datavorous/entrance-exam-dataset huggingface dataset:
The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is

Options:
A $1: \sqrt{2}$
B $1: 3$
C $2: 1$
D $\sqrt{5}: \sqrt{6}$

The correct answer is:
$\sqrt{5}: \sqrt{6}$

Solution:
$\begin{aligned} & \mathrm{I}_{\text {disk }}=\frac{5}{4} \mathrm{MR}^2=\mathrm{MK}_{\text {disk }}^2 \\ & \mathrm{I}_{\text {ring }}=\frac{3}{2} \mathrm{MR}^2=\mathrm{MK}_{\text {ring }}^2 \\ & =\frac{\mathrm{K}_{\text {disk }}}{\mathrm{K}_{\text {ring }}}=\sqrt{\frac{\overline{5}}{6}}\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axes is

Options:
A $\sqrt{2}: \sqrt{3}$
B $\sqrt{3}: \sqrt{2}$
C $1: \sqrt{2}$
D $\sqrt{2}: 1$

The correct answer is:
$1: \sqrt{2}$

Solution:
$$
\begin{aligned}
& \frac{l_{\text {diso }}}{l_{\text {ring }}}=\frac{M R^2 / 2}{M R^2}=\frac{M K_{\text {disc }}^2}{M K_{\text {ring }}^2} \\
& \Rightarrow \frac{K_{\text {dise }}}{K_{\text {ring }}}=\frac{1}{\sqrt{2}}
\end{aligned}
$$

Topic: Physics, Rotational Motion, NEET, NEET 2008 (Mains)
The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The spectral series are:

Options:
A Paschen and Pfund
B Balmer and Brackett
C Lyman and Paschen
D Brackett and Pfund

The correct answer is:
Lyman and Paschen

Solution:
1λ2=RH1n12-1n221λ1=RH1m12-1m22
As for shortest wavelengths, n2 and m2 is ∞.
∴ λ1λ2=91=m12n12
Now if m1=3 and n1=1 it will justify the statement hence Lyman and Paschen B is correct.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The ratio of the slopes of \( p-V \) graphs of adiabatic and isothermal is

Options:
A \( \frac{\gamma-1}{\gamma} \)
B \( \gamma-1 \)
C \( \gamma \)
D \( \gamma+1 \)

The correct answer is:
\( \gamma \)

Solution:
Slope of p-V graph of adiabatics
Slope of p-V graph of isothermal
Required ratio =γ

Topic: Physics, Thermodynamics, JEE Main
The ratio of the speed of sound in nitrogen gas to that in helium gas, at 300 K is

Options:
A 27
B 17
C 35
D 65

The correct answer is:
35

Solution:
Speed of sound in an ideal gas is given by
v=γRTM
∴ v∝γM [T is same for both the gases]
γ N 2 = 7 5 (Diatomic)
γ He = 5 3 (Monoatomic)

∴ vN2vHe=γN2γHe·MHeMN2=7/55/3428
=35

Topic: Physics, Thermodynamics, JEE Main
The ratio of the variance of first n positive integral multiples of 4 to the variance of first n positive odd numbers is

Options:

The correct answer is:
4

Solution:
Let the variance of first n natural numbers is σ2 then the variance of first n integral multiple of 4 is 16σ2
and the variance of first n odd natural numbers is 4σ2
Then, the required ratio is 16σ24σ2=4

Topic: Mathematics, Statistics, JEE Main
The ratio of the velocity of sound in hydrogen $\left(\gamma=\frac{7}{5}\right)$ to that in helium $\left(\gamma=\frac{5}{3}\right)$ at the same temperature is

Options:
A $\sqrt{\frac{5}{42}}$
B $\sqrt{\frac{5}{21}}$
C $\frac{\sqrt{42}}{5}$
D $\frac{\sqrt{21}}{5}$

