datavorous/entrance-exam-dataset huggingface dataset:
The differential rate law for the reaction $\mathrm{H}_2+\mathrm{I}_2 \rightarrow 2 \mathrm{HI}$ is

Options:
A
$-\frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=-\frac{\mathrm{d}\left[\mathrm{I}_2\right]}{\mathrm{dt}}=-\frac{\mathrm{d}[\mathrm{HI}]}{\mathrm{dt}}$

B
$\frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{I}_2\right]}{\mathrm{dt}}=\frac{1}{2} \frac{1 \mathrm{~d}[\mathrm{Hl}]}{\mathrm{dt}}$

C
$\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{I}_2\right]}{\mathrm{dt}}=-\frac{\mathrm{d}[\mathrm{HI}]}{\mathrm{dt}}$

D
$-2 \frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=-2 \frac{\mathrm{d}\left[\mathrm{I}_2\right]}{\mathrm{dt}}=\frac{\mathrm{d}[\mathrm{HI}]}{\mathrm{dt}}$

The correct answer is:
$-2 \frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=-2 \frac{\mathrm{d}\left[\mathrm{I}_2\right]}{\mathrm{dt}}=\frac{\mathrm{d}[\mathrm{HI}]}{\mathrm{dt}}$

Solution:
No Solution Available

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2002
The diffraction pattern of a light of wavelength 400 nm diffracting from a slit of width 0.2 mm is focused on the focal plane of a convex lens of focal length 100 cm. The width of the 1st secondary maxima will be :

Options:
A 2 mm
B 2 cm
C 0.02 mm
D 0.2 mm

The correct answer is:
2 mm

Solution:
Width of 1st secondary maxima =λa·D

Here

a=0.2×10-3 m

λ=400×10-9 m

D=100×10-2=1 m

Therefore, width of 1st secondary maxima

=400×10-90.2×10-3×1

=2×10-3 m=2 mm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (30 Jan Shift 1)
The diffrerential equation \(\frac{d y}{d x}+\frac{1}{x} \sin 2 y=x^3 \cos ^2 y\) represents a family of curves given by the equation

Options:
A \(\mathrm{x}^6+6 \mathrm{x}^2=\mathrm{C} \tan \mathrm{y}\)
B \(6 \mathrm{x}^2 \tan \mathrm{y}=\mathrm{x}^6+\mathrm{C}\)
C \(\sin 2 \mathrm{y}=\mathrm{x}^3 \cos ^2 \mathrm{y}+\mathrm{C}\)
D none of these

The correct answer is:
\(6 \mathrm{x}^2 \tan \mathrm{y}=\mathrm{x}^6+\mathrm{C}\)

Solution:
The given equation can be converted to linear form by dividing both the sides by \(\cos ^2 \mathrm{y}\). We get
\(\sec ^2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{1}{\mathrm{x}} 2 \tan \mathrm{y}=\mathrm{x}^3\)
Put tan \(y=z \Rightarrow \sec ^2 y \frac{d y}{d x}=\frac{d z}{d x}\)
The equation becomes \(\frac{d z}{d x}+\frac{2}{x} z=x^3\), which is linear in \(z\) The integrating factor is
\(\text {I.F. }=e^{\int \frac{2}{x} d x}=e^{2 \log x}=e^{\log x^2}=x^2\)
Hence, the solution is
\(z\left(x^2\right)=\int x^3\left(x^2\right) d x+a \Rightarrow x^2=\frac{x^6}{6}+a,\)
\(a\) is constant of integration.
\(\begin{aligned}
& \therefore(6 \tan y) x^2=x^6+6 a \Rightarrow 6 x^2 \tan y=x^6+c, \\
& \quad[c=6 a]
\end{aligned}\)

Topic: Mathematics, Differential Equations, JEE Main
The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times. As a result, the diffusion coefficient of this gas increases x times. The value of x is

Options:

The correct answer is:
4

Solution:
Diffusion coefficient ∝ mean free path × mean speed
D∝ī×c̄
But ī=12πσ2RTPN0
or ī∝TP and c̄=8RTπM
or c̄∝TM
So D∝TP.TM or D∝T32PM
or D1D2=T4T322PP=1432×2
=123×2=14
or D2=4D1

Topic: Chemistry, States of Matter, JEE Main
The dimension of B22μ0, where B is magnetic filed and μ0 is the magnetic permeability of vacuum, is:

Options:
A MLT-2
B ML2T-1
C ML2T-2
D ML-1T-2

The correct answer is:
ML-1T-2

Solution:
Energy density in magnetic filed =B22μ0
=Force×displacementdisplacement3=MLT-2.LL3=ML-1T-2

Topic: Physics, Units and Dimensions, JEE Main
The dimension of magnetic field in $\mathrm{M}, \mathrm{L}, \mathrm{T}$ and $\mathrm{C}$ (Coulomb) is given as

Options:
A
$\mathrm{MLT}^{-1} \mathrm{C}^{-1}$

B
$\mathrm{MT}^2 \mathrm{C}^{-2}$

C
$\mathrm{MT}^{-1} \mathrm{C}^{-1}$

D
$\mathrm{MT}^{-2} \mathrm{C}^{-1}$

The correct answer is:
$\mathrm{MT}^{-1} \mathrm{C}^{-1}$

Solution:
$$
\begin{aligned}
& F=q v B \\
& B=F / q v=M C^{-1} T^{-1}
\end{aligned}
$$

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2008
The dimension of mutual inductance is

Options:
A ML2T-2A-1
B ML2T-2A-2
C ML2T-3A-1
D ML2T-3A-2

The correct answer is:
ML2T-2A-2

Solution:
The formula for mutual inductance is given by, ε=MdIdt ⇒M=εdIdt

Dimension of emf or voltage is M L2 T-3 A-1,
Dimension of current is A,
Dimension of time T.

Therefore, the dimension of M is
M=ML2T-3A-1AT-1
=ML2T-2A-2

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2022 (26 Jun Shift 2)
The dimensional formula for areal velocity is

Options:
A [M0L-2T]
B [M0L-2T-1]
C [M0L2T-1]
D [M0L2T]

The correct answer is:
[M0L2T-1]

Solution:
Areal velocity is the area covered per unit time.
∵ Area =[L2]
∴Areal velocity=[L2T-1]

Topic: Physics, Units and Dimensions, JEE Main
The dimensional formula for entropy is

Options:
A [MLT-2K-1]
B [ML2T-2]
C [ML2T-2K-1]
D [ML-2T-2K-1]

The correct answer is:
[ML2T-2K-1]

Solution:
Entropy=heattemperature ( ΔQ=TΔS )
[Entropy]=QT=[ML2T-2][K]=[ML2T-2K-1]

Topic: Physics, Units and Dimensions, JEE Main
The dimensional formula for magnetic flux is

Options:
A [M L 2 â T â1 â A 2 ]
B [ M 0 L −2   T 2 A −2 ]
C [M L 3   T −2   A −2 ]
D [M L 2 â T â2 â A â1 ]

The correct answer is:
[M L 2   T −2   A −1 ]

Solution:
According to Faraday’s law of electromagnetic induction, E.M.F. = -dϕdt
so ϕ=[EMF][T]=Wq[T]
ϕ=ML 2T−2AT[T]=[ML 2T−2A−1]

Topic: Physics, Units and Dimensions, JEE Main
The dimensional formula for the capacitance of a capacitor is

Options:
A A2M-1L-2T4
B AM-1L-2T4
C A2M-2L-2T4
D A0M0L-2T4

The correct answer is:
A2M-1L-2T4

Solution:
U=q22C
∴ C=q22U
∴ C=AT2ML2T-2
=A2M-1L-2T4

Topic: Physics, Units and Dimensions, JEE Main
The dimensional formula of angular impulse is :

Options:
A [M L–2T –1]
B [M L2 T –2 ]
C [M L T –1]
D [M L2 T –1 ]

The correct answer is:
[M L2 T –1 ]

Solution:
Angular impulse = change in angular momentum.

[Angular impulse] = [Angular momentum] =[mvr]

=[M L2 T –1 ]

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2024 (01 Feb Shift 1)
The dimensional formula of electron mobility in semiconductor is ______.

Options:
A M-1L1T2A1
B M1L0T-2A-1
C M1L-1T-2A-1
D M-1L0T2A1

The correct answer is:
M-1L0T2A1

Solution:
Mobility =VdE=Drift velocityElectric field
=LT-1MLT-3A-1
=M-1L0T2A1

Topic: Physics, Current Electricity, JEE Main
The dimensional formula of gravitational constant is

Options:
A

[ML2 T−2]

B

[ML−1 T−1]

C

[M−1 L3 T−2]

D

None of these

The correct answer is:

[M−1 L3 T−2]

Solution:
From Newton’s law of gravitation, we have:

F=G m 1 m 2 r 2

or G= F r 2 m 1 m 2

Substituting the dimensions of all physical quantities:

[G]=[MLT−2] [L2][M][M]

or [G]=[M−1 L3 T−2]

Its SI unit is m3  kg-1 s-2  or  N m2 kg-2.

