datavorous/entrance-exam-dataset huggingface dataset:
The correct order of \(\mathrm{C}=\mathrm{O}\) bond length in ethyl propanoate (I), ethyl propenoate (II) and ethenyl propanoate (III) is

Options:
A I \( > \) II \( > \) III
B III \( > \) II \( > \) I
C I \( > \) III \( > \) II
D II \( > \) I \( > \) III

The correct answer is:
II \( > \) I \( > \) III

Solution:
Hint :
\(\mathrm{C}=\mathrm{O}\) bond length
\(\text {II } > \text { I } > \text { III }\)
bond has the most single bond character in compound II and the least single bond character in compound III.

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2023
The correct order of $\mathrm{O}-\mathrm{O}$ bond length in $\mathrm{O}_{2}, \mathrm{H}_{2} \mathrm{O}_{2}$ and $\mathrm{O}_{3}$ is

Options:
A $\mathrm{O}_{2}>\mathrm{O}_{3}>\mathrm{H}_{2} \mathrm{O}_{2}$
B $\mathrm{H}_{2} \mathrm{O}_{2}>\mathrm{O}_{3}>\mathrm{O}_{2}$
C $\mathrm{O}_{3}>\mathrm{O}_{2}>\mathrm{H}_{2} \mathrm{O}_{2}$
D $\mathrm{O}_{3}>\mathrm{H}_{2} \mathrm{O}_{2}>\mathrm{O}_{2}$

The correct answer is:
$\mathrm{H}_{2} \mathrm{O}_{2}>\mathrm{O}_{3}>\mathrm{O}_{2}$

Solution:
Correct order of $O-O$ bond length is
$$
\mathrm{H}_{2} \mathrm{O}_{2}>\mathrm{O}_{3}>\mathrm{O}_{2}
$$
(i) Between $\mathrm{O}_{3}$ and $\mathrm{O}_{2}$
$\because$ Bond order of $\mathrm{O}_{2}>\mathrm{O}_{3}$
Thus, bond length of $\mathrm{O}_{3}>\mathrm{O}_{2}$
$$
\left ( \because \text { Bond order } \propto \frac{1}{\text { Bond length }}\right)
$$
(ii) Between $\mathrm{O}_{3}$ and $\mathrm{H}_{2} \mathrm{O}_{2}$
$\because \mathrm{O}_{3}$ has $\pi$ -bonds in resonance so, has shorter bond length than $O-O$ in $H_{2} O_{2}$
$\therefore$ Hence, (b) is the answer.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2016
The correct order of reactivity for the addition reaction of the following carbonyl compounds with ethylmagnesium iodide is

Options:
A I > III > II > IV
B IV > III > II > I
C I > II > IV > III
D III > II > I > IV

The correct answer is:
I > III > II > IV

Solution:
The correct order of reactivity for the addition reaction of the given carbonyl compounds with ethylmagnesium iodide is $1>111>11>\mathrm{IV}$

This order can be explained on the basis of following two factors:
(i) Inductive effect Greater the number of alkyl (electron releasing) groups attached to carbonyl group, greater will be the electron density on carbonyl carbon. Thus, it lowers the attack of nucleophile and hence, reactivity decreases.
(ii) Steric effect As the number of alkyl group attached to carbonyl carbon increases, the attack of nucleophile on carbonyl group becomes more and more difficult due to steric hinderance.

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2018
The correct order of relative stability for the given free radicals is

Options:
A II > I > III
B II > III > I
C III > I > II
D III > II > I

The correct answer is:
III > II > I

Solution:
III>II>I
Adjacent functional group appear to weaken $\mathrm{C}-\mathrm{H}$ Bonds, hence making free radical more stable
( All the three are molecules - not radicals. We have assumed them as radicals)

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2022
The correct order of relative stability of the given conformers of $n$-butane is

Options:
A II > I = III
B II > III > I
C II > I > III
D I = III > II

The correct answer is:
II > I = III

Solution:
II > I = III
Anti form is more stable than Gauche form.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2022
The correct order of the addition reaction rates of halogen acids with ethylene is

Options:
A hydrogen chloride $>$ hydrogen bromide > hydrogen iodide
B hydrogen iodide $>$ hydrogen bromide > hydrogen chloride
C hydrogen bromide $>$ hydrogen chloride > hydrogen iodide
D hydrogen iodide $>$ hydrogen chloride > hydrogen bromide

