datavorous/entrance-exam-dataset huggingface dataset:
The order of melting point of ortho, para, meta-nitrophenol is

Options:
A $0 \gt m \gt p$
B $p \gt m \gt 0$
C $m \gt p \gt 0$
D $p \gt 0 \gt m$

The correct answer is:
$p \gt m \gt 0$

Solution:
Due to intramolecular hydrogen bonding in ortho-isomer, it has least melting point. Due to effective intermolecular hydrogen bonding and close packed structure in para-isomer, it has highest melting point among the isomers. So, the order is para
$\left(114^{\circ} \mathrm{C}\right) \gt$ meta $\left(97^{\circ} \mathrm{C}\right) \gt \operatorname{ortho}\left(45^{\circ} \mathrm{C}\right)$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The order of reactivity of phenyl magnesium bromide (PhMgBr) with the following compounds

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

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

Solution:
Since alkyl group has $+I$-effect and aryl group has $+R$-effect, thus greater the number of alkyl and aryl groups attached to the carbonyl group, its reactivity towards nucleophilic addition reaction. Secondly, as the steric crowding on carbonyl group increases, the reactivity decreases accordingly.
$\therefore$ Correct reactivity order for reaction with $\mathrm{PhMgBr}$ is

Topic: Chemistry, Aldehydes and Ketones, NEET, NEET 2011 (Mains)
The order of reactivity of the given haloalkanes towards nucleophile is :

Options:
A
$\mathrm{RI}>\mathrm{RBr}>\mathrm{RCl}$

B
$\mathrm{RCl}>\mathrm{RBr}>\mathrm{RI}$

C
$\mathrm{RBr}>\mathrm{RCl}>\mathrm{RI}$

D
$\mathrm{RBr}>\mathrm{RI}>\mathrm{RCl}$

The correct answer is:
$\mathrm{RI}>\mathrm{RBr}>\mathrm{RCl}$

Solution:
For a given alkyl group, the order of reactivity is
$\frac{\mathrm{R}-\mathrm{I}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F}}{\text { increasing bond energy }}$
decreasing halogen reactivity.
This order depends on the carbon-halogen bond energy; the carbon-fluorine bond energy is maximum and thus fluorides are least reactive while carbon-iodine bond energy is minimum hence iodides are most reactive.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2013 (23 Apr Online)
The order of relative stability of the contributing structure is:

Choose the correct answer from the options given below:

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

The correct answer is:
I>II>III

Solution:
The factors that affect the stability of a species are:
Number of Covalent Bonds: More the number of covalent bonds, more will be the atoms with octet, more will be the stability.
Formal Charge: The species should have the least formal charge to be stable.
Charge Separation: The species should have least charge separation to be stable.
Charge on Atoms: The species with negative charge on electropositive atom and positive charge on electronegative atom will be least stable.

I>II, since neutral resonating structures are more stable than charged resonating structures.II>III, since the stability of the structure with –ve charge on more electronegative atom is higher.

Also, in structure I, the octet of every element is complete.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The order of $S_{N} 1$ reactivity in aqueous acetic acid solution for the compounds
is

Options:
A $1>2>3$
B $1>3>2$
C $3>2>1$
D $3>1>2$

The correct answer is:
$3>2>1$

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
The order of stability of the following carbocations:

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

The correct answer is:
III > I > II

Solution:
CH 2 = CH - C ⊕ H 2 ↔ C ⊕ H 2 - CH = CH 2
(I) (Stabilised by delocalisation ( 2 eq. R.S.) )
CH 3 - CH 2 - C ⊕ H 2
(II) (Stabilised by only + I & hyperconjugation)(2 conjugated C-H)
III is stabilised by maximum delocalisation ( 4 R.S.)
The order of stability is III > I > II
Order of stability is:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2013 (07 Apr)
The order of stability of the following tautomeric compound is

CH2=C(OH)-CH2-CO-CH3I→CH3-CO-CH2-CO-CH3II→CH3-C(OH)=CH-CO-CH3III

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

The correct answer is:
III > II > I

Solution:
The enol of beta-dicarbonyl compounds are more stable because of conjugation and intramolecular H-bonding. Thus, the order of stability is,
CH 3 − C | = OH CH− C || O − CH 3 III > CH 3 −CO− CH 2 −CO− CH 3 II > CH 2 =C( OH )− CH 2 −CO− CH 3 I
I is less stable as (=) bond is not in conjugation with carbonly group.
III is most stable due to conjugation of (=) bond and intramolecular hydrogen bonding.

