datavorous/entrance-exam-dataset huggingface dataset:
For a reaction, A+B2+⟶B+A2+, at 250C E0=0.2955V. The value of Keq is

Options:
A 10
B 1010
C -10
D 10-10

The correct answer is:
1010

Solution:
Given that,

For a reaction,

A+B2+⟶B+A2+,E°=0.2955V

T=25°C, ⇒T=298 K n=2

Using Nernst equation,

E°=2.303RTnFlogkeq

E°=0.0591nlogkeqat 25oC

⇒ 0.2955=0.05912logkeq

keq=1010

Therefore, the value of Keq is 1010.

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2018
For a reversible reaction, A⇌B, which one of the following statement is wrong from the given energy profile diagram?

Options:
A Activation energy of forward reaction is greater than backward reaction
B The forward reaction is endothermic
C The threshold energy is less than that of activation energy
D The energy of activation of forward reaction is equal to the sum of heat of reaction and the energy of activation of backward reaction

The correct answer is:
The threshold energy is less than that of activation energy

Solution:
where, Ea= activation energy of forward reaction

Ea'= activation energy of backward reaction

The above energy profile diagram shows that Ea>Ea'

The potential energy of the product is greater than that of the reactant, so the reaction is endothermic.

Ea=Ea'+ΔE

Et=Ea or Et>Ea'

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2019
For a sparingly soluble salt $A_{p} B_{q}$, the relationship of its solubility product $L s \rightarrow K_{s p}$ with its solubility (S) is

Options:
A $\quad \mathrm{Ls} \rightarrow \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{\mathrm{pq}}(\mathrm{pq})^{\mathrm{P}+\mathrm{q}}$
B $\mathrm{Ls}=\mathrm{S}^{\mathrm{p}+\mathrm{q}} \cdot \mathrm{p}^{\mathrm{p}} \mathrm{q}^{\mathrm{q}}$
C $\quad \mathrm{Ls} \rightarrow \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{\mathrm{p}+\mathrm{q}} \cdot \mathrm{p}^{\mathrm{q}} \mathrm{q}^{\mathrm{p}}$
D $\quad \mathrm{Ls} \rightarrow \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{\mathrm{pq}} \mathrm{p}^{\mathrm{P}} \mathrm{q}^{\mathrm{q}}$

The correct answer is:
$\mathrm{Ls}=\mathrm{S}^{\mathrm{p}+\mathrm{q}} \cdot \mathrm{p}^{\mathrm{p}} \mathrm{q}^{\mathrm{q}}$

Solution:
$\mathrm{ApBq} \rightleftharpoons \mathrm{pA}_{\mathrm{Sp}}^{+}+\mathrm{qB}_{\mathrm{Sq}}$

Let the solubility be $\mathrm{S} \mathrm{mol} /$ liter Thus,

\begin{aligned}

\mathrm{K}_{\mathrm{sp}} &=\left[\mathrm{A}^{+}\right]^{\mathrm{p}}\left[\mathrm{B}^{-}\right]^{\mathrm{q}} \\

&=[\mathrm{Sp}]^{\mathrm{p}}[\mathrm{Sq}]^{\mathrm{q}}=\mathrm{p}^{\mathrm{p} \cdot} \mathrm{q}^{\mathrm{q}}(\mathrm{S})^{\mathrm{p}+\mathrm{q}}

\end{aligned}

Topic: Chemistry, Ionic Equilibrium, BITSAT, BITSAT 2016
For adsorption of a gas on a solid, the plot of log $x / m$ vs log $P$ is linear with slope equal to ( $n$ being whole number)

Options:
A $k$
B $\log k$
C $n$
D $\frac{1}{n}$

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

Solution:
According to Freundlich adsorption isotherm, at intermediate pressure, extent of adsorption
$$
\frac{x}{m}=k \mathrm{P}^{1 / n} \text { or } \log \frac{x}{m}=\log k+\frac{1}{n} \log \mathrm{P}
$$
plot of $\log \frac{x}{m}$ vs log $\mathrm{P}$ is linear with slope
$=\frac{1}{n}$

