datavorous/entrance-exam-dataset huggingface dataset:
Let $\mathrm{T}(k)$ be the statement $1+3+5+\ldots+$ $(2 k-1)=k^{2}+10$
Which of the following is correct?

Options:
A $\mathrm{T}(1)$ is true
B $\mathrm{T}(k)$ is true $\Rightarrow \mathrm{T}(k+1)$ is true
C $\mathrm{T}(n)$ is true for all $n \in \mathbf{N}$
D All above are correct

The correct answer is:
$\mathrm{T}(k)$ is true $\Rightarrow \mathrm{T}(k+1)$ is true

Solution:
When $\mathrm{k}=1, \mathrm{LHS}=1$ but $\mathrm{RHS}=1+10=11$
$\therefore \mathrm{T}(1)$ is not true Let $\mathrm{T}(k)$ is true. That is
$\begin{array}{l}
1+3+5+\ldots .+(2 k-1)=k^{2}+10 \\
\text { Now, } 1+3+5+\ldots . .+(2 k-1)+(2 k+1) \\
=k^{2}+10+2 k+1=(k+1)^{2}+10
\end{array}$
$\therefore \mathrm{T}(k+1)$ is true.
That is $\mathrm{T}(k)$ is true $\Rightarrow \mathrm{T}(k+1)$ is true.
But $\mathrm{T}(n)$ is not true for all $n \in \mathbf{N},$ as $\mathrm{T}(1)$ is not true.

Topic: Mathematics, Mathematical Induction, BITSAT, BITSAT 2012
Let $x$ and $y$ be two natural numbers such that $x y=12(x+y)$ and $x \leq y$. Then the total number of pairs $(x, y)$ is

Options:
A 8
B 6
C 4
D 16

The correct answer is:
8

Solution:
$x y-12 x-12 y=0 \Rightarrow(x-12)(y-12)=144$
Now 144 can be factorised into two factors $x$ and $y$ where $x \leq y$ and the factors are (1,144),(2,72),(3,48),(4,36),(6,24),(8,18)
(9,16),(12,12)
Thus there are eight solutions.

Topic: Mathematics, Basic of Mathematics, BITSAT, BITSAT 2012
Let $[x]$ denote the greatest integer $\leq x$. If $f(x)=$ $[\mathrm{x}]$ and $g(x)=|x|,$ then the value of $\mathrm{f}\left(\mathrm{g}\left(\frac{8}{5}\right)\right)-\mathrm{g}\left(\mathrm{f}\left(-\frac{8}{5}\right)\right)$ is

Options:
A 2
B -2
C 1
D -1

The correct answer is:
-1

Solution:
$\begin{aligned}

&\text { Given that, } f(x)=[x] \text { and } g(x)=|x|\\

&\text { Now, } \mathrm{f}\left(\mathrm{g}\left(\frac{8}{5}\right)\right)=\mathrm{g}\left(\frac{8}{5}\right)=\left[\frac{8}{5}\right]=1

\end{aligned}$

and $\mathrm{g}\left(\mathrm{f}\left(-\frac{8}{5}\right)\right)=\mathrm{g}\left(\left[-\frac{8}{5}\right]\right)=\mathrm{g}(-2)=|-2|=2$

$\therefore \mathrm{f}\left(\mathrm{g}\left(\frac{8}{5}\right)\right)-\mathrm{g}\left(\mathrm{f}\left(-\frac{8}{5}\right)\right)=1-2=-1$

Topic: Mathematics, Functions, BITSAT, BITSAT 2020
Let $x+y=3-\cos 4 \theta$ and $x-y=4 \sin 2 \theta$ then the greatest of $x y$ is

Options:
A $\frac{3}{4}$
B 1
C $\frac{1}{2}$
D 2

The correct answer is:
1

Solution:
$x=\frac{3-\cos 4 \theta+4 \sin 2 \theta}{2}$

$=\frac{3-\left(1-\sin ^{2} 2 \theta\right)+4 \sin 2 \theta}{2}=(1+\sin 2 \theta)^{2}$

$y=\frac{3-\cos 4 \theta-4 \sin 2 \theta}{2}$

$=\frac{3-\left(1-\sin ^{2} 2 \theta\right)+4 \sin 2 \theta}{2}=(1-\sin 2 \theta)^{2}$

$\therefore x y=\left(1-\sin ^{2} 2 \theta\right)^{2}=\cos ^{4} 2 \theta \leq 1$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2020
Liebig's method is used for the estimation of

Options:
A Nitrogen
B Sulphur
C Carbon and hydrogen
D Halogens

The correct answer is:
Carbon and hydrogen

Solution:
No Solution available

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2010
Light of wavelength $180 \mathrm{nm}$ ejects photoelectron from a plate of a metal whose work function is $2 \mathrm{eV}$. If a uniform magnetic field of $5 \times 10^{-5} \mathrm{~T}$ is applied parallel to plate, what would be theradius of the path followed by electrons ejected normally from the plate with maximum energy?

