datavorous/entrance-exam-dataset huggingface dataset:
The equations of the lines through $(1,1)$ and making angles of $45^{\circ}$ with the line $x+y=0$ are

Options:
A $x-1=0, x-y=0$
B $x-y=0, y-1=0$
C $x+y-2=0, y-1=0$
D $\mathrm{x}-1=0, \mathrm{y}-1=0$

The correct answer is:
$\mathrm{x}-1=0, \mathrm{y}-1=0$

Solution:
Hints : $m=1, y-1=\frac{m \pm \tan 45}{1 \mp m \tan 45}(x-1), y-1=\frac{(-1) \pm 1}{1 \pm 1}(x-1)$
$y=1, x=1$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2010
The equations $x^{2}+x+a=0 \quad$ and
$x^{2}+a x+1=0$ have a common real root

Options:
A for no value of $a$
B for exactly one value of $a$
C for exactly two values of $a$
D for exactly three values of $a$

The correct answer is:
for exactly one value of $a$

Solution:
Let \(\alpha\) be the common roots
\(\begin{array}{l}
\therefore \alpha^{2}+\alpha+a=0 \ldots \text { (i) } \\
\text { and } \alpha^{2}+a \alpha+1=0 \ldots \text { (ii) } \\
\frac{\alpha^{2}}{1-a^{2}}=\frac{\alpha}{a-1}=\frac{1}{a-1}[a \neq 1]
\end{array}\)
Eliminating \(\alpha\), we get
\(\begin{array}{l}
(a-1)^{2}=\left(1-a^{2}\right)(a-1) \\
\Rightarrow a-1)=1-a^{2} \\
\Rightarrow a^{2}+a-2=0 \\
\Rightarrow a^{2}+2 a-a-2=0 \\
\Rightarrow a(a+2)-1(a+2)=0 \\
\Rightarrow(a+2)(a-1)=0 \\
\Rightarrow a=-2 \cdot[\because a \neq 1]
\end{array}\)

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2012
The equations $\mathrm{y}= \pm \sqrt{3 x}, \mathrm{y}=1$ are the sides of

Options:
A an equilateral triangle
B a right angled triangle
C an isosceles triangle
D an obtuse angled triangle

The correct answer is:
an equilateral triangle

Solution:
Hints: $y=\tan 60^{\circ} x, y=-\tan 60^{\circ} x$
$y=1$, equilateral

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2010
The equilibrium constant for the following reactions are given at $25^{\circ} \mathrm{C}$
$\begin{array}{l}
2 \mathrm{~A} \rightleftharpoons \mathrm{B}+\mathrm{C}, \mathrm{K}_{1}=1.0 \\
2 \mathrm{~B} \rightleftharpoons \mathrm{C}+\mathrm{D}, \mathrm{K}_{2}=16 \\
2 \mathrm{C}+\mathrm{D} \rightleftharpoons 2 \mathrm{P}, \mathrm{K}_{3}=25
\end{array}$
The equilibrium constant for the reaction $\mathrm{P} \rightleftharpoons \mathrm{A}+\frac{1}{2} \mathrm{~B}$ at $25^{\circ} \mathrm{C}$ is

Options:
A $\frac{1}{20}$
B 20
C $\frac{1}{42}$
D 21

The correct answer is:
$\frac{1}{20}$

Solution:
Hint:
We can manipulate the given equations as follows
$\frac{1}{2} \mathrm{~B}+\frac{1}{2} \mathrm{C} \rightleftharpoons \mathrm{A}, \quad \mathrm{K}_{1}^{\prime}=1$
$\frac{1}{2} \mathrm{C}+\frac{1}{2} \mathrm{D} \rightleftharpoons \mathrm{B}, \quad \mathrm{K}_{2}^{\prime}=\frac{1}{\sqrt{16}}$
$\mathrm{P} \rightleftharpoons \mathrm{C}+\frac{1}{2} \mathrm{D}, \quad \mathrm{K}_{\mathrm{3}}^{\prime}=\frac{1}{\sqrt{25}}$
$\mathrm{P} \rightleftharpoons \mathrm{A}+\frac{1}{2} \mathrm{~B}, \quad \mathrm{~K}_{\text {final }}=1 \times \frac{1}{4} \times \frac{1}{5}=\frac{1}{20}$

