datavorous/entrance-exam-dataset huggingface dataset:
The correct order of increasing $\mathrm{C}-\mathrm{O}$ bond length $\mathrm{CO}, \mathrm{CO}_{3}^{2-}, \mathrm{CO}_{2}$ is:

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

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

Solution:
$\mathrm{C} \equiv \mathrm{Ohas}$ lowest bond length due to highest bond order. $\mathrm{O}=\mathrm{C}=\mathrm{O}$ has second lowest bond length due to double bond.

has highest has bond length due to lowest bond order which is due to

resonance.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The correct order of increasing $\mathrm{C}-\mathrm{O}$ bond length $\mathrm{CO}, \mathrm{CO}_{3}^{2-}, \mathrm{CO}_{2}$ is:

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

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

Solution:
$\mathrm{C} \equiv \mathrm{Ohas}$ lowest bond length due to highest bond order. $\mathrm{O}=\mathrm{C}=\mathrm{O}$ has second lowest bond length due to double bond.

has highest has bond length due to lowest bond order which is due to

resonance.

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2014
The correct order of magnetic moments (spin only values in B.M.) is:

Options:
A $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}>\left[\mathrm{MnCl}_{4}\right]^{2-}>\left[\mathrm{CoCl}_{4}\right]^{2-}$
B $\left[\mathrm{MnCl}_{4}\right]^{2-}>\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}>\left[\mathrm{CoCl}_{4}\right]^{2-}$
C $\left[\mathrm{MnCl}_{4}\right]^{2-}>\left[\mathrm{CoCl}_{4}\right]^{2-}>\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$
D $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}>\left[\mathrm{CoCl}_{4}\right]^{2-}>\left[\mathrm{MnCl}_{4}\right]^{2-}$

(Atomic nos. : $\mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27)$

The correct answer is:
$\left[\mathrm{MnCl}_{4}\right]^{2-}>\left[\mathrm{CoCl}_{4}\right]^{2-}>\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$

Solution:
Number of unpaired electrons in central atom

$$

\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-},\left[\mathrm{CoCl}_{4}\right]^{2-},\left[\mathrm{MnCl}_{4}\right]^{2-}

$$

zer three five The greater the number of unpaired electrons, the higher the value of magnetic

moment

Topic: Chemistry, Coordination Compounds, BITSAT, BITSAT 2014
The correct order of mobility of alkali metal ions in aqueous solution is-

Options:
A Li+>Na+>K+>Rb+
B Na+>K+>Rb+>Li+
C K+>Rb+>Na+>Li+
D Rb+>K+>Na+>Li+

The correct answer is:
Rb+>K+>Na+>Li+

Solution:
Mobility of alkali ions in aqueous solution ∝1 size of hydrated ion

The mobility of alkali ions is inversely proportional to the size of ion in hydrated state, because bigger ion can't move faster. Also, small ion get hydrolysed to large exent due to which its size get increased. In other words, smaller ion will form hydrated ion of larger size.

Hence, correct order of mobility is

Rb+>K+>Na+>Li+

Topic: Chemistry, s Block Elements, BITSAT, BITSAT 2019
The correct order of radii is

Options:
A $\quad \mathrm{N}<\mathrm{Be}<\mathrm{B}$
B $\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{N}^{3-}$
C $\quad \mathrm{Na}<\mathrm{L}_{1}<\mathrm{K}$
D $\mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}<\mathrm{Fe}^{4+}$

The correct answer is:
$\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{N}^{3-}$

Solution:
Effective nuclear charge (i,e. Z/e ratio) decreases from $\mathrm{F}^{-}$ to $\mathrm{N}^{3-}$ hence the radii follows the order: $\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{N}^{3-} . \mathrm{Z} / \mathrm{e}$ for
$\mathrm{F}^{-}=9 / 10=0.9,$ for $\mathrm{O}^{2-}=8 / 10=.8,$ for $\mathrm{N}^{3-}=0.7$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, BITSAT, BITSAT 2011
The correct order of reactivity of halogens with alkalies is

Options:
A $\quad \mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$
B $\quad \mathrm{F}<\mathrm{Cl}>\mathrm{Br}<\mathrm{I}$
C $\quad \mathrm{F}<\mathrm{Cl}<\mathrm{Br}<\mathrm{I}$
D $\quad \mathrm{F}<\mathrm{Cl}<\mathrm{Br}>\mathrm{I}$

