datavorous/entrance-exam-dataset huggingface dataset:
Consider the greatest integer function, defined by f(x)=[x],0≤x<2. Then

Options:
A f is derivable at x=1
B f is not derivable at x=1
C f is derivable at x=2
D None of these

The correct answer is:
f is not derivable at x=1

Solution:
The given function is

fx=0,0≤x<11,1≤x<2

Now, LHD=limx→1-f(x)-f(1)x-1=limx→1-0-1x-1

Which does not exist

∴ f is not derivable at x=1.

Topic: Mathematics, Continuity and Differentiability, BITSAT, BITSAT 2017
Consider the optical system shown in Fig. The point source of light $S$ is having wavelength equal to $\lambda$. The light is reaching screen only after reflection. For point $P$ to besecond maxima, the value of $\lambda$ would be $(\mathrm{D}>>d$ and $d>>\lambda)$

Options:
A $\frac{12 d^{2}}{\mathrm{D}}$
B $\frac{6 d^{2}}{\mathrm{D}}$
C $\frac{3 d^{2}}{\mathrm{D}}$
D $\frac{24 d^{2}}{\mathrm{D}}$

The correct answer is:
$\frac{12 d^{2}}{\mathrm{D}}$

Solution:
$\text { At } \mathrm{P}, \Delta x=\frac{(8 d) \times 3 d}{\mathrm{D}}$

For 2 nd maxima, $\Delta x=2 \lambda$

$\begin{array}{l}

\Rightarrow \frac{24 d^{2}}{\mathrm{D}}=2 \lambda \\

\Rightarrow \lambda=\frac{12 d^{2}}{\mathrm{D}}

\end{array}$

Topic: Physics, Wave Optics, BITSAT, BITSAT 2021
Consider the reaction :

$\mathrm{Cl}_{2}(a q)+\mathrm{H}_{2} \mathrm{~S}(a q) \rightarrow$

$\mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q)$

The rate equation for this reaction is rate $=k\left[\mathrm{Cl}_{2}\right]\left[\mathrm{H}_{2} \mathrm{~S}\right]$

Which of these mechanisms is/are consistent with this rate equation?

A. $\mathrm{Cl}_{2}+\mathrm{H}_{2} \mathrm{~S} \rightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}+\mathrm{Cl}^{+}+\mathrm{HS}^{-}$ (slow)

$\mathrm{Cl}^{+}+\mathrm{HS}^{-} \rightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}+\mathrm{S}$

B. $\mathrm{H}_{2} \mathrm{~S} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HS}^{-}$ (fast equilibrium)

$\mathrm{Cl}_{2}+\mathrm{HS}^{-} \rightarrow 2 \mathrm{Cl}^{-}+\mathrm{H}^{+}+\mathrm{S}$ (Slow)

Options:
A $\mathrm{B}$ only
B $\quad$ Both $\mathrm{A}$ and $\mathrm{B}$
C Neither A nor B
D A only

The correct answer is:
A only

Solution:
No Solution

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2016
Coolant used in car radiator is aqueous solution of ethylene glycol. In order to prevent the solution from freezing at $-0.3^{\circ} \mathrm{C}$. How much ethylene glycol must be added to $5 \mathrm{~kg}$ of water ? $\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{Kg} \mathrm{mol}^{-1}\right)$

Options:
A $50 \mathrm{~g}$
B $55 \mathrm{~g}$
C $45 \mathrm{~g}$
D $40 \mathrm{~g}$

The correct answer is:
$50 \mathrm{~g}$

Solution:
$\Delta \mathrm{T}_{\mathrm{f}}=0.3^{\circ} \mathrm{C}$
$$
\begin{array}{l}
\Delta \mathrm{T}_{\mathrm{f}}=\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}} \\
0.3=\frac{1.86 \times \mathrm{W}_{\mathrm{B}} \times 1000}{62 \times 5000}
\end{array}
$$
$\therefore \mathrm{W}_{\mathrm{B}}=50 \mathrm{~g}$
The amount used should be more than $50 \mathrm{~g}$.

