datavorous/entrance-exam-dataset huggingface dataset:
In the electrical circuit shown in the figure, the current through the $4 \Omega$ resistor is

Options:
A $1 \mathrm{A}$
B $0.5 \mathrm{A}$
C $0.25 \mathrm{A}$
D $0.1 \mathrm{A}$

The correct answer is:
$0.5 \mathrm{A}$

Solution:
The current through the various branches of the circuit will be shown as

According to Kirchhoff's second law in closed circuit $B C D E B$
$$
\begin{array}{r}
2 I_{1}+4 l_{1}+2 l_{1}-8\left(1-l_{1}\right)=0 \\
\Rightarrow \quad 16 l_{1}-81=0
\end{array}
$$
$\Rightarrow \quad l_{1}=\frac{8 l}{16} \Rightarrow l_{1}=\frac{1}{2} l$
In closed circuit ABEFA
$$
-9+3 l+8\left(1-l_{1}\right)+2 l=0
$$
$\Rightarrow \quad 13l-8 I_{1}=9 \Rightarrow 13 /-8\left(\frac{1}{2} I\right)=9$
$\Rightarrow \quad I=\frac{9}{9}=1 \mathrm{A}$
So current through $4 \Omega$ resistor $l_{1}=\frac{1}{2} \times 1=0.5 \mathrm{A}$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2013
In the expansion of $(x-1)(x-2) \ldots(x-18)$ the coefficient of $x^{17}$ is

Options:
A 684
B -171
C 171
D -342

The correct answer is:
-171

Solution:
Coefficient of $x^{17}$ in $(x-1)(x-2) \ldots(x-18)$
$$
\begin{array}{l}
=-(1+2+3+\ldots+16) \\
=-(171)=-171
\end{array}
$$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2016
In the extraction of Ca by electro reduction of molten $\mathrm{CaCl}_{2}$ some $\mathrm{CaF}_{2}$ is added to the electrolyte for the following reason:

Options:
A To keep the electrolyte in liquid state at temperature lower than the m. p. of $\mathrm{CaCl}_{2}$
B To effect precipitation of Ca
C To effect the electrolysis at lower voltage
D To increase the current efficiency

The correct answer is:
To keep the electrolyte in liquid state at temperature lower than the m. p. of $\mathrm{CaCl}_{2}$

Solution:
Hint:
$\mathrm{CaF}_{2}$ decreases the melting point of $\mathrm{CaCl}_{2}$

Topic: Chemistry, s Block Elements, WBJEE, WBJEE 2020
In the face-centred cubic lattice structure of gold the closest distance between gold atoms is ('a' being the edge length of the cubic unit cell)

Options:
A $a \sqrt{2}$
B $\frac{a}{\sqrt{2}}$
C $\frac{a}{2 \sqrt{2}}$
D $2 \sqrt{2}$ a

The correct answer is:
$\frac{a}{\sqrt{2}}$

Solution:
Hint:
Closest distance between two gold atoms in fcc lattice of gold is $\frac{a}{\sqrt{2}} .$ This is the distance between the corner atom and the closest face centre atom.

Topic: Chemistry, Solid State, WBJEE, WBJEE 2020
$$
\text { In the figure below, the capacitance of each capacitor is } 3 \mu \mathrm{F} \text {. The effective capacitance between } \mathrm{A} \text { and } \mathrm{B} \text { is : }
$$

Options:
A $\frac{3}{4} \mu \mathrm{F}$
B $3 \mu \mathrm{F}$
C $6 \mu \mathrm{F}$
D $5 \mu \mathrm{F}$

The correct answer is:
$5 \mu \mathrm{F}$

Solution:
Hints:
$$
\frac{2 \mathrm{C}}{3}+\mathrm{C}=2+3=5 \mu \mathrm{F}
$$

Topic: Physics, Capacitance, WBJEE, WBJEE 2010
In the following compound, the number of sp-bybridised carbons are