Topic: Chemistry, General Organic Chemistry, JEE Main
The order of the differential equation of the family of circles touching the y-axis at the origin is k, then the maximum value of y=kcosâ¡x âxâR is

Options:

The correct answer is:
1

Solution:
The equation of the family is x-a2+y2=a2, which has only one arbitrary constant
∴ the order will be 1.
Hence, the maximum value of y=kcos⁡x=cos⁡x is ‘1’

Topic: Mathematics, Differential Equations, JEE Main
The order of the differential equation of the family of curves y=a3bx+c+dsinx+e is (where, a, b, c, d, e are arbitrary constants)

Options:
A 5
B 4
C 3
D 2

The correct answer is:
4

Solution:
We can re-write the equation of the curve as
y=a⋅3c⋅3bx+dsinx+e
or y=k⋅3bx+dsinx+e (where, k is an arbitrary constant)
As there are ‘4’ such constants, hence the order of the differential equation will be ‘4’

Topic: Mathematics, Differential Equations, JEE Main
The order of the differential equation of the family of curves y=acsin⁡bx+3dx where a,b,c,d are arbitrary constants is

Options:
A 4
B 3
C 2
D 1

The correct answer is:
3

Solution:
Given equation is
y=Asin⁡bx+3dx
which has 3 arbitrary constants, hence the order of the differential equation will be 3.

Topic: Mathematics, Differential Equations, JEE Main
The order of the differential equation of the family of curves y=k12k2x+k33x+k4 is (where, k1,k2,k3,k4 are arbitrary constants)

Options:
A 4
B 5
C 3
D 6

The correct answer is:
3

Solution:
The given family of curves is y=k12k2x+k33k43x=k12k2x+k53x
which has ‘3’ arbitrary constants.
Hence, the order will be ‘three’

Topic: Mathematics, Differential Equations, JEE Main
The order of the differential equation of the family of parabolas symmetric about y=1 and tangent to x=2 is

Options:
A 2
B 1
C 3
D 4

The correct answer is:
1

Solution:
The equation of the family is which has one arbitrary constant.
Hence, the order of the differential equation will be 1

Topic: Mathematics, Differential Equations, JEE Main
The order of the differential equation whose general solution is given by y = c 1 + c 2 cos x + c 3 - c 4 e x + c 5 where c1, c2, c3, c4 & c5 are arbitrary constants, is

Options:
A 5
B 4
C 3
D 2

The correct answer is:
3

Solution:
The given equation can be written as y=Câ¡cos(x+c3)- c4ec5ex (where C =c1+c2)
â´ 3 arbitary constants

Topic: Mathematics, Differential Equations, JEE Main
The order of the reaction
$\mathrm{N}_{2} \mathrm{O}_{5} \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \mathrm{is}$

Options:
A 3
B 2
C 1
D 0

The correct answer is:
1

Solution:
It is a first order reaction because rate of reaction $\propto\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]$

Topic: Chemistry, Chemical Kinetics, JEE Main
The ordered pair a,b, for which the system of linear equations
3x-2y+z=b

5x-8y+9z=3

2x+y+az=-1
has no solution, is

Options:
A 3,13
B -3,13
C -3,-13
D 3,-13

The correct answer is:
-3,-13

Solution:
Given system of linear equations are
3x-2y+z=b

5x-8y+9z=3

2x+y+az=-1

For system to have no solution,

∆=3-215-8921a=0

⇒3-8a-9+25a-18+121=0

⇒a=-3

Also, ∆2=3-2b5-8321-1=35+2-11+b21=21b-7

If b=13, ∆2=0

So b must be equal to -13

Topic: Mathematics, Determinants, JEE Main, JEE Main 2022 (26 Jun Shift 1)
The ordinates of the points P and Q on the parabola y2=12x are in the ratio 1:2. The locus of the point of intersection of the normals to the parabola at P and Q is

