datavorous/entrance-exam-dataset huggingface dataset:
In a third order matrix $A$, $a_{0}$ denotes the element in the ith row and $j$ th column. If $a_{i j}=0$ for $i=j,1$ for $i>j,-1$ for $i < j$ Then the matrix is

Options:
A skew symmetric
B symmatric
C not invertible
D non-singular

The correct answers are:
skew symmetric, not invertible

Solution:
$$
A=\left[\begin{array}{ccc}
0 & -1 & -1 \\
1 & 0 & -1 \\
1 & 1 & 0
\end{array}\right]
$$
$\Rightarrow$
$A^{\prime}=\left[\begin{array}{ccc}0 & 1 & 1 \\ -1 & 0 & 1 \\ -1 & -1 & 0\end{array}\right]=-A$
$\Rightarrow A$ is a skew-symmetric matrix. $\begin{aligned}|A| &=0+1(0+1)-1(-0) \\ &=0+1-1=0=\text { singular } \end{aligned}$
$\Rightarrow A$ is not invertible.

Topic: Mathematics, Matrices, WBJEE, WBJEE 2018
In a transistor output characteristicy commonly used in common emitter configuration, the base current $I_{B}$, the collector current $I_{C}$ and the collector-emitter voltage $V_{C E}$ have values of the following orders of magnitude in the active region

Options:
A $I_{B}$ and $I_C$ both are in $\mu \mathrm{A}$ and $V_{C E}$ in volt
B $I_{B}$ is in $\mathrm{\mu A}$ and $\mathrm{I_C}$ is in $\mathrm{mA}$ and $\mathrm{V}_{\mathrm{CE}}$ in volt
C $I_{B}$ is in $\mathrm{mA}$ and $I_C$ is in $\mu \mathrm{A}$ and $\mathrm{V}_{C E}$ in $\mathrm{mV}$
D $I_{B}$ is in $\mathrm{mA}$ and $I_C$ is in $\mathrm{m} \mathrm{A}$ and $\mathrm{V}_{C E}$ in $\mathrm{mV}$

The correct answer is:
$I_{B}$ is in $\mathrm{\mu A}$ and $\mathrm{I_C}$ is in $\mathrm{mA}$ and $\mathrm{V}_{\mathrm{CE}}$ in volt

Solution:
$I_{g}$ is in $\mu$ amp. $I_{c}$ is in $\mathrm{m}$ amp and $V_{CE}$ in volt.

Topic: Physics, Semiconductors, WBJEE, WBJEE 2014
In a triangle $A B C$, the sides $A B$ and $A C$ are represented by the vectors $3 \mathrm{i}+\mathrm{j}+\mathrm{k}$ and $\mathrm{i}+2 \mathrm{j}+\mathrm{k}$, respectively. Calculate the angle $\angle A B C$

Options:
A $\cos ^{-1} \sqrt{\frac{5}{11}}$
B $\cos ^{-1} \sqrt{\frac{6}{11}}$
C $\left(90^{\circ}-\cos ^{-1} \sqrt{\frac{5}{11}}\right)$
D $\left(180^{\circ}-\cos ^{-1} \sqrt{\frac{5}{11}}\right)$

The correct answer is:
$\cos ^{-1} \sqrt{\frac{5}{11}}$

Solution:
$A B=3 i+j+\hat{k}$ and $A C=i+2 j+i$
$8 A=-(3 i+j+k)$
$\angle A B C$ is angle between $B A$ and $B C$
$B C=A C+B A$
$=A C-A B$
$=\hat{i}+2 \hat{j}+\hat{k}-(3 \hat{i}+\hat{j}+\hat{k})=-2 \hat{i}+\hat{j}$
$B A \cdot B C=|B A \| B C| \cos \theta$
$-(3 \hat{i}+\hat{j}+\hat{k}) \cdot(-2 \hat{i}+\hat{j})$
$=\left(\sqrt{3^{2}+1^{2}+1^{2}}\right) \cdot\left(\sqrt{2^{2}+1^{2}}\right) \cos \theta$
$\Rightarrow \quad 6-1=\sqrt{11 \times 5} \cdot \cos \theta$
$\cos \theta=\frac{5}{\sqrt{55}}=\sqrt{\frac{5}{11}}=\theta=\cos ^{-1}=\sqrt{\frac{5}{11}}$

