datavorous/entrance-exam-dataset huggingface dataset:
When light of frequency $v_{1}$ is incident on a metal with work function $W$ (where $h v_{1}>w$ ), then photocurrent falls to zero at a stopping potential of $V_{1}$. If the frequency of light is increased to $v_{2},$ the stopping potential changes to $V_{2}$. Therefore, the charge of an electron is given by

Options:
A $\frac{W\left(v_{2}+v_{1}\right)}{v_{1} V_{2}+v_{2} V_{1}}$
B $\frac{W\left(v_{2}+v_{1}\right)}{v_{1} V_{1}+v_{2} V_{2}}$
C $\frac{W\left(v_{2}-v_{1}\right)}{v_{1} V_{2}-v_{2} V_{1}}$
D $\frac{W\left(v_{2}-v_{1}\right)}{v_{2} V_{2}-v_{1} V_{1}}$

The correct answer is:
$\frac{W\left(v_{2}-v_{1}\right)}{v_{1} V_{2}-v_{2} V_{1}}$

Solution:
From Einstein's photoelectric equation,
$$
\begin{array}{l}
h v=W+e V \\
W=\text { Work function }
\end{array}
$$
$v=$ Frequency of incident photon
$V=$ Stopping potential
For a photon of frequency $v_{1},$
$$
h v_{1}=W+e V_{1}
$$
For photon of frequency $v_{2}$
$$
h v_{2}=W+e V_{2}
$$
Divided Eq. (i) by Eq. (ii),
$$
\begin{aligned}
\frac{h v_{1}}{h v_{2}} &=\frac{W+e V_{1}}{W+e V_{2}} \\
\frac{v_{1}}{v_{2}} &=\frac{W+e V_{1}}{W+e V_{2}}
\end{aligned}
$$
$\Rightarrow W v_{1}+e V_{2} v_{1}=W v_{2}+e V_{1} v_{2}$
Solving above equation, $\Rightarrow \quad e=\frac{W\left(v_{2}-v_{1}\right)}{V_{2} v_{1}-V_{1} v_{2}}$

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2017
When phenol is treated with $\mathrm{D}_{2} \mathrm{SO}_{4} / \mathrm{D}_{2} \mathrm{O},$ some of the hydrogens get exchanged. The final product in this exchange reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
In acidic medium, phenol exists in following resonating structures.

Since, o/p positions are electron rich sites, so electrophile will attack on these site, i.e. hydrogen of these sites get exchanged by D (deuterium). Hence, the final product of the

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2014
When the frequency of the $A C$ voltage applied to a series LCR circuit is gradually increased from a low value, the impedance of the circuit

Options:
A monotonically increases
B first increases and then decreases
C first decreases and then increases
D monotonically decreases

The correct answer is:
first increases and then decreases

Solution:
The impedance of LCR circuit $Z=\sqrt{R^{2}+\left(\omega_{c}-\frac{1}{\omega_{c}}\right)^{2}}$
Where, $R=$ Resistance $L=$ Inductance
$C=$ Capacitance
When frequency of AC voltage applied to a series LCR circuit, it will first increase and resonance state $\omega_{i}=\frac{1}{\omega_{c}},$ it will decrease. In that state $(Z=R),$ it will be minimum. So, it will first increases and then decreases.

Topic: Physics, Alternating Current, WBJEE, WBJEE 2017
When the frequency of the light used is changed from $4 \times 10^{14} \mathrm{s}^{-1}$ to $5 \times 10^{14} \mathrm{s}^{-1}$. the angular width of the principal (central) maximum in a single slit Fraunhoffer diffraction pattern changes by 0.6 radian. What is the width of the slit (assume that the experiment is performed in vacuum

Options:
A $1.5 \times 10^{-7} \mathrm{m}$
B $3 \times 10^{-7} \mathrm{m}$
C $5 \times 10^{-7} \mathrm{m}$
D $6 \times 10^{-7} \mathrm{m}$