Topic: Chemistry, Solutions, BITSAT, BITSAT 2010
$\cos ^{2}\left(\frac{\pi}{6}+\theta\right)-\sin ^{2}\left(\frac{\pi}{6}-\theta\right)=$

Options:
A $\frac{1}{2} \cos 2 \theta$
B 0
C $-\frac{1}{2} \cos 2 \theta$
D $\frac{1}{2}$

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

Solution:
$\begin{aligned} & \cos ^{2}\left(\frac{\pi}{6}+\theta\right)-\sin ^{2}\left(\frac{\pi}{6}-\theta\right) \\ &=\cos \left(\frac{\pi}{6}+\theta+\frac{\pi}{6}-\theta\right) \cos \left(\frac{\pi}{6}+\theta-\frac{\pi}{6}+\theta\right) \\ &=\cos \frac{2 \pi}{6} \cos 2 \theta=\frac{1}{2} \cos 2 \theta \end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, BITSAT, BITSAT 2014
$\mathrm{Cu}_{\mathrm{aq}}^{+}$ is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction :

$2 \mathrm{Cu}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$

choose correct $\mathrm{E}^{\circ}$ for above reaction if

$\mathrm{E}^{\circ} \mathrm{Cu}^{2+}=0.34 \mathrm{~V}$ and $\mathrm{E}^{\circ} \mathrm{Cu}^{2+}=0.15 \mathrm{~V}$

Options:
A $-0.38 \mathrm{~V}$
B $+0.49 \mathrm{~V}$
C $+0.38 \mathrm{~V}$
D $-0.19 \mathrm{~V}$

The correct answer is:
$+0.38 \mathrm{~V}$

Solution:
$2 \mathrm{Cu}^{+} \longrightarrow \mathrm{Cu}^{+2}+\mathrm{Cu}$

$2 \mathrm{e}^{-}+\mathrm{Cu}^{+2} \longrightarrow \mathrm{Cu} ; \mathrm{E}_{1}^{\circ}=0.34 \mathrm{~V} ; \ldots$ (i)

$\mathrm{e}^{-}+\mathrm{Cu}^{+2} \longrightarrow \mathrm{Cu}^{+} ; \mathrm{E}_{2}^{\circ}=0.15 \mathrm{~V} ; \ldots$ (ii)

$\mathrm{Cu}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Cu} ; \mathrm{E}_{3}^{\circ}=?$ ...(iii)

Now, $\Delta \mathrm{G}_{1}^{\circ}=-\mathrm{nFE}_{1}^{\circ}=-2 \times 0.34 \mathrm{~F}$

$\Delta \mathrm{G}_{2}^{\circ}=-1 \times 0.15 \mathrm{~F}, \Delta \mathrm{G}_{3}^{\circ}=-1 \times \mathrm{E}_{3}^{\circ} \mathrm{F}$

Again, $\Delta \mathrm{G}_{1}^{\circ}=\Delta \mathrm{G}_{2}^{\circ}+\Delta \mathrm{G}_{3}^{\circ}$

$\Rightarrow-0.68 \mathrm{~F}=-0.15 \mathrm{~F}-\mathrm{E}_{3} \mathrm{~F}$

$0 \Rightarrow \mathrm{E}_{3}^{\circ}=0.68-0.15=0.53 \mathrm{~V}$

$\mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\text {cathode }}^{\circ}\left(\mathrm{Cu}^{+} / \mathrm{Cu}\right)$

$=0.53-0.15=0.38 \mathrm{~V}$

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2014
$\frac{d}{d x}\left(\tan ^{-1} \sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}\right)$ is equal to

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

The correct answer is:
$-\frac{1}{4}$

Solution:
We have $\frac{d}{d x}\left(\tan ^{-1} \sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}\right)$

$=\frac{d}{d x}\left[\tan ^{-1} \sqrt{\frac{1+\left(2 \cos ^{2} \frac{x}{4}-1\right)}{1-\left(1-2 \sin ^{2} \frac{x}{4}\right)}}\right]$

$=\frac{d}{d x}\left(\tan ^{-1} \sqrt{\left.\frac{2 \cos ^{2} \frac{x}{4}}{2 \sin ^{2} \frac{x}{4}}\right)}\right.$

$=\frac{d}{d x}\left(\tan ^{-1} \sqrt{\cot ^{2} \frac{x}{4}}\right)=\frac{d}{d x}\left(\tan ^{-1} \cot \frac{x}{4}\right)$