Topic: Physics, Mathematics in Physics, WBJEE, WBJEE 2018
In a triangle $\mathrm{ABC}$ if $\sin \mathrm{A} \sin \mathrm{B}=\frac{\mathrm{ab}}{\mathrm{c}^2}$, then the triangle is

Options:
A equilateral
B isosceles
C right angled
D obtuse angled

The correct answer is:
right angled

Solution:
$$
\begin{aligned}
& \text { Hints }: \sin A \sin B=\frac{a b}{c^2} \\
& \Rightarrow c^2=\frac{a b}{\sin A \sin B}=\left(\frac{a}{\sin A}\right)\left(\frac{b}{\sin B}\right) \\
& \Rightarrow c^2=\left(\frac{c}{\sin C}\right)^2 \Rightarrow \sin ^2 C=1 \Rightarrow \sin C=1 \Rightarrow C=90^{\circ}
\end{aligned}
$$

Topic: Mathematics, Properties of Triangles, WBJEE, WBJEE 2009
In a triangle $\mathrm{PQR}, \angle \mathrm{R}=\pi / 2$. If $\tan \left(\frac{\mathrm{p}}{2}\right)$ and $\tan \left(\frac{\mathrm{Q}}{2}\right)$ are roots of $\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0$, where $\mathrm{a} \neq 0$, then which one is true ?

Options:
A $c=a+b$
B $\mathrm{a}=\mathrm{b}+\mathrm{c}$
C $\mathrm{b}=\mathrm{a}+\mathrm{c}$
D $\mathrm{b}=\mathrm{c}$

The correct answer is:
$c=a+b$

Solution:
Hints: $\frac{\mathrm{P}}{2}+\frac{\mathrm{Q}}{2}=\frac{\pi}{2}-\frac{\mathrm{P}}{2}=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$
$$
\begin{aligned}
& \tan \left(\frac{\rho}{2}+\frac{Q}{2}\right)=1, \frac{-b / a}{1-c / a}=1 \Rightarrow \frac{-b}{a-c}=1 \\
& -b=a-c \Rightarrow a+b=c
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2010
In a Young's double slit experiment, the intensity of light at a point on the screen where the path difference between the interfering waves is $\lambda$, ( $\lambda$ being the wavelength of light used) is $\mathrm{I}$. The intensity at a point where the path difference is $\frac{\lambda}{4}$ will be (assume two waves have same amplitude)

Options:
A zero
B I
C 0.5I
D 0.25I

The correct answer is:
0.5I

Solution:
$\mathrm{I}^{\prime}=\mathrm{I} \cos ^2 \phi / 2$
$\phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}, \quad \phi=\frac{2 \pi}{\lambda} \cdot \frac{\lambda}{4}=\frac{\pi}{2}$
$\frac{\phi}{2}=\frac{\lambda}{4}, \quad \mathrm{I}^{\prime}=\mathrm{I}^2 \cos ^2 \frac{\pi}{4}=\frac{\mathrm{I}}{2}$

Topic: Physics, Wave Optics, WBJEE, WBJEE 2022
In aluminium extraction by the Bayer process, alumina is extracted from bauxite by sodium hydroxide at high temperature and pressures :
$\mathrm{Al}_2 \mathrm{O}_3(\mathrm{~s})+2 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{Al}_2 \mathrm{O}_2(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(1)$
Solid impurities such as $\mathrm{Fe}_2 \mathrm{O}_3$ and $\mathrm{SiO}_2$ are removed and then $\mathrm{Al}(\mathrm{OH})_4^{-}$is reprecipitated : $2 \mathrm{Al}(\mathrm{OH})_4^{-} \rightarrow \mathrm{Al}_2 \mathrm{O}_3 \cdot 3 \mathrm{H}_2 \mathrm{O}(\mathrm{s})+2 \mathrm{OH}^{-}$(aq) . In the industrial world :

