datavorous/entrance-exam-dataset huggingface dataset:
The speed of sound in oxygen $\left(\mathrm{O}_2\right)$ at a certain temperature is $460 \mathrm{~ms}^{-1}$. The speed of sound in helium (He) at the same temperature will be (assumed both gases to be ideal)

Options:
A
$460 \mathrm{~ms}^{-1}$

B
$500 \mathrm{~ms}^{-1}$

C
$650 \mathrm{~ms}^{-1}$

D
None of these

The correct answer is:
None of these

Solution:
$$
\begin{aligned}
& v=\sqrt{\frac{\gamma R T}{M}} \\
& \frac{V_1}{V_2}=\sqrt{\frac{\gamma_1 M_2}{\gamma_2 M_1}}=\sqrt{\frac{\frac{7}{5} \times 4}{\frac{5}{3} \times 32}} \\
& \frac{460}{V_2}=\sqrt{\frac{21}{25 \times 8}} \Rightarrow \quad v_2=\frac{460 \times 5 \times 2 \sqrt{2}}{\sqrt{21}}=1420
\end{aligned}
$$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2008
The speed of the earth's rotation about its axis is \( \omega \). Its speed is increased to \( x \) times to make the effective acceleration due to gravity equal to zero at the equator, then \( x \) is around \( \left(g=10 \mathrm{~m} \mathrm{~s}^{-2}, R=6400 \mathrm{~km}\right) \)

Options:
A \( 1 \)
B \( 8.5 \)
C \( 17 \)
D \( 34 \)

The correct answer is:
\( 17 \)

Solution:
At the equator, g′=g-Rω2

When angular velocity be ω′=xω

then, 0=g-Rω′2 or
ω=gR

present angular velocity of the earth ω0=2π24×60×60

ωω0=17

Topic: Physics, Gravitation, JEE Main
The speed of the particle is increasing at the rate of a $\mathrm{m} / \mathrm{s}^2$ when it is rotating in the circular path of radius $R$. Find the acceleration of the particle when its speed is $v$.

Options:
A a
B $v^2 / R$
C $\sqrt{a^2+\frac{v^2}{r}}$
D $\sqrt{a^2+\frac{v^4}{r^2}}$

The correct answer is:
$\sqrt{a^2+\frac{v^4}{r^2}}$

Solution:
$\sqrt{a^2+\frac{v^4}{r^2}}$
Given $R=$ radius of the path
- The tangential acceleration is given as,
$\Rightarrow a_t=a$ ...(1)
- The centripetal acceleration is given as,
$\Rightarrow a_c=\frac{v^2}{R} \quad-----(2)$
When two vectors act perpendicular to each other, the resultant vector is given as,
$\Rightarrow R=\sqrt{P^2+Q^2}$
Where $\mathrm{P}$ and $\mathrm{Q}$ are the two vectors
- Since acceleration is a vector quantity so the resultant acceleration of the rotating particle is given as,
$\begin{aligned} & \Rightarrow a_R=\sqrt{a^2+\left(\frac{v^2}{R}\right)^2} \\ & \Rightarrow a_R=\sqrt{a^2+\frac{v^4}{R^2}}\end{aligned}$
- Hence, option 4 is correct.

Topic: Physics, Laws of Motion, JEE Main
The speed verses time graph for a particle is shown in the figure. The distance travelled (in m) by the particle during the time interval t=0 to t=5 s will be __________

Options:

The correct answer is:
20

Solution:
Distance = Area of |v| – t graph

=1/2×8×5=20 m

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The 'Spin only’ Magnetic moment for [Ni(NH3)6] 2+ is______× 10–1 BM. (given = Atomic number of Ni : 28)

Round off your answer to the nearest integer.

Options:

The correct answer is:
28

Solution:
In the presence of ligand NH3, pairing does not occur in Ni2+ because pairing of electrons cannot create two vacant d-orbitals. Hence, nickel ion undergoes sp3d2 hybridisation and it has two unpaired electrons.

Ni2+=3d8

No. of unpaired electron =2

The spin only magnetic moment

μ=nn+2=8=2.82 BM

=28.2×10-1 BM

x=28

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (31 Jan Shift 1)
The spin only magnetic moment of a divalent ion in aqueous solution (atomic number 29) is ___ BM.

