datavorous/entrance-exam-dataset huggingface dataset:
Two vectors \(\vec{A}\) and \(\vec{B}\) are such that \(|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|\). The angle between the two vectors will be-

Options:
A \(60^{\circ}\)
B \(90^{\circ}\)
C \(180^{\circ}\)
D \(0^{\circ}\)

The correct answer is:
\(90^{\circ}\)

Solution:
\(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=| \overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}} \mid\)
\(\Rightarrow \sqrt{\mathrm{A}^2+\mathrm{B}^2+2 \mathrm{AB} \cos \theta}=\sqrt{\mathrm{A}^2+\mathrm{B}^2-2 \mathrm{AB} \cos \theta}\)
Squaring both the sides, we get
\(\overrightarrow{\mathrm{A}}^2+\overrightarrow{\mathrm{B}}^2+2 \overrightarrow{\mathrm{A}} \overrightarrow{\mathrm{B}} \cos \theta=\overrightarrow{\mathrm{A}}^2+\overrightarrow{\mathrm{B}}^2-2 \overrightarrow{\mathrm{A}} \overrightarrow{\mathrm{B}} \cos \theta\)
or \(4 \overrightarrow{\mathrm{A}} \overrightarrow{\mathrm{B}} \cos \theta=0\) or \(\cos \theta=0\) (since the scalar or dot product is zero). Therefore angle between \(\overrightarrow{\mathrm{A}}\) to \(\overrightarrow{\mathrm{B}}\) is \(90^{\circ}\)

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2011
Two vessels of volumes $16.4 \mathrm{~L}$ and $5 \mathrm{~L}$ contain two ideal gases of molecular existence at the respective temperature of $27^{\circ} \mathrm{C}$ and $227^{\circ} \mathrm{C}$ and exert 1.5 and 4.1 atmospheres respectively. The ratio of the number of molecules of the former to that of the later is

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

The correct answer is:
2

Solution:
Given conditions $\mathrm{V}_{1}=16.4 \mathrm{~L}, \mathrm{~V}_{2}=5 \mathrm{~L}$

$\mathrm{P}_{1}=1.5 \mathrm{~atm}, \mathrm{P}_{2}=4.1 \mathrm{~atm}$

$\mathrm{T}_{1}=273+27=300 \mathrm{~K}$

$\mathrm{T}_{2}=273+227=500 \mathrm{~K}$

Applying gas equation, $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{P}_{2} \mathrm{~V}_{2}}=\frac{\mathrm{n}_{1} \mathrm{~T}_{1}}{\mathrm{n}_{2} \mathrm{~T}_{2}}$

$\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{\mathrm{P}_{1} \mathrm{~V}_{1} \mathrm{~T}_{1}}{\mathrm{P}_{2} \mathrm{~V}_{2} \mathrm{~T}_{2}}$

$\therefore \frac{1.5 \times 16.4 \times 500}{4.1 \times 5 \times 300}=\frac{2}{1}$

Topic: Chemistry, States of Matter, BITSAT, BITSAT 2020
Two waves of wavelengths $99 \mathrm{~cm}$ and $100 \mathrm{~cm}$ both travelling with velocity $396 \mathrm{~m} / \mathrm{s}$ are made to interfere. The number of beats produced by them per second is

Options:
A 1
B 2
C 4
D 8

The correct answer is:
4

Solution:
Velocity of wave $\mathrm{v}=\mathrm{n} \lambda$ where $n=$ frequency of wave $\Rightarrow n=\frac{v}{\lambda}$

$\mathrm{n}_{2}=\frac{\mathrm{v}_{2}}{\lambda_{2}}=\frac{396}{100 \times 10^{-2}}=396 \mathrm{~Hz}$

no. of beats $=\mathrm{n}_{1}-\mathrm{n}_{2}=4$

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2013
Two wires $\mathrm{A}$ and $\mathrm{B}$ of the same material, having radii in the ratio 1: 2 and carry currents in the ratio 4: 1 . The ratio of drift speed of electrons in $\mathrm{A}$ and $\mathrm{B}$ is