Options:
A Carbon dioxide is added to precipitate the alumina
B Temperature and pressure are dropped and the supersaturated solution seeded
C Both (1) and (2) are practised
D The water is evaporated

The correct answer is:
Temperature and pressure are dropped and the supersaturated solution seeded

Solution:
No Solution Available

Topic: Chemistry, General Principles and Processes of Isolation of Metals, WBJEE, WBJEE 2009
In an experiment, the length of an object is measured to be \(6.50 \mathrm{~cm}\). This measured value can be written as \(0.0650 \mathrm{~m}\). The number of significant figures on \(0.0650 \mathrm{~m}\) is

Options:
A 3
B 4
C 2
D 5

The correct answer is:
3

Solution:
Hint :

Topic: Physics, Mathematics in Physics, WBJEE, WBJEE 2023
In an inelastic collision an electron excites as hydrogen atom from its ground state to a M-shell state. A second electron collides instantaneously with the excited hydrogen atom in the M-State and ionizes it. At least how much energy the second electron transfers to the atom in the M-state?

Options:
A \(+3.4 \mathrm{eV}\)
B \(+1.51 \mathrm{eV}\)
C \(-3.4 \mathrm{eV}\)
D \(-1.51 \mathrm{eV}\)

The correct answer is:
\(+1.51 \mathrm{eV}\)

Solution:
Hints: \(E_m=-\frac{13.6}{(3)^2}=-1.51\)
Minimum energy required by electron should be \(+1.51 \mathrm{eV}\)

Topic: Physics, Atomic Physics, WBJEE, WBJEE 2011
In an $n-p-n$ transistor

Options:
A the emitter has higher degree of doping compared to that of the collector
B the collector has higher degree of doping compared to that of the emitter
C both the emitter and collector have same degree of doping
D the base region is most heavily doped

The correct answer is:
the emitter has higher degree of doping compared to that of the collector

Solution:
In an $n-p-n$ or $p-n-p$ transistor, the left hand side thick layer of the transistor is heavily doped known as emitter and right hand side thick layer of the transistor is moderately doped known as collector.

Topic: Physics, Semiconductors, WBJEE, WBJEE 2013
In aqueous alkaline solution, two electrons reduction of $\mathrm{HO}_{2}^{-}$ gives

Options:
A $\mathrm{HO}^{-}$
B $\mathrm{H}_{2} \mathrm{O}$
C $\mathrm{O}_{2}$
D $O_{2}^{-}$

The correct answer is:
$\mathrm{HO}^{-}$

Solution:
In aqueous alkaline solution, two electron reduction of $\mathrm{HO}_{2}^{-}$ gives OH The reaction is as follows:
$\begin{array}{l}
\mathrm{H}_{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HO}_{2}^{-} \\
\mathrm{HO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O}+2 e^{-} \rightleftharpoons 3 \mathrm{OH}
\end{array}$

Topic: Chemistry, Redox Reactions, WBJEE, WBJEE 2015
In basic medium the amount of $\mathrm{Ni}^{2+}$ in a solution can be estimated with the dimethylglyoxime reagent. The correct statement(s) about the reaction and the product is (are)

Options:
A in a ammoniacal solution $\mathrm{Ni}^{2+}$ salts give cherry-red precipitate of nickel(il) dimethylglyoximate
B two dimethylglyoximate units are bound to one $\mathrm{Ni}^{2 *}$
C in the complex two dimethylglyoximate units are hydrogen bonded to each other
D each dimethylglyoximate unit forms a six membered chelate ring with $\mathrm{Ni}^{2+}$

The correct answers are:
in a ammoniacal solution $\mathrm{Ni}^{2+}$ salts give cherry-red precipitate of nickel(il) dimethylglyoximate, two dimethylglyoximate units are bound to one $\mathrm{Ni}^{2 *}$, in the complex two dimethylglyoximate units are hydrogen bonded to each other, each dimethylglyoximate unit forms a six membered chelate ring with $\mathrm{Ni}^{2+}$