Topic: Physics, Units and Dimensions, JEE Main
The dimensional formula of latent heat is :

Options:
A $\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$
B $\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]$
C $\left[\mathrm{MLT}^{-2}\right]$
D $\left[\mathrm{M}^{\circ} \mathrm{LT}^{-2}\right]$

The correct answer is:
$\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]$

Solution:
Latent heat is specific heat
$\Rightarrow \frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{M}}=\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-2}$

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2024 (09 Apr Shift 1)
The dimensions of $\left(\mu_0 \varepsilon_0\right)^{-1 / 2}$ are

Options:
A $\left[\mathrm{L}^{1 / 2} \mathrm{~T}^{-1 / 2}\right]$
B $\left[\mathrm{L}^{-1} \mathrm{~T}\right]$
C $\left[\mathrm{LT}^{-1}\right]$
D $\left[\mathrm{L}^{1 / 2} \mathrm{~T}^{1 / 2}\right]$

The correct answer is:
$\left[\mathrm{LT}^{-1}\right]$

Solution:
$\left(\mu_0 \varepsilon_0\right)^{-1 / 2}$ is velocity of light
Hence $\quad c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}=\left[\mathrm{LT}^{-1}\right]$

Topic: Physics, Units and Dimensions, NEET, NEET 2012 (Mains)
The dimensions of ab, in the equation P=a-t2bx, where P is pressure, x is distance and t is time, are

Options:
A M2LT-3
B MT-2
C ML3T-1
D LT-3

The correct answer is:
MT-2

Solution:
P=a-t2bx
â Pbx=a-t2
â Pbx=a=t2
Or b= t2P x=T2ML-1T-2 L=M-1T4
â´ ab= T2M-1T4=MT-2

Topic: Physics, Units and Dimensions, JEE Main
The dimensions of angular momentum, latent heat and capacitance are, respectively.

Options:
A
$\mathrm{ML}^2 \mathrm{~T}^1 \mathrm{~A}^2, \mathrm{~L}^2 \mathrm{~T}^{-2}, \mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2$

B
$\mathrm{ML}^2 \mathrm{~T}^{-2}, \mathrm{~L}^2 \mathrm{~T}^2, \mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{~A}^2$

C
$\mathrm{ML}^2 \mathrm{~T}^{-1}, \mathrm{~L}^2 \mathrm{~T}^{-2}, \mathrm{ML}^2 \mathrm{TA}^2$

D
$\mathrm{ML}^2 \mathrm{~T}^{-1}, \mathrm{~L}^2 \mathrm{~T}^{-2}, \mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{~A}^2$

The correct answer is:
$\mathrm{ML}^2 \mathrm{~T}^{-1}, \mathrm{~L}^2 \mathrm{~T}^{-2}, \mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{~A}^2$

Solution:
Angular momentum $=\mathrm{m} \times \mathrm{v} \times \mathrm{r}$
$$
=\mathrm{ML}^2 \mathrm{~T}^{-1}
$$
Latent heat $\mathrm{L}=\frac{\mathrm{Q}}{\mathrm{m}}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{M}}=\mathrm{L}^2 \mathrm{~T}^{-2}$
Capacitance $\mathrm{C}=\frac{\text { Charge }}{\text { P.d. }}=\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{~A}^2$

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2013 (22 Apr Online)
The dimensions of B2μ0 will be

(if μ0: permeability of free space and B :magnetic field)

Options:
A ML2 T-2
B MLT-2
C ML-1T-2
D MLL-2A-1

The correct answer is:
ML-1T-2

Solution:
Energy density in magnetic field is given by u=B22μ0. Therefore, dimensions of B2μ0 will be equal to the dimensions of energy per unit volume.

B2μ0=u=ML2T-2L3=ML-1T-2

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2022 (28 Jul Shift 1)
The dimensions of four wires of the same material are given below. In which wire the increase in length will be maximum when the same tension is applied?

