datavorous/entrance-exam-dataset huggingface dataset:
BF3 is an acid, according to

Options:
A Lewis
B Arrhenius
C Bronsted and Lowery
D All of the above

The correct answer is:
Lewis

Solution:
BF3 is a Lewis acid, as it accepts lone pair of electrons.

It is not an Arrhenius acid because it does not furnish H+ ion in its aqueous solution.

It is also not a Bronsted Lowry acid as if does not give H+ ions in any solvent.

Topic: Chemistry, Ionic Equilibrium, BITSAT, BITSAT 2017
Biochemical oxygen demand(BOD) value can be a measure of water pollution caused by the organic matter. Which of the following statements is correct?

Options:
A Polluted water has BOD value higher than $10 \mathrm{ppm}$
B Aerobic bacteria decreases the BOD value
C Anaerobic bacteria increases the BOD value
D Clean water has $B O D$ value higher than $10 \mathrm{ppm}$

The correct answer is:
Polluted water has BOD value higher than $10 \mathrm{ppm}$

Solution:
Clean water has BOD value less than $5 \mathrm{ppm}$. Polluted water has BOD value higher than $10 \mathrm{ppm}$.

Topic: Chemistry, Environmental Chemistry, BITSAT, BITSAT 2021
Block A of mass $\mathrm{m}$ and block $\mathrm{B}$ of mass $2 \mathrm{~m}$ are placed on a fixed triangular wedge by means of a massless, inextensible string and a frictionless pulley as shown in figure.

The wedge is inclined at $45^{\circ}$ to the horizontal on both the sides. If the coefficient of friction between the block $\mathrm{A}$ and the wedge is $2 / 3$ and that between the block $\mathrm{B}$ and the wedge is $1 / 3$ and both the blocks $\mathrm{A}$ and $\mathrm{B}$ are released from rest, the acceleration of A will be

Options:
A $-1 \mathrm{~ms}^{-2}$
B $1.2 \mathrm{~ms}^{-2}$
C $0.2 \mathrm{~ms}^{-2}$
D zero

The correct answer is:
zero

Solution:
Total maximum friction force,\(f_{t}=f_{A}+f_{B}\)

\(\mu_{\mathrm{A}} \mathrm{mg} \cos 45^{\circ}+\mu_{\mathrm{B}} 2 \mathrm{mg} \cos 45^{\circ}\)

or, \(f_{t}=m g\left(\frac{2}{3} \times \frac{1}{\sqrt{2}}+\frac{1}{3} \times 2 \times \frac{1}{\sqrt{2}}\right)\)

\(=\frac{4}{3 \sqrt{2}} \mathrm{mg}\)

The net pulling force, \(F=2 m g \sin 45^{\circ}-m g \sin 45^{\circ}\) \(=\mathrm{mg} \sin 45^{\circ}\)

\(=\frac{m g}{\sqrt{2}}\)

\(\because \mathrm{F}<\mathrm{f}_{\mathrm{t}}\)

\(\therefore\) The blocks will not move.

So, acceleration of block A is Zero.

\(\therefore\) option (D) is correct.

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2016
Blocks of magnesium are fixed to the bottom of a ship to

Options:
A block hole in the ship
B acidity of sea water
C make the ship lighter
D prevent the action of water and salt

The correct answer is:
prevent the action of water and salt

Solution:
Block of magnesium prevents the action of water and salt of iron (of ship) by sacrificing itself.

This is method of protection of ship (i.e. of iron), known as sacrificial protection.

It prevents corrosive action of water and salt by providing cathodic protection.

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2017
Bond enthalpies of $\mathrm{H}_{2}, \mathrm{X}_{2}$ and $\mathrm{HX}$ are in the ratio 2: 1: 2 . If enthalpy of formation of $\mathrm{HX}$ is $-50 \mathrm{~kJ}$ $\mathrm{mol}^{-1},$ the bond enthalpy of $\mathrm{X}_{2}$ is