Solution:
Estimation of $\mathrm{Ni}^{2+}$ is carried out as.
Filtrate of group III $+\mathrm{NH}_{4} \mathrm{OH}+\mathrm{NH}_{4} \mathrm{Cl} \longrightarrow$
passing $\mathrm{H}_{2} \mathrm{S}$ gas $\rightarrow$ black pPt of NiS.
This black ppt of NiS is soluble in conc. HCl in presence of oxidising agent like $\mathrm{KClO}_{3}$ $\mathrm{NiS}+\underset{\mathrm{COnc}}{2 \mathrm{HCl}}+\mathrm{O} \longrightarrow \mathrm{NiCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{S}$
Now this $\mathrm{NiCl}_{2}$, in basic medium, treated with dimethyl glyoxime, cherry red ppt of nickel (II) dimethyl glyoximate is obtained, in which two dimethyl glyoximate units are hydrogen bonded to each other and each unit forms a six-membered chelate ring with $\mathrm{Ni}^{2+}$

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2013
In Bohr model of atom, radius of hydrogen atom in ground state is $\mathrm{r}_1$ and radius of $\mathrm{He}^{+}$ion in ground state is $\mathrm{r}_2$. Which of the following is correct?

Options:
A $\frac{r_1}{r_2}=4$
B $\frac{r_1}{r_2}=\frac{1}{2}$
C $\frac{r_2}{r_1}=\frac{1}{4}$
D $\frac{r_2}{r_1}=\frac{1}{2}$

The correct answer is:
$\frac{r_2}{r_1}=\frac{1}{2}$

Solution:
$r_n=a_0 \times \frac{n^2}{z}$
$\underline{\text {For H-atom}}$
$r_1=a_0$
$\underline{\text {Also, For}}$ $\underline {\mathrm{He}^{+}}$ - $\underline{\text { ion }}$
$r_2=a_0 \times \frac{1^2}{2}=\frac{r_1}{2}$
$\therefore \frac{r_2}{r_1}=\frac{1}{2}$

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2022
In borax the number of B $-\mathrm{O}-\mathrm{B}$ links and B - OH bonds present are, respectively

Options:
A five and four
B four and five
C three and four
D five and five

The correct answer is:
five and four

Solution:
$5 B-O-B bonds
4 B-O H bonds

Topic: Chemistry, p Block Elements (Group 13 & 14), WBJEE, WBJEE 2013
In case of a bipolar transistor $\beta=45$. The potential drop across the collector resistance $\begin{array}{llllll}\text { of } 1 \mathrm{k} \Omega & \text { is } 5 \mathrm{V} . & \text { The base } & \text { current } & \text { is }\end{array}$ approximately

Options:
A $222 \mu \mathrm{A}$
B $55 \mu \mathrm{A}$
C $111 \mu \mathrm{A}$
D $45 \mu \mathrm{A}$

The correct answer is:
$111 \mu \mathrm{A}$

Solution:
Current gain $\beta=45$
We know that, $\quad \beta=\frac{\Delta I_{c}}{\Delta I_{b}}$
$\therefore$
$$
\frac{\Delta I_{c}}{\Delta I_{b}}=45
$$
Potential drop across $1 \mathrm{k} \Omega$ or $1 \times 10^{3} \Omega$
Collector resistor is $5 \mathrm{V}$.
$$
\begin{aligned}
V &=I_{C} R \\
I_{C} &=\frac{V}{R}=\frac{5}{1 \times 10^{3}} \\
&=5 \times 10^{-3} \mathrm{A}
\end{aligned}
$$
From Eqs. (i) and (ii).
$$
\begin{aligned}
\Delta I_{B} &-\frac{\Delta I_{C}}{45} \\
&=\frac{5 \times 10^{-3}}{45}=\frac{1}{9} \times 10^{-3} \\
&=0.111 \times 10^{-3} \mathrm{A} \\
&=111 \times 10^{-6} \mathrm{A}=111 \mu \mathrm{A}
\end{aligned}
$$

Topic: Physics, Semiconductors, WBJEE, WBJEE 2016
In case of a simple harmonic motion, if the velocity is plotted along the $X$ -axis and the displacement (from the equilibrium position) is plotted along the $Y$ -axis, the resultant curve happens to be an ellipse with the ratio:
major axis (along X) $=20 \pi$ minor axis (along $Y)$
What is the frequency of the simple harmonic motion?

