datavorous/entrance-exam-dataset huggingface dataset:
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, BITSAT, BITSAT 2010
Negation of "Paris in France and London is in England" is

Options:
A Paris is in England and London is in France
B Paris is not in France or London is not in England
C Paris is in England or London is in France
D None of these

The correct answer is:
Paris is not in France or London is not in England

Solution:
Let \(p\) : Paris is in France, \(q\) : London is in England
\(\therefore \quad\) we have \(p \wedge q\)
Its negation is \(\sim(p \wedge q)=\sim p \vee \sim q\)
i.e., Paris is not in France or London is not in England.

Topic: Mathematics, Mathematical Reasoning, BITSAT, BITSAT 2011
Negation of the Boolean expression $\mathrm{p} \Leftrightarrow(\mathrm{q} \Rightarrow \mathrm{p})$ is

Options:
A $(\sim p) \wedge q$
B $(\mathrm{p}) \wedge(\sim \mathrm{q})$
C $(\sim \mathrm{p}) \vee(\sim \mathrm{q})$
D $(\sim p) \wedge(\sim q)$

The correct answer is:
$(\sim p) \wedge(\sim q)$

Solution:
Given expression is $p \Leftrightarrow(q \Rightarrow p)$
$\sim(\mathrm{p} \leftrightarrow(\mathrm{q} \rightarrow \mathrm{p}))$
$\sim(\mathrm{p} \leftrightarrow \mathrm{q})=(\mathrm{p} \wedge \sim \mathrm{q}) \vee(\mathrm{q} \wedge \sim \mathrm{p})$
$\sim(\mathrm{p} \leftrightarrow(\mathrm{q} \rightarrow \mathrm{p}))$ $=(\mathrm{p} \wedge \sim(\mathrm{q} \rightarrow \mathrm{p})) \vee((\mathrm{q} \rightarrow \mathrm{p}) \wedge \sim \mathrm{p})$
$(\mathrm{p} \wedge \sim(\mathrm{q} \rightarrow \mathrm{p}))=\mathrm{p} \wedge(\mathrm{q} \wedge \sim \mathrm{p})$ $=(p \wedge \sim p) \wedge q=c$
$(q \rightarrow p) \wedge \sim p=(\sim q \vee p) \wedge \sim p$ $=\sim p \wedge(\sim q \vee p)$
$=(\sim \mathrm{p} \wedge \sim \mathrm{q}) \vee(\sim \mathrm{p} \wedge \mathrm{p})=\sim \mathrm{p} \wedge \sim \mathrm{q}$
$\sim(p \leftrightarrow(\dot{q} \rightarrow p))=c \vee(\sim p \wedge \sim q)=\sim p \wedge \sim q$

Topic: Mathematics, Mathematical Reasoning, BITSAT, BITSAT 2023 (Memory Based Paper 1)
Neoprene is a

Options:
A monomer of rubber
B synthetic rubber
C a natural rubber
D vulcanised rubber

The correct answer is:
synthetic rubber

Solution:
Neoprene is a synthetic rubber. It is a polymer of chloroprene (2-chlorobuta,1,3 diene).

Topic: Chemistry, Polymers, BITSAT, BITSAT 2017
Ninhydrin gives yellow colour in paper chromatography with which amino acid?

Options:
A
Tryptophan

B
Proline

C
Alanine

D
Tyrosine

The correct answer is:

Proline

Solution:
The amino acid which gives yellow colour with ninhydrin in paper chromatography is proline.
Ninhydrin + Proline $\longrightarrow$ Yellow-orange product

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2022
Nitrogen is obtained by the thermal decomposition of:

Options:
A $\mathrm{NH}_{4} \mathrm{NO}_{2}$
B $\mathrm{NH}_{4} \mathrm{NO}_{3}$
C $\mathrm{AgNO}_{3}$
D $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$

