyoxime is added to ammoniacal $\mathrm{Ni}(\mathrm{II})$. Which of the following statements is not true?

Options:
A Red complex has a square planar geometry
B Complex has symmetrical $\mathrm{H}$-bonding
C Red complex has a tetrahedral geometry
D Dimethylglyoxime functions as bindentate ligand

The correct answer is:
Red complex has a tetrahedral geometry

Solution:

Topic: Chemistry, Coordination Compounds, NEET, NEET 2012 (Mains)
Reducing smog is a mixture of:

Options:
A Smoke, fog and O3
B Smoke, fog and SO2
C Smoke, fog and CH2=CH-CHO
D Smoke, fog and N2O3

The correct answer is:
Smoke, fog and SO2

Solution:
Reducing or classical smog is the combination of smoke, fog and SO2.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2021 (17 Mar Shift 1)
Reduction in vascular tissue, mechanical tissue and cuticle is characteristic of :

Options:
A Hydrophytes
B Xerophytes
C Mesophytes
D Epiphytes

The correct answer is:
Hydrophytes

Solution:
No Solution Available

Topic: Biology, Ecosystem, NEET, NEET 2009 (Mains)
Reduction of acetyl chloride with $\mathrm{H}_2$ in presence of $\mathrm{Pd}$ gives

Options:
A $\mathrm{CH}_3 \mathrm{COCH}_3$
B $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
C $\mathrm{CH}_3 \mathrm{COOH}$
D $\mathrm{CH}_3 \mathrm{CHO}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{CHO}$

Solution:
$\mathrm{CH}_3 \mathrm{COCl}+\mathrm{H}_2 \xrightarrow[\text { Reduction }]{\mathrm{Pd}} \mathrm{CH}_3 \mathrm{CHO}+\mathrm{HCl}$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Reduction of Aldehydes and Ketones to hydrocarbon take place in the presence of

Options:
A $\mathrm{Zn}$ amalgam and $\mathrm{HCl}$ acid
B $\mathrm{Pd} / \mathrm{BaSO}_4$
C Anhydrous $\mathrm{AlCl}_3$
D $\mathrm{Ni} / \mathrm{Pt}$

The correct answer is:
$\mathrm{Zn}$ amalgam and $\mathrm{HCl}$ acid

Solution:
Hydrocarbons are formed when aldehydes and ketones are reacted with amalgamated zinc and conc. $\mathrm{HCl}$

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Reduction of hexose A (molecular formula C6H12O6 ) with sodium borohydride gives compound B and C. Compound B is optically inactive, whereas compound C is optically active. Which of the following is compound A?

Options:
A D-fructose
B D-glucose
C D-mannose
D D-galactose

The correct answer is:
D-fructose

Solution:

Topic: Chemistry, Biomolecules, JEE Main
Reduction of metal centre in aqueous permanganate ion involves

(1) 5 electrons in neutral medium
(2) 5 electrons in acidic medium
(3) 3 electrons in neutral medium
(4) 3 electrons in alkaline medium

Options:
A 1,2
B 2,3
C 1,3
D 2,3,4

The correct answer is:
2,3

Solution:
In acidic medium
MnO4-+8H++5e-1→Mn2++4H2O
In neutral medium
MnO4-+2H2O+3e-→MnO2+4OH-
Hence, number of electrons gained in acidic and neutral mediums are 5 and 3 electrons respectively.

Topic: Chemistry, d and f Block Elements, JEE Main
Reduction of nitroalkanes yields

Options:
A Acid
B Alcohol
C Amine
D Diazo compounds

The correct answer is:
Amine

Solution:

Topic: Chemistry, Amines, JEE Main
Reduction of nitrobenzene with $\mathrm{Zn}$ and alcoholic $\mathrm{KOH}$ solution results in the formation of the following compound

Options:
A hydrazobenzene
B azobenzene
C aniline
D phenyl hydroxyl amine

The correct answer is:
hydrazobenzene

Solution:
Reduction of nitrobenzene with $\mathrm{Zn}$ and alcoholic $\mathrm{KOH}$ results in the formation of hydrazobenzene.

Topic: Chemistry, Amines, JEE Main
Reduction potential of four elements $P, Q, R, S$ is $-2.90,+0.34,+1.20$ and -0.76. Reactivity decreases in the order

Options:
A $P \gt Q \gt R \gt S$
B $Q \gt P \gt R \gt S$
C $R \gt Q \gt S \gt P$
D $P \gt S \gt Q \gt R$

The correct answer is:
$P \gt S \gt Q \gt R$

Solution:
Reducing power i.e. the tendency to lose electrons increases as the reduction potential decreases.