The correct answer is:
$\frac{\sqrt{42}}{5}$

Solution:
Velocity of sound in a gas
$v=\sqrt{\frac{\gamma p}{d}}$
$\frac{v_{\mathrm{H}_{2}}}{V_{\mathrm{He}}}=\sqrt{\frac{\gamma_{\mathrm{H}_{2}} \times d_{\mathrm{He}}}{d_{\mathrm{H}_{2}} \times \gamma_{\mathrm{He}}}}$
$\frac{\mathrm{H}_{2}}{\mathrm{He}}=\sqrt{\frac{7 \times 3 \times 2}{5 \times 5}}$
$\begin{array}{ll}\text { As } \quad & \frac{d_{\mathrm{He}}}{d_{\mathrm{H}_{2}}}=2 \\ \therefore & \frac{v_{\mathrm{H}_{2}}}{v_{\mathrm{He}}}=\frac{\sqrt{42}}{5}\end{array}$

Topic: Physics, Waves and Sound, JEE Main
The ratio of thermal conductivity of two rods of different material is $5: 4$. The two rods of same area of cross-section and same thermal resistance will have the lengths in the ratio

Options:
A $4: 5$
B $9: 1$
C $1: 9$
D $5: 4$

The correct answer is:
$5: 4$

Solution:
The thermal resistance of a rod is given by the formula
$\mathrm{R}=\mathrm{L} / \mathrm{kA}$
Given that the two rods have the same thermal resistance and the same cross-sectional area, we can set their thermal resistances equal to each other and solve for the ratio of their lengths
$\mathrm{L} 1 / \mathrm{k} 1 \mathrm{~A}=\mathrm{L} 2 / \mathrm{k} 2 \mathrm{~A}$
Solving for L1/L2 gives:
L1/L2=k1/k2
So, the ratio of the lengths of the two rods is equal to the ratio of their thermal conductivities. Therefore, if the ratio of the thermal conductivities of the two rods is $5: 4$, then the ratio of their lengths will also be $5: 4$

Topic: Physics, Thermal Properties of Matter, JEE Main
The ratio of volumes of $\mathrm{CH}_{3} \mathrm{COOH } \text{ 0.1}$ (N) to CH COONa 0.1 (N) required to prepare a buffer solution of $\mathrm{pH} 5.74$ is (Given, $p K_{a}$ of $C H_{3} C O O H$ is $\left.4.74\right)$

Options:
A $10 : 1$
B $5 : 1$
C $1 : 5$
D $1 : 10$

The correct answer is:
$1 : 10$

Solution:
Given, that $\mathrm{pH}=5.74, \mathrm{pK}_{a}=4.74$
Suppose that volume of acid solution $=x \mathrm{L}$ Volume of salt solution $=y \mathrm{L}$
From Henderson equation,
$$
\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}
$$
or, $\mathrm{pH}-\mathrm{p} K_{\mathrm{a}}=\log \frac{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}$
or, $5.74-4.74=1=\log \frac{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}$
or $\frac{\left.\mathrm{CH}_{3} \mathrm{COONa}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}=10$
or $\frac{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}=\frac{1}{10}=\frac{\frac{x+y}{0.1 \mathrm{y}}}{\mathrm{x}+\mathrm{y}}$
Thus, $\frac{x}{y}=\frac{1}{10}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The ratio of wavelength of spectral lines Hα and Hβ in the Balmer series is x20. The value of x is _____.

Options:

The correct answer is:
27

Solution:
The formula to calculate the wavelength of Hα-line in Balmer series is given by

hcλα=13.614-19 ...1

The formula to calculate the wavelength of Hβ-line in Balmer series is given by

hcλβ=13.614-116 ...2

Divide equation (2) by equation (1)
⇒λαλβ=14-11614-19
=3664×125
=2720

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The ratio of wavelengths of proton and deuteron accelerated by potential Vp and Vd is 1:2. Then, the ratio of Vp to Vd will be

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

The correct answer is:
4:1

Solution:
Kinetic energy gained by a charged particle accelerated by a potential V is qV

KE=qV

Using relation between momentum and kinetic energy

⇒p22 m=qV⇒p=2mqV

We know that momentum of charged particle is p=hλ.