Topic: Chemistry, Surface Chemistry, BITSAT, BITSAT 2011
For all $n \in N$, the sum of $\frac{n^5}{5}+\frac{n^3}{3}+\frac{7 n}{15}$ is

Options:
A a negative integer
B a whole number
C a real number
D a natural number

The correct answer is:
a natural number

Solution:
Let the statement $\mathrm{P}(\mathrm{n})$ be defined as $P(n): \frac{n^5}{5}+\frac{n^3}{3}+\frac{7 n}{15}$ is a natural number for all $n \in N$ :
Step I : For $\mathrm{n}=1, \mathrm{P}(1): \frac{1}{5}+\frac{1}{3}+\frac{7}{15}=1 \in \mathrm{N}$
Hence, it is true for $\mathrm{n}=1$.
Step II : Let it is true for $n=k$,
i.e. $\frac{k^5}{5}+\frac{k^3}{3}+\frac{7 k}{15}=\lambda \in N$ $\ldots$ (i)
Step III : For $\mathrm{n}=\mathrm{k}+1$,
$\frac{(\mathrm{k}+1)^5}{5}+\frac{(\mathrm{k}+1)^3}{3}+\frac{7(\mathrm{k}+1)}{15}$
$=\frac{1}{5}\left(k^5+5 k^4+10 k^3+10 k^2+3 k+1\right)$ $+\frac{1}{3}\left(k^3+3 k^2+3 k+1\right)+\frac{7}{15} k+\frac{7}{15}$
$=\left(\frac{k^5}{5}+\frac{k^3}{3}+\frac{7}{15} k\right)+\left(k^4+2 k^3+3 k^2+2 k\right)$ $+\frac{1}{5}+\frac{1}{3}+\frac{7}{15}$
$=\lambda+\mathrm{k}^4+2 \mathrm{k}^3+3 \mathrm{k}^2+2 \mathrm{k}+1$ [using equation (i)]
which is a natural number, since $\lambda \mathrm{k} \in \mathrm{N}$. Therefore, $P(k+1)$ is true, when $P(k)$ is true. Hence, from the principle of mathematical induction, the statement is true for all natural numbers $\mathrm{n}$

Topic: Mathematics, Mathematical Induction, BITSAT, BITSAT 2023 (Memory Based Paper 1)
For an endothermic reaction, where ΔH represent the enthalpy of reaction in kJ mol-1, the minimum value for energy of activation (for forward reaction) will be

Options:
A less than ΔH
B zero
C more than ΔH
D equal to ΔH

The correct answer is:
more than ΔH

Solution:
We know that, ΔH=Ef-Eb

where, Eb and Ef= activation energy for backward and forward respectively.

Thus,Ef=ΔH+Eb, means Ef>ΔH.

ΔH=heat of reaction (endothermic)

Ea=energy of activation

Thus, Ea>ΔH

Heat is absorbed in an endothermic reaction. Thus, for an endothermic reaction, where ΔH represent the enthalpy of reaction in kJ mol-1, the minimum value for energy of activation (for forward reaction) will be more than ΔH.

Topic: Chemistry, Chemical Kinetics, JEE Main
For an endothermic reaction, where ΔH represent the enthalpy of reaction in kJ mol-1, the minimum value for energy of activation (for forward reaction) will be

Options:
A less than ΔH
B zero
C more than ΔH
D equal to ΔH

The correct answer is:
more than ΔH

Solution:
We know that, ΔH=Ef-Eb

where, Eb and Ef= activation energy for backward and forward respectively.

Thus,Ef=ΔH+Eb, means Ef>ΔH.

ΔH=heat of reaction (endothermic)

Ea=energy of activation

Thus, Ea>ΔH

Heat is absorbed in an endothermic reaction. Thus, for an endothermic reaction, where ΔH represent the enthalpy of reaction in kJ mol-1, the minimum value for energy of activation (for forward reaction) will be more than ΔH.