The correct answer is:
$\quad \mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$

Solution:
Reactivity follows the order $\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), BITSAT, BITSAT 2010
The correct sequence of reagents for the following conversion will be

Options:
A O3/Red P, AlCl3, MeCOOH
B H2SO4+HgSO4,H2O/Heat
C O3/Zn-AcOH, H2SO4+HgSO4/H2O/Heat
D CH3COOH, H2O2+OH/H2O

The correct answer is:
H2SO4+HgSO4,H2O/Heat

Solution:
R-C≡CH is converted to ketone by catalytic hydration with reagents.

Hydration of alkynes is carried out in the presence of a warm 40% sulfuric acid in the presence of 1% mercuric sulfate.

In this reaction, first a molecule of water is added to form enol which then tautomerizes to keto form.

Topic: Chemistry, Hydrocarbons, BITSAT, BITSAT 2018
The correct stability order of following species is -

Options:
A $x>y>w>z$
B $y>x>w>z$
C $x>w>z>y$
D $z>x>y>w$

The correct answer is:
$x>w>z>y$

Solution:
$x$ has a conjugated diene system, $\mathrm{w}$ an isolated diene system, $\mathrm{z}$ a cumulated diene system, and y an antiaromatic system.

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2021
The correct statement about silicone is

Options:
A They are ketones with silyl group SiH3 similar to alkyl, SiH32CO
B They are synthetic polymer containing repeated R2SiO2 units.
C They are formed by hydrolysis of R2SiCl2.
D All of the above

The correct answer is:
They are formed by hydrolysis of R2SiCl2.

Solution:
Silicones are synthetic organo-silicon polymers containing repeated R2SiO units. Since the empirical formula is same as that of a ketone R2CO, the name silicon has been given to these materials. They can be formed by hydrolysis of dichlorosilanes R2SiCl2

R2SiCl2+2H2O⟶R2SiOH2+2HCl

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The correct statement about silicone is

Options:
A They are ketones with silyl group SiH3 similar to alkyl, SiH32CO
B They are synthetic polymer containing repeated R2SiO2 units.
C They are formed by hydrolysis of R2SiCl2.
D All of the above

The correct answer is:
They are formed by hydrolysis of R2SiCl2.

Solution:
Silicones are synthetic organo-silicon polymers containing repeated R2SiO units. Since the empirical formula is same as that of a ketone R2CO, the name silicon has been given to these materials. They can be formed by hydrolysis of dichlorosilanes R2SiCl2

R2SiCl2+2H2O⟶R2SiOH2+2HCl

Topic: Chemistry, p Block Elements (Group 13 & 14), BITSAT, BITSAT 2018
The correct statement about the following disaccharide is-

Options:
A Ring I is pyranose with α-glycosidic link
B Ring I is furanose with α-glycosidic link
C Ring II is furanose with α-glycosidic link
D Ring II is pyranose with α-glycosidic link

The correct answer is:
Ring I is pyranose with α-glycosidic link

Solution:
In the given structure, ring one I has one O-atom and is a 6 membered ring, thus is called pyranose ring, whereas ring II has one O-atom and is a five membered ring, thus is called furanose.

The disaccharide is linked by a glycosidic bond between C-1 of glucose (in the α -position, α linkage) and C-2 of fructose (in the β-position, β-linkage).

Topic: Chemistry, Biomolecules, BITSAT, BITSAT 2019
The curve given by $x+y=e^{x y}$ has a tangent parallel to the Y-axis at the point

Options:
A $(0,1)$
B $(1,0)$
C $(1,1)$
D None of these

The correct answer is:
$(1,0)$

Solution:
$\because x+y=e^{x y}$
Differentiating w.r.t. $x$, we get
$1+\frac{d y}{d x}=e^{x y}\left[y+x \frac{d y}{d x}\right]$
$\Rightarrow \quad \frac{d y}{d x}\left(1-x e^{x y}\right)=y e^{x y}-1$
$\Rightarrow \frac{d y}{d x}=\frac{y e^{x y}-1}{1-x e^{x y}} \because \frac{d y}{d x}=\infty$,
as tangent is parallel to $\mathrm{Y}$-axis
$\Rightarrow 1-x e^{x y}=0$
$\therefore \mathrm{xe}^{\mathrm{xy}}=1$
This holds, when $x=1$ and $y=0$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The curve $y=x e^{x}$ has minimum value equal to

