datavorous/entrance-exam-dataset huggingface dataset:
According to Faraday's law of electromagnetic induction

Options:
A electric field is produced by time varying magnetic flux.
B magnetic field is produced by time varying electric flux.
C magnetic field is associated with a moving charge.
D None of these.

The correct answer is:
electric field is produced by time varying magnetic flux.

Solution:
Faraday's law states that when the magnetic flux through a loop of the conductor will change with respect to time, then the EMF will be generated across the ends of the conductor.

According to Faraday's law of electromagnetic induction,

ε=-dϕdt-E→·dr→=-AdBdt

where, A is the area of loop and Bis the magnetic field.

Clearly, electric field is produced by time varying magnetic field.

Topic: Physics, Electromagnetic Induction, JEE Main
According to Freundlich adsorption isotherm, which of the following is correct ?

Options:
A
$\frac{x}{m} \propto P^0$

B
$\frac{x}{m} \propto p^1$

C
$\frac{x}{m} \propto p^{1 / n}$

D
All the above are correct for different ranges of pressure

The correct answer is:
All the above are correct for different ranges of pressure

Solution:
$\frac{\mathrm{x}}{\mathrm{m}} \propto \mathrm{P}^0$ is true at extremely high pressures
$\frac{x}{m} \propto p^1 ; \frac{x}{m} \propto p^{1 / n}$ are true at low and moderate pressures.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2012 (Offline)
According to Gauss' Theorem, electric field of an infinitely long straight wire is proportional to

Options:
A $r$
B $\frac{1}{r^2}$
C $\frac{1}{r^3}$
D $\frac{1}{r}$

The correct answer is:
$\frac{1}{r}$

Solution:
$e=\frac{\lambda}{2 \pi \varepsilon_0 r}$ $\Rightarrow E \propto \frac{1}{r}$

Topic: Physics, Electrostatics, JEE Main
According to Graham's law at a given temperature, the ratio of the rates of diffusion $r_A / r_B$ of gases $A$ and $B$ is given by (where $P$ and $M$ are the pressures and molecular weights of gases $A$ and $B$ respectively)

Options:
A $\left(P_A / P_B\right)\left(M_A / M_B\right)^{1 / 2}$
B $\left(M_A / M_B\right)\left(P_A / P_B\right)^{1 / 2}$
C $\left(P_A / P_B\right)\left(M_B / M_A\right)^{1 / 2}$
D $\left(M_A / M_B\right)\left(P_B / P_A\right)^{1 / 2}$

The correct answer is:
$\left(P_A / P_B\right)\left(M_B / M_A\right)^{1 / 2}$

Solution:
$\therefore(c)\left(P_A / P_B\right)\left(M_B / M_A\right)^{1 / 2}$

Topic: Chemistry, States of Matter, JEE Main
According to Hardy-Weinberg’s principle, if allele one is denoted as ‘A’ and allele two as ‘a’ and their frequencies are denoted by p and q, and if random mating occurs. The frequency of heterozygous individual would be:

Options:
A

2pq

B

q 2

C

pq

D

p 2

The correct answer is:

2pq

Solution:
Hardy-Weinbergâs principle states that the allele frequency of a population remains constant from generation to generation unless specific disturbing influences are introduced. This is called genotype equilibrium.

A single locus with two alleles: allele one is denoted as âAâ and allele two as âaâ and their frequencies are denoted by p and q;

frequency ( A ) = p
frequency ( a ) = q

And, p + q = 1 .

If mating is random, then new individuals will have:

frequency ( AA ) = p 2 for the AA homozygotes in the population,

frequency ( aa ) = q 2 for the aa homozygotes, and

frequency ( Aa ) = 2pq for the Aa heterozygotes.

Topic: Biology, Evolution, NEET
According to Heisenberg's uncertainty principle, the product of uncertainties in position and velocities for an electron of mass $9.1 \times 10^{-31} \mathrm{~kg}$ is

Options:
A $2.8 \times 10^{-3} \mathrm{~m}^2 \mathrm{~s}^{-1}$
B $3.8 \times 10^{-5} \mathrm{~m}^2 \mathrm{~s}^{-1}$
C $5.8 \times 10^{-5} \mathrm{~m}^2 \mathrm{~s}^{-1}$
D $6.8 \times 10^{-6} \mathrm{~m}^2 \mathrm{~s}^{-1}$

