datavorous/entrance-exam-dataset huggingface dataset:
$\int_{0}^{\pi / 2} \frac{2^{\sin x}}{2^{\sin x}+2^{\cos x}} d x$ equals

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

The correct answer is:
$\pi / 4$

Solution:
$\mathrm{I}=\int_{0}^{\pi / 2} \frac{2^{\sin \mathrm{x}}}{2^{\sin x}+2^{\cos \mathrm{x}} \mathrm{dx}}$
$\mathrm{I}=\int_{0}^{\pi / 2} \frac{2^{\sin (\pi / 2-\mathrm{x})}}{2^{\sin (\pi / 2-\mathrm{x})}+2^{\cos (\pi / 2-\mathrm{x})} \mathrm{dx}}$
$=\int \frac{2^{\cos \mathrm{x}}}{2^{\cos \mathrm{x}}+2^{\sin \mathrm{x}}} \mathrm{dx} \Rightarrow 21=\int_{0}^{\pi / 2} \mathrm{~d} \mathrm{x}=\frac{\pi}{2} \Rightarrow 1=\frac{\pi}{4}$

Topic: Mathematics, Definite Integration, BITSAT, BITSAT 2012
$\int_0^{\pi / 4} \cos x e^{\sin x} d x$ is equal to

Options:
A $e+1$
B $e-1$
C $e$
D $-e$

The correct answer is:
$e-1$

Solution:
Suppose, $I=\int_0^{\pi / 2} \cos x e^{\sin x} d x$
Let $\sin x=t \Rightarrow \cos x d x=d t$
$x \rightarrow 0 \Rightarrow t \rightarrow 0$
and $x \rightarrow \frac{\pi}{2} \Rightarrow t \rightarrow 1$
So $I=\int_0^1 e^t d t=\left[e^t\right]_0^1=e-1$

Topic: Mathematics, Definite Integration, BITSAT, BITSAT 2023 (Memory Based Paper 2)
0.5F of electricity is passed through 500 mL of copper sulphate solution. The amount of copper which can be deposited will be

Options:
A 63.5 g
B 31.75 g
C 15.80 g
D unpredictable

The correct answer is:
15.80 g

Solution:
Cu2++2e-⟶Cu ...i

According to Faraday's first law, when an electric current is passed through an electrolyte, amount of substance deposited is directly proportional to the quantity of electric charge passed through the electrolyte.

∵1 mol of 63.5 g of Cu is obtained by passing 2F of electricity.

Mass of Cu obtained by passing 1F of electricity =63.52g

∴ Mass of Cu obtained by passing 0.5F of electricity
=63.52×0.5=15.80 g

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2018
$\int_{0}^{10} \frac{x^{10}}{(10-x)^{10}+x^{10}} d x$ is equal to

Options:
A 10
B 5
C 2
D $\frac{1}{0}$

The correct answer is:
5

Solution:
Let $I=\int_{0}^{10} \frac{x^{10}}{(10-x)^{10}+x^{10}} d x$...(1)

$I=\int_{0}^{10} \frac{(10-x)^{10}}{(10-x)^{10}+x^{10}} d x$...(2)

Adding

(1) and (2), we get

$2 I=\int_{0}^{10} d x \Rightarrow 2 I=10 \Rightarrow I=5$

Topic: Mathematics, Definite Integration, BITSAT, BITSAT 2014
0.45 g acid of molecular mass 90 was neutralise by 20 ml of 0.5 N KOH. The basicity of the acid is

Options:
A 2
B 4
C 1
D 3

The correct answer is:
2

Solution:
Given that 0.45 g of the acid of molecular mass 90 was neutralized by 20 ml of 0.5 N caustic potash.