Options:
A 16: 1
B 1: 16
C 1: 4
D 4: 1

The correct answer is:
16: 1

Solution:
Current flowing through the conductor, $\mathrm{I}=\mathrm{n} \mathrm{e} \mathrm{v} \mathrm{A}$. Hence

$\frac{4}{1}=\frac{\text { nev }_{\mathrm{d}_{1}} \pi(1)^{2}}{\operatorname{nev}_{\mathrm{d}_{2}} \pi(2)^{2}}$ or $\frac{v_{\mathrm{d}_{1}}}{\mathrm{v}_{\mathrm{d}_{2}}}=\frac{4 \times 1}{1}=\frac{16}{1}$

Topic: Physics, Current Electricity, BITSAT, BITSAT 2020
Two wires are made of the same material and have the same volume. However wire 1 has crosssectional area $A$ and wire 2 has cross-sectional area $3 A$. If the length of wire 1 increases by $\Delta x$ on applying force $F$, how much force is needed to stretch wire 2 by the same amount?

Options:
A $4 F$
B $6 \mathrm{~F}$
C $9 F$
D $F$

The correct answer is:
$9 F$

Solution:
As shown in the figure, the wires will have the same Young's modulus (same material) and the length of the wire of area of crosssection $3 A$ will be $\ell / 3$ (same volume as wire 1 ). For wire $1, \quad Y=\frac{F / A}{\Delta x / \ell}$

For wire $2, Y=\frac{F^{\prime} / 3 A}{\Delta x /(\ell / 3)}$

From (i) and

(ii) $, \frac{F}{A} \times \frac{\ell}{\Delta x}=\frac{F^{\prime}}{3 A} \times \frac{\ell}{3 \Delta x}$

$\Rightarrow F^{\prime}=9 F$

Topic: Physics, Mechanical Properties of Solids, BITSAT, BITSAT 2016
Two wires are made of the same material and have the same volume. However wire 1 has crosssectional area $A$ and wire 2 has cross-sectional area $3 A$. If the length of wire 1 increases by $\Delta x$ on applying force $F,$ how much force is needed to stretch wire 2 by the same amount?

Options:
A $4 F$
B $6 F$
C $9 F$
D $F$

The correct answer is:
$9 F$

Solution:
As shown in the figure, the wires will have the same Young's modulus (same material) and the length of the wire of area of crosssection $3 A$ will be $\ell / 3$ (samevolume as wire 1 ). For wire $1, \quad Y=\frac{F / A}{\Delta x / \ell}$
For wire $2, Y=\frac{F^{\prime} / 3 A}{\Delta x /(\ell / 3)}$
From (i) and (ii) $, \frac{F}{A} \times \frac{\ell}{\Delta x}=\frac{F^{\prime}}{3 A} \times \frac{\ell}{3 \Delta x}$
$\Rightarrow F^{\prime}=9 F$

Topic: Physics, Mechanical Properties of Solids, JEE Main
Two wires are made of the same material and have the same volume. However wire 1 has crosssectional area $A$ and wire 2 has cross-sectional area $3 A$. If the length of wire 1 increases by $\Delta x$ on applying force $F$, how much force is needed to stretch wire 2 by the same amount?

Options:
A $4 F$
B $6 \mathrm{~F}$
C $9 F$
D $F$

The correct answer is:
$9 F$

Solution:
As shown in the figure, the wires will have the same Young's modulus (same material) and the length of the wire of area of crosssection $3 A$ will be $\ell / 3$ (same volume as wire 1 ). For wire $1, \quad Y=\frac{F / A}{\Delta x / \ell}$

For wire $2, Y=\frac{F^{\prime} / 3 A}{\Delta x /(\ell / 3)}$

From (i) and

(ii) $, \frac{F}{A} \times \frac{\ell}{\Delta x}=\frac{F^{\prime}}{3 A} \times \frac{\ell}{3 \Delta x}$

$\Rightarrow F^{\prime}=9 F$

Topic: Physics, Mechanical Properties of Solids, JEE Main
Two wires are made of the same material and have the same volume. However wire 1 has crosssectional area $A$ and wire 2 has cross-sectional area $3 A$. If the length of wire 1 increases by $\Delta x$ on applying force $F,$ how much force is needed to stretch wire 2 by the same amount?