$=\frac{d}{d x}\left(\tan ^{-1} \tan \left(\frac{\pi}{2}-\frac{x}{4}\right)\right)=\frac{d}{d x}\left(\frac{\pi}{2}-\frac{x}{4}\right)=0-\frac{1}{4}=-\frac{1}{4}$

Topic: Mathematics, Differentiation, BITSAT, BITSAT 2021
$\left(\frac{d y}{d x}\right) \tan x=y \sec ^2 x+\sin x$, find general solution

Options:
A
$y=\tan x(\log |\operatorname{cosec} x-\cot x|+\cos x+c)$

B
$y=\sec ^2 x+\tan x+c$

C
$y=\log |\sec x+\tan x|+\operatorname{cosec} x+c$

D
$y=\tan ^2 x+\sin x+c$

The correct answer is:

$y=\tan x(\log |\operatorname{cosec} x-\cot x|+\cos x+c)$

Solution:
Given that
$$
\begin{aligned}
& \left(\frac{d y}{d x}\right) \tan x=y \sec ^2 x+\sin x \\
& \tan x\left(\frac{d y}{d x}\right)-y \sec ^2 x=\sin x \\
& \frac{d y}{d x}+\left(\frac{-\sec ^2 x}{\tan x}\right) y=\frac{\sin x}{\tan x} \\
& \text { Here, } P=-\frac{\sec ^2 x}{\tan x} \text { and } Q=\cos x \\
& \text { IF }=e^{\int P d x}=e^{\int \frac{-\sec ^ 2 x}{\tan x} d x} \\
& \text { IF }=e^{-\log (\tan x)}=e^{\log \cot x}=\cot x \\
& \text { Now, } y \times \mathrm{IF}=\int(Q \times \mathrm{IF}) d x+C \\
& y \cdot \cot x=\int \cos x \cdot \cot x+C \\
& y \cdot \cot x=\int \frac{1-\sin ^2 x}{\sin x}+C \\
& y \cdot \cot x=\log |\operatorname{cosec} x-\cot x|+\cos x+C
\end{aligned}
$$

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2022
Determine the current in $2 \Omega$ resistor.

Options:
A $1 \mathrm{~A}$
B $1.5 \mathrm{~A}$
C $0.9 \mathrm{~A}$
D $0.6 \mathrm{~A}$

The correct answer is:
$0.9 \mathrm{~A}$

Solution:
At steady state the capacitor will be fully charged and thus there will be no current in the $1 \Omega$ resistance. So the effective circuit becomes

Net current from the $6 \mathrm{~V}$ battery,

$I=\frac{6}{\left(\frac{2 \times 3}{2+3}\right)+\frac{2.8}{1}}=\frac{6}{1.2+2.8}=\frac{3}{2}=1.5 \mathrm{~A}$

Between $\mathrm{A}$ and $\mathrm{B}$, voltage is same in both resistances, $2 \mathrm{I}_{1}=3 \mathrm{I}_{2}$ where

$\mathrm{I}_{1}+\mathrm{I}_{2}=\mathrm{I}=1.5$

$\Rightarrow 2 \mathrm{I}_{1}=3\left(1.5-\mathrm{I}_{1}\right) \Rightarrow \mathrm{I}_{1}=0.9 \mathrm{~A}$

Topic: Physics, Current Electricity, BITSAT, BITSAT 2020
Diborane $\left(\mathrm{B}_{2} \mathrm{H}_{6}\right)$ reacts independently with $\mathrm{O}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$ to produce, respectively:

Options:
A $\mathrm{B}_{2} \mathrm{O}_{3}$ and $\mathrm{H}_{3} \mathrm{BO}_{3}$
B $\mathrm{B}_{2} \mathrm{O}_{3}$ and $\left[\mathrm{BH}_{4}\right]^{-}$
C $\mathrm{H}_{3} \mathrm{BO}_{3}$ and $\mathrm{B}_{2} \mathrm{O}_{3}$
D $\mathrm{HBO}_{2}$ and $\mathrm{H}_{3} \mathrm{BO}_{3}$

The correct answer is:
$\mathrm{B}_{2} \mathrm{O}_{3}$ and $\mathrm{H}_{3} \mathrm{BO}_{3}$

Solution:
$\mathrm{B}_{2} \mathrm{H}_{6}+3 \mathrm{O}_{2} \rightarrow \mathrm{B}_{2} \mathrm{O}_{3}+3 \mathrm{H}_{2} \mathrm{O}$

$\mathrm{B}_{2} \mathrm{H}_{6}+6 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{H}_{3} \mathrm{BO}_{3}+6 \mathrm{H}_{2}$

Topic: Chemistry, p Block Elements (Group 13 & 14), BITSAT, BITSAT 2021
Dimension of which base quantity corresponds to that Ghc3=?