Options:

The correct answer is:
2

Solution:
Z=29Cu

Cu2+ form CuH2O42+ complex ion with H2O

CuH2O42+⇒Cu2+Ar3 d9,H2O→WFL

number of unpaired e-=1

μ=11+2B.M

μ=3⇒1.73 B.M.⇒round off ans. ⇒2

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The spin only magnetic moment of MnH2O62+ complexes is B.M. (Nearest integer)

(Given atomic number of Mn=25)

Options:

The correct answer is:
6

Solution:
A spin-only magnetic moment occurs when an electron rotates on its own axis and generates a magnetic field.
μ = n ( n + 2 )BM
Here, n is the number of unpaired electrons.
The unit for a spin-only magnetic moment is the Bohr magneton (BM).

MnH2O62+

Mn2+=3d5

μ=5(5+2)=5.91BM

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (01 Feb Shift 2)
The spin only magnetic moment of the complex present in Fehling's reagent is____B.M. (Round off your answer to the nearest integer)

Options:

The correct answer is:
2

Solution:
Fehling's solution consists of two parts A and B. Fehling’s solution A is deep blue colour due to the bis(tartrate) complex of Cu2+. Fehling's solution B is the Rochelle salt (potassium sodium tartrate tetrahydrate) serves as a chelating agent in the solution.

Fehling solution is a complex of Cu++

Cu++=3 d9

Number of unpaired e-=1

Magnetic Moment=11+2=3=1.73BM

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2022 (27 Jul Shift 2)
The spin only magnetic moment of transition metal ion found to be 5.92 BM . The number of unpaired electrons present in the species is:

Options:

The correct answer is:
5

Solution:
Spin only magnetic moment, Î¼â=â n(n+2) âBM
where n = no. of unpaired electrons there; n(n + 2)  = 5.92
â´nâ=â5

Topic: Chemistry, Coordination Compounds, JEE Main
The spin only magnetic moments (in BM) for free Ti3+,V2+ and Sc3+ ions respectively are

(At.No. Sc : 21,Ti:22, V:23 )

Options:
A 3.87,1.73,0
B 1.73,3.87,0
C 1.73,0,3.87
D 0,3.87,1.73

The correct answer is:
1.73,3.87,0

Solution:
μ=n(n+2)BM

Ti+3=[Ar]3 d1 n=1 μ=1.73BM

V+2=[Ar]3 d3 n=3 μ=3.87BM

Sc+3=[Ar]3 d04 s0 n=0 μ=0

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2021 (25 Jul Shift 2)
The spinal cord passes through which of the following vertebral structures?

Options:
A Neural canal
B Diocoel
C Central canal
D Ventricular system

The correct answer is:
Neural canal

Solution:
The vertebra extends from the base of the skull and forms the main framework of the trunk.
Neural canal (or vertebral canal): It is the central hollow portion of the vertebra through which the spinal cord passes. Neurocoel or central canal is filled with CSF.

Topic: Biology, Neural Control and Coordination, NEET
The spinel structure consists of an array of O2- ions in fcc arrangement. General formula of spinel is AB2O4 . Cations of A occupy 1/8th the tetrahedral voids and cations of B ions occupy half of the octahedral voids. If oxide ions are replaced by X-8/3 ions then number of an ionic vacancy per unit cell will be

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
These are four O2- ions in the unit cell; let these ions are replaced by n, X-8/3 ions.
Charge of O2- ions = Charge of X-8/3 ions
4(-2) = n (-8/3)
n = 3 (i.e., there will one anionic vacancy)
Thus, there will be one an ionic vacancy per unit cell.

Topic: Chemistry, Solid State, JEE Main
The "spin-only" magnetic moment [in units of Bohr magneton, $\left.\left(\mu_{\mathrm{B}}\right)\right]$ of $\mathrm{Ni}^{2+}$ in aqueous solution would be (Atomic number of $\mathrm{Ni}=28$ )

Options:
A
2.84

B
4.90

C
0

D
1.73

The correct answer is:
2.84

Solution:
No Solution Available

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2006
The spin-only magnetic moment value for the complex Co(CN)64- is...... BM. (nearest integer value)
[At. no. of Co=27]

Options:

The correct answer is:
2

Solution:
Co(CN)64-
x+6×(-1)=-4
x=+2
Co2+:[Ar]3 d7 and CN- is a strong field ligand which can pair electron of central atom.