Options:
A $100 \mathrm{Hz}$
B $20 \mathrm{Hz}$
C $10 \mathrm{Hz}$
D $\frac{1}{10} \mathrm{Hz}$

The correct answer is:
$10 \mathrm{Hz}$

Solution:
In representation of simple harmonic motion in ellipse as velocity along $X$ -axis and displacement along $Y$ -axis. $\frac{\text { Major axis (along } X \text { -axis) }}{\text { Minor axis (along } X \text { -axis) }}=20 \pi$
$\Rightarrow \quad \frac{\omega A}{A}=20 \pi \Rightarrow \omega=20 \pi$
The frequency of the simple harmonic motion

Topic: Physics, Oscillations, WBJEE, WBJEE 2018
In case of heteronuclear diatomics of the type $A B,$ where $A$ is more electronegative than $B$, bonding molecular orbital resembles the character of $A$ more than that of $B$. The statement

Options:
A is false
B is true
C cannot be evaluated since data is not sufficient
D is true only for centain systems

The correct answer is:
is true

Solution:
The given statement is true as electrons are shifted more towards the more electronegative atom.

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2014
In Cu-ammonia complex, the state of hybridization of $\mathrm{Cu}^{+2}$ is

Options:
A $\mathrm{sp}^3$
B $\mathrm{d}^3 \mathrm{~s}$
C $\mathrm{sp}^2 \mathrm{f}$
D $\mathrm{dsp}^2$

The correct answer is:
$\mathrm{dsp}^2$

Solution:
In $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{+}$
$\mathrm{Cu}^{+2}$ is in a state of $d s p^2$ hybridization and shape of the complex is square planar. (One $e^{-}$is excited from $3 d$ to $4 p$ during complex formation)

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2009
In diborane, the number of electrons that accounts for bonding in the bridges is

Options:
A six
B two
C eight
D four

The correct answer is:
four

Solution:
In diborone, odd-electron bonds are found in the bridges. From figure, it is clear that 4 electrons are present for bonding in the bridges.

Topic: Chemistry, p Block Elements (Group 13 & 14), WBJEE, WBJEE 2013
In DNA, the consecutive deoxynucleotides are connected via

Options:
A phosphodiester linkage
B phosphomonoester linkage
C phosphotriester linkage
D amide linkage

The correct answer is:
phosphodiester linkage

Solution:
In DNA, the consecutive deoxynucleotides are linked through phosphodiester linkage.

Topic: Chemistry, Biomolecules, WBJEE, WBJEE 2014
In hemoglobin the metal ion present is

Options:
A $\mathrm{Fe}^{2+}$
B $\mathrm{Zn}^{2+}$
C $\mathrm{Co}^{2+}$
D $\mathrm{Cu}^{2+}$

The correct answer is:
$\mathrm{Fe}^{2+}$

Solution:
No Solution Available

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2010
In $\mathrm{O}_{2}$ and $\mathrm{H}_{2} \mathrm{O}_{2}$, the $\mathrm{O}-\mathrm{O}$ bond lengths are 1.21 Ã… and 1.48 Ã… respectively. In ozone, the average $O-O$ bond length is

Options:
A $1.28 Ã…$
B $1.18 Ã…$
C $1.44 Ã…$
D $1.52 Ã…$

The correct answer is:
$1.28 Ã…$

Solution:
Ozone shows following resonance structure

It contains bond angle $116.8^{\circ}$ and $O-O$ bond length $1.28 Ã…$

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2013
In open interval $\left(0, \frac{\pi}{2}\right)$,