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2017
For an insulator, the forbidden energy gap is

Options:
A
Zero

B
$1 \mathrm{eV}$

C
$2 \mathrm{eV}$

D
$5 \mathrm{eV}$

The correct answer is:

$5 \mathrm{eV}$

Solution:
For an insulator, $E_{\mathrm{g}}>3 \mathrm{eV}$, that is why electron transition from valence band to conduction band is not possible. For semiconductor $E_{\mathrm{g}}$ is $0.2 \mathrm{eV}$ to $0.3 \mathrm{eV}$, while for metals $E_g>5 \mathrm{eV}$.

Topic: Physics, Semiconductors, BITSAT, BITSAT 2022
For an isomerisation reaction $A \rightleftharpoons B$, the temperature dependence of equilibrium constant is given by
$\log _e K=4.0-\frac{2000}{T}$
The value of $\Delta S^{\circ}$ at Hook is, therefore

Options:
A
$4 R$

B
$5 R$

C
$400 R$

D
$2000 R$

The correct answer is:

$4 R$

Solution:
van't Hoff equation giving variation of equilibrium constant, $K$ with temperature $T$ is
$\log _e K=-\frac{\Delta H^{\circ}}{R T}+\frac{\Delta S^{\circ}}{R}$
Given, $\quad \log _e K=-\frac{2000}{T}+4.0$
$\therefore \quad \Delta S^{\circ} / R=4 \Rightarrow \Delta S^{\circ}=4 R$

Topic: Chemistry, Chemical Equilibrium, BITSAT, BITSAT 2022
For an isothermal expansion of a perfect gas, the value of $\frac{\Delta P}{P}$ is equal to

Options:
A $-\gamma^{1 / 2} \frac{\Delta V}{V}$
B $-\frac{\Delta V}{V}$
C $-\gamma \frac{\Delta V}{V}$
D $-\gamma^{2} \frac{\Delta V}{V}$

The correct answer is:
$-\frac{\Delta V}{V}$

Solution:
Differentiate $P V=$ constant w.r.t $V$
$\Rightarrow P \Delta V+V \Delta P=0 \quad \Rightarrow \frac{\Delta P}{P}=-\frac{\Delta V}{V}$

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2012
For an RLC circuit driven with voltage of amplitude $V_{m}$ and frequency $\omega_{0}=\frac{1}{\sqrt{\mathrm{LC}}}$ the current exhibits resonance the quality factor, $Q$ is given by

Options:
A $\frac{\omega_{0}}{\mathrm{~L}}$
B $\frac{\omega_{0} \mathrm{~L}}{\mathrm{R}}$
C $\frac{\mathrm{R}}{\omega_{0} \mathrm{C}}$
D $\frac{\mathrm{CR}}{\omega_{0}}$

The correct answer is:
$\frac{\omega_{0} \mathrm{~L}}{\mathrm{R}}$

Solution:
Total angular momentum of the system initially \(\mathrm{L}_{\mathrm{i}}=\mathrm{I}_{1} \mathrm{w}+\mathrm{I}_{2}(0)=\mathrm{I}_{1} \mathrm{w}\) Total angular momentum of the system finally \(\mathrm{L}_{\mathrm{f}}=\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) \mathrm{w}_{2}\)

According to conservation of angular momentum i.e. \(\mathrm{L}_{\mathrm{i}}=\mathrm{L}_{\mathrm{f}}\)

\(\begin{aligned}

&\Rightarrow \mathrm{I}_{1} \omega=\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) \omega_{2} \\

&\Rightarrow \omega_{2}=\frac{\mathrm{I}_{1} \omega}{\mathrm{I}_{1}+\mathrm{I}_{2}}

\end{aligned}\)

Topic: Physics, Alternating Current, BITSAT, BITSAT 2021
For any differentiable function $y$ of $x$, $\frac{d^{2} x}{d y^{2}}\left(\frac{d y}{d x}\right)^{3}+\frac{d^{2} y}{d x^{2}}=$