The correct answer is:
$5.8 \times 10^{-5} \mathrm{~m}^2 \mathrm{~s}^{-1}$

Solution:
Given that mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$
Planck's constant $=6.63 \times 10^{-34} \mathrm{kgm}^2 \mathrm{~s}^{-1}$
By using $\Delta x \times \Delta p=\frac{h}{4 \pi} ; \quad \Delta x \times \Delta v \times m=\frac{h}{4 \pi}$
where: $\Delta x=$ uncertainty in position
$\Delta v=$ uncertainty in velocity
$\Delta x \times \Delta v=\frac{h}{4 \pi \times m}$
$=\frac{6.63 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31}}=5.8 \times 10^{-5} \mathrm{~m}^2 \mathrm{~s}^{-1}$.

Topic: Chemistry, Structure of Atom, JEE Main
According to Hess's law, the heat of reaction depends upon

Options:
A Initial condition of reactants
B Initial and final conditions of reactants
C Intermediate path of the reaction
D End conditions of reactants

The correct answer is:
End conditions of reactants

Solution:
According to Hess's law, the heat of reaction depends upon Initial and final conditions of reactants. The the heat of reaction is independant of the intermediate path of the reaction. For example, the heat of the reaction for the reaction \(\mathrm{A} \rightarrow \mathrm{D}\) is equal to the sum of the heats of the reactions
\(\begin{aligned}
& \mathrm{A} \rightarrow \mathrm{B} \\
& \mathrm{B} \rightarrow \mathrm{C} \\
& \mathrm{C} \rightarrow \mathrm{D}
\end{aligned}\)

Topic: Chemistry, Thermodynamics (C), JEE Main
According to Hooke's law of elasticity, if stress is increased, then the ratio of stress to strain

Options:
A becomes zero
B remains constant
C decreases
D increases

The correct answer is:
remains constant

Solution:
Hooke in 1679 , showed experimentally that if strain is small, then the stress is proportional to strain. The ratio of stress to strain is constant for material of the given body and is called modulus of elasticity $E$. Thus, $E=\frac{\text { stress }}{\text { strain }}=$ constant
Hence, if stress is increased, then the ratio of stress to strain remains constant.

Topic: Physics, Mechanical Properties of Fluids, MHT CET, MHT CET 2007
According to Hugo deVries, the main cause of evolution is

Options:
A Saltation
B Inheritance of acquired characters
C Natural Selection
D All of these

The correct answer is:
Saltation

Solution:
The two key concepts of Darwin's theory of evolution are:

Natural selection explains the mechanism of evolution.

Branching descent explains the pattern of evolution.

According to Hugo deVries, a single step large mutation, called as saltation is the ultimate source of variation which causes speciation and hence evolution.
Inheritance of acquired characters was given by Lamarck.

Topic: Biology, Evolution, NEET
According to IUPAC nomenclature sodium nitroprusside is named as:

Options:
A sodium nitroferricyanide
B sodium nitroferrocyanide
C sodium pentacyanonitrosyl ferrate (II)
D sodium pentacyanonitrosyl ferrate (III)

The correct answer is:
sodium pentacyanonitrosyl ferrate (II)

Solution:
Sodium nitroprusside
$$
\mathrm{Na}_2\left[\mathrm{Fe}\left(\mathrm{CN}_5\right) \mathrm{NO}\right]
$$
Let oxidation state of $\mathrm{Fe}$ is $x$ in this compound.
$$
\begin{aligned}
& +2+x+(-1) 5+1=0 \\
& \text { or }+2+x-5+1=0 \text { or } x=+2
\end{aligned}
$$
Thus IUPAC name of sodium nitroprusside is sodium pentacyanonitrosyl ferrate (II)

Topic: Chemistry, Coordination Compounds, JEE Main
According to IUPAC nomenclature sodium nitroprussied is named is

Options:
A Sodium pentacyanonitrosyl ferrate (III)
B Sodium nitroferricyanide
C Sodium nitroferrocyanide
D Sodium pentacyanonitrosyl ferrate (II)

The correct answer is:
Sodium pentacyanonitrosyl ferrate (III)

Solution:
Sodium nitroprusside has the formula \(\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]\). It is named based on the ligands attached to the iron center. The nitrosyl group ($\mathrm{NO}$) and cyanide ($\mathrm{CN}$) ligands are named accordingly, and the oxidation state of iron in this complex is +3. Therefore, the correct name is Sodium pentacyanonitrosyl ferrate (III).