Options:
A $-\frac{1}{\mathrm{e}}$
B $\frac{1}{\mathrm{e}}$
C $-\mathrm{e}$
D e

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

Solution:
Let $y=x e^{x}$. Differentiate both side w.r.t. 'x'.

$$

\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}+\mathrm{xe}^{\mathrm{x}}=\mathrm{e}^{\mathrm{x}}(1+\mathrm{x})

$$

Put $\frac{\mathrm{dy}}{\mathrm{dx}}=0$

$$

\begin{array}{l}

\Rightarrow \mathrm{e}^{\mathrm{x}}(1+\mathrm{x})=0 \\

\Rightarrow \mathrm{x}=-1

\end{array}

$$

Now, $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{\mathrm{x}}(1+\mathrm{x})=\mathrm{e}^{\mathrm{x}}(\mathrm{x}+2)$

$$

\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)_{(\mathrm{x}=-1)}=\frac{1}{\mathrm{e}}+0>0

$$

Hence, $\mathrm{y}=\mathrm{xe}^{\mathrm{x}}$ is minimum function and

$$

\mathrm{y}_{\min }=-\frac{1}{\mathrm{e}} \text { . }

$$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2016
The curve \(y-e^{x y}+x=0\) has a vertical tangent at

Options:
A \((1,1)\)
B \((0,1)\)
C \((1,0)\)
D no point

The correct answer is:
\((1,0)\)

Solution:
\(\begin{aligned}
& y-e^{x y}+x=0 \\
& \therefore \frac{d y}{d x}-e^{x y}\left(y+x \frac{d y}{d x}\right)+1=0 \\
& \text {i.e., } \frac{d y}{d x}-y(x+y)-x(x+y) \frac{d y}{d x}+1=0 \\
& \text {i.e., }[1-x(x+y)] \frac{d y}{d x}=y(x+y)-1
\end{aligned}\)
for the vertical tangents
\(1-x(x+y)=0\) i.e., \(\quad y=\frac{1-x^2}{x}\)
\(\therefore \mathrm{x}=1\) and \(\mathrm{y}=0\)

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2011
The $d$ -electron configurations of $\mathrm{Cr}^{2+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{2+}$ and $\mathrm{Co}^{2+}$ are $d^{4}, d^{5}, d^{6}$ and $d^{7}$ respectively. Which one of the following will exhibit the lowest paramagnetic behaviour? (Atomic no. $\mathrm{Cr}=24, \mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27)$

Options:
A $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$
B $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$
C $\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$
D $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$

The correct answer is:
$\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$

Solution:
$\therefore$ Since $\mathrm{Co}^{2+}$ has lowest no. of unpaired electrons hence lowest paramagnetic behaviour is shown by $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$

Topic: Chemistry, Coordination Compounds, BITSAT, BITSAT 2020
The de-Broglie wavelength of a proton (charge =1.6×10-19 C,m=1.6×10-27 kg) accelerated through a potential difference of 1 kV is

Options:
A 600 Å
B 0.9×10-12 m
C 7 Å
D 0.9 nm

The correct answer is:
0.9×10-12 m

Solution:
de-Broglie wavelength,

λ=hp=h2mE=h2mqV

⇒ λ=6.6×10-342×1.6×10-27×1.6×10-19×1000

⇒ λ=6.6×10-347.16×10-22=0.9×10-12 m

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2017
The de-Broglie wavelength of an electron moving with a velocity $\frac{c}{3}\left(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)$ is equal to the wavelength of photon. The ratio of the kinetic energies of electron and photon is

Options:
A
$1: 4$

B
$1: 3$

C
$1: 2$

D
$2: 1$

The correct answer is:

$1: 3$

Solution:
The de-Broglie wavelength, $\lambda=\frac{h}{m v}$
Here, $\quad \lambda_e=\frac{h}{m_e \frac{c}{3}}$ and $\lambda_p=\frac{h}{m_p c}$
Given, $\quad \lambda_e=\lambda_p$
So, $\quad \frac{h}{m_e \frac{c}{3}}=\frac{h}{m_p c}$
$\Rightarrow \quad \frac{m_e}{m_p}=3$
Ratio of KE $=\frac{K_e}{K_p}=\frac{\frac{1}{2} m_e v_e^2}{\frac{1}{2} m_p v_p^2}=\frac{1}{3}$
$\left(\because \frac{m_e}{m_p}=3, v_e=\frac{c}{3}\right.$ and $\left.v_p=c\right)$