The correct answer is:
$\mathrm{NH}_{4} \mathrm{NO}_{2}$

Solution:
$\mathrm{NH}_{4} \mathrm{NO}_{2} \stackrel{\Delta}{\rightarrow} \mathrm{N}_{2}+2 \mathrm{H}_{2} \mathrm{O}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), BITSAT, BITSAT 2021
Normals drawn to y2=4ax at the points where it is intersected by the line y=mx+c intersected at P. Coordinates of foot of the another normal drawn to the parabola from the point P is

Options:
A am2,-2am
B 9m,-6am
C am2,-2am
D 4am2,-4am

The correct answer is:
4am2,-4am

Solution:
Let y=mx+c, intersect y2=4ax at Aat12,2at1 and Bat22,2at2.

Then, 2t1+t2=m

⇒t1+t2=2m

Let the foot of another normal be Cat32,2at3

Then, t1+t2+t3=0

t3=t1+t2=-2m

Thus, other foot is 4am2,-4am

Hence, option (d) is correct.

Topic: Mathematics, Parabola, BITSAT, BITSAT 2018
Nuclear energy is released in fission since binding energy per nucleon is

Options:
A sometimes larger and sometimes smaller
B larger for fission fragments than for parent nucleus
C same for fission fragments and nucleus
D smaller for fission fragments than for parent nucleus

The correct answer is:
larger for fission fragments than for parent nucleus

Solution:
Nuclear energy is relased in fission because $\mathrm{BE} /$ nucleon is larger for fission fragments than for parent nucleus.

Topic: Physics, Nuclear Physics, BITSAT, BITSAT 2009
Nucleophilic addition reaction will be most favoured in

Options:
A $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{O}$
B $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}$
C $\mathrm{CH}_{3} \mathrm{CHO}$
D

The correct answer is:
$\mathrm{CH}_{3} \mathrm{CHO}$

Solution:
Aldehydes are more reactive than ketones due to $+\mathrm{I}$ effect of $-\mathrm{CH}_{3}$ group. There are two $-\mathrm{CH}_{3}$ group in acetone which reduces +ve charge density on carbon atom of carbonyl group. More hindered carbonyl group becomes less reactive. So in the given case $\mathrm{CH}_{3} \mathrm{CHO}$ is the right choice.

Topic: Chemistry, Aldehydes and Ketones, BITSAT, BITSAT 2013
Number of atoms of He in 100 amu of He(atomic wt. of He is 4) are:

Options:
A 25
B 100
C 50
D $100 \times 6 \times 10^{-23}$

The correct answer is:
25

Solution:
$\begin{array}{l}
100 \text { amu of He }=\frac{100}{4} \text { atoms of He } \\
=25 \text { atoms. }
\end{array}$
$[1$ a.m.u. $=$ mass of one proton (approx.)]

Topic: Chemistry, Some Basic Concepts of Chemistry, BITSAT, BITSAT 2012
Number of solution of the equation |cosx|=2[x] are (where |x|,[x] are modulus and greatest integer function respectively).

Options:
A 0
B 2
C 1
D infinitely many

The correct answer is:
0

Solution:
Given, |cosx|=2[x]

0≤|cosx|≤1

∴ 0≤2[x]≤1

0≤[x]≤12

x=0

⇒x∈[0,1)

But |cosx|=0

⇒x=nπ+π2 for any value of n∈Z

Hence, the equation has no solution.

Topic: Mathematics, Functions, BITSAT, BITSAT 2019
Number of words from the letters of the words BHARAT in which $B$ and $\mathrm{H}$ will never come together is

Options:
A 210
B 240
C 422
D 400

The correct answer is:
240

Solution:
There are 6 letters in the word BHARAT, 2 of them are identical. Hence total number of words $=6 ! / 2 !=360$
Number of words in which $B$ and $\mathrm{H}$ come together
$=\frac{5 ! 2 !}{2 !}=120$
final ans $=360-120=240$

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2023 (Memory Based Paper 1)
Nylon- 6,6 is obtained from

Options:
A
adipic acid and hexamethylene diamine

B
tetrafluoroethylene

C
vinyl cyanide

D
vinyl benzene

The correct answer is:

adipic acid and hexamethylene diamine

Solution:
The monomer units of nylon- 6,6 are hexamethylene diamine and adipic acid.