The correct answer is:
hydrogen iodide $>$ hydrogen bromide > hydrogen chloride

Solution:
In halogen acids, as the size of halogen atom increases, the bond between halogen and hydrogen atom weakens. Therefore, the case with which the bond can be broken increases.
The correct order of the addition reaction rates of halogen acids with ethylene is hydrogen iodide (HI) $>$ Hydrogen bromide (HBr)
Hydrogen chloride (HCl) So, the option (b) is correct answer.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2019
The correct pair of electron affinity order is

Options:
A $\mathrm{O}>\mathrm{S}, \mathrm{F}>\mathrm{Cl}$
B $\mathrm{O} < \mathrm{S}, \mathrm{Cl}>\mathrm{F}$
C $\mathrm{S}>\mathrm{O}, \mathrm{F}>\mathrm{Cl}$
D $\mathrm{S} < \mathrm{O}, \mathrm{Cl}>\mathrm{F}$

The correct answer is:
$\mathrm{O} < \mathrm{S}, \mathrm{Cl}>\mathrm{F}$

Solution:
Second period elements of p-block have lower values of electron affinity than expected due to unusually smaller size.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, WBJEE, WBJEE 2022
The correct relationship between molecules I and II is

Options:
A Enantiomer
B Homomer
C Diastereomer
D Constitutional isomer

The correct answer is:
Homomer

Solution:
They are homomers

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2022
The correct set(s) of reactions to synthesize benzoic acid starting from benzene is/are

Options:
A (i) \(\mathrm{Br}_2 / \mathrm{Fe}\)
(ii) Mg / dry ether
(iii) \(\mathrm{CO}_2\)
(iv) \(\mathrm{H}_3 \mathrm{O}^{\oplus}\)
B (i) \(\mathrm{Br}_2 / \mathrm{Fe}\)
(ii) \(\mathrm{NH}_3, 25^{\circ} \mathrm{C}\)
(iii) \(\mathrm{NaNO}_2\), dil. \(\mathrm{HCl}, 0^{\circ}\) to \(5^{\circ} \mathrm{C}\)
(iv) \(\mathrm{CuCN} / \mathrm{KCN}\)
(v) dil. \(\mathrm{HCl}, \Delta\)
C (i) \(\mathrm{CH}_3 \mathrm{Cl}\), Anhydrous \(\mathrm{AlCl}_3\)
(ii) \(\mathrm{KMnO}_4 \mid \mathrm{OH}^{\ominus}, \Delta\)
(iii) \(\mathrm{H}_3 \mathrm{O}^{\oplus}\)
D (i) \(\mathrm{CH}_3 \mathrm{COCl}\), Anhydrous \(\mathrm{AlCl}_3\)
(ii) \(\mathrm{Br}_2, \mathrm{NaOH}\),
(iii) \(\mathrm{H}_3 \mathrm{O}^{\oplus}\)

The correct answers are:
(i) \(\mathrm{Br}_2 / \mathrm{Fe}\)
(ii) Mg / dry ether
(iii) \(\mathrm{CO}_2\)
(iv) \(\mathrm{H}_3 \mathrm{O}^{\oplus}\), (i) \(\mathrm{CH}_3 \mathrm{Cl}\), Anhydrous \(\mathrm{AlCl}_3\)
(ii) \(\mathrm{KMnO}_4 \mid \mathrm{OH}^{\ominus}, \Delta\)
(iii) \(\mathrm{H}_3 \mathrm{O}^{\oplus}\), (i) \(\mathrm{CH}_3 \mathrm{COCl}\), Anhydrous \(\mathrm{AlCl}_3\)
(ii) \(\mathrm{Br}_2, \mathrm{NaOH}\),
(iii) \(\mathrm{H}_3 \mathrm{O}^{\oplus}\)

Solution:
Hint : (A)
(C)
(D)

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2023
The correct stability order of the following carbocations is
\(\underset{\text{I}}{\mathrm{H}_2 \stackrel{\oplus}{\mathrm{C}}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3}\)
\(\underset{\text{II}}{\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_2-\mathrm{CH}=\mathrm{CH}-\mathrm{BMe}_2}\)
\(\underset{\text{III}}{\mathrm{H}_2 \stackrel{\oplus}{\mathrm{C}}-\mathrm{CH}=\mathrm{CH}-\mathrm{NMe}_2}\)
\(\underset{\text{Iv}}{\mathrm{H}_2 \stackrel{\oplus}{\mathrm{C}}-\mathrm{CH}=\mathrm{CH}-\mathrm{OMe}}\)