Thus, λ=h2mqV

Now, λpλd=mdVdmpVp

⇒12=2VdVp⇒VpVd=4

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2022 (25 Jul Shift 2)
The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is

Options:
A 25
B 32
C 35
D 23

The correct answer is:
25

Solution:
Work done by an ideal daitomic gas is given by, W=∆Q-∆U∆Q=nCp∆T where Cp for diatomic gas is 52R∆U=nCv∆T where Cv for diatomic gas is 32Rhence W=n52R∆T-n32R∆T=nR∆T52-32=nR∆T

And heat supplies is given by, (∆Q)p=nCp∆T=n52R∆T

Hence the ratio is given by∆W∆QP=nR∆Tn52R∆T
∆W∆QP=25

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2016 (09 Apr Online)
The ratio (R) of output resistance r0, and the input resistance ri in measurements of input and output characteristics of a transistor is typically in the range :

Options:
A R~102-103
B R~1-10
C R~0.1-1.0
D R~0.1-0.01

The correct answer is:
R~102-103

Solution:
The input resistance ri of a transistor is in the range of 1-10 kΩ.

The output resistance r0 of a transistor is in the order of 100 kΩ.

So, the ratio of output resistance and input resistance,

R=r0ri=100 kΩ1-10 kΩ

⇒ R~102-103

Topic: Physics, Semiconductors, JEE Main, JEE Main 2016 (10 Apr Online)
The raw material for Raschig process is

Options:
A chlorobenzene
B phenol
C benzene
D anisole

The correct answer is:
benzene

Solution:
Raschig process is the commercial method for the preparation of chlorobenzene from benzene.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The ray diagram could be correct

Options:
A If $n_1=n_2=n_g$
B If $n_1=n_2$ and $n_1 \lt n_g$
C If $n_1=n_2$ and $n_1\gt n_g$
D Under no circumstances

The correct answer is:
If $n_1=n_2$ and $n_1\gt n_g$

Solution:
If $n_l\gt n_g$ then the lens will be in more denser medium. Hence its nature will change and the convex lens will behave like a concave lens.

Topic: Physics, Ray Optics, JEE Main
The reactance of a coil when used in the domestic ac power supply ( $220 \mathrm{volt}, 50 \mathrm{cycles}$ ) is $100 \mathrm{ohm}$. The self inductance of the coil is nearly

Options:
A 3.2 henry
B 0.32 henry
C 2.2 henry
D 0.22 henry

The correct answer is:
0.32 henry

Solution:
$\begin{aligned} & \text { Reactance }=2 \pi v \Rightarrow 100 \Omega=2 \times \frac{22}{7} \times 50 \times L \\ & \therefore L=0.32 \text { henry }\end{aligned}$

Topic: Physics, Alternating Current, JEE Main
The reactant A is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
The reaction

Options:
A Wurtz reactions
B Koch reaction
C Clemenson's reduction
D Kolbe's reaction

The correct answer is:
Koch reaction

Solution:
The given reaction is koch reaction.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The reaction $2 \mathrm{~A} \rightarrow 2 \mathrm{~B}+\mathrm{C}$ has a rate constant of $1.2 \times 10^{-2} \mathrm{~s}^{-1}$. Which of the following is correct?

Options:
A Plot of $[\mathrm{A}]$ vs $" \frac{1}{t}$ " will be straight line
B Plot of $\frac{1}{[\mathrm{~A}]}$ vs $t^2$ will be a straight line
C Plot of $\ln [\mathrm{A}]$ vs $t$ will be a straight line
D Plot of [A] vs $t^2$ will be a straight line

The correct answer is:
Plot of $\ln [\mathrm{A}]$ vs $t$ will be a straight line

Solution:
$\mathrm{K}=1.2 \times 10^{-2} \mathrm{~s}^{-1}$; This is a 1 st order reaction. Hence, $[\mathrm{A}]=[\mathrm{A}]_0 \mathrm{e}^{-\mathrm{kt}} ; \ln [\mathrm{A}]=\ln [\mathrm{A}]_0-\mathrm{kt}$ $\therefore \quad \ln [\mathrm{A}]$ vs $\mathrm{t}$ is a straight line with negative slope $(\mathrm{k})$.