Topic: Chemistry, Polymers, BITSAT, BITSAT 2022
OPQR is a square and M, N are the middle points of the sides PQ and QR respectively, then the ratio of the areas of the square and ΔOMN is

Options:
A 4:1
B 2:1
C 8:3
D 4:3

The correct answer is:
8:3

Solution:
Taking the coordinates of vertices O, P, Q, R as (0,0), (a,0), (a,a), (0,a), respectively.

∴ The coordinates of M is a,a2 and N is a2,a

∴ Area of ΔOMN=12001aa/21a/2a1=3a28

and area of the square =a2

∴ The required ratio is 8:3.

Hence, option (c) is correct.

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2018
Observe the given boundary surface diagrams of two orbitals I and II and choose the correct option.

Options:
A $\mathrm{I}-d_{x^2-y^2}, \mathrm{II}-d_{y z}$
B $\mathrm{I}-d_{y z}, \mathrm{II}-d_{x^2-y^2}$
C $\mathrm{I}-d_{x z}, \mathrm{II}-d_{z^2}$
D $\mathrm{I}-d_{x y}, \mathrm{II}-d_{x z}$

The correct answer is:
$\mathrm{I}-d_{y z}, \mathrm{II}-d_{x^2-y^2}$

Solution:
In boundary surface diagram (1) the four lobes lie between $y$ and $z$-axis $\left(d_{y z}\right)$ whereas, in boundary surface diagram (2) the four lobes lie on the $x$ and $y$-axis $\left(d_{x^2-y^2}\right)$.

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2023 (Memory Based Paper 2)
$\mathrm{OF}$ and $\mathrm{F}_2$ can be compared in terms of

Options:
A
$\mathrm{OF}$ is paramagnetic while $\mathrm{F}_2$ is diamagnetic

B
OF is more stable towards dissociation into atoms

C
Both (a) and (b) are correct

D
None of the above is correct

The correct answer is:

Both (a) and (b) are correct

Solution:
This is one unpaired electron in $(\mathrm{O}-\mathrm{F})$, hence paramagnetic, thus (a) is correct.
As bond order of $\mathrm{O}-\mathrm{F}>\mathrm{F}-\mathrm{F}$ thus, bond energy of $\mathrm{O}-\mathrm{F}>\mathrm{F}-\mathrm{F}$.
Hence, (a) and (b) both are correct.

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2022
On a temperature scale ' $X$ '. The boiling point of water is $65^{\circ} \mathrm{X}$ and the freezing point is $-15^{\circ} \mathrm{X}$. Assume that the $\mathrm{X}$ scale is linear. The equivalent temperature corresponding to $-95^{\circ} \mathrm{X}$ on the Farenheit scale would be:

Options:
A $-63^{\circ} \mathrm{F}$
B $-112^{\circ} \mathrm{F}$
C $-48^{\circ} \mathrm{F}$
D $-148^{\circ} \mathrm{F}$

The correct answer is:
$-148^{\circ} \mathrm{F}$

Solution:
Boiling point of water, $=$ UFP $65^{\circ}$
Freezing point of water, $=\mathrm{LFP}-15^{\circ}$
$\frac{\mathrm{X}-\mathrm{LFP}}{\mathrm{UFP}-\mathrm{LFP}}=\frac{\mathrm{T}_{\mathrm{F}}-32}{212-32}$
$\Rightarrow \frac{-95-(-15)}{65-(-15)}=\frac{\mathrm{T}_{\mathrm{F}}-32}{180}$
$\Rightarrow \frac{-80}{80}=\frac{T_F-32}{180} \Rightarrow T_F=-180+32$
$\Rightarrow \mathrm{T}_{\mathrm{F}}=-148^{\circ} \mathrm{F}$