Options:
A Time
B Length
C Mass
D Temperalure

The correct answer is:
Length

Solution:
As, gravitational constant,

G=M-1 L3 T-2

Planck's constant, h=ML2 T-1

Hence, Gh=M-1L3T-2ML2T-1

=M0L5 T-3

Velocity of light, [c]=LT-1

Now, Ghc31/2=L5T-31/2L3T-31/2=L21/2=L

Hence [L]= length

Topic: Physics, Units and Dimensions, BITSAT, BITSAT 2018
Due to Frenkel defect, the density of ionic solids

Options:
A decreases
B increases
C does not change
D changes

The correct answer is:
does not change

Solution:
No change in density.

Topic: Chemistry, Solid State, BITSAT, BITSAT 2013
During the kinetic study of the reaction, $2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}$, following results were obtained:

$$
\begin{array}{|c|c|c|c|}
\hline \text { Run } & {[\mathrm{A}] / \mathrm{mol} \mathrm{L}^{-1}} & {[\mathrm{~B}] / \mathrm{mol} \mathrm{L}^{-1}} & \begin{array}{l}
\text { Initial rate of } \\
\text { formation of } \\
\mathrm{D} / \mathrm{mol} \mathrm{L}^{-1} \mathrm{~min}^{-1}
\end{array} \\
\hline \text { I } & 0.1 & 0.1 & 6.0 \times 10^{-1} \\
\hline \text { II } & 0.3 & 0.2 & 7.2 \times 10^{-1} \\
\hline \text { III } & 0.3 & 0.4 & 28.8 \times 10^{-1} \\
\hline \text { IV } & 0.4 & 0.1 & 24.0 \times 10^{-1} \\
\hline
\end{array}
$$

Based on the above data which one of the following is correct?

Options:
A rate $=k[\mathrm{~A}]^{2}[\mathrm{~B}]$
B rate $=k[\mathrm{~A}][\mathrm{B}]$
C rate $=k[\mathrm{~A}]^{2}[\mathrm{~B}]^{2}$
D rate $=k[\mathrm{~A}][\mathrm{B}]^{2}$

The correct answer is:
rate $=k[\mathrm{~A}][\mathrm{B}]^{2}$

Solution:
In case of (II) and (III) Keeping concentration of [A] constant, when the concentration of $[\mathrm{B}]$ is doubled, the rate quadruples. Hence it is second order with respect to $\mathrm{B}$. In case of $\mathrm{I} \&$ IV Keeping the concentration of [B] constant, when the concentration of $[\mathrm{A}]$ is increased four times, rate also increases four times. Hence, the order with respect to $\mathrm{A}$ is one. Hence Rate $=k[\mathrm{~A}][\mathrm{B}]^{2}$

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2011
E1, E2 and E3 are the emf of the following three galvanic cells respectively

(i) ZnsZn2+0.1 MCu2+1 MCus

(ii) ZnsZn2+1 MCu2+1 MCus

(iii) ZnsZn2+1 MCu2+0.1 MCus

Which of the following is true?

Options:
A E2>E1>E3
B E1>E2>E3
C E3>E1>E2
D E3>E2>E1

The correct answer is:
E1>E2>E3

Solution:
Key-idea

Use Nernst equation, i.e.

Eemf=E°cell-0.05912logZn2+Cu2+

For reaction i

E1=E°cell-0.05912log0.11

E1=E°cell+0.05912

For reaction ii

E2=E°cell-0.05912log11=0

E2=E°cell

For reaction iii

E3=E°cell-0.05912log10.1

E3=E°cell-0.05912

Hence, E1>E2>E3.