Options:
A $\cos x+x \sin x < 1$
B $\cos x+x \sin x>1$
C no specific order relation can be ascertained between $\cos x+x \sin x$ and 1
D $\cos x+x \sin x < \frac{1}{2}$

The correct answer is:
$\cos x+x \sin x>1$

Solution:
Hint:
$f(x)=\cos x+x \sin x-1$
$\Rightarrow f^{\prime}(x)=-\sin x+\sin x+x \cos x>0 ; x \in\left(0, \frac{\pi}{2}\right)$
$\Rightarrow f(x)$ is increasing function
$\begin{array}{l}
\Rightarrow f(x)>f(0) \\
\cos x+x \sin x-1>0 \\
\cos x+x \sin x>1
\end{array}$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2020
In order to get a head at least once with probability $\geq 0.9$, the minimum number of times a unbiased coin needs to be tossed is

Options:
A 3
B 4
C 5
D 6

The correct answer is:
4

Solution:
We have, $P(H)=\frac{1}{2}, P(T)=\frac{1}{2}$
Now, $P(X \geq 1)=1-P(X=0$
$=1-{ }^{n} C_{o}(p)^{0}(q)^{n}=1-\left(\frac{1}{2}\right)^{n}$
$1-\frac{1}{2^{n}} \geq 0.9$
$\Rightarrow$
$0.1 \geq \frac{1}{2 n}$
$\Rightarrow$
$\frac{1}{10} \geq \frac{1}{2^{n}}$
$\Rightarrow$
$z^{n} \geq 10$
$\Rightarrow$
$n \geq 4$
Minimum number of tossed = 4

Topic: Mathematics, Probability, WBJEE, WBJEE 2018
In $\mathrm{SOCl}_{2}$, the $\mathrm{Cl} - \mathrm{S}-\mathrm{Cl}$ and $\mathrm{Cl}-\mathrm{S}-\mathrm{O}$
bond angles are

Options:
A 130 and $115^{\circ}$
B 106 and $96^{\circ}$
C 107 and $108^{\circ}$
D $96^{\circ}$ and $106^{\circ}$

The correct answer is:
$96^{\circ}$ and $106^{\circ}$

Solution:
In $\mathrm{SOCl}_{2}$, the $\mathrm{Cl}-\mathrm{S}-\mathrm{Cl}$ and $\mathrm{Cl}-\mathrm{S}-\mathrm{O}$ bond
angles are respectively $96^{\circ}$ and $106^{\circ}$

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2013
In Sommerfeld's modification of Bohr's theory, the trajectory of an electron in a hydrogen atom is

Options:
A a perfect ellipse
B a closed ellipse - like curve, narrower at the perihelion position and flatter at the aphelion position
C a closed loop on spherical surface
D a rosette

The correct answer is:
a closed loop on spherical surface

Solution:
No Solution Available

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2010
In the adjoining figure, the potential difference between $X$ and $Y$ is 60 V. The potential difference between the points $M$ and $N$ will be

Options:
A $10 \mathrm{V}$
B $15 \mathrm{V}$
C $20 \mathrm{V}$
D $30 \mathrm{V}$

The correct answer is:
$30 \mathrm{V}$

Solution:
Capacitors in series
\(2 C_{1} \times V_{1}=C \times V_{2}\)
\(2 V_{1}=V_{2} \ldots\) (i)
and \(V_{1}+V_{2}+V_{1}=60 \ldots\) (ii)
Solving Eqs. (i) and (ii), we get
\(V_{2}=30 \mathrm{~V}\)

Topic: Physics, Capacitance, WBJEE, WBJEE 2012
In the argand plane, the distinct roots of $1+z+z^{3}+z^{4}=0(z$ is a complex number) represent vertices of

Options:
A a square
B an equilateral triangle
C a thombus
D a rectangle

The correct answer is:
an equilateral triangle

Solution:
Given equation is
$1+z+z^{3}+z^{4}=0$
$\Rightarrow \quad(1+z)+z^{3}(1+z)=0$
$\Rightarrow \quad(1+z)\left(1+z^{3}\right)=1$
$\Rightarrow \quad z=-1, z^{3}=-1$
$\Rightarrow \quad z=-1, z=-1,-\omega-\omega^{2}$
Hence, these roots are the vertices of an equilateral triangle.