Options:
A II \( > \) I \( > \) III \( > \) IV
B III \( > \) I \( > \) II \( > \) IV
C III \( > \) IV \( > \) I \( > \) II
D IV \( > \) III \( > \) II \( > \) I

The correct answer is:
III \( > \) IV \( > \) I \( > \) II

Solution:
Hint: \(\quad\) III \( > \) IV \( > | > | \mid\)
\(\left.+\mathrm{R} \text { effect : }-\ddot{\mathrm{N}}\mathrm{Me}_2\right\rangle-\underset{\cdot\cdot}{\ddot{\mathrm{O}}} \mathrm{CH}_3\)
\(-\mathrm{BMe}_2\) can show \(-\mathrm{R}\) effect

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2023
The correct statement about the magnetic properties of \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}\) and \(\left[\mathrm{FeF}_6\right]^{3-}\) is

Options:
A Both are paramagnetic
B Both are diamagnetic
C \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}\) is diamagnetic, \(\left[\mathrm{FeF}_6\right]^{3-}\) is paramagnetic
D \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}\) is paramagnetic, \(\left[\mathrm{FeF}_6\right]^{3-}\) is diamagnetic

The correct answer is:
Both are paramagnetic

Solution:
Hint : \(\mathrm{Fe}^{+3}=[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^0\)
For \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}\)
Pairing of \(\mathrm{e}^{-}\)takes place as \(\mathrm{CN}^{-}\)is strong field ligand but has one unpaired electron thus paramagnetic.
For \(\left[\mathrm{FeF}_6\right]^{-3}\)
As \(\mathrm{F}^{\text {- }}\) is weak field ligand, so no pairing of electron, thus it has five unpaired electron. Therefore paramagnetic.

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2023
The correct statement regarding the following compounds is

Options:
A all three compounds are chiral
B only I and II are chiral
C I and III are diastereomers
D only I and III are chiral

The correct answer is:
only I and III are chiral

Solution:
$2-R, 3-R$ configuration, s0 it is a chiral compound

$2-R, 3-S$ configuration has plane of symmetry or in other words, rotation of hall part is cancelled by the other half, so it is not a chiral compound.

$2-S, 3-S$ configuration, so it is also a chiral compound.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2014
The correct statement regarding the following energy diagrams is

Options:
A Reaction $M$ is faster and less exothermic than reaction $N$
B Reaction $M$ is slower and less exothermic than reaction $N$
C Reaction $M$ is faster and more exothermic than reaction $N$
D Reaction $M$ is slower and more exothermic than reaction $N$

The correct answer is:
Reaction $M$ is faster and more exothermic than reaction $N$

Solution:
In reaction $M$ there is only an intermediate step. $\$ 0$ it is a fast reaction. Further, in reaction $M$ energy of product is much lesser than that of the reactant, as compared to as shown in reaction $N$. Thus reaction $M$ is faster and more exothermic than reaction $N$.

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2014
The correct statement(s) about $\mathrm{B}_2 \mathrm{H}_6$ is / are :

Options:
A All B atoms are $\mathrm{sp}^3$ hybridised
B It is paramagnetic
C It contains 3C - 4e bonding
D There are two types of $\mathrm{H}$ present

The correct answers are:
All B atoms are $\mathrm{sp}^3$ hybridised, There are two types of $\mathrm{H}$ present

Solution:
In diborane, all boron atoms are $\mathrm{sp}^3$ hybridised. It is diamagnetic and contains $3 \mathrm{C}-2 \mathrm{e}$ bonds. There are two types of hydrogens $-$ terminal and bridging.

Topic: Chemistry, p Block Elements (Group 13 & 14), WBJEE, WBJEE 2022
The correct structure of the dipeptide gly-ala is

Options:
A
B
C
D

The correct answer is:

Solution:
The structure of dipeptide gly-ala is

Topic: Chemistry, Biomolecules, WBJEE, WBJEE 2013
The correct structure of the drug paracetamol is

Options:
A
B
C
D

The correct answer is:

Solution:
Paracetamol is 4 -acetamidophenol, i.e. N-(4-hydroxy phenyl) ethanamide (IUPAC Name).