Options:
A 2
B 3
C 4
D 5

The correct answer is:
4

Solution:
$$
\text { Thus, number of } s p \text { -hybridised } C=4
$$

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2015
In the following electron - dot structure, calculate the formal charge from left to right nitrogen atom; \(\ddot{\mathrm{N}}=\mathrm{N}=\ddot{\mathrm{N}}\)

Options:
A \(-1,-1,+1\)
B \(-1,+1,-1\)
C \(+1,-1,-1\)
D \(+1,-1,+1\)

The correct answer is:
\(-1,+1,-1\)

Solution:
Hints: Formal chargl= Number of electrons in
Valence shell \(-\left(\frac{1}{2} \times\right.\) numbers of electrons as bond pair + numbers of electrons as lone pair)
\(\begin{gathered}
1 \quad~~ 2 ~~\quad 3\\
: \ddot{\mathrm{N}}=\mathrm{N}=\ddot{\mathrm{N}}:
\end{gathered}\)
For \(\mathrm{N}_1 ~\&~ \mathrm{~N}_3\)
Formal charge \(=5-\left(\frac{4}{2}+4\right)=5-(6)=-1\)
For \(\mathrm{N}_2=5-\frac{1}{2} \times 8-0=5-4=+1\)

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2011
In the following reaction, the product $E$ is

Options:
A
B
C
D

The correct answer is:

Solution:
Aldehydes lacking $\alpha-\mathrm{H}$ atom, undergo Cannizzaro reaction if treated with concentrated alkali (NaOH), i.e., one mole of the aldehyde is reduced to alcohol and other mole is oxidised to salt of acid. Thus, the given aldehyde (glyoxal) undergoes intramolecular Cannizzaro reaction, ie. its half part is oxidised and half is reduced.

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2014
In the following species, the one which is likely to be the intermediate during benzoin condensation of benzaldehyde is

Options:
A
B
C
D

The correct answer is:

Solution:
When benzaldehyde is heated with aqueous ethanolic NaCN or $\mathrm{KCN},$ it dimerises to form an $\alpha$ -hydroxy ketone called benzoin, and this reaction is formed as benzoin condensation. It involves self condensation of an aromatic aldehyde in the presence of $\mathrm{CN}^{-}$ as catalyst.

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2013
In the given circuit, find the voltage drop \(V_L\) in the load resistance \(R_L\).

Options:
A \(5 \mathrm{~V}\)
B \(3 \mathrm{~V}\)
C \(9 \mathrm{~V}\)
D \(6 \mathrm{~V}\)

The correct answer is:
\(3 \mathrm{~V}\)

Solution:
Hint :
\(\mathrm{i}=\frac{9}{300} \mathrm{~A}\)
So, \(V_a-V_a=\frac{9}{300} \times 100=3 V\)
So, diode is not activated,
So, voltage across load is \(3 \mathrm{~V}\).

Topic: Physics, Semiconductors, WBJEE, WBJEE 2023
In the given circuit, the binary inputs at $A$ and $B$ are both 1 in one case and both 0 in the next case. The respective outputs at $Y$ in these two cases will be

Options:
A 1,1
B 0,0
C 0,1
D 1,0

The correct answer is:
0,0

Solution:
The output of logic gate circuit
for
$$
\begin{array}{l}
Y=\frac{}{A \cdot B+\bar{A} \bar{B}} \\
A=1
\end{array}
$$
$$
B=1 \Rightarrow Y=0
$$
and for $\quad A=0, B=0$
$\Rightarrow \quad Y=0$

Topic: Physics, Semiconductors, WBJEE, WBJEE 2017
In the given figure, 1 represents isobaric, 2 represents isothermal and 3 represents adiabatic processes of an ideal gas. If $\Delta \mathrm{U}_{1}, \Delta \mathrm{U}_{2}, \Delta \mathrm{U}_{3}$ be the changes in internal energy in these processes respectively, then