The correct answer is:
$5 \times 10^{-7} \mathrm{m}$

Solution:
Given, initial frequency of light,
$$
f_{1}=4 \times 10^{14} \mathrm{s}^{-1}
$$
Final frequency of light, $f_{2}=5 \times 10^{14} \mathrm{s}^{-1}$
Change in wavelength $, \Delta \lambda=\lambda_{1}-\lambda_{2}$
or
$$
\Delta \lambda=\frac{c}{f_{1}}-\frac{c}{f_{2}}
$$
$$
\begin{array}{l}
=\frac{3 \times 10^{8}}{4 \times 10^{14}}-\frac{3 \times 10^{8}}{5 \times 10^{14}} \\
=\frac{3 \times 10^{8}}{10^{14}}\left(\frac{1}{4}-\frac{1}{5}\right) \\
=\frac{3 \times 10^{8}}{10^{14}} \times \frac{1}{20} \\
=1.5 \times 10^{-7} \mathrm{m}
\end{array}
$$
Now, we know that, Angular width of central maxima,
$$
\begin{array}{l}
\theta=\frac{2 \lambda}{d} \text { or } \Delta \theta=\frac{2 \Delta \lambda}{d} \\
d=\frac{2 \Delta \lambda}{\Delta \theta}
\end{array}
$$
Here, $\Delta \theta=0.6$ radian
$$
d=\frac{2 \times 1.5 \times 10^{-7}}{0.6}=5 \times 10^{-7} \mathrm{m}
$$

Topic: Physics, Wave Optics, WBJEE, WBJEE 2019
Which atomic species cannot be used as a nuclear fuel?

Options:
A ${ }_{92}^{233} U$
B ${ }_{92}^{235} U$
C ${ }_{94}^{239} \mathrm{Pu}$
D ${ }_{92}^{238} U$

The correct answer is:
${ }_{92}^{238} U$

Solution:
Among the given options, ${ }_{92} \mathrm{U}^{238}$ is an isotope of uranium but cannot be used as a nuclear-fuel. U- 238 is non-fissile, i.e., it does not undergo fission by thermal neutrons. The energy released when U-238 absorbs a neutron is insufficient to carry out nuclear fission. Hence, (d) is the correct option.

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2019
Which information below regarding this reaction is applicable?

Options:
A It follows $\mathrm{S}_{\mathrm{N}} 2$ pathway, because it is a primary alkyl chloride.
B It follows $S_{N} 1$ pathway, because the intermediate carbocation is resonance stabilized.
C $\mathrm{S}_{\mathrm{N}} 1$ pathway is not followed, because the intermediate carbocation is destabilised by $-1$ effect of oxygen
D A mixed $\mathrm{S}_{\mathrm{N}} 1$ and $\mathrm{S}_{\mathrm{N}} 2$ pathway is followed.

The correct answer is:
A mixed $\mathrm{S}_{\mathrm{N}} 1$ and $\mathrm{S}_{\mathrm{N}} 2$ pathway is followed.

Solution:
Hint:

$\mathrm{S}_{\mathrm{N}} 2$ is also facilitated by OH due to resonance stablization of transition state.

Topic: Chemistry, Haloalkanes and Haloarenes, WBJEE, WBJEE 2020
Which of the following atoms should have the highest $1^{\prime \prime}$ electron affinity?

Options:
A F
B 0
C $\mathrm{N}$
D C

The correct answer is:
F

Solution:
$\because$ F-atom has 2 nd highest electron affinity
(Cl has highest) in the periodic table
$\therefore$ Among the given options F-has highest electron gain affinity

Topic: Chemistry, Classification of Elements and Periodicity in Properties, WBJEE, WBJEE 2017
Which of the following carbocations will be most stable?

Options:
A $\mathrm{Ph}_3 \mathrm{C}^{++}$
B $\mathrm{CH}_3-\stackrel{+}{\mathrm{C}} \mathrm{H}_3$
C $\left(\mathrm{CH}_3\right)_2 \stackrel{+}{\mathrm{C}} \mathrm{H}$
D $\mathrm{CH}_2=\mathrm{CH}-\stackrel{+}{\mathrm{C}} \mathrm{H}_2$

The correct answer is:
$\mathrm{Ph}_3 \mathrm{C}^{++}$

Solution:
Hints:
(Highly resonance stabilized)

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2009
Which of the following chemicals may be used to identify three unlabelled beakers containing conc.NaOH, conc.$\mathrm{H}_{2}\mathrm{SO}_{4}$ and water?