Options:
A y2=12343x-63
B y2=12343x+63
C y-62=12x343
D y+62=12x343

The correct answer is:
y2=12343x-63

Solution:
Here, a=3. So, let P3t12,6t1 and Q3t22,6t2 be two points on the parabola y2=12x. Then,
6t16t2=12ât2=2t1.
Let, h,k be the point of intersection of the normals at P and Q. Then,
k=-t1h+6t1+3t13 and k=-t2h+6t2+3t23
âh=6+3t12+t22+t1t2 and k=-3t1t2t1+t2
âh=6+3t12+4t12+2t12 and k=-6t123t1
âh-6=21t12 and k=-18t13
âh-6213=t16 and -k182=t16
âh-6213=k2324
âk2=12343h-63
Hence, the locus of h,k is
y2=12343x-63

Topic: Mathematics, Parabola, JEE Main
The ore that contains both iron and copper is:

Options:
A Azurite
B Malachite
C Dolomite
D Copper pyrites

The correct answer is:
Copper pyrites

Solution:
Azurite -2CuCO3.CuOH2
Dolomite - CaCO3. MgCO3 ,
Malachite - CuCO3.Cu(OH)2
Copper pyrites - CuFeS2,
So both copper and Iron are present in copper pyrite.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2019 (09 Jan Shift 1)
The ore that contains the metal in the form of fluoride is known as which of the following?

Options:
A cryolite
B magnetite
C malachite
D sphalerite

The correct answer is:
cryolite

Solution:
Cryolite ⇒Na3[AlF6]
Sphalerite ⇒ZnS
Magnetite ⇒Fe3O4
Malachite ⇒CuCO3.CuOH2

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The ore which is concentrated by leaching

Options:
A PbS
B Al2O3·2H2O
C SnO2
D Fe2O3

The correct answer is:
Al2O3·2H2O

Solution:
Leaching: It involves the treatment of the ore with a suitable reagent as to make it soluble while impurity remains insoluble and after this, the ore is recovered from the solution by suitable methods.

The principal ore of aluminium and bauxite usually contain SiO2, iron oxides and titanium oxides as impurities.

Al2O3 s+2NaOH aq+3H2O→2NaAlOH4 aq2NaAlOH4 aq+CO2 g→Al2O3·xH2O s+2NaHCO3 aq

The sodium silicate remains in the solution and hydrated alumina is filtered, dried and heated to give pure Al2O3.

Al2O3·xH2O s→1470 KAl2O3 s+xH2O g

Topic: Chemistry, General Principles and Processes of Isolation of Metals, AP EAMCET, AP EAMCET 2018 (25 Apr Shift 1)
The organic chloro compound, which shows complete stereochemical inversion during a $\mathrm{S}_{\mathrm{N}} 2$ reaction, is

Options:
A
$\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{CHCl}$

B
$\left(\mathrm{CH}_3\right)_3 \mathrm{CCl}$

C
$\left(\mathrm{CH}_3\right)_2 \mathrm{CHCl}$

D
$\mathrm{CH}_3 \mathrm{Cl}$

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

Solution:
For $\mathrm{S}_{\mathrm{N}} 2$ reaction, the $\mathrm{C}$ atom is least hindered towards the attack of nucleophile in the case of $\left(\mathrm{CH}_3 \mathrm{Cl}\right)$.
Hence, (D) is the correct answer.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2008
The organic compound obtained during the addition of HBr to propene in the presence of peroxide catalyst is

Options:
A 2-bromopropane
B 2-bromopropene
C 1-bromopropane
D 1-bromopropene

The correct answer is:
1-bromopropane

Solution:
The addition of $\mathrm{HBr}$ to propene in presence of peroxide follows Kharash effect.