Options:
A 0
B $y$
C $-y$
D $x$

The correct answer is:
0

Solution:
$\frac{d y}{d x}=\left(\frac{d x}{d y}\right)^{-1}$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=-1\left(\frac{d x}{d y}\right)^{-2}\left\{\frac{d}{d x}\left(\frac{d x}{d y}\right)\right\}$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=(-1)\left(\frac{d x}{d y}\right)^{-2}\left\{\frac{d}{d y}\left(\frac{d x}{d y}\right) \frac{d y}{d x}\right\}$

$=(-1)\left(\frac{d y}{d x}\right)^{2}\left\{\frac{d^{2} x}{d y^{2}} \cdot \frac{d y}{d x}\right\}=-\left(\frac{d y}{d x}\right)^{3}\left\{\frac{d^{2} x}{d y^{2}}\right\}$

$\Rightarrow \frac{d^{2} x}{d y^{2}}\left(\frac{d y}{d x}\right)^{3}+\frac{d^{2} y}{d x^{2}}=0$

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2016
For each parabola $y=x^2+p x+q$, meeting coordinate axes at 3-distinct points, if circles are drawn through these points, then the family of circles must pass through

Options:
A
$(1,0)$

B
$(0,1)$

C
$(1,1)$

D
$(p, q)$

The correct answer is:

$(0,1)$

Solution:
Suppose the parabola $y=x^2+p x+q$ cuts $X$-axis at $A(\alpha, 0)$ and $B(\beta, 0)$.
Then, $\alpha, \beta$ are roots of the equation $x^2+p x+q=0$
$$
\therefore \quad \alpha+\beta=-p \text { and } \alpha \beta=q
$$
The parabola $y=x^2+p x+q$ cuts $Y$-axis at $(0, q)$.
Let the equation of the circle passing through $A, B$ and $C$ be
$$
\begin{aligned}
& x^2+y^2+2 g x+2 f y+c =0 &...(i)\\
\therefore & \alpha^2+2 g \alpha+c =0 &...(ii)\\
& \beta^2+2 g \beta+c =0 &...(iii)\\
\text { and } & q^2+2 f q+c =0 &...(iv)
\end{aligned}
$$
Subtracting Eq. (iii) from Eq. (ii), we get
$$
\begin{array}{rlrl}
\alpha+\beta+2 g & =0 \\
\Rightarrow & g = p / 2
\end{array}
$$
Adding Eqs. (ii) and (iii), we get
$$
\begin{array}{r}
\alpha^2+\beta^2+2 g(\alpha+\beta)+2 c=0 \\
(\alpha+\beta)^2-2 \alpha \beta+2 g(\alpha+\beta)+2 c=0 \\
p^2-2 q-p^2+2 c=0
\end{array}
$$
$$
\left[\because \alpha+\beta=p \text { and } g=\frac{p}{2}\right]
$$
Putting $c=q$ in Eq. (iv), we get $f=-\left(\frac{q+1}{2}\right)$
Substituting the values of $g, f$ and $c$ in Eq. (i), we obtain the equation of family of circles passing through $A, B$ and $C$ as
$$
x^2+y^2+p x-(q+1) y+q=0
$$
Clearly, it passes through $(0,1)$.

Topic: Mathematics, Circle, BITSAT, BITSAT 2022
$\text { If }\left[\begin{array}{cc}
1 & -\tan \theta \\
\tan \theta & 1
\end{array}\right]\left[\begin{array}{cc}
1 & \tan \theta \\
-\tan \theta & 1
\end{array}\right]^{-1}=\left[\begin{array}{cc}
a & -b \\
b & a
\end{array}\right],$
then

Options:
A
$a=1, b=1$

B
$a=\sin 2 \theta, b=\cos 2 \theta$

C
$a=\cos 2 \theta, b=\sin 2 \theta$

D
None of these

The correct answer is:

$a=\cos 2 \theta, b=\sin 2 \theta$

Solution:
Given,
$\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]\left[\begin{array}{cc}1 & \tan \theta \\ -\tan \theta & 1\end{array}\right]^{-1}=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right] \cdot \frac{1}{1+\tan ^2 \theta}\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]$
$=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$
$\Rightarrow \frac{1}{1+\tan ^2 \theta}\left[\begin{array}{cc}1-\tan ^2 \theta & -2 \tan \theta \\ 2 \tan \theta & 1-\tan ^2 \theta\end{array}\right]=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$
$\Rightarrow \quad\left[\begin{array}{cc}\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta} & \frac{2 \tan \theta}{1+\tan ^2 \theta} \\ \frac{2 \tan \theta}{1+\tan ^2 \theta} & \frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\end{array}\right]=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$
$\Rightarrow \quad a=\cos 2 \theta, b=\sin 2 \theta$