Options:
A Length $100 \mathrm{~cm}$, diameter $1 \mathrm{~mm}$
B Length $200 \mathrm{~cm}$, diameter $2 \mathrm{~mm}$
C Length $300 \mathrm{~cm}$, diameter $3 \mathrm{~mm}$
D Length $50 \mathrm{~cm}$, diameter $0.5 \mathrm{~mm}$

The correct answer is:
Length $50 \mathrm{~cm}$, diameter $0.5 \mathrm{~mm}$

Solution:
We know that, Young's modulus of elasticity,
$$
\begin{array}{ll}
& Y=\frac{F L}{\pi r^{2} l} \quad \text { [where, } l \text { is change in length] } \\
\Rightarrow \quad & l=\frac{F L}{\pi r^{2} Y} \\
\Rightarrow \quad & l \propto \frac{L}{r^{2}} \propto \frac{L}{d^{2}} \quad \text { [where, } d=\text { diameter] } \\
\Rightarrow \quad & l \propto \frac{L}{d^{2}}
\end{array}
$$
Among the given options, $\frac{L}{d^{2}}$ is maximum for the values of $L=50 \mathrm{~cm}$ and $d=0.5 \mathrm{~mm}$.

Topic: Physics, Mechanical Properties of Solids, JEE Main
The dimensions of magnetic flux are

Options:
A $M L T^{-2} A^{-2}$
B $M L^2 T^{-2} A^{-2}$
C $M L^2 T^{-1} A^{-2}$
D $M L^2 T^{-2} A^{-1}$

The correct answer is:
$M L^2 T^{-2} A^{-1}$

Solution:
The force experienced by a moving charge in a magnetic field is given by $F=q(\vec{v} \times \vec{B})$
Here, $\vec{B}$ is the magnetic field or magnetic flux density or magnetic flux per unit area.
Thus, dimensions of magnetic flux
$[\Phi]=[\mathrm{B}] \times[$ Area $]=\frac{[\mathrm{F}][\mathrm{A}]}{[\mathrm{q}][\mathrm{v}]}$
We know the dimensions of the following as-
$[\mathrm{F}]=\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-2}\right]$
$[\mathrm{A}]=\left[\mathrm{L}^2\right]$
$[q]=\left[\mathrm{A}^1 \mathrm{~T}^1\right]$
$[\mathrm{V}]=\left[\mathrm{L}^1 \mathrm{~T}^{-1}\right]$
Thus, $[\Phi]=$ $\frac{\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-2}\right]\left[\mathrm{L}^2\right]}{\left[\mathrm{A}^1 \mathrm{~T}^1\right]\left[\mathrm{L}^1 \mathrm{~T}^{-1}\right]}=\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]$

Topic: Physics, Electromagnetic Induction, JEE Main
The dimensions of Planck's constant is the same as the product of

Options:
A force and time
B force, displacement and time
C force and distance
D time and displacement

The correct answer is:
force, displacement and time

Solution:
The dimensions of Planck constant, $[h]=\left[\frac{E}{v}\right]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]$
Force, $[F]=\left[\mathrm{MLT}^{-2}\right]$
Displacement, $[d]=[\mathrm{L}]$
Time $[t]=[\mathrm{T}]$
$\therefore$
$$
[F][d][t]=\left[\mathrm{MLT}^{-2}\right][\mathrm{L}][\mathrm{T}]
$$
$$
=\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]
$$
Hence the dimensions of Planck's constant is the same as the product of force, displacement and time.

Topic: Physics, Units and Dimensions, JEE Main
The dimensions of resistance are the same as (where h is the Plank's constant and e is the charge)

Options:
A h2e2
B h2e
C he2
D he

The correct answer is:
he2

Solution:
Resistance, R=Vi=Wqi
=ML2T-2A2T
R=ML2T-3A-2
he2=ML2T-1AT2
=ML2T-3A-2

Topic: Physics, Units and Dimensions, JEE Main
The dimensions of self or mutual inductance are given as

Options:
A L-2M1T-2I-2
B L2M-2T-2I-2
C L2M1T-2I-2
D L2M2T-2I-2

The correct answer is:
L2M1T-2I-2

Solution:
The self or mutual inductance of a coil is the flux change due to change in current, i.e.
L or M = ϕI
So, dimension of Mutual Inductance or Self-Inductance can be given by
⇒L=ϕI
==ML2T-2I-1M0L0I1=ML2T-2I-2