Topic: Chemistry, Hydrocarbons, JEE Main
The organic compound that gives following qualitative analysis is:

Test	Inference

(a)	Dil.HCl	Insoluble

(b)	NaOH solution	Soluble

(c)	Br2/water	Decolourization

Options:
A
B
C
D

The correct answer is:

Solution:
(i) Aromatic compounds having +M group can decolourise bromine water.
(ii) Since the compound is insoluble in HCl, it shows that the compound is acidic.
(iii) As the compound is soluble in NaOH, this again shows that the compound is acidic.
The above three conclusions indicate that the compound is aromatic having +M group and acidic.
Hence, the answer is a phenol as it is aromatic and slightly acidic.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The orthocentre of the triangle whose vertices are 1,1,5,1 and 4,5 is

Options:
A 94,-13
B 4,13
C 4,94
D 4,74

The correct answer is:
4,74

Solution:
Consider the triangle with the given information,

The slope of AB=0, so the equation of altitude from C is x=4
The slope of BC=-4, so the equation of altitude from A is x-4y+3=0

Solving these two equations, we get,
orthocentre of the triangle is 4,74

Topic: Mathematics, Straight Lines, JEE Main
The orthogonal trajectory of \( x^{2}-y^{2}=a^{2} \), where \( a \) is an arbitrary constant, is

Options:
A A parabola
B A circle
C An ellipse
D A hyperbola

The correct answer is:
A hyperbola

Solution:
x2 – y2 = a2
⇒ 2 x - 2 y dy dx = 0
dy dx = x y
Product of slope of x 2 − y 2 = a 2 and slope of orthogonal trajectory of x 2 − y 2 = a 2 is −1
∴ Slope of orthogonal trajectory of x 2 − y 2 = a 2 is
dy dx =− y x
∫ dx x = ∫ - dy y
⇒ logx = – logy + logc
logxy = logc
⇒ xy = c
which is a rectangular hyperbola.

Topic: Mathematics, Differential Equations, JEE Main
The oscillation of a body on a smooth horizontal surface is represented by the equation, X=Acos⁡ωt, where X= displacement at time t, ω= frequency of oscillation, a= acceleration at time t and T= time period.
Which one of the following graph shows correctly the variation a with t ?

Options:
A
B
C
D

The correct answer is:

Solution:
The equation of displacement, X=Acosωt then the speed and acceleration will be in terms of amplitude of the wave and angular acceleration,

v=dxdt=Aωsinωt

a=d2xdt2=-Aω2cosωt, so the graph of acceleration and displacement.

Topic: Physics, Oscillations, JEE Main
The osmotic pressure exerted by a solution prepared by dissolving 2.0 g of protein of molar mass 60 kg mol-1 in 200 mL of water at 27 °C is____Pa. [integer value] (use R=0.083 L bar mol-1 K-1)

Options:

The correct answer is:
415

Solution:
Osmotic pressure is the minimum pressure which needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane.It is a colligative property.

π=CRTWhere,π=Osmotic pressureC=MolarityT=Temperature.

π=CRT=2×10360×103×200×8.31×300 × 1000 =415Pa

Topic: Chemistry, Solutions, JEE Main, JEE Main 2022 (26 Jun Shift 2)
The osmotic pressure of a dilute solution of an ionic compound XY in water is four times that of a solution 0.01 M BaCl2 in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of XY ( in mol L-1 ) in solution is

Options:
A 4×10-2
B 16×10-4
C 4×10-4
D 6×10-2

The correct answer is:
6×10-2

Solution:
Osmotic pressure π=i×CRT

i=Van't hoff factor

According to question,

for XY i=2, BaCl2 i=3

πXY=4×πBaCl2

2×C×RT=4×3×0.01×RT

C=6×0.01=0.06 M

Topic: Chemistry, Solutions, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The osmotic pressure of a solution of NaCl is 0.10 atm and that of a glucose solution is 0.20 atm. The osmotic pressure of a solution formed by mixing 1 L of the sodium chloride solution with 2 L of the glucose solution is x×10-3 atm. x is ________ (nearest integer)