Options:
A $4 F$
B $6 F$
C $9 F$
D $F$

The correct answer is:
$9 F$

Solution:
As shown in the figure, the wires will have the same Young's modulus (same material) and the length of the wire of area of crosssection $3 A$ will be $\ell / 3$ (samevolume as wire 1 ). For wire $1, \quad Y=\frac{F / A}{\Delta x / \ell}$
For wire $2, Y=\frac{F^{\prime} / 3 A}{\Delta x /(\ell / 3)}$
From (i) and (ii) $, \frac{F}{A} \times \frac{\ell}{\Delta x}=\frac{F^{\prime}}{3 A} \times \frac{\ell}{3 \Delta x}$
$\Rightarrow F^{\prime}=9 F$

Topic: Physics, Mechanical Properties of Solids, BITSAT, BITSAT 2011
Under the same reaction conditions, initial concentration of 1.386 mol dm-3 of a substance becomes half in 40 s and 20 s through first order and zero order kinetics, respectively. Ratio k1k0 of the rate of constants for first order k1 and zero order k0 of the reaction is

Options:
A 0.5 mol-1dm3
B 1.0 mol dm-3
C 1.5 moldm-3
D 2.0 mol-1dm3

The correct answer is:
0.5 mol-1dm3

Solution:
Given that,

Under the same reaction conditions, initial concentration of 1.386 mol dm-3 of a substance becomes half in 40 s and 20 sfor first order and zero order kinetics, respectively.

For first order reaction,

t1/2=ln2k1=40 s...i

For zero order reaction, t1/2=A02k0=20 s....ii

Divide eqn. (i) by (ii), we get

12=[A]02k0×k1ln2

k1k0=ln2[A]0=0.6931.386=0.5 mol-1dm3

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2018
Universal set,
$U=\left\{x \mid x^{5}-6 x^{4}+11 x^{3}-6 x^{2}=0\right\}$
$A=\left\{x \mid x^{2}-5 x+6=0\right\}$ $\mathrm{B}=\left\{\mathrm{x} \mid \mathrm{x}^{2}-3\mathrm{x}+2=0\right\}$

What is $(A \cap B)^{\prime}$ equal to ?

Options:
A {1,3}
B {1,2,3}
C {0,1,3}
D {0,1,2,3}

The correct answer is:
{0,1,3}

Solution:
$U=\left\{x: x^{5}-6 x^{4}+11 x^{3}-6 x^{2}=0\right\}$

Solving for values of $x$, we get $\mathrm{U}=\{0,1,2,3\}$

$\mathrm{A}=\left\{\mathrm{x}: \mathrm{x}^{2}-5 \mathrm{x}+6=0\right\}$

Solving for values of $x$, we get $\mathrm{A}=\{2,3\}$

and $\mathrm{B}=\left\{\mathrm{x}: \mathrm{x}^{2}-3 \mathrm{x}+2=0\right\}$

Solving for values of $x$, we get $\mathrm{B}=\{2,1\}$

$A \cap B=\{2\}$

$\therefore(A \cap B)^{\prime}=U-(A \cap B)$

$=\{0,1,2,3\}-\{2\}=\{0,1,3\}$

Topic: Mathematics, Sets and Relations, BITSAT, BITSAT 2015
Using MOT, which of the following pairs denote paramagnetic species?

Options:
A $\mathrm{B}_{2}$ and $\mathrm{C}_{2}$
B $\mathrm{B}_{2}$ and $\mathrm{O}_{2}$
C $\mathrm{N}_{2}$ and $\mathrm{C}_{2}$
D $\mathrm{O}_{2}$ and $\mathrm{O}_{2}^{2-}$

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

Solution:
$\mathrm{B}_{2}$ and $\mathrm{O}_{2}$ are paramagnetic due to presence of unpaired electron. MO electronic configuration of $\mathrm{B}_{2}$ is:

$$

\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi 2 p_{x}^{1}=\pi 2 p_{y}^{1}

$$

MO electorn i.e. configuration of $\mathrm{O}_{2}$ is:

$$

\begin{array}{l}

\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}^{2} \pi 2 p_{x}^{2} \\

=\pi 2 p_{y}^{2} \pi * 2 p_{x}^{1}=\pi * 2 p_{y}^{1}

\end{array}

$$

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2013
Using the standard electrode potential, find out the pair between which redox reaction is not feasible.
$E^{\ominus}$ values $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=+0.77 ; \mathrm{I}_2 / \mathrm{I}^{-}=+054$
$\mathrm{Cu}^{2+} / \mathrm{Cu}=+0.34 ; \mathrm{Ag}^{+} / \mathrm{Ag}=0.80 \mathrm{~V}$

