datavorous/entrance-exam-dataset huggingface dataset:
Spin only magnetic moment 2.84 BM is given by:

(Atomic Number of Ni=28,Ti=22,Cr=24,Co=27 )

Options:
A Ni 2+
B Ti 3+
C Cr 2+
D Co 2+

The correct answer is:
Ni 2+

Solution:
Î¼= n( n+2 ) =2.84ân=2
electronic configuration of Ni +2 =1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 8 e â having two unpaired electrons
electron configuration of T i 3+ =1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 1 only one unpaired electron
electron configuration of C r 2+ =1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 4 having four unpaired electrons
electron configuration of C o +2 =1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 7 having three unpaired electrons.

Topic: Chemistry, Structure of Atom, JEE Main
Spin only magnetic moment in BM of FeCO4C2O4+ is :

Options:
A 5.92
B 1
C 0
D 1.73

The correct answer is:
1.73

Solution:
In Fe(CO)4C2O4+

Oxidation no. of Fe=+3

26Fe+3=Ar183 d54 s0

CO is a Strong field ligand, so paring occurs against the Hund's rule.

Fe+3=t2g2,2,1 eg0,0

Number of unpaired electron =1

so magnetic moment μ=nn+2 B.M. =11+2=3=1.73 BM

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (31 Aug Shift 2)
Spin only magnetic moment of Mn4+ ion is nearly

Options:
A 3 BM
B 6 BM
C 4 BM
D 5 BM

The correct answer is:
4 BM

Solution:
Electronic configuration of Mn is Z=25=Ar3d54s2

∴ Electronic configuration of Mn4+ is =Ar3d3

Thus, Mn4+ has 3-unpaired electron and its spin only magnetic moment μ is, μ=nn+2 BM

where, n= number of unpaired electrons.

μ=3(3+2). BM⇒μ=15=3.87BM≈4BM

Topic: Chemistry, d and f Block Elements, JEE Main
Spin only magnetic moment of MnBr64- is _____ B.M. (round off to the closest integer)

Options:

The correct answer is:
6

Solution:
MnBr64-

Mn2+: [Ar]3d5

Number of unpaired electrons =5

Magnetic moment of the complex,

μ=n(n+2) B.M=35=5.91 B.M.

5.91 is rounded off to the nearest integer 6.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Spin only magnetic moment of the compound HgCoSCN4 is x B.M. The value of ' x ' is

Options:

The correct answer is:
15

Solution:
HgCoSCN4 :
n=3
μs=nn+2=3×5=15

Topic: Chemistry, Coordination Compounds, JEE Main
Spirometer is useful in assessment of all the following pulmonary functions except

Options:
A Vital Capacity (VC)
B Functional Residual Capacity (FRC)
C Inspiratory Capacity (IC)
D Expiratory Capacity (EC)

The correct answer is:
Functional Residual Capacity (FRC)

Solution:
Spirometer is an instrument used to measure lung volumes. It can measure tidal volume (TV), inspiratory reserve volume (IRV) and expiratory reserve volume (ERV). It cannot measure residual volume (RV) which is the volume of air remaining in the lungs even after a forcible expiration. Inspiratory capacity is the sum of IRV and TV. Expiratory capacity is the sum of TV and ERV. Vital capacity is the sum of TV, ERV and IRV. These three capacities can be measured using spirometer as they involve TV, IRV and ERV. Functional residual capacity (FRC) is the sum of ERV and RV. Since Residual capacity cannot be measured by spirometer, FRC cannot be calculated using spirometer.

Topic: Biology, Breathing and Exchange of Gases, NEET
Splitting of spectral lines under the influence of magnetic field is called

Options:
A Zeeman effect
B Stark effect
C Photoelectric effect
D None of these

The correct answer is:
Zeeman effect

Solution:
The spliting of spectral line by the magnetic field is called Zeeman effect.

Topic: Chemistry, Structure of Atom, JEE Main
Spraying of juvenile conifers with which hormone hastens the maturity period and thus leading to early seed production?