It has one unpaired electron (n) in 4 d-subshell. So spin only magnetic moment (μ)=
n(n+2)B·M
where n= number of unpaired electrons.
μ=3 B.M
μ=1.73 BM ≃2 BM

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The spin-only magnetic moment value of an octahedral complex among CoCl3·4NH3, NiCl2·6H2O and PtCl4·2HCl, which upon reaction with excess of AgNO3 gives 2 moles of AgCl is_____B.M. (Nearest Integer)

Options:

The correct answer is:
3

Solution:
(I) H2PtCl6→ no ppt with AgNO3

Pt+4, CN=6

(II) NiH2O6Cl2 AgNO3excess 2 mole of AgCl↓,

Ni+2,CN=6,n=2, μm=nn+2 BM=8=2.82BM≃3

Aqua complexes, generally have CN=6

(III) CoCl2NH34Cl⟶AgNO3excess1 Mole of AgCl↓,

Co+3,CN=6

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2022 (26 Jun Shift 1)
The spin-only magnetic moment value of B2+ species is ×10-2BM (Nearest integer)

[ Given : 3=1.73]

Options:

The correct answer is:
173

Solution:
In B2number of electron +=9

B2=+(σ1 s)2σ*1 s2(σ2 s)2σ⋆2 s2π2px=1π2py0,

so unpaired electron =1

μ=n(n+2)B

μ=3 B.M. =1.73=173×10-2 B.M

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (01 Sep Shift 2)
The spin-only magnetic moment value of the most basic oxide of vanadium among V2O3' V2O4 and V2O5 is____B.M. (Nearest integer)

Options:

The correct answer is:
3

Solution:
V2O3 is the most basic oxide among the given oxides.

In V2O3, Vanadium is in +3 oxidation state

V3+⇒Ar3d24s0

So, 2 unpaired electrons are there,

Its magnetic moment is equal to μ=nn+2 BM=22+2 BM

=2.83 BM ≃ 3 BM

(n= number of unpaired electrons)

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (26 Jun Shift 1)
The spring balance $A$ reads $2 \mathrm{~kg}$ with a block $m$ suspended from it. $A$ balance $B$ reads $5 \mathrm{~kg}$ when a beaker filled with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging mass is inside the liquid as shown in figure. In this situation

Options:
A The balance $A$ will read more than $2 \mathrm{~kg}$
B The balance $B$ will read more than $5 \mathrm{~kg}$
C The balance $A$ will read less than $2 \mathrm{~kg}$ and $B$ will read more than $5 \mathrm{~kg}$
D The balances $A$ and $B$ will read $2 \mathrm{~kg}$ and $5 \mathrm{~kg}$ respectively

The correct answers are:
The balance $B$ will read more than $5 \mathrm{~kg}$, The balance $A$ will read less than $2 \mathrm{~kg}$ and $B$ will read more than $5 \mathrm{~kg}$

Solution:
Force of upthrust will be there on mass $m$ shown in figure, so $A$ weighs less than $2 \mathrm{~kg}$. Balance will show sum of load of beaker and reaction of upthrust so it reads more than $5 \mathrm{~kg}$.

Topic: Physics, Laws of Motion, JEE Main
The spring block system lies on a smooth horizontal surface. The free end of the spring is being pulled towards the right with constant speed V0=2 m s-1. At t=0 s, the spring of constant k=100 N cm-1 is unstretched and the block has a speed 1 m s-1 to left. The maximum extension of the spring will be

Options:
A 2 cm
B 4Â cm
C 6 cm
D 8Â cm

The correct answer is:
6 cm

Solution:
In the frame (inertial w.r.t. earth) of the free end of spring, the initial velocity of the block is 3 m/s to left and the spring is unstretched.
Applying conservation of energy between initial and maximum extension state,
12mV2=12kA2
A=mk V=410000Ã3=6 cm

Topic: Physics, Oscillations, JEE Main
The square of the distance of the image of the point $(6,1,5)$ in the line $\frac{x-1}{3}=\frac{y}{2}=\frac{z-2}{4}$, from the origin is _________

Options:

The correct answer is:
62

Solution:
Let $\mathrm{M}(3 \lambda+1,2 \lambda, 4 \lambda+2)$
$\overrightarrow{\mathrm{AM}} \cdot \overrightarrow{\mathrm{b}}=0$
$\Rightarrow \quad 9 \lambda-15+4 \lambda-2+16 \lambda-12=0$
$\Rightarrow \quad 29 \lambda=29$
$\Rightarrow \quad \lambda=1$
$\mathrm{M}(4,2,6), \mathrm{I}=(2,3,7)$
Required Distance $=\sqrt{4+9+49}=\sqrt{62}$
Ans. 62

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (09 Apr Shift 2)
The square of the distance of the point of intersection of the line x-12=y-23=z+16 and the plane 2 x-y+z=6 from the point (-1,-1,2) is

Options:

The correct answer is:
61

Solution:
Equation of line

x-12=y-23=z+16

Let x-12=y-23=z+16=λ

⇒x = 2λ+1, y = 3λ+2, z=6λ-1

⇒ Let P(2λ+1,3λ+2,6λ-1) is point of intersection of line and plane

So it should satisfy the given plane.

⇒ 2(2λ+1)-3λ-2+6λ-1=6

7λ=7⇒λ=1

Point P(3,5,5)

Square of distance from point (-1,-1,2) and P(3,5,5)

=x2-x12 +y2-y12 +z2-z12

=42+62+32=61

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (31 Aug Shift 1)
The square root of the product of inductance and capacitance has the dimension of

Options:
A Length
B Mass
C Time
D Frequency

The correct answer is:
Time

Solution:
We know, f=12ÏLC
Or LC=12Ïf=time
Thus, LC has the dimension of time.

Topic: Physics, Units and Dimensions, JEE Main
The stability of +1 oxidation state among Al, Ga, In and Tl increases in the sequence:

Options:
A Ga<In<Al<Tl
B Al<Ga<In<Tl
C Tl<In<Ga<Al
D In<Tl<Ga<Al

The correct answer is:
Al<Ga<In<Tl

Solution:
The inert-pair effect is the tendency of the two electrons in the outermost atomic-s orbital to remain unshared in compounds of post-transition elements.

Inert-pair effect is seen in the elements of 13th,14th, 15th and 16th groups.

Due to the inert-pair effect elements down the group exhibit oxidation state which is 2 units lesser than the elements above them.

Stability of +1 oxidation state due to inert pair effect is in the order Al<Ga<In<Tl

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET, NEET 2015 (Phase 2)
The stability of +1 oxidation state among Al, Ga, In and Tl increases in the sequence:

Options:
A Ga<In<Al<Tl
B Al<Ga<In<Tl
C Tl<In<Ga<Al
D In<Tl<Ga<Al

The correct answer is:
Al<Ga<In<Tl

Solution:
The inert-pair effect is the tendency of the two electrons in the outermost atomic-s orbital to remain unshared in compounds of post-transition elements.

Inert-pair effect is seen in the elements of 13th,14th, 15th and 16th groups.

Due to the inert-pair effect elements down the group exhibit oxidation state which is 2 units lesser than the elements above them.

Stability of +1 oxidation state due to inert pair effect is in the order Al<Ga<In<Tl

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The stability of +1 oxidation state increases in the sequence :

Options:
A $\mathrm{Ga} < $ In $ < \mathrm{Al} < \mathrm{Tl}$
B $\mathrm{Al} < \mathrm{Ga} < \mathrm{In} < \mathrm{Tl}$
C $\mathrm{Tl} < $ In $ < \mathrm{Ga} < \mathrm{Al}$
D In $ < \mathrm{Tl} < \mathrm{Ga} < \mathrm{Al}$

The correct answer is:
$\mathrm{Al} < \mathrm{Ga} < \mathrm{In} < \mathrm{Tl}$

Solution:
$+\mathrm{I}$ stability down the group increase due to inert pair effect
$$
\mathrm{Al} < \mathrm{Ga} < \mathrm{In} < \mathrm{Tl}
$$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET, NEET 2009 (Mains)
The stability of +1 oxidation state increases in the sequence

Options:
A $\mathrm{Al} < \mathrm{Ga} < \mathrm{In} < \mathrm{Tl}$
B $\mathrm{Tl} < $ In $ < \mathrm{Ga} < \mathrm{Al}$
C $\mathrm{In} < \mathrm{Tl} < \mathrm{Ga} < \mathrm{Al}$
D $\mathrm{Ga} < \mathrm{In} < \mathrm{Al} < \mathrm{Tl}$