Options:
A
$\mathrm{Fe}^{3+}$ and $\mathrm{I}^{-}$

B
$\mathrm{Ag}^{+}$and $\mathrm{Cu}$

C
$\mathrm{Fe}^{3+}$ and $\mathrm{Cu}$

D
Ag and $\mathrm{Fe}^{3+}$

The correct answer is:

Ag and $\mathrm{Fe}^{3+}$

Solution:
(a)

This reaction is feasible, since $E_{\text {cell }}^{\circ}$ is positive.
(b)

This reaction is feasible, since $E_{\text {cell }}^{\circ}$ is positive.
(c)

This reaction is feasible, since $E_{\text {cell }}^{\circ}$ is positive.
(d)

This reaction is not feasible, since $E_{\text {cell }}^{\circ}$ is negative.

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2022
Valence electrons in the element A are 3 and that in element B are 6. Most probable compound formed from A and B is

Options:
A A2B
B AB2
C A6B3
D A2B3

The correct answer is:
A2B3

Solution:
Valence electrons in A=3

Valence electron in B=6

Thus, A is electropositive and B is electronegative, A can lose three electrons and B can gain two electrons to attain stable configuration.

Hence, A exist as A3+ and B as B2-

∴ Compound formed is A2B3

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2018
Value of $\lim _{x \rightarrow 0} \frac{a^{\sin x}-1}{\sin x}$ is

Options:
A $\log \mathrm{a}$
B $\sin x$
C $\log (\sin x)$
D $\cos x$

The correct answer is:
$\log \mathrm{a}$

Solution:
We have, $\lim _{x \rightarrow 0} \frac{a^{\sin x}-1}{\sin x}=\lim _{y \rightarrow 0} \frac{a^{y}-1}{y}$ $=\log \mathrm{a}$, where $\mathrm{y}=\sin \mathrm{x}$

$\because \mathrm{x} \rightarrow 0 \Rightarrow \mathrm{y}=\sin \mathrm{x} \rightarrow 0$

Topic: Mathematics, Limits, BITSAT, BITSAT 2021
Value of $\int \frac{d x}{\sqrt{x(a-x)}}$ is

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

The correct answer is:
$2 \sin ^{-1} \sqrt{\frac{x}{a}}+c$

Solution:
Let $\mathrm{x}=\mathrm{a} \sin ^2 \theta$
then $d x=2 a \sin \theta \cos \theta d \theta$
$\begin{aligned}
\therefore \quad I & =\int \frac{2 a \sin \theta \cdot \cos \theta}{\sqrt{a \sin ^2 \theta \cdot a \cos ^2 \theta}} d \theta \\
& =2 \int d \theta=2 \theta+c \\
& =2 \sin ^{-1}(\sqrt{x / a})+c
\end{aligned}$

Topic: Mathematics, Indefinite Integration, BITSAT, BITSAT 2023 (Memory Based Paper 2)
Velocity constant of a reaction at $290 \mathrm{~K}$ was found to be $3.2 \times 10^{-3}$. At $300 \mathrm{~K}$ it will be

Options:
A $1.28 \times 10^{-2}$
B $9.6 \times 10^{-3}$
C $6.4 \times 10^{-3}$
D $3.2 \times 10^{-4}$

The correct answer is:
$6.4 \times 10^{-3}$

Solution:
The velocity constant doubles for every $10^{\circ} \mathrm{C}$ rise in temperature.