Options:
A The hormone which also causes bolting in some plants
B The hormone which is activated during deficiency of water
C The hormone which is responsible for apical dominance
D The hormone which prevents apical dominance is plants

The correct answer is:
The hormone which also causes bolting in some plants

Solution:
The hormone which also causes bolting in some plants is Gibberellin. Gibberellins are growth promoters. They were first discovered in a fungus Gibberella fujikuroi. There are many gibberellins and they are denoted as GA1, GA2, GA3 and so on. However, GA3 was one of the first gibberellins to be discovered and remains the most intensively studied form. Spraying juvenile conifers with GAs hastens the maturity period, thus leading to early seed production. Gibberellins also promotes bolting (internode elongation just prior to flowering) in beet, cabbages and many plants with rosette habit.The hormone which is activated during deficiency of water is absicisic acid, the hormone which is responsible for apical dominance is auxin and the hormone which prevents apical dominance is plants is cytokinin.

Topic: Biology, Plant - Growth and Development, NEET
\( \sqrt{-1-\sqrt{-1-\sqrt{-1 \ldots \infty}}} \) is equal to, where \( \omega \) is the imaginary cube of root of unity and \( i=\sqrt{-1} . \)

Options:
A w and \( \omega^{2} \)
B \( -\omega \) or \( -\omega^{2} \)
C \( 1+i \) or \( 1-i \)
D \( -1+i \) or \( -1-i \)

The correct answer is:
w and \( \omega^{2} \)

Solution:
Let z=-1--1--1.............∞

or z=-1-z

z2=-1-z

z2+z+1=0

z=-1±i32

or z=ω and ω2.

Topic: Mathematics, Complex Number, JEE Main
Square matrix $\left[a_{i j}\right]_{n \times n}$ will be an upper triangular matrix, if

Options:
A $a_{i j} \neq 0$, for $i \gt j$
B $a_{i j}=0$, for $i \gt j$
C $a_{i j}=0$, for $i \lt j$
D None of these

The correct answer is:
$a_{i j}=0$, for $i \gt j$

Solution:
$\left[a_{i j}\right]_{n \times n}$ square matrix is a upper triangular matrix for $a_{i j}=0, i \gtj$

Topic: Mathematics, Matrices, JEE Main
Square of either of the two imaginary cube roots of unity will be

Options:
A Real root of unity
B Other imaginary cube root of unity
C Sum of two imaginary roots of unity
D None of these

The correct answer is:
Other imaginary cube root of unity

Solution:
Since imaginary cube root of unity are square of each other.

Topic: Mathematics, Complex Number, JEE Main
Square-planar geometry is shown by

Options:
A
$\left[\mathrm{PtCl}_2\left(\mathrm{NH}_3\right)_2\right]$

B
$\left[\mathrm{NiCl}_4\right]^{2-}$

C
$\mathrm{MnO}_4^{-}$

D
$\mathrm{CrO}_4^{2-}$

The correct answer is:
$\left[\mathrm{PtCl}_2\left(\mathrm{NH}_3\right)_2\right]$

Solution:
$$
\text { }\left[\mathrm{PtCl}_2\left(\mathrm{NH}_3\right)_2\right] \text { has square planar geometry. }
$$

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2012 (19 May Online)
'Stability of a crystal is reflected in the magnitude of its melting points'. Comment.
Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules?

Options:

None

Solution:
Higher the melting point, greater are the forces holding the constituent particles together and thus greater is the stability of a crystal. Melting points of given substances are following.
Solid water \(=273 \mathrm{~K}\), Ethyl alcohol \(=155.7 \mathrm{~K}\), Diethylether \(=156.8 \mathrm{~K}\), Methane \(=90.5 \mathrm{~K}\).
The intermolecular forces present in case of water and ethyl alcohol are mainly due to the hydrogen bonding which is responsible for their high melting points. Hydrogen
bonding is stronger in case of water than ethyl alcohol and hence water has higher melting point then ethyl alcohol.
Dipole-dipole interactions are present in case of diethylether. The only forces present in case of methane is the weak van der Waal's forces (or London dispersion forces).