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2019
Eddy currents are produced when

Options:
A a metal is kept in varying magnetic field
B a metal is kept in steady magnetic field
C a circular coil is placed in a magnetic field
D through a circular coil, current is passed

The correct answer is:
a metal is kept in varying magnetic field

Solution:
Eddy can be created when a conductor is moving through a magnetic field, or when the magnetic field surrounding a stationary conductor is varying. Option A

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2013
$[\mathrm{EDTA}]^{4-}$ is a :

Options:
A monodentate ligand
B bidentate ligand
C quadridentate ligand
D hexadentate ligand

The correct answer is:
hexadentate ligand

Solution:
It can act as hexadentate ligand as it has six donar atoms ( 2 nitrogen atoms and 4 oxygen atom).

Topic: Chemistry, Coordination Compounds, BITSAT, BITSAT 2009
Eight equal drops of water are falling through air with a steady speed of $10 \mathrm{~cm} / \mathrm{s}$. If the drops coalesce, the new velocity is:

Options:
A $10 \mathrm{~cm} / \mathrm{s}$
B $40 \mathrm{~cm} / \mathrm{s}$
C $16 \mathrm{~cm} / \mathrm{s}$
D $5 \mathrm{~cm} / \mathrm{s}$

The correct answer is:
$40 \mathrm{~cm} / \mathrm{s}$

Solution:
Let $\mathrm{r}$ be the radius of small drops of water.
$\mathrm{R}=$ radius of big drop formed as volume remain same.
$\therefore 8 . \frac{4}{3} \pi \mathrm{r}^3-\frac{4}{3} \pi \mathrm{R}^3 \Rightarrow \mathrm{R}=2 \mathrm{r}$
Terminal velocity,
$\begin{aligned}
& \mathrm{v}_{\mathrm{T}}=\frac{2}{9 \eta}(\rho-\sigma) \mathrm{r}^2 \mathrm{~g} \\
& \therefore \mathrm{v}_{\mathrm{T}} \propto \mathrm{r}^2 \\
& \therefore \frac{\mathrm{v}_1}{\mathrm{v}_2}=\left(\frac{\mathrm{r}}{\mathrm{R}}\right)^2 \\
& \Rightarrow \frac{10}{\mathrm{v}_2}=\left(\frac{1}{2}\right)^2\left(\because \mathrm{v}_1=10 \mathrm{~cm} / \mathrm{s} \text { given }\right) \\
& \quad \Rightarrow \mathrm{v}_2=40 \mathrm{~cm} / \mathrm{s}
\end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2023 (Memory Based Paper 2)
Electric field in a certain region is given by $\vec{E}=\left(\frac{A}{x^2} \hat{i}+\frac{B}{y^3} \hat{j}\right)$. The SI unit of $A$ and $B$ are:

Options:
A $\mathrm{Nm}^3 \mathrm{C}^{-1} ; \mathrm{Nm}^2 \mathrm{C}^{-1}$
B $\mathrm{Nm}^2 \mathrm{C}^{-1} ; \mathrm{Nm}^3 \mathrm{C}^{-1}$
C $\mathrm{Nm}^3 \mathrm{C} ; \mathrm{Nm}^2 \mathrm{C}$
D $\mathrm{Nm}^2 \mathrm{C} ; \mathrm{Nm}^3 \mathrm{C}$

The correct answer is:
$\mathrm{Nm}^2 \mathrm{C}^{-1} ; \mathrm{Nm}^3 \mathrm{C}^{-1}$

Solution:
Electric field in a certain region is given by,
$\vec{E}=\frac{A}{x^2} \hat{i}+\frac{B}{y^3} \hat{j}$
$\left[\frac{\mathrm{A}}{\mathrm{x}^2}\right]=\mathrm{NC}^{-1} \Rightarrow[\mathrm{A}]=\mathrm{Nm}^2 \mathrm{C}^{-1}$
$\left[\frac{\mathrm{B}}{\mathrm{y}^3}\right]=\mathrm{NC}^{-1} \Rightarrow[\mathrm{B}]=\mathrm{Nm}^3 \mathrm{C}^{-1}$

Topic: Physics, Units and Dimensions, BITSAT, BITSAT 2023 (Memory Based Paper 1)
Electric field in theregion is given by $\mathrm{E}=\left(\frac{\mathrm{M}}{\mathrm{x}^{3}}\right) \hat{\mathrm{i}}$, then the correct expression for the potential in the region is [assume potential at infinity is zero]