Options:
A $1.239 \mathrm{~m}$
B $0.149 \mathrm{~m}$
C $3.182 \mathrm{~m}$
D $2.33 \mathrm{~m}$

The correct answer is:
$0.149 \mathrm{~m}$

Solution:
If $\mathrm{v}_{\max }$ is the speed of the fastest electron emitted from the metal surface, then
$$
\begin{aligned}
& \frac{h c}{\lambda}=W_{0}+\frac{1}{2} m v_{\max }^{2} \frac{\left(6.63 \times 10^{-34}\right) \times\left(3 \times 10^{8}\right)}{\left(180 \times 10^{-9}\right)} \\
&=2 \times\left(1.6 \times 10^{-19}\right)+\frac{1}{2}\left(9.1 \times 10^{-31}\right) v_{\max }^{2} \\
\therefore \quad & v=1.31 \times 10^{6} \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
The radius of the electron is given by
$$
r=\frac{m v}{q B}=\frac{\left(9.1 \times 10^{-31}\right) \times\left(1.31 \times 10^{6}\right)}{\left(1.6 \times 10^{-19}\right) \times\left(5 \times 10^{-9}\right)}=0.149 \mathrm{~m}
$$

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2011
$\lim _{n \rightarrow \infty} \frac{a^{n}+b^{n}}{a^{n}-b^{n}},$ where $a>b>1,$ is equal to

Options:
A -1
B 1
C 0
D None

The correct answer is:
1

Solution:
limit $=\lim _{n \rightarrow \infty} \frac{1+\left(\frac{b}{a}\right)^{n}}{1-\left(\frac{b}{a}\right)^{n}}=1,$

because $0<\frac{b}{a}<1$ implies

$\left(\frac{b}{a}\right)^{n} \rightarrow 0$ as $n \rightarrow \infty$

Topic: Mathematics, Limits, BITSAT, BITSAT 2016
limn→∞sinπn2+1 is equal to

Options:
A ∞
B 0
C does not exist
D None of these

The correct answer is:
0

Solution:
limn→∞sinnπ1+1n21/2

=limn→∞sinnπ1+12n2-18n4+…

=limn→∞sinnπ+π2n-π8n3+…

=limn→∞-1nsinπ12n-18n3+…=0

Topic: Mathematics, Limits, BITSAT, BITSAT 2017
$\lim _{x \rightarrow 0}(\operatorname{cosec} x)^{1 / \log x}$ is equal to :

Options:
A 0
B 1
C $\frac{1}{e}$
D None of these

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

Solution:
Let $y=\lim _{x \rightarrow 0}(\operatorname{cosec} x)^{1 / \log x}$

Taking $\log$ on both sides, we get

$\log y=\lim _{x \rightarrow 0} \frac{\log \operatorname{cosec} x}{\log x}\left[\frac{\infty}{\infty}\right.$ form $]$

$=\lim _{x \rightarrow 0} \frac{-\cot x}{1 / x} \quad$ (ByL' Hospital rule)

$=-\lim _{x \rightarrow 0} \frac{x}{\tan x} \quad\left(\because \cot x=\frac{1}{\tan x}\right)$

$\Rightarrow \log y=-1$

$\Rightarrow y=e^{-1}=\frac{1}{e}$

Hence, required limit $=\frac{1}{e}$

Topic: Mathematics, Limits, BITSAT, BITSAT 2014
$\lim _{x \rightarrow 0} \frac{2 \sin ^{2} 3 x}{x^{2}}$ is equal to