Topic: Physics, Units and Dimensions, BITSAT, BITSAT 2023 (Memory Based Paper 2)
On electrolysis of water, a total of 1 mole of gases is evolved. The amount of water decomposed is

Options:
A 1 mol
B 2 mol
C 13 mol
D 23 mol

The correct answer is:
23 mol

Solution:
∵ Only one mole of total gases are evolved and H2O decomposes as

H2O→decompositionH2+12O2

i.e. total moles =1+12=32 moles by one mole of H2O

Now,

∵ 32 moles of gases are evolved by =1 mole of H2O

∴ 1 mole of gases are evolved by =13/2=23

Hence, 1 mole of total gases are evolved by 23 mole of H2O.

Topic: Chemistry, Some Basic Concepts of Chemistry, BITSAT, BITSAT 2017
On reaction with sodium, 1 mol of a compound $\mathrm{X}$ gives $1 \mathrm{~mol}$ of $\mathrm{H}_{2} .$ Which one of the following compounds might be $X ?$

Options:
A $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}$
B $\mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}$
C $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$
D $\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$

The correct answer is:
$\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$

Solution:
Since 1 mole of compound Xreacts with Na to evolve 1 mole of $\mathrm{H}_{2}$ gas, therefore the compound should have 2 active hydrogen atoms per mole which is possible only in option $d$.

$\begin{array}{l}

\mathrm{CH}_{2} \mathrm{OHCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}+2 \mathrm{Na} \\

\longrightarrow \mathrm{NaOCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{ONa}+2 \mathrm{H}

\end{array}$

Topic: Chemistry, Alcohols Phenols and Ethers, BITSAT, BITSAT 2020
One end of a long metallic wire of length $\mathrm{L}$ tied to the ceiling. The other end is tied with a massless spring of spring constant $\mathrm{K}$. Amass hangs freely from the free end of the spring. The area of cross section and the young's modulus of the wire are A and Y respectively. If the mass slightly pulled down and released, it will oscillate with a time period T equal to:

Options:
A $2 \pi \sqrt{(\mathrm{m} / \mathrm{K})}$
B $2 \pi \sqrt{\mathrm{m}(\mathrm{YA}+\mathrm{KL}) /(\mathrm{YAK})}$
C $2 \pi \sqrt{(\mathrm{m} \mathrm{YA} / \mathrm{KL})}$
D $2 \pi \sqrt{(\mathrm{mL} / \mathrm{YA})}$

The correct answer is:
$2 \pi \sqrt{\mathrm{m}(\mathrm{YA}+\mathrm{KL}) /(\mathrm{YAK})}$