Topic: Chemistry, Coordination Compounds, JEE Main
According to IUPAC system, the compound is named as:

Options:
A Cyclohex-1-en-2-ol
B 1-Hydroxyhex-2-ene
C Cyclohex-1-en-3-ol
D Cyclohex-2-en-1-ol

The correct answer is:
Cyclohex-2-en-1-ol

Solution:
The structure of the IUPAC compound is

Secondary prefix+ primary prefix + root word + Primary suffix + Secondary suffix

It is a cyclic compound, with one OH group on position 1 and double bond at position 2.

There is no secondary prefix in the above compound. primary prefix is cyclo. Root word is hex, primary suffix is en and secondary suffix is ol.

Hence, the name is Cyclohex-2-en-1-ol.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (29 Jan Shift 2)
According to Kepler's law, the areal velocity of the radius vector drawn from the Sun to any planet always

Options:
A decreases.
B first increases and then decreases.
C remains constant.
D increases.

The correct answer is:
remains constant.

Solution:
Kepler's law of planetary motion

All planets move around the sun in elliptical path, having the sun at one of its foci

Radial vector joining the sun and the planet is sweeps equal area in equal interval of time.

The square of time period of revolution of planet is directly proportional to mean distance of planet from the Sun.

Topic: Physics, Gravitation, JEE Main
According to kinetic theory of gases

Options:
A There are intermolecular attractions
B Molecules have considerable volume
C No intermolecular attractions
D The velocity of molecules decreases after each collision

The correct answer is:
No intermolecular attractions

Solution:
No intermolecular attractions

Topic: Chemistry, States of Matter, JEE Main
According to kinetic theory of gases,

A. The motion of the gas molecules freezes at 0°C.

B. The mean free path of gas molecules decreases if the density of molecules is increased.

C. The mean free path of gas molecules increases if temperature is increased keeping pressure constant.

D. Average kinetic energy per molecule per degree of freedom is 32kBT (for monoatomic gases).

Choose the most appropriate answer from the options given below

Options:
A A and C only
B B and C only
C A and B only
D C and D only

The correct answer is:
B and C only

Solution:
According to kinetic theory of gases,

A. The motion of the gas molecules freezes at 0 K.

B. The mean free path decreases on increasing the number density of the molecules.

C. The mean free path increases on increasing the volume. Now if temperature is increased by keeping the pressure constant the volume should increase that is mean free path increases.

Mean free path λ=12πd2n, here, n=NV

⇒λ=12πd2NV

Using, PV=NKT

⇒λ=KT2πd2P

D. Average kinetic energy per molecule per degree of freedom is 12kBT.

Thus, B and C only.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2022 (27 Jun Shift 2)
According to Langmuir adsorption isotherm, the amount of gas adsorbed at very high pressure

Options:
A goes on increasing with pressure
B goes on decreasing with pressure
C increases first and decreases later with pressure
D reaches a constant limiting value

The correct answer is:
reaches a constant limiting value

Solution:
According to Langmuir adsorption isotherm the amount of gas adsorbed at very high pressures reaches a constant limiting volume.

Topic: Chemistry, Surface Chemistry, JEE Main
According to Langmuir adsorption isotherm, the amount of gas adsorbed at very high pressures

Options:
A Reaches a constant limiting value
B Goes on increasing with pressure
C Goes on decreasing with pressure
D Increases first and decreases later with pressure

The correct answer is:
Reaches a constant limiting value

Solution:
According to langmuir Adsorption isotherm the amount of gas adsorbed at very high pressures reaches a constant limiting volume.

Topic: Chemistry, Surface Chemistry, JEE Main
According to law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is :-

Options:
A 92R
B 52R
C 32R
D 72R

The correct answer is:
72R

Solution:
A diatomic gas has 5 degrees of freedom apart from vibrational degree of freedom. For each additional vibrational mode, degree of freedom is increased by 2. So new degree of freedom,

f=5+2=7

Therefore, we require molar specific heat capacity at constant volume, which is given by

Cv=f2R=72R

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2023 (25 Jan Shift 2)
According to law of mass action rate of a chemical reaction is proportional to

Options:
A Concentration of reactants
B Molar concentration of reactants
C Concentration of products
D Molar concentration of products

The correct answer is:
Molar concentration of reactants

Solution:
According to law of mass-action, "at a given temperature, the rate of a reaction at a particular instant is proportional to the product of the active masses of the reactants at that instant raised to powers which are numerically equal to the numbers of their respective molecules in the stoichiometric equation describing the reaction".

