datavorous/entrance-exam-dataset huggingface dataset:
In terms of resistance \( \mathrm{R} \) and time \( \mathrm{T} \), the dimensions of ratio \( \frac{\mu}{\epsilon} \) the permeability \( \mu \) and permittivity \( \epsilon \) is :

Options:
A \( \left[\mathrm{RT}^{-2}\right] \)
B \( \left[\mathrm{R}^{2} \mathrm{~T}^{2}\right] \)
C \( \left[\mathrm{R}^{2}\right] \)
D \( \left[\mathrm{R}^{2} \mathrm{~T}^{-1}\right] \)

The correct answer is:
\( \left[\mathrm{R}^{2}\right] \)

Solution:
μ ∈ = MLT - 2 A - 2 M - 1 L - 3 T 4 A 2
= M 2 L 4 T - 6 A - 4
= M L 2 T - 3 A - 2 2
= R2

Topic: Physics, Units and Dimensions, JEE Main
In terms of resistance $R$ and time $T$, the dimensions of ratio $\frac{\mu}{\varepsilon}$ of the permeability $\mu$ and permittivity $\varepsilon$ is:

Options:
A
$\left[\mathrm{RT}^{-2}\right]$

B
$\left[\mathrm{R}^2 \mathrm{~T}^{-1}\right]$

C
$\left[\mathrm{R}^2\right]$

D
$\left[\mathrm{R}^2 \mathrm{~T}^2\right]$

The correct answer is:
$\left[\mathrm{R}^2\right]$

Solution:
Dimensions of $\mu=\left[\right.$ MLT $\left.^{-2} \mathrm{~A}^{-2}\right]$
Dimensions of $\in=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]$
Dimensions of $\mathrm{R}=\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]$
$$
\begin{aligned}
&\therefore \frac{\text { Dimensions of } \mu}{\text { Dimensions of } e}=\frac{\left[\mathrm{MLT}^{-2} \mathrm{~A}^{-2}\right]}{\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]} \\
&=\left[\mathrm{M}^2 \mathrm{~L}^4 \mathrm{~T}^{-6} \mathrm{~A}^{-4}\right]=\left[\mathrm{R}^2\right]
\end{aligned}
$$

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2014 (11 Apr Online)
In the above arrangement, each side of the cube has 1 Ω resistance and it is known that the effective resistance between A and B is 59 Ω. Now, if the resistor between A and B is removed, then find the new effective resistance (in Ω) between the same two points. If your answer is K, mark 4K as answer after rounding off to nearest integer.

Options:

The correct answer is:
5

Solution:
i. For a cube of resistors each of resistance R the equivalent resistance between two edges on the same face is Req=5R9

ii. If AB is removed, assuming the equivalent resistance between remaining resistors is x,

1Req=1R+1x

Req=RxR+x

∴ 59=1(x)1+x=x1+x ....∵ R=1Ω

∴ 5(1+x)=9x

∴ 5+5x=9x

∴ 5=4x

∴ x=54Ω

∴ x=1.25Ω

Topic: Physics, Current Electricity, JEE Main
In the above arrangement, each side of the cube has 1 Ω resistance and it is known that the effective resistance between A and B is 59 Ω. Now, if the resistor between A and B is removed, then find the new effective resistance (in Ω) between the same two points.

Options:

The correct answer is:
01.25

Solution:
i. For a cube of resistors each of resistance R the equivalent resistance between two edges on the same face is Req=5R9

ii. If AB is removed, assuming the equivalent resistance between remaining resistors is x,

1Req=1R+1x

Req=RxR+x

∴ 59=1(x)1+x=x1+x ....∵ R=1Ω

∴ 5(1+x)=9x

∴ 5+5x=9x

∴ 5=4x

∴ x=54Ω

∴ x=1.25Ω

Topic: Physics, Current Electricity, JEE Main
In the above chemical reaction, intermediate "X" and reagent/condition "A" are :

Options:
A
B
C

D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (16 Mar Shift 1)
In the above chemical reaction sequence " \(\mathrm{A}\) " and " \(\mathrm{B}\) " respectively are