Topic: Chemistry, Solid State
Stability of α - Helix structure of proteins depends upon

Options:
A H-bonding interaction
B dipolar interaction
C van der Waals forces
D π-stacking interaction

The correct answer is:
H-bonding interaction

Solution:
The secondary structure of protein refers to the shape in which a long polypeptide chain can exist. They are found to exist in two different types of structures viz. α-helix and β-pleated sheet structure. These structures arise due to the regular folding of the backbone of the polypeptide chain due to hydrogen bonding between and -NH- groups of the peptide bond.

α-helix Is one of the most common ways in which a polypeptide chain forms all possible hydrogen bonds by twisting into a right handed screw (helix).

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2022 (28 Jun Shift 1)
Stability of the species Li2,Li2- and Li2+ increases in the order of

Options:
A Li2-<Li2<Li2+
B Li2-<Li2+<Li2
C Li2<Li2-<Li2+
D Li2<Li2+<Li2-

The correct answer is:
Li2-<Li2+<Li2

Solution:
Li2:σ1s2 σ1s*2 σ1s2 Bond order =1
Li2+:σ1s2 σ1s*2 σ2s1 Bond order =0.5
Li2-:σ1s2 σ1s*2 σ2s2 σ2s*1 Bond order =0.5
Li2- is less stable than Li2+ because the incoming electron goes to anti bonding molecular orbital. So, the stability order is Li2>Li2+>Li2- .

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Stability of the species Li 2 , Li 2 - and Li 2 + increases in the order of

Options:
A Li 2 < Li 2 - < Li 2 +
B Li 2 - < Li 2 < Li 2 +
C Li 2 < Li 2 + < Li 2 -
D Li 2 - < Li 2 + < Li 2

The correct answer is:
Li 2 - < Li 2 + < Li 2

Solution:
Li 2 - < Li 2 + < Li 2
Bond order ∝ stability of species
Li2=σ1s2<σ*1s2<σ2s2                      ;  B.O=4−22=1
Li2−=σ1s2<σ*1s2<σ2s2<σ*2s1                      ;  B.O=4−32=0.5
Li2+=σ1s2<σ*1s2<σ2s1                      ;  B.O=3−22=0.5
Li2+ and Li2- both have same bond order, but Li2- has more number of electron in anti-bonding molecular orbitals (experience more inter electronic repulsion) and hence least stable among three species.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2013 (07 Apr)
Stability of the species Li 2 , Li 2 - and Li 2 + increases in the order of

Options:
A Li 2 < Li 2 - < Li 2 +
B Li 2 - < Li 2 < Li 2 +
C Li 2 < Li 2 + < Li 2 -
D Li 2 - < Li 2 + < Li 2

The correct answer is:
Li 2 - < Li 2 + < Li 2

Solution:
Li 2 - < Li 2 + < Li 2
Bond order ∝ stability of species
Li2=σ1s2<σ*1s2<σ2s2                      ;  B.O=4−22=1
Li2−=σ1s2<σ*1s2<σ2s2<σ*2s1                      ;  B.O=4−32=0.5
Li2+=σ1s2<σ*1s2<σ2s1                      ;  B.O=3−22=0.5
Li2+ and Li2- both have same bond order, but Li2- has more number of electron in anti-bonding molecular orbitals (experience more inter electronic repulsion) and hence least stable among three species.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Stability of the species Li 2 , Li 2 - and Li 2 + increases in the order of

Options:
A Li 2 < Li 2 - < Li 2 +
B Li 2 - < Li 2 < Li 2 +
C Li 2 < Li 2 + < Li 2 -
D Li 2 - < Li 2 + < Li 2

The correct answer is:
Li 2 - < Li 2 + < Li 2

Solution:
Li 2 - < Li 2 + < Li 2
Bond order â stability of species
Li2=Ï1s2<Ï*1s2<Ï2s2ââââââââââââââââââââââ;â B.O=4â22=1
Li2â=Ï1s2<Ï*1s2<Ï2s2<Ï*2s1ââââââââââââââââââââââ;â B.O=4â32=0.5
Li2+=Ï1s2<Ï*1s2<Ï2s1ââââââââââââââââââââââ; âB.O=3â22=0.5
Li2+ and Li2- both have same bond order,but Li2- has more number of electron in antibonding molecular orbitals (experience more inter electronic repulsions) and hence least stable among three species.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Staggered and eclipsed conformers of ethane are:

Options:
A Polymers
B Rotamers
C Enantiomers
D Mirror images

The correct answer is:
Rotamers

Solution:
Staggered and eclipsed conformers of ethane also known as rotamers.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Stalk with which ovules remain attached to placenta is called:

Options:
A Funicle
B Integuments
C Hilum
D Chalaza

The correct answer is:
Funicle

Solution:
The ovule is a small structure attached to the placenta by means of a stalk called funicle. The body of the ovule fuses with funicle in the region called hilum. Thus, hilum represents the junction between ovule and funicle. Each ovule has one or two protective envelopes called integuments. Integuments encircle the nucellus except at the tip where a small opening called the micropyle is organised. Opposite the micropylar end, is the chalaza, representing the basal part of the ovule.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Stalk with which ovules remain attached to the placenta is called

Options:
A funicle.
B raphe.
C hilum.
D chalaza.

The correct answer is:
funicle.

Solution:
Funicle is the stalk of an ovule by which it is attached to the placenta. Raphe is an elongated mass of tissue, containing vascular bundle, and lying on the side of an anatropous ovule, between the chalaza and the attachment to the placenta. Helium is the scar on the testa where it was attached to the funicle. Chalaza is the opposite side of the side bearing micropyle of an ovule.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Standard deviation of $n$ observations $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ is ${\sigma} .$ Then, the standard deviation of the observations $\lambda a_{1}, \lambda a_{2}, \ldots, \lambda a_{n}$ is

Options:
A $\lambda {\sigma}$
B $-\lambda \sigma$
C $|\lambda| \sigma$
D $\lambda^{n} \boldsymbol{\sigma}$

The correct answer is:
$|\lambda| \sigma$

Solution:
Let $x_{i}=a_{1}, a_{2}, a_{3}, \ldots, a_{n}$
and $y_{i}=\lambda a_{1}, \lambda a_{2}, \lambda a_{3}, \ldots, \lambda a_{n-}$
$$
y_{i}=\lambda x_{i}
$$
$\because$
$\begin{aligned} \bar{y} &=\frac{\Sigma y_{i}}{N} \\ &=\frac{\Sigma \lambda x_{i}}{n} \end{aligned}$
$$
\begin{aligned}
&=\lambda \frac{\Sigma x_{i}}{n} \\
&=\lambda \bar{x} \\
\sigma_{y} &=\sqrt{\frac{\Sigma\left(y_{i}-\bar{y}\right)^{2}}{N}} \\
&=\sqrt{\frac{\Sigma(\lambda x_{i}-\lambda \bar{x})^{2}}{n}} \\
&=\sqrt{\frac{\lambda^{2} \Sigma\left(x_{i}-\bar{x}\right)^{2}}{n}} \\
&=\sqrt{\lambda^{2} \sigma_{x}^{2}}=|\lambda| \sigma_{x}=|\lambda| \sigma
\end{aligned}
$$

Topic: Mathematics, Statistics, JEE Main
Standard electrode potentials of $Z n$ and $F e$ are known to be (i) $-0.76 \mathrm{~V}$ and (ii) $-0.44 \mathrm{~V}$ respectively. How does it explain that galvanization prevents rusting of iron while zinc slowly dissolves away

Options:
A Since (i) is less than (ii), zinc becomes the cathode and iron the anode
B Since (i) is less than (ii), zinc becomes the anode and iron the cathode
C Since (i) is more than (ii), zinc becomes the anode and iron the cathode
D Since (i) is more than (ii), zinc becomes the cathode and iron the anode

The correct answer is:
Since (i) is less than (ii), zinc becomes the anode and iron the cathode

Solution:
Correct option is (2). Since (i) is less than (ii), zinc between the anode and iron the cathode