Topic: Chemistry, Chemical Equilibrium, JEE Main
According to Le Chatelier's principle, how does a system respond to a change in equilibrium conditions?

Options:
A By ignoring the change and remaining in equilibrium.
B By shifting the equilibrium in the direction that exacerbates the change.
C By changing in a manner that reduces or counteracts the effect of the change.
D By accelerating the rate of the reaction.

The correct answer is:
By changing in a manner that reduces or counteracts the effect of the change.

Solution:
Le Chatelier's principle states that a system responds to a change in equilibrium conditions by changing in a manner that reduces or counteracts the effect of the change.

Topic: Chemistry, Chemical Equilibrium, NEET
According to Le-chatelier principle, if heat is given to solid-liquid system, then

Options:
A Quantity of solid will reduce
B Quantity of liquid will reduce
C Increase in temperature
D Decrease in temperature

The correct answer is:
Quantity of solid will reduce

Solution:
On increasing temperature equilibrium will shift in forward direction due to decrease in intermolecular forces of solid

Topic: Chemistry, Chemical Equilibrium, JEE Main
According to Le-chatelier's principle, an increase in the temperature of the following reaction will $\mathrm{N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}-43,200 \mathrm{kcal}$

Options:
A Increase the yield of NO
B Decrease the yield of NO
C Not effect the yield of NO
D Not help the reaction to proceed in forward direction

The correct answer is:
Increase the yield of NO

Solution:
Being endothermic, the forward reaction is faxeured by high temperature.

Topic: Chemistry, Chemical Equilibrium, JEE Main
According to Maxwell's law of distribution of velocities of molecules, the most probable velocity is

Options:
A Greater than the mean velocity
B Equal to the mean velocity
C Equal to the root mean square velocity
D Less than the root mean square velocity

The correct answer is:
Less than the root mean square velocity

Solution:
From Maxwell's velocity distribution law, we infer that
vrms>v>vmp
ie, most probable velocity is less than the root mean square velocity.

Topic: Physics, Kinetic Theory of Gases, JEE Main
According to MO theory, number of species/ions from the following having identical bond order is____

CN-,NO+,O2,O2+,O22+

Options:

The correct answer is:
3

Solution:
The bond order of a diatomic species can be calculated using molecular orbital theory as follows,

Bond order=No.of BMO electrons-No.of ABMO electrons2

BMO = bonding molecular orbital

ABMO = Anti bonding molecular orbital.

Based on the total number electrons in bond orders are given in the following table.

Number of e-	Bond order

13	2.5

14	3

15	2.5

16	2

CN- , NO+ and O22+ are having 14 electrons each, hence the bond order is 3.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2022 (27 Jul Shift 1)
According to MO theory the bond orders for O22-, CO and NO+ respectively, are

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

The correct answer is:
1,3 and 3

Solution:
(i) Molecular orbital configuration of O22- is KK (σ2s)2 (σ*2s)2 𝜋(2px)2 𝜋(2py)2 (σ2pz)2 𝜋*(2px)2 𝜋*(2py)2

Bond order = 12 (Nb - Na) = 12 8 - 6 = 1

(ii) Molecular orbital configuration of CO is KK (σ2s)2 (σ*2s)2 𝜋(2px)2 𝜋(2py)2 (σ2pz)2

Bond order = 12 (Nb - Na) = 12 8 - 2 = 3

(iii) Molecular orbital configuration of NO+ is KK (σ2s)2 (σ*2s)2 𝜋(2px)2 𝜋(2py)2 (σ2pz)2

Bond order = 12 (Nb - Na) = 12 8 - 2 = 3

So, according to molecular orbital theory, the bond orders respectively are 1, 3 and 3.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (29 Jan Shift 2)
According to MO theory which of the following lists ranks the nitrogen species in terms of increasing bond order ?