Topic: Chemistry, Chemical Equilibrium, WBJEE, WBJEE 2020
The equilibrium constant $(\mathrm{K})$ of a reaction may be written as :

Options:
A $K=e^{-\Delta G / R T}$
B $K=e^{-\Delta \mathrm{G}^0 / R T}$
C $K=e^{-\Delta H / R T}$
D $K=e^{-\Delta H^0 / R T}$

The correct answer is:
$K=e^{-\Delta \mathrm{G}^0 / R T}$

Solution:
$\Delta \mathrm{G}^{\mathrm{o}}=-\mathrm{RT} \ln \mathrm{K}$
$\Rightarrow \frac{\Delta \mathrm{G}^{\mathrm{o}}}{-\mathrm{RT}}=\ln \mathrm{K}$
$\mathrm{K}=e^{-\Delta \mathrm{G}^{\circ} / \mathrm{RT}}$

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2009
The equivalent conductance of \(\mathrm{NaCl}, \mathrm{HCl}\) and \(\mathrm{CH}_3 \mathrm{COONa}\) at infinite dilution are \(126.45,426.16\) and 91 \(\mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{eq}^{-1}\) respectively at \(25^{\circ} \mathrm{C}\). The equivalent conductance of acetic acid (at infinite dilution) would be

Options:
A \(461.61 \mathrm{~ohm}^{-1} \mathrm{~cm}^2 \mathrm{~eq}^{-1}\)
B \(390.71 \mathrm{~ohm}^{-1} \mathrm{~cm}^2 \mathrm{~eq}^{-1}\)
C cannot be determined from the given data
D \(208.71 \mathrm{~ohm}^{-1} \mathrm{~cm}^2 \mathrm{~eq}^{-1}\)

The correct answer is:
\(390.71 \mathrm{~ohm}^{-1} \mathrm{~cm}^2 \mathrm{~eq}^{-1}\)

Solution:
Hint : According to Kohlrausch's law
\(\begin{aligned}
& \wedge_{\mathrm{CH}_3 \mathrm{COOH}}^0=\wedge_{\mathrm{CH}_{3}\mathrm{COONa}}^0 +\wedge_{\mathrm{HCl}}^0-\wedge_{\mathrm{NaCl}}^0 \\
& \wedge_{\mathrm{CH}_3 \mathrm{COOH}}^0=(91+426.16-126.45) \mathrm{~ohm}^{-1} \mathrm{~cm}^2 \mathrm{~eq}^{-1} \\
& \wedge_{\mathrm{CH}_3 \mathrm{COOH}}^0=390.71 \mathrm{~ohm}^{-1} \mathrm{~cm}^2 \mathrm{~eq}^{-1}
\end{aligned}\)

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2023
The equivalent resistance between the points $a$ and $b$ of the electrical network shown in the figure is

Options:
A $6 r$
B $4 r$
C $2 r$
D $r$

The correct answer is:
$r$

Solution:
The equlvalent circuit for the given electrical network is

This is a balanced Wheatstone bridge, so the arm containing $2 r$ not in use. So, we have

So, the equivalent resistance between a and bis
$$
\frac{1}{r^{\prime}}=\frac{1}{2 r}+\frac{1}{2 r} \Rightarrow r^{\prime}=\frac{2 r}{2}=r
$$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2013
The equivalent weight of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ in acidic medium is expressed in terms of its molecular weight (M) as

Options:
A $M / 3$
B $M / 4$
C $M / 6$
D $M / 7$

The correct answer is:
$M / 6$

Solution:
In acidic medium,
$$
\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \longrightarrow \mathrm{Cr}^{3+}
$$
Equivalent weight of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}=\frac{\text { Molecular weight }}{\text { Change in oxidation state of } \mathrm{c}}$
$$
=\frac{M}{2(6-3)}=\frac{M}{6}
$$

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2012
The equivalent weight of \(\mathrm{KIO}_3\) in the given reaction is (\(\mathrm{M}=\) molecular mass) :
\(2 \mathrm{Cr}(\mathrm{OH})_3+4 \mathrm{OH}^{-}+\mathrm{KIO}_3 \rightarrow 2 \mathrm{CrO}_4{ }^{2-}+5 \mathrm{H}_2 \mathrm{O}+\mathrm{KI}\)