Options:
A $\mathrm{NH}_{4} \mathrm{NO}_{3}$
B $\mathrm{NaCl}$
C $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}$
D HCOONa

The correct answers are:
$\mathrm{NH}_{4} \mathrm{NO}_{3}$, $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}$

Solution:
$\mathrm{NH}_{4} \mathrm{NO}_{3}$ will evolve $\mathrm{NH}_{3}$ pungent smelling gas with $\mathrm{NaOH}$.
- $\mathrm{NO}_{2}$ brown gas with $\operatorname{con} . \mathrm{H}_{2} \mathrm{SO}_{4}$
- No reaction with water
Thus, $\mathrm{NH}_{4} \mathrm{NO}_{3}$ can be used to label all three beakers.
(c) $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}$ will evolve pungent smelling gas $\mathrm{NH}_{3}$ with $\mathrm{NaOH}$
- Effervescence of $\mathrm{CO}_{2}$ gas with con. $\mathrm{H}_{2} \mathrm{SO}_{4}$
- No reaction with water.
Thus, $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}$ can be used to label all three beakers.
Thus, options $(a, c)$ are correct.

Topic: Chemistry, Practical Chemistry, WBJEE, WBJEE 2019
Which of the following compounds does not exhibit optical isomerism?
en = ethylenediamine).

Options:
A $\operatorname{cis}-\left[\mathrm{CrCl}_{2}(\mathrm{Ox})_{2}\right]^{3-}$
B $\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}$
C $\operatorname{trans}-\left[\mathrm{CrCl}_{2}(\mathrm{ox})_{2}\right]^{3-}$
D $\left[\mathrm{Co}(\mathrm{O} \mathrm{x})(\mathrm{en})_{2}\right]^{+}$

The correct answers are:
$\operatorname{cis}-\left[\mathrm{CrCl}_{2}(\mathrm{Ox})_{2}\right]^{3-}$, $\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}$, $\left[\mathrm{Co}(\mathrm{O} \mathrm{x})(\mathrm{en})_{2}\right]^{+}$

Solution:
trans $\left[\mathrm{CrCl}_{2}(\mathrm{ox})_{2}\right]^{3-} \quad$ isomer-optically inactive (superimposable mirror images and plane of symmetry).
$\left.\mathrm{cis}-\mathrm{CrCl}_{2}(\mathrm{ox})_{2}\right]^{3-},\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}$ and $\left[\mathrm{Co}(\mathrm{ox})(\mathrm{en})_{2}\right]^{+}$
exhibited optical isomerism.

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2015
Which of the following compounds has maximum volatility?

Options:
A

B

C

D

The correct answer is:

Solution:
Hints: Due to intramolecular \(\mathrm{H}\) - bonding

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2011
Which of the following compounds have magnetic moment identical with $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$

Options:
A $\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$
B $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$
C $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$
D $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{4+}$

The correct answer is:
$\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{4+}$

Solution:
$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$
$\mathrm{Cr} \Rightarrow(\mathrm{Ar}) 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{1}$
$\mathrm{Cr}^{+3} \Rightarrow(\mathrm{Ar}) 3 \mathrm{~d}^{3} 4 \mathrm{~s}^{0}$

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2021
Which of the following compounds is asymmetric?

Options:
A
B
C
D

The correct answer is:

Solution:
Hint: No plane of symmetry as well as centre of symmetry is present

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2020
Which of the following compounds is not formed in iodoform reaction of acetone

Options:
A \(\mathrm{CH}_3 \mathrm{COCH}_2 \mathrm{I}\)
B \(\mathrm{ICH}_2 \mathrm{COCH}_2 \mathrm{I}\)
C \(\mathrm{CH}_3 \mathrm{COCH}_2\)
D \(\mathrm{CH}_3 \mathrm{COCI}_3\)

The correct answer is:
\(\mathrm{ICH}_2 \mathrm{COCH}_2 \mathrm{I}\)

Solution:
Hints:

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2011
Which of the following compounds shows evidence of the strongest hydrogen bonding?