Topic: Chemistry, Chemistry in Everyday Life, WBJEE, WBJEE 2016
The cosine of the angle between any two diagonals of a cube is

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

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

Solution:
Let the direction ratio's of two diagonals of a cube are
(1,1,1) and (-1,1,1) $\begin{aligned} \therefore \cos \theta &=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{f} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ &=\frac{(1) \times(-1)+(1) \times(1)+(1) \times(1)}{\sqrt{(1)^{2}+(1)^{2}+(1)^{2}} \sqrt{(-1)^{2}+(1)^{2}+(1)^{2}}} \\ &=\frac{-1+1+1}{\sqrt{3} \sqrt{3}}=\frac{1}{3} \end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, WBJEE, WBJEE 2016
The cross-section of a reflecting surface is represented by the equation $x^{2}+y^{2}=R^{2}$ as shown in the figure. A ray travelling in the positive $x$ direction is directed toward positive y direction after reflection from the surface at point $M$. The coordinate of the point $\mathrm{M}$ on the reflecting surface is

Options:
A $\left(\frac{\mathrm{R}}{\sqrt{2}}, \frac{\mathrm{R}}{\sqrt{2}}\right)$
B $\left(-\frac{\mathrm{R}}{2},-\frac{\mathrm{R}}{2}\right)$
C $\left(-\frac{R}{\sqrt{2}}, \frac{R}{\sqrt{2}}\right)$
D $\left(\frac{R}{\sqrt{2}},-\frac{R}{\sqrt{2}}\right)$

The correct answer is:
$\left(-\frac{R}{\sqrt{2}}, \frac{R}{\sqrt{2}}\right)$

Solution:
$x^{2}+y^{2}=R^{2}$
$2 x+2 y \frac{d y}{d x}=0$
$\frac{d y}{d x}=-\frac{x}{y}=\tan 45^{\circ}=1$
$\therefore x=-\frac{R}{\sqrt{2}}$
$y=+\frac{R}{\sqrt{2}}$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2021
The current flowing through a wire depends on time as $I=3 t^2+2 t+5$. The charge flowing through the cross section of the wire in time from $t=0$ to $t=2 \mathrm{sec}$. is

Options:
A $22 \mathrm{C}$
B $20 \mathrm{C}$
C $18 \mathrm{C}$
D $5 \mathrm{C}$

The correct answer is:
$22 \mathrm{C}$

Solution:
$Q=\int_0^2\left(3 t^2+2 t+5\right) d t=22 C$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2009
The current $I$ is in the circuit shown is

Options:
A $1.33 \mathrm{A}$
B zero
C $2.00 \mathrm{A}$
D $1.00 \mathrm{A}$

The correct answer is:
$1.33 \mathrm{A}$

Solution:
The circuit diagram can be redrawn as

For the loop $A B C D A$,
$$
+2-2 I^{\prime}+2-2 I=0
$$
For the loop $A B F E A$,
$$
\begin{array}{r}
2-2 I+2-2\left(I-I^{\prime}\right)=0 \\
4-2 I-2 I+2 I^{\prime}=0
\end{array}
$$
$$
\begin{aligned}
4-4 I+2 I^{\prime} &=0 \\
2 &=2 I-I^{\prime}
\end{aligned}
$$
From Eqs (i) and (in), we get
$$
2=2 I-I^{\prime} \cdot 2=I+I^{\prime}
$$
$$
4=3 I, I=4 / 3=1.33 \mathrm{A}
$$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2015
The current through a coil of self inductance $\mathrm{L}=2 m \mathrm{H}$ is given by $\mathrm{I}=t^2 e^{-t}$ at time $t$. How long it will take to make the e $\mathrm{m} . f$ zero?