Options:
A \(\mathrm{M}\)
B \(\mathrm{M} / 2\)
C \(\mathrm{M} / 6\)
D \(\mathrm{M} / 8\)

The correct answer is:
\(\mathrm{M} / 6\)

Solution:
Hint : \(\mathrm{Cr}(\mathrm{OH})_3+4 \overline{\mathrm{O}} \mathrm{H}+\mathrm{K~} \mathrm{I}^{+5} \mathrm{O}_3 \longrightarrow 2 \mathrm{CrO}_4^{-2}+5 \mathrm{H}_2 \mathrm{O}+\mathrm{K~} \mathrm{I}^{-1}\)
Change in oxidation state of iodine \(=6\)
\(\therefore\) Equivalent weight of \(\mathrm{KIO}_3=\mathrm{M} / 6\)

Topic: Chemistry, Redox Reactions, WBJEE, WBJEE 2023
The even function of the following is

Options:
A \(f(x)=\frac{a^x+a^{-x}}{a^x-a^{-x}}\)
B \(f(x)=\frac{a^x+1}{a^x-1}\)
C \(f(x)=x \cdot \frac{a^x-1}{a^x+1}\)
D \(\mathrm{f}(\mathrm{x})=\log _2\left(\mathrm{x}+\sqrt{\mathrm{x}^2+1}\right)\)

The correct answer is:
\(f(x)=x \cdot \frac{a^x-1}{a^x+1}\)

Solution:
Hints: \(f(-x)=(-x) \frac{a^{-x}-1}{a^{-x}+1}\)
\(\begin{aligned}
& =(-n) \frac{1-a^x}{1+a^x} \\
& =n \frac{\left(a^x-1\right)}{\left(a^x+1\right)}=f(x)
\end{aligned}\)

Topic: Mathematics, Functions, WBJEE, WBJEE 2011
The exact order of acidity of the compounds p-nitrophenol, acetic acid, acetylene and ethanol is

Options:
A p-nitrophenol $ < $ acetic acid $ < $ acetylene $ < $ ethanol
B acetic acid $ < \mathrm{p}-$ nitrophenol $ < $ acetylene $ < $ ethanol
C acetylene $ < \mathrm{p}$-nitrophenol $ < $ ethanol $ < $ acetic acid
D acetylene $ < $ ethanol $ < $ p-nitrophenol $ < $ acetic acid

The correct answer is:
acetylene $ < $ ethanol $ < $ p-nitrophenol $ < $ acetic acid

Solution:
Acetic acid is most acidic as its conjugte base has equivalent Resonating structure followed by para-nitrophenol. Ethanol is less acidic then para-nitrophenol as conjugate base of p-nitrophenol is resonance stablized but not ethoxide $\left(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{O}\right)$ Also electronegativity of carbon is lesser than oxygen, hence acetylene is less acidic.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2021
The exact order of boiling points of the compounds $\mathrm{n}$-pentane, isopentane, butanone and 1 -butanol is

Options:
A n-pentane $ < $ isopentane $ < $ butanone $ < 1$-butanol
B isopentane $ < \mathrm{n}$-pentane $ < $ butanone $ < 1$-butanol
C butanone $ < \mathrm{n}$-pentane $ < $ isopentane $ < 1$-butanol
D 1-butanol $ < $ butanone $ < \mathrm{n}$-pentane $ < $ isopentane

The correct answer is:
isopentane $ < \mathrm{n}$-pentane $ < $ butanone $ < 1$-butanol

Solution:
As intermolecular force varies as:
H bonding > dipole dipole force > vander waals force
$\therefore$ boiling point order
$\mathrm{n}$-butanol $>$ butanone $>\mathrm{n}$-Pentane $>$ iso pentane

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2021
The expression $\frac{(1+i)^{n}}{(1-i)^{n-2}}$ equals