Topic: Chemistry, Electrochemistry, JEE Main
Standard enthalpy and standard entropy changes for the oxidation of ammonia at \( 298 \mathrm{~K} \) are \( -382.64 \)
\( k J \mathrm{~mol}^{-1} \) and \( -145.6 J K^{-1} \mathrm{~mol}^{-1} \), respectively. Standard Gibb's energy change for the same reaction at \( 298 \mathrm{~K} \) is

Options:
A \( -339.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \)
B \( -22.1 k J m o l^{-1} \)
C \( -523.2 k J m o l^{-1} \)
D \( -439.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \)

The correct answer is:
\( -339.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \)

Solution:
Using equation.

ΔG∘=ΔH∘-TΔS∘         =−382.64+(298×145.6×10−3)         =−339.3 kJ mol−1

Topic: Chemistry, Thermodynamics (C), JEE Main
Standard enthalpy of vapourisation for CCl4 is 30.5 kJ mol-1. Heat required for vapourisation of 284 g of CCl4 at constant temperature is __________ kJ. (Given molar mass in gmol-1;C=12,Cl=35.5)

Round off your answer to the nearest integer.

Options:

The correct answer is:
56

Solution:
The enthalpy of vapourisation ∆H of CCl4=30.5 kJ/mol

qp=∆H=30.5 kJ/mol

No. of moles of CCl4=given massmolar mass

=284154=1.844 mol

Hence, the heat required for the vapourisation of 1.844 mol of CCl4

=1.844×30.5Heat required=56.242 kJ/mol≈56 kJ/mol

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2024 (29 Jan Shift 2)
Standard entropies of x2,y2 and xy3 are 70, 50 and 60 J K-1mol-1 respectively. For the reaction 12x2+32y2⇌xy3.ΔH=-30kJ to be at equilibrium, the temperature should be

Options:
A 450 K
B 600 K
C 1200 K
D 300 K

The correct answer is:
600 K

Solution:
ÎS=Sxy3-12Sx2-32Sy2
ÎS=60-12Ã70-32Ã50
=-50 JK-1 mol-1
At equilibrium ÎG=0
ÎG=ÎH-TÎS
0=-30-(-50Ã10-3ÃT)
T=600 K

Topic: Chemistry, Thermodynamics (C), JEE Main
Standard entropies of X 2 ,  Y 2 and XY 3 are 60, 30 and 50  JK −1   mol −1 respectively. For the reaction 1 2 X 2  +  3 2 Y 2 ⇌ XY 3 , ΔH=−30 kJ to be at equilibrium, the temperature should be:

Options:
A 1200âK
B 1000 K
C 750âK
D 500 K

The correct answer is:
1200 K

Solution:
ΔS for the reaction 1 2 X 2 + 3 2 Y 2 ⇌ XY 3
ΔS = 50− 1 2  × 60 +  3 2  × 30 =−25 J
For equilibrium ΔG = 0 = ΔH−TΔS
T =  ΔH ΔS  =  −30000 −25  = 1200 K

Topic: Chemistry, Thermodynamics (C), JEE Main
Standard entropy of $X_2, Y_2$ and $X Y_3$ are 60,40 and $50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, respectively. For the reaction, $\frac{1}{2} X_2+\frac{3}{2} Y_2 \rightarrow X Y_3, \Delta \mathrm{H}=-30 \mathrm{~kJ}$, to be at equilibrium, the temperature will be