Options:
A 12
B 18
C 0
D 6

The correct answer is:
18

Solution:
Consider $\lim _{x \rightarrow 0} \frac{2 \sin ^{2} 3 x}{x^{2}}$

$\begin{array}{l}

=2 \cdot \lim _{x \rightarrow 0}\left[\frac{\sin 3 x}{x}\right]^{2}=2 \cdot \lim _{x \rightarrow 0}\left[3 \frac{\sin 3 x}{3 x}\right]^{2} \\

=2.9 \lim _{x \rightarrow 0}\left(\frac{\sin 3 x}{3 x}\right)^{2}=18 \times 1=18

\end{array}$

Topic: Mathematics, Limits, BITSAT, BITSAT 2021
$\lim _{x \rightarrow 0} \sqrt{\frac{x-\sin x}{x+\sin ^{2} x}}$ is equal to

Options:
A 1
B 0
C $\infty$
D None of these

The correct answer is:
0

Solution:
$\lim _{x \rightarrow 0} \sqrt{\frac{x-\sin x}{x+\sin ^{2} x}=\lim _{x \rightarrow 0} \sqrt{\frac{1-\frac{\sin x}{x}}{1+\frac{\sin ^{2} x}{x}}}}$

$=\lim _{x \rightarrow 0} \sqrt{\frac{1-\frac{\sin x}{x}}{1+\left(\frac{\sin x}{x}\right) \sin x}}=\sqrt{\frac{1-1}{1+1 \times 0}}=0$

Topic: Mathematics, Limits, BITSAT, BITSAT 2021
$\lim _{x \rightarrow 0} \frac{|\sin x|}{x}$ is equal to

Options:
A 1
B -1
C Does not exist
D None of these

The correct answer is:
Does not exist

Solution:
Let $l=\lim _{x \rightarrow 0} \frac{[\sin x]}{x} ; \mathrm{RHL}=\lim _{x \rightarrow 0^{+}} \frac{\sin x}{x}=1$
$\mathrm{LHL}=\lim _{x \rightarrow 0^{-}}\left(\frac{-\sin x}{x}\right)=-\left(\lim _{x \rightarrow 0^{-}} \frac{\sin x}{x}\right)=-1$
$\left[|x|=\begin{array}{c}x \text { if } x \geq 0 \\ -x \text { if } x \leq 0\end{array}\right]$
As, LHL $\neq$ RHL Hence, limit $l$.does not exist.

Topic: Mathematics, Limits, BITSAT, BITSAT 2023 (Memory Based Paper 1)
$\lim _{x \rightarrow \pi / 2} \frac{\left(1-\tan \left(\frac{x}{2}\right)\right)(1-\sin x)}{\left(1+\tan \left(\frac{x}{2}\right)\right)(\pi-2 x)^{3}}=?$

Options:
A $1 / 8$
B 0
C $1 / 32$
D $\infty$

The correct answer is:
$1 / 32$

Solution:
Put $x=\frac{\pi}{2}-h$ as $x \rightarrow \frac{\pi}{2}, h \rightarrow 0$
$\therefore$ Given limit
$=\lim _{h \rightarrow 0} \frac{1-\tan \left(\frac{\pi}{4}-\frac{\mathrm{h}}{2}\right)}{1+\tan \left(\frac{\pi}{4}-\frac{\mathrm{h}}{2}\right)} \cdot \frac{(1-\cosh )}{(2 \mathrm{~h})^{3}}$
$=\lim _{\mathrm{h} \rightarrow 0} \tan \frac{\mathrm{h}}{2} \frac{2 \sin ^{2} \frac{\mathrm{h}}{2}}{8 \mathrm{~h}^{3}}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{1}{4} \cdot \frac{\tan \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2} \times 2}\left(\frac{\sin \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2}}\right)^{2} \times \frac{1}{4}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{1}{32} \cdot\left(\frac{\tan \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2} \times 2}\right)\left(\frac{\sin \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2}}\right)^{2}=\frac{1}{32}$

Topic: Mathematics, Limits, BITSAT, BITSAT 2012
$$\lim _{x \rightarrow \infty} \frac{\int_{0}^{2 x} x e^{x^{2}} d x}{e^{4 x^{2}}} \text { equals }$$

Options:
A $\underline{0}$
B $\infty$
C 2
D $\frac{1}{2}$

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

Solution:
Consider $\lim _{x \rightarrow \infty} \frac{\int_{0}^{2 x} x e^{x^{2}} d x}{e^{4 x^{2}}}$

$$

\begin{array}{l}

=\lim _{x \rightarrow \infty} \frac{2 \int_{0}^{2 x} x e^{x^{2}} d x}{2 e^{4 x^{2}}} \\