Options:
A $\Delta \mathrm{U}_{1} < \Delta \mathrm{U}_{2} < \Delta \mathrm{U}_{3}$
B $\Delta \mathrm{U}_{1}>\Delta \mathrm{U}_{3} < \Delta \mathrm{U}_{2}$
C $\Delta \mathrm{U}_{1}=\Delta \mathrm{U}_{2}>\Delta \mathrm{U}_{3}$
D $\Delta \mathrm{U}_{1}>\Delta \mathrm{U}_{2}>\Delta \mathrm{U}_{3}$

The correct answer is:
$\Delta \mathrm{U}_{1}>\Delta \mathrm{U}_{2}>\Delta \mathrm{U}_{3}$

Solution:
$\Delta \mathrm{T}_{1}=\mathrm{T}_{0}$
$\Delta \mathrm{T}_{2}=0$
$\Delta \mathrm{T}_{3}=-\mathrm{ve}$
$\Rightarrow \Delta \mathrm{U}_{1}>\Delta \mathrm{U}_{2}>\Delta \mathrm{U}_{3}$

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2021
In the hydrolysis of an organic chloride in presence of large excess of water; $\mathrm{RCI}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{ROH}+\mathrm{HCl}$

Options:
A Molecularity and order of reaction both are 2
B Molecularity is 2 but order of reaction is 1
C Molecularity is 1 but order of reaction is 2
D Molecularity is 1 and order of reaction is also 1

The correct answer is:
Molecularity is 2 but order of reaction is 1

Solution:
Hints : As water used is in large excess.

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2010
In the interval \((-2 \pi, 0)\), the function \(f(x)=\sin \left(\frac{1}{x^3}\right)\)

Options:
A never changes sign
B changes sign only once
C changes sign more than once but finitely many times
D changes sign infinitely many times

The correct answer is:
changes sign infinitely many times

Solution:
Hint : \(x \in(-2 \pi, 0), f(x)=\sin \left(\frac{1}{x^3}\right)\)
\(\begin{aligned}
& \because-2 \pi < x < 0 \\
& \Rightarrow-8 \pi^3 < x^3 < 0 \\
& \Rightarrow-\infty < \frac{1}{x^3} < -\frac{1}{8 \pi^3}
\end{aligned}\)
Hence, \(\sin \left(\frac{1}{x^3}\right)\) will take all values from -1 to 1 , and will change its sign infinitely many times since it is a periodic function

Topic: Mathematics, Functions, WBJEE, WBJEE 2023
In the IUPAC system, $PhCH_{2}CH_{2}\mathrm{CO}_{2} \mathrm{H}$ is named as

Options:
A 3 -phenylpropanoic acid
B benzylacetic acid
C carboxyethylbenzene
D 2 -phenylpropanoic acid

The correct answer is:
3 -phenylpropanoic acid

Solution:
IUPAC name of $\mathrm{Ph} \cdot \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$ is,
3-phenylpropanoic acid

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2017
In the Lassaigne's test for the detection of nitrogen in an onganic compound. the appearance of blue coloured compound is due to

Options:
A ferric ferricyanide
B ferrous ferricyanide
C ferric ferrocyanide
D ferrous ferrocyanide

The correct answer is:
ferric ferrocyanide

Solution:
To a part of sodium extract, FeSO $_{4}$ solution is added and the contents are warmed. A few drops of $\mathrm{FeCl}_{3}$ solution are then added and resulting $\begin{array}{lllll}\text { solution } & \text { is } & \text { acidified } & \text { with } & \text { conc. } & \text { HCl. } & \text { The }\end{array}$ appearance of bluish green or Prussian blue colouration confirms the presence of nitrogen. The reactions that occur during this test are as

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2015
In the nuclear reaction \({ }_7^{14} \mathrm{~N}+\mathrm{X} \rightarrow{ }_6^{14} \mathrm{C}+{ }_1^1 \mathrm{H}\) the \(\mathrm{X}\) will be