Topic: Chemistry, Electrochemistry, JEE Main
Reduction potential of ions are given below:

ClO4 - IO4- BrO4-

E°=1.19 V ; E°=1.65 V; E°=1.74 V

The correct order of their oxidising power is:

Options:
A ClO4->IO4->BrO4-
B BrO4->IO4->ClO4-
C BrO4->ClO4->IO4-
D IO4->BrO4->ClO4-

The correct answer is:
BrO4->IO4->ClO4-

Solution:
The reduction potential of the substance is the ability of the substance to be reduced. So as the reduction potential increases, the reducing ability of the substance increases. It means the oxidising power of the substance (The ability of the substance to make the other substance to lose the electrons increases, which nothing but oxidising power) increases.

So the decreasing order of oxidising power is BrO4->IO4->ClO4-.

Hence option B is correct.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Reduction potentials of four elements P, Q, R, S is −2.90 V , 0.34 V , 1.2 V and −0.76 V . The decreasing order of reducing power is

Options:
A P>Q>R>S
B S>R>Q>P
C P>S>Q>R
D Q>S>R>P

The correct answer is:
P>S>Q>R

Solution:
Reducing power, i.e. the tendency to lose electrons increases as the reduction potential decreases

Topic: Chemistry, Electrochemistry, JEE Main
Reena was experiencing headache and dizziness for the last few days. When visited a clinic, the doctor informed that her blood pressure is high. Which of the following hormones will be secreted in the body to reduce the blood pressure?

I. Aldosterone

II. Cortisol

III. Atrial Natriuretic factor

IV. Erythropoietin

Options:
A I only
B I and II
C III only
D I and III

The correct answer is:
I and III

Solution:
Atrial Natriuretic factor is released by atrial walls of heart in response to high blood pressure. It dilates the blood vessel thus resulting in reducing blood pressure. Aldosterone is a mineralocorticoid secreted by adrenal cortex. It reabsorbs water and sodium by acting on renal tubules thus maintaining body fluid volume and blood pressure.

Topic: Biology, Excretory Products and their Elimination, NEET
Refer to the given figure and choose the organism not related to the condition shown here.

Options:
A Microsporum
B Trichophyton
C W. malayi
D Epidermophyton

The correct answer is:
W. malayi

Solution:
The person shown in the given figure is suffering from ringworm infection. Many fungi belonging to the genera Microsporum, Trichophyton and Epidermophyton are responsible for ringworm disease. Appearance of dry, scaly lesions on skin, nails and scalp are the main symptoms of this disease.

Topic: Biology, Human Health and Diseases, AIIMS, AIIMS 2017
Refer to the plot of temperature versus time (figure) showing the changes in the state if ice on heating (not to scale). Which of the following is correct?

Options:
A
The region $A B$ represent ice and water in thermal equilibrium

B
At $B$ water starts boiling

C
At $C$ all the water gets converted into steam

D
$C$ to $D$ represents water and steam in equilibrium at boiling point

The correct answers are:
The region $A B$ represent ice and water in thermal equilibrium
,
$C$ to $D$ represents water and steam in equilibrium at boiling point

Solution:
When a substance heat continuously but its remperature does not change, then its state changes, state may be solid, liquid or gas.
So, during the process $A B$ temperature of the system is $0^{\circ} \mathrm{C}$. Hence, it represents phase change i.e. transformation of ice into water while temperature remains $0^{\circ} \mathrm{C}$.
$B C$ represents rise in temperature of water from $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$ (at $C$ ). Now, water starts converting into steam which is represent by $C D$.

Topic: Physics, Thermal Properties of Matter
Refer to this figure. Which symbol is correct for an LED?