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2014
In the bandgap between valence band and conduction band in a material is $5.0 \mathrm{eV},$ then the material is

Options:
A semiconductor
B good conductor
C superconductor
D insulator

The correct answer is:
insulator

Solution:
The band gap of $5 \mathrm{eV}$ corresponds to that of an
insulator.

Topic: Physics, Semiconductors, WBJEE, WBJEE 2014
In the Bayer's process, the leaching of alumina is done by using

Options:
A $\mathrm{Na}_{2} \mathrm{CO}_{3}$
B NaOH
C $\mathrm{SiO}_{2}$
D $\mathrm{CaO}$

The correct answer is:
NaOH

Solution:
In the Bayer's process, the leaching of alumina is done by using $\mathrm{NaOH}$.
$\mathrm{Al}_{2} \mathrm{O}_{3}(s)+2 \mathrm{NaOH}(a q) \stackrel{150^{\circ} \mathrm{C}, 35 \mathrm{atm}}{\longrightarrow} 2 \mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_{4}\right]+3 \mathrm{H}_{2} \mathrm{O}$
$2 \mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_{4}\right]+\mathrm{CO}_{2} \rightarrow \mathrm{Al}_{2} \mathrm{O}_{3} \cdot x \mathrm{H}_{2} \mathrm{O}+2 \mathrm{NaHCO}_{3}$
$\mathrm{Al}_{2} \mathrm{O}_{3} \cdot x \mathrm{H}_{2} \mathrm{O}(s) \underset{\longrightarrow}{\text{1470K}} \mathrm{Al}_{2} \mathrm{O}_{3}(s)+x \mathrm{H}_{2} \mathrm{O}(g)$
So, the option (b) is correct answer.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, WBJEE, WBJEE 2019
In the brown ring complex $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{NO})\right] \mathrm{SO}_{4}$
nitric oxide behaves as

Options:
A $\mathrm{NO}^{+}$
B neutral NO molecule
C $\mathrm{NO}^{-}$
D $\mathrm{NO}^{2-}$

The correct answer is:
$\mathrm{NO}^{+}$

Solution:
In Fe-complexes, NO behaves as $\mathrm{NO}^{+}$.

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2012
In the circuit shown assume the diode to be ideal. When $V_{i}$ increases from $2 \mathrm{V}$ to $6 \mathrm{V},$ the change in the current is (in mA)

Options:
A zero
B 20
C $80 / 3$
D 40

The correct answer is:
20

Solution:
$I_{\text {intitial }}=0$,

$I_{\text {final }}=\frac{3}{150}=0.02 \mathrm{A}$
So, change in $I=0.02 \mathrm{A}=20 \mathrm{mA}$

Topic: Physics, Semiconductors, WBJEE, WBJEE 2014
In the circuit shown below, the switch is kept in position $a$ for a long time and is then thrown to position $b .$ The amplitude of the resulting oscillating current is given by

Options:
A $E \sqrt{L/C}$
B $E / R$
C intinity
D $E \sqrt{C/L}$

The correct answer is:
$E \sqrt{C/L}$

Solution:
When switch is in position $a$, then capacitor will be charged, such that Charge on capacitor, $q=C E$ Energy stored in capacitor, $U=q^{2} / 2 C$ When switch is in position $b$, then circuit becomes an $L C$ oscillator. Let amplitude of current in $L C$ circuit be $I_{o}$. From conservation of energy, maximum electrical energy = maximum magnetic energy $\frac{q^{2}}{2 C}=\frac{1}{2} L I_{0}^{2}$
$[\because q=C E]$
$\Rightarrow \quad \frac{C^{2} E^{2}}{C}=L I_{0}^{2}$
$\therefore \quad I_{0}=E \sqrt{\frac{C}{L}}$