Options:
A \({ }_{-1}^0 \mathrm{e}\)
B \({ }_1^1 \mathrm{H}\)
C \({ }_1^2 \mathrm{H}\)
D \({ }_0^1 \mathrm{n}\)

The correct answer is:
\({ }_0^1 \mathrm{n}\)

Solution:
\(\text {Hints: } \mathrm{X} \rightarrow{ }_0^1 \mathrm{n}\)

Topic: Physics, Nuclear Physics, WBJEE, WBJEE 2011
In the reaction of sodium thiosulphate with $\mathrm{I}_2$ in aqueous medium the equivalent weight of sodium thiosulphate is equal to

Options:
A molar mass of sodium thiosulphate
B the average of molr masses of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ and $\mathrm{I}_2$
C half the molar mass of sodium thiosulphate
D molar mass of sodium thiosulphate $\times 2$

The correct answer is:
molar mass of sodium thiosulphate

Solution:
$\begin{gathered}
\text {Hints : } 2 \mathrm{Na}_2 \stackrel{+2}{\mathrm{~S}_2} \mathrm{O}_3+\mathrm{I}_2 \longrightarrow \mathrm{Na}_2 \stackrel{+25}{\mathrm{~S}_4} \mathrm{O}_6+2 \mathrm{NaI} \\
\text {n-factor }=1 \\
\mathrm{E}=\frac{\mathrm{M}}{1}=\mathrm{M}
\end{gathered}$

Topic: Chemistry, Redox Reactions, WBJEE, WBJEE 2010
In the reaction $R \mathrm{MgBr}+\mathrm{HC}(\mathrm{OEt})_{3} \stackrel{\text { Ether } \mathrm{H}_{3} \mathrm{O}^{+}}{\longrightarrow} P$ the proxtuct $P$ is

Options:
A $\mathrm{ACHO}$
B $R_{2}$ CHOEt
C $R_{3} \mathrm{CH}$
D $R \mathrm{CH}(\mathrm{OEt})_{2}$

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

Solution:

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2015
In the set of all $3 \times 3$ real matrices a relation is defined as follows. A matrix $A$ is related to a matrix $B,$ if and only if there is a non-singular $3 \times 3$ matrix $P,$ such that $B=P^{-1} A P .$ This relation is

Options:
A reflexive, symmetric but not transitive
B reflexive, transitive but not symmetric
C symmetric, transitive but not reflexive
D an equivalence relation

The correct answer is:
an equivalence relation

Solution:
Let the relation defined as $R=\left\{(A, B) \mid B=P^{-1} A P\right\}$
For reflexive, $A=I^{-1} A l$
$\Rightarrow (A, A) \in R$
$\Rightarrow R$ is reflexive
For symmetric $\operatorname{Let}(A B) \in R$
$\because$
$B=P^{-1} A P$
$\Rightarrow P B=A P \Rightarrow P B P^{-1}=A$
$\Rightarrow$
$A=\left(P^{-1}\right)^{-1} B\left(P^{-1}\right)$
$\Rightarrow (B, A) \in R \Rightarrow R$ is symmetric.
For transitive
Let $(A, B) \in R,(B, C) \in R$
$\because A=P^{-1} B P$ and $B=Q^{-} C Q$
$\Rightarrow A=P^{-} Q^{-1} C Q P=(Q P)^{-1} C(Q P)$
$\Rightarrow (A, C) \in R$
$\Rightarrow R$ is transitive. since, $A$ is reflexive, symmetric and transitive. So, $R$ is an equivalence relation.