=\lim _{x \rightarrow \infty} \frac{2 \int_{0}^{2 x} e^{x^{2}} d\left(x^{2}\right)}{2 e^{4 x^{2}}} \\

=\lim _{x \rightarrow \infty} \frac{\left[e^{x^{2}}\right]_{0}^{2 x}}{e^{4 x^{2}}}=\lim _{x \rightarrow \infty} \frac{e^{4 x^{2}}-1}{2 e^{4 x^{2}}} \\

=\lim _{x \rightarrow \infty}\left(\frac{1}{2}-\frac{1}{2 e^{4 x^{2}}}\right)=\frac{1}{2}

\end{array}

$$

Topic: Mathematics, Limits, BITSAT, BITSAT 2015
Line joining the points (0,3) and (5,-2) is a tangent to the curve y=ax1+x, then

Options:
A a=1±3
B a∈ϕ
C a=-1±3
D a=-2±23

The correct answer is:
a∈ϕ

Solution:
Equation of line joining the points (0,3) and (5,-2) is y=3-x.

If this line is tangent to y=ax(x+1), then

(3-x)(x+1)=ax should have equal roots.

Thus, (a-2)2+12=0

⇒ No value of a⇒a∈ϕ

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2017
Line x+1λ=y-1=z+2-4 is perpendicular to 2x+2y-8z+5=0, then λ is

Options:
A 1
B -4
C -5
D -3

The correct answer is:
1

Solution:
When line x-x1a1=y-y1b1=z-z1c1 is perpendicular to plane ax+by+cz+d=0, then

a1a=b1b=c1c

∴ λ2=12=-4-8

⇒λ=1

Hence, option (a) is correct.

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2018
Liquids A and B form an ideal solution in the entire composition range. At $350 \mathrm{~K}$, the vapor pressures of pure A and pure B are $7 \times 10^{3} \mathrm{~Pa}$ and $12 \times 10^{3} \mathrm{~Pa}$, respectively. The composition of the vapour is in equilibrium with a solution containing 40 mole percent of $\mathrm{A}$ at this temperature is:

Options:
A $x_{\mathrm{A}}=0.37 ; x_{\mathrm{B}}=0.63$
B $x_{\mathrm{A}}=0.28 ; x_{\mathrm{B}}=0.72$
C $x_{\mathrm{A}}=0.4 ; x_{\mathrm{B}}=0.6$
D $x_{\mathrm{A}}=0.76 ; x_{\mathrm{B}}=0.24$

The correct answer is:
$x_{\mathrm{A}}=0.28 ; x_{\mathrm{B}}=0.72$

Solution:
$\mathrm{P}_{\mathrm{A}}^{\circ}=7 \times 10^{3}$

$\mathrm{P}_{\mathrm{B}}^{\circ}=12 \times 10^{3}$

$x_{\mathrm{A}}^{\prime}=0.4 ; x_{\mathrm{B}}^{\prime}=1-0.4$

$x_{\mathrm{B}}^{\prime}=0.6$

$\mathrm{P}_{\text {total }}=\mathrm{P}_{\mathrm{A}^{\circ}}^{x_{\mathrm{A}}^{\prime}}+\mathrm{P}_{\mathrm{B}}^{\circ} x_{\mathrm{B}}^{\prime}$

$=7 \times 10^{3} \times 0.4+12 \times 10^{3} \times 0.6$

$=(7 \times 0.4+12 \times 0.6) \times 10^{3}=10^{4}$

$x_{\mathrm{A}}=\frac{\mathrm{P}_{\mathrm{A}^{\circ}}^{x^{\prime}} \mathrm{A}}{\mathrm{P}_{\text {total }}}=\frac{7 \times 10^{3} \times 0.4}{10^{4}}$