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2022
The decreasing order of the ionization potential of the following elements is

Options:
A $\quad \mathrm{Ne}>\mathrm{Cl}>\mathrm{P}>\mathrm{S}>\mathrm{Al}>\mathrm{Mg}$
B $\quad \mathrm{Ne}>\mathrm{Cl}>\mathrm{P}>\mathrm{S}>\mathrm{Mg}>\mathrm{Al}$
C $\quad \mathrm{Ne}>\mathrm{Cl}>\mathrm{S}>\mathrm{P}>\mathrm{Mg}>\mathrm{Al}$
D $\mathrm{Ne}>\mathrm{Cl}>\mathrm{S}>\mathrm{P}>\mathrm{Al}>\mathrm{Mg}$

The correct answer is:
$\quad \mathrm{Ne}>\mathrm{Cl}>\mathrm{P}>\mathrm{S}>\mathrm{Mg}>\mathrm{Al}$

Solution:
Closed shell (Ne), half filled (P) and completely filled configuration (Mg) are the cause of higher value of I.E.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, BITSAT, BITSAT 2016
The degree of dissociation of $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$ in dilute aq. solution containing $7.0 \mathrm{~g}$ of salt per $100 \mathrm{~g}$ of water at $100^{\circ} \mathrm{C}$ is $70 \%$. If vapour pressure of water at $100^{\circ} \mathrm{C}$ is $760 \mathrm{~mm} \mathrm{Hg}$. The vapour pressure of solution is

Options:
A 735
B $\quad 730$
C 760
D $\quad 746$

The correct answer is:
$\quad 746$

Solution:
$\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \rightleftharpoons \mathrm{Ca}^{2+}+2 \mathrm{NO}_{3}^{-}$
$1-0.71 \quad 0.7 \quad 2 \times 0.7$
$\quad\left(\because \alpha=\frac{70}{100}=0.7\right)$
$\quad \mathrm{n}_{1}=\frac{100}{18}=5.55$
$\frac{\mathrm{p}^{0}-\mathrm{p}_{\mathrm{s}}}{\mathrm{p}^{0}}=\frac{\mathrm{n}_{2} \times \mathrm{i}}{\mathrm{n}_{1}+\mathrm{n}_{2}}, \frac{760-\mathrm{p}_{\mathrm{s}}}{760}$
$=\frac{0.042 \times 2.4}{5.55+0.042}$
$\therefore \mathrm{p}_{\mathrm{s}}-746 \mathrm{~mm} \mathrm{Hg} .$

Topic: Chemistry, Solutions, BITSAT, BITSAT 2009
The degree of dissociation of $\mathrm{PCl}_{5}(\alpha)$ obeying

the equilibrium $\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{2}$ is related to the equilibrium pressure by

Options:
A $\alpha \propto \frac{1}{\mathrm{P}^{4}}$
B $\alpha \propto \frac{1}{\sqrt{\mathrm{P}}}$
C $\alpha \propto \frac{1}{\mathrm{P}^{2}}$
D $\alpha \propto \mathrm{P}$

The correct answer is:
$\alpha \propto \frac{1}{\sqrt{\mathrm{P}}}$

Solution:
$\therefore \mathrm{K}_{\mathrm{p}}=\frac{\frac{\alpha}{1+\alpha} \mathrm{P} \times \frac{\alpha}{1+\alpha} \mathrm{P}}{\frac{1-\alpha}{1+\alpha} \mathrm{P}}=\frac{\alpha^{2} \mathrm{P}}{1-\alpha^{2}}$

or, $\mathrm{K}_{\mathrm{p}}=\alpha^{2} \mathrm{P}$

$\therefore \alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{P}}}$ when $1-\alpha^{2}=1$

Topic: Chemistry, Chemical Equilibrium, JEE Main
The degree of dissociation of $\mathrm{PCl}_{5}(\alpha)$ obeying

the equilibrium $\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{2}$ is related to the equilibrium pressure by