Options:
A $1 \mathrm{~s}$
B $2 \mathrm{~s}$
C $3 \mathrm{~s}$
D $4 \mathrm{~s}$

The correct answer is:
$2 \mathrm{~s}$

Solution:
Hints: $\mathrm{I}=t^2 e^{-t}$
$$
\begin{aligned}
& \frac{d \mathrm{I}}{d t}=2 t e^{-t}-e^{-t} t^2=e^{-t} t(2-t) \\
& e=-\mathrm{L} \frac{d \mathrm{I}}{d t} \\
& \Rightarrow \frac{d \mathrm{I}}{d t}=0 \Rightarrow e^{-t} t(2-t)=0 \\
& t=2 \mathrm{sec}
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, WBJEE, WBJEE 2010
The curve $y=(\cos x+y)^{1 / 2}$ satisfies the differential equation

Options:
A $(2 y-1) \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}+\cos x=0$
B $\frac{d^{2} y}{d x^{2}}-2 y\left(\frac{d y}{d x}\right)^{2}+\cos x=0$
C $(2 y-1) \frac{d^{2} y}{d x^{2}}-2\left(\frac{d y}{d x}\right)^{2}+\cos x=0$
D $(2 y-1) \frac{d^{2} y}{d x^{2}}-\left(\frac{d y}{d x}\right)^{2}+\cos x=0$

The correct answer is:
$(2 y-1) \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}+\cos x=0$

Solution:
Given curve is
$$
y=(\cos x+y)^{1 / 2}
$$
On differentiating both sides w.r.t. $x,$ we get
$$
\begin{aligned}
\frac{d y}{d x} &=\frac{1}{2(\cos x+y)^{1 / 2}} \times\left(-\sin x+\frac{d y}{d x}\right) \\
&=\frac{\left(-\sin x+\frac{d y}{d x}\right)}{2 y} \\
\Rightarrow \quad & 2 y \frac{d y}{d x}=-\sin x+\frac{d y}{d x} \\
\Rightarrow \quad(2 y-1) \frac{d y}{d x} &=-\sin x
\end{aligned}
$$
Again, differentiating both sides w.r.t. $x$, we get
$$
\begin{array}{r}
(2 y-1) \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}\left(2 \frac{d y}{d x}\right)=-\cos x \\
\Rightarrow \quad(2 y-1) \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}\left(2 \frac{d y}{d x}\right)+\cos x=0
\end{array}
$$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2014
The de-Broglie wavelength $(\lambda)$ for electron (e), proton $(p)$ and $\mathrm{He}^{2+}$ ion $(\alpha)$ are in the following order. (Speed of e, p and $\alpha$ are the same)

Options:
A $\alpha>p>e$
B $e>p>\alpha$
C $e>\alpha>p$
D $\alpha < p>e$

The correct answer is:
$e>p>\alpha$

Solution:
$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$
$\because$ speed are same
$\begin{aligned}
&\therefore \lambda \propto \frac{1}{\mathrm{~m}} \\
&\therefore \lambda_e>\lambda_p>\lambda_\alpha \text { as } \quad \mathrm{m}_\alpha>\mathrm{m}_{\mathrm{p}}>\mathrm{m}_e
\end{aligned}$

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2022
The de-Broglie wavelength of an electron is $0.4 \times 10^{-10} \mathrm{m}$ when its kinetic energy is $1.0 \mathrm{keV} .$ Its wavelength will be $1.0 \times 10^{-10} \mathrm{m}$
when its kinetic energy is

Options:
A $0.2 \mathrm{keV}$
B $0.8 \mathrm{keV}$
C $0.63 \mathrm{keV}$
D $0.16 \mathrm{keV}$

The correct answer is:
$0.16 \mathrm{keV}$

Solution:
De-Broglie wavelength, $\lambda=\frac{h}{\sqrt{2 m E}}$
$\therefore \quad \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{E_{2}}{E_{1}}}$
$\lambda_{1}=$ Wave length of electron
$\begin{aligned} &=0.4 \times 10^{-10} \mathrm{m} \\ E_{1} &=1 \cdot \text { OkeV } \\ \lambda_{2} &=1 \cdot 0 \times 10^{-10} \mathrm{m} \\ E_{2} &=? \\ \frac{0 \cdot 4 \times 10^{-10}}{1 \times 10^{-10}} &=\sqrt{\frac{E_{2}}{1 \cdot 0 \mathrm{keV}}} \\ \frac{4}{10}=& \sqrt{\frac{E_{2}}{1}} \\ \frac{16}{100}=& \frac{E_{2}}{1} \\ E_{2}=& 0 \cdot 16 \mathrm{keV} \end{aligned}$