We know that,

Normality N = Weight of the base ×1000Eq. weight×volume

∵ Gram equivalents of an acid

Gram equivalents of KOH i.e. 0.45E=20×0.51000

where, E= Equivalent mass E=0.45×100020×0.5=45

Thus, basicity =9045=2 ∵Basicity=MolecularmassEquivalentmass

Topic: Chemistry, Redox Reactions, BITSAT, BITSAT 2019
0.765 g of an acid gives 0.535 g of CO2 and 0.138 g of H2O. Then, the ratio of percentage of carbon and hydrogen is

Options:
A 19:2
B 18:17
C 20:17
D 1:7

The correct answer is:
19:2

Solution:
% for CO2

∵ 0.765 g of acid gives CO2=0.535 g

∴100 g of acid gives CO2=0.535×1000.765

% of CO2=70.00

% for carbon

∵ 44 g of CO2 gives, C=12 g

∴ 70 g of CO2 gives, C=12×7044

=19.09 g

∴ % of carbon =19.00

Similarly,

(i) for hydrogen

% of H2O

⇒ 0.138×1000.765

=18.03=18.00

% of Hydrogen

⇒ 2×1818=2.00

Hence ratio of carbon and hydrogen is 19:2

Topic: Chemistry, Some Basic Concepts of Chemistry, BITSAT, BITSAT 2017
$\int_{0}^{\infty} \frac{\mathrm{dx}}{\left(\mathrm{x}^{2}+\mathrm{a}^{2}\right)\left(\mathrm{x}^{2}+\mathrm{b}^{2}\right)}$ is

Options:
A $\frac{\pi a b}{a+b}$
B $\frac{\pi}{2(a+b)}$
C $\frac{\pi}{2 a b(a+b)}$
D $\frac{\pi(a+b)}{2 a b}$

The correct answer is:
$\frac{\pi}{2 a b(a+b)}$

Solution:
$\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}=\frac{1}{b^{2}-a^{2}}$

$\int_{0}^{\infty} \frac{\left(x^{2}+b^{2}\right)-\left(x^{2}+a^{2}\right)}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}$

$=\frac{1}{b^{2}-a^{2}} \int_{0}^{\infty}\left[\frac{1}{x^{2}+a^{2}}-\frac{1}{x^{2}+b^{2}}\right] d x$

$=\frac{1}{b^{2}-a^{2}}\left[\frac{1}{a} \tan ^{-1} \frac{x}{a}-\frac{1}{b} \tan ^{-1} \frac{x}{b}\right]_{0}^{\infty}$

$=\frac{1}{b^{2}-a^{2}}\left[\frac{\pi}{2 a}-\frac{\pi}{2 b}\right]=\frac{\pi}{2 a b(a+b)}$

Topic: Mathematics, Definite Integration, BITSAT, BITSAT 2020
$2^{3 n}-7 n-1$ is divisible by

Options:
A 64
B 36
C 49
D 25

The correct answer is:
49

Solution:
$2^{3 n}-7 n-1 \quad$ Taking $n=2 ;$

$$

\begin{array}{l}

2^{6}-7 \times 2-1 \\

=64-15=49

\end{array}

$$

Therefore this is divisible by 49 .

Topic: Mathematics, Basic of Mathematics, BITSAT, BITSAT 2014
$2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightleftharpoons 4 \mathrm{NO}_{2}+\mathrm{O}_{2}$
If rate and rate constant for above reaction are $2.40 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ and $3 \times 10^{-5} \mathrm{~s}^{-1}$ respectively, then calculate the concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$

Options:
A 1.4
B 1.2
C 0.04
D 0.8

The correct answer is:
0.8

Solution:
The reaction is of first and for a first order reaction, rate, $R=k\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]$ $2.4 \times 10^{-5}=3 \times 10^{-5} \times\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]$
$\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=\frac{2.4 \times 10^{-5}}{3 \times 10^{-5}}=0.8 \mathrm{~mol} \mathrm{~L}^{-1}$

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2012
$\int_{\pi / 3}^{\pi / 2} x \sin (\pi[x]-x) \mathrm{dx}$ is equal to

Options:
A $\frac{1}{2}+\frac{\pi}{6}$
B $1-\frac{\sqrt{3}}{2}+\frac{\pi}{6}$
C $-\frac{1}{2}-\frac{\pi}{6}$
D $\frac{\sqrt{3}}{2}-1-\frac{\pi}{6}$

The correct answer is:
$1-\frac{\sqrt{3}}{2}+\frac{\pi}{6}$

Solution:
In the interval $\frac{\pi}{3}$ to $\frac{\pi}{2},[\mathrm{x}]=1$ $\begin{aligned} \therefore \mathrm{I} &=\int_{\pi / 3}^{\pi / 2} \mathrm{x} \sin (\pi-\mathrm{x}) \mathrm{dx}=\int_{\pi / 3}^{\pi / 2} \mathrm{x} \sin \mathrm{x} \mathrm{d} \mathrm{x} \\ &=[-\mathrm{x} \cos \mathrm{x}+\sin \mathrm{x}]_{\pi / 3}^{\pi / 2}=1-\frac{\sqrt{3}}{2}+\frac{\pi}{6} \end{aligned}$