Options:
A $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$
B $300 \mathrm{~kJ} \mathrm{~mol}^{-1}$
C $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$
D $400 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$100 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
$\frac{1}{2} \mathrm{H}_{2}+\frac{1}{2} \mathrm{X}_{2} \longrightarrow \mathrm{HX}$
Let the bond enthalpy of $\mathrm{X}-\mathrm{X}$ bond be $x$.
$\Delta \mathrm{H}_{\mathrm{f}}(\mathrm{HX})=-50$
$=\frac{1}{2} \Delta \mathrm{H}_{\mathrm{H}-\mathrm{H}}+\frac{1}{2} \Delta \mathrm{H}_{\mathrm{X}-\mathrm{X}}-\Delta \mathrm{H}_{\mathrm{H}-\mathrm{X}}$
$=\frac{1}{2} 2 x+\frac{1}{2} x-2 x=\frac{-x}{2}$
$\therefore x=50 \times 2=100 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2012
$\mathrm{Br}^{\huge\cdot}$ will abstract which of the hydrogen most readily?

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

The correct answer is:
a

Solution:
Bromine is more selective

$\therefore$ abstract that hydrogen which forms stable free- radical

$\therefore$

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2020
Breaking stress of a steel wire is p and the density of steel is ρ. The greatest length of steel wire that can hang vertically without breaking is

Options:
A pρg
B p2ρg
C 2pρg
D None of these

The correct answer is:
pρg

Solution:
Maximum stress = Maximum weight Cross-sectional area

Now, maximum weight of steel wire

= Volume × Density ×g=Alρg

where, l is the maximum length of steel wire that can hang vertically without breaking. ρ is the density of steel and A is the cross-sectional area of steel wire.
∴ Maximum stress, p=AlρgA=lρg

∴ l=pρg

Topic: Physics, Mechanical Properties of Solids, BITSAT, BITSAT 2018
$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C} \equiv \mathrm{N}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N} \stackrel{\rightarrow}{=} \mathrm{C}$ exhibit which
type of isomerism?

Options:
A Position
B Functional
C Demotropism
D Position isomerism

The correct answer is:
Functional

Solution:
No Solution available

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2011
$\int \frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)} \mathrm{dx}$ equals

Options:
A $-\cot \left(e x^{x}\right)+\mathrm{C}$
B $\tan \left(x e^{x}\right)+\mathrm{C}$
C $\tan \left(e^{x}\right)+\mathrm{C}$
D $\cot \left(e^{x}\right)+\mathrm{C}$

The correct answer is:
$\tan \left(x e^{x}\right)+\mathrm{C}$

Solution:
$\int \frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)} \mathrm{dx}$

Let $x e^{x}=t$

$\Rightarrow\left(x e^{x}+e^{x}\right)=\frac{\mathrm{dt}}{\mathrm{dx}}$

$\Rightarrow d x=\frac{\mathrm{dt}}{e^{x}(x+1)}$

$\therefore \int \frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)} \mathrm{dx}=\int \frac{e^{x}(1+x)}{\cos ^{2} t} \times \frac{d t}{e^{x}(1+x)}=\int \frac{1}{\cos ^{2} t} \mathrm{dt}=\int \sec ^{2} t \mathrm{dt}=$$\tan t+\mathrm{C}=\tan \left(x e^{x}\right)+\mathrm{C}$

Topic: Mathematics, Indefinite Integration, BITSAT, BITSAT 2021
Calcination is used in metallurgy for removal of?

Options:
A Water and sulphide
B Water and $\mathrm{CO}_{2}$
C $\mathrm{CO}_{2}$ and $\mathrm{H}_{2} \mathrm{~S}$
D $\mathrm{H}_{2} \mathrm{O}$ and $\mathrm{H}_{2} \mathrm{~S}$

The correct answer is:
Water and $\mathrm{CO}_{2}$

Solution:
Calcination is used for removal of volatile impurities and decompose carbonates.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, BITSAT, BITSAT 2011
Calculate the area of the plates of a one farad parallel plate capacitor if separation between plates is $1 \mathrm{~mm}$ and plates are in vacuum

Options:
A $18 \times 10^{8} \mathrm{~m}^{2}$
B $0.3 \times 10^{8} \mathrm{~m}^{2}$
C $1.3 \times 10^{8} \mathrm{~m}^{2}$
D $1.13 \times 10^{8} \mathrm{~m}^{2}$

The correct answer is:
$1.13 \times 10^{8} \mathrm{~m}^{2}$

Solution:
For a parallel plate capacitor $\mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$
$$
\begin{aligned}
\therefore \quad & \mathrm{A}=\frac{\mathrm{Cd}}{\varepsilon_{0}}=\frac{1 \times 10^{-3}}{8.85 \times 10^{-12}} \\
&=1.13 \times 10^{8} \mathrm{~m}^{2}
\end{aligned}
$$
This corresponds to area of square of side $10.6 \mathrm{~km}$ which shows that one farad is very large unit of capacitance.