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2013
Vertical displacement of a plank with a body of mass \(m\) on it is varying according to law \(y=\sin \omega t+\sqrt{3} \cos \omega t\). The minimum value of \(\omega\) for which the mass just breaks off the plank and the moment it occurs first after \(\mathrm{t}=0\), are given by

Options:
A \(\sqrt{\mathrm{g} / 2}, \frac{\sqrt{2}}{6} \frac{\pi}{\sqrt{\mathrm{g}}}\)
B \(\frac{\mathrm{g}}{\sqrt{2}}, \frac{2}{3} \sqrt{\pi / \mathrm{g}}\)
C \(\sqrt{\mathrm{g} / 2}, \frac{\pi}{3} \sqrt{2 / \mathrm{g}}\)
D \(\sqrt{2 g}, \sqrt{2 \pi / 3 g}\)

The correct answer is:
\(\sqrt{\mathrm{g} / 2}, \frac{\sqrt{2}}{6} \frac{\pi}{\sqrt{\mathrm{g}}}\)

Solution:
From, figure,

$$

\begin{array}{ll}

& \Lambda_{\mathrm{R}}=\sqrt{(\sqrt{3})^{2}+(1)^{2}}-2 \\

& \theta=\tan ^{-1}\left(\frac{\sqrt{3}}{1}\right)=\frac{\pi}{3} \\

\therefore \quad & \mathrm{y}=2 \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right)

\end{array}

$$

$$

\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}^{2}}=\mathrm{a}=-2 \omega^{2} \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right)

$$

$a_{\max }=-2 \omega^{2}=g$

For which mass just breaks off the plank $\omega=\sqrt{\mathrm{g} / 2}$

This will be happen for the first time when

$$

\begin{array}{l}

\omega \mathrm{t}+\frac{\pi}{3}+\frac{\pi}{2} \text { or } \omega \mathrm{t}=\frac{\pi}{6} \\

\therefore \quad \mathrm{t}=\frac{\pi}{6 \omega}=\frac{\pi}{6} \sqrt{\frac{2}{\mathrm{~g}}}

\end{array}

$$

Topic: Physics, Oscillations, BITSAT, BITSAT 2015
Viscosity is the property of a liquid due to which it :

Options:
A occupies minimum surface area
B opposes relative motion between its adjacent layers
C becomes spherical in shape
D tends to regain its deformed position

The correct answer is:
opposes relative motion between its adjacent layers

Solution:
No Solution available

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2011
Voltage $V$ and current $i$ in $A C$ circuit are given by $\mathrm{V}=50 \sin (50 \mathrm{t})$ volt, $\mathrm{i}=50 \sin \left(50 \mathrm{t}+\frac{\pi}{3}\right) \mathrm{mA}$

The power dissipated in the circuit is

Options:
A $5.0 \mathrm{~W}$
B $2.5 \mathrm{~W}$
C $1.25 \mathrm{~W}$
D zero

The correct answer is:
$1.25 \mathrm{~W}$

Solution:
Given $V=50 \sin (50$ t) $V$ Maximum voltage, $\mathrm{V}_{0}=50 \mathrm{~V}$

$$

i=\left(50 t+\frac{\pi}{3}\right) \mathrm{mA}

$$

Maximum current, $i_{0}=50 \mathrm{~mA}=50 \times 10^{-3} \mathrm{~A}$

Power dissipated, $\mathrm{P}=\frac{\mathrm{i}_{0}}{\sqrt{2}} \times \frac{\mathrm{V}_{0}}{\sqrt{2}}$

$=\frac{50 \times 50 \times 10^{-3}}{2}=\frac{2500 \times 10^{-3}}{2}=1.25 \mathrm{~W}$

Topic: Physics, Alternating Current, BITSAT, BITSAT 2014
Volume of $3 \%$ solution of sodium carbonate necessary to neutralise a litre of $0.1 \mathrm{~N}$ sulphuric acid

Options:
A $176.66 \mathrm{ml}$
B $156.6 \mathrm{ml}$
C $116.0 \mathrm{ml}$
D $196.1 \mathrm{ml}$

The correct answer is:
$176.66 \mathrm{ml}$

Solution:
Normality of $3 \% \mathrm{Na}_{2} \mathrm{CO}_{3}$

$\mathrm{N}=\frac{3 \times 1000}{53 \times 100}=0.566 \mathrm{~N}$

For $\mathrm{H}_{2} \mathrm{SO}_{4}$ sol. $\mathrm{N}_{1}=0.1, \mathrm{~V}_{1}=100 \mathrm{~mL}$