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2017
The de-Broglie wavelength of an electron is the same as that of a $50 \mathrm{keV}$ X-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is $0.5 \mathrm{MeV}$ )

Options:
A 01:50
B 01:20
C 20: 1
D 50: 1

The correct answer is:
20: 1

Solution:
de-Broglie wavelength $\lambda=\frac{h}{\sqrt{2 m K}}$
The kinetic energy of the electron
$$
K_{\text {electron }}=\frac{1}{2 m} \cdot \frac{h^{2}}{\lambda^{2}}
$$
Where $h=$ Planck constant $\lambda=$ wavelength
The photon energy
$$
E_{\text {phocton }}=\frac{h c}{\lambda}
$$
From Eqs. (ii) and (i), we get
$\frac{E_{\text {photon }}}{K_{\text {electron }}}=\frac{h c / \lambda}{h^{2} / 2 m \cdot \lambda^{2}}=\frac{h c \cdot \lambda^{2} \times 2 m}{h^{2} \cdot \lambda}$
where $m=0.5 \mathrm{MeV}=5 \times 10^{5} \mathrm{eV}$
$\begin{aligned} \frac{h}{\lambda c}=50 \times 10^{3} \mathrm{eV}=\frac{2 \mathrm{m} \lambda c}{h} \\=\frac{2 \mathrm{m}}{h / \lambda c} &=\frac{2 \times 5 \times 10^{5}}{50 \times 10^{3}} \quad\left(\because \mathrm{m}=\frac{h}{c \lambda}\right) \\ &=20: 1 \end{aligned}$

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2014
The de-Broglie wavelength of an electron (mass $=1 \times 10^{-30} \mathrm{kg},$ charge $\left.=16 \times 10^{-19} \mathrm{C}\right)$ with a kinetic energy of $200 \mathrm{eV}$ is (Planck's constant $=6.6 \times 10^{-34} \mathrm{Js}$

Options:
A $9.60 \times 10^{-11} \mathrm{m}$
B $8.25 \times 10^{-11} \mathrm{m}$
C $6.25 \times 10^{-11} \mathrm{m}$
D $5.00 \times 10^{-11} \mathrm{m}$

The correct answer is:
$8.25 \times 10^{-11} \mathrm{m}$

Solution:
The de-Broglie wavelength, $\lambda=\frac{h}{\sqrt{2 m k}}$
Given,
$$
h=6.6 \times 10^{-34} \mathrm{J}-\mathrm{s}
$$
$$
\begin{aligned}
m &=1 \times 10^{-30} \mathrm{kg} \\
K=200 \mathrm{eV} &=200 \times 1.6 \times 10^{-19} \mathrm{J}
\end{aligned}
$$
Substituting all these values
$$
\begin{aligned}
\lambda &=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 1 \times 10^{-30} \times 200 \times 1.6 \times 10^{-19}}} \\
&=0.825 \times 10^{-10}=8.25 \times 10^{-11} \mathrm{m}
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2013
The de-Broglie wavelength of an electron moving with a velocity $c / 2$ ( $c=$ velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is

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

The correct answer is:
$1: 2$

Solution:
de-Broglie wavelength
$$
\lambda=\frac{h}{m v}
$$
Here, $\lambda_{e}=\frac{h}{m_{e} \frac{c}{2}}$ and $\lambda_{p}=\frac{h}{m_{p} c}$
Given, $\quad \lambda_{e}=\lambda_{B}$
So, $\quad \frac{h}{m_{e} \frac{c}{2}}=\frac{h}{m_{p} c}$
$\frac{m_{e}}{m_{p}}=2$
Ratio of KE
$$
\begin{array}{l}
\frac{K_{e}}{K_{p}}=\frac{\frac{1}{2} m_{e} v_{e}^{2}}{\frac{1}{2} m_{p} v_{p}^{2}} \\
\frac{K_{e}}{K_{p}}=\frac{1}{2}
\end{array}
$$

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2012
The decimal equivalent of the binary number $(11010.101)_2$ is

Options:
A 9.625
B 25.265
C 26.625
D 26.265

The correct answer is:
26.625

Solution:
$$
\text { Hints : }(11010.101)=0 \times 2^0+1 \times 2^1+0 \times 2^2+1 \times 2^3+1 \times 2^4+1 \times 2^{-1}+0 \times 2^{-2}+1 \times 2^{-3}=2+8+16+\frac{1}{2}+\frac{1}{8}=26.625
$$

