datavorous/entrance-exam-dataset huggingface dataset:
The general solution of the different equation $100 \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}-20 \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0$ is

Options:
A $y=\left(c_1+c_2 x\right) e^x$
B $y=\left(c_1+c_2 x\right) e^{-x}$
C $y=\left(c_1+c_2 x\right) e^{\frac{x}{10}}$
D $y=c_1 e^x+c_2 e^{-x}$

The correct answer is:
$y=\left(c_1+c_2 x\right) e^{\frac{x}{10}}$

Solution:
Hints: $100 p^2-20 p+1=$
$$
\begin{aligned}
& (10 P-1)^2=0, P=\frac{1}{10} \\
& y=\left(c_1+c_2 x\right) e^{\frac{x}{10}}
\end{aligned}
$$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2010
The general solution of the differential equation $\left(1+e^{\frac{x}{y}}\right) d x+\left(1-\frac{x}{y}\right) e^{x / y} d y=0$ is$(C$ is an arbitrary constant $)$

Options:
A $x-y e^{\frac{x}{y}}=C$
B $y-x e^{\frac{x}{y}}=C$
C $x+y e^{\frac{x}{y}}=C$
D $y+x e^{\frac{x}{y}}=C$

The correct answer is:
$x+y e^{\frac{x}{y}}=C$

Solution:
$\left(1+e^{\frac{x}{y}}\right) d x+\left(1-\frac{x}{y}\right) e^{x / y} d y=0$
$\Rightarrow \quad\left(1+e^{x / y}\right) d x=-e^{x / y}\left(1-\frac{x}{y}\right) d y$
$\Rightarrow \quad \frac{d x}{d y}=\frac{-e^{x / y}(1-x / y)}{\left(1+e^{x / y}\right)}$ ...(i)
This is homogeneous differential equation.
So, put $\quad x=v y$
$\frac{d x}{d y}=v+y \frac{d v}{d y}$ ....(ii)
Therefore, $v+y \frac{d v}{d y}=\frac{-e^{v} (1-v)}{1+e^{v}}$
[from Eqs.(i) and (ii)]
$\Rightarrow \quad y \frac{d v}{d y}=\frac{-e^{v}+ve^{v}}{1+e^{v}}-v$
$=\frac{-e^{v}+ve^{v}-v\left(1+e^{v}\right)}{1+e^{v}}$
$=-e^{v}+v e^{v}-v-v e^{v}$
$\Rightarrow \quad y \frac{d v}{d y}=\frac{-\left(e^{v}+v\right)}{1+e^{v}}$
$\Rightarrow \quad \frac{1+e^{v}}{v+e^{r}} d v=-\frac{d y}{y} \quad$ [variable separation $]$
$\Rightarrow \quad \int \frac{1+e^{v}}{v+e^{v}} d v=-\int \frac{d y}{y} \quad$ [by integration $] \ldots$ (iii)
Let $\left.v+e^{v}=t \Rightarrow (1+e^{v}\right) d v=d t$
$\int \frac{1+e^{v}}{v+e^{v}} d v=\int \frac{d t}{t}=\log t=\log \left(v+e^{v}\right)$ ...(iv)
$\Rightarrow \log \left(v+e^{v}\right)=-\log y+\log C$
[from Eqs. (iii) and (iv)] $\Rightarrow \log \left(v+e^{v}\right)+\log y=\log C$
$\Rightarrow \quad y\left(v+e^{v}\right)=C$
$y\left(\frac{x}{y}+e^{x / y}\right)=C$
$y\left(\frac{x+ye^{x / y}}{y}\right)=C$
$\Rightarrow \quad x+y e^{x / y}=C$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2019
The general solution of the differential equation $\frac{d y}{d x}=\frac{x+y+1}{2 x+2 y+1}$ is

Options:
A $\log _{e}|3 x+3 y+2|+3 x+6 y=C$
B $\log _{e}|3 x+3 y+2|-3 x+6 y=C$
C $\log _{e}|3 x+3 y+2|-3 x-6 y=C$
D $\log _{e}|3 x+3 y+2|+3 x-6 y=C$