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2013
In the solid state, $\mathrm{PCl}_{5}$ exists as

Options:
A $\left[\mathrm{PCl}_{4}\right]^{-}$ and $\left[\mathrm{PCl}_{6}\right]^{+}$ ions
B covalent $\mathrm{PCl}_{5}$ molecules only
C $\left[\mathrm{PCl}_{4}\right]^{+}$ and $\left[\mathrm{PCl}_{6}\right]^{-}$ ions
D covalent $\mathrm{P}_{2} \mathrm{Cl}_{10}$ molecules only

The correct answer is:
$\left[\mathrm{PCl}_{4}\right]^{+}$ and $\left[\mathrm{PCl}_{6}\right]^{-}$ ions

Solution:
$\mathrm{PCl}_{5}$ exists as $\left[\mathrm{PCl}_{4}\right]^{+}$ and $\left[\mathrm{PCl}_{6}\right]^{-}$ in solid state.
$$
2 \mathrm{PCl}_{5} \longrightarrow\left[\mathrm{PCl}_{4}\right]^{+}\left[\mathrm{PCl}_{6}]^{-}\right.
$$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2016
In triangle $\mathrm{ABC}, \mathrm{a}=2, \mathrm{~b}=3$ and $\sin \mathrm{A}=\frac{2}{3}$, then $\mathrm{B}$ is equal to

Options:
A $30^{\circ}$
B $60^{\circ}$
C $90^{\circ}$
D $120^{\circ}$

The correct answer is:
$90^{\circ}$

Solution:
Hints: $\frac{a}{\sin A}=\frac{b}{\sin B}$
$\sin B=\frac{b}{a} \cdot \sin A=\frac{3}{2} \cdot \frac{2}{3}=1$
$$
\mathrm{B}=\frac{\pi}{2}
$$

Topic: Mathematics, Properties of Triangles, WBJEE, WBJEE 2009
In which of the following functions, Rolle's theorem is applicable?

Options:
A $f(x)=|x|$ in $-2 \leq x \leq 2$
B $\mathrm{f}(\mathrm{x})=\tan \mathrm{x}$ in $0 \leq \mathrm{x} \leq \pi$
C $\mathrm{f}(\mathrm{x})=1+(\mathrm{x}-2)^{\frac{2}{3}}$ in $1 \leq \mathrm{x} \leq 3$
D $\mathrm{f}(\mathrm{x})=\mathrm{x}(\mathrm{x}-2)^2$ in $0 \leq \mathrm{x} \leq 2$

The correct answer is:
$\mathrm{f}(\mathrm{x})=\mathrm{x}(\mathrm{x}-2)^2$ in $0 \leq \mathrm{x} \leq 2$

Solution:
Hints: $(\mathrm{A}) \mathrm{f}(\mathrm{x})=|\mathrm{x}|$ not differentiable at $\mathrm{x}=0$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2010
In which of the following mixed aqueous solutions, $\mathrm{pH}=\mathrm{p} K_{a}$ at equilibrium?
(1) $100 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}+100 \mathrm{mL}$ of $0.1 \mathrm{MCH}_{3} \mathrm{COONa}$
(2) $100 \mathrm{mL}$ of $0.1 \mathrm{MCH}_{3} \mathrm{COOH}+50 \mathrm{mL}$ of 0.1 $\mathrm{M} \mathrm{NaOH}$
(3) $100 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}+100 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$
(4) $100 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}+100 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{NH}_{3}$

Options:
A (1) is correct
B (2) is correct
C (3) is correct
D Both (1) and (2) are correct

The correct answers are:
(1) is correct, (2) is correct, Both (1) and (2) are correct

Solution:
pH of buffer solutions is given by
$$
\mathrm{pH}=\mathrm{pK}_{a}+\mathrm{log} \frac{[\mathrm{salt}]}{[\mathrm{acid}]}
$$
(1) $\mathrm{CH}_{3} \mathrm{COOH}$ and $\mathrm{CH}_{3} \mathrm{COONa}$
$\quad \quad$ $100 \times 0.1$ $\quad \quad$$100 \times 0.1$
No. of milimoles $=10 \quad=10$
$\because[$ salt $]=\mid a c i d]=10, p H=p K_{a}+\log \frac{10}{10}$
$\therefore \mathrm{pH}=\mathrm{pK_{a}}$