Topic: Physics, Mathematics in Physics, WBJEE, WBJEE 2010
The decimal number equivalent to a binary number 1011001 is

Options:
A 13
B 17
C 89
D 178

The correct answer is:
89

Solution:
Given, binary number $=1011001$ Its equivalent decimal number
$\begin{array}{l}
=2^{\circ}+0+0+2^{3}+2^{4}+2^{6} \\
=1+8+16+64=89
\end{array}$

Topic: Physics, Mathematics in Physics, WBJEE, WBJEE 2012
$$
\begin{aligned}
&\text { The decreasing order of basic character of }\\
&\mathrm{K}_{2} \mathrm{O}, \mathrm{BaO}, \mathrm{CaO} \text { and } \mathrm{MgO} \text { is }
\end{aligned}
$$

Options:
A $\mathrm{K}_{2} \mathrm{O}>\mathrm{BaO}>\mathrm{CaO}>\mathrm{Mg} \mathrm{O}$
B $\mathrm{K}_{2} \mathrm{O}>\mathrm{CaO}>\mathrm{BaO}>\mathrm{MgO}$
C $\mathrm{MgO}>\mathrm{BaO}>\mathrm{CaO}>\mathrm{K}_{2} \mathrm{O}$
D $\mathrm{MgO}>\mathrm{CaO}>\mathrm{BaO}>\mathrm{K}_{2} \mathrm{O}$

The correct answer is:
$\mathrm{K}_{2} \mathrm{O}>\mathrm{BaO}>\mathrm{CaO}>\mathrm{Mg} \mathrm{O}$

Solution:
On the basis of electropositivity of metal, the order of basic character is
$$
\mathrm{K}_{2} \mathrm{O}>\mathrm{BaO}>\mathrm{CaO}>\mathrm{MgO}
$$

Topic: Chemistry, s Block Elements, WBJEE, WBJEE 2015
The degree and order of the differential equation \(y=x\left(\frac{d y}{d x}\right)^2+\frac{d y}{d x}\) are respectively

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

The correct answer is:
4,1

Solution:
Hints: \(y\left(\frac{d y}{d x}\right)^2=x\left(\frac{d y}{d x}\right)^4+1\)

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2011
The degree of the differential equation $x=1+\left(\frac{d y}{d x}\right)+\frac{1}{2 !}\left(\frac{d y}{d x}\right)^2+\frac{1}{3 !}\left(\frac{d y}{d x}\right)^3+\ldots \ldots \ldots$

Options:
A 3
B 2
C 1
D not defined

The correct answer is:
1

Solution:
Hints: $x=e \frac{d y}{d x}, \frac{d y}{d x}=\log _e x$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2010
The density of the material of a cube can be estimated by measuring its mass and the length of one of its sides. If the maximum error in the measurement of mass and length are $0.3 \%$ and $0.2 \%$ respectively, the maximum error in the estimation of the density of the cube is approximately.

Options:
A $1.1 \%$
B $0.5 \%$
C $0.9 \%$
D $0.7 \%$

The correct answer is:
$0.9 \%$

Solution:
Given,
Maximum error in the measurement of mass $=0.3 \%$
Maximum error in the measurement of length $=0.2 \%$
We know that,
Error in density is given as,
$\text {Density, } \rho=\frac{\text { mass }(m)}{\text { volume }(V)}=\frac{m}{L^{3}}$
where, $L=$ side of cube
Error in density is given as,
$\left(\frac{\Delta \rho}{\rho}\right)=\frac{\Delta m}{m}+\frac{3 \Delta L}{L}$
or $\left(\frac{\Delta \rho}{\rho}\right) \times 100=\left(\frac{\Delta m}{m}+\frac{3 \Delta L}{L}\right) \times 100$
Substituting the given values, we get
$\left(\frac{\Delta \rho}{\rho}\right)_{\max }=(0.3 \%+3(0.2) \%=0.3 \%+0.6 \%$
$\therefore$ Maximum percentage error measurement of density $\left(\frac{\Delta \rho}{\rho}\right)_{\max }=0.9 \%$

Topic: Physics, Mathematics in Physics, WBJEE, WBJEE 2019