Topic: Physics, Units and Dimensions, JEE Main
The dimensions of stefan-Boltzmann constant $\sigma$ can be written in terms of Planck's constanth' Boltzmann constant $\mathrm{K}_{\mathrm{B}}$ and the speed of light $\mathrm{c}$ as $\sigma=\mathrm{h}^{\alpha} \mathrm{K}_{\mathrm{B}}{ }^{\beta} \mathrm{C}^{\gamma} .$ Here

Options:
A $\alpha=3, \beta=4$ and $\gamma=-3$
B $\alpha=3, \beta=-4$ and $\gamma=2$
C $\alpha=-3, \beta=4$ and $\gamma=-2$
D $\alpha=2, \beta=-3$ and $\gamma=-1$

The correct answer is:
$\alpha=-3, \beta=4$ and $\gamma=-2$

Solution:
$\quad\left(\mathrm{ML}^{\circ} \mathrm{T}^{-2} \theta^{-4}\right)=\left(\mathrm{ML}^{2} \mathrm{~T}^{-1}\right)^{\gamma}\left(\mathrm{ML}^{2} \mathrm{~T}^{-2} \theta^{-1}\right)^{\beta}\left(\mathrm{LT}^{-1}\right)^{\gamma}$
Solving we get

Topic: Physics, Units and Dimensions, JEE Main
The dimensions of universal gravitational constant are:

Options:
A $\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]$
B $\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]$
C $\left[\mathrm{ML}^{-2} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]$
D $\left[\mathrm{ML}^{-2} \mathrm{~L}^2 \mathrm{~T}^{-1}\right]$

The correct answer is:
$\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]$

Solution:
$\begin{aligned}
F & =\frac{\mathrm{GMm}}{R^2} \\
\Rightarrow \mathrm{G}=\frac{F R^2}{M m} & =\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^2\right]}{\left[\mathrm{M}^2\right]} \\
& =\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]
\end{aligned}$

Topic: Physics, Units and Dimensions, JEE Main
The dimensions of universal gravitational constant are

Options:
A $M^2 L^2 T^{-2}$
B $M^{-1} L^3 T^{-2}$
C $M {L}^{-1} T^{-2}$
D $M {L}^2 T^{-2}$

The correct answer is:
$M^{-1} L^3 T^{-2}$

Solution:
$\begin{aligned} & F=\frac{G m_1 m_2}{d^2} \Rightarrow G=\frac{F d^2}{m_1 m_2} \\ & \therefore[G]=\frac{\left[M L T^{-2}\right]\left[L^2\right]}{\left[M^2\right]}=\left[M^{-1} L^3 T^{-2}\right]\end{aligned}$

Topic: Physics, Units and Dimensions, JEE Main
The diode shown in the circuit is a silicon diode. The potential difference between the points $A$ and $\mathcal{B}$ will be

Options:
A $6 \mathrm{~V}$
B $0.6 \mathrm{~V}$
C $0.7 \mathrm{~V}$
D $O V$

The correct answer is:
$6 \mathrm{~V}$

Solution:
In the given condition diode is in reverse biasing so it acts as open circuit. Hence potential difference between $A$ and $B$ is $6 \mathrm{~V}$

Topic: Physics, Semiconductors, JEE Main
The dipeptide, Gln-Gly, on treatment with $\mathrm{CH}_3 \mathrm{COCl}$ followed by aqueous work up gives.

Options:
A

B

C

D

The correct answer is:

Solution:
Amino group of glutamine is acetylated. Amide group of glutamine is not acetylated. Note :
Acetylation of amide requires activation of amides and /or acyl donors, since the nitrogen atom of amides is less basic than that of the corresponding amines due to amide resonance.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B1 . When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is B2 . The ratio B1B2 is:

Options:
A 12
B 2
C 3
D 2

The correct answer is:
2

Solution:
m=magnetic dipole moment=IA. For Loop1, m=Iπr2, B1=μ0i2r
For Loop2. If current is constant m is doubled. r 1 = 2 r . B 2 = μ0 i 2 2 r .B1B2=2

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2018 (08 Apr)
The dipole moment of AX3, BX3 and CX3 are 1.5 D, 0.5 D and 0 D respectively. The possible shapes of molecules may be (consider C has no Lone pair) (A, B and C are are more electronegative than X)

Options:
A Pyramidal, T-shape, Trigonal planar respectively.
B T-shape, Pyramidal, Square planar respectively.
C T-shape, Pyramidal, Trigonal planar respectively.
D Pyramidal, T-shape, Square planar respectively.