The correct answer is:
$\log _{e}|3 x+3 y+2|+3 x-6 y=C$

Solution:
Given differential equation, $\frac{d y}{d x}=\frac{x+y+1}{2 x+2 y+1}$
Put $x+y=v$ $\Rightarrow \quad 1+\frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{d y}{d x}=\frac{d v}{d x}-1$
$\therefore$ Eq.
(i) becomes, $\frac{d y}{d x}=\frac{v+1}{2 v+1}$
$\Rightarrow \quad \frac{d v}{d x}-1=\frac{v+1}{2 v+1} \Rightarrow \frac{d v}{d x}=\frac{v+1}{2 v+1}+1$
$\Rightarrow \quad \frac{d v}{d x}=\frac{v+1+2 v+1}{2 v+1} \Rightarrow \frac{d v}{d x}=\frac{3 v+2}{2 v+1}$
$\Rightarrow \frac{2 v+1}{3 v+2} d v=d x \Rightarrow \frac{\left[\frac{2}{3}(3 v+2)-\frac{1}{3}\right]}{3 v+2} d v=d x$
$\Rightarrow \quad\left[\frac{2}{3}-\frac{1}{3} \times \frac{1}{(3 v+2)}\right] d v=d x$
Integrating both sides, we get $\Rightarrow \quad \int\left(\frac{2}{3}-\frac{1}{3} \times \frac{1}{(3 v+2)}\right) d v=\int d x+C^{\prime}$
Where $c$ is a constant of integration.
$\Rightarrow \quad \frac{2}{3} v-\frac{1}{3} \frac{\log (3 v+2)}{3}=x+C^{\prime}$
$\Rightarrow \quad \frac{2}{3}(x+y)-\frac{1}{9} \log (3 x+3 y+2)=x+C$
$\Rightarrow \quad 6 x+6 y-\log (3 x+3 y+2)=9 x+C$
$\Rightarrow \quad 3 x-6 y+\log (3 x+3 y+2)=C$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2012
The general solution of the differential equation $\frac{d y}{d x}=e^{y+x}+e^{y-x}$ is
where $c$ is an arbitrary constant

Options:
A $\mathrm{e}^{-\mathrm{y}}=\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}+\mathrm{c}$
B $\mathrm{e}^{-\mathrm{y}}=\mathrm{e}^{-\mathrm{x}}-\mathrm{e}^{\mathrm{x}}+\mathrm{c}$
C $e^{-y}=e^x+e^{-x}+c$
D $e^y=e^x+e^{-x}+c$

The correct answer is:
$\mathrm{e}^{-\mathrm{y}}=\mathrm{e}^{-\mathrm{x}}-\mathrm{e}^{\mathrm{x}}+\mathrm{c}$

Solution:
Hints: $e^{-y} d y=\left(e^x+e^{-x}\right) d x$ Integrate
$-\mathrm{e}^{-\mathrm{y}}=e^x-e^{-x}+c, \quad \mathrm{e}^{-\mathrm{y}}=e^{-x}-e^{+x}+c$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2009
The general solution of the differential equation \(\frac{d^2 y}{d x^2}+8 \frac{d y}{d x}+16 y=0\) is

Options:
A \((A+B x) e^{5 x}\)
B \((\mathrm{A}+\mathrm{Bx}) \mathrm{e}^{-4 x}\)
C \(\left(\mathrm{A}+\mathrm{Bx}^2\right) \mathrm{e}^{4 \mathrm{x}}\)
D \(\left(A+B x^4\right) \mathrm{e}^{4 \mathrm{x}}\)

The correct answer is:
\((\mathrm{A}+\mathrm{Bx}) \mathrm{e}^{-4 x}\)

Solution:
Hints: \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}+8 \frac{\mathrm{dy}}{\mathrm{dx}}+16 \mathrm{y}=0\)
auxilary equation \(\mathrm{m}^2+8 \mathrm{~m}+16=0 \Rightarrow \mathrm{m}=-4\)
Solution \(y=(a x+b) e^{-4 x}\)

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2011
The general solution of the differential equation \(\log _{\mathrm{e}}\left(\frac{d y}{d x}\right)=x+y\) is

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

The correct answer is:
\(e^x+e^{-y}=C\)

Solution:
Hints: \(\frac{d y}{d x}=e^x \cdot e^y \Rightarrow \int e^{-y} d y .=\int e^x d x \Rightarrow e^x+e^{-y}=c\)

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2011
The general value of the real angle $\theta$, which satisfies the equation. $(\cos \theta+i \sin \theta)(\cos 2 \theta+i \sin 2 \theta) \ldots$$(\cos n \theta+t \sin n \theta)=1$ is given by, (assuming $k$ is an integer)