Options:

The correct answer is:
167

Solution:
π=iCRT

On mixing two solution πfinal=π1V1+π2V2V1+V2

πfinal=0.1×1+0.2×23

=0.1+0.43=0.53=5003×100-3atm

So, X=167

Topic: Chemistry, Solutions, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The osmotic pressure of blood is 7.47 bar at 300 K. To inject glucose to a patient intravenously, it has to be isotonic with blood. The concentration of glucose solution in gL-1 is (Molar mass of glucose =180 g mol-1, R=0.083 Lbar-1 mol-1) (Nearest integer)

Options:

The correct answer is:
54

Solution:
Osmotic pressure, π=iCRT

i= van't hoff factor

C= concentration in molarity

R= universal gas constant

T= temperature

C= Molarity

=molesLitre

=ωGMω/Litre

π=iCRT

i=1 for glucose as it neither associates nor dissociates in solution.

7.47=1×C×0.083×300

7.47=1×ω180×0.083×300

ω=7.47×1800.083×300=54 g/Lit

Topic: Chemistry, Solutions, JEE Main, JEE Main 2022 (24 Jun Shift 1)
The osmotic pressure of equimolar solutions of urea, BaCl 2 and AlCl 3 will be in the order of

Options:
A AlCl 3  > Urea >  BaCl 2
B Urea >  BaCl 2  >  AlCl 3
C AlCl3>BaCl2>Ureaâ
D BaCl 2  >  AlCl 3  > Urea

The correct answer is:
AlCl3>BaCl2>Urea

Solution:
Since, osmotic pressure is a colligative property, therefore
Osmotic pressure â number of solute particle in the solution.( i -factor)
âµ Dissociation of AlCl 3 is maximum and of urea is minimum.
Hence, order of osmotic pressure will be
AlCl 3  >  BaCl 2  > Urea

Topic: Chemistry, Solutions, JEE Main
The osmotic pressure of one molar solution at \( 27^{\circ} \mathrm{C} \) is

Options:
A \( 2.46 \mathrm{~atm} \)
B \( 24.6 \) atm
C 1. \( 21 \) atm
D \( 12.1 \mathrm{~atm} \)

The correct answer is:
\( 24.6 \) atm

Solution:
Osmotic pressure of solution is given by

π=nVRT
π=MRT Molarity (M) = n V
π=1×0.0821×300
=24.63 atm

Topic: Chemistry, Solutions, JEE Main
The osmotic pressure of solutions of PVC in cyclohexanone at 300 K are plotted on the graph. The molar mass of PVC is g mol-1 (Nearest integer)

(Given : R=0.083 L atm K-1 mol-1 )

Options:

The correct answer is:
41500

Solution:
Using osmotic pressure in ideal gas equation:
π=M'RT=W/MVRT

⇒π=WV1MRT=CRTM

⇒πC=RTM

If we assume given graph between πC and C

Slope =RTM=0.083×300M=6×10-4

∴M=0.083×3006×10-4=830×3006

=41,500gm/mole

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (25 Jan Shift 1)
The outcome of each of 30 items was observed; 10 items

gave an outcome $\frac{1}{2}-\mathrm{d}$ each, 10 items gave outcome $\frac{1}{2}$

each and the remaining 10 items gave outcome $\frac{1}{2}+\mathrm{d}$ each.