Options:
A \(\mathrm{H}_2 \mathrm{O}, \mathrm{H}^{+}\)and \(\mathrm{KMnO}_4\)
B \(\mathrm{O}_3, \mathrm{Zn} / \mathrm{H}_2 \mathrm{O}\) and \(\mathrm{NaOH}_{(\text {alc) }} / \mathrm{I}_2\)
C \(\mathrm{O}_3, \mathrm{Zn} / \mathrm{H}_2 \mathrm{O}\) and \(\mathrm{KMnO}_4\)
D \(\mathrm{H}_2 \mathrm{O}, \mathrm{H}^{+}\)and \(\mathrm{NaOH}_{(\mathrm{alc})} / \mathrm{I}_2\)

The correct answer is:
\(\mathrm{O}_3, \mathrm{Zn} / \mathrm{H}_2 \mathrm{O}\) and \(\mathrm{NaOH}_{(\text {alc) }} / \mathrm{I}_2\)

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (04 Apr Shift 2)
In the above circuit the current in each resistance is:

Options:
A 0 A
B 1 A
C 0.25 A
D 0.5 A

The correct answer is:
0 A

Solution:
Potential difference across each resistor is zero. Hence current is zero in each resistor.

E1-E2=0, E3-E4=0, E4-E5=0

Hence, current through all the 1 Ω ressitor is zero.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2017 (02 Apr)
In the above compound, the bond orders of bonds I and II are:

Options:
A I<2, II>2
B I>2, II>2
C I>2, II<2
D I<2, II<2

The correct answer is:
I<2, II>2

Solution:
Azide ion N3-
Bond order: I<2
Bond Order: II>2
Due to the delocalisation in I double character increase in single bond and II triple bond character decrease in triple bond (double bond character comes).

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2018 (15 Apr)
In the above conversion the correct sequence of reagents to be added is

Options:
A i KMnO4, ii Br2/Fe, iii Fe/H+, iv Cl2
B i Br2/Fe, ii Fe/H+, iii KMnO4, iv Cl2
C i Fe/H+, ii HONO, iii CuCl, iv KMnO4 v Br2
D i Br2/Fe, ii Fe/H+, iii HONO, iv CuCl v KMnO4

The correct answer is:
i Br2/Fe, ii Fe/H+, iii HONO, iv CuCl v KMnO4

Solution:
The required product can be obtained by the following method. In this method nitro toluene is subjected to bromination, bromo group is substituted at ortho position to the methyl as it is more activating group. Then nitro group is reduced to amine group with iron metal in acid. The amine group undergoes diazotisation with nitrous acid. The diazonium group is replaced with chloro group in the presence of cuprous chloride. Finally, methyl group is oxidised in the presence of potassium permanganate.

Topic: Chemistry, Amines, JEE Main, JEE Main 2023 (15 Apr Shift 1)
In the above problem if the lift moves up with an acceleration equal to the acceleration due to gravity, the reading on the spring balance will be

Options:
A $2 \mathrm{~kg}$
B $(2 \times g) \mathrm{kg}$
C $(4 \times g) \mathrm{kg}$
D $4 \mathrm{~kg}$

The correct answer is:
$4 \mathrm{~kg}$

Solution:
When lift moves upward then reading of the spring balance, $R=m(g+a)=2(g+g)=4 g N=4 \mathrm{~kg}$ [As $a=g$ ]

Topic: Physics, Laws of Motion, JEE Main
âââââââ

In the above reaction, 5 g of toluene is converted into benzaldehyde with 92% yield. The amount of benzaldehyde produced is____ ×10-2 g

Options:

The correct answer is:
530

Solution:
Molar mass of benzaldehyde=106 g/mol

Number of moles of toluene =592

Number of moles of Benzaldehyde =0.92×592

Mass of Benzaldehyde =0.92×592×106=5.30 g=530×10-2 g

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (27 Jul Shift 2)
In the above reaction, left hand side and right hand side rings are named as ‘A’ and ‘B’ respectively. They undergo ring expansion. The correct statement for this process is:

Options:
A

Ring expansion can go upto seven membered rings

B

Finally both rings will become six membered each.