Topic: Physics, Capacitance, BITSAT, BITSAT 2009
Calculate the effective resistance between A and B in following network.

Options:
A $5 \Omega$
B $10 \Omega$
C $20 \Omega$
D $30 \Omega$

The correct answer is:
$20 \Omega$

Solution:
Equivalent resistance
$$
\begin{array}{l}
=(5+10+15) \|(10+20+30) \\
\text { So, } R_{\text {eq }}=\frac{30 \times 60}{30+60}=20 \Omega
\end{array}
$$

Topic: Physics, Current Electricity, JEE Main
Calculate the effective resistance between A and B in following network.

Options:
A $5 \Omega$
B $10 \Omega$
C $20 \Omega$
D $30 \Omega$

The correct answer is:
$20 \Omega$

Solution:
Equivalent resistance
$$
\begin{array}{l}
=(5+10+15) \|(10+20+30) \\
\text { So, } R_{\text {eq }}=\frac{30 \times 60}{30+60}=20 \Omega
\end{array}
$$

Topic: Physics, Current Electricity, BITSAT, BITSAT 2011
Calculate the wavelength (in nanometer) associated with a proton moving at $1.0 \times 10^{3} \mathrm{~ms}^{-1}$.
(Mass of proton $=1.67 \times 10^{-27} \mathrm{~kg}$ and $\left.\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}\right)$

Options:
A $0.40 \mathrm{~nm}$
B $2.5 \mathrm{~nm}$
C $14.0 \mathrm{~nm}$
D $0.32 \mathrm{~nm}$

The correct answer is:
$0.40 \mathrm{~nm}$

Solution:
$\begin{aligned} \lambda &=\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{1.67 \times 10^{-27} \times 1 \times 10^{3}} \\ &=3.97 \times 10^{-10} \text { meter }=0.397 \text { nanometer } \approx 0.40 \mathrm{~nm} \end{aligned}$

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2021
Calculate the wavelength of light used in an interference experiment from the following data:

Fringe width $=0.03 \mathrm{~cm}$. Distance between the slits and eyepiece through which the interference pattern is observed is $1 \mathrm{~m}$. Distance between the images of the virtual source when a convex lens of focal length $16 \mathrm{~cm}$ is used at a distance of $80 \mathrm{~cm}$ from the eyepiece is $0.8 \mathrm{~cm}$.

Options:
A $0.0006 Ã…$
B $0.0006 \mathrm{~m}$
C $600 \mathrm{~cm}$
D $6000 Ã…$

The correct answer is:
$6000 Ã…$

Solution:
Given: fringe with $\beta=0.03 \mathrm{~cm}, \mathrm{D}=1 \mathrm{~m}=100$

$\mathrm{cm}$

Distance between images of the source $=$ $0.8 \mathrm{~cm} .$

Image distance $v=80 \mathrm{~cm}$ Object distance $=u$ Using mirror formula,

$$

\begin{array}{l}

\qquad \begin{array}{l}

\frac{1}{v}+\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{60}+\frac{1}{u}=\frac{1}{16} \\

\Rightarrow \mathrm{u}=20 \mathrm{~cm}

\end{array} \\

\text { Magnification, } \mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{80}{20}=4 \\

\text { Magnification } \begin{aligned}

{\text { distances between images of slits }}{\text { distance between slits }} \\

=\frac{0.8}{\mathrm{~d}}=\frac{0.8}{\mathrm{~d}}=4 \Rightarrow \mathrm{d}=0.2 \mathrm{~cm}