$\therefore x_{\mathrm{A}}=0.28, x_{\mathrm{B}}=1-0.28=0.72$

Topic: Chemistry, Solutions, BITSAT, BITSAT 2021
∑m=1ntan-12mm4+m2+2 is equal to

Options:
A tan-1n2+nn2+n+2
B tan-1n2-nn2-n+2
C tan-1n2+n+2n2+n
D None of these

The correct answer is:
tan-1n2+nn2+n+2

Solution:
We have, ∑m=1ntan-12mm4+m2+2

=∑m=1ntan-1m2+m+1-m2-m+11+m2+m+1m2-m+1

=∑m=1ntan-1m2+m+1-tan-1m2-m+1

=tan-13-tan-11+tan-17-tan-13+tan-1(13)-tan-17+…+tan-1n2+n+1-tan-1n2-n+1

=tan-1n2+n+1-tan-11

=tan-1n2+n+1-11+n2+n+1=tan-1n2+nn2+n+2

Topic: Mathematics, Inverse Trigonometric Functions, BITSAT, BITSAT 2019
Magnesium burns in $\mathrm{CO}_2$ to form

Options:
A $\mathrm{MgO}+\mathrm{C}$
B $\mathrm{MgO}+\mathrm{CO}$
C $\mathrm{MgCO}_3$
D
MgO

The correct answer is:
$\mathrm{MgO}+\mathrm{C}$

Solution:
$\mathrm{Mg}$ burns in $\mathrm{CO}_2$ to give $\mathrm{MgO}$ and $\mathrm{C}$.

Topic: Chemistry, s Block Elements, BITSAT, BITSAT 2023 (Memory Based Paper 2)
Magnetic moment of bar magnet is $\mathrm{M}$. The work done to turn the magnet by $90^{\circ}$ of magnet in direction of magnetic field $\mathrm{B}$ will be

Options:
A zero
B $\frac{1}{2} \mathrm{MB}$
C $2 \mathrm{MB}$
D $\mathrm{MB}$

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

Solution:
Work done, $\mathrm{W}=\mathrm{MB}(1-\cos \theta)$
$\theta=90^{\circ}$
$\mathrm{W}=\mathrm{MB}$

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2012
Maleic acid and fumaric acids are

Options:
A Chain isomers
B Functional isomers
C Tautomers
D Geometrical isomers

The correct answer is:
Geometrical isomers

Solution:
Maleic acid and fumaric acids are geometrical isomers.

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2020
Match List I (Reaction) with List II (Reagent) and select the correct answer using the codes given below the lists:

Options:
A $\quad \mathrm{I}-\mathrm{A}, \mathrm{II}-\mathrm{B}, \mathrm{III}-\mathrm{C}, \mathrm{IV}-\mathrm{B}$
B $\quad \mathrm{I}-\mathrm{D}, \mathrm{II}-\mathrm{C}, \mathrm{III}-\mathrm{A}, \mathrm{IV}-\mathrm{B}$
C $\quad \mathrm{I}-\mathrm{C}, \mathrm{II}-\mathrm{D}, \mathrm{III}-\mathrm{A}, \mathrm{IV}-\mathrm{B}$
D $\quad \mathrm{I}-\mathrm{B}, \mathrm{II}-\mathrm{A}, \mathrm{III}-\mathrm{D}, \mathrm{IV}-\mathrm{C}$

The correct answer is:
$\quad \mathrm{I}-\mathrm{C}, \mathrm{II}-\mathrm{D}, \mathrm{III}-\mathrm{A}, \mathrm{IV}-\mathrm{B}$

Solution:
No Solution available

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Match List I (Reaction) with List II (Reagent) and select the correct answer using the codes given below the lists:

Options:
A $\quad \mathrm{I}-\mathrm{A}, \mathrm{II}-\mathrm{B}, \mathrm{III}-\mathrm{C}, \mathrm{IV}-\mathrm{B}$
B $\quad \mathrm{I}-\mathrm{D}, \mathrm{II}-\mathrm{C}, \mathrm{III}-\mathrm{A}, \mathrm{IV}-\mathrm{B}$
C $\quad \mathrm{I}-\mathrm{C}, \mathrm{II}-\mathrm{D}, \mathrm{III}-\mathrm{A}, \mathrm{IV}-\mathrm{B}$
D $\quad \mathrm{I}-\mathrm{B}, \mathrm{II}-\mathrm{A}, \mathrm{III}-\mathrm{D}, \mathrm{IV}-\mathrm{C}$

The correct answer is:
$\quad \mathrm{I}-\mathrm{C}, \mathrm{II}-\mathrm{D}, \mathrm{III}-\mathrm{A}, \mathrm{IV}-\mathrm{B}$