Options:
A $-i^{n+1}$
B $i^{n+1}$
C $-2 i^{n+1}$
D 1

The correct answer is:
$-2 i^{n+1}$

Solution:
$(\mathrm{c}) \frac{(1+i)^{n}}{(1-i)^{n-2}}=\frac{(1+i)^{n}}{(1-i)^{n}(1-i)^{-2}}$
$$
\begin{aligned}
&=\left(\frac{1+i}{1-i}\right)^{n}(1-i)^{2} \\
=&\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{n}\left(1+i^{2}-2 i\right) \\
=&\left(\frac{1+2 i+i^{2}}{1-i^{2}}\right)^{n}\left(1+i^{2}-2 i\right) \\
=&\left(\frac{1+2 i-1}{1-(-1)}\right)^{n}(1-1-2 i) \\
=&\left(\frac{2 i}{2}\right)^{n}(-2 i) \\
=& i^{n}(-2 i)=-2^{n+1}
\end{aligned}
$$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2017
The expression $\bar{A}(A+B)+(B+A A)(A+\bar{B})$ simplifies to

Options:
A $A+B$
B $A B$
C $\overline{\mathrm{A}+\mathrm{B}}$
D $\bar{A}+\bar{B}$

The correct answer is:
$A+B$

Solution:
$\begin{aligned} & \bar{A}(A+B)+(B+A A)(A+\bar{B}) \\=& \bar{A} A+\bar{A} B+(B+A)(A+\bar{B}) \\=& O+\bar{A} B+A B+A+B \bar{B}+A \bar{B} \\=&(\bar{A}+A) B+A+O+A \bar{B} \\=& 1 . B+A(1+\bar{B}) \\=& A+B \end{aligned}$

Topic: Physics, Mathematics in Physics, WBJEE, WBJEE 2022
The expression $a x^{2}+b x+c(a, b$ and $c$ are real $)$ has the same sign as that of a for all $x$ if

Options:
A $b^{2}-4 a c>0$
B $b^{2}-4 a c \neq 0$
C $b^{2}-4 a c \leq 0$
D $\mathrm{b}$ and $\mathrm{c}$ have the same sign as that of $\mathrm{a}$

The correct answer is:
$b^{2}-4 a c \leq 0$

Solution:
Hint:
C-l: If $a>0, a x^{2}+b x+c>0, b^{2}-4 a c < 0, \quad C-I I:$ If $a \leq 0, a x^{2}+b x+c \leq 0, D \leq 0$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2020
The expression \(\frac{\int_0^n[x] d x}{\int_0^n\{x\} d x}\), where \([x]\) and \(\{x\}\) are respectively integral and fractional part of \(x\) and \(n \in \mathbb{N}\), is equal to

Options:
A \(\frac{1}{n-1}\)
B \(\frac{1}{n}\)
C \(n\)
D \(n-1\)

The correct answer is:
\(n-1\)

Solution:
Hint : \(\frac{\int_0^n[x] d x}{\int_0^n\{x\} d x}=\frac{I_1}{I_2}\)
\(\begin{aligned} & I_1=\int_0^1[x] d x+\int_1^2[x] d x+\ldots+\int_{n-1}^n[x] d x \\ & =0+1+\ldots .+(n-1)=\frac{n(n-1)}{2} \\ & I_2=\int_0^n\{x\} d x=n \int_0^1\{x\} d x=n\left[\frac{x^2}{2}\right]_0^1=\frac{n}{2} \\ & \therefore \frac{I_1}{I_2}=\frac{n(n-1)}{n}=n-1\end{aligned}\)

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2023
The family of curves \(y=e^{a \sin x}\), where ' \(a\) ' is arbitrary constant, is represented by the differential equation

Options:
A \(y \log y=\tan x \frac{d y}{d x}\)
B \(y \log x=\cot x \frac{d y}{d x}\)
C \(\log y=\tan x \frac{d y}{d x}\)
D \(\log y=\cot x \frac{d y}{d x}\)

The correct answer is:
\(y \log y=\tan x \frac{d y}{d x}\)

Solution:
\(\begin{aligned} & \Rightarrow \log y=a \sin x \\ & \Rightarrow \frac{d}{d x}\left(\frac{\log y}{\sin x}\right)=\frac{d}{d x}(a) \Rightarrow \frac{\sin x\left(\frac{d y}{d x}\right)}{y}-\log y \times \cos x=0 \\ & \Rightarrow y \log y=\tan x \times \frac{d y}{d x}\end{aligned}\)