C

Finally both rings will become five membered each.

D

Only A will become 6 membered.

The correct answer is:

Finally both rings will become six membered each.

Solution:
The reaction proceeds through a carbocation intermediate. The carbocation formed undergo rearrangement by ring expansion. Four membered ring undergo ring expansion to five membered ring and five membered ring undergo ring expansion to six membered ring and finally it converts to alkene by losing proton.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2023 (13 Apr Shift 1)
In the above reaction, number of possible alkenes (products) are

Options:

The correct answer is:
3

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
In the above reaction product B is

Options:
A

B

C

D

The correct answer is:

Solution:
Alcohols are more reactive than phenol towards the reaction with hydrogen chloride. The preparation of alkyl iodide from alkyl bromide or chloride with potassium or sodium iodide in acetone is generally known as the Finkelstein reaction.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2022 (27 Jul Shift 1)
In the above reaction product ' \(\mathrm{P}\) ' is

Options:
A

B

C

D

The correct answer is:

Solution:
Due to NGP effect of phenyl ring Nucleophilic substitution of Br will occurs.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2024 (09 Apr Shift 2)
In the above reaction, the reagent "A" is :

Options:
A NaBH4,H3O+
B LiAlH4
C Alkaline KMnO4,H+
D HCl,Zn-Hg

The correct answer is:
Alkaline KMnO4,H+

Solution:
KMnO4 in presence of acidic medium shows oxidation reaction.

It converts carbon into acid if carbon is connected to benzene group directly and having at least one hydrogen with it then it can be converted into acid otherwise not. As shown below:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2021 (16 Mar Shift 2)
In the above reaction, the structural formula of (A), "X" and "Y" respectively are :

Options:
A
B
C
D

The correct answer is:

Solution:
It is nucleophilic substitution reaction.

Here nucleophile ethyl alcohol behaves as nucleophile and it attacks on electrophilic site to form methoxy benzene, acetaldehyde, hydrochloric acid ans water as shown below:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2021 (17 Mar Shift 2)
In the above reactions, the enzyme A and enzyme B respectively are :-

Options:
A Amylase and Invertase
B Invertase and Amylase
C Invertase and Zymase
D Zymase and Invertase

The correct answer is:
Invertase and Zymase

Solution:
Informative

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2021 (17 Mar Shift 2)
In the above sequence, C is

Options:
A 1, 2, 3-Tribromobenzene
B 2, 4, 6-Tribromoaniline
C A mixture of o- and p-Bromoaniline
D 1, 3, 5-Tribromoaniline

The correct answer is:
2, 4, 6-Tribromoaniline

Solution:

Topic: Chemistry, Amines, JEE Main
In the above sequence of reaction, (Y) is

Options:
A Isobutyl alcohol
B n-Butyl alcohol
C Tertiary butyl alcohol
D Isobutylene

The correct answer is:
Tertiary butyl alcohol

Solution:
Here tertiary butanol is formed as product Y.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
In the absence of applied potential, the electric current flowing through a metallic wire is zero because

Options:
A the electrons remain stationary
B the electrons are drifted in random direction with a speed of the order of $10^{-2} \mathrm{~cm} / \mathrm{s}$
C the electrons move in random direction with a speed of the order close to that of velocity of light
D electrons and ions move in opposite direction

The correct answer is:
the electrons are drifted in random direction with a speed of the order of $10^{-2} \mathrm{~cm} / \mathrm{s}$

Solution:
If there is no potential difference through a metallic wire, the current is zero because the electrons drift in a random direction with a speed of the order of $10^{-2} \mathrm{~cm} / \mathrm{s}$ so that the net charge crossing a particular cross-section in a given time is zero.

Topic: Physics, Work Power Energy, VITEEE, VITEEE 2008
In the adiabatic compression, the decrease in volume is associated with

Options:
A increase in temperature and decrease in pressure
B decrease in temperature and increase in pressure
C decrease in temperature and decrease in pressure
D increase in temperature and increase in pressure

The correct answer is:
increase in temperature and increase in pressure

Solution:
In adiabatic compression temperature and hence internal energy of the gas increases. In compression pressure will increase.