Solution:
No Solution available

Topic: Chemistry, Haloalkanes and Haloarenes, BITSAT, BITSAT 2010
Match List I with List II and select the correct answer using the code given below the lists:
\(\begin{array}{l|l} \text{List I} & \text{List II} \\ \hline \text{(A)} f(x)=\cos x & \text{1. The graph cuts y-axis in infinite number of points} \\ \text{(B)} f(x)=\ln x & \text{2. The graph cuts x-axis in two points} \\ \text{(C)} f(x)=x^2-5 x+43. & \text{3. The graph cuts y-axis in only one point} \\ \text{(D)} f(x)=e^x & \text{4. The graph cuts x-axis in only one point} \\ & \text{5. The graph cuts x-axis in infinite number of points} \end{array}\)Codes:

Options:
A (A) - 1, (B) - 4, (C) - 5, (D) - 3
B (A) - 1, (B) - 3, (C) - 5, (D) - 4
C (A) - 5, (B) - 4, (C) - 2, (D) - 3
D (A) - 5, (B) - 3, (C) - 2, (D) - 4

The correct answer is:
(A) - 5, (B) - 4, (C) - 2, (D) - 3

Solution:
(A) Graph of $f(x)=\cos x$ cuts $x$ -axis at infinite number of points. ( 5 of list II)

(B) Graph of $f(x)=\operatorname{In} x$ cuts $x$ -axis in only one point. (4 of list II)

(C) Graph of $f(x)=x^{2}-5 x+4$ cuts $x$ axis in two points ( 2 of list II)

(D) Graph of $\mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}$ cuts $\mathrm{y}$ -axis in only one point. (3 of list II)

Topic: Mathematics, Functions, BITSAT, BITSAT 2016
Match the polymer given in column I with correct monomer of column II and choose the correct option.

Column I	Column II

A.	Neoprene	I.	Isoprene

B.	Natural rubber	II.	Tetrafluoro ethane

C.	Teflon	III.	Chloroprene

D.	Acrilan	IV.	Acryl nitrile

Options:
A ABCDIVIIIIII
B ABCDIIIIIIIV
C ABCDIIIIIIIV
D ABCDIIIVIIII

The correct answer is:
ABCDIIIIIIIV

Solution:
Monomers basically create blocks for molecules, including proteins, starch and many other polymers. Four big monomers are found: amino acids, nucleotides, monosaccharides, and fatty acids. The main forms of macromolecules are those monomers: proteins, nucleic acids, carbohydrates, and lipids.

Topic: Chemistry, Polymers, BITSAT, BITSAT 2018
Match the species in Column I with their types in Column II.

Codes

Options:
A
$\mathrm{A} \rightarrow 4, \mathrm{~B} \rightarrow 3, \mathrm{C} \rightarrow 2, \mathrm{D} \rightarrow 1$

B
$\mathrm{A} \rightarrow 1, \mathrm{~B} \rightarrow 2, \mathrm{C} \rightarrow 3, \mathrm{D} \rightarrow 1$

C
$\mathrm{A} \rightarrow 2, \mathrm{~B} \rightarrow 3, \mathrm{C} \rightarrow 1, \mathrm{D} \rightarrow 4$

D
$\mathrm{A} \rightarrow 3, \mathrm{~B} \rightarrow 1, \mathrm{C} \rightarrow 2, \mathrm{D} \rightarrow 4$

The correct answer is:

$\mathrm{A} \rightarrow 4, \mathrm{~B} \rightarrow 3, \mathrm{C} \rightarrow 2, \mathrm{D} \rightarrow 1$

Solution:
The correct match is
$\mathrm{A} \rightarrow(4) ; \mathrm{B} \rightarrow(3) ; \mathrm{C} \rightarrow(2) ; \mathrm{D} \rightarrow(1)$
A. DDT $\quad \rightarrow$ Pesticides
B. $\mathrm{NaClO}_3 \rightarrow$ Herbicides
C. $\mathrm{Cl}_2 \rightarrow$ Disinfectant
D. PAN $\quad \rightarrow$ Photochemical smog

Topic: Chemistry, Coordination Compounds, BITSAT, BITSAT 2022
Maximum pKb value is of

Options:
A

B CH3CH22NH
C CH32NH
D

The correct answer is:

Solution:
Maximum pKb (least basic) is (d) since, lone pair on
N-atom is used to delocalise π -electrons on two benzene
rings.

Least basic has maximum value of pKb.