Topic: Physics, Alternating Current, WBJEE, WBJEE 2015
In the circuit shown, inputs $A$ and $B$ are in states 1 and 0 respectively. What is the only possible stable state of the outputs $X$ and $Y ?$

Options:
A $X=1, Y=1$
B $X=1, Y=0$
C $X=0, Y=1$
D $X=0, Y=0$

The correct answer is:
$X=0, Y=1$

Solution:
$X=\overline{A Y}=\bar{Y}=0$
$\therefore$
$$
Y=\overline{B X}=1
$$

Topic: Physics, Semiconductors, WBJEE, WBJEE 2018
In the circuit shown, the value of $\beta$ of the transistor is 48 . If the base current supplied is $200 \mu \mathrm{A}$, what is the voltage at the terminal $\mathrm{Y}$?

Options:
A $0.2 \mathrm{~V}$
B $0.5 \mathrm{~V}$
C $4 \mathrm{~V}$
D $4.8 \mathrm{~V}$

The correct answer is:
$0.2 \mathrm{~V}$

Solution:
Hint:
$\beta=48$
$l_{B}=200 \mu \mathrm{A}$
$I_{C}=\beta I_{B}=48 \times 200 \times 10^{-6}$
$V_{\mathrm{CC}}=I_{C} R_{C}+V_{C E} \therefore V_{C E}=5-\left(96 \times 10^{-4}\right) \times 500=5-4.8=0.2$ volt

Topic: Physics, Semiconductors, WBJEE, WBJEE 2020
In the circuit shown, what will be the current through the $6 \mathrm{V}$ zener?

Options:
A $6 \mathrm{mA}$, from $\mathrm{A}$ to $\mathrm{B}$
B $2 \mathrm{mA},$ from $\mathrm{A}$ to $\mathrm{B}$
C $2 \mathrm{mA},$ from $B$ to $\mathrm{A}$
D Zero

The correct answer is:
Zero

Solution:
According to the question
Input voltage $V_{s}=10 \mathrm{V}$
Source resistance $R_{S}=1 \mathrm{k} \Omega$
Zener diode voltage $V_{Z}=6 \mathrm{V}$
$\because$ Breakdown voltage of zener diode is $=6 \mathrm{V},$ and the potential difference across the zener diode $5 \mathrm{V}$.
$\therefore$ Current flow in zener diode $I_{z}=0$

Topic: Physics, Semiconductors, WBJEE, WBJEE 2019
In the compound $\mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_2$ the hybridization of $\mathrm{C}-2$ and $\mathrm{C}-3$ carbons are respectively:

Options:
A $\mathrm{sp}^3 \& \mathrm{sp}^3$
B $\mathrm{sp}^2 \& \mathrm{sp}^3$
C $\mathrm{sp}^2 \& \mathrm{sp}$
D $\mathrm{sp}^3 \& \mathrm{sp}$

The correct answer is:
$\mathrm{sp}^2 \& \mathrm{sp}$

Solution:
Hints:
(Double bond is preferred)

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2009
In the crystalline solid $M \mathrm{SO}_{4} \cdot n \mathrm{H}_{2} \mathrm{O}$ of molar mass 250 g $mol^{-1}$ , the percentage of anhydrous salt is 64 by weight. The value of $n$ is

Options:
A 2
B 3
C 5
D 7

The correct answer is:
5

Solution:
$$
\begin{aligned}
&\text {Mass of } \mathrm{H}_{2} \mathrm{O}=\frac{36}{100} \times 250=90 \mathrm{g}\\
&\text { Moles of } \mathrm{H}_{2} \mathrm{O}=\frac{90}{18}=5 \mathrm{mol}
\end{aligned}
$$
In the crystalline solid $M \mathrm{SO}_{4} \cdot n \mathrm{H}_{2} \mathrm{O},$ the value of $n$ is 5
Hence, option (c) is correct answer.

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2019