$\because \quad$ (salt) $=[\mathrm{acid}]=5, \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{5}{5}$
$\therefore \quad \mathrm{pH}=\mathrm{p} K_{\mathrm{a}}$
$[$ salt $]=10$ and $[a c i d]=0$
$p H=p K_{a}+\log \frac{10}{0}$
$\therefore \quad \mathrm{pH} \neq \mathrm{pK}_{a}$
(4) $\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NH}_{3} \longrightarrow \mathrm{CH}_{3} \mathrm{COONH}_{4}$
initial $10 \quad \quad \quad \quad 10 \rightarrow$ No buffer action
Final Due to weak $\mathrm{CH}_{3} \mathrm{COONH}_{4}$ salt
$\therefore \quad p H \neq p K_{a}$

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2016
In which of the following molecules is hydrogen bridge bond present?

Options:
A Water
B Inorganic benzene
C Diborane
D Methanol

The correct answer is:
Diborane

Solution:
Hints:

Topic: Chemistry, Hydrogen, WBJEE, WBJEE 2010
In which of the following pairs, the two physical quantities have different dimensions?

Options:
A Planck's constant and angular momentum
B impulse and linear momentum
C Moment of inertia and moment of a force
D Energy and torque

The correct answer is:
Moment of inertia and moment of a force

Solution:
(a) Planck's constant = Energy
$=\left[\mathrm{ML}^{2} \mathrm{T}^{-1}\right]$
Angular momentum = Moment of inertia Angular velocity
$=\left[\mathrm{ML}^{2}\right] \times\left[\mathrm{T}^{-1}\right)=\left[\mathrm{ML}^{2} \mathrm{T}^{-1}\right]$
(b) Impulse = Force x Time
$=\left[\mathrm{MLT}^{-2}\right] \mathrm{IT}=\left[\mathrm{MLT}^{-1}\right]$
and linear momentum = Mass $\times$ Velocity
$=\left[\mathrm{M}\left[\mathrm{LT}^{-1}\right]=\left[\mathrm{MLT}^{-1}\right)\right.$
(c) Moment of inertia = Mass $\times(\text { Distance })^{2}$
$=\left[M L^{2}\right]$
and moment of force = Force x Distance
$=\left[\mathrm{MLT}^{-2}\right] \mathrm{U}=\left[\mathrm{M}^{2} \mathrm{T}^{-2}\right]$
(d) Energy $=\left[\mathrm{ML}^{2} \mathrm{T}^{-2} \mathrm{l}\right.$ and torque $=\left[\mathrm{ML}^{2} \mathrm{T}^{-2} 1\right.$
So, (c) option has different dimensions.

Topic: Physics, Units and Dimensions, WBJEE, WBJEE 2014
In which of the following phenomena, the heat waves travel along straight lines with the speed of light?

Options:
A Thermal conduction
B Forced convection
C Natural convection
D Thermal radiation

The correct answer is:
Thermal radiation

Solution:
Thermal radiation is the heat waves travel along straight lines with the speed of light.

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2014
In which of the following reactions new carbon-carbon bond is not formed :

Options:
A Cannizaro reaction
B Wurtz reaction
C Aldol condensation
D Friedel-Craft reaction

The correct answer is:
Aldol condensation

Solution:
In cannizaro's reaction no new $\mathrm{C}-\mathrm{C}$ bond is formed

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2009
In Young's double slit experiment, light of wavelength $\lambda$ passes through the double-slit and forms interference fringes on a screen $1.2 \mathrm{~m}$ away. If the difference between 3 rd order maximum and 3 rd order minimum is $0.18 \mathrm{~cm}$ and the slits are $0.02 \mathrm{~cm}$ apart, then $\lambda$ is

Options:
A $1200 \mathrm{~nm}$
B $450 \mathrm{~nm}$
C $600 \mathrm{~nm}$
D $300 \mathrm{~nm}$