Options:
A $\alpha \propto \frac{1}{\mathrm{P}^{4}}$
B $\alpha \propto \frac{1}{\sqrt{\mathrm{P}}}$
C $\alpha \propto \frac{1}{\mathrm{P}^{2}}$
D $\alpha \propto \mathrm{P}$

The correct answer is:
$\alpha \propto \frac{1}{\sqrt{\mathrm{P}}}$

Solution:
$\therefore \mathrm{K}_{\mathrm{p}}=\frac{\frac{\alpha}{1+\alpha} \mathrm{P} \times \frac{\alpha}{1+\alpha} \mathrm{P}}{\frac{1-\alpha}{1+\alpha} \mathrm{P}}=\frac{\alpha^{2} \mathrm{P}}{1-\alpha^{2}}$

or, $\mathrm{K}_{\mathrm{p}}=\alpha^{2} \mathrm{P}$

$\therefore \alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{P}}}$ when $1-\alpha^{2}=1$

Topic: Chemistry, Chemical Equilibrium, BITSAT, BITSAT 2016
The degree of the differential equation \(\left(\frac{d^3 y}{d x^3}\right)^{2 / 3}+4-3 \frac{d^2 y}{d x^2}+5 \frac{d y}{d x}=0\) is

Options:
A 1
B 2
C 3
D None of these

The correct answer is:
2

Solution:
\(\begin{aligned}
& \left(\frac{d^3 y}{d x^3}\right)^{2 / 3}+4-3 \frac{d^2 y}{d x^2}+5 \frac{d y}{d x}=0 \\
& \Rightarrow\left(\frac{d^3 y}{d x^3}\right)^2=\left[3 \frac{d^2 y}{d x^2}-5 \frac{d y}{d x}-4\right]^3
\end{aligned}\)
It is a differential equation of degree 2.

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2011
The degree of the differential equation satisfying

$\sqrt{1-x^{2}}+\sqrt{1+y^{2}}=a(x-y)$ is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
$\quad$ Put $x=\sin \theta$ and $y=\sin \phi$

$\Rightarrow \cos \theta+\cos \phi=a(\sin \theta-\sin \phi)$

$\Rightarrow 2 \cos$

$\begin{aligned} & \frac{\theta+\phi}{2} \cos \frac{\theta-\phi}{2}=2 a \cos \frac{\theta+\phi}{2} \sin \frac{\theta-\phi}{2} \\ \Rightarrow \cot \frac{\theta-\phi}{2}=a \Rightarrow \theta-\phi=2 \cot ^{-1} \mathrm{a} \\ \Rightarrow \sin ^{-1} \mathrm{x}-\sin ^{-1} \mathrm{y}=2 \cot ^{-1} \mathrm{a} \\ \text { Differentiate } \frac{1}{\sqrt{1-\mathrm{x}^{2}}}-\frac{1}{\sqrt{1-\mathrm{y}^{2}}} \frac{\mathrm{dy}}{\mathrm{dx}}=0 \end{aligned}$

so the degree is one

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2015
The density of a certain gas at 300C and 768 torr is 1.35 kg/m3, then density of the gas at STP is

Options:
A 1.48 kg/m3
B 1.27 kg/m3
C 1.35 kg/m3
D 1.00 kg/m3

The correct answer is:
1.48 kg/m3

Solution:
Using PM=dRT

P1MP2M = d1RT1d2RT2768760 = 1.35×303d2×273d2 = 1.35×303×760273×768d2 = 1.48 kg/m3

Topic: Chemistry, States of Matter, BITSAT, BITSAT 2019
The derivative of $e^{x^3}$ with respect to $\log x$ is

Options:
A $\mathrm{e}^{\mathrm{x}^3}$
B $3 \mathrm{x}^2 2 \mathrm{e}^{\mathrm{x}^3}$
C $3 \mathrm{x}^3 \mathrm{e}^{\mathrm{x}^3}$
D $3 \mathrm{x}^3 \mathrm{e}^{\mathrm{x}^3}+3 \mathrm{x}^2$