Options:
A
$1250 \mathrm{~K}$

B
$500 \mathrm{~K}$

C
$750 \mathrm{~K}$

D
$1000 \mathrm{~K}$

The correct answer is:
$750 \mathrm{~K}$

Solution:
$$
\begin{aligned}
& \frac{1}{2} X_2+\frac{3}{2} Y_2 \longrightarrow X Y_3 \\
& \Delta S_{\text {reaction }}=50-\left(\frac{3}{2} \times 40+\frac{1}{2} \times 60\right)=-40 \mathrm{Jmol}^{-1} \\
& \Delta G=\Delta H-T \Delta S \\
& \text { at equilibrium } \Delta G=0
\end{aligned}
$$
$$
\begin{aligned}
& \Delta \mathrm{H}=\mathrm{T} \Delta \mathrm{S} \\
& 30 \times 10^3=\mathrm{T} \times 40 \\
& \Rightarrow \mathrm{T}=750 \mathrm{~K}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2008
Standard free energies of formation (in $\mathrm{kJ} / \mathrm{mol}$ ) at $298 \mathrm{~K}$ are $-237.2,-394.4$ and -8.2 for $\mathrm{H}_2 \mathrm{O}(\mathrm{l}), \mathrm{CO}_2(\mathrm{~g})$ and pentane $(\mathrm{g})$ respectively. The value of $\mathrm{E}_{\text {cell }}^{\circ}$ for the pentaneoxygen fuel cell is

Options:
A $0.0968 \mathrm{~V}$
B $1.968 \mathrm{~V}$
C $2.0968 \mathrm{~V}$
D $1.0968 \mathrm{~V}$

The correct answer is:
$1.0968 \mathrm{~V}$

Solution:
After calculation $\Delta \mathrm{G}^{\circ}$, use the formula, $\Delta \mathrm{G}^{\circ}=-n F E^{\circ}$
Here, $n=32$ is taken because balanced equation is
$$
\mathrm{C}_5 \mathrm{H}_{12}+8 \mathrm{O}_2 \rightarrow 5 \mathrm{CO}_2+6 \mathrm{H}_2 \mathrm{O}
$$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2008 (Mains)
Standard reduction electrode potentials of three metals A,B\&C are respectively $+0.5 \mathrm{~V},-3.0 \mathrm{~V} \&-1.2 \mathrm{~V}$. The reducing, powers of these metals are

Options:
A
$\mathrm{A}>\mathrm{B}>\mathrm{C}$

B
$\mathrm{C}>\mathrm{B}>\mathrm{A}$

C
$\mathrm{A}>\mathrm{C}>\mathrm{B}$

D
$\mathrm{B}>\mathrm{C}>\mathrm{A}$

The correct answer is:
$\mathrm{B}>\mathrm{C}>\mathrm{A}$

Solution:
No Solution Available

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2003
Standard reduction potentials of the half reactions are given below
$$
\begin{array}{ll}
\mathrm{F}_2(g)+2 e^{-} \longrightarrow 2 \mathrm{~F}^{-}(a q) ; & E^{\circ}=+2.85 \mathrm{~V} \\
\mathrm{Cl}_2(g)+2 e^{-} \longrightarrow 2 \mathrm{Cl}^{-}(a q) ; & E^{\circ}=+1.36 \mathrm{~V} \\
\mathrm{Br}_2(l)+2 e^{-} \longrightarrow 2 \mathrm{Br}^{-}(a q) ; & E^{\circ}=+1.06 \mathrm{~V} \\
\mathrm{I}_2(s)+2 e^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) ; & E^{\circ}=+0.53 \mathrm{~V}
\end{array}
$$
The strongest oxidising and reducing agents respectively are

Options:
A $\mathrm{F}_2$ and $\mathrm{I}^{-}$
B $\mathrm{Br}_2$ and $\mathrm{Cl}^{-}$
C $\mathrm{Cl}_2$ and $\mathrm{Br}^{-}$
D $\mathrm{Cl}_2$ and $\mathrm{I}_2$

The correct answer is:
$\mathrm{F}_2$ and $\mathrm{I}^{-}$

Solution:
Higher the value of standard reduction potential, stronger will be the oxidising agent. Therefore, $\mathrm{F}_2$ will act as strongest oxidising agent.
Similarly lower the value of standard reduction potential, stronger will be the reducing agent. Therefore, $\mathrm{I}^{-}$will act as strongest reducing agent.