Topic: Physics, Thermodynamics, JEE Main
In the adjacent circuit, the instantaneous current equation is

Options:
A 2sin⁡100t-π4
B 2sin⁡100t-π4
C 2sin⁡200t-π4
D 2100t+π4

The correct answer is:
2sin⁡100t-π4

Solution:
Let Ï is the phase difference between V and i
tanâ¡Ï=XLR=ÏLR=100Ã1100=1
âÎ¸= Ï 4
Since this is an LR circuit V will lead i
â´ Instantaneous current equation will be
i=i0 sinâ¡100 t-Ï where Ï=Ï4 and i0=V0Z
i0=V0R2+ÏL2=2001002+100Ã12=2
â´i=2sinâ¡100 t-Ï4

Topic: Physics, Alternating Current, JEE Main
In the adjacent diagram, CP represents wavefront and AO and BP corresponding two rays. One particular condition on Î¸ for destructive interference at P between the ray BP and reflected ray OP is cosÎ¸=Î»nd, where n is

Options:

The correct answer is:
2

Solution:
Path difference =CO+OP+λ2
=OPcos2θ+OP+λ2
=OP(1+cos2θ)+λ2
OPcosθ = d

Path difference =dcosθ(1+cos2θ)+λ2
=dcosθ⋅2cos2θ+λ2=2dcosθ+λ2
⇒2dcosθ=kλ (k∈N)

For k=1, 2dcosθ=λ
cosθ=λ2d

Topic: Physics, Wave Optics, JEE Main
In the adjoining circuit diagram each resistance is of $10 \Omega$. The current in the arm AD will be

Options:
A $\frac{2 i}{5}$
B $\frac{3 i}{5}$
C $\frac{4 i}{5}$
D $\frac{i}{5}$

The correct answer is:
$\frac{2 i}{5}$

Solution:
Applying Kirchoffs law in mesh $A B C D A$
$-10\left(i-i_1\right)-10 i_2+20 i_1=0 \quad \Rightarrow 3 i_1-i_2=i$ ....(i)
and in mesh BEFCB
$-20\left(i-i_1-i_2\right)+10\left(i_1+i_2\right)+10 i_2=0$
$\Rightarrow \quad 3 i_1+4 i_2=2 i$ ...(ii)
From equation (i) and (ii) $i_1=\frac{2 i}{5}, i_2=\frac{i}{5} \Rightarrow$
$i_{A D}=\frac{2 i}{5}$

Topic: Physics, Current Electricity, JEE Main
In the Alkaline earth metals, the element forming predominantly covalent compound is

Options:
A $\mathrm{Be}$
B $M g$
C $\mathrm{Sr}$
D $\mathrm{Ca}$

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

Solution:
Be to $B a$ ionic character increasing.

Topic: Chemistry, s Block Elements, JEE Main
In the anion $\mathrm{HCOO}^{-}$the two carbon-oxygen bonds are found to be of equal length. What is the reason for it?

Options:
A
The $\mathrm{C}=\mathrm{O}$ bond is weaker than the $\mathrm{C}-\mathrm{O}$ bond

B
The anion $\mathrm{HCOO}^{-}$has two resonating structures

C
The anion is obtained by removal of a proton from the acid molecule

D
Electronic orbitals of carbon atom are hybridised

The correct answer is:
The anion $\mathrm{HCOO}^{-}$has two resonating structures

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2003
In the Arcitc region hemispherical houses called Igloos are made of ice. It is possible to maintain a temperature inside an Igloo as high as $20^{\circ} \mathrm{C}$ because-

Options:
A Ice has high thermal conductivity
B Ice has low thermal conductivity
C Ice has high specific heat
D Ice has higher density than water

The correct answer is:
Ice has low thermal conductivity

Solution:
Ice has low thermal conductivity So no exchange of heat outside surrounding.