Topic: Mathematics, Definite Integration, BITSAT, BITSAT 2020
3 moles of an ideal gas at a temperature of $27^{\circ} \mathrm{C}$ are mixed with 2 moles of an ideal gas at a temperature $227^{\circ} \mathrm{C}$, determine the equilibrium temperature of the mixture, assuming no loss of energy.

Options:
A $327^{\circ} \mathrm{C}$
B $107^{\circ} \mathrm{C}$
C $318^{\circ} \mathrm{C}$
D $410^{\circ} \mathrm{C}$

The correct answer is:
$107^{\circ} \mathrm{C}$

Solution:
Energy possessed by the ideal gas at $27^{\circ} \mathrm{C}$ is
$$
E_{1}=3\left(\frac{3}{2} R \times 300\right)=\frac{2700 R}{2}
$$
Energy possessed by the ideal gas at $227^{\circ} \mathrm{C}$ is
$$
\mathrm{E}_{2}=2\left(\frac{3 \mathrm{R}}{2} \times 500\right)=1500 \mathrm{R}
$$
If $\mathrm{T}$ be the equilibrium temperature, of the mixture, then its energy will be
$$
\mathrm{E}_{\mathrm{m}}=5\left(\frac{3 \mathrm{R} \mathrm{T}}{2}\right)
$$
Since, energy remains conserved,
$$
\mathrm{E}_{\mathrm{m}}=\mathrm{E}_{1}+\mathrm{E}_{2}
$$
or $\quad 5\left(\frac{3 R T}{2}\right)=\frac{2700 R}{2}+1500 R$
or $\quad \mathrm{T}=380 \mathrm{~K}$ or $107^{\circ} \mathrm{C}$

Topic: Physics, Kinetic Theory of Gases, BITSAT, BITSAT 2010
4 g of copper was dissolved in conc.HNO3. The copper nitrate thus obtained gave 5 g of its oxide on strong heating the equivalent weight of copper is

Options:
A 23
B 32
C 12
D 20

The correct answer is:
32

Solution:
4 g of copper gave 5 g of its oxide means,

1 g of oxygen combine with 4 g of copper.

∵ Eq. wt of oxygen =8,

Therefore,

8 g of oxygen combine with =4×8 g of copper =32 g

Hence,

Equivalent weight of copper =32

Topic: Chemistry, Redox Reactions, BITSAT, BITSAT 2017
4 moles of A are mixed with 4 moles of B. At equilibrium for the reaction A+B⇌C+D, 2 moles of C and D are formed, the equilibrium constant for the reaction will be,

Options:
A 1
B 1/2
C 4
D 1/4

The correct answer is:
1

Solution:
The initial concentrations of A, B, C, and D are 4M, 4M, 0M, and 0M respectively.

The equilibrium concentrations of A,B,C and D are 4−2=2M, 4−2=2M, 2M and 2M respectively.

A + B ⇌ C + Dinitial conc 4 4 0 0at equilibrium 4-2 4-2 2 2

The expression for the equilibrium constant is =[C][D]​[A][B]

∴ Keq=2×22×2=1

Topic: Chemistry, Chemical Equilibrium, BITSAT, BITSAT 2019
0.5 mole of an ideal gas at constant temperature $27^{\circ} \mathrm{C}$ kept inside a cylinder of length $L$ and crosssection area A closed by a massless piston.