Solution:
Step 1: Calculation of spring constant of wire
Young's modulus of wire, \(\mathrm{Y}=\frac{\Delta \mathrm{F} / \mathrm{A}}{\Delta \mathrm{L} / \mathrm{L}}=\frac{\Delta \mathrm{F}}{\Delta \mathrm{L}} \times \frac{\mathrm{L}}{\mathrm{A}}\)
Due to elasticity of the wire, it behaves as equivalent spring \(\left(\mathrm{K}_{\text {spring }}=\mathrm{F} / \Delta \mathrm{x}\right)\).
Hence, Spring constant for Wire, \(\mathrm{K}_{\mathrm{w}}=\frac{\Delta \mathrm{F}}{\Delta \mathrm{L}}=\mathrm{Y} \frac{\mathrm{A}}{\mathrm{L}}\)
Step 2: Calculation of Equivalent spring constant Let a force \(\mathrm{F}\) be applied to the end of the spring. The wire will extend.
Total extension \(\Delta \mathrm{S}=\mathrm{x}_{\mathrm{wire}}+\mathrm{x}_{\text {spring }}=\frac{\mathrm{F}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{w}}}+\frac{\mathrm{F}_{\mathrm{S}}}{\mathrm{K}}\)
But \(\mathrm{F}_{\mathrm{S}}=\mathrm{F}_{\mathrm{w}} \quad\) (By Newton's, third Law, Both will apply equal and opposite forces at their joint A )
\(\therefore \Delta \mathrm{S}=\mathrm{F}_{\mathrm{w}}\left(\frac{1}{\mathrm{~K}_{\mathrm{w}}}+\frac{1}{\mathrm{~K}}\right)\)
So, Equivalent spring constant \(\left(\mathrm{K}_{\text {eq }}\right)\) is given by:
\(\mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{F}_{\mathrm{eq}}}{\mathrm{x}_{\mathrm{eq}}}=\frac{\mathrm{F}_{\mathrm{w}}}{\Delta \mathrm{S}}=\frac{1}{\frac{1}{\mathrm{~K}_{\mathrm{w}}}+\frac{1}{\mathrm{~K}}}\)
\(\mathrm{K}_{\mathrm{eq}}=\frac{(\mathrm{K})\left(\mathrm{K}_{\mathrm{w}}\right)}{\mathrm{K}_{\mathrm{w}}+\mathrm{K}}=\frac{\mathrm{KY} \frac{\mathrm{A}}{\mathrm{L}}}{\mathrm{Y} \frac{\mathrm{A}}{\mathrm{L}}+\mathrm{K}}=\frac{\mathrm{KYA}}{\mathrm{YA}+\mathrm{KL}}\)
Step 3: Calculation of Time period
Angular frequency \(\omega\) for spring block system is given by:
\(\omega=\sqrt{\frac{\mathrm{K}_{\text {eq }}}{\mathrm{m}}}=\sqrt{\frac{\mathrm{KYA}}{\mathrm{m}(\mathrm{YA}+\mathrm{KL})}}\)
\(\therefore\) Time period, \(\mathrm{T}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{(\mathrm{YA}+\mathrm{KL}) \mathrm{m}}{\mathrm{KYA}}}\)
Hence, Option B is correct

Topic: Physics, Mechanical Properties of Solids, JEE Main
One end of a long metallic wire of length $\mathrm{L}$ tied to the ceiling. The other end is tied with a massless spring of spring constant $\mathrm{K}$. Amass hangs freely from the free end of the spring. The area of cross section and the young's modulus of the wire are A and Y respectively. If the mass slightly pulled down and released, it will oscillate with a time period T equal to:

Options:
A $2 \pi \sqrt{(\mathrm{m} / \mathrm{K})}$
B $2 \pi \sqrt{\mathrm{m}(\mathrm{YA}+\mathrm{KL}) /(\mathrm{YAK})}$
C $2 \pi \sqrt{(\mathrm{m} \mathrm{YA} / \mathrm{KL})}$
D $2 \pi \sqrt{(\mathrm{mL} / \mathrm{YA})}$

The correct answer is:
$2 \pi \sqrt{\mathrm{m}(\mathrm{YA}+\mathrm{KL}) /(\mathrm{YAK})}$