Topic: Physics, Thermal Properties of Matter, KVPY, KVPY 2012 (SA)
In the arrangement shown, by what acceleration the boy must go up so that 100 kg block remains stationary on the wedge. The wedge is fixed and friction is absent everywhere. (Take g=10 m s-2)

Options:
A 2 m s-2

B 4 m s-2

C 8 m s-2

D 6 m s-2

The correct answer is:
6 m s-2

Solution:
For block to be stationary T= 800 N
If man moves up by acceleration a
T-mg=ma
800-500=50 a,
a=6 m s-2

Topic: Physics, Laws of Motion, NEET
In the arrangement shown in the diagram, pulleys are small and springs are ideal. k1=k2=k3=k4=10 N m-1are force constants of the springs and mass m=10 kg. If the time period of small vertical oscillations of the block of mass m is given by 2Ïx seconds, then find the value of x.

Options:

The correct answer is:
4

Solution:
T=2ÏmKeq
where Keq=14 1k1+1k2+1k3+1k4=1016
T=2ÏmKeq=2Ïx
So, x = 1 0 1 0 / 1 6 = 4 , x = 4

Topic: Physics, Oscillations, JEE Main
In the arrangement shown in the figure, friction exists only between the two blocks, A and B. The coefficient of static friction Î¼s=0.6 and coefficient of kinetic friction Î¼k=0.4, the masses of the blocks A and B are m1=20 kg and m2=30 kg, respectively. Find the acceleration (in m s-2) of m1, if a force F=150 N is applied, as shown in the figure. [Assume that string and pulleys are massless]

Options:

The correct answer is:
1.5

Solution:
Let us assume that the two blocks move together, without slipping, relative to each other. The acceleration of the system in that case is

a=Fcos60Â°m1+m2

a=150Ã1220+30=1.5 m s-2

In this case, if the frictional force acting between the two blocks is f, then writing the Newton's second law of motion, for the block of mass m1, we get

T-f=m1a

150-f=20Ã1.5=30

f=120 N

fmax=0.6Ã200=120 N

Since fâ¤fmax, our assumption about the two blocks moving together is correct and hence the acceleration of the blocks is 1.5 m s-2

Topic: Physics, Laws of Motion, JEE Main
In the arrangement shown in the figure, slits S3 and S4 are having a variable separation Z. Point O on the screen is at the common perpendicular bisector of S1S2 and S3S4. (Assume D>>d)

The minimum value of Z for which the intensity at O is zero is

Options:
A 3λD2d
B λD2d
C λD3d
D λDd

The correct answer is:
λDd

Solution:
Intensity at O is due to S3 and S4.
∴ Intensity at S3 = zero and Intensity at S3 is given by
I=K cos 2 ( ϕ 2 )
Intensity at O is zero if ϕ2=π2
ϕ=2πβ×Z2=π ⇒Z=β ;
Z=β=λDd
ϕ=phase difference between waves at S3 and S4

Topic: Physics, Wave Optics, JEE Main
In the arrangement shown in the figure, the coefficient of friction between two blocks is 0.5. The force of friction between the two blocks is (Assume that the \(4 \mathrm{~kg}\) block is placed on a smooth horizontal surface.) (Acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\).)

Options:
A \(8 \mathrm{~N}\)
B \(10 \mathrm{~N}\)
C \(6 \mathrm{~N}\)
D \(4 \mathrm{~N}\)

The correct answer is:
\(8 \mathrm{~N}\)

Solution:
According to the question, the given condition is shown in the figure,
If \(f\) be the friction force between \(2 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) block then the static friction force on \(2 \mathrm{~kg}\) block.
\(\begin{aligned}
& \left.f^{\prime}=\mu R_1=0.5 \times 2 \mathrm{~g}=0.5 \times 2 \times 10 \quad \text { [Given, } \mu=0.5\right] \\
& f^{\prime}=10 \mathrm{~N}
\end{aligned}\)
Since, the static friction on \(2 \mathrm{~kg}\) block is more than force applied on it.
\(\text {i.e, } \quad f^{\prime} > 2 \mathrm{~N}\)
Hence, \(2 \mathrm{~kg}\) body will move along the direction of \(4 \mathrm{~kg}\) body.
Hence, net friction force on the block of \(2 \mathrm{~kg}\),
\(\begin{aligned}
f & =f^{\prime}-2 \mathrm{~N} \\
& =10-2=8 \mathrm{~N}
\end{aligned}\)