The correct answer is:
$3 \mathrm{x}^3 \mathrm{e}^{\mathrm{x}^3}$

Solution:
Let $y=\mathrm{e}^{\mathrm{x}^3}, \mathrm{z}=\log \mathrm{x}$
On differentiating w.r.t.x, we get
$\frac{d y}{d x}=e^{x^3}\left(3 x^2\right)=3 x^2 e^{x^3} \text { and } \frac{d z}{d x}=\frac{1}{x}$
$\therefore \frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}=\frac{3 x^2 e^{x^3}}{\left(\frac{1}{x}\right)}=3 x^3 e^{x^3}$

Topic: Mathematics, Differentiation, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The determinant $\left|\begin{array}{ccc}1 & (x-3) & (x-3)^{2} \\ 1 & (x-4) & (x-4)^{2} \\ 1 & (x-5) & (x-5)^{2}\end{array}\right|$

Options:
A 3 values of $x$
B 2 values of $x$
C 1 values of $x$
D No value of $x$

The correct answer is:
No value of $x$

Solution:
The given determinant vanishes, i.e.,

$$

\left|\begin{array}{ccc}

1 & x-3 & (x-3)^{2} \\

1 & x-4 & (x-4)^{2} \\

1 & x-5 & (x-5)^{2}

\end{array}\right|=0

$$

Expanding along $\mathrm{C}_{1},$ we get

$$

\begin{array}{l}

(x-4)(x-5)^{2}-(x-5)(x-4)^{2}-\left\{(x-3)(x-5)^{2}\right. \\

\left.-(x-5)(x-3)^{2}\right\}+(x-3)(x-4)^{2} \\

-(x-4)(x-3)^{2}=0 \\

\Rightarrow(x-4)(x-5)(x-5-x+4) \\

-(x-3)(x-5)(x-5-x+3) \\

+(x-3)(x-4)(x-4-x+3)=0

\end{array}

$$

$$

\Rightarrow-(x-4)(x-5)+2(x-3)(x-5)-(x-3)

$$

$(x-4)=0$

$$

\Rightarrow-x^{2}+9 x-20+2 x^{2}-16 x+30-x^{2}+7 x-12=0

$$

$\Rightarrow-32+30=0 \Rightarrow-2=0$

Which is not possible, hence no value of $\mathrm{x}$ satisfies the given condition.

Topic: Mathematics, Determinants, BITSAT, BITSAT 2013
The diagonal of a square is changing at the rate of $0.5 \mathrm{~cm} / \mathrm{sec}$. Then the rate of change of area, when the area is $400 \mathrm{~cm}^{2}$, is equal to

Options:
A $20 \sqrt{2} \mathrm{~cm}^{2} / \mathrm{sec}$
B $10 \sqrt{2} \mathrm{~cm}^{2} / \mathrm{sec}$
C $\frac{1}{10 \sqrt{2}} \mathrm{~cm}^{2} / \mathrm{sec}$
D $\frac{10}{\sqrt{2}} \mathrm{~cm}^{2} / \mathrm{sec}$

The correct answer is:
$10 \sqrt{2} \mathrm{~cm}^{2} / \mathrm{sec}$

Solution:
Diagonal $\mathrm{D}=\sqrt{2}$ a
Differentiating w.r.t.
t: $\frac{\mathrm{dD}}{\mathrm{dt}}=\sqrt{2} \frac{\mathrm{da}}{\text { at }}$
or $\frac{\mathrm{da}}{\mathrm{dt}}=\frac{1}{\sqrt{2}} \frac{\mathrm{da}}{\mathrm{dt}}=\frac{1}{\sqrt{2}} \times 0.5 \mathrm{~cm} / \mathrm{s}$
Let Area is denoted by $\underline{A}$ $\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=2 \mathrm{a} \frac{\mathrm{da}}{\mathrm{dt}}$
when area $A$ is $400 \mathrm{~cm}^{2}$ then $\mathrm{a}=20$ $\therefore \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=2 \times 20 \times \frac{0.5}{\sqrt{2}}=10 \sqrt{2} \mathrm{~cm}^{2} / \mathrm{sec}$

Topic: Mathematics, Differentiation, BITSAT, BITSAT 2012
The diameter of the objective of a telescope is a, its magnifying power is $\mathrm{m}$ and wavelength of light is $\lambda,$ The resolving power of the telescope is:

Options:
A $(1.22 \lambda) / \mathrm{a}$
B $(1.22 \mathrm{a}) / \lambda$
C $\lambda \mathrm{m} /(1.22 \mathrm{a})$
D a/(1.22\lambda)