Topic: Chemistry, Chemical Kinetics, JEE Main
The reaction 2 A+B2→2AB is an elementary reaction.

For a certain quantity of reactants, if the volume of the reaction vessel is reduced by a factor of 3, the rate of the reaction increases by a factor of _____ . (Round off to the Nearest Integer).

Options:

The correct answer is:
27

Solution:
Reaction : 2 A+B2⟶2AB

As the reaction is elementary, the rate of reaction is

r=K.[A]2 B2

on reducing the volume by a factor of 3, the concentrations of A and B2 will become 3 times and hence, the rate becomes 32×3=27 times of initial rate.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2021 (17 Mar Shift 2)
The reaction $2 \mathrm{~A} \rightarrow \mathrm{B}+\mathrm{C}$, would be a zero-order reaction when:

Options:
A the rate of the reaction is proportional to the square of concentration of A
B the rate of reaction remains same at any concentration of A
C the rate remains unchanged at any concentration of $\mathrm{B}$ and $\mathrm{C}$.
D the rate of reaction doubles if concentration of $\mathrm{B}$ is increased to double.

The correct answer is:
the rate of reaction remains same at any concentration of A

Solution:
The characteristic of zero-order reaction is Rate of reaction depends only on concentration of reactants.
A reaction is zero-order if concentration data is plotted versus time and the result is a straight line.

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2002
The reaction,
$$
2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}(\mathrm{g}) \rightleftharpoons 3 \mathrm{C}(\mathrm{g})+\mathrm{D}(\mathrm{g})
$$
is begun with the concentrations of $A$ and $B$ both at an initial value of $1.00 \mathrm{M}$. When equilibrium is reached, the concentration of $D$ is measured and found to be $0.25 \mathrm{M}$. The value for the equilibrium constant for this reaction is given by the expression

Options:
A $\left[(0.75)^{3}(0.25)\right] \div\left[(1.00)^{2}(1.00)\right]$
B $\left[(0.75)^{3}(0.25)\right] \div\left[(0.50)^{2}(0.75)\right]$
C $\left[(0.75)^{3}(0.25)\right] \div\left[(0.50)^{2}(0.25)\right]$
D $\left[(0.75)^{3}(0.25)\right] \div\left[(0.75)^{2}(0.25)\right]$

The correct answer is:
$\left[(0.75)^{3}(0.25)\right] \div\left[(0.50)^{2}(0.75)\right]$

Solution:
Thereaction-
$\quad 2 A(\mathrm{~g})+B(\mathrm{~g}) \rightleftharpoons 3 C(\mathrm{~g})+D(\mathrm{~g})$
Initial $\quad 1 \quad 1$
At equil $1-0.501-0.25 \quad 0.75 \quad 0$
$\mathrm{~K}=\frac{(0.75)^{3}(0.25)}{(0.50)^{2}(0.75)}$

Topic: Chemistry, Chemical Equilibrium, JEE Main
The reaction,
\(2 \mathrm{~A}_{(\mathrm{g})}+\mathrm{B}_{(\mathrm{g})} \leftrightharpoons 3 \mathrm{C}_{(\mathrm{g})}+\mathrm{D}_{(\mathrm{g})}\)
is begun with the concentrations of \(A\) and \(B\) both at an initial value of 1.00 M When equilibrium is reached, the concentration of \(D\) is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression

Options:
A \(\left[(0.75)^3(0.25)\right] \div\left[(1.00)^2(1.00)\right]\)
B \(\left[(0.75)^3(0.25)\right] \div\left[(0.50)^2(0.75)\right]\)
C \(\left[(0.75)^3(0.25)\right] \div\left[(0.50)^2(0.25)\right]\)
D \(\left[(0.75)^3(0.25)\right] \div\left[(0.75)^2(0.25)\right]\)