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2023
The figure represents two equipotential lines in \(\mathrm{x}-\mathrm{y}\) plane for an electric field. The \(\mathrm{x}\)-component \(\mathrm{E}_{\mathrm{x}}\) of the electric field in space between these equipotential lines is,

Options:
A \(100 \mathrm{~V} / \mathrm{m}\)
B \(-100 \mathrm{~V} / \mathrm{m}\)
C \(200 \mathrm{~V} / \mathrm{m}\)
D \(-200 \mathrm{~V} / \mathrm{m}\)

The correct answer is:
\(-100 \mathrm{~V} / \mathrm{m}\)

Solution:
Hint : \(E_{\mathrm{x}}=-\frac{d v}{d x}=-\left(\frac{4-2}{2}\right)=-1 \mathrm{~V} / \mathrm{cm}=-100 \mathrm{~V} / \mathrm{m}\)

Topic: Physics, Electrostatics, WBJEE, WBJEE 2023
The figure shows two identical parallel plate capacitors \(A\) and \(B\) of capacitances \(C\) connected to a battery. The key \(\mathrm{K}\) is initially closed. The switch is now opened and the free spaces between the plates of the capacitors are filled with a dielectric constant 3. Then which of the following statement (s) is/are true?

Options:
A Which the switch is closed, total energy stored in the two capacitors is \(\mathrm{CV}^2\)
B When the switch is opened, no charge is stored in the capacitor B
C When the switch is opened, energy stored in the capacitor \(\mathrm{B}\) is \(\frac{3}{2} \mathrm{CV}^2\)
D When the switch is opened, total energy stored in two capacitors is \(\frac{5}{2} \mathrm{CV}^2\)

The correct answers are:
Which the switch is closed, total energy stored in the two capacitors is \(\mathrm{CV}^2\), When the switch is opened, total energy stored in two capacitors is \(\frac{5}{2} \mathrm{CV}^2\)

Solution:
Hint :
\(\begin{aligned} & \mathrm{U}=\frac{1}{2} \times 2 \mathrm{CV}^2 \\ & =\mathrm{CV}^2\end{aligned}\)
\(\begin{aligned} & \mathrm{U}_{\mathrm{A}}=\frac{1}{2}(\mathrm{KC}) \mathrm{V}^2=\frac{3}{2} \mathrm{CV}^2 \\ & \mathrm{U}_B=\frac{\mathrm{q}^2}{2 \mathrm{KC}}=\frac{[\mathrm{CV}]^2}{2 \mathrm{KC}}=\frac{\mathrm{CV}^2}{2 \mathrm{~K}}=\frac{1}{6} \mathrm{CV}^2\end{aligned}\)
\(\underline{\text{Total energy when switch is open}}\)
\(\begin{aligned}
& \mathrm{U}=\frac{1}{2} \mathrm{KCV}^2+\frac{1}{6} \mathrm{CV}^2 \\
& =\frac{3}{2} \mathrm{CV}^2+\frac{1}{6} \mathrm{CV}^2 \\
& =\frac{10}{6} \mathrm{CV}^2=\frac{5}{3} \mathrm{CV}^2
\end{aligned}\)

Topic: Physics, Capacitance, WBJEE, WBJEE 2023
The first electron affinity of $\mathrm{C}, \mathrm{N}$ and $\mathrm{O}$ will be of the order

Options:
A $C < N < O$
B $N < C < O$
C $C < O < N$
D $\mathrm{O} < \mathrm{N} < \mathrm{C}$