Topic: Mathematics, Matrices, BITSAT, BITSAT 2022
If 1+x+x220=∑r=040ar·xr, then ∑r=039-1r·ar·ar+1 equal to

Options:
A 79
B 239·C3978
C 339·C3978
D 0

The correct answer is:
0

Solution:
We have, 1+x+x220=∑r=040arxr

x is replaced by -1x, we get

1-1x+1x220=∑r=040(-1)rar1xr

1-x+x220=∑r=040(-1)rarx40-r

∑r=039-1rar·ar+1 is coefficient of x in product of 1+x+x2201-1x+1x220

Now, 1+x+x2201-1x+1x220=1+x2+1x220

Here, there will be no term of x.

Hence, the coefficient of x will be 0.

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2019
If 0.2 gram of an organic compound containing carbon, hydrogen and oxygen on combustion, yielded 0.147 gram carbon dioxide and 0.12 gram water. What will be the content of oxygen in the substance?

Options:
A $73.29 \%$
B $78.45 \%$
C $83.23 \%$
D $89.50 \%$

The correct answer is:
$73.29 \%$

Solution:
$\%$ of $\mathrm{C}=\frac{12}{44} \times \frac{0.147}{0.2} \times 100=20.045 \%$

$\%$ of $\mathrm{H}=\frac{2}{18} \times \frac{0.12}{0.2} \times 100=6.666 \%$

$\%$ of $\mathrm{O}=100-(20.045+6.666)=73.29 \%$

Topic: Chemistry, Some Basic Concepts of Chemistry, BITSAT, BITSAT 2020
$\text { If } 2^{x}+2^{y}=2^{x+y} \text {, then } \frac{d y}{d x}=$

Options:
A $2^{x-y} \frac{2^{y}-1}{2^{x}-1}$
B $2^{x-y} \frac{2^{y}-1}{1-2^{x}}$
C $\frac{2^{x}+2^{y}}{2^{x}-2^{y}}$
D None of these

The correct answer is:
None of these

Solution:
On differentiating

$\begin{array}{l}

2^{x} \log 2+2^{y} \log 2 \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \\

=2^{x} \cdot 2^{y} \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \log 2+2^{y} \cdot 2^{x} \log 2 \\

\Rightarrow 2^{x}+2^{y} \frac{\mathrm{dy}}{\mathrm{dx}}=2^{x+y} \frac{\mathrm{dy}}{\mathrm{dx}}+2^{x+y} \\

\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2^{x+y}-2^{x}}{2^{y}-2^{x+y}} \\

\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2^{x}+2^{y}-2^{x}}{2^{y}-2^{x}-2^{y}}=-2^{y-x}

\end{array}$

Topic: Mathematics, Differentiation, BITSAT, BITSAT 2021
If $(-4,5)$ is one vertex and $7 x-y+8=0$ is one diagonal of a square, then the equation of second diagonal is

Options:
A $x+3 y=21$
B $2 x-3 y=7$
C $x+7 y=31$
D $2 x+3 y=21$

The correct answer is:
$x+7 y=31$

Solution:
One vertex of square is $(-4,5)$ and equation of one diagonal is $7 x-y+8=0$ Diagonal of a square are perpendicular and bisect each other
Let the equation of the other diagonal be $y=m x+$ $c$ where $m$ is the slope of the line and $c$ is the $\mathrm{y}$-intercept.
Since this line passes through $(-4,5)$
Since this line is at right angle to the line
$7 x-y+8=0$ or $y=7 x+8$, having slope $=7$,
$\therefore 7 \times \mathrm{m}=-1$ or $\mathrm{m}=\frac{-1}{7}$
Putting this value of $m$ in equation (i) we get
$5=-4 \times\left(\frac{-1}{7}\right)+c$
or $5=\frac{4}{7}+c \quad$ or $\quad c=5-\frac{4}{7}=\frac{31}{7}$
Hence equation of the other diagonal is
$y=-\frac{1}{7} x+\frac{31}{7}$ or $7 y=-x+31$ or $x+7 y-31=0$ or $x+7 y=31$