The correct answer is:
\(\left[(0.75)^3(0.25)\right] \div\left[(0.50)^2(0.75)\right]\)

Solution:
\(\begin{array}{lcccc}
& 2 A_{(g)} & + & B_{(g)} & \rightleftharpoons & 3 C_{(g)} & + & D_{(g)} \\
\text {Initial moles : } & 1 & & 1 & & 0 & & 0 \\
\text {Moles at eq. : } & 1-(2 \times 0.25) & & 1-0.25 & & 3 \times 0.25 & & 0.25 \\
& =0.5 & & =0.75 & & =0.75 & & =0.25
\end{array}\)
Equilibrium constant, \(K=\frac{[C]^3[D]}{[A]^2[B]}\)
\(\therefore \quad K=\frac{(0.75)^3(0.25)}{(0.5)^2(0.75)}\)

Topic: Chemistry, Chemical Equilibrium, NEET, NEET 2010 (Mains)
The reaction 2N2O5g→4NO2g+O2(g) follows first order kinetics. The pressure of a vessel containing only N2O5 was found to increase from 50 mm Hg to 87.5 mm Hg in 30 min. The pressure exerted by the gases after 60 min. Will be (Assume temperature remains constant) :

Options:
A 106.25 mm Hg
B 125 mm Hg
C 116.25 mm Hg
D 150 mm Hg

The correct answer is:
106.25 mm Hg

Solution:
2N2O5g â4NO2g+O2g
at t=0, P0 0 0
at t=30 min. (P0-x) 2x x 2
at t=60 min. (P0-y) 2y y 2
âµP0=50 mmHg
âµafter 30min. â¹ P0-x+2x+x2=87.5
P0+32x=87.5
32x=37.5 ; x=25mmHg
P0=50 ; P30min=25mmHg
That mean 30 min. is Half life of reaction.
P0â30 min.P02â30 min.P04
at t= 60 min
PN2O5 =P04 â¹504=12.5mmHg
P0-y=12.5 â¹50-y=12.5
y=37.5mmHg
Total P after 60 min. â¹P0-y+2y+y2â¹50Ã32 Ã37.5
=106.25 mmHg

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2015 (10 Apr Online)
The reaction 2NO+Br2→2NOBr takes place through the mechanism given below

NO+Br2⇌NOBr2 fast

NOBr2+NO→2NOBr slow

The overall order of the reaction is _____.

Options:

The correct answer is:
3

Solution:
On applying Law of mass action over slowest step,

R=k[NOBr2][NO]

The slowest step is the rate determining step.

On applying Law of mass action on equilibrium of Step I

k=NOBr2NOBr2

NOBr2=kNOBr2

Hence, overall rate equation will be

R=kNO2Br2

=k'NO2Br2

The overall order is equal to sum of the powers of the concentration terms in the rate law.

The overall order of the reaction is 3.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2023 (12 Apr Shift 1)
The reaction $2 \mathrm{X} \rightarrow \mathrm{B}$ is a zeroth order reaction. If the

initial concentration of $\mathrm{X}$ is $0.2 \mathrm{M}$, the half-life is $6 \mathrm{~h}$. When the initial concentration of $X$ is $0.5 \mathrm{M}$, the time required to reach its final concentration of $0.2 \mathrm{M}$ will be

Options:
A $9.0 \mathrm{~h}$
B $12.0 \mathrm{~h}$
C $18.0 \mathrm{~h}$
D $7.2 \mathrm{~h}$

The correct answer is:
$18.0 \mathrm{~h}$

Solution:
For the reaction $2 \mathrm{X} \rightarrow \mathrm{B}$, follow zeroth order Rate equation is $\mathrm{Kt}=[\mathrm{A}]_{0}-[\mathrm{A}]$