The correct answer is:
$N < C < O$

Solution:
Key Point
(i) Half-filled/fully-filled configuration have (-) ve or low electron affinity i.e., (+) ve or high electron gain enthalpy.
(ii) Smaller the size, more is value of electron affinity, i.e. more $(-)$ ve will be the electron gain enthalpy.
On moving across a period, the size of atoms decreases due to increase in nuclear charge.
(i) Electron configuration of the elements are shown below.
$C(Z=6)=1 s^{2} 2 s^{2} 2 p^{2}$
$\mathrm{N}(Z=7)=1 s^{2} 2 s^{2} 2 p^{3}$
$\mathrm{O}(Z=8)=1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 p^{4}$
(ii) Due to half-filled electronic configuration of nitrogen (i.e. 3 electrons in $3 p$ orbitals), it is highly stable and has $(+)$ ve value of electron gain enthalpy (i.e. about $30.9 \mathrm{kJ} / \mathrm{mol}$ ).
(iii) As size of ' $\mathrm{O}^{\prime},$ is smaller than that of ${ }^{\prime} \mathrm{C}^{\prime},{ }^{\prime} \mathrm{O}^{\prime}$
atom has higher $(-)$ ve electron gain enthalpy (about $-141.1 \mathrm{kJ} / \mathrm{mol}$ ) than that of carbon (C-atom) [about $-122.3 \mathrm{kJ} / \mathrm{mol}]$.
Hence, correct order is $\mathrm{O}>\mathrm{C}>\mathrm{N}$ and $(\mathrm{b})$ is the correct option.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, WBJEE, WBJEE 2019
The focus of the conic $x^{2}-6 x+4 y+1=0$ is

Options:
A (2,3)
B (3,2)
C (3,1)
D (1,4)

The correct answers are:
(3,1)

Solution:
We have,
$$
x^{2}-6 x+4 y+1=0
$$
$\Rightarrow \quad(x-3)^{2}-9+4 y+1=0$
$\Rightarrow \quad(x-3)^{2}+4 y-8=0$
$\Rightarrow \quad(x-3)^{2}=-4(y-2)$
It represents parabola whose vertex is (3,2)
$\therefore$ Focus $=(3,-1+2)=(3,1)$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2017
The following equilibrium constants are given
$\begin{array}{l}
\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} ; K_{1} \\
\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO} ; K_{2}
\end{array}$
$\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O} ; K_{3}$
The equilibrium constant for the oxidation of 2 mole of $\mathrm{NH}_{3}$ to give NO is

Options:
A $K_{1} \cdot \frac{K_{2}}{K_{3}}$
B $K_{2} \cdot \frac{K_{3}^{3}}{K_{1}}$
C $K_{2}, \frac{K_{2}^{2}}{K_{1}}$
D $K_{2}^{2} \frac{K_{3}}{K_{1}}$

The correct answer is:
$K_{2} \cdot \frac{K_{3}^{3}}{K_{1}}$

Solution:
$\begin{array}{l}
\text {Given, } \\
\begin{aligned}
\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} ; K_{1} &=\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}} \\
\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons & 2 \mathrm{NO} ; K_{2}=\frac{[\mathrm{NO}]^{2}}{\left[\mathrm{N}_{2}\right]\left[\mathrm{O}_{2}\right]} \\
\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons & \mathrm{H}_{2} \mathrm{O} ; K_{3}=\frac{\left[\mathrm{H}_{2} \mathrm{O}\right]}{\left[\mathrm{H}_{2}\right]\left[\mathrm{O}_{2}\right]^{1 / 2}}
\end{aligned}
\end{array}$
The chemical equation for the oxidation of 2 mol of
$\mathrm{NH}_{3}$ to give NO is
$2 \mathrm{NH}_{3}+\frac{5}{2} \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}+3 \mathrm{H}_{2} \mathrm{O}$
(ii) and To get the equation (iv) from equation (i),
(iii) following steps are followed:
Reversing equation (i), we get
$2 \mathrm{NH}_{3} \rightleftharpoons \mathrm{N}_{2}+3 \mathrm{H}_{2} \text { so } K^{\prime}=\frac{1}{K_{1}}$
Multiplying cquation (iii) by 3, we get
$3 \mathrm{H}_{2}+\frac{3}{2} \mathrm{O}_{2} \rightleftharpoons 3 \mathrm{H}_{2} \mathrm{O} \mathrm{so} \mathrm{K}^{\prime}=K_{3}^{3}$
Adding equation (ii) and (vi)
$\mathrm{N}_{2}+3 \mathrm{H}_{2}+\frac{5}{2} \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}+3 \mathrm{H}_{2} \mathrm{O}$
so, $K^{\prime}=K_{2} \cdot K_{3}^{3}$
On combining (v) and (vii), we get the required equation having equilibrium constant $K^{\prime}=K_{2} \cdot \frac{K_{3}^{3}}{K_{1}}$