\end{aligned} \\

\text { Fringe width } \beta=\frac{\mathrm{D} \lambda}{\mathrm{d}}

\end{array}

$$

or, $\beta=\frac{100 \lambda}{0.2}=0.03 \times 10^{-2}$

Therefore, wavelength of light used $\lambda=6000$ $Ã…$

Topic: Physics, Wave Optics, BITSAT, BITSAT 2015
Cards are drawn one by one at random from a well shuffled pack of 52 cards until 2 aces are obtained for the first time. The probability that 18 draws are required for this is

Options:
A 334
B 17455
C 56115925
D None of these

The correct answer is:
56115925

Solution:
18 draws are required for 2 aces means in the first 17 draws, there is one ace and 16 non-ace and 18th draw 2nd ace.

Required probability =C1648×C14C1752×335=56115925

Topic: Mathematics, Probability, BITSAT, BITSAT 2019
Cetyltrimethyl ammonium bromide is a popular

Options:
A anionic detergent
B cationic detergent
C non-ionic detergent
D sweetener

The correct answer is:
cationic detergent

Solution:
Cetyltrimethyl ammonium bromide which is a germicide, is a popular cationic detergent.

Topic: Chemistry, Chemistry in Everyday Life, BITSAT, BITSAT 2021
$\quad \mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{3} \stackrel{\mathrm{H}_{2} / \mathrm{Pt}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{D}_{2} / \mathrm{Pt}}{\longrightarrow} \mathrm{B}$

The compounds $A$ and $B$, respectively are

Options:
A cis-butene-2 and $\mathrm{rac}-2,3$ -dideuterobutane
B trans-butene-2 and $\mathrm{rac}-2$, 3-dideuterobutane
C cis-butene- 2 and meso-2, 3-dideuterobutane
D trans-butene-2 and meso-2, 3-dideuterobutane

The correct answer is:
cis-butene- 2 and meso-2, 3-dideuterobutane

Solution:
Catalytic hydrogenation of alkynes gives cis-alkene which in turn adds deuterium atoms in presence of $\mathrm{H}_{2}$ again in $\mathrm{ci}$ -manner forming meso -2,3 -dideuterobutane.

Topic: Chemistry, Hydrocarbons, BITSAT, BITSAT 2016
Chief source of soil and water pollution is:

Options:
A Mining
B Agro industry
C Thermal power plant
D All of the above

The correct answer is:
All of the above

Solution:
No Solution available

Topic: Chemistry, Environmental Chemistry, BITSAT, BITSAT 2009
Chloride ion and potassium ion are isoelectronic. Then :

Options:
A Their sizes are same
B $\mathrm{Cl}^{-}$ ion is bigger than $\mathrm{K}^{+}$ ion
C $\mathrm{K}^{+}$ ion is relatively bigger
D Their sizes depend on other cation and anion

The correct answer is:
$\mathrm{Cl}^{-}$ ion is bigger than $\mathrm{K}^{+}$ ion

Solution:
No Solution available

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2010
Chlorobenzene→Reaction-(X)Phenol→Reaction-(Y)Salicylaldehyde.

The reactions(s) X'' andY'' respectively are

Options:
A Fries rearrangement and Kolbe
B Cumene and Reimer-Tiemann
C Dow and Reimer-Tiemann
D Dow and Sandmeyer.

The correct answer is:
Dow and Reimer-Tiemann

Solution:
Phenol is manufactured by heating chlorobenzene with 10% aqueous NaOH solution at about 623K under 300 atm and in the presence of copper salt acting as a catalyst to form sodium phenoxide. The sodium salt when treated with dil. HCl, gives phenol. This method is known as Dow's process.

When phenol is heated with CHCl3 in the presence of aqueous NaOH, an aldehydic group (−CHO) get introduced in the ring at a position ortho to the phenolic group. This reaction is known as the Reimer-Tiemann reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Chlorobenzene→Reaction-(X)Phenol→Reaction-(Y)Salicylaldehyde.

The reactions(s) X'' andY'' respectively are

Options:
A Fries rearrangement and Kolbe
B Cumene and Reimer-Tiemann
C Dow and Reimer-Tiemann
D Dow and Sandmeyer.