Options:
A
B
C
D

The correct answer is:

Solution:
The correct respresentation is in option A

Topic: Physics, Semiconductors, JEE Main
Referring to the following reactions the missing products A, B, C, and D respectively are

NH4Cl (aq) + NaNO2 (aq) â [A] + H2O (l)+ NaCl (aq)

NH 4 2 Cr 2 O 7 â Î B + H 2 O l + Cr 2 O 3 s

Cu + HNO3 (dilute) â Cu(NO3)2 + [C] + H2O(l)

Cu + HNO3 (concentrate) â Cu(NO3)2 + [D] + H2O(l)

Options:
A N 2 ,  N 2 , NO,  NO 2
B N 2 ,  NH 3 ,  N 2 , NO
C N 2 ,  N 2 ,  NO 2 ,  NO 2
D N 2 ,  NH 3 ,  NO 2 ,  N 2 O 4

The correct answer is:
N 2 ,  N 2 , NO,  NO 2

Solution:
(i) In the laboratory, dinitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite.
NH 4 Cl aq + NaNO 2 aq â N 2 g +2 H 2 O l +NaCl aq
Small amounts of NO and HNO 3 are also formed as impurities, which can be removed by passing the gas through aqueous sulphuric acid containing potassium dichromate.
(ii) Dinitrogen can also be prepared by heating ammonium dichromate
NH 4 2 Cr 2 O 7 s â¶ Î N 2 (g) + 4H 2 O lâ¡ + Cr 2 O 3 s
(iii) Copper metal reacts with dilute HNO 3 to form nitrogen (II) oxide NO
3Cus+8HNO3 diluteâ 3CuNO32aq +2NOg+4H2Ol
(iv) Copper metal reacts with conc. HNO 3 to form nitrogen (IV) oxide or nitrogen dioxide NO 2 .
Cu( s ) +4HNO 3 ( conc. )âCu ( NO 3 ) 2 ( aq ) +2NO 2 ( g ) +2H 2 O( l )

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Refining using liquation method is the most suitable for metals with

Options:
A Low melting point
B High boiling point
C High electrical conductivity
D Less tendency to be soluble in melts than impurities

The correct answer is:
Low melting point

Solution:
This process is used for refining easily fusible metals like lead and tin. The impure metal is heated on the sloped hearth of a reverberatory furnace. The metal melts and flows down leaving the impurities.
Liquation is a metallurgical method for separating metals from an ore or alloy. The material must be heated until one of the metals starts to melt and drain away from the other and can be collected. This method was largely used to remove lead containing silver from copper, but it can also be used to remove antimony minerals from ore, and refine tin.

Liquation is used to purify metals having lower melting point than impurities present in them. For example, Pb, Sn, Sb, Bi and Hg.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2022 (26 Jul Shift 1)
Reflection property of ellipse is applicable in which of the following?

Options:
A Medical procedure for treating kidney stones.
B Rare View mirror
C The planets travel around the sun in an elliptical orbit with the sun as its foci.
D Catheral/ whispering gallery

The correct answer is:
The planets travel around the sun in an elliptical orbit with the sun as its foci.

Solution:
Reflection properly of ellipse is applicable to
Planets travel around the sun in an elliptical orbit with the Sun at its foci.

Topic: Mathematics, Ellipse, JEE Main
Reflexion of the point \( (\alpha, \beta, \gamma) \) in \( X Y \) plane is

Options:
A \( (\alpha, \beta, 0) \)
B \( (0,0, y) \)
C \( (-\alpha,-\beta, \gamma) \)
D \( (\alpha, \beta,-\gamma) \)

The correct answer is:
\( (\alpha, \beta,-\gamma) \)

Solution:
Given that, point \( (\alpha, \beta, y) \) has reflection in \( x y \)-plane, that is, only \( z \)-axis coordinate change, \( x \) and \( y \) coordinates remain
same.
So, \( (\alpha, \beta,-\gamma) \)

Topic: Mathematics, Area Under Curves, KCET, KCET 2017
Refractive index of glass is $\frac{3}{2}$ and refractive index of water is $\frac{4}{3}$. If the speed of light in glass is $2.00 \times 10^8 \mathrm{~m} / \mathrm{s}$, the speed in water will be

Options:
A $2.67 \times 10^8 \mathrm{~m} / \mathrm{s}$
B $2.25 \times 10^8 \mathrm{~m} / \mathrm{s}$
C $1.78 \times 10^8 \mathrm{~m} / \mathrm{s}$
D $1.50 \times 10^8 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$2.25 \times 10^8 \mathrm{~m} / \mathrm{s}$

Solution:
$\begin{aligned} \mu \propto \frac{1}{v} & \Rightarrow \frac{\mu_g}{\mu_w}=\frac{v_w}{v_g} \Rightarrow \frac{3 / 2}{4 / 3}=\frac{v_w}{2 \times 10^8} \\ & \Rightarrow v_w=2.25 \times 10^8 \mathrm{~m} / \mathrm{s}\end{aligned}$