The cylinder is attached with a conducting rod of length $L,$ cross-section area $(1 / 9) \mathrm{m}^{2}$ and thermal conductivity $\mathrm{k}$, whose other end is maintained at $0^{\circ} \mathrm{C}$. If piston is moved such that rate of heat flow through the conducing rod is constant then velocity of piston when it is at height $L / 2$ from the bottom of cylinder is :

[Neglect any kind of heat loss from system

Options:
A $\left(\frac{\mathrm{k}}{R}\right) \mathrm{m} / \mathrm{sec}$
B $\left(\frac{\mathrm{k}}{10 R}\right) \mathrm{m} / \mathrm{sec}$
C $\left(\frac{\mathrm{k}}{100 R}\right) \mathrm{m} / \mathrm{sec}$
D $\left(\frac{\mathrm{k}}{1000 R}\right) \mathrm{m} / \mathrm{sec}$

The correct answer is:
$\left(\frac{\mathrm{k}}{100 R}\right) \mathrm{m} / \mathrm{sec}$

Solution:
$\frac{\Delta Q}{\Delta t}=\frac{\Delta W}{\Delta t}=$ work done per unit time = $\frac{\mathrm{k} a \theta}{L}$

$$

\frac{d W}{d t}=P \frac{d v}{d t}=\mathrm{k} \frac{a \theta}{L}, P=\frac{n R T}{V}

$$

$\Rightarrow \quad \frac{0.5 R(300)}{V} A \cdot \frac{d \ell}{d t}=\frac{\mathrm{k} a \theta}{L}$

$$

\Rightarrow \frac{0.5 R(300)}{A \cdot \frac{L}{2}} A \cdot v=\frac{\mathrm{k} a \theta}{L}

$$

$\Rightarrow \quad v=\frac{\mathrm{k} a}{R}\left(\frac{27}{300}\right)=\frac{\mathrm{k}}{100 R}$

Topic: Physics, Thermal Properties of Matter, BITSAT, BITSAT 2015
\(10^n+3\left(4^{n+2}\right)+5\) is divisible by \((n \in N)\)

Options:
A 7
B 5
C 9
D 17

The correct answer is:
9

Solution:
\(\begin{aligned}
& 10^n+3\left(4^{n+2}\right)+5 \text { Taking } n=2, \\
& 10^2+3 \times 4^4+5=100+768+5=873
\end{aligned}\)
Therefore this is divisible by 9.

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2009
12 eV energy is given to electron in third orbit of H-atom, then final energy of electron when it ionise from third orbit, is

Options:
A 5.25 eV
B 10.49 eV
C 12 eV
D 11.15 eV

The correct answer is:
10.49 eV

Solution:
Given, total energy given to electron in third orbit,

E=12 eV

Energy of third orbit of H-atom, E'=-13.6n2

=-13.632=-1.51 eV

lonisation energy of electron of third orbit, E''=-E'

=--1.51 eV⇒E''=1.51 eV

Hence, final energy of electron when it comes out of H-atom from third orbit

Ef=E-E''

=12-1.51=10.49 eV

Topic: Physics, Atomic Physics, BITSAT, BITSAT 2019
$0.20 \mathrm{~g}$ of an organic compound gave $0.12 \mathrm{~g}$ of AgBr By using Carius method, the percentage of bromine in the compound will be

Options:
A
$34.06 \%$

B
$44.04 \%$

C
$54 \%$

D
$25 \%$

The correct answer is:

$25 \%$

Solution:
Given,
Mass of an organic compound $=0.20 \mathrm{~g}$
Mass of $\mathrm{AgBr}=0.12 \mathrm{~g}$
Molecular mass of $\mathrm{AgBr}=188 \mathrm{~g} \mathrm{~mol}^{-1}$
$188 \mathrm{~g}$ of AgBr contains $80 \mathrm{~g}$ of bromine
$\therefore 0.12 \mathrm{~g}$ of $\mathrm{AgBr}$ will contain $=\frac{80}{188} \times 0.12$
$=0.05 \mathrm{~g}$ of bromine
$\therefore$ Percentage of bromine $=\frac{0.05}{0.20} \times 100$
$=25 \%$

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2022
2,4 -Dichlorophenoxyacetic add is used as

Options:
A Fungicide
B Insecticide
C Herbicide
D Moth repellant

The correct answer is:
Herbicide

Solution:
2,4 -dichlorophenoxyacetic acid is used as a herbicide.