Options:
A $\frac{2 k \pi}{n+2}$
B $\frac{4 k \pi}{n(n+1)}$
C $\frac{4 x \pi}{n+1}$
D $\frac{6 k \pi}{n(n+1)}$

The correct answer is:
$\frac{4 k \pi}{n(n+1)}$

Solution:
We have,
$(\cos\theta+i\sin\theta)(cos2\theta+isin2\theta)...(cosn\theta+isinn\theta) = 1$
$\Rightarrow e^{i\theta}.e^{i(2\theta)}.e^{i(3\theta)}. ...e^{i(n\theta)} = 1$
$\Rightarrow e^{i\theta(1+2+3+...+n}=1$
$\Rightarrow e^{\frac{in(n+1)\theta}{2}}=1$
$\Rightarrow cos(\frac{n(n+1)}{2}\theta)+isin(\frac{n(n+1)}{2}\theta = 1 + 0i$
$\Rightarrow cos(\frac{n(n+1)}{2}\theta)=1$
$\Rightarrow \frac{n(n+1)}{2}\theta = 2k\pi$
$\Rightarrow \theta=\frac{4k}{n(n+1)}\pi$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2019
The glass prisms $P_{1}$ and $P_{2}$ are to be combined together to produce dispersion without deviation. The angles of the prisms $P_{1}$ and $P_{2}$ are selected as $4^{\circ}$ and $3^{\circ}$ respectively. If the refractive index of prism $P_{1}$ is $1.54,$ then that of $P_{2}$ will be

Options:
A 1.48
B 1.58
C 1.62
D 1.72

The correct answer is:
1.72

Solution:
Given for prism $\mathrm{P}_{1}$
$$
\mu=1.54,A=4^{\circ}
$$
For prism $\mathrm{P}_{2}$
$$
\mu^{\prime}=?, A^{\prime}=3^{\circ}
$$
For no deviation $=\delta +\delta ^{\prime}=0$
$\Rightarrow \quad(\mu-1) A=\left(\mu^{\prime}-1\right) A^{\prime}$
$\Rightarrow \quad(1.54-1) 4=\left(\mu^{\prime}-1\right) 3$
On solving we get $\mu^{\prime}=1.72$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2013
The graphs of the polynomial $x^{2}-1$ and $\cos x$ intersect

Options:
A at exactly two points
B at exactly 3 points
C at least 4 but at finitely many points.
D at infinitely many points.

The correct answer is:
at exactly two points

Solution:
It is clear from the graph that it intersects at exactly two points.

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2019
The greatest and least values of $f(x)=\tan ^{-1} x-\frac{1}{2} \ln x$ on $\left[\frac{1}{\sqrt{3}}, \sqrt{3}\right]$ are

Options:
A $\mathrm{f}_{\min }=\sqrt{3}-1$
B $\mathrm{f}_{\max }=\pi / 6+\frac{1}{4} \ell \mathrm{n} 3$
C $\mathrm{f}_{\min }=\pi / 3-\frac{1}{4} \ell \mathrm{n} 3$
D $\mathrm{f}_{\max }=\pi / 12+\ell \mathrm{n} 5$

The correct answers are:
$\mathrm{f}_{\max }=\pi / 6+\frac{1}{4} \ell \mathrm{n} 3$, $\mathrm{f}_{\min }=\pi / 3-\frac{1}{4} \ell \mathrm{n} 3$

Solution:
$f(x)=\tan ^{-1} x-1 / 2 \ln x, x \in[1 / \sqrt{3}, \sqrt{3}]$
$f^{\prime}(x)=\frac{1}{1+x^{2}}-\frac{1}{2 x}=\frac{2 x-\left(1+x^{2}\right)}{2 x\left(1+x^{2}\right)}=\frac{-(x-1)^{2}}{2 x\left(1+x^{2}\right)}$
when $x>0, f^{\prime}(x) < 0$
$f(x)$ is decreasing function.
$\begin{array}{l}
f_{\min }=f(\sqrt{3})=\tan ^{-1}(\sqrt{3})-1 / 2 \ln (\sqrt{3})=\pi / 3-\frac{1}{4} \ln 3 \\
f_{\max }=f(1 / \sqrt{3})=\pi / 6+1 / 4 \ln 3
\end{array}$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2021
The greatest integer which divides $(p+1)(p+2)(p+3) \ldots(p+q)$ for all
$p \in N$ and fixed $q \in N$ is