Topic: Chemistry, Electrochemistry, NEET, NEET 2012 (Mains)
Standing waves are produced by the superposition of two waves y1=0.05sin⁡3πt-2x and y2=0.05sin⁡3πt+2x, where x and y are in metre and t is in second. The amplitude of the particle at x=0.5 m is [Given cos57.3°=0.54]

Options:
A 2.7 cm
B 5.4 cm
C 8.1 cm
D 10.8 cm

The correct answer is:
5.4 cm

Solution:
Here, y1=0.05sin⁡(3πt-2x)
y2=0.05sin⁡(3πt+2x)
According to superposition principle, the resultant displacement is
y=y1+y2
=0.05sin⁡(3πt-2x)+sin⁡(3πt+2x)
y=0.05×2sin⁡3πtcos⁡2x
y=0.01cos⁡2xsin⁡2πt=Rsin⁡3πt
Where R=0.1 cos⁡2 x= amplitude of the resultant standing wave.
At x=0.5 m
R=0.1cos⁡2x=0.1cos⁡2×0.5
=0.1cos⁡1radian=0.1cos⁡180°π
=0.1cos⁡57.3°
R=0.1×0.54m=0.054m=5.4cm.

Topic: Physics, Waves and Sound, JEE Main
Starch can be used as an indicator for the detection of traces of

Options:
A glucose in aqueous solution
B proteins in blood
C iodine in aqueous solution
D urea in blood

The correct answer is:
iodine in aqueous solution

Solution:
$\mathrm{I}_{2}$ gives blue colour with starch.

Topic: Chemistry, Biomolecules, JEE Main
Starch is a polymer of

Options:
A Sucrose
B Maltose
C Glucose
D Hexose

The correct answer is:
Glucose

Solution:
Starch is a polymer of Glucose.

Topic: Chemistry, Biomolecules, JEE Main
Stars are not visible in the day time because

Options:
A Stars hide behind the sun
B Stars do not reflect sun rays during day
C Stars vanish during the day
D Atmosphere scatters sunlight into a blanket of extreme brightness through which faint stars cannot be visible

The correct answer is:
Atmosphere scatters sunlight into a blanket of extreme brightness through which faint stars cannot be visible

Solution:
Correct option is 4. Atmosphere scatters sunlight into a blanket of extreme
brightness through which faint stars cannot be visible
During the night time the sunlight doesn't complete reach the part of earth we are residing in. Hence, the sky is darker compared to day time. So, the visibility of the stars increases and we can see them properly.
During the day time, the sunlight reaches the part of earth we are residing in directly. The particles present in the atmosphere causes the phenomenon of atmospheric scattering which causes the sunlight to scatter brightly across the sky. This brightness are so high that the light emitted by the stars are completely covered by the sunlight. Hence, we cannot see the stars in daytime.

Topic: Physics, Ray Optics, JEE Main
Starting from rest, acceleration of a particle is $a=2(t-1)$. The velocity of the particle at $t=5 \mathrm{~s}$ is

Options:
A 15 m/sec
B 25 m/sec
C 5 m/sec
D None of these

The correct answer is:
15 m/sec

Solution:
$\begin{aligned} & \because a=\frac{d v}{d t}=2(t-1) \Rightarrow d v=2(t-1) d t \\ & \Rightarrow v=\int_0^5 2(t-1) d t=2\left[\frac{t^2}{2}-t\right]_0^5=2\left[\frac{25}{2}-5\right]=15 \mathrm{~m} / \mathrm{s}\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main
Starting from the centre of the earth having radius $R$, the variation of $g$ (acceleration due to gravity) is shown by

Options:
A

B

C

D

The correct answer is:

Solution:
Acceleration due to gravity at a depth $d$ below the surface of the earth is given by
$\begin{aligned}
g_{\text {depth }} & =g_{\text {surface }}\left(1-\frac{d}{R}\right) \\
& =g_{\text {surface }}\left[\frac{R-d}{R}\right]=d_{\text {surface }}\left(\frac{r}{R}\right)
\end{aligned}$
Also, for a point at height $h$ above surface,
$g_{\text {height }}=g_{\text {surface }}\left[\frac{R^2}{(R+h)^2}\right]$
Therefore, we can say that value of $g$ increases from centre to maximum at the surface and then decreases as depicted in graph (b).

Topic: Physics, Gravitation, JEE Main