Options:
A $\frac{\mathrm{M}}{2 \mathrm{x}^{2}}$
B $\mathrm{Mx}^{2}$
C $\frac{\mathrm{M}}{3 \mathrm{x}^{4}}$
D $\frac{\mathrm{M}}{\mathrm{x}^{2}}$

The correct answer is:
$\frac{\mathrm{M}}{2 \mathrm{x}^{2}}$

Solution:
$V=-\int_{\infty}^{x} E \cdot d r=-\int_{\infty}^{x} \frac{M}{x^{3}} d x=\frac{M}{2 x^{2}}$

Topic: Physics, Electrostatics, BITSAT, BITSAT 2014
Electrode potential data are given below:
$\mathrm{Fe}^{+3}(\mathrm{aq})+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{+2}(\mathrm{aq}) ; \mathrm{E}^{\circ}=+0.77 \mathrm{~V}$
$\mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}_{(\mathrm{s})} ; \mathrm{E}^{\circ}=-1.66 \mathrm{~V}$
$\mathrm{Br}_2(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Br}^{-}(\mathrm{aq}) ; \mathrm{E}^{\circ}=+1.08 \mathrm{~V}$
Based on the data, the reducing power of $\mathrm{Fe}^{2+}$, $\mathrm{Al}$ and $\mathrm{Br}^{-}$will increase in the order

Options:
A $\mathrm{Br}^{-} < \mathrm{Fe}^{2+} < \mathrm{Al}$
B $\mathrm{Fe}^{2+} < \mathrm{Al} < \mathrm{Br}^{-}$
C $\mathrm{Al} < \mathrm{Br}^{-} < \mathrm{Fe}^{2+}$
D $\mathrm{Al} < \mathrm{Fe}^{2+} < \mathrm{Br}^{-}$

The correct answer is:
$\mathrm{Br}^{-} < \mathrm{Fe}^{2+} < \mathrm{Al}$

Solution:
$\begin{array}{rrrr} & \text { Fe } & \text { A1 } & \text { B } \\ E_{\text {Red }}^{\circ} & 0.77 & -1.66 & 1.08 \\ E_{\text {Oxi }}^{\circ} & -0.77 & 1.66 & -1.08\end{array}$
Hence, reducing power $\mathrm{Al}>\mathrm{Fe}^{2+}>\mathrm{Br}$

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2023 (Memory Based Paper 1)
Electrolysis is carried out in three cells

(A) $1.0 \mathrm{M} \mathrm{OuSO}_{4}$ Pt electrode

(B) $1.0 \mathrm{M} \mathrm{CuSO}_{4}$ copper electrodes

(C) $1.0 \mathrm{M} \mathrm{KCl}$ Pt electrodes If volume of electrolytic solution is maintained constant in each of the cell, which is correct set of pH changes in $(\mathrm{A}),(\mathrm{B})$ and $(\mathrm{C})$ cell respectively?

Options:
A decrease in all the three
B increase in all the three
C decrease, constant, increase
D increase, constant, increase

The correct answer is:
decrease, constant, increase

Solution:
No Solution

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2016
Electrons are accelerated through a potential difference $\mathrm{V}$ and protons are accelerated through a potential difference $4 \mathrm{~V}$. The de - Broglie wavelengths are $\lambda_{e}$ and $\lambda_{p}$ for electrons and protons respectively. The ratio of $\frac{\lambda_{e}}{\lambda_{p}}$ is given by: (given $m_{e}$ is mass of electron and $m_{p}$ is mass of proton).

Options:
A $\sqrt{\frac{m_{p}}{m_{e}}}$
B $\sqrt{\frac{m_{e}}{m_{p}}}$
C $\frac{1}{2} \sqrt{\frac{m_{e}}{m_{p}}}$
D $2 \sqrt{\frac{m_{p}}{m_{e}}}$

The correct answer is:
$2 \sqrt{\frac{m_{p}}{m_{e}}}$

Solution:
Energy $=$ charge $\times$ potential diff.