Options:
A Propan-1-ol
B Propan-2-ol
C Propan-1,2-diol
D Propan-1,2,3-triol

The correct answer is:
Propan-1,2,3-triol

Solution:
$$
\text { Hints : Propan-1,2,3 triol have three polar }-\mathrm{OH} \text { group. }
$$

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2010
Which of the following compounds would not react with Lucas reagent at room temperature?

Options:
A $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}_{2} \mathrm{OH}$
B $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}$
C $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$
D $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}$

The correct answer is:
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$

Solution:
Primary ( $1^{\circ}$ )-alcohols do not give Lucas reagent test at room temperature. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$ does not undergo $S_{N}1 or S_{N}2$ at room temperature.

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2016
Which of the following contributes to the double helical structure of DNA

Options:
A hydrogen bond
B covalent bond
C disulphide bond
D van-der Waal's force

The correct answer is:
hydrogen bond

Solution:
No Solution Available

Topic: Chemistry, Biomolecules, WBJEE, WBJEE 2009
Which of the following diagrams correctly shows the relation between the terminal velocity $\mathrm{v}_{\mathrm{T}}$ of a spherical body falling in a liquid and viscosity $\eta$ of the liquid?

Options:
A
B
C
D

The correct answer is:

Solution:
Hint : At terminal speed.
$\mathrm{V}_{\mathrm{T}}=\frac{2 \mathrm{r}^{2}}{9 \eta}\left(\rho_{\mathrm{s}}-\rho_{\mathrm{L}}\right) \mathrm{g} \quad$ (Assuming other factor constant)
So, $\mathrm{V}_{\mathrm{T}} \propto \frac{1}{\eta}$ (rectangular hyperbola)

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2020
Which of the following does not represent the mathematical expression for the Heisenberg uncertainty principle?

Options:
A $\Delta x \cdot \Delta P \geq h /(4 \pi)$
B $\Delta x \cdot \Delta v \geq h /(4 \pi m)$
C $\Delta E \cdot \Delta t \geq h /(4 \pi)$
D $\Delta E \cdot \Delta x \geq h /(4 \pi)$

The correct answer is:
$\Delta E \cdot \Delta x \geq h /(4 \pi)$

Solution:
From Heisenberg uncertainty principle, $\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$
or $\quad \Delta x \cdot m \Delta v \geq \frac{h}{4 \pi}$
or $\quad \Delta x \cdot \Delta v \geq \frac{h}{4 \pi m}$
This principle is also applicable for pairs like energy-time $(\Delta E \cdot \Delta t) \quad$ and $\quad$ angular $\begin{array}{llll}\text { moment-angle } & (\Delta w \cdot \Delta \theta) & \text { along } & \text { with }\end{array}$
position-moment $(\Delta x \cdot \Delta p)$
Thus,
$$
\Delta E \cdot \Delta t \geq \frac{h}{4 \pi}
$$

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2012
Which of the following electronic configuration is not possible?

Options:
A $n=3 l=0 m=0$
B $n=3 l=1 m=-1$
C $n=2.l=0 \mathrm{mm}-1$
D $n=2, l=1 m=0$

The correct answer is:
$n=2.l=0 \mathrm{mm}-1$

Solution:
$n=2,1=0, m=-1$ is not a possible electronie configuration. With $n=2$ and $/-0$, the orbital is having only one $m_{1}$ value i.e. $0 .$ As the value of $m_{1}=-1 t 0+l$
Other options are correct.

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2018
Which of the following has the dimension if $\left[\mathrm{ML}^{2} \mathrm{T}^{-2}\right] ?$

Options:
A Coefficient of viscosity
B Surface tension
C Vapour pressure
D Kinetic energy

The correct answer is:
Surface tension

Solution:
$\because$ Surface tension $(T)=$ Work done per unit area
$$
\begin{aligned}
\therefore \quad T &=\frac{d W}{d A} \\
&=\frac{F \cdot d x}{d A} \quad\left(\mathrm{FL}^{-1}=\mathrm{MT}^{-2}\right.
\end{aligned}
$$
$\therefore$ Dimension of surface tension $=\mathrm{ML}^{2} \mathrm{T}^{-2}$

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2017
Which of the following has the largest number of atoms?