The correct answer is:
$\mathrm{Al} < \mathrm{Ga} < \mathrm{In} < \mathrm{Tl}$

Solution:
Key Idea Stability of lower oxidation states increases on moving down a group due to inert pair effect.
The given elements belong to third group. These elements mainly exhibit +3 and +1 oxidation state. As we know the stability of lower oxidation state, $i e,+1$ state, increases on moving down a group, the sequence of stability is
$$
\mathrm{Al} < \mathrm{Ga} < \mathrm{In} < \mathrm{Tl}
$$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET, NEET 2009 (Screening)
The stability of +1 oxidation state increases in the sequence

Options:
A $\mathrm{Al} < \mathrm{Ga} < \mathrm{In} < \mathrm{Tl}$
B $\mathrm{Tl} < $ In $ < \mathrm{Ga} < \mathrm{Al}$
C $\mathrm{In} < \mathrm{Tl} < \mathrm{Ga} < \mathrm{Al}$
D $\mathrm{Ga} < \mathrm{In} < \mathrm{Al} < \mathrm{Tl}$

The correct answer is:
$\mathrm{Al} < \mathrm{Ga} < \mathrm{In} < \mathrm{Tl}$

Solution:
Key Idea Stability of lower oxidation states increases on moving down a group due to inert pair effect.
The given elements belong to third group. These elements mainly exhibit +3 and +1 oxidation state. As we know the stability of lower oxidation state, $i e,+1$ state, increases on moving down a group, the sequence of stability is
$$
\mathrm{Al} < \mathrm{Ga} < \mathrm{In} < \mathrm{Tl}
$$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The stability of carbanions in the following
is in the order of

Options:
A $a>c>b>d$
B $a>b>c>d$
C b>c $>$ d $>$ a
D d $>$ b $>$ c $>$ a

The correct answer is:
$a>b>c>d$

Solution:
Higher is the electronegativity of atom higher will be stability of carbanion on it. Higher is the s-character, higher will be the electronegativity of atom.

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2008 (Mains)
The stability of dihalides of $\mathrm{Si}, \mathrm{Ge}, \mathrm{Sn}$ and $\mathrm{Pb}$ increases steadily in the sequence

Options:
A
$\mathrm{GeX}_2 \ll \mathrm{SiX}_2 \ll \mathrm{SnX_2}
\ll \mathrm{PbX}_2$

B
$\mathrm{SiX}_2 \ll \mathrm{GeX}_2 \ll \mathrm{PbX}_2 \ll \mathrm{SnX}{ }_2$

C
$\mathrm{SiX}_2 \ll \mathrm{GeX}_2 \ll \mathrm{SnX}_2 \ll \mathrm{PbX}_2$

D
$\mathrm{PbX}_2 \ll \mathrm{SnX}_2 \ll \mathrm{GeX}_2 \ll \mathrm{SiX}_2$

The correct answer is:
$\mathrm{SiX}_2 \ll \mathrm{GeX}_2 \ll \mathrm{SnX}_2 \ll \mathrm{PbX}_2$

Solution:
Due to inert pair effect, the stability of $+2$ oxidation state increases as we move down this group.
$\therefore \mathrm{SiX}_2 \ll \mathrm{GeX}_2 \ll \mathrm{SnX}{ }_2 \ll \mathrm{PbX}_2$
Hence, (C) is correct.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The stability of dihalides of $\mathrm{Si}, \mathrm{Ge}, \mathrm{Sn}$ and $\mathrm{Pb}$ increases steadily in the sequence

Options:
A
$\mathrm{GeX}_2 \ll \mathrm{SiX}_2 \ll \mathrm{SnX_2}
\ll \mathrm{PbX}_2$

B
$\mathrm{SiX}_2 \ll \mathrm{GeX}_2 \ll \mathrm{PbX}_2 \ll \mathrm{SnX}{ }_2$

C
$\mathrm{SiX}_2 \ll \mathrm{GeX}_2 \ll \mathrm{SnX}_2 \ll \mathrm{PbX}_2$

D
$\mathrm{PbX}_2 \ll \mathrm{SnX}_2 \ll \mathrm{GeX}_2 \ll \mathrm{SiX}_2$

The correct answer is:
$\mathrm{SiX}_2 \ll \mathrm{GeX}_2 \ll \mathrm{SnX}_2 \ll \mathrm{PbX}_2$