Topic: Chemistry, Chemistry in Everyday Life, BITSAT, BITSAT 2012
$0.45 \mathrm{~g}$ of acid molecular weight 90 is neutralised by $20 \mathrm{ml}$ of $0.5 \mathrm{~N}$ caustic potash. The basicity of acid is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
Eq. of acid = Eq of base,
$\begin{array}{l}
\therefore \frac{0.45}{\mathrm{E} \cdot \mathrm{wt}}=\frac{20 \times 0.5}{1000}=\mathrm{E} \mathrm{wt}=45 \\
\text { Basicity }=\frac{\mathrm{M} \cdot \mathrm{wt}}{\mathrm{E} \cdot \mathrm{wt}}=\frac{90}{45}=2
\end{array}$

Topic: Chemistry, Redox Reactions, BITSAT, BITSAT 2012
$1.00 \mathrm{~g}$ of a non-electrolyte solute (molar mass $\left.250 \mathrm{~g} \mathrm{~mol}^{-1}\right)$ was dissolved in $51.2 \mathrm{~g}$ of benzene. If the freezing point depression constant, $\mathrm{K}_{\mathrm{f}}$ of benzene is $5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, the freezing point of benzene will be lowered by

Options:
A $0.3 \mathrm{~K}$
B $0.5 \mathrm{~K}$
C $0.4 \mathrm{~K}$
D 0.2

The correct answer is:
$0.4 \mathrm{~K}$

Solution:
$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \mathrm{m}=5.12 \times \frac{1}{250} \times \frac{1000}{51.2}=0.4 \mathrm{~K}$

Topic: Chemistry, Solutions, BITSAT, BITSAT 2011
$100 \mathrm{~mL}$ of $2 \mathrm{M}$ of formic acid $\left(\mathrm{p} K_a=3.74\right)$ is neutralise by $\mathrm{NaOH}$, at the equivalence point $\mathrm{pH}$ is

Options:
A
$7$

B
$6$

C
$9.5$

D
$8.87$

The correct answer is:

$8.87$

Solution:
Sodium formate is present at the equivalence point. It is the salt of weak acid $+$ strong base. So, final solution will be basic in nature.
As we know
$\mathrm{pH}=7+\frac{\mathrm{p} K_a}{2}+\frac{\log C}{2}$
where, $C$ is concentration of salt.
Total volume of solution $=100+100=200 \mathrm{~mL}$
Concentration of salt $(C)=2 \times \frac{100}{200}=1 \mathrm{M}$
$\mathrm{pH}=7+\frac{3.74}{2}+\frac{\log [1]}{2}$
$\mathrm{pH}=7+1.87+0$
$\mathrm{pH}=8.87$

Topic: Chemistry, Ionic Equilibrium, BITSAT, BITSAT 2022
107. $\hat{\mathbf{u}}$ and $\hat{\mathbf{v}}$ are two non-collinear unit vectors such that $\left|\frac{\hat{\mathbf{u}}+\hat{\mathbf{v}}}{\mathbf{2}}+\hat{\mathbf{u}} \times \hat{\mathbf{v}}\right|=1$. Then the value of $|\hat{\mathbf{u}} \times \hat{\mathbf{v}}|$ is equal to

Options:
A
$\left|\frac{\hat{\mathbf{u}}+\hat{\mathbf{v}}}{2}\right|$

B
$|\hat{\mathbf{u}}+\hat{\mathbf{v}}|$

C
$|\hat{\mathbf{u}}-\hat{\mathbf{v}}|$

D
$\left|\frac{\hat{\mathbf{u}}-\hat{\mathbf{v}}}{2}\right|$

The correct answer is:

$\left|\frac{\hat{\mathbf{u}}-\hat{\mathbf{v}}}{2}\right|$

Solution:
Given that, $\left|\frac{\hat{\mathbf{u}}+\hat{\mathbf{v}}}{2}+\hat{\mathbf{u}} \times \hat{\mathbf{v}}\right|=1$
$$
\begin{array}{ll}
\Rightarrow & \left|\frac{\hat{\mathbf{u}}+\hat{\mathbf{v}}}{2}+\hat{\mathbf{u}} \times \hat{\mathbf{v}}\right|^2=1 \\
\Rightarrow & \frac{2+2 \cos \theta}{4}+\sin ^2 \theta=1 \\
& \quad[\because \hat{\mathbf{u}} \cdot(\hat{\mathbf{u}} \times \hat{\mathbf{v}})=\hat{\mathbf{v}}(\hat{\mathbf{u}} \times \hat{\mathbf{v}})=0] \\
\Rightarrow & \cos ^2 \frac{\theta}{2}=\cos 2 \theta \\
\Rightarrow & \theta=n \pi \pm \frac{\theta}{2}, n \in z \\
\Rightarrow & \theta=\frac{2 \pi}{3}
\end{array}
$$
$\Rightarrow \quad|\hat{\mathbf{u}} \times \hat{\mathbf{v}}|=\sin \frac{2 \pi}{3}=\sin \frac{\pi}{3}=\left|\frac{\hat{\mathbf{u}}-\hat{\mathbf{v}}}{2}\right|$