If the variance of this outcome data is $\frac{4}{3}$ then $|\mathrm{d}|$ equals:

Options:
A $\frac{2}{3}$
B 2
C $\frac{\sqrt{5}}{2}$
D $\sqrt{2}$

The correct answer is:
$\sqrt{2}$

Solution:
Outcomes are $\left(\frac{1}{2}-d\right),\left(\frac{1}{2}-d\right), 0 \ldots, 10$ times, $\frac{1}{2}, \frac{1}{2}$,

$\ldots, 10$ times, $\frac{1}{2}+d, \frac{1}{2}+d, \ldots, 10$ times

$$

\text { Mean }=\frac{1}{30}\left(\frac{1}{2} \times 30\right)=\frac{1}{2}

$$

Variance of the outcomes is,

$$

\begin{aligned}

\sigma^{2} &=\frac{1}{30} \Sigma x_{i}^{2}-(\bar{x})^{2} \\

&=\frac{1}{30}\left[\left(\frac{1}{2}-d\right)^{2} \times 10+\left(\frac{1}{2}\right)^{2} \times 10+\left(\frac{1}{2}+d\right)^{2} \times 10\right]-\frac{1}{4} \\

\Rightarrow & \frac{4}{3}=\frac{1}{30}\left[30 \times \frac{1}{4}+20 d^{2}\right]-\frac{1}{4} \\

\Rightarrow & \frac{4}{3}=\frac{1}{4}+\frac{2}{3} d^{2}-\frac{1}{4} \\

\Rightarrow & d^{2}=2 \Rightarrow|d|=\sqrt{2}

\end{aligned}

$$

Topic: Mathematics, Statistics, JEE Main
The outcome of each of 30 items was observed; 10 items

gave an outcome $\frac{1}{2}-\mathrm{d}$ each, 10 items gave outcome $\frac{1}{2}$

each and the remaining 10 items gave outcome $\frac{1}{2}+\mathrm{d}$ each.

If the variance of this outcome data is $\frac{4}{3}$ then $|\mathrm{d}|$ equals:

Options:
A $\frac{2}{3}$
B 2
C $\frac{\sqrt{5}}{2}$
D $\sqrt{2}$

The correct answer is:
$\sqrt{2}$

Solution:
Outcomes are $\left(\frac{1}{2}-d\right),\left(\frac{1}{2}-d\right), 0 \ldots, 10$ times, $\frac{1}{2}, \frac{1}{2}$,

$\ldots, 10$ times, $\frac{1}{2}+d, \frac{1}{2}+d, \ldots, 10$ times

$$

\text { Mean }=\frac{1}{30}\left(\frac{1}{2} \times 30\right)=\frac{1}{2}

$$

Variance of the outcomes is,

$$

\begin{aligned}

\sigma^{2} &=\frac{1}{30} \Sigma x_{i}^{2}-(\bar{x})^{2} \\

&=\frac{1}{30}\left[\left(\frac{1}{2}-d\right)^{2} \times 10+\left(\frac{1}{2}\right)^{2} \times 10+\left(\frac{1}{2}+d\right)^{2} \times 10\right]-\frac{1}{4} \\

\Rightarrow & \frac{4}{3}=\frac{1}{30}\left[30 \times \frac{1}{4}+20 d^{2}\right]-\frac{1}{4} \\

\Rightarrow & \frac{4}{3}=\frac{1}{4}+\frac{2}{3} d^{2}-\frac{1}{4} \\

\Rightarrow & d^{2}=2 \Rightarrow|d|=\sqrt{2}

\end{aligned}

$$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2019 (11 Jan Shift 1)
The outcomes of 5 tosses of a coin are recorded in a single sequence as $\mathrm{H}$ (head) and T (tail) for each toss. What is the number of elementary events in the sample space?

Options:
A 5
B 10
C 25
D 32

The correct answer is:
10

Solution:
A coin has two faces and is tossed 5 times. So, number of elements in the sample space $=10$.

Topic: Mathematics, Probability, JEE Main
The outcomes of an experiment classified as success $A$ or failure $A$ will followa binomial distribution, if

Options:
A $P(A)=1 / 2$
B $P(A)=0$
C $P(A)=1$
D $P(A)$ remains constant in all trials

The correct answer is:
$P(A)$ remains constant in all trials

Solution:
Given, the outcomes of an experiment classified as success A will follow a binomial distribution if $\mathrm{P}(\mathrm{A})$ remains constant in all trials.