Since, in (d), lone pair of electrons over N-atom are delocalised to two benzene rings, thus is least basic.

Topic: Chemistry, Amines, BITSAT, BITSAT 2017
Mean of 25 observations was found to be 78.4. But later on it was found that 96 was misread 69 . The correct mean is

Options:
A 79.24
B 79.48
C 80.10
D None of these

The correct answer is:
79.48

Solution:
Mean \(\bar{x}=\frac{\sum x}{n}\) or \(\Sigma x=n \bar{x}\)
\(\Sigma \mathrm{x}=25 \times 78.4=1960\)
But this \(\Sigma \mathrm{x}\) is incorrect as 96 was misread as 69 .
\(\begin{aligned}
& \therefore \text { correct } \Sigma \mathrm{x}=1960+(96-69)=1987 \\
& \therefore \text { correct mean }=\frac{1987}{25}=79.48
\end{aligned}\)

Topic: Mathematics, Statistics, BITSAT, BITSAT 2010
Melamine plastic crockery is a copolymer of

Options:
A $\mathrm{HCHO}$ and melamine
B $\mathrm{HCHO}$ and ethylene
C melamine and ethylene
D None of these

The correct answer is:
$\mathrm{HCHO}$ and melamine

Solution:
Melamine plastic crockery is a copolymer of $\mathrm{HCHO}$ and Melamine.

Topic: Chemistry, Polymers, BITSAT, BITSAT 2023 (Memory Based Paper 1)
Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at $25^{\circ} \mathrm{C}$. For this process, which ofthe following statement is correct?

Options:
A The adsorption requires activation at $25^{\circ} \mathrm{C}$
B The adsorption is accompanied by a decrease in enthalpy
C The adsorption increases with increase of temperature
D The adsorption is irreversible

The correct answer is:
The adsorption is accompanied by a decrease in enthalpy

Solution:
The adsorption of methylene blue on activated charcoal is an example of physiosorption which is exothermic, multilayer and does not have energy barrier.

Topic: Chemistry, Surface Chemistry, BITSAT, BITSAT 2020
M.I of a circular loop of radius $\mathrm{R}$ about the axis in figure is

Options:
A $\mathrm{MR}^{2}$
B $(3 / 4) \mathrm{MR}^{2}$
C $\mathrm{MR}^{2} / 2$
D $2 \mathrm{MR}^{2}$

The correct answer is:
$(3 / 4) \mathrm{MR}^{2}$

Solution:
Use theorem of parallel axes.

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2020
Minamata disease is due to pollution of

Options:
A aresenic into the atmosphere
B organic waste into drinking water
C oil spill in water
D industrial waste mercury into fishing water

The correct answer is:
industrial waste mercury into fishing water

Solution:
Minamata is caused by Hg poisoning.

Topic: Chemistry, Environmental Chemistry, BITSAT, BITSAT 2010
Minimise $Z=\sum_{j=1}^{n} \sum_{i=1}^{m} c_{i j} x_{i j}$

Subject to $\sum_{i=1}^{m} x_{i j}=b_{j}, j=1,2, \ldots, n$

$\sum_{j=1}^{n} x_{i j}=b_{j}, i=1,2, \ldots, m$ is a $L P P$ with number of constr

Options:
A $m-n$
B $m$
C $\mathrm{m}+\mathrm{n}$
D $\frac{\mathrm{m}}{\mathrm{n}}$

The correct answer is:
$\mathrm{m}+\mathrm{n}$

Solution:
Constraints will be

$$

\mathrm{x}_{11}+\mathrm{x}_{21}+\ldots .+\mathrm{x}_{\mathrm{ml}}=\mathrm{b}_{1}

$$

$\mathrm{x}_{12}+\mathrm{x}_{22}+\ldots \ldots \mathrm{x}_{\mathrm{m} 2}=\mathrm{b}_{2}$

$\mathrm{x}_{\ln }+\mathrm{x}_{2 \mathrm{n}}+\ldots .+\mathrm{x}_{\mathrm{mn}}=\mathrm{b}_{\mathrm{n}}$

$\mathrm{x}_{11}+\mathrm{x}_{12}+\ldots+\mathrm{x}_{1 \mathrm{n}}=\mathrm{b}_{1}$

$\mathrm{x}_{21}+\mathrm{x}_{22}+\ldots+\mathrm{x}_{2 \mathrm{n}}=\mathrm{b}_{2}$