Topic: Chemistry, Chemical Equilibrium, WBJEE, WBJEE 2018
The following graph shows how $\mathrm{T}_{1 / 2}$ (half-life) of a reactant $\mathrm{R}$ changes with the initial reactant concentration $a_0$.
The order of the reaction will be :

Options:
A $0$
B $1$
C $2$
D $3$

The correct answer is:
$2$

Solution:
$t_{1 / 2} \propto \frac{1}{a^{n-1}}$
Hence (C)

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2009
The foot of the perpendicular drawn from the point (1,8,4) on the line joining the point (0,-11,4) and (2,-3,1) is

Options:
A (4,5,2)
B (-4,5,2)
C (4,-5,2)
D (4,5,-2)

The correct answer is:
(4,5,-2)

Solution:
Equation of line joining points (0,-11,4) and (2,-3,1) is
$$
\frac{x-2}{2}=\frac{y+3}{8}=\frac{z-1}{-3}=\lambda
$$
Let $P$ is any point of the above line then coordinate of $P$ is $(2 \lambda+2,8 \lambda-3,-3 \lambda+1)$
$\therefore \mathrm{DR}^{\prime} \mathrm{s}$ of $P Q$ is $(2 \lambda+1,8 \lambda-11,-3 \lambda-3$
Now, $(2 \lambda+1)(4+(8 \lambda-11)(8)+(-3 \lambda-3) (-3)=0$
$\Rightarrow \quad 4 \lambda+2+64 \lambda-88+9 \lambda+9=0$
$77 \lambda-77=0$

$\therefore$ Required foot of perpendicular.
$$
P(4,5,-2)
$$

Topic: Mathematics, Three Dimensional Geometry, WBJEE, WBJEE 2018
The force $\mathrm{F}$ acting on a particle moving in a straight line is shown below. What is the work done by the force on the particle in the $1^{\text {st }}$ meter of the trajectory?

Options:
A $5 \mathrm{~J}$
B $10 \mathrm{~J}$
C $15 \mathrm{~J}$
D $2.5 \mathrm{~J}$

The correct answer is:
$2.5 \mathrm{~J}$

Solution:
Work done in 1 meter $=$ area of shaded curve $=1 / 2 \times 1 \times 5=2.5 \mathrm{~J}$

Topic: Physics, Work Power Energy, WBJEE, WBJEE 2009
The frequency of the first overtone of a closed pipe of length $l_{1}$ is equal to that of the first. overtone of an open pipe of length $l_{2}$. The ratio of their lengths $\left(l_{1}: l_{2}\right)$ is

Options:
A $2: 3$
B $4: 5$
C $3: 5$
D $3: 4$

The correct answer is:
$3: 4$

Solution:
Here, $\quad v_{1}=\frac{(2 n-1)}{4 l_{1}} v$
$$
v_{2}=\frac{n}{2 l_{2}} v
$$
Hence, $y_{1}=v_{2}$
$$
\frac{l_{1}}{l_{2}}=\frac{3}{4}
$$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2012
The frequency $v$ of the radiation emitted by an atom when an electron jumps from one orbit to another is given by $=\mathrm{k} \delta \mathrm{E}$, where $\mathrm{k}$ is a constant and $\delta \mathrm{E}$ is the change in energy level due to the transition. Then dimension of $\mathrm{k}$ is

Options:
A $\mathrm{ML}^{2} \mathrm{~T}^{-2}$
B the same dimension of angular momentum
C $\mathrm{ML}^{2} \mathrm{~T}^{-1}$
D $\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}$

The correct answer is:
$\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}$

Solution:
Hint:
$v=k \delta E$
$[k]=\left[\frac{v}{\delta E}\right]=\frac{\left[T^{-1}\right]}{\left[M L^{2} T^{-2}\right]}$
$[k]=\left[M^{-} L^{-2} T^{1}\right]$

Topic: Physics, Units and Dimensions, WBJEE, WBJEE 2020
The function $f(x)=\frac{\tan \left\{\pi\left[x-\frac{\pi}{2}\right]\right\}}{2+[x]^{2}},$ where $[x]$ denotes the greatest integer $\leq x$, is

Options:
A continuous for all values of $x$
B discontinuous at $x=\frac{\pi}{2}$
C not differentiable for some values of $x$
D discontinuous at $x=-2$