The correct answer is:
$600 \mathrm{~nm}$

Solution:
Distance between 3 rd order minima and 3 rd order maxima is $\frac{\beta}{2}$
$
\begin{array}{l}
\frac{\beta}{2}=0.18 \mathrm{~cm} \\
\beta=0.36 \mathrm{~cm} \\
\beta=\frac{\mathrm{D} \lambda}{\mathrm{d}} \\
\therefore \lambda=600 \mathrm{~nm}
\end{array}
$

Topic: Physics, Wave Optics, WBJEE, WBJEE 2021
In Young's double slit experiment, the fringe width is $\beta$. If the entire arrangement is placed in a liquid of refractive index $n$, the fringe width becomes :

Options:
A $n \beta$
B $\frac{\beta}{n+1}$
C $\frac{\beta}{n-1}$
D $\frac{\beta}{n}$

The correct answer is:
$\frac{\beta}{n}$

Solution:
No Solution Available

Topic: Physics, Wave Optics, WBJEE, WBJEE 2010
In Young's double slit experiment the two slits are d distance apart. Interference pattern is observed on a screen at a distance D from the slits. A dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light is

Options:
A \(\frac{\mathrm{D}^2}{2 \mathrm{~d}}\)
B \(\frac{\mathrm{d}^2}{2 \mathrm{D}}\)
C \(\frac{\mathrm{D}^2}{\mathrm{~d}}\)
D \(\frac{\mathrm{d}^2}{\mathrm{D}}\)

The correct answer is:
\(\frac{\mathrm{d}^2}{\mathrm{D}}\)

Solution:
Hints:
\(\mathrm{n}^{\text {th }}\) Dark fringe
\(\begin{aligned}
& (2 n-1) \frac{D \lambda}{2 d}=\frac{d}{2} \\
& \lambda=\frac{d^2}{(2 n-1) D}=\frac{d^2}{D} \quad[\text { for } n=1]
\end{aligned}\)

Topic: Physics, Wave Optics, WBJEE, WBJEE 2011
In Young's double slit experiment with a monochromatic light, maximum intensity is 4 times the minimum intensity in the interference pattern. What is the ratio of the intensities of the two interfering waves?

Options:
A $1 / 9$
B $1 / 3$
C $1 / 16$
D $1 / 2$

The correct answer is:
$1 / 9$

Solution:
$\left(\sqrt{I_1}+\sqrt{I_2}\right)^2=4\left(\sqrt{I_1}-\sqrt{I_2}\right)^2$
$\sqrt{l_1}+\sqrt{I_2}=2\left(\sqrt{l_1}-\sqrt{I_2}\right), 3 \sqrt{I_2}=\sqrt{I_1}, \frac{l_2}{I_1}=\frac{1}{9}$

Topic: Physics, Wave Optics, WBJEE, WBJEE 2022
In Young's experiment for the interference of light, the separation between the slits is $d$ and the distance of the screen from the slits is $D$. If $D$ is increased by $0.5 \%$ and $d$ is decreased by $0.3 \%$ then for the light of a given wavelength, which one of the following is true? "The fringe width.........

Options:
A increases by $0.8 \%$
B decreases by $0.8 \%$
C increases by $0.2 \%$
D decreases by $0.2 \%$

The correct answer is:
increases by $0.8 \%$

Solution:
As we know that, fringe width, $\beta=\frac{\lambda D}{d}$
where,
$\lambda=$ wavelength of light
$D=$ Distance of the screen from the slits
$d=$ separation between the slits
Now, $\frac{\Delta \beta}{\beta} \times 100=\frac{\Delta D}{D} \times 100-\frac{\Delta d}{d} \times 100$
According to the question.
$$
\frac{\Delta \beta}{\beta} \times 100=0.5-(-0.3)=0.8 \%
$$
Hence, the fringe width increases by $0.2 \%$

Topic: Physics, Wave Optics, WBJEE, WBJEE 2019
Indicate the correct IUPAC name of the co-ordination compound shown in the figure.