Topic: Physics, Ray Optics, JEE Main
Refractive index of material is equal to tangent of polarising angle. It is called

Options:
A Brewster's law
B Lambert's law
C Malus's law
D Bragg's law

The correct answer is:
Brewster's law

Solution:
Brewster's law

Topic: Physics, Wave Optics, JEE Main
Regarding the oxidation states of elements of transition element the incorrect statement is

Options:
A Mo+6 is more stable than Cr+6
B W+6 is more stable than Cr+6
C Oxoanion of Cr+6 in acidic medium is better oxidizing agent than oxides of Mo and W in +6 oxidation state.
D Higher oxidation states are shown by metals when they are attached to π - acceptor ligands.

The correct answer is:
Higher oxidation states are shown by metals when they are attached to π - acceptor ligands.

Solution:
Lower oxidation states are shown by metals when they are attached to π - acceptors ligands.

Topic: Chemistry, d and f Block Elements, JEE Main
Regarding the structure of SO2 and SeO2 which of the following is true -

Options:
A The gaseous SO2 and SeO2 have same V -shaped molecule both in solid and gas phase
B At room temperature both SO2 and SeO2 are solids.
C In the solid phase the structure of SO2 is V -shaped discrete molecule but SeO2 is cyclic trimeric
D In solid phase the structure of SO2 is V -shaped discrete molecule but SeO2 has linear polymeric chain

The correct answer is:
In solid phase the structure of SO2 is V -shaped discrete molecule but SeO2 has linear polymeric chain

Solution:
The gaseous SO2 and solid SO2 both are V-shaped discrete molecule. SeO2 is solid at room temperature and solid SeO2 is polymerised by infinite chain.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Region I and II are separated by a spherical surface of radius 25 cm. An object is kept in region I at a distance of 40 cm from the surface. The distance of
the image from the surface is:

Options:
A 55.44 cm
B 9.52 cm
C 18.23 cm
D 37.58 cm

The correct answer is:
37.58 cm

Solution:
μ2v-μ1u=μ2-μ1R
1.4v-1.25-40=1.4-1.25-25
1.4v=-0.1525-1.2540
v=-37.58 cm
Hence option (4).

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (20 Jul Shift 1)
Regular use of which of the following fertilizer increases the acidity of soil?

Options:
A
Potassium nitrate

B
Urea

C
Superphosphate of lime

D
Ammonium sulphate

The correct answer is:
Ammonium sulphate

Solution:
$\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4$ is a salt of strong acid and weak base, on hydrolysis it ill produce $\mathrm{H}^{+}$ion. This will increase the acidity of soil.
$\begin{aligned}
& \left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \longrightarrow 2 \mathrm{NH}_4^{+}+\mathrm{SO}_4^{2-} \\
& \mathrm{NH}_4^{+}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{NH}_4 \mathrm{OH}+\mathrm{H}^{+}
\end{aligned}$
Hence, (D) is correct answer.

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2007
Relation between rate constant and temperature by Arrhenius equation is

Options:
A $\log _e A=\log _e K+\frac{E_a}{R T}$
B $\log K=A \frac{E_a}{R T}$
C $\log _e K=\log _e A-\frac{E_a}{R T^2}$
D $\log A=R T \ln E_a-\ln K$

The correct answer is:
$\log _e A=\log _e K+\frac{E_a}{R T}$

Solution:
Arrhenius equation is $\quad \log k=\log A-\frac{E_a}{R T}$

Topic: Chemistry, Chemical Kinetics, JEE Main
Relationship between given pair is:

Options:
A Identical
B Enantiomer
C Diastereomer
D Consitutional isomer

The correct answer is:
Diastereomer

Solution:
Diastereomers are stereoisomers that are not mirror images of one another and are non-superimposable on one another.
These two are diastereomers of each other.