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2023 (Memory Based Paper 2)
If $(-4,5)$ is one vertex and $7 x-y+8=0$ is one diagonal of a square, then the equation of second diagonal is

Options:
A $x+3 y=21$
B $2 x-3 y=7$
C $x+7 y=31$
D $2 x+3 y=21$

The correct answer is:
$x+7 y=31$

Solution:
One vertex of square is $(-4,5)$ and equation of one diagonal is $7 x-y+8=0$ Diagonal of a square are perpendicular and bisect each other
Let the equation of the other diagonal be $y=m x+$ $c$ where $m$ is the slope of the line and $c$ is the $\mathrm{y}$-intercept.
Since this line passes through $(-4,5)$
Since this line is at right angle to the line
$7 x-y+8=0$ or $y=7 x+8$, having slope $=7$,
$\therefore 7 \times \mathrm{m}=-1$ or $\mathrm{m}=\frac{-1}{7}$
Putting this value of $m$ in equation (i) we get
$5=-4 \times\left(\frac{-1}{7}\right)+c$
or $5=\frac{4}{7}+c \quad$ or $\quad c=5-\frac{4}{7}=\frac{31}{7}$
Hence equation of the other diagonal is
$y=-\frac{1}{7} x+\frac{31}{7}$ or $7 y=-x+31$ or $x+7 y-31=0$ or $x+7 y=31$

Topic: Mathematics, Straight Lines, JEE Main
$\quad$ If $4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c=0,$ the family of lines $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ is concurrent at one or the other of the two points-

Options:
A $\left(-1,-\frac{1}{2}\right),(-2,-1)$
B $(-1,-1),\left(-2,-\frac{1}{2}\right)$
C $(-1,2),\left(\frac{1}{2},-1\right)$
D $(1,2),\left(\frac{1}{2},-1\right)$

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

Solution:
$$

\text { } \begin{array}{l}

4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c \\

\quad \equiv(2 a+b)^{2}-3(2 a+b) c+2 c^{2}=0 \\

\quad \Rightarrow(2 a+b-2 c)(2 a+b-c)=0 \Rightarrow c=2 a+b \\

\quad \text { or } c=a+\frac{1}{2} b

\end{array}

$$

The equation of the family of lines is

$$

a(x+2)+b(y+1)=0 \text { or } a(x+1)+b\left(y+\frac{1}{2}\right)=0

$$

giving the point of concurrence (-2,-1) or

$$

\left(-1,-\frac{1}{2}\right)

$$

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2013
If 1.0 mole of $I_{2}$ is introduced into 1.0 litre flask at $1000 \mathrm{~K},$ at equilibrium $\left(\mathrm{K}_{\mathrm{c}}=10^{-6}\right),$ which one is correct?

Options:
A $\left[\mathrm{I}_{2}(\mathrm{~g})\right]>\left[\mathrm{I}^{-}(\mathrm{g})\right]$
B $\left[\mathrm{I}_{2}(\mathrm{~g})\right]<\left[\mathrm{I}^{-}(\mathrm{g})\right]$
C $\left[\mathrm{I}_{2}(\mathrm{~g})\right]=\left[\mathrm{I}^{-}(\mathrm{g})\right]$
D $\left[\mathrm{I}_{2}(\mathrm{~g})\right]=\frac{1}{2}\left[\mathrm{I}^{-}(\mathrm{g})\right]$

The correct answer is:
$\left[\mathrm{I}_{2}(\mathrm{~g})\right]<\left[\mathrm{I}^{-}(\mathrm{g})\right]$

Solution:
$K_{C}=\frac{(2 x)^{2}}{(1-x)}=10^{-6}$

It shows that $(1-x)<2 x$

Topic: Chemistry, Chemical Equilibrium, JEE Main
If 1.0 mole of $I_{2}$ is introduced into 1.0 litre flask at $1000 \mathrm{~K},$ at equilibrium $\left(\mathrm{K}_{\mathrm{c}}=10^{-6}\right),$ which one is correct?