Solution:
Step 1: Calculation of spring constant of wire
Young's modulus of wire, \(\mathrm{Y}=\frac{\Delta \mathrm{F} / \mathrm{A}}{\Delta \mathrm{L} / \mathrm{L}}=\frac{\Delta \mathrm{F}}{\Delta \mathrm{L}} \times \frac{\mathrm{L}}{\mathrm{A}}\)
Due to elasticity of the wire, it behaves as equivalent spring \(\left(\mathrm{K}_{\text {spring }}=\mathrm{F} / \Delta \mathrm{x}\right)\).
Hence, Spring constant for Wire, \(\mathrm{K}_{\mathrm{w}}=\frac{\Delta \mathrm{F}}{\Delta \mathrm{L}}=\mathrm{Y} \frac{\mathrm{A}}{\mathrm{L}}\)
Step 2: Calculation of Equivalent spring constant Let a force \(\mathrm{F}\) be applied to the end of the spring. The wire will extend.
Total extension \(\Delta \mathrm{S}=\mathrm{x}_{\mathrm{wire}}+\mathrm{x}_{\text {spring }}=\frac{\mathrm{F}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{w}}}+\frac{\mathrm{F}_{\mathrm{S}}}{\mathrm{K}}\)
But \(\mathrm{F}_{\mathrm{S}}=\mathrm{F}_{\mathrm{w}} \quad\) (By Newton's, third Law, Both will apply equal and opposite forces at their joint A )
\(\therefore \Delta \mathrm{S}=\mathrm{F}_{\mathrm{w}}\left(\frac{1}{\mathrm{~K}_{\mathrm{w}}}+\frac{1}{\mathrm{~K}}\right)\)
So, Equivalent spring constant \(\left(\mathrm{K}_{\text {eq }}\right)\) is given by:
\(\mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{F}_{\mathrm{eq}}}{\mathrm{x}_{\mathrm{eq}}}=\frac{\mathrm{F}_{\mathrm{w}}}{\Delta \mathrm{S}}=\frac{1}{\frac{1}{\mathrm{~K}_{\mathrm{w}}}+\frac{1}{\mathrm{~K}}}\)
\(\mathrm{K}_{\mathrm{eq}}=\frac{(\mathrm{K})\left(\mathrm{K}_{\mathrm{w}}\right)}{\mathrm{K}_{\mathrm{w}}+\mathrm{K}}=\frac{\mathrm{KY} \frac{\mathrm{A}}{\mathrm{L}}}{\mathrm{Y} \frac{\mathrm{A}}{\mathrm{L}}+\mathrm{K}}=\frac{\mathrm{KYA}}{\mathrm{YA}+\mathrm{KL}}\)
Step 3: Calculation of Time period
Angular frequency \(\omega\) for spring block system is given by:
\(\omega=\sqrt{\frac{\mathrm{K}_{\text {eq }}}{\mathrm{m}}}=\sqrt{\frac{\mathrm{KYA}}{\mathrm{m}(\mathrm{YA}+\mathrm{KL})}}\)
\(\therefore\) Time period, \(\mathrm{T}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{(\mathrm{YA}+\mathrm{KL}) \mathrm{m}}{\mathrm{KYA}}}\)
Hence, Option B is correct

Topic: Physics, Mechanical Properties of Solids, BITSAT, BITSAT 2012
One kilogram of ice at $0^{\circ} \mathrm{C}$ is mixed with one kilogram of water at $80^{\circ} \mathrm{C}$. The final temperature of the mixture is (Take specific heat of water $=4200 \mathrm{~kJ} / \mathrm{kg}-{ }^{\circ} \mathrm{C},$ Latent heat of ice $\left.=336 \mathrm{~kJ} / \mathrm{kg}\right)$

Options:
A $0^{\rho} \mathrm{C}$
B $40^{\circ} \mathrm{C}$
C $50^{\circ} \mathrm{C}$
D $60^{\circ} \mathrm{C}$

The correct answer is:
$0^{\rho} \mathrm{C}$

Solution:
No Solution available

Topic: Physics, Thermal Properties of Matter, BITSAT, BITSAT 2010
One mole of an ideal diatomic gas undergoes a process as shown in the figure. The molar specific heat of the gas in the process is

Options:
A
$3 R / 2$

B
$R / 2$

C
$5 R / 2$

D
$7 R / 2$

The correct answer is:

$3 R / 2$

Solution:
In the curve, $V \propto \frac{1}{T}$
$$
\begin{aligned}
& \Rightarrow V=m \frac{1}{T} \text { where, } m=\text { slope of curve. } \\
& \Rightarrow V T=\text { constant } \\
& \therefore \quad T d V+V d T=0 \\
& \Rightarrow \quad d V=-\frac{V}{T} d T \\
& \text { As, } Q=\Delta U+W \\
& C d T=C_V d T+p d V \\
& C d T=C_V d T-\frac{p}{T} V d T \\
& \Rightarrow \quad C=C_V-\frac{p V}{T} \\
& =C_V-R \quad\left(\because \frac{p V}{T}=R\right) \\
&
\end{aligned}
$$
For diatomic gas, $C_V=\frac{5}{2} R$
$$
\therefore \quad C=C_V-R=\frac{5}{2} R-R=\frac{3 R}{2}
$$