For the half-life; $t=t_{1 / 2}$ and $[A]=0.1$ $\mathrm{Kt}_{12}=0.2-0.1$

$\frac{0.2-0.1}{6}=\frac{0.1}{6} \mathrm{Mhr}^{-1}$

$\therefore$ Time required to reach from $0.5 \mathrm{M}$ to $0.2 \mathrm{M}$ $\mathrm{Kt}=[\mathrm{A}]_{0}-[\mathrm{A}]$

$\frac{0.1}{6} \times \mathrm{t}=(0.5-0.2) ; \mathrm{t}=18$ hour

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The reaction $A+B \rightleftharpoons C+D+$ heat has reached equilibrium. The reaction may be made to proceed forward by

Options:
A Adding more $\mathrm{C}$
B Adding more $\mathrm{D}$
C Decreasing the temperature
D Increasing the temperature

The correct answer is:
Decreasing the temperature

Solution:
Exothermic reaction is favoured by low temperature to proceed in forward direction.

Topic: Chemistry, Chemical Equilibrium, JEE Main
The reaction A+B+C→ products is formed to obey the rate law, r=dAdt=KA2B32C-12 . The overall order of the reaction is

Options:

The correct answer is:
3

Solution:
Overall order =2+32+-12=3

Topic: Chemistry, Chemical Kinetics, JEE Main
The reaction between an alcohol and an acid with the elimination of water molecule is called

Options:
A Esterification
B Saponification
C Etherification
D Elimination

The correct answer is:
Esterification

Solution:
CH 3 COOH+CH 3 OH⇌ CH 3 COOCH 3 +H 2 O                                                                          Esterification
When a carboxylic acid is treated with an alcohol and an acid catalyst, an ester is formed (along with water). This reaction is called the Fischer esterification.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The reaction between an ester and excess of Grignard reagent shall finally result in a

Options:
A Primary alcohol
B Secondary alcohol
C Tertiary alcohol
D Ketone

The correct answer is:
Tertiary alcohol

Solution:
The reaction between an ester and excess of Grignard reagent gives tertiary alcohol. Hence, option 3 is correct.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The reaction between phosphorus (P4 ) and hydrogen (H2 ) can result in the formation of phosphine (PH3 ) as shown :

P 4 s + 6 H 2 g â 4 PH 3 g

The graph shows the change in concentration of hydrogen for this reaction in which the system was disturbed after four hours.

Which of the following could explain the change in hydrogen concentration at time = 4 hours?

Options:
A The volume on the reaction vessel was decreased
B A catalyst was added
C The pressure on the reaction mixture was decreased
D More phosphorus was added

The correct answer is:
The volume on the reaction vessel was decreased

Solution:
Volume of the vessel was decreased, which led to sudden increase in [H2 ]. The reaction then moves in direction to achieve a state of new equilibrium.
A catalyst does not affect the equilibrium concentration of reactants and products.
Increase in pressure would decrease concentration as reaction favoured forward.
P4(s) will have no effect on kP and concentration of H2(g)
( because active mass of solid is taken to be unit )

Topic: Chemistry, Chemical Equilibrium, JEE Main
The reaction
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2+\mathrm{CHCl}_3+3 \mathrm{KOH} \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{NC}+3 \mathrm{KCl}+2 \mathrm{H}_2 \mathrm{O}$
is known as

Options:
A Carbylamine reaction
B Reimer-Tiemann reaction
C Kolbe reaction
D Hofmann's degradation

The correct answer is:
Carbylamine reaction

Solution:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2+\mathrm{CHCl}_3+3 \mathrm{KOH} \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{NC}+3 \mathrm{KCl}+2 \mathrm{H}_2 \mathrm{O}$ this is the Carbylamine reaction.