The correct answer is:
a/(1.22\lambda)

Solution:
No Solution available

Topic: Physics, Ray Optics, BITSAT, BITSAT 2010
The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is:

Options:
A $\mathrm{Ni}^{2+}$
B $\mathrm{Fe}^{2+}$
C $\mathrm{Co}^{2+}$
D $\mathrm{Mn}^{2+}$

The correct answer is:
$\mathrm{Co}^{2+}$

Solution:

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2021
The difference of maximum and minimum values of f(x)=x2e-x is

Options:
A e
B 1/e
C 1-16
D 1+16

The correct answer is:
1/e

Solution:
We have,

fx=x2e-x2

f'x=2xe-x2-2x3e-x2

f'x=2xe-x21-x2

f'x=0,x=0,-1,1

f'-h<0,f'h>0, So minimum at x=0

f'1-h>0,f'1+h<0, So maximum at x=1

f'-1-h>0,f'-1+h<0, So maximum at x=-1

Minimum fx=f0=0

Maximum fx=e-1=1e

Since, fx≥0∀x

So, difference between maximum and minimum values

=1e-0=1e

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2019
The differential equation whose solution is $A x^{2}+B y^{2}=1$ where $A$ and $B$ are arbitrary constants is of

Options:
A second order and second degree
B first order and second degree
C first order and first degree
D second order and first degree

The correct answer is:
second order and first degree

Solution:
$A x^{2}+B y^{2}=1$ ...(1)
$A x+B y \frac{d y}{d x}=0$ ...(2)
$A+B y \frac{d^{2} y}{d x^{2}}+B\left(\frac{d y}{d x}\right)^{2}=0$ ...(3)
$x\left\{-B y \frac{d^{2} y}{d x^{2}}-B\left(\frac{d y}{d x}\right)^{2}\right\}+B y \frac{d y}{d x}=0$
Dividing both sides by $-\mathrm{B}$, we get $x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0$
Which is a DE of order 2 and degree 1

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2012
The digits \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are such that the three digit numbers \(A 88,6 \mathrm{~B} 8,86 \mathrm{C}\) are divisible by 72 then the determinant \(\left|\begin{array}{ccc}\text { A } & 6 & 8 \\ 8 & B & 6 \\ 8 & 8 & C\end{array}\right|\) is divisible by

Options:
A 72
B 144
C 288
D 216

The correct answer is:
72

Solution:
\(R_3 \rightarrow 100 R_1+10 R_2+R_3\)
\(\Rightarrow\left|\begin{array}{ccc}
A & 6 & 8 \\
8 & B & 6 \\
8 & 8 & C
\end{array}\right|=\left|\begin{array}{ccc}
A & 6 & 8 \\
8 & B & 6 \\
A 88 & 6 B C & 86 C
\end{array}\right|\)
which is divisible by 72.

Topic: Mathematics, Determinants, BITSAT, BITSAT 2009
The dimensional formula of μ0ε0 is

Options:
A ML2 T-3A2
B M0LT-1A0
C ML2 T-3A-2
D M-1 L-2 T3 A2

The correct answer is:
ML2 T-3A-2

Solution:
We know that, velocity of light c is terms of ϵ0 and μ0 is given by

c=1μ0ε0⇒c2=1μ0ε0

c2=1μ02μ0ε0⇒c=1μ0μ0ε0⇒μ0ε0=cμ0 ...i

Dimension of the magnetic field, B=τNIA=ML2 T-2AL2

=MA-1T-2

Again, B=μ0nI

∴ μ0=Bnl=MA-1T-2L-1A=MLA-2T-2

μ0=MLA-2T-2 ...ii

and C=LT-1 ...iii

From Equations. (i), (ii) and (iii), we get

μ0ϵ0=LT-1MLA-2T-2=ML2T-3A-2

Topic: Physics, Units and Dimensions, BITSAT, BITSAT 2019
The disease caused the high concentration of hydrocarbon pollutants in atmosphere is/are

Options:
A silicosis
B $\mathrm{TB}$
C cancer
D asthma

The correct answer is:
cancer

Solution:
The high concentration of hydrocarbon pollutants in atmosphere causes cancer.

Topic: Chemistry, Environmental Chemistry, BITSAT, BITSAT 2015