The correct answer is:
Dow and Reimer-Tiemann

Solution:
Phenol is manufactured by heating chlorobenzene with 10% aqueous NaOH solution at about 623K under 300 atm and in the presence of copper salt acting as a catalyst to form sodium phenoxide. The sodium salt when treated with dil. HCl, gives phenol. This method is known as Dow's process.

When phenol is heated with CHCl3 in the presence of aqueous NaOH, an aldehydic group (−CHO) get introduced in the ring at a position ortho to the phenolic group. This reaction is known as the Reimer-Tiemann reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, BITSAT, BITSAT 2017
Choose the incorrect statement.

Options:
A Sodium borohydride reacts very slowly with cold water
B Sodium borohydride reacts violently with cold water to give H2(g)
C Solubility of sodium borohydride in water at 250C is 10.05 g/ml
D Melting point of sodium borohydride is 5000C

The correct answer is:
Sodium borohydride reacts violently with cold water to give H2(g)

Solution:
Option (b) is the incorrect statement as reaction of NaBH4 with cold water is very slow. It is not violent. All other are correct statements.

The reaction of sodium borohydride (NaBH4â) with cold water is very slow.

Topic: Chemistry, s Block Elements, BITSAT, BITSAT 2017
Cinnabar is an ore of

Options:
A $\mathrm{Hg}$
B $\mathrm{Cu}$
C $\mathrm{Pb}$
D $\mathrm{Zn}$

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

Solution:
Cinnabar (HgS) is an ore of Hg.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, BITSAT, BITSAT 2009
$\mathrm{CO}_{2}$ goes to air, causes green house effect and gets dissolved in water. What will be the effect on soil fertility and $\mathrm{pH}$ of the water?

Options:
A Increases
B Decreases
C Remain same
D None of these

The correct answer is:
Decreases

Solution:
$\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3} \rightarrow \mathrm{H}^{+}+\mathrm{HCO}_{3}^{-}$
Here $\left[\mathrm{H}^{+}\right]$ increases hence, $\mathrm{pH}$ decreases due to which soil fertility will also decreases.

Topic: Chemistry, Ionic Equilibrium, BITSAT, BITSAT 2012
Cobalt (60) isotope is used in the treatment of :

Options:
A Heart diseases
B Skin diseases
C Diabetes
D Cancer

The correct answer is:
Cancer

Solution:
Cobalt ( 60 ) isotope is used in the treatment of cancer.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, BITSAT, BITSAT 2015
Coefficient of variation of two distributions are 50 and 60 and their arithmetic means are 30 and 25, respectively. Difference of their standard deviation is

Options:
A 0
B 1
C 1.3
D 2.5

The correct answer is:
0

Solution:
Given, coefficient of variation C1=50

and coefficient of variation C2=60.

We have, x¯1=30 and x¯2=25

∵ C=σx¯×100

50=σ130×100

⇒ σ1=15

and 60=σ225×100

⇒ σ2=15

∴ Required difference σ1-σ2=15-15=0

Topic: Mathematics, Statistics, BITSAT, BITSAT 2017
Coefficient of $x^{13}$ in the expansion of $(1-x)^{5}\left(1+x+x^{2}+x^{3}\right)^{4}$ is

Options:
A 4
B 6
C 32
D 5

The correct answer is:
4

Solution:
Expression $=(1-x)^{5} \cdot(1+x)^{4}\left(1+x^{2}\right)^{4}$

$\begin{array}{l}

=(1-x)\left(1-x^{2}\right)^{4}\left(1+x^{2}\right)^{4} \\

=(1-x)\left(1-x^{4}\right)^{4}

\end{array}$

$\therefore$ Coefficient of $x^{13}=-{ }^{4} \mathrm{C}_{3}(-1)^{3}=4$

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2021
Complete hydrolysis of cellulose gives

Options:
A D-ribose
B $\mathrm{D}$ -glucose
C $\quad \mathrm{L}$ -glucose
D D-fructose

The correct answer is:
$\mathrm{D}$ -glucose

Solution:

Topic: Chemistry, Biomolecules, BITSAT, BITSAT 2012
Complete hydrolysis of cellulose gives

Options:
A D-ribose
B $\mathrm{D}$ -glucose
C $\quad \mathrm{L}$ -glucose
D D-fructose

The correct answer is:
$\mathrm{D}$ -glucose

Solution:

Topic: Chemistry, Biomolecules, JEE Main
Compounds I and II may be grouped as

Options:
A diastereomers
B epimers
C $\mathrm{C}-2$ epimers
D all of the three.