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2022
$1.25 \mathrm{~g}$ of a sample of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ and $\mathrm{Na}_{2} \mathrm{SO}_{4}$ is

dissolved in $250 \mathrm{ml}$ solution. $25 \mathrm{ml}$ of this solution neutralises $20 \mathrm{ml}$ of $0.1 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}$. The $\%$ of $\mathrm{Na}_{2} \mathrm{CO}_{3}$

in this sample is

Options:
A $84.8 \%$
B $8.48 \%$
C $15.2 \%$
D $42.4 \%$

The correct answer is:
$84.8 \%$

Solution:
Let the amount of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ present in the mixture be $\mathrm{xg} . \mathrm{Na}_{2} \mathrm{SO}_{4}$ will not react with $\mathrm{H}_{2} \mathrm{SO}_{4} .$ Then

$\frac{x}{53}=\frac{20 \times 0.1 \times 10}{1000} \therefore x=1.06 g$

$\therefore$ Percentage of $\mathrm{Na}_{2} \mathrm{CO}_{3}=$

$\frac{1.06 \times 100}{1.25}=84.8 \%$

Topic: Chemistry, Some Basic Concepts of Chemistry, BITSAT, BITSAT 2016
$3.92 \mathrm{~g}$ of ferrous ammonium sulphate react completely with $50 \mathrm{ml} \frac{\mathrm{N}}{10} \mathrm{KMnO}_{4}$ solution. The percentage purity of the sample is

Options:
A 50
B 78.4
C 80
D 39.2

The correct answer is:
50

Solution:
Eq of $\mathrm{KMnO}_{4}$ used $=\frac{50 \times 1}{1000 \times 10}=0.005$

$\therefore$ Eq of FAS reacted $=0.005$

$\therefore$ weight of FAS needed

$$

=0.005 \times 392=1.96 \mathrm{~g}

$$

Thus percentage purity of FAS is $50 \%$

Topic: Chemistry, Some Basic Concepts of Chemistry, BITSAT, BITSAT 2014
$1.520 \mathrm{~g}$ of hydroxide of a metal on ignition gave $0.995 \mathrm{~g}$ of oxide. The equivalent weight of metal is:

Options:
A 1.52
B 0.995
C 190
D 9

The correct answer is:
9

Solution:
Let $E$ be the equivalent weight of the metal
$$
\text { So, } \frac{E+17}{E+8}=\frac{1.52}{0.995}
$$
[17 is equivalent weight of $\mathrm{OH}$ and 8 is equivalent weight of oxygen]
$$
\begin{array}{l}
\Rightarrow 0.995 \mathrm{E}+17 \times 0.995=\mathrm{E} \times 1.52+8 \times 1.52 \\
\Rightarrow 0.525 \mathrm{E}=16.915-12.16=4.755 \\
\therefore \quad \mathrm{E}=\frac{4.755}{0.525}=9
\end{array}
$$

Topic: Chemistry, Redox Reactions, BITSAT, BITSAT 2011
A $1 \mathrm{~kg}$ mass is attached to a spring of force constant $600 \mathrm{~N} / \mathrm{m}$ and rests on a smooth horizontal surface with other end of the spring tied to wall as shown in figure. A second mass of $0.5 \mathrm{~kg}$ slides along the surface towards the first at $3 \mathrm{~m} / \mathrm{s}$. If the masses make a perfectly inelastic collision, then find amplitude and time period of oscillation of combined mass.