Options:
A $p !$
B $q !$
C $p$
D $q$

The correct answer is:
$q !$

Solution:
$(p+1)(p+2)(p+3) \ldots(p+q)$ is the product of $q$ consecutive natural number ( $R$ q e $N$ ) The product of $q$ consecutive natural number is always divisible by $q!$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2017
The ground state electronic configuration of CO molecule is

Options:
A $1 \sigma^{2} 2 \sigma^{2} 1 \pi^{4} 3 \sigma^{2}$
B $1 \sigma^{2} 2 \sigma^{2} 3 \sigma^{2} 1 \pi^{2} 2 \pi^{2}$
C $1 \sigma^{2} 2 \sigma^{2} 1 \pi^{2} 3 \sigma^{2} 2 \pi^{2}$
D $1 \sigma^{2} 1 \pi^{2} 2 \sigma^{2} 2 \sigma^{2}$

The correct answer is:
$1 \sigma^{2} 2 \sigma^{2} 1 \pi^{4} 3 \sigma^{2}$

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2013
The ground state magnetic property of $\mathrm{B}_{2}$ and
$\mathrm{C}_{2}$ molecules will be

Options:
A $\mathrm{B}_{2}$ paramagnetic and $\mathrm{C}_{2}$ diamagnetic
B $\mathrm{B}_{2}$ diamagnetic and $\mathrm{C}_{2}$ paramagnetic
C Both are diamagnetic
D Both are paramagnetic

The correct answer is:
$\mathrm{B}_{2}$ paramagnetic and $\mathrm{C}_{2}$ diamagnetic

Solution:
According to molecular orbital theory, outer configuration for
$$
\left[\mathrm{C}_{2}\right]=\mathrm{C}_{12} \rightarrow \pi p_{x}^{2}=\pi p_{y}^{2}
$$
Thus, is diamagnetic in nature.
$$
\left[\mathrm{B}_{2}\right]=\mathrm{B}_{10}->\pi p_{x}^{1}=\pi p_{y}^{1}
$$
Thus, is paramagnetic in nature.

Topic: Chemistry, Solid State, WBJEE, WBJEE 2017
The $\mathrm{H}_{3} \mathrm{O}^{+}$ion has the following shape

Options:
A Tetrahedral
B Pyramidal
C Triangular planar
D "T" shaped

The correct answer is:
Pyramidal

Solution:
$\mathrm{H}_{3} \mathrm{O}^{+}$
total valence $e^{-1}$ count
$\begin{array}{l}
\mathrm{O}+\mathrm{H}+\mathrm{H}+\mathrm{H}-\mathrm{e}^{-1}=8 \\
2,6 \quad 1 \quad 1 \quad 1
\end{array}$
using Lever rule
$2|8| 4 \leadsto$ sp $^{3}$ hybrid $(0)$
$\frac{8}{x}$

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2021
The H $-N-H$ angle in ammonia is $107.6^{\circ}$ while the $\mathrm{H}-\mathrm{P} -\mathrm{H}$ angle in phosphine is $93.5^{\circ} .$ Relative to phosphine, the $p$ -character of the lone-pair on ammonia is expected to be

Options:
A Less
B More
C Same
D Cannot be predicted

The correct answer is:
Less

Solution:
Key point As the percentage of s character in a bond increases, the bond angle also increases. Therefore.
as the bond angle increases, the percentage of $p$ -character decreases.
In the given options, Ammonia ( $\mathrm{NH}_{3}$ ) has bond angle $=107.6^{\circ}$ and Phosphine $\left(\mathrm{PH}_{3}\right)$ has bond angle $=93.5^{\circ}$
$\because$ Bond angle $\propto \frac{1}{p-\text { character }}$ and $\propto s$ -character
Bond angle of $\mathrm{NH}_{3}>\mathrm{PH}_{3}$ due to lone-pair lone-pair repulsion. Thus, the lone pair on $\mathrm{NH}_{3}$ has less $p$ -character. Hence, option (a) is the correct answer.