The correct answer is:
all of the three.

Solution:
When structures I and II are $\mathrm{C}-2$ epimers, it implies that these are epimers and diastereomers too.

Topic: Chemistry, Biomolecules, BITSAT, BITSAT 2020
Consider $\frac{x}{2}+\frac{y}{4} \geq 1$ and $\frac{x}{3}+\frac{y}{2} \leq 1, x, y \geq 0$. Then

number of possible solutions are :

Options:
A Zero
B Unique
C Infinite
D None of these

The correct answer is:
Infinite

Solution:
Consider $\frac{x}{2}+\frac{y}{4} \geq 1, \frac{x}{3}+\frac{y}{2} \leq 1$,

x, $y \geq 0$ convert them into equation and solve them and draw the graph of these equations we get $y=1$ and $x=3 / 2$

From graph region is finite but numbers of possible solutions are infinite because for different values of $x$ and $y$ we have different or infinite no. of solutions.

Topic: Mathematics, Linear Programming, BITSAT, BITSAT 2014
Consider an infinite geometric series with first term $a$ and common ratio $r$. If its sum is 4 and the second term is $\frac{3}{4}$, then:

Options:
A $a=\frac{4}{7}, r=\frac{3}{7}$
B $a=2, r=\frac{3}{8}$
C $a=\frac{3}{2}, r=\frac{1}{2}$
D $a=3, r=\frac{1}{4}$

The correct answer is:
$a=3, r=\frac{1}{4}$

Solution:
Since, sum $=4$ and second term $=\frac{3}{4}$ $\Rightarrow \frac{a}{1-r}=4$, and $a r=\frac{3}{4}$ $\Rightarrow \frac{a}{1-\frac{3}{4 a}}=4$ $\Rightarrow(a-1)(a-3)=0$ $\Rightarrow a=1$ or $a=3$

$\begin{array}{l}

\Rightarrow \frac{a}{1-\frac{3}{4 a}}=4 \\

\Rightarrow(a-1)(a-3)=0 \\

\Rightarrow a=1 \text { or } a=3

\end{array}$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2021
Consider elastic collision of a particle of mass m moving with a velocity u with another particle of the same mass at rest. After the collision the projectile and the struck particle move in directions making angles $\theta_{1}$ and $\theta_{2}$ respectively with the initial direction of motion. The sum of the angles $\theta_{1}+\theta_{2},$ is :

Options:
A $45^{\circ}$
B $90^{\circ}$
C $135^{\circ}$
D $180^{\circ}$

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

Solution:
No Solution

Topic: Physics, Center of Mass Momentum and Collision, BITSAT, BITSAT 2016
Consider the acceleration, velocity and displacement of a tennis ball as its falls to the ground and bounces back. Directions of which of these change in the process?

Options:
A Velocity only
B Displacement and velocity
C Acceleration, velocity and displacement
D Displacement and acceleration

The correct answer is:
Displacement and velocity

Solution:
When a tennis ball falls on the ground and bounces back, its velocity and displacement changes in reverse direction while direction of acceleration remains same.

Topic: Physics, Motion In One Dimension, BITSAT, BITSAT 2018
Consider the following phenols:

The decreasing order of acidity of the above phenols is

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

The correct answer is:
$\quad \mathrm{III}>\mathrm{IV}>\mathrm{II}>\mathrm{I}$

Solution:
Electron withdrawing group $\left(-\mathrm{NO}_{2}\right)$ increases the acidity while electron releasing group $\left(-\mathrm{CH}_{3},-\mathrm{H}\right)$ decreases acidity. Also effect will be more if functional group is present at para position than ortho and meta position.

Topic: Chemistry, Alcohols Phenols and Ethers, BITSAT, BITSAT 2014
Consider the following radioactive decays

I. U92⟶-α90Th and

II. 90Th⟶-αRa88

In which case group of parent and daughter elements remains unchanged.