Options:
A $1 \mathrm{~g}$ of $\mathrm{Ag}$
B $1 \mathrm{~g}$ of $\mathrm{Fe}$
C $1 \mathrm{~g}$ of $\mathrm{Cl}_{2}$
D $1 \mathrm{~g}$ of $\mathrm{Mg}$

The correct answer is:
$1 \mathrm{~g}$ of $\mathrm{Mg}$

Solution:
Hint:
Number of atoms of $\mathrm{Ag}=\frac{1}{108} \times \mathrm{N}_{\mathrm{A}}=0.009 \mathrm{~N}_{\mathrm{A}}$
Number of atoms of $\mathrm{Fe}=\frac{1}{56} \times \mathrm{N}_{\mathrm{A}}=0.0178 \mathrm{~N}_{\mathrm{A}}$
Number of atoms of $\mathrm{Cl}_{2}=\frac{1}{71} \times 2 \times \mathrm{N}_{\mathrm{A}}=0.028 \mathrm{~N}_{\mathrm{A}}$
Number of atoms of $\mathrm{Mg}=\frac{1}{24} \mathrm{~N}_{\mathrm{A}}=0.041 \mathrm{~N}_{\mathrm{A}}$ [largest]

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2020
Which of the following has the strongest H-bond?

Options:
A $0-H-S$
B $\mathrm{S}-\mathrm{H}-\mathrm{O}$
C $F-H$. F
D $F-H-O$

The correct answer is:
$F-H$. F

Solution:
The strongest H-bond is $\mathrm{F}-\mathrm{H}-\mathrm{F}$ because in this case, a hydrogen is bonded to a most. electronegative atom i.e. fluorine. In all other options, hydrogen is bonded to oxygen and sulphur that are less electronegative atom than fluorine.
The correct electronegativity order of elements are as follows: $\mathrm{F}>\mathrm{O}>\mathrm{S}$

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2018
Which of the following is correct?

Options:
A Radius of $\mathrm{Ca}^{2+} < \mathrm{Cl}^{-} < \mathrm{S}^{2-}$
B Radius of $\mathrm{Cl}^{-} < \mathrm{S}^{2-} < \mathrm{Ca}^{2+}$
C Radius of $\mathrm{S}^{2-}=\mathrm{Cl}^{-}=\mathrm{Ca}^{2+}$
D Radius of $\mathrm{S}^{2-} < \mathrm{Cl}^{-} < \mathrm{Ca}^{2+}$

The correct answer is:
Radius of $\mathrm{Ca}^{2+} < \mathrm{Cl}^{-} < \mathrm{S}^{2-}$

Solution:
\begin{array}{c|c|c}
\hline Species & Atomic number & Number of electrons \\
\hline \mathrm{Ca}^{2+} & 20 & 20-2=18 \\
\mathrm{Cl}^{-} & 17 & 17+1=18 \\
\mathrm{s}^{2-} & 16 & 16+2=18 \\
\hline
\end{array}
For isoelectronic species, an increase in atomic number, result in decreased size. Thus, order of radius will be
$$
\mathrm{Ca}^{2+} < \mathrm{Cl}^{-} < \mathrm{S}^{2-}
$$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, WBJEE, WBJEE 2012
Which of the following is correct?

Options:
A Evaporation of water causes an increase in disorder of the system
B Melting of ice causes a decrease in randomness of the system
C Condensation of steam causes an increase in disorder of the system
D There is practically no change in the randomness of the system when water is evaporated

The correct answer is:
Evaporation of water causes an increase in disorder of the system

Solution:
Water evaporates into steam. In this physical state its disorder or randomness increases.

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2012
Which of the following is least thermally stable?