For $\mathrm{Na}_{2} \mathrm{CO}_{3}$ sol. $\mathrm{N}_{2}=0.566$

Now apply $\mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2}$

$\mathrm{V}_{2}=\frac{\mathrm{N}_{1} \mathrm{~V}_{1}}{\mathrm{~N}_{2}}=\frac{0.1 \times 1000 \mathrm{~mL}}{0.566}=176.66 \mathrm{~mL}$

Topic: Chemistry, Redox Reactions, BITSAT, BITSAT 2013
Water is flowing on a horizontal fixed surface, such that its flow velocity varies with $y$ (vertical direction) as $v=k\left(\frac{2 y^{2}}{a^{2}}-\frac{y^{3}}{a^{3}}\right) .$ If coefficient of viscosity for water is $\eta$, what will be shear stress between layers of water at $y=a$.

Options:
A $\frac{\eta k}{a}$
B $\frac{\eta}{k a}$
C $\frac{\eta a}{k}$
D None of these

The correct answer is:
$\frac{\eta k}{a}$

Solution:
Newton's law of viscosity, $F=\eta A \frac{d v}{d y}$

$$

\begin{array}{l}

\text { Stress }=\frac{F}{A}=\eta\left(\frac{d v}{d y}\right)=\eta k\left(\frac{4 y}{a^{2}}-\frac{3 y^{2}}{a^{3}}\right) \\

\text { At } y=a, \text { stress }=\eta k\left(\frac{4}{a}-\frac{3}{a}\right)=\frac{\eta k}{a}

\end{array}

$$

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2015
Wave pulse can travel along a tense string like a violin spring. Aseries of experiments showed that the wave velocity $\mathrm{V}$ of a pulse depends on the following quantities, the tension T of the string, the cross-section area A of the string and then as per unit volume $\rho$ of the string. Obtain an expression for $V$ in terms of the $\mathrm{T}, \mathrm{A}$ and $\rho$ using dimensional analysis.

Options:
A $\mathrm{V}=\mathrm{k} \sqrt{\frac{\mathrm{T}}{\mathrm{A} \rho}}$
B $\mathrm{V}=\mathrm{k} \sqrt{\frac{\mathrm{T}}{\mathrm{A}}}$
C $V=k \sqrt{\frac{A \rho}{T}}$
D None of these

The correct answer is:
$\mathrm{V}=\mathrm{k} \sqrt{\frac{\mathrm{T}}{\mathrm{A} \rho}}$

Solution:
Let $\mathrm{V}=\mathrm{kT}^{\mathrm{a}} \mathrm{A}^{\mathrm{b}} \rho^{\mathrm{c}}$,

$\mathrm{k}=$ dimensional constant

Writing dimension on both we side

$$

\begin{aligned}

\left[\mathrm{LT}^{-1}\right] &=\left[\mathrm{MLT}^{-2}\right]^{a}\left[\mathrm{~L}^{2}\right]^{\mathrm{b}}\left[\mathrm{ML}^{-3}\right]^{\mathrm{c}} \\

&=\left[\mathrm{M}^{\mathrm{a}+\mathrm{c}} \mathrm{L}^{\mathrm{a}+2 \mathrm{~b}-3 \mathrm{c}} \mathrm{T}^{-2 \mathrm{a}}\right]

\end{aligned}

$$

Comparing power on both sides we have

$$

\begin{array}{l}

a+c=0, a+2 b-3 c=1, \quad-2 a=-1 \\

\therefore \quad a=\frac{1}{2}, c=-\frac{1}{2} \Rightarrow b=-\frac{1}{2} \therefore V=k \sqrt{\frac{T}{A \rho}}

\end{array}

$$

Topic: Physics, Units and Dimensions, BITSAT, BITSAT 2013
What is $\left[\mathrm{NH}_4^{+}\right]$in a solution that is $0.02 \mathrm{M} \mathrm{NH}_3$ $0.01 \mathrm{M} \mathrm{KOH} ?\left[K_b\left(\mathrm{NH}_3\right)=1.8 \times 10^{-5}\right]$