Options:
A $\mathrm{N}_2^{-} < \mathrm{N}_2 < \mathrm{N}_2^{2-}$
B $\mathrm{N}_2^{2-} < \mathrm{N}_2^{-} < \mathrm{N}_2$
C $\mathrm{N}_2 < \mathrm{N}_2^{2-} < \mathrm{N}_2^{-}$
D $\mathrm{N}_2^{-} < \mathrm{N}_2^{2-} < \mathrm{N}_2$

The correct answer is:
$\mathrm{N}_2^{2-} < \mathrm{N}_2^{-} < \mathrm{N}_2$

Solution:
Key Idea Bond order $=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}$
where, $\mathrm{N}_{\mathrm{b}}=$ number of electrons in bonding $M O$
$\mathrm{N}_{\mathrm{a}}=$ number of electrons in anti bonding $M O$
$$
\begin{aligned}
& \mathrm{N}_2(7+7=14)=\sigma 1 \mathrm{~s}^2, \dot{*} 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \stackrel{\dot{*}}{\sigma} 2 \mathrm{~s}^2, \pi 2 \mathrm{p}_{\mathrm{x}}^2 \\
& \approx \pi 2 \mathrm{p}_{\mathrm{y}}^2, \sigma 2 \mathrm{p}_{\mathrm{z}}^2
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{BO}= \frac{10-4}{2}=3 \\
& \mathrm{~N}_2(7+7+1=15) \\
&= \sigma 1 \mathrm{~s}^2, \dot{*} 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \\
& \stackrel{*}{\sigma} 2 \mathrm{~s}^2, \sigma 2 \mathrm{p}_z^2, \pi 2 \mathrm{p}_{\mathrm{x}}^2 \approx \pi 2 \mathrm{p}_{\mathrm{y}}^2, \pi 2 \mathrm{p}_{\mathrm{x}}^1
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{BO}=\frac{10-5}{2}=2.5 \\
& \mathrm{~N}_2^{2-}(7+7+2=16)
\end{aligned}
$$
$\begin{aligned} & =\sigma 1 \mathrm{~s}^2, \star 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \star 2 \mathrm{~s}^2, \sigma 2 \mathrm{p}_{\mathrm{z}}^2, \pi 2 \mathrm{p}_{\mathrm{z}}^2 \\ & \approx \pi 2 \mathrm{p}_{\mathrm{y}}^2, \pi 2 \mathrm{p}_{\mathrm{x}}^1 \approx \pi 2 \mathrm{p}_{\mathrm{y}}^1 \\ & \end{aligned}$
$$
\mathrm{BO}=\frac{10-6}{2}=2
$$
Hence, the increasing order of B.O is,
$$
\mathrm{N}_2^{2-} < \mathrm{N}_2^{-} < \mathrm{N}_2
$$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
According to MO theory which of the following lists ranks the nitrogen species in terms of increasing bond order :

Options:
A $\mathrm{N}_2^{-} < \mathrm{N}_2^{2-} < \mathrm{N}_2$
B $\mathrm{N}_2^{-} < \mathrm{N}_2 < \mathrm{N}_2^{2-}$
C $\mathrm{N}_2^{2-} < \mathrm{N}_2^{-} < \mathrm{N}_2$
D $\mathrm{N}_2 < \mathrm{N}_2^{2-} < \mathrm{N}_2^{-}$

The correct answer is:
$\mathrm{N}_2^{2-} < \mathrm{N}_2^{-} < \mathrm{N}_2$

Solution:
$$
\begin{aligned}
& \mathrm{N}_2=14 \mathrm{e}=\mathrm{B} \cdot \mathrm{O} \cdot=3 \\
& \mathrm{~N}_2^{-}=15 \mathrm{e}=\mathrm{B} \cdot O \cdot=2.5 \\
& \mathrm{~N}_2^{2-}=16 \mathrm{e}=\mathrm{B} \cdot \mathrm{O} \cdot=2
\end{aligned}
$$

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2009 (Mains)
According to molecular orbital theory, antibonding molecular orbitals of $\mathrm{O}_{2}$ contain

Options:
A $4$ electrons
B $6$ electrons
C $10$ electrons
D $8$ electrons

The correct answer is:
$4$ electrons

Solution:
$\left(\begin{array}{l}\sigma^{*} 2 \mathrm{~s}=2 \mathrm{e}^{-} \\ \pi * 2 \mathrm{p}_{\mathrm{x}}=1 \mathrm{e}^{-} \\ \frac{\pi^{*} 2 \mathrm{p}_{\mathrm{y}}=1 \mathrm{e}^{-}}{\text {Total }=4 \text { electrons }}\end{array}\right)$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
According to molecular orbital theory, the number of unpaired electron(s) in O22- is _________ .