Topic: Physics, Kinetic Theory of Gases, BITSAT, BITSAT 2022
One mole of an ideal gas at $300 \mathrm{~K}$ is expanded isothermally from an initial volume of 1 litre to 10 litres. The $\Delta \mathrm{E}$ for this process is $\left(\mathrm{R}=2\right.$ cal $\left.\mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)$

Options:
A $163.7 \mathrm{cal}$
B zero
C $1381.1 \mathrm{cal}$
D 9 lit atom

The correct answer is:
zero

Solution:
For isothermal process, $\Delta \mathrm{E}=0$

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2009
One mole of $\mathrm{O}_{2}$ gas having a volume equal to 22.4 Litres at $0^{\circ} \mathrm{C}$ and 1 atmospheric pressure in compressed isothermally so that its volume reduces to 11.2 litres. The work done in this process is-

Options:
A $1672.5 \mathrm{~J}$
B $1728 \mathrm{~J}$
C $-1728 \mathrm{~J}$
D $-1572.5 \mathrm{~J}$

The correct answer is:
$-1572.5 \mathrm{~J}$

Solution:
Work done in adiabatic process, \(\mathrm{W}=\mu \mathrm{RT} \log _{\mathrm{e}}\left(\frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{\mathrm{i}}}\right)\)

\(\mathrm{W}=1 \times 8.314 \times 273 \times \log _{\mathrm{e}}\left(\frac{11.2}{22.4}\right)\)

\(\mathrm{W}=8.314 \times 273 \times \log _{\mathrm{e}}\left(\frac{1}{2}\right)\)

\(\mathrm{W}=8.314 \times 273 \times\left[\log _{\mathrm{e}}(1)-\log _{\mathrm{e}}(2)\right]\)

\(\mathrm{W}=8.314 \times 273 \times(0-0.69)\)

\(\mathrm{W}=-1566.1 \mathrm{~J}\)

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2014
One moles each of four ideal gases are kept as follows.

I. 5 L of gas (A) at 2 atm pressure

II. 2.5 L of gas (B) at 2 atm pressure

III. 1.25 L of gas (C) at 2 atm pressure

IV. 2.5 L of gas (D) at 2.5 atm pressure

Which of the above gases is kept at highest temperature.

Options:
A Gas (A)
B Gas (B)
C Gas (C)
D Gas (D)

The correct answer is:
Gas (A)

Solution:
∵ For ideals gas

pV=nRT

or T=pVnR

∵ n=1= constant

R=0.083= constant

∴ T∝PV

Thus,

The gas that shows highest product of p×V, has highest temperature.

for
(A) p×V=2×5=10 atm L

(B) p×V=2×2.5=5 atmL

(C) p×V=2×1.25=2.5 atmL

(D) p×V=2.5×2.5=6.25 atm L

Hence gas (A) show highest temperature.

Topic: Chemistry, States of Matter, BITSAT, BITSAT 2017
One of the lines in the emission spectrum of $\mathrm{Li}^{2+}$ has the same wavelength as that of the $2^{\text {nd }}$ line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is $\mathrm{n}=12 \rightarrow \mathrm{n}=\mathrm{x} .$ Find the value of $\mathrm{x}$

Options:
A 8
B 6
C 7
D 5

The correct answer is:
6

Solution:
For $2^{\text {nd line of Balmer series in hydrogen }}$ spectrum

$$

\frac{1}{\lambda}=\mathrm{R}(1)\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{3}{16} \mathrm{R}

$$

$$

\text { For } \mathrm{Li}^{2+}\left[\frac{1}{\lambda}=\mathrm{R} \times 9\left(\frac{1}{\mathrm{x}^{2}}-\frac{1}{12^{2}}\right)=\frac{3 \mathrm{R}}{16}\right]

$$

which is satisfied by $n=12 \rightarrow n=6$

Topic: Physics, Atomic Physics, BITSAT, BITSAT 2015
Ordinary glass is:

Options:
A Sodium silicate
B Calcium silicate
C Sodium and calcium silicate
D Mixed salt of $\mathrm{Na}$ and $\mathrm{Ca}$

The correct answer is:
Sodium silicate

Solution:
Normal glass is calcium alkali silicate glass made by fusing the alkali metal carbonate, $\mathrm{CaCO}_{3}$ and $\mathrm{SiO}_{2}$

Topic: Chemistry, p Block Elements (Group 13 & 14), BITSAT, BITSAT 2015
ortho and para form of hydrogen have

Options:
A
different physical and chemical properties

B
identical physical properties but different chemical properties

C
identical chemical properties but different physical properties

D
identical chemical and physical properties

The correct answer is:

identical chemical properties but different physical properties

Solution:
Ortho and para hydrogen similar to each other in their chemical properties but they have different physical properties like boiling point, thermal conductivity, due to their difference in overall spins.

Topic: Chemistry, Hydrogen, BITSAT, BITSAT 2022
$\mathrm{P}$ and $\mathrm{Q}$ are isomers. Identify $\mathrm{Q}$

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
$\mathrm{P}$ and $\mathrm{Q}$ are isomers. Identify $\mathrm{Q}$

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, BITSAT, BITSAT 2014
$\mathrm{PCl}_{5}$ is dissociating $50 \%$ at $250^{\circ} \mathrm{C}$ at a total pressure of $\mathrm{P}$ atm. If equilibrium constant is $\mathrm{K}_{p}$, then which of the following relation is numerically correct -

Options:
A $\mathrm{K}_{p}=3 \mathrm{P}$
B $\mathrm{P}=3 \mathrm{~K}_{p}$
C $\mathrm{P}=\frac{2 K_{\mathrm{P}}}{3}$
D $\mathrm{~K}_{p}=\frac{2 \mathrm{P}}{3}$

The correct answer is:
$\mathrm{P}=3 \mathrm{~K}_{p}$

Solution:
$\mathrm{K}_{\mathrm{p}}=\frac{\frac{\mathrm{P}}{3} \times \frac{\mathrm{P}}{3}}{\frac{\mathrm{P}}{3}}=\frac{\mathrm{P}}{3}$

Topic: Chemistry, Chemical Equilibrium, BITSAT, BITSAT 2021
Period of \(\frac{\sin \theta+\sin 2 \theta}{\cos \theta+\cos 2 \theta}\) is

Options:
A \(2 \pi\)
B \(\pi\)
C \(\frac{2 \pi}{3}\)
D \(\frac{\pi}{3}\)

The correct answer is:
\(\frac{2 \pi}{3}\)

Solution:
\(\begin{aligned}
& \frac{\sin \theta+\sin 2 \theta}{\cos \theta+\cos 2 \theta}=\frac{2 \sin \left(\frac{3 \theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}{2 \cos \left(\frac{3 \theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}=\tan \left(\frac{3 \theta}{2}\right) \\
& \text {Hence period }=\frac{2 \pi}{3}
\end{aligned}\)

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2009
pH of a saturated solution of BaOH2 is 12 . The value of its Ksp is

Options:
A 3.3×10-7 M3
B 5.0×10-7 M3
C 4.0×10-6 M3
D 5.0×10-6 M

The correct answer is:
5.0×10-7 M3

Solution:
Given that,

pH of BaOH2=12

∵ pH+pOH=14

∴ pOH=14-pH=14-12=2

Now, pOH=-logOH-

⇒ 2=-logOH

OH-=10-2 M

BaOH2sS mol L-1⇌Ba2+S+2OH-2S

OH-=2S=10-2 M

∴ S=0.5×10-2=5×10-3 M

Ksp=Ba2+OH-2

⇒(5×10-3)(10-2)2=5×10-7M3

Therefore, the value of Ksp is 5.0×10-7 M3.

Topic: Chemistry, Ionic Equilibrium, BITSAT, BITSAT 2018