Options:
A $5 \mathrm{~cm}, \frac{\pi}{10} \mathrm{~s}$
B $5 \mathrm{~cm}, \frac{\pi}{5} \mathrm{~s}$
C $4 \mathrm{~cm}, \frac{2 \pi}{5} \mathrm{~s}$
D $4 \mathrm{~cm}, \frac{\pi}{3} \mathrm{~s}$

The correct answer is:
$5 \mathrm{~cm}, \frac{\pi}{10} \mathrm{~s}$

Solution:
No Solution

Topic: Physics, Oscillations, BITSAT, BITSAT 2016
A $2 \mathrm{~m}$ wide truck is moving with a uniform speed $v_{0}=8 \mathrm{~m} / \mathrm{s}$ along a straight horizontal road. $\mathrm{A}$ pedestrain starts to cross the road with a uniform speed $v$ when the truck is 4 m away from him. The minimum value of $v$ so that he can cross the road safely is

Options:
A $2.62 \mathrm{~m} / \mathrm{s}$
B $4.6 \mathrm{~m} / \mathrm{s}$
C $3.57 \mathrm{~m} / \mathrm{s}$
D $1.414 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$3.57 \mathrm{~m} / \mathrm{s}$

Solution:
Let the man starts crossing the road at an angle $\theta$ as shown in figure. For safe crossing the condition is that the man must cross the road by the time the truck describes the distance 4 । $A C$ or 4 । $2 \cot \theta$. $\therefore \frac{4+2 \cot \theta}{8}=\frac{2 / \sin \theta}{v}$ or $v=\frac{8}{2 \sin \theta+\cos \theta}$

For minimum $v, \frac{d v}{d \theta}=0$

or $\frac{-8(2 \cos \theta-\sin \theta)}{(2 \sin \theta+\cos \theta)^{2}}=0$ or $2 \cos \theta-$

$\sin \theta=0$

or $\tan \theta=2$

From equation (i),

$$

v_{\min }=\frac{8}{2\left(\frac{2}{\sqrt{5}}\right)+\frac{1}{\sqrt{5}}}=\frac{8}{\sqrt{5}}=3.57 \mathrm{~m} / \mathrm{s}

$$

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2015
$\int_{-a}^{4}\left(x^{8}-x^{4}+x^{2}+1\right) d x=2 \int_{0}^{4}\left(x^{8}-x^{4}+x^{2}+1\right) d x$ then $a=$

Options:
A 3
B 4
C 6
D 8

The correct answer is:
4

Solution:
If $\mathrm{f}(x)$ is an even function then

$\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x$

Here $\mathrm{f}(x)=x^{8}-x^{4}+x^{2}+1$ is an even function, therefore $a=4$.

Topic: Mathematics, Definite Integration, BITSAT, BITSAT 2021
A 0.5 molal solution of ethylene glycol in water is used as coolant in a car. If the freezing point constant of water be $1.86^{\circ} \mathrm{C}$ per mole, the mixture shall freeze at

Options:
A $0.93^{\circ} \mathrm{C}$
B $-0.93^{\circ} \mathrm{C}$
C $1.86^{\circ} \mathrm{C}$
D $-1.86^{\circ} \mathrm{C}$

The correct answer is:
$-0.93^{\circ} \mathrm{C}$

Solution:
$\begin{aligned} &\Delta T_f=K_f \times m=1.86 \times 0.5=0.93^{\circ} \mathrm{C} \text {; } \\ & T_f=-0.93^{\circ} \mathrm{C}\end{aligned}$

Topic: Chemistry, Solutions, BITSAT, BITSAT 2023 (Memory Based Paper 2)
A $12 \mathrm{~V}$ battery connected to a coil of resistance $6 \Omega$ through a switch, drives a constant current in the circuit. The switch is opened in $1 \mathrm{~ms}$. The emf induced across the coil is $20 \mathrm{~V}$. The inductance of the coil is :

Options:
A $5 \mathrm{mH}$
B $12 \mathrm{mH}$
C $8 \mathrm{mH}$
D $10 \mathrm{mH}$

The correct answer is:
$10 \mathrm{mH}$

Solution:
Induced emf, $\varepsilon=-\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}} \Rightarrow 20=-\mathrm{L} \frac{(0-2)}{10^{-3}}$ Hence, inductance of the coil.
$\mathrm{L}=10 \mathrm{mH}$