$\mathrm{x}_{\mathrm{ml}}+\mathrm{x}_{\mathrm{m} 2}+\ldots+\mathrm{x}_{\mathrm{mn}}=\mathrm{b}_{\mathrm{n}}$

So, total number of constraints $=\mathrm{m}+\mathrm{n}$

Topic: Mathematics, Linear Programming, BITSAT, BITSAT 2015
Minimum value of $Z=3 x+5 y$ subject to constraints $x+y \geq 2, x+3 y \geq 3, x, y \geq 0$

Options:
A
6
B
7
C
8
D
9

The correct answer is:
7

Solution:
$\mathrm{Z}$ is 7 minimum at $\left(\frac{3}{2}, \frac{1}{2}\right)$

Topic: Mathematics, Linear Programming, BITSAT, BITSAT 2023 (Memory Based Paper 2)
Monochromatic light of wavelength $667 \mathrm{~nm}$ is produced by a helium neon laser. The power emitted is $9 \mathrm{~mW}$. The number of photons arriving per sec on the average at a target irradiated by this beam is:

Options:
A $3 \times 10^{16}$
B $9 \times 10^{15}$
C $3 \times 10^{19}$
D $9 \times 10^{17}$

The correct answer is:
$3 \times 10^{16}$

Solution:
$\lambda=667 \times 10^{-9} m, \mathrm{P}=9 \times 10^{-3} \mathrm{~W}$

$\mathrm{P}=\frac{\text { Nhc }}{\lambda}, \mathrm{N}=$ No. of photons emitted/sec.

$\mathrm{N}=\frac{9 \times 10^{-3} \times 667 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^{8}}=3 \times 10^{16} / \mathrm{sec}$

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2021
Monochromatic radiation of wavelength λ is incident on a hydrogen sample in ground state. Hydrogen atom absorbs a fraction of light and subsequently emits radiations of six different wavelengths. The wavelength λ is

Options:
A 97.5 nm
B 121.6 nm
C 110.3 nm
D 45.2 nm

The correct answer is:
97.5 nm

Solution:
As H-atom emits 6 spectral lines nn-12=6

∴ n=4

ΔE=E4-E1=13.6-0.85

=12.75 eV

λ=hc∆E=1242 nm12.75

∴λ=97.5 nm

Topic: Physics, Atomic Physics, BITSAT, BITSAT 2017
$\mathrm{N}_{2}$ and $\mathrm{O}_{2}$ are converted into mono anions, $\mathrm{N}_{2}^{-}$ and $\mathrm{O}_{2}^{-}$ respectively. Which of the following statements is wrong ?

Options:
A $\operatorname{In} \mathrm{N}_{2}$, the $\mathrm{N}-\mathrm{N}$ bond weakens
B In $\mathrm{O}_{2}$, the $\mathrm{O}-\mathrm{O}$ bond order increases
C In $\mathrm{O}_{2},$ bond length decreases
D $\mathrm{N}_{2}^{-}$ becomes diamagnetic

The correct answer is:
In $\mathrm{O}_{2}$, the $\mathrm{O}-\mathrm{O}$ bond order increases

Solution:
We know that in $\mathrm{O}_{2}$ bond, the order is 2 and in $\mathrm{O}_{2}^{-}$ bond, the order is 1.5 . Therefore the wrong statements is (b).

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2014
$\mathrm{n}\left(\mathrm{CF}_{2}=\mathrm{CF}_{2}\right) \frac{\text { Benzoyl peroxide or }}{\left(\mathrm{NH}_{4}\right)_{2} \mathrm{~S}_{2} \mathrm{O}_{8}} \mathrm{X}$

Here, $\mathrm{X}$ is:

Options:
A RMMA
B $\mathrm{PVC}$
C $\mathrm{PAN}$
D None of these

The correct answer is:
None of these

Solution:
Here $X$ is polytetrafluoroethylene. So, none of these i.e., option (d) is correct choice.

Topic: Chemistry, Polymers, BITSAT, BITSAT 2020
Name of following reaction is

Options:
A Claisen Condensation
B Diel's Alder reaction
C Dieckmann cyclisation
D Michael addition reaction

The correct answer is:
Diel's Alder reaction

Solution:
The given reaction is Diel's Alder reaction.

Topic: Chemistry, Hydrocarbons, JEE Main