Options:
A $\left[\mathrm{I}_{2}(\mathrm{~g})\right]>\left[\mathrm{I}^{-}(\mathrm{g})\right]$
B $\left[\mathrm{I}_{2}(\mathrm{~g})\right]<\left[\mathrm{I}^{-}(\mathrm{g})\right]$
C $\left[\mathrm{I}_{2}(\mathrm{~g})\right]=\left[\mathrm{I}^{-}(\mathrm{g})\right]$
D $\left[\mathrm{I}_{2}(\mathrm{~g})\right]=\frac{1}{2}\left[\mathrm{I}^{-}(\mathrm{g})\right]$

The correct answer is:
$\left[\mathrm{I}_{2}(\mathrm{~g})\right]<\left[\mathrm{I}^{-}(\mathrm{g})\right]$

Solution:
$K_{C}=\frac{(2 x)^{2}}{(1-x)}=10^{-6}$

It shows that $(1-x)<2 x$

Topic: Chemistry, Chemical Equilibrium, BITSAT, BITSAT 2013
If \(12 \cot ^2 \theta-31 \operatorname{coses} \theta+32=0\), then the value of \(\sin \theta\) is

Options:
A \(\frac{3}{5}\) or 1
B \(\frac{2}{3}\) or \(\frac{-2}{3}\)
C \(\frac{4}{5}\) or \(\frac{3}{4}\)
D \(\pm \frac{1}{2}\)

The correct answer is:
\(\frac{3}{5}\) or 1

Solution:
\(\begin{aligned}
& 12 \cot ^2 \theta-31 \operatorname{cosec} \theta+32=0 \\
& \Rightarrow 12\left(\operatorname{cosec}^2 \theta-1\right)-31 \operatorname{cosec} \theta+32=0 \\
& \Rightarrow \quad 12 \operatorname{cosec}^2 \theta-31 \operatorname{cosec} \theta+20=0 \\
& \Rightarrow \quad 12 \operatorname{cosec}^2 \theta-16 \operatorname{cosec} \theta-15 \operatorname{cosec} \theta+20=0 \\
& \Rightarrow \quad(4 \operatorname{cosec} \theta-5)(3 \operatorname{cosec} \theta-4)=0 \\
& \Rightarrow \quad \operatorname{cosec} \theta=\frac{5}{4}, \frac{4}{3} ; \therefore \sin \theta=\frac{4}{5}, \frac{3}{4}
\end{aligned}\)

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2010
If 16 identical pencils are distributed among 4 children such that each gets at least 3 pencils. The number of ways of distributing the pencils is

Options:
A 15
B 25
C 35
D 40

The correct answer is:
35

Solution:
Required number of ways

$=\text { coeff. of } x^{16} \text { in }\left(x^{3}+x^{4}+x^{5}+x^{6}+x^{7}\right)^{4}$

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2020
If \({ }^{20} \mathrm{C}_{\mathrm{r}}={ }^{20} \mathrm{C}_{\mathrm{r}-10}\) then \({ }^{18} \mathrm{C}_{\mathrm{r}}\) is equal to

Options:
A 4896
B 816
C 1632
D None of these

The correct answer is:
816

Solution:
\({ }^{20} \mathrm{C}_{\mathrm{r}}={ }^{20} \mathrm{C}_{\mathrm{r}-10} \Rightarrow \mathrm{r}+(\mathrm{r}-10)=20 \Rightarrow \mathrm{r}=15\)
\(\therefore \quad{ }^{18} \mathrm{C}_{\mathrm{r}}={ }^{18} \mathrm{C}_{15}={ }^{18} \mathrm{C}_3=\frac{18 \cdot 17 \cdot 16}{1 \cdot 2 \cdot 3}=816\)