Options:
A
$3.6 \times 10^{-5} \mathrm{M}$

B
$1.8 \times 10^{-5} \mathrm{M}$

C
$0.9 \times 10^{-5} \mathrm{M}$

D
$7.2 \times 10^{-5} \mathrm{M}$

The correct answer is:

$3.6 \times 10^{-5} \mathrm{M}$

Solution:
$\mathrm{NH}_4 \mathrm{OH} \rightleftharpoons \mathrm{NH}_4^{+}+\mathrm{OH}^{-}$
$K_b=\frac{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_4 \mathrm{OH}\right]}$
$1.8 \times 10^{-5}=\frac{\left[\mathrm{NH}_4^{+}\right][0.01]}{[0.02]}$
$\therefore \quad\left[\mathrm{NH}_4^{+}\right]=3.6 \times 10^{-5} \mathrm{M}$

Topic: Chemistry, Ionic Equilibrium, BITSAT, BITSAT 2022
What is the angle between the two straight lines $y=(2-\sqrt{3}) x+5$ and $y=(2+\sqrt{3}) x-7 ?$

Options:
A $60^{\circ}$
B $45^{\circ}$
C $30^{\circ}$
D $15^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
The given lines are:
$y=(2-\sqrt{3}) x+5 \text { and } y=(2+\sqrt{3}) x-7$
Therefore, slope of first line $=m_1=2-\sqrt{3}$ and slope of second line $=m_2=2+\sqrt{3}$
$\therefore \quad \tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|=\left|\frac{2+\sqrt{3}-2+\sqrt{3}}{1+(4-3)}\right|$
$=\left|\frac{2 \sqrt{3}}{2}\right|=\sqrt{3}=\tan \frac{\pi}{3} \Rightarrow \theta=\frac{\pi}{3}=60^{\circ}$

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2023 (Memory Based Paper 1)
What is the area bounded by \(y=\tan x, y=0\) and \(x=\frac{\pi}{4} ?\)

Options:
A \(\ell n ~2~ \mathrm{sq}\). units
B \(\frac{\ell n 2}{2} ~\mathrm{sq}\). units
C \(2~(\ell n ~2)~ \mathrm{sq}\). units
D None of these

The correct answer is:
\(\frac{\ell n 2}{2} ~\mathrm{sq}\). units

Solution:
Required area \(=\int_0^{\frac{\pi}{4}} \tan x d x\)
\(=\ln |\sec 1|_0^{\frac{\pi}{4}}=\ln \sqrt{2}=\frac{\ln 2}{2}\)

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2011
What is the correct increasing order of Bronsted bases?

Options:
A $\mathrm{ClO}_{4}^{-}<\mathrm{ClO}_{3}^{-}<\mathrm{ClO}_{2}^{-}<\mathrm{ClO}^{-}$
B $\mathrm{ClO}_{4}^{-}>\mathrm{ClO}_{3}^{-}>\mathrm{ClO}_{2}^{-}>\mathrm{ClO}^{-}$
C $\mathrm{ClO}_{3}^{-}<\mathrm{ClO}_{4}^{-}<\mathrm{ClO}_{2}^{-}<\mathrm{ClO}^{-}$
D $\mathrm{ClO}^{-}>\mathrm{ClO}_{3}^{-}>\mathrm{ClO}_{2}^{-}<\mathrm{ClO}_{4}^{-}$

The correct answer is:
$\mathrm{ClO}_{4}^{-}<\mathrm{ClO}_{3}^{-}<\mathrm{ClO}_{2}^{-}<\mathrm{ClO}^{-}$

Solution:
$\mathrm{ClO}_{4}^{-}<\mathrm{ClO}_{3}^{-}<\mathrm{ClO}_{2}^{-}<\mathrm{ClO}^{-}$ is the correct

increasing order of Bronsted base. With increase in the number of oxygen atoms in the conjugate bases, the delocalisation of the $\pi$ bond becomes more and more extended. This results in decrease in the electron density. Consequently basicity also decreases.

Topic: Chemistry, Ionic Equilibrium, BITSAT, BITSAT 2015
What is the free energy change, $\Delta \mathrm{G}$, when 1.0 mole of water at $100^{\circ} \mathrm{C}$ and 1 atm pressure is converted into steam at $100^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$.
pressure?