Topic: Mathematics, Probability, NDA, NDA 2008 (Phase 2)
The outer electron configuration of Gd (Atomic No : 64 is :

Options:
A
$4 f^8 5 d^0 6 s^2$

B
$4 f^4 5 d^4 6 s^2$

C
$4 f^7 5 d^1 6 s^2$

D
$4 f^3 4 d^5 6 s^2$

The correct answer is:
$4 f^7 5 d^1 6 s^2$

Solution:
No Solution Available

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2011
The outer electronic configuration of alkaline earth metal is

Options:
A $n s^2$
B $n s^1$
C $n p^6$
D $n d^{10}$

The correct answer is:
$n s^2$

Solution:
The electronic configuration of the outermost orbit in the case of alkaline earth metals is : $\mathrm{ns}_2$.

Topic: Chemistry, s Block Elements, JEE Main
The outermost electronic configuration of the most electronegative element is

Options:
A $n s^2 n p^3$
B $n s^2 n p^4$
C $n s^2 n p^5$
D $n s^2 n p^6$

The correct answer is:
$n s^2 n p^5$

Solution:
Halogens are most electronegative.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The output characteristics of a transistor is shown in the figure. When VCE is 10V and IC=4.0 mA, then value of βac is

Options:

The correct answer is:
150

Solution:
From the formula of current gain, β=ΔiCΔiBVCE=10

ΔiC=4.5-3 mA

⇒ΔiC=1.5 mA

⇒ΔiC=1.5×10-3 A

ΔiB=30-20 μA

⇒ΔiB=10 μA

⇒ΔiB=10×10-6 A

⇒ΔiB=10-5 A

⇒β=ΔiCΔiBVCE=10=1.5×10-310-5

⇒β=1.5×10-3×105

⇒β=150

Topic: Physics, Semiconductors, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The output characteristics of an $n-p-n$ transistor represent, ( $I_C=$ collector current, $V_{C E}=$ potential difference between collector and emitter, $I_B=$ base current, $V_{B B}=$ voltage given to base, $V_{B E}=$ the potential difference between base and emitter)

Options:
A changes in $I_C$ as $I_B$ and $V_{B B}$ are changed
B changes in $I_C$ with changes in $V_{C E}\left(I_B=\right.$ constant $)$
C changes in $I_B$ with changes in $V_{C E}$
D change in $I_C$ as $V_{B E}$ is changed

The correct answer is:
changes in $I_C$ with changes in $V_{C E}\left(I_B=\right.$ constant $)$

Solution:
The potential difference between base and emitter changes in $I_C$ with changes in $V_{C E}$ (when $I_B=$ constant $)$

Topic: Physics, Semiconductors, JEE Main
The output from a NAND gate having inputs A and B given below will be,

Options:
A

B

C

D

The correct answer is:

Solution:
The output of NAND gate is

Y=(AB)'.

The truth table is

ABY001011101110

Y would be zero only when both A and B are 1 . Clearly, for all other input conditions the output will be 1. Hence, this is the right option.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2023 (13 Apr Shift 2)
The output of a NAND gate is $O$

Options:
A If both inputs are $O$
B If one input is $O$ and the other input is 1.
C If both inputs are 1
D Either if both inputs are 1 or if one of the inputs is 1 and the other 0

The correct answer is:
If both inputs are 1

Solution:
If inputs are $A$ and $B$ then output for NAND gate is $Y=\overline{A B}$
$\Rightarrow$ If $A=B=1, Y=\overline{1.1}=\overline{1}=0$

Topic: Physics, Semiconductors, JEE Main
The output of a step down transformers is measured to be 48 V when connected to a 12 W bulb. The value of peak current is

Options:
A 12A
B 2 A
C 122 A
D 14 A

The correct answer is:
122 A

Solution:
Given,
Output of step down transformer Vs=48 V
Power associated with secondary coil Ps=12 W
Is=PsVs=1248=14=0.25 A
Amplitude of current I0=Is2
=0.25 2
=24
=122A

Topic: Physics, Alternating Current, JEE Main