Options:
A In (I)
B In (II)
C Both in (I) and (II)
D None of the above

The correct answer is:
In (I)

Solution:
For the following radioactive decays

I. U92⟶-α90Th and
II. Th90⟶-α88Ra

In both Uranium and Thorium belongs to IIIB group.

∵ Elements with atomic number 90 to 103 belongs to same group, i.e. III-B 3rd group (i.e. actinoids).

Topic: Chemistry, Classification of Elements and Periodicity in Properties, BITSAT, BITSAT 2017
Consider the following sequence of reactions.

Z→PCl5X→Alc.KOHY→2H2O/ΔConc.H2SO4Z

'Z' is

Options:
A CH3CH2CH2OH
B (CH3)2C=O
C

D None of these

The correct answer is:

Solution:
When CH3CHOHCH3 is reacted with PCl5, 2-Chloropropane is formed. Thus, X is 2-Chloropropane CH3CHClCH3.

When CH3CHOHCH3 is reacted with alc. KOH , Propene is formed. Thus, Y is CH3-CH=CH2 .

When CH3-CH=CH2 is treated with conc. H2SO4 and H2O, 2-Propanol CH3CHOHCH3 is formed. Thus Z is CH3CHOHCH3.

âââ

Topic: Chemistry, Hydrocarbons, BITSAT, BITSAT 2018
Consider the following statements in respect of the function $f(x)=x^{3}-1, x \in[-1,1]$

I. $f(x)$ is increasing in [-1,1]

II. $f(x)$ has no root in (-1,1) .

Which of the statements given above is/are correct?

Options:
A Only I
B Only II
C Both I and II
D Neither I nor II

The correct answer is:
Only I

Solution:
Since $f(x)$ is an increasing function in [-1,1] and it has a root in (-1,1) $\therefore \quad$ Only statement $I$ is correct.

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2016
Consider the function $\mathrm{f}$ in $\mathrm{A}=\mathrm{R}-\left\{\frac{2}{3}\right\}$ defined as $f(x)=\frac{4 x+3}{6 x-4}$, then $f^{-1}$ is equal to

Options:
A $\frac{3+4 x}{6 x-4}$
B $\frac{6 x-4}{3+4 x}$
C $\frac{3-4 x}{6 x-4}$
D $\frac{9+2 x}{6 x-4}$

The correct answer is:
$\frac{3+4 x}{6 x-4}$

Solution:
Given $f(x)=\frac{4 x+3}{6 x-4}$
Let $y=\frac{4 x+3}{6 x-4}$,
$\begin{aligned}
& \Rightarrow 6 x y-4 y=4 x+3 \Rightarrow x(6 y-4)=3+4 y \\
& \Rightarrow x=\frac{3+4 y}{6 y-4} \\
& f^{-1}(x)=\frac{3+4 x}{6 x-4}
\end{aligned}$

Topic: Mathematics, Functions, BITSAT, BITSAT 2023 (Memory Based Paper 2)
Consider the function $f(x)=|x-1| /x^{2},$ then $f(x)$ is

Options:
A increasing in (0,1)$\cup(2, \infty)$
B increasing in $(-\infty, 0) \cup(1,2)$
C decreasing in (0,1)$\cup(2, \infty)$
D decreasing in (0,1)$\cup(2, \infty)$

The correct answer is:
decreasing in (0,1)$\cup(2, \infty)$

Solution:
$f(x)=\frac{|x-1|}{x^{2}}=\left\{\begin{array}{l}\frac{1-x}{x^{2}}, x<1, x \neq 0 \\ \frac{x-1}{x^{2}}, x>1\end{array}\right.$

Clearly, $f(x)$ is continuous for all $x \in R$ except at $x=0$

$f^{\prime}(x)=\left\{\begin{array}{l}

\frac{x-2}{x^{3}}, x<1, x \pm 0 \\

\frac{2-x}{x^{3}}, x>1

\end{array}\right.$

$f^{\prime}(x)>0 \Rightarrow x<0$ or $1
$f^{\prime}(x)<0 \Rightarrow 02$

Hence, $f(x)$ is increasing in $(-\infty, 0) \cup(1,2)$ and

decreasing in (0,1)$\cup(2, \infty)$.

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2020