Options:
A $540 \mathrm{cal}$
B -9800 cal
C $9800 \mathrm{cal}$
D $0 \mathrm{cal}$

The correct answer is:
$0 \mathrm{cal}$

Solution:
Condition of equilibrium, hence $\Delta \mathrm{G}=0$.

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2011
What is the hybridisation of $\left[\mathrm{CrF}_6\right]^{3-}$ ?

Options:
A
$s p^3 d$

B
$s p^3 d^2$

C
$d^2 s p^3$

D
$d^2 s p$

The correct answer is:

$d^2 s p^3$

Solution:
The electronic configuration of $\mathrm{Cr}$ is $[\mathrm{Ar}] 3 d^5 4 s^1$.

$\left[\mathrm{CrF}_6\right]^{3-}$ has $d^2 s p^3$-hybridisation and octahedral geometry.

Topic: Chemistry, Coordination Compounds, BITSAT, BITSAT 2022
What is the magnetic field at the centre of arc in the figure below?

Options:
A μ04π·2Ir2+π
B μ04π·2Ir2+π4
C μ04π·Ir2+π
D μ04π·Ir2+π4

The correct answer is:
μ04π·2Ir2+π4

Solution:
Distance of straight path from O

=r2

Hence, B=B1+B2

=μ04π·2Ir2+μ04π2πIr14

=μ04π·2Ir2+π4

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2017
What is the maximum wavelength line in the Lyman series of $\mathrm{He}^{+}$ ion?

Options:
A $3 R$
B $\frac{1}{\mathrm{R}}$
C $\frac{1}{3 R}$
D None of these

The correct answer is:
$\frac{1}{3 R}$

Solution:
$\frac{1}{\lambda}=\mathrm{RZ}^{2}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]=\mathrm{R} \times 2^{2}\left[\frac{1}{\mathrm{l}^{2}}-\frac{1}{2^{2}}\right]$

$\Rightarrow 3 \mathrm{R} ; \lambda=\frac{1}{3 \mathrm{R}}$

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2020
What is the moment of inertia of a solid sphere of density $\rho$ and radius $\mathrm{R}$ about its diameter?

Options:
A $\frac{105}{176} \mathrm{R}^{5} \rho$
B $\frac{105}{176} \mathrm{R}^{2} \rho$
C $\frac{176}{105} \mathrm{R}^{5} \rho$
D $\frac{176}{105} \mathrm{R}^{2} \rho$

The correct answer is:
$\frac{176}{105} \mathrm{R}^{5} \rho$

Solution:
For solid sphere
$\begin{array}{l}
I=\frac{2}{5} M R^{2}=\frac{2}{5}\left(\frac{4}{3} \pi R^{3} \rho\right) R^{2} \\
\rho=\frac{176}{105} R^{5} \rho
\end{array}$

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2012
What is the product of the following reaction?

Options:
A
B
C
D

The correct answer is:

Solution:
Only-CHO group reacts with CN $^{-}$ ion and the reaction is

Topic: Chemistry, Aldehydes and Ketones, BITSAT, BITSAT 2020
What is the solution of \(\frac{d y}{d x}+2 y=1\) satisfying \(y(0)=0\) ?

Options:
A \(y=\frac{1-e^{-2 x}}{2}\)
B \(y=\frac{1+e^{-2 x}}{2}\)
C \(y=1+e^x\)
D \(y=\frac{1+e^x}{2}\)

The correct answer is:
\(y=\frac{1-e^{-2 x}}{2}\)

Solution:
\(\begin{aligned}
& \frac{d y}{d x}+2 y=1 \Rightarrow \frac{d y}{d x}=1-2 y \\
& \int \frac{d y}{1-2 y}=\int d x-\frac{1}{2} \log |1-2 y|=x+C \\
& \text { at } x=0, y=0 ;-\frac{1}{2} \log 1=0+C \Rightarrow C=0 \\
& 1-2 y=e^{-2 x} \Rightarrow y=\frac{1-e^{-2 x}}{2}
\end{aligned}\)

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2010