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2023 (Memory Based Paper 2)
A $100 \mathrm{~m}$ long wire having cross-sectional area $6.25 \times 10^{-4} \mathrm{~m}^2$ and Young's modulus is $10^{10}$ $\mathrm{Nm}^{-2}$ is subjected to a load of $250 \mathrm{~N}$, then the elongation in the wire will be :

Options:
A $6.25 \times 10^{-3} \mathrm{~m}$
B $4 \times 10^{-4} \mathrm{~m}$
C $6.25 \times 10^{-6} \mathrm{~m}$
D $4 \times 10^{-3} \mathrm{~m}$

The correct answer is:
$4 \times 10^{-3} \mathrm{~m}$

Solution:
$\Delta \ell=\frac{F \ell}{Y A}=\frac{250 \times 100}{10^{10} \times 6.25 \times 10^{-4}}=40 \times 10^{-4} \mathrm{~m}$
$=4 \times 10^{-3} \mathrm{~m}$

Topic: Physics, Mechanical Properties of Solids, BITSAT, BITSAT 2023 (Memory Based Paper 1)
A $12.5 \mathrm{eV}$ electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be:

Options:
A 2
B 1
C 3
D 4

The correct answer is:
3

Solution:
If we assume electron in hydrogen atom takes energy $12.09 \mathrm{eV}$ from the incoming radiation, the maximum excited state of electron will be $n=3$. So, number of spectral lines is $\frac{3(3-1)}{2}=3$.
Here we assume some part of energy $12.5 \mathrm{eV}$ $-12.09 \mathrm{eV}=0.41 \mathrm{eV}$ get lost due to collision.

Topic: Physics, Atomic Physics, BITSAT, BITSAT 2023 (Memory Based Paper 1)
A $3.628 \mathrm{~kg}$ freight car moving along a horizontal rail road spur track at $7.2 \mathrm{~km} /$ hour strikes a bumper whose coil springs experiences a maximum compression of $30 \mathrm{~cm}$ in stopping the car. The elastic potential energy of the springs at the instant when they are compressed $15 \mathrm{~cm}$ is

Options:
A $12.1 \times 10^{4} \mathrm{~J}$
B $121 \times 10^{4} \mathrm{~J}$
C $1.21 \times 10^{4} \mathrm{~J}$
D $1.21 \times 10^{6} \mathrm{~J}$

The correct answer is:
$1.21 \times 10^{4} \mathrm{~J}$

Solution:
\(\begin{aligned} 7.2 \mathrm{~km} / \mathrm{ram} &=\frac{7.2 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}}{7 / 2^{2}} \\ &=\frac{72 \times 5}{10 \times 18}=2 \mathrm{~m} / \mathrm{s} \end{aligned}\)

firt case EC \(\frac{1}{2} k x^{2}=\frac{1}{2} m v^{2}\) \(\Rightarrow \frac{1}{7} k\left(30 \times 10^{-2}\right)^{2}=\frac{1}{2} \times 3.628 \times(2)^{2}\) \(k=\frac{4 \times 3.628}{9 \times 10^{-2}}\)

\(\begin{aligned} \text { EC } & \frac{1}{2} k x^{2}=\frac{1}{2} m v^{2} \\ \Rightarrow & \frac{1}{2} k\left(30 \times 10^{-2}\right)^{2}=\frac{1}{2} \times 3.628 \times(2)^{2} \\ k &=\frac{4 \times 3.628}{9 \times 10^{-2}} \end{aligned}\)

energy potential enrgy when conpressed at \(15 \mathrm{~cm}\)

\(\begin{aligned}\Rightarrow & \frac{1}{2} k\left(15 \times 10^{-2}\right)^{2} \\=& \frac{1}{2} \times \frac{4 \times 3.628}{9 \times 10^{-2}} \times 225 \times 10^{-4} \\=& \frac{1639 \times 10^{-2}}{9} \\=& 182.206 \times 10^{-2} \end{aligned}\)

\(\begin{aligned} &\Rightarrow \frac{1}{2} k\left(15 \times 10^{2}\right. \\ &=\frac{1}{2} \times \frac{\alpha \times 3.628}{9 \times 10^{-2}} \times 225 \times 10^{-4} \\ &=\frac{1639 \times 10^{-2}}{9} \\ &=182.206 \times 10^{-2} \end{aligned}\)

Topic: Physics, Work Power Energy, BITSAT, BITSAT 2013