Topic: Physics, Laws of Motion, JEE Main
In the arrangement shown, the equivalent resistance between points \( \mathrm{A} \) and \( \mathrm{B} \) is :

Options:
A \( \frac{3 \mathrm{R}}{4} \)
B \( \frac{\mathrm{R}}{2} \)
C \( \frac{5 \mathrm{R}}{8} \)
D \( 2 \mathrm{R} \)

The correct answer is:
\( \frac{5 \mathrm{R}}{8} \)

Solution:
The circuit shown in diagram (1) can be redrawn as shown in diagram (2).
â´ R AB = 5 R 8

Topic: Physics, Current Electricity, JEE Main
In the balanced chemical reaction,
$1 \mathrm{O}_3^{-}+a \mathrm{I}^{-}+b \mathrm{H}^{+} \rightarrow \mathrm{cH}_2 \mathrm{O}+d \mathrm{I}_2$
$a, b, c$ and $d$ respectively correspond to

Options:
A $5,6,3,3$
B $5,3,6,3$
C $3,5,3,6$
D $5,6,5,5$

The correct answer is:
$5,6,3,3$

Solution:
\(1 \mathrm{O}_3^{-}+a \mathrm{I}^{-}+b \mathrm{H} \rightarrow \mathrm{CH}_2 \mathrm{O}+\mathrm{dl}_2\)
Step 1: \(r^1 \rightarrow I_2\) (oxidation)
\(10_3^{-} \rightarrow I_2\) (reduction)
Step 2: \(21 \mathrm{O}_3^{-}+12 \mathrm{H} \rightarrow \mathrm{I}_2+6 \mathrm{H}_2 \mathrm{O}\)
Step \(3: 21 \mathrm{O}_3^{-}+12 \mathrm{H}+10 e \rightarrow \mathrm{I}_2+6 \mathrm{H}_2 \mathrm{O}\)
\(.2 r \rightarrow I_2+2 e\)
Step 4: \(210_3^{-}+12 \mathrm{H}+10 e^{-} \rightarrow 12+6 \mathrm{H}_2 \mathrm{O}\)
\(\left[2 \Gamma \rightarrow I_2+2 e\right] 5\)
Step 5: \(210_3^{-}+10 r+12 \mathrm{H} \rightarrow 6 / 2+6 \mathrm{H}_2 \mathrm{O}\)
\(1 \mathrm{O}_3^{-}+5 \Gamma+6 \mathrm{H} \rightarrow 3 \mathrm{I}_2+3 \mathrm{H}_2 \mathrm{O}\)
On comparing, \(a=5, b=6, c=3, d=3\)

Topic: Chemistry, Redox Reactions, JEE Main
In the bandgap between valence band and conduction band in a material is $5.0 \mathrm{eV},$ then the material is

Options:
A semiconductor
B good conductor
C superconductor
D insulator

The correct answer is:
insulator

Solution:
The band gap of $5 \mathrm{eV}$ corresponds to that of an
insulator.

Topic: Physics, Semiconductors, JEE Main
In the basic structural unit of silicates, the number of oxygen atoms bonded to each Si atom is

Options:

The correct answer is:
4

Solution:
The basic structural unit of silicates is $\mathrm{SiO}_{4}^{4-}$
Each Si atom is bonded to 4 oxygen atoms.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In the below mentioned compounds the decreasing order of reactivity towards electrophilic substitution is
(i)
(ii)
(iii)
(iv)

Options:
A
(iv) $>$ (i) $>$ (ii) $>$ (iii)

B
(ii) $>$ (iii) $>$ (i) $>$ (iv)

C
(iii) $>$ (i) $>$ (iv) $>$ (ii)

D
(i) $>$ (ii) $>$ (iii) $>$ (iv)

The correct answer is:
(i) $>$ (ii) $>$ (iii) $>$ (iv)