Options:
A Cis-dichlorotetraminochromium (III) chloride
B Trans-dichlorotetraminochromium (III) chloride
C Trans-tetraminedichlorochromium (III) chloride
D Cis-tetraamminedichlorochromium (III) chloride

The correct answer is:
Cis-tetraamminedichlorochromium (III) chloride

Solution:
Hint:

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2020
Indicate the number of unpaired electrons in $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ and $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$

Options:
A $1 \quad 0$
B $5 \quad 6$
C $6 \quad 5$
D $0\quad 1$

The correct answer is:
$1 \quad 0$

Solution:
$\begin{array}{l}
\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \quad \mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \\
\mathrm{Fe}=[\mathrm{Ar}] 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{2} \\
\mathrm{Fe}^{+3} \rightarrow(\mathrm{Ar}) 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{0} 4 \mathrm{p}^{0} \quad \mathrm{Fe}^{+2} \rightarrow(\mathrm{Ar}) 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{0} 4 \mathrm{p}^{0}
\end{array}$
As $\mathrm{CN}$ is the strong field ligand; we pair up

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2021
Indicate the products $(\mathrm{X})$ and $(\mathrm{Y})$ in the following reactions:
$\begin{array}{lr}
\mathrm{Na}_{2} \mathrm{~S}+\mathrm{nS}(\mathrm{n}=1-8) & \rightarrow(\mathrm{X}) \\
\mathrm{Na}_{2} \mathrm{SO}_{3}+\mathrm{S} & \rightarrow(\mathrm{Y})
\end{array}$
$\begin{array}{ll}(\mathrm{X}) & (\mathrm{Y})\end{array}$

Options:
A $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \quad \mathrm{Na}_{2} \mathrm{~S}_{2}$
B $\mathrm{Na}_{2} \mathrm{~S}_{(n+1)} \quad \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$
C $\mathrm{Na}_{2} \mathrm{~S}_{\mathrm{n}} \quad \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$
D $\mathrm{Na}_{2} \mathrm{~S}_{5} \quad \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{4}$

The correct answer is:
$\mathrm{Na}_{2} \mathrm{~S}_{(n+1)} \quad \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$

Solution:
$\begin{array}{l}
\mathrm{Na}_{2} \mathrm{~S}+\mathrm{nS} \longrightarrow \mathrm{Na}_{2} \mathrm{~S}_{(\mathrm{n}+1)} \\
\mathrm{Na}_{2} \mathrm{SO}_{3}+\mathrm{S} \longrightarrow \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}
\end{array}$

Topic: Chemistry, Redox Reactions, WBJEE, WBJEE 2021
\(\int 2^x\left(f^{\prime}(x)+f(x) \log 2\right) d x\) is

Options:
A \(2^x \mathrm{f}^{\prime}(\mathrm{x})+\mathrm{C}\)
B \(2^x f(x)+C\)
C \(2^x(\log 2) f(x)+C\)
D \((\log 2) f(x)+C\)

The correct answer is:
\(2^x f(x)+C\)

Solution:
Hints: \(I=\int 2^x f^{\prime}(x) d x+\int 2^x f(x) \log 2 d x\) \(=2^x f(x)\)

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2011
\(\int \frac{\cos 2 x}{\cos x} d x=\)

Options:
A \(2 \sin x+\log |\sec x+\tan x|+C\)
B \(2 \sin x-\log |\sec x-\tan x|+c\)
C \(2 \sin x-\log |\sec x+\tan x|+C\)
D \(2 \sin x+\log |\sec x-\tan x|+C\)

The correct answer is:
\(2 \sin x-\log |\sec x+\tan x|+C\)

Solution:
Hints: \(\int \frac{2 \cos ^2 x-1}{\cos x} d x=2 \sin x-\log |\sec x+\tan |\)

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2011