The correct answer is:
continuous for all values of $x$

Solution:
Given, $f(x)=\frac{\tan \left\{\pi\left[x-\frac{\pi}{2}\right]\right\}}{2+[x]^{2}}$
Since. $\left[x-\frac{\pi}{2}\right]$ is an integer for all $x$, therefore $\pi\left[x-\frac{\pi}{2}\right]$ is an integral multiple of $\pi$ for all $x$. Hence, $\tan \left\{\pi\left[x-\frac{\pi}{2}\right]\right\}=0$ for all $x$
Also, $2+[x]^{2} \neq 0$ for all $x$
Hence, $f(x)=0$ for all $x$ Hence, $f(x)$ is continuous and derivable for all $x$.

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2014
The function $f(x)=a \sin |x|+b e^{| x \mid} \quad$ is differentiable at $x=0$ when

Options:
A $3 a+b=0$
B $3 a-b=0$
C $a+b=0$
D $a-b=0$

The correct answer is:
$a+b=0$

Solution:
Given, $f(x)=a \sin |x|+b e^{|x|}$
We know that $\sin |x|$ and $e^{|x|}$ is not
differentiable at $x=0$.
Therefore, for $f(x)$ to differentiable at $x=0$, we
must have $a=b=0$.
$\therefore$
$$
a+b=0
$$

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2014
The function $\mathrm{f}(\mathrm{x})=\sec \left[\log \left(\mathrm{x}+\sqrt{1+\mathrm{x}^2}\right)\right]$ is

Options:
A odd
B even
C neither odd nor even
D constant

The correct answer is:
even

Solution:
Hints : $\mathrm{f}(\mathrm{x})=\sec \left(\ell \ln \left(\mathrm{x}+\sqrt{1+\mathrm{x}^2}\right)\right)=\sec ($ odd function $)=$ even function
$\because$ sec is an even function

Topic: Mathematics, Functions, WBJEE, WBJEE 2010
The function $f(x)=x^{2}+b x+c,$ where $b$ and $c$ real constants, describes

Options:
A one-to-one mapping
B onto mapping
C not one-to-one but onto mapping
D neither one-to-one nor onto mapping

The correct answer is:
neither one-to-one nor onto mapping

Solution:
Given function is
$$
f(x)=x^{2}+b x+c
$$
It is a quadratic equation in $x$. So, we will get a parabola either downward or upward.

Hence, it is a many one mapping and not onto mapping. Hence, it is neither one-to-one nor onto mapping.

Topic: Mathematics, Functions, WBJEE, WBJEE 2014
The function \(f(x)=a x+b\) is strictly increasing for all real \(x\) if

Options:
A a \( > 0\)
B a \( < 0\)
C \(a=0\)
D \(\mathrm{a} \leq 0\)

The correct answer is:
a \( > 0\)

Solution:
Hints : \(f^{\prime}(x)=a\)
\(f^{\prime}(x) > 0 \Rightarrow a > 0\)

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2011
The function \(y=e^{k x}\) satisfies \(\left(\frac{d^2 y}{d x^2}+\frac{d y}{d x}\right)\left(\frac{d y}{d x}-y\right)=y \frac{d y}{d x}\). It is valid for

Options:
A exactly one value of \(k\)
B two distinct values of \(k\)
C three distinct values of \(k\)
D infinitely many values of \(k\)

The correct answer is:
three distinct values of \(k\)

Solution:
Hint : \(y=e^{k x}\)
\(\Rightarrow \frac{d y}{d x}=k e^{k x}=k y \Rightarrow \frac{d^2 y}{d x^2}=k^2 e^{k x}=k^2 y\)
Now LHS
\(=\left(\frac{d^2 y}{d x^2}+\frac{d y}{d x}\right)\left(\frac{d y}{d x}-y\right)=\left(k^2 y+k y\right)(k y-y)=k(k+1)(k-1) y^2=k\left(k^2-1\right) y^2\)
Now RHS \(y \frac{d y}{d x}=k y^2, A / q~ k\left(k^2-1\right) y^2=k y^2 \Rightarrow k\left[k^2-2\right]=0 \Rightarrow k=0, k= \pm \sqrt{2}\)

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2023