Solution:
$-\mathrm{OCH}_3$ and $-\mathrm{CH}_3$ groups are activating group while $-\mathrm{CF}_3$ is a deactivating group.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2012 (12 May Online)
In the binomial expansion of 23+133n, the ratio of the 7th term from the beginning to the 7th term from the end is 1:6, then the value of n is

Options:
A 13
B 16
C 9
D 23

The correct answer is:
9

Solution:
7th term from the beginning is C6n2n-631363
7th term from the end is C6n13n-63263
⇒nC62n3-2132nC62213n3-2=16
⇒2n3-4134-n3=16
⇒2×3n3-4=2×3-1⇒n=9

Topic: Mathematics, Binomial Theorem, JEE Main
In the binomial expansion of ab3+ba321, the term containing same powers of a and b is

Options:
A 11th
B 13th
C 12th
D 6th

The correct answer is:
13th

Solution:
tr+1=C21rab21-r3br3ar6
=C21ra21-r3-r6br3-21-r3

Since, power of a and b are the same
⇒21-r3-r6=r3-21-r3
⇒221-r3=r3+r6=r2
84-4r=3r⇒7r=84⇒r=12
Hence, the 13th term contains the same powers of a and b .

Topic: Mathematics, Binomial Theorem, JEE Main
In the binomial expansion of $(a-b)^n, n \geq 5$, the sum of $5^{\text {th }}$ and $6^{\text {th }}$ terms is zero, then $\frac{a}{b}$ equals

Options:
A
$\frac{5}{n-4}$

B
$\frac{6}{n-5}$

C
$\frac{n-5}{6}$

D
$\frac{n-4}{5}$

The correct answer is:
$\frac{n-4}{5}$

Solution:
${ }^n C_4 a^{n-4}(-b)^4+{ }^n C_5 a^{n-5}(-b)^5=0$
$\Rightarrow\left(\frac{a}{b}\right)=\frac{n-5+1}{5}$.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2007
In the binomial expansion of \( \left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{n} \), the ratio of the \( 7^{\text {th }} \) term from the beginning to the \( 7^{\text {th }} \) term from the end is \( 1: 6 \), then the value of \( n \) is

Options:
A \( 13 \)
B \( 16 \)
C \( 9 \)
D \( 23 \)

The correct answer is:
\( 9 \)

Solution:
7th term from the beginning is C6n2n-631363
7th term from the end is C6n13n-63263
⇒nC62n3-2132nC62213n3-2=16
⇒2n3-4134-n3=16
⇒2×3n3-4=2×3-1⇒n=9

Topic: Mathematics, Binomial Theorem, JEE Main
In the binomial \(\left(2^{1 / 3}+3^{-1 / 3}\right)^{\mathrm{n}}\), if the ratio of the seventh term from the beginning of the expansion to the seventh term from its end is \(1 / 6\), then \(n\) equal to

Options:
A 6
B 9
C 12
D 15

The correct answer is:
9

Solution:
\(\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} a^{\mathrm{n}-\mathrm{r}} \cdot \mathrm{b}^{\mathrm{r}}\) where \(\mathrm{a}=2^{1 / 3}\) and \(\mathrm{b}=3^{-1 / 3}\)
\(\mathrm{T}_7\) from beginning \(={ }^n \mathrm{C}_6 a^{\mathrm{n}-6} \mathrm{~b}^6\) and \(\mathrm{T}_7\) from end \(={ }^{\mathrm{n}} \mathrm{C}_6 \quad \mathrm{~b}^{\mathrm{n}-6} \quad \mathrm{a}^6\)
\(\begin{aligned}
& \Rightarrow \frac{\mathrm{a}^{\mathrm{n}-12}}{\mathrm{~b}^{\mathrm{n}-12}}=\frac{1}{6} \Rightarrow 2^{\frac{\mathrm{n}-12}{3}} \cdot 3^{\frac{\mathrm{n}-12}{3}}=6^{-1} \\
& \Rightarrow \mathrm{n}-12=-3 \Rightarrow \mathrm{n}=9
\end{aligned}\)

Topic: Mathematics, Binomial Theorem, JEE Main