The correct answer is:
Pyramidal, T-shape, Trigonal planar respectively.

Solution:
Maximum dipole moment for MX3 type structures would be in pyramidal while least would be in trigonal planar.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The dipole moment of $H B r$ is $1.6 \times 10^{-30} \mathrm{~cm}$ and interatomic spacing is $1 A$. The $\%$ ionic character of $\mathrm{HBr}$ is

Options:
A 7
B 10
C 15
D 27

The correct answer is:
10

Solution:
Charge of $e^{-}=1.6 \times 10^{-19}$
Dipole moment of $\mathrm{HBr}=1.6 \times 10^{-30}$
Inter atomic spacing $=1 \mathbb{A}=1 \times 10^{-10} \mathrm{~m}$
$\%$ of ionic character in
$\mathrm{HBr}=\frac{\text { dipole moment of } \mathrm{HBr} \times 100}{\text { inter spacing distance } \times q}$
$=\frac{1.6 \times 10^{-30}}{1.6 \times 10^{-19} \times 10^{-10}} \times 100$
$=10^{-30} \times 10^{29} \times 100=10^{-1} \times 100=0.1 \times 100=10 \%$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The dipole moment of HBr is 1.6×10-30 Coloumb-metre and inter-atomic spacing is 1  A o . The % ionic character of HBr is

Options:

The correct answer is:
10

Solution:
Charge of electron =1.6Ã10-19 C
Dipole moment of HBr=1.6 Ã 10 â30 âCÃm
Inter-atomic space (distance) = 1 A o
=1Ã10-10 m
Percentage of ionic character in HBr
=Dipole moment of HBr Ã 100inter-atomic distance Ã q
=1.6Ã10-301.6Ã10-19Ã10-10Ã100
=10-30 Ã1029Ã100
=10-1Ã100
=0.1Ã100
=10%

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The dipole moment of HBr is 2.6Ã10-30 esv.cm and the interatomic spacing is 1.41Ã . The percentage of ionic character in HBr is

Options:
A 10.5
B 11.5
C 12.5
D 13.5

The correct answer is:
11.5

Solution:
Percentage ionic character =Observed dipole momentTheoretical dipole momentÃ100
Theoretical dipole moment of a 100% ionic character
=eÃd=(1.6Ã10-19C)Ã(1.41Ã10-10m)
=2.256Ã10-29

% ionic character =2.6Ã10-302.256Ã6Ã10-29Ã100=11.5%

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The dipole moment of is 1.5 D . The dipole Moment ofin Debye

Options:

The correct answer is:
1.5

Solution:
At Para positions Î¼â values cancel each other
So, net dipole moment is due to ortho positions Î¼â
Î¼=Î¼2+Î¼2+2Î¼2cos 120
Î¼â=â1.5 D

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The dipole moments of CCl4,CHCl3 and CH4 are in the order

Options:
A CHCl3<CH4=CCl4
B CCl4<CH4<CHCl3
C CH4<CCl4<CHCl3
D CH4=CCl4<CHCl3

The correct answer is:
CH4=CCl4<CHCl3

Solution:
In CHCl3, dipole moment of C-H is added to the resultant dipole moment of three C-Cl bonds, hence compound has non-zero value of dipole moment.

Electronegativity of carbon =2.5
Electronegativity of hydrogen=2.1.
Elelctrongegativity of C=2.5
Electronegativity of Cl=3

Due to symmetry and similar bonds, vector summation of net dipole moments of both CH4, CCl4 are zero, hence equal.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2020 (07 Jan Shift 1)
The dipole moments of CCl4,CHCl3 and CH4 are in the order

Options:
A CH4<CCl4<CHCl3
B CHCl3<CH4=CCl4
C CH4=CCl4<CHCl3
D CCl4<CH4<CHCl3

The correct answer is:
CH4=CCl4<CHCl3

Solution:
The dipole moments of CCl4,CHCl3 and CH4 are in the order CH4=CCl4<CHCl3

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The direct conversion of A to B is difficult, hence it is carried out by the following shown path
CA↑→DB↓
Given
ΔS(A→C)=50 e.u.
ΔS(C→D)=30 e.u.
ΔS(B→D)=20 e.u.
where e.u. is entropy unit then ΔS(A→B) in e.u. is

Options:

The correct answer is:
60

Solution:
ΔS(A→B)=ΔS(A→C)+ΔS(C→D)-ΔS(B→D)
=50+30-20=60 e.u.