Options:
A $\mathrm{MgCO}_{3}$
B $\mathrm{CaCO}_{3}$
C $\mathrm{SrCO}_{3}$
D $\mathrm{BeCO}_{3}$

The correct answer is:
$\mathrm{BeCO}_{3}$

Solution:
BeCO, is least thermally stable. The thermal stability of carbonates increases down the group
i.e from Be to Ba. $\mathrm{BeCO}_{3} < \mathrm{MgCO}_{3} < \mathrm{CaCO}_{3} < \mathrm{SrCO}_{3} < \mathrm{BaCO}_{3}$
$(523 \mathrm{K}) \quad(813 \mathrm{K})$
$(1173 \mathrm{K}) \quad(1562 \mathrm{K}) \quad(1633 \mathrm{K})$
BeCO, is unstable to the extent that it is stable only in atmosphere of $\mathrm{CO}_{y}$. These carbonates show reversible decomposition in closed container.
$M \mathrm{CO}, \quad \rightarrow \mathrm{MO}+\mathrm{CO}_{2}$
More stable the oxide is formed, lesser will be stability of carbonates.

Topic: Chemistry, Practical Chemistry, WBJEE, WBJEE 2018
Which of the following is not always true?

Options:
A $\mid a+b\mid^{2}=|a|^2+|b|^2. \text{if a and b are perpendicular to each other}$
B $\mid a+\lambda b| \geq|a|$ for all $\lambda \in R$ if a and b are perpendicular to each other
C $\mid a+b \mid^{2}+\mid a-b \mid^{2}=2(|a|^2+|b|^2)$
D $\mid a+\lambda b| \geq|a|$ for all $\lambda \in R$ if a is parallel to b

The correct answer is:
$\mid a+\lambda b| \geq|a|$ for all $\lambda \in R$ if a is parallel to b

Solution:
(a) If a and b are perpendicular to each other, then
$\mathbf{a} \cdot \mathbf{b}=0$
Now consider,
$\begin{aligned}
|\mathbf{a}+\mathbf{b}|^{2} &=(\mathbf{a}+\mathbf{b}) \cdot(\mathbf{a}+\mathbf{b}) \\
&=|\mathbf{a}|^{2}+|\mathbf{b}|^{2}
\end{aligned}$
So, option (a) is always true.
(b) If $\mathbf{a}$ and $\mathbf{b}$ are perpendicular to each other, then
$\mathbf{a} \cdot \mathbf{b}=0$
Now consider, $\begin{aligned}|\mathbf{a}+\lambda \mathbf{b}|^{2} &=(\mathbf{a}+\lambda \mathbf{b}) \cdot(\mathbf{a}+\lambda \mathbf{b}) \\ &=|\mathbf{a}|^{2}+\lambda^{2}|\mathbf{b}|^{2} \\ \Rightarrow|\mathbf{a}+\lambda \mathbf{b}| &=\sqrt{|\mathbf{a}|^{2}+\lambda^{2}|\mathbf{b}|^{2}} \end{aligned}$
$\geq|\mathbf{a}|$ for all $\lambda \in R$
So, option (b) is always true.
(c) Consider,
$\left|a+\left.b\right|^{2}+|a-b|^{2}=(a+b) \cdot(a+b) + (a-b) \cdot (a-b)\right.$
$=|\mathbf{a}|^{2}+\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{a}+|\mathbf{b}|^{2}+|\mathbf{a}|^{2}-\mathbf{a} \cdot \mathbf{b} -b \cdot a + |b|^2$
$=2\left(|\mathbf{a}|^{2}+|\mathbf{b}|^{2}\right)$
So, option (c) is always true.
(d) Consider.
$\mathbf{a}=-\mathbf{b} \text { and } \mathbf{b} \neq 0$
Then, $|\mathbf{a}+\lambda \mathbf{b}| \geq|\mathbf{a}|$
$\Rightarrow|-\mathbf{b}+\lambda \mathbf{b}| \geq|-\mathbf{b}|$
$\Rightarrow \quad|\mathbf{b}||\lambda-1| \geq|\mathbf{b}| \Rightarrow|\lambda-1| \geq 1$
which is not true for all $\lambda,$ as we consider $\lambda=\frac{1}{2}$, then it is not true
Hence, option (d) is not always true.

Topic: Mathematics, Vector Algebra, WBJEE, WBJEE 2015
Which of the following is present in maximum amount in 'acid rain'?