Solution:
Due to inert pair effect, the stability of $+2$ oxidation state increases as we move down this group.
$\therefore \mathrm{SiX}_2 \ll \mathrm{GeX}_2 \ll \mathrm{SnX}{ }_2 \ll \mathrm{PbX}_2$
Hence, (C) is correct.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2007
The stability of Me $\mathcal{C}=\mathrm{CH}_{2}$ is more than that of $\mathrm{MeCH}_{2} \mathrm{CH}=\mathrm{CH}_{2}$ due to

Options:
A inductive effect of the Me groups
B resonance effect of the Me groups
C hyperconjugative effect of the Me groups
D resonance as well as inductive effect of the Me groups

The correct answer is:
hyperconjugative effect of the Me groups

Solution:
2-hyperconltjugate forms. Therefore, $\mathrm{Me} \underline{\mathcal{C}}=\mathrm{CH}_{2}$ is more stable than
$\mathrm{MeCH}_{2} \mathrm{CH}=\mathrm{CH}_{2}$

Topic: Chemistry, General Organic Chemistry, JEE Main
The standard deviation of 5 scores $1,2,3,4,5$ is $\sqrt{a}$. The value of ' $a$ ' is

Options:
A
2
B
3
C
5
D
1

The correct answer is:
2

Solution:
$\begin{aligned} & \text { Mean }(\bar{x})=\frac{1+2+3+4+5}{5}=3 \\ & \text { S.D }=\sigma=\sqrt{\frac{1}{5}(1+4+9+16+25)-9}=\sqrt{11-9} \\ & =\sqrt{2}\end{aligned}$

Topic: Mathematics, Statistics, JEE Main
The standard deviation of a distribution is 30. If each observation is increased by 5, then the new standard deviation will be

Options:
A 32
B 28
C 27
D None of these

The correct answer is:
None of these

Solution:
S.D. of a series is unaltered if each observation is increased (or reduced) by the same scalar quantity, S.D. is independent of change of origin.
Hence, S.D. will be the same.
∴ S.D. =30

Topic: Mathematics, Statistics, JEE Main
The standard deviation of the following distribution
\begin{array}{|c|c|c|c|}
\hline C.I. & 0-6 & 6-12 & 12-18 \\
\hline \mathrm{f}_{\mathrm{i}} & 2 & 4 & 6 \\
\hline
\end{array}
is

Options:
A $5 \sqrt{2}$
B $\sqrt {5}$
C $2\sqrt {5}$
D $20$

The correct answer is:
$2\sqrt {5}$

Solution:
\begin{array}{|c|c|c|c|c|c|}
\hline C.I. & \mathbf{f}_{\mathbf{i}} & \boldsymbol{x}_{\mathbf{i}} & \boldsymbol{x}_{\mathbf{i}}^{\mathbf{2}} & \mathbf{f}_{\mathbf{i}} \boldsymbol{x}_{\mathbf{i}} & \mathbf{f}_{\mathbf{i}} \boldsymbol{x}_{\mathbf{i}}^{\mathbf{2}} \\
\hline 0-6 & 2 & 3 & 9 & 6 & 18 \\
\hline 6-12 & 4 & 9 & 81 & 36 & 324 \\
\hline 12-18 & 6 & 15 & 225 & 90 & 1350 \\
\hline Total & \mathbf{1 2} & & & \mathbf{1 3 2} & \mathbf{1 6 9 2} \\
\hline
\end{array}
Here $\sum \mathrm{f}_{\mathrm{i}}=12, \sum \mathrm{f}_{\mathrm{i}} x_{\mathrm{i}}=132, \sum \mathrm{f}_{\mathrm{i}} x_{\mathrm{i}}^2=1692$
$\begin{aligned}
\therefore \quad \mathrm{V}(\mathrm{X}) & =\frac{1692}{12}-\left(\frac{132}{12}\right)^2 \\
& =141-121 \\
& =20
\end{aligned}$
$\therefore \quad$ Standard deviation $=\sqrt{20}=2 \sqrt{5}$