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2011
If 1000 droplets of water of surface tension 0.07 $\mathrm{N} / \mathrm{m}$. having same radius $1 \mathrm{~mm}$ each, combine to from a single drop. In the process the released surface energy is( Take $\left.\pi=\frac{22}{7}\right)$

Options:
A $7.92 \times 10^{-6} \mathrm{~J}$
B $7.92 \times 10^{-4} \mathrm{~J}$
C $9.68 \times 10^{-4} \mathrm{~J}$
D $8.8 \times 10^{-5} \mathrm{~J}$

The correct answer is:
$7.92 \times 10^{-4} \mathrm{~J}$

Solution:
We have $V_f=V_i$
$\Rightarrow \frac{4}{3} \pi r_{\mathrm{f}}^3=1000 \times \frac{4}{3} \pi \mathrm{r}_{\mathrm{i}}^3 \Rightarrow \mathrm{r}_{\mathrm{f}}^3=1000 \mathrm{r}_{\mathrm{i}}^3$
$\Rightarrow r_f=10 r_i$
So, released energy $=$ Initial surface energy - final surface energy
$=1000 \times \mathrm{T} \times 4 \pi \mathrm{r}_{\mathrm{i}}^2-\mathrm{T} \times 4 \pi \mathrm{r}_{\mathrm{f}}^2$
$=4 \pi \mathrm{T}\left(1000 \mathrm{r}_{\mathrm{i}}^2-\mathrm{r}_{\mathrm{f}}^2\right)$
$=4 \pi \times 0.07\left(1000 \mathrm{r}_{\mathrm{i}}^2-100 \mathrm{r}_{\mathrm{i}}^2\right)$
$=4 \pi \times 0.07 \times 900 r_i^2$
$=4 \pi \times 63 \times 10^{-6}=7.92 \times 10^{-4} \mathrm{~J}$

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2023 (Memory Based Paper 1)
If $\mathrm{A}=\left[\begin{array}{cc}0 & 2 \\ 3 & -4\end{array}\right]$ and $k \mathrm{~A}=\left[\begin{array}{cc}0 & 3 a \\ 2 b & 24\end{array}\right]$, then the values of $k, a$ and $b$ respectively are:

Options:
A $-6,-12,-18$
B $-6,-4,-9$
C $-6,4,9$
D $-6,12,18$

The correct answer is:
$-6,-4,-9$

Solution:
$k A=\left[\begin{array}{cc}0 & 2 k \\ 3 k & -4 k\end{array}\right]=\left[\begin{array}{cc}0 & 3 a \\ 2 b & 24\end{array}\right]$
$\Rightarrow \quad k=-6, a=-4$ and $b=-9$

Topic: Mathematics, Matrices, BITSAT, BITSAT 2023 (Memory Based Paper 1)
If \(a > 0, a \in R, z=a+2 i\) and \(z|z|-a z+1=0\) then

Options:
A \(z\) is always a positive real number
B \(z\) is always a negative real number
C \(z\) is purely imaginary number
D such a complex z does not exist

The correct answer is:
such a complex z does not exist

Solution:
Putting \(z=a+2 i\) in the given equation and comparing imaginary parts, we get \(\mathrm{a}^2+4=\mathrm{a}^2\), which is not possible.

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2011
If $a>0, b>0, c>0$ and $a, b, c$ are distinct, then $(a+$b) $(b+c)(c+a)$ is greater than

Options:
A $2(a+b+c)$
B $3(a+b+c)$
C $6 \mathrm{abc}$
D $8 \mathrm{abc}$

The correct answer is:
$8 \mathrm{abc}$

Solution:
$\mathrm{AM}>\mathrm{GM}$

$\Rightarrow \frac{\mathrm{a}+\mathrm{b}}{2}>\sqrt{\mathrm{ab}}, \frac{\mathrm{b}+\mathrm{c}}{2}>\sqrt{\mathrm{bc}}, \frac{\mathrm{c}+\mathrm{a}}{2}>\sqrt{\mathrm{ca}}$

So, $(a+b)(b+c)(c+a)>8 a b c$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2020