Options:

The correct answer is:
0

Solution:
O2-2=18e-, there is no unpaired electron
σ1s2σ*1s2σ2s2σ*2s2σ2pz2π2p2x=π2p2yπ*2px2=π*2p2y

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (31 Aug Shift 2)
According to molecular orbital theory, the species among the following that does not exist is:

Options:
A He2-
B O22-
C He2+
D Be2

The correct answer is:
Be2

Solution:
Chemical Species	Bond Order

He2+	0.5

He2-	0.5

Be2	0

O22-	1

According to M.O.T. If bond order of chemical species is zero then that chemical species does not exist.

The bond order (half the difference in the numbers of bonding and antibonding electrons) for such a diatomic entity is zero.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
According to molecular orbital theory, the species among the following that does not exist is:

Options:
A He2-
B O22-
C He2+
D Be2

The correct answer is:
Be2

Solution:
Chemical Species	Bond Order

He2+	0.5

He2-	0.5

Be2	0

O22-	1

According to M.O.T. If bond order of chemical species is zero then that chemical species does not exist.

The bond order (half the difference in the numbers of bonding and antibonding electrons) for such a diatomic entity is zero.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (25 Feb Shift 1)
According to molecular orbital theory, which of the following molecule will not be available?

Options:
A H22-
B He22+
C He2+
D H2-

The correct answer is:
H22-

Solution:
The electronic configuration of H22⊖ is ⇒ σ1s2, σ*1s2
Bond order of H22⊝=Nb-Na2=2-22=0
Hence, H22⊖ does not exist, due to zero bond order.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2018 (08 Apr)
According to Newton's law of cooling, the rate of cooling of a body is proportional to $(\Delta \theta)^{\mathrm{n}}$, where $\Delta \theta$ is the difference of the temperature of the body and the surroundings, and $n$ is equal to

Options:
A
two

B
three

C
four

D
one

The correct answer is:
one

Solution:
$\frac{\mathrm{dH}}{\mathrm{dt}} \propto\left(\theta_2-\theta_1\right)=(\Delta \theta)^{\mathrm{n}} \Rightarrow \mathrm{n}=1$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2003
According to recent views which is the correct representation of hydrated proton in aqueous solution?

Options:
A H+
B H9O5+
C H9O4+
D H3O+

The correct answer is:
H9O4+

Solution:
the correct representation of hydrated proton in aqueous solution is H9O4+ as H+ remains surrounded by four H2O molecules (H9O4+) .

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?

Options:
A $n=6$ to $n=1$
B $n=5$ to $n=4$
C $n=6$ to $n=5$
D $n=5$ to $n=3$

The correct answer is:
$n=6$ to $n=5$

Solution:
$\because \Delta E \propto\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$, where $n_2>n_1$ Energy of photon obtained from the transition $n=6$ to $n=5$ will have least energy.

Topic: Chemistry, Structure of Atom, JEE Main
According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?

Options:
A $n=6$ to $n=1$
B $n=5$ to $n=4$
C $n=6$ to $n=5$
D $n=5$ to $n=3$

The correct answer is:
$n=6$ to $n=5$

Solution:
$\because \Delta E \propto\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$, where $n_2>n_1$ Energy of photon obtained from the transition $n=6$ to $n=5$ will have least energy.

Topic: Chemistry, Structure of Atom, NEET, NEET 2011 (Mains)
According to the equation:
4NH3g + 5 O2gâ4NOg + 6H2Ol ;
when 1 mole of O2 and 1 mole of ammonia are mixed, then

Options:
A 0.2 mole of H 2 O is produced
B 0.1 mole of NO is produced
C all the oxygen will be consumed
D all the ammonia will be consumed in order to form 1 mole NO

The correct answer is:
all the oxygen will be consumed

Solution:
4NH3(g)4 moles+5O2g5 molesâ4NO(g)4 moles+6H2O(l)6 moles
âµ 4 moles of NH3â 5 moles of O2
1 moles of NH3â54 moles of O2â 1.25 moles of O2
thus, all volume of oxygen will be consumed.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
According to the first law of Faraday, the weight of a substance discharge at the electrode is

Options:
A $W=Z Q$
B $W=e F$
C $W=\frac{Z}{F} I t$
D $W=Z I$

The correct answer is:
$W=Z Q$

Solution:
$W=Z Q ; W=Zit$

Topic: Chemistry, Electrochemistry, JEE Main