Options:
A $\mathrm{HNO}_{3}$
B $\mathrm{H}_{2} \mathrm{SO}_{4}$
C HCI
D $\mathrm{H}_{2} \mathrm{CO}_{3}$

The correct answer is:
$\mathrm{H}_{2} \mathrm{SO}_{4}$

Solution:
$\mathrm{H}_{2} \mathrm{SO}_{4}$ (sulphuric acid) is present in maximu $\mathrm{h}_{4}$ amount in 'acid rain'. Oxides of nitrogen and sulphur, released into the atmosphere from industries and automobiles thermal power plants, are the miain sources of acid rain. These oxides on oxidation followed by hydrolysis give sulphuric acid and nitric acid that alongwith $\mathrm{HCl}$ are responsible for the acidity of rain. The oxidation reaction is catalysed by particulate matter present in the polluted atmosphere. $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)+2 \mathrm{H}, \mathrm{O}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)$
$4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}\left(h \longrightarrow 4 \mathrm{HNO}_{3}(a q)\right.$

Topic: Chemistry, Environmental Chemistry, WBJEE, WBJEE 2018
Which of the following is radioactive?

Options:
A Hydrogen
B Deuterium
C Tritium
D none

The correct answer is:
Tritium

Solution:
Fact

Topic: Chemistry, Hydrogen, WBJEE, WBJEE 2022
Which of the following is used to prepare \(\mathrm{Cl}_2\) gas at room temperature from concentrated \(\mathrm{HCl}\) ?

Options:
A \(\mathrm{MnO}_2\)
B \(\mathrm{H}_2 \mathrm{~S}\)
C \(\mathrm{KMnO}_4\)
D \(\mathrm{Cr}_2 \mathrm{O}_3\)

The correct answer is:
\(\mathrm{KMnO}_4\)

Solution:
Hints : \(2 \mathrm{MnO}_4^{-}+16 \mathrm{H}^{+}+10 \mathrm{Cl}^{-} \rightarrow 2 \mathrm{Mn}^{2+}+5 \mathrm{Cl}_2+8 \mathrm{H}_2 \mathrm{O}\)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2011
Which of the following is/are always false?

Options:
A A quadratic equation with rational coefficients hes zero or two irrational roots
B A quadratic equation with real coefficients has zero or two non-real roots
C A quadratic equation with irrational coeflicients has zero or two irrational roots
D A quadratic equation with integer coefficients has zero or two irrational roots

The correct answers are:
A quadratic equation with irrational coeflicients has zero or two irrational roots

Solution:
If a quadratic equation has irrational coefficients, then roots are also inational.
So, option (c) is always false.

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2015
Which of the following materials is the best conductor of electricity?

Options:
A Platinum
B Gold
C Silicon
D Copper

The correct answer is:
Copper

Solution:
No Solution Available

Topic: Physics, Semiconductors, WBJEE, WBJEE 2009
Which of the following mixtures act(s) as buffer solution?

Options:
A \(\mathrm{NaOH}+\mathrm{CH}_3 \mathrm{COOH}\) (1:1 mole ratio)
B \(\mathrm{NH}_4 \mathrm{OH}+\mathrm{HCl}(2: 1\) mole ratio)
C \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaOH}(2: 1\) more ratio)
D \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaOH}(1: 2\) mole ratio)

The correct answers are:
\(\mathrm{NH}_4 \mathrm{OH}+\mathrm{HCl}(2: 1\) mole ratio), \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaOH}(2: 1\) more ratio)

Solution:
Hint : (B) \(\mathrm{NH}_4 \mathrm{OH}+\mathrm{HCl}~(2: 1 \text { mole ratio}) \rightarrow \underset{\large 1 \quad:\quad 1 \atop
\large \text{Buffer} \quad \text{buffer}}{\mathrm{NH}_4 \mathrm{Cl}+\mathrm{NH}_4 \mathrm{OH}}\)
(C) \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaOH}(2: 1\) mole ratio \() \rightarrow \underset{\large 1\quad : \quad 1 \atop \large \text{Acidic buffer}}{\mathrm{CH}_3 \mathrm{COONa}+\mathrm{CH}_3 \mathrm{COOH}}\)

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2023