Topic: Chemistry, General Organic Chemistry, JEE Main
Relationship between van’t Hoff’s factor (i) and degree of dissociation α is

Options:
A i=α-1n'-1
B i=α-11-n'
C α=1-in'-1
D α=i-1n'-1

The correct answer is:
α=i-1n'-1

Solution:
Relationship between vant Hoff factor (i) and degree of dissociation α is given by
α=1-in'-1
Where, n is the number of ions formed after dissociation. The relationship can be obtained as follows: For the reaction, A⇌n'B
Initially 1 mole 0
After dissociation 1-α mole n’ α
Total number of moles present in the solution
=1-α+n'α=1+n'-1α
Van’t Hoff factor, i=1+n'-1,α> 1 if n'≥2
∴                α=i-1n'-1

Topic: Chemistry, Solutions, JEE Main
Relative abundance (in percentage) of ${ }^{14} \mathrm{C}$ isotope is

Options:
A 1.1
B $2 \times 10^{-10}$
C $2 \times 10^{-4}$
D $2 \times 10^{-5}$

The correct answer is:
$2 \times 10^{-10}$

Solution:
Carbon has mainly three isotopes. C-12, C-13 and $\mathrm{C}-14 . \mathrm{C}-14$ is a radioactive isotope of carbon and present in least quantity, i.e. near about $\left(2 \times 10^{-10}\right)$.
$$
\begin{array}{lll}
{ }^{12} \mathrm{C} & { }^{13} \mathrm{C} & { }^{14} \mathrm{C} \\
98.9 \% & 1.08 \% & 2 \times 10{ }^{10}
\end{array}
$$

Topic: Chemistry, p Block Elements (Group 13 & 14), TS EAMCET, TS EAMCET 2018 (04 May Shift 2)
Relative permeability of an iron is 5500 , then its magnetic susceptibility will he

Options:
A $5500 \times 10^{-3}$
B $5500 \times 10^{3}$
C 5499
D 5501

The correct answer is:
5499

Solution:
Relative permeability $\mu=(1+x)$
$\therefore$ susceptibility $\chi=\mu-1=5500-1=5499$

Topic: Physics, Magnetic Properties of Matter, JEE Main
Relative permeability of diamagnetic material is slightly:

Options:
A More than two
B More than unity
C Less than unity
D More than one of the above

The correct answer is:
Less than unity

Solution:
- Relative increase or decrease in the resultant magnetic field inside a material compared with the magnetizing field in which the given material is located.
- It is the property of a material that is equal to the magnetic flux density $\mathrm{B}$ established within the material by a magnetizing field divided by the magnetic field strength $\mathrm{H}$ of the magnetizing field.
- Magnetic permeability is thus defined as $\mu=B / H$.

Topic: Physics, Magnetic Properties of Matter, JEE Main
Relative permittivity and permeability of a material are $\varepsilon_{\mathrm{r}}$ and $\mu_{\mathrm{r}}$, respectively. Which of the following values of these quantities are allowed for a diamagnetic material?

Options:
A
$\varepsilon_{\mathrm{r}}=0.5, \mu_{\mathrm{r}}=1.5$

B
$\varepsilon_{\mathrm{r}}=1.5, \mu_{\mathrm{r}}=0.5$

C
$\varepsilon_r=0.5, \mu_r=0.5$

D
$\varepsilon_{\mathrm{r}}=1.5, \mu_{\mathrm{r}}=1.5$

The correct answer is:
$\varepsilon_{\mathrm{r}}=1.5, \mu_{\mathrm{r}}=0.5$

Solution:
No Solution Available

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2008
Relative stability of the contributing structures is :

Options:
A (I) \( > \) (II) \( > \) (III)
B (I) \( > \) (III) \( > \) (II)
C \(\text { (II) } > \text { (I) } > \text { (III) }\)
D \(\text { (III) } > \text { (II) } > \text { (I) }\)

The correct answer is:
(I) \( > \) (II) \( > \) (III)

Solution:
(1) Neutral structures are more stable than charged ones. Therefore I is more stable than II and III.
(2) +ve charge on less electronegative atom is more stable i.e., $\mathrm{C}^{\oplus}$ is more stable than $\mathrm{O}^{\oplus}$ Order is I $>$ III $>$ III

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (09 Apr Shift 1)
Remaining quantity (in\%) of radioactive element after 5 half lives is

Options:
A $4.125 \%$
B $3.125 \%$
C $31.1 \%$
D $42.125 \%$

The correct answer is:
$3.125 \%$

Solution:
Remaining quantity
$N=N_0\left(\frac{1}{2}\right)^n=N_0\left(\frac{1}{2}\right)^5=\frac{N_0}{32}$
In percentage, $N=\frac{N_0}{30 \times N_0} \times 100=3.125 \%$

Topic: Physics, Nuclear Physics, JIPMER, JIPMER 2014