$\mathrm{E}_{\text {electron }}=\mathrm{q}_{\mathrm{e}} \mathrm{V}$ and $\mathrm{E}_{\text {proton }}=\mathrm{q}_{\mathrm{p}} 4 \mathrm{~V}$

de $-$ Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$

$\lambda_{e}=\frac{h}{\sqrt{2 m_{e} \mathrm{eV}}}$ and $\lambda_{\mathrm{P}}=\frac{h}{\sqrt{2 m_{\mathrm{p}} e 4 \mathrm{~V}}} \because\left(q_{e}=q_{p}\right)$

$\therefore \frac{\lambda_{e}}{\lambda_{\mathrm{P}}}=\frac{\frac{h}{\sqrt{2 m_{e} \mathrm{eV}}}}{\frac{h}{\sqrt{2 m_{\mathrm{p}} e 4 \mathrm{~V}}}}=2 \sqrt{\frac{m_{\mathrm{P}}}{m_{e}}}$

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2021
Element 'B' forms ccp structure and 'A' 'occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is :

Options:
A $\mathrm{A}_{2} \mathrm{BO}_{4}$
B $\mathrm{AB}_{2} \mathrm{O}_{4}$
C $\mathrm{A}_{2} \mathrm{~B}_{2} \mathrm{O}$
D $\mathrm{A}_{4} \mathrm{~B}_{2} \mathrm{O}$

The correct answer is:
$\mathrm{AB}_{2} \mathrm{O}_{4}$

Solution:
No. of lattice points $=$ No. of octahedral voids $=\frac{1}{2} \times$ No. of tetrahedral

voids in $c c p$ structure

$\therefore$ No. of atoms of $B=4$

No. of atoms of $A=\frac{1}{2} \times$ No. of octahedral voids

$=\frac{1}{2} \times 4=2$

No. of atoms of $O=$ All tetrahedral voids $=2 \times$ No. of lattice points

$=2 \times 4=8$

Hence, $A: B: O=1: 2: 4$

Therefore, the formula of the compound is $A B_{2} O_{4}$

Topic: Chemistry, Solid State, BITSAT, BITSAT 2021
Element with electronic configuration $1 s^2 2 s^2 2 p^6$ $3 s^2 3 p^6 3 d^{10} 4 s^2 4 p^6 4 d^{10} 5 s^2 5 p^3$ belongs to the following group of the periodic table

Options:
A 5th
B 15th
C $3 \mathrm{rd}$
D 17th

The correct answer is:
15th

Solution:
Its valence shell has 5 electrons $\left(n s^2, n p^3\right)$ It belongs to 15 th group of the periodic table.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, BITSAT, BITSAT 2023 (Memory Based Paper 1)
Elvaluate $\int_{0}^{\pi / 2} \frac{\sin x}{1+\cos ^{2} x} d x$

Options:
A $\pi / 2$
B $\pi / 4$
C $\pi / 3$
D $\pi$

The correct answer is:
$\pi / 4$

Solution:
Let $\mathrm{I}=\int_{0}^{\pi / 2} \frac{\sin \mathrm{x}}{1+\cos ^{2} \mathrm{x}} \mathrm{dx}$

Let $\cos \mathrm{x}=\mathrm{t}$ and $-\sin \mathrm{x} \mathrm{d} \mathrm{x}=\mathrm{dt}$

Now, $\mathrm{x}=0 \Rightarrow \mathrm{t}=\cos 0=1$ and

$\mathrm{x}=\frac{\pi}{2} \Rightarrow \mathrm{t}=\cos \frac{\pi}{2}=0$

$\therefore \quad \mathrm{I}=\int_{1}^{0} \frac{\sin \mathrm{x}}{1+\mathrm{t}^{2}}\left(\frac{-\mathrm{dt}}{\sin \mathrm{x}}\right)=-\int_{1}^{0} \frac{\mathrm{dt}}{1+\mathrm{t}^{2}}$

$=\left[\tan ^{-1} \mathrm{t}\right]_{1}^{0}=-\left[0-\frac{\pi}{4}\right]=\frac{\pi}{4}$

Topic: Mathematics, Definite Integration, BITSAT, BITSAT 2013
Energy required to completely separate one mole of a solid ionic compound into gaseous. constituent ions is called

Options:
A
Interstitial defect
B
Valency defect
C
Frenkel defect
D
Schottky defect

The correct answer is:
Frenkel defect

Solution:
Frenkel defect is due to dislocation of ion from its usual lattice site to interstitial position.