Topic: Mathematics, Statistics, JEE Main
The standard deviation of the following distribution is
$\begin{array}{|l|c|c|c|c|}
\hline \begin{array}{l}
\text { Class interval } \\
\end{array} & 0-10 & 10-20 & 20-30 & 30-40 \\
\hline \begin{array}{l}
\text { Frequency } \\
\end{array} & 1 & 3 & 4 & 2 \\
\hline
\end{array}$

Options:
A 9
B 8
C 7
D 6

The correct answer is:
9

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Statistics, JEE Main
The standard deviation of the wages of \( 85 \) employees is \( ₹ 15.40 \). After one year, each of them is given an increment of \( ₹ 25 \) The standard deviation of new wages (in ₹) is

Options:
A \( 15.40 \)
B \( 40.40 \)
C \( 20.40 \)
D \( 10.40 \)

The correct answer is:
\( 15.40 \)

Solution:
The arithmetic mean of the square of deviation of the variable values from its actual arithmetic mean is known as variance and its square root is known as standard deviation.

Standard deviation is given by formula

σ2=∑xi-x¯2n

If data is changed from xi to xi+k then new standard deviation is given by σ12=∑xi+k-x¯+k2n=σ2

since, SD is independent of change of origin. Therefore, the standard deviation remains same i.e. 15.40.

Topic: Mathematics, Statistics, JEE Main
The standard deviations of xii=1,2,….10 and yii=1,…10 are respectively a and b. x¯, y¯ are the means of these two sets of observations respectively. If zi=xi-x¯yi-y¯ and ∑i=110zi=c, then the standard deviation of the observations xi-yi, i=1,2,…10 is

Options:
A a2+b2+c5
B a2+b2-c5
C a2+b2-c25
D a2+b2+c25

The correct answer is:
a2+b2-c5

Solution:
It is given that, zi=xi-x¯yi-y¯

And ∑i=110zi=c

Rewrite the above equation.

∑i=110xi-x¯yi-y¯=c

⇒ ∑i=110xiyi-xiy¯-x¯yi+x¯·y¯=c

⇒ ∑i=110xiyi-∑i=110xiy¯-∑i=110x¯yi+∑i=110x¯·y¯=c

⇒ ∑i=110xiyi-10x¯·y¯-10x¯·y¯+10x¯·y¯=c

⇒ ∑i=110xiyi-10x¯·y¯=c

Now,

σx1-y1=110∑xi-yi2-x¯-y¯2

=110∑xi2+110∑yi2-15∑xiyi-x¯2-y¯2+2xy¯

=110∑xi2-x¯2+110∑yi2-y¯2-15∑xiyi-10xy¯

=a2+b2-c5

Topic: Mathematics, Statistics, JEE Main
The standard electrode potential Eo and its temperature coefficient dEdT for a cell are 2 V and -5×10-4  V K-1 at 300 K, respectively. The reaction is Zn s+Cu2+ aq→Zn2+ aq+Cu s. The standard reaction enthalpy ΔrH- at 300 K in mol-1 is

[Use R=8 J K-1 mol-1 and F=96,500 C mol-1]

Options:
A -412.8
B 206.4
C -384.0
D 192.0

The correct answer is:
-412.8

Solution:
ΔG=-nFEcell=-2×96500×2=-386 kJ
ΔS=nFdEdT=2×96500×-5×10-4 J degree celcious-1
=-96.5 J
At 298 K,
TΔS=298×-96.5 J=-28.757 kJ
At constant temperature T=248 K and pressure,
ΔG=ΔH-TΔS
ΔH=ΔG+TΔS
=-386-28.757=-414.75≃-412.8

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2019 (12 Jan Shift 1)
The standard electrode potential (M3+/M2+) for V, Cr, Mn & Co are -0.26 V, - 0.41 V, +1.57 V and +1.97 V, respectively. The metal ions which can liberate H2 from a dilute acid are

Options:
A V2+ and Mn2+
B Cr2+ and Co2+
C V2+ and Cr2+
D Mn2+ and Co2+

The correct answer is:
V2+ and Cr2+

Solution:
The E⊝ values for the redox couple (M3+/M2+) shows that Mn2+ and Co2+ ions are the strongest oxidising agents in aqueous solution.

The ions Ti2+, V2+ and Cr2+ are strong reducing agents and will liberate hydrogen from a dilute acid.

So, the Correct answer is V2+ and Cr2+.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2023 (29 Jan Shift 1)