Topic: Chemistry, Amines, JEE Main
The reaction

Can be classified as:

Options:
A Williamson ether synthesis reaction
B Alcohol formation reaction
C Dehydration reaction
D Williamson alcohol synthesis reaction

The correct answer is:
Williamson ether synthesis reaction

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The reaction: $\left(\mathrm{CH}_3\right)_3 \mathrm{C}-\mathrm{Br} \stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow}\left(\mathrm{CH}_3\right)_3-\mathrm{C}-\mathrm{OH}$

Options:
A
elimination reaction

B
substitution reaction

C
free radical reaction

D
displacement reaction

The correct answer is:
substitution reaction

Solution:
No Solution Available

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The reaction: $\left(\mathrm{CH}_3\right)_3 \mathrm{C}-\mathrm{Br} \stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow}\left(\mathrm{CH}_3\right)_3-\mathrm{C}-\mathrm{OH}$

Options:
A
elimination reaction

B
substitution reaction

C
free radical reaction

D
displacement reaction

The correct answer is:
substitution reaction

Solution:
No Solution Available

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2002
The reaction, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}+\mathrm{KOH}(a q) \longrightarrow$ $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{KI}$ is classified as

Options:
A electrophilic substitution
B nucleophilic substitution
C electrophilic addition
D nucleophilic addition

The correct answer is:
nucleophilic substitution

Solution:
Alkyl halides are hydrolysed to corresponding alcohols by boiling with aqueous alkali solution ( $\mathrm{NaOH}$ or $\mathrm{KOH}$ ).
$$
\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}+\mathrm{KOH} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{KI}
$$
This is nucleophilic substitution reaction in which the attacking nucleophile is $\mathrm{OH}^{-}$.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The reaction, $\mathrm{CH}_3 \mathrm{CHO} \frac{[\mathrm{H}]}{\mathrm{Zn}(\mathrm{Hg}) / \text { Conc. } \mathrm{HCl}} \mathrm{CH}_3 \mathrm{CH}_3$

Options:
A
Cannizaro's reaction

B
Rosenmund reduction

C
Wolf-Kishner reduction

D
Clemmenson reduction

The correct answer is:
Clemmenson reduction

Solution:
The reaction given is a Clemenson reduction.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2012 (12 May Online)
The reaction $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{Cl}_2 \xrightarrow{\text{Phosphorus, Water}} \mathrm{ClCH}_2 \mathrm{COOH}+\mathrm{HCl}$ is called

Options:
A Hell-Yolhard-Zelinsky reaction
B Birch reaction
C Rosenmund reaction
D Hunsdieckert reaction

The correct answer is:
Hell-Yolhard-Zelinsky reaction

Solution:
Hell-Yolhard-Zelinsky reaction

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The reaction condition leading to the best yield of C 2 H 5 Cl are

Options:
A C 2 H 6 ( excess )+C l 2 ââ â UVââlight
B C 2 H 6 +C l 2 ( excess ) → room  temp. dark
C C 2 H 6 +C l 2    → UV  light
D C 2 H 6 +C l 2 ( excess )ââ â UVââlight

The correct answer is:
C 2 H 6 ( excess )+C l 2    → UV  light

Solution:
During chlorination of alkane, if excess of alkane is treated with Cl 2 ( g ) in presence of light or heat, chance of monoâchlorination predominant.

C 2 H 6 ( excess )+C l 2 ââ â UVââlight CH 3 âC H 2 âClâ+HCl

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The reaction conditions used for converting 1, 2-dibromo ethane to ethylene are

Options:
A Zn, alcohol, $\Delta$
B $\mathrm{KOH}$, alcohol, $\Delta$
C $\mathrm{KOH}$, water, $\Delta$
D $\mathrm{Na}$, alcohol, $\Delta$

The correct answer is:
Zn, alcohol, $\Delta$

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
The reaction given below is known as
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{ONa}+I \mathrm{C}_2 \mathrm{H}_5 \longrightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5+\mathrm{Nal}$

Options:
A Kolbe's synthesis
B Wurtz's synthesis
C Williamson's synthesis
D Grignard's synthesis

The correct answer is:
Williamson's synthesis

Solution:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{ONa}+\mathrm{IC}_2 \mathrm{H}_5 \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5+\mathrm{NaI}$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main