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2023 (Memory Based Paper 2)
Enthalpy of combustion of methane and ethane are -210 k cal/mol and -368 k cal/mol respectively. The enthalpy of combustion of decane is

Options:
A -1582 kcal
B -1632 kcal
C -1700 kcal
D -1480 kcal

The correct answer is:
-1632 kcal

Solution:
Given,

CH4+2O2⟶CO2+2H2O; ΔH=-210 kcal/mol ...(i)

C2H6+72O2⟶2CO2+3H2O; ΔH=-368 kcal/mol ...(ii)

On subtracting Eq. (i) from Eq (ii), we get

CH2+32O2⟶CO2+H2O; ΔH=-158 kcal/mol

∴ Enthalpy of combustion of one CH2 unit

=-158 kcal/mol

ΔHcombC10H22=ΔHcombCH4+9×ΔHcombCH2

=-210+9×-158

=-1632 kcal

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2018
Equal weights of $\mathrm{NaCl}$ and $\mathrm{KCl}$ are dissolved separately in equal volumes of solutions. Molarity of the two solutions will be :

Options:
A Equal
B That of $\mathrm{NaCl}$ will be less than that of $\mathrm{KCl}$
C That of $\mathrm{NaCl}$ will be more than that of $\mathrm{KCl}$ solution
D That of $\mathrm{NaCl}$ will be about half of that of $\mathrm{KCl}$ solution

The correct answer is:
That of $\mathrm{NaCl}$ will be more than that of $\mathrm{KCl}$ solution

Solution:
When equal weights of different solutes are present in equal volumes of solution the molarity is inversely related to molecular mass of the solute. Mol. mass of $\mathrm{NaCl}$ is less than $\mathrm{KCl}$. Hence, molarity of $\mathrm{NaCl}$ solution will be more.

Topic: Chemistry, Some Basic Concepts of Chemistry, BITSAT, BITSAT 2013
Equation sinx+cost+x+cost-x=2 has real solution, then sint can be

Options:
A 12
B 15
C 34
D -34

The correct answer is:
12

Solution:
Given, sinx+cost+x+cost-x=2

⇒sinx+2cost·cosx=2

For real solution

1+4cos2t≥2

⇒cos2t≥34

⇒-12≤sint≤12

Hence, option (a) is correct.

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2018
Eutrophication causes reduction in

Options:
A Dissolved oxygen
B Nutrients
C Dissolved salts
D All of these

The correct answer is:
Dissolved oxygen

Solution:
Eutrophication causes reduction in dissolved oxygen.

Topic: Chemistry, Environmental Chemistry, BITSAT, BITSAT 2010
Evaluate: $\int_{0}^{\pi} \frac{1}{5+4 \cos x} d x$

Options:
A $\frac{\pi}{3}$
B $\frac{\pi}{2}$
C $\frac{\pi}{4}$
D $\frac{\pi}{6}$

The correct answer is:
$\frac{\pi}{3}$

Solution:
We have, $\int_{0}^{\pi} \frac{1}{5+4 \cos x} d x$

$=\int_{0}^{\pi} \frac{1}{5+4\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x$

$=\int_{0}^{\pi} \frac{1+\tan ^{2} \frac{x}{2}}{5\left(1+\tan ^{2} \frac{x}{2}\right)+4\left(1-\tan ^{2} \frac{x}{2}\right)} d x$

$=\int_{0}^{\pi} \frac{1+\tan ^{2} \frac{x}{2}}{9+\tan ^{2} \frac{x}{2}} \mathrm{dx}=\int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{9+\tan ^{2} \frac{x}{2}} \mathrm{dx}$

Let $\tan \frac{x}{2}=t \Rightarrow \frac{1}{2} \sec ^{2} \frac{x}{2} d x=d t$

Also, $x=0 \Rightarrow t=0$ and $x=\pi \Rightarrow t=\infty$

$\therefore \mathrm{I}=\int_{0}^{\infty} \frac{\mathrm{dt}}{9+t^{2}}$

$\therefore \mathrm{I}=2 \int_{0}^{\infty} \frac{\mathrm{dt}}{9+t^{2}}$

$\therefore \mathrm{I}=\frac{2}{3}\left[\tan ^{-1} \frac{t}{3}\right]_{0}^{\infty}=\frac{2}{3}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right]$

$=\frac{2}{3}\left(\frac{\pi}{2}-0\right)=\frac{\pi}{3}$

Topic: Mathematics, Definite Integration, BITSAT, BITSAT 2021