Topic: Chemistry, Thermodynamics (C), JEE Main
The direction cosines l, m and n of two lines are connected by the relations l+m+n=0 and lm=0, then the angle between the lines is

Options:
A π3
B π4
C π2
D 0

The correct answer is:
π3

Solution:
Given, l+m+n=0 ...(i)
and lm=0
i.e. either m=0 or l=0
If l=0 , then putting in equation (i), we get m=-n
â´ Direction ratios are 0,-n,n i.e. 0,-1,1
If m=0 , then putting in equation (i), we get l=-n
â´ Direction ratios are -n,0,n i.e. -1,0,1
â´ cosâ¡Î¸=0Ã-1+-1Ã0+1Ã102+-12+12-12+02+12=12
â Î¸=Ï3

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The direction cosines of a line equally inclined to three mutually perpendicular lines having direction cosines as $l_1, m_1, n_1 ; l_2, m_2, n_2$ and $l_3, m_3, n_3$ are

Options:
A $l_1+l_2+l_3, m_1+m_2+m_3, n_1+n_2+n_3$
B $\frac{l_1+l_2+l_3}{\sqrt{3}}, \frac{m_1+m_2+m_3}{\sqrt{3}}, \frac{n_1+n_2+n_3}{\sqrt{3}}$
C $\frac{l_1+I_2+l_3}{3}, \frac{m_1+m_2+m_3}{3}, \frac{n_1+n_2+n_3}{3}$
D None of these

The correct answer is:
$\frac{l_1+l_2+l_3}{\sqrt{3}}, \frac{m_1+m_2+m_3}{\sqrt{3}}, \frac{n_1+n_2+n_3}{\sqrt{3}}$

Solution:
Since the three lines are mutually perpendicular,
$\therefore \quad \begin{aligned}
& l_1 l_2+m_1 m_2+n_1 n_2=0 \\
& l_2 l_3+m_2 m_3+n_2 n_3=0 \\
& l_3 l_1+m_3 m_1+n_3 n_1=0
\end{aligned}$
Also,
$l_1^2+m_1^2+n_1^2=1, l_2^2+m_2^2+n_2^2=1, l_3^2+m_3^2+n_3^2=1$
Now, $\left(l_1+l_2+l_3\right)^2+\left(m_1+m_2+m_3\right)^2+\left(n_1+n_2+n_3\right)^2$
$=\left(l_1^2+m_1^2+n_1^2\right)+\left(l_2^2+m_2^2+n_2^2\right)+\left(l_3^2+m_3^2+n_3^2\right)$
$\begin{array}{r}+2\left(l_1 l_2+m_1 m_2+n_1 n_2\right)+2\left(l_2 l_3+m_2 m_3+n_2 n_3\right) \\ +2\left(l_3 l_1+m_3 m_1+n_3 n_1\right)=3\end{array}$
$\Rightarrow\left(l_1+l_2+l_3\right)^2+\left(m_1+m_2+m_3\right)^2+\left(n_1+n_2+n_3\right)^2=3$
Hence, direction cosines of required line are :
$\left(\frac{l_1+l_2+l_3}{\sqrt{3}}, \frac{m_1+m_2+m_3}{\sqrt{3}}, \frac{n_1+n_2+n_3}{\sqrt{3}}\right)$
Note: Students should remember it as a fact.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The direction cosines of the line $\frac{3 x+1}{-3}=\frac{3 y+2}{6}=\frac{z}{-1}$ are

Options:
A $\left(\frac{1}{3}, \frac{2}{3}, 0\right)$
B $\left(-1, \frac{2}{3}, 1\right)$
C $\left(-\frac{1}{2}, 1,-\frac{1}{2}\right)$
D $\left(-\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}}\right)$

The correct answer is:
$\left(-\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}}\right)$

Solution:
Change the given equation in standard form, we get,
$\frac{x+\frac{1}{3}}{-1}=\frac{y+\frac{2}{3}}{2}=\frac{z}{-1}$
So direction cosine are, $\left(\frac{-1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}\right)$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The direction cosines of the line $x=y=z$ are

Options:
A $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$
B $\frac{1}{3}, \frac{1}{3}, \frac{1}{3}$
C $1,1,1$
D None of these

The correct answer is:
$\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$

Solution:
Direction cosines $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main