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2019
The half-life for decay of ${ }^{14} \mathrm{~C, by~} \beta$ -emission is $5730 \mathrm{yr}$. The fraction of ${ }^{14} \mathrm{C}$ decays, in a sample that is 22920 yr old, would be

Options:
A $\frac{1}{8}$
B $\frac{1}{16}$
C $\frac{7}{8}$
D $\frac{15}{16}$

The correct answer is:
$\frac{15}{16}$

Solution:
$n=\frac{\text { total time }(t)}{\text { half -lite }(t / 2)}=\frac{22920}{5730}=4$
$$
\begin{aligned}
\text { Left amount, } N=& N_{0}\left(\frac{1}{2}\right)^{n} \\
&=N_{0}\left(\frac{1}{2}\right)^{4}=\frac{N_{0}}{16} \\
\therefore \text { Decayed fraction } &=N_{0}-\frac{N_{0}}{16} \\
&=\frac{16 N_{0}-N_{0}}{16}=\frac{15 N_{0}}{16}
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2013
The half-life of a radioactive element is 10 hours. How much will be left after 4 hours in $1 \mathrm{~g}$ atom sample?

Options:
A $45.6 \times 10^{23}$ atoms
B $4.56 \times 10^{23}$ atoms
C $4.56 \times 10^{21}$ atoms
D $4.56 \times 10^{20}$ atoms

The correct answer is:
$4.56 \times 10^{23}$ atoms

Solution:
$t_{1 / 2}=10 \mathrm{hr} . \quad \mathrm{K}=\frac{0.693}{10}$
$4=\frac{2.303 \times 10}{0.693} \log \frac{1}{\mathrm{~N}}$
$\log \frac{1}{\mathrm{~N}}=\frac{4 \times 0.693}{2.303 \times 10}=0.12036$
$\log \mathrm{N}=-0.12036=\overline{1} .87964$
$\mathrm{N}=7.575 \times 10^{-1} \mathrm{~g}$ atoms
$\therefore$ No. of atoms $=7.575 \times 10^{-1} \times 6.023 \times 10^{23}$ atoms $=4.56 \times 10^{23}$ atoms

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2009
The half-life of $C^{14}$ is 5760 years. For a $200 \mathrm{mg}$ sample of $\mathrm{C}^{14}$, the time taken to change to $25 \mathrm{mg}$ is

Options:
A 11520 years
B 23040 years
C 5760 years
D 17280 years

The correct answer is:
17280 years

Solution:
Given. Half-life of $\mathrm{C}^{4}, t_{1 / 2}=5760$ years Initial concentration of sample of $\mathrm{C}^{4}, \mathrm{N}_{0}=200 \mathrm{mg}$ Final concentration of sample of $\mathrm{C}^{4}, \mathrm{N}_{1}=2 \mathrm{S} \mathrm{mg}$ The given decay is radioactive and all radiouctive decay follows first order kinetics.
$$
t_{1 / 2}=\frac{0.693}{\lambda}
$$
$\therefore$
$$
\lambda=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{5760} \mathrm{yr}^{-1}
$$
We know that, for first order reaction
$\begin{aligned} t &=\frac{2.303}{\lambda} \log \frac{N_{0}}{N_{1}}=\frac{2303}{0.693} \log \frac{[200]}{125} \\ &=\frac{2303 \times 5760}{0.693} \log 8 \\ &=17,28678 \mathrm{yrs} \approx 17.280 \mathrm{yrs} \end{aligned}$
Therefore, the time taken to change 200 me to $25 \mathrm{mg}$ is $17,286.78 \mathrm{yr}$, which is very close to option (d)

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2018
The half-life period of ${ }_{53} I^{125}$ is 60 days. The radioactivity after 180 days will be

Options:
A $25 \%$
B $12.5 \%$
C $33.3 \%$
D $3.0 \%$

The correct answer is:
$12.5 \%$

Solution:
Let, initial concentration $(a)=100$
Given, half-life $\left(t_{1 / 2}\right)=60$ days
To find, radioactivity,
i.e. $(a-x)$ after time $T$ ( 180 days)
$$
\because \quad T=n \times t_{1 / 2}
$$
where, $n=$ no. of half-lives
$$
\begin{aligned}
180 &=n \times 60 \\
n &=\frac{180}{60}=3
\end{aligned}
$$
and $(a-x)=\frac{a}{2^{n}}=\frac{100}{2^{3}}=\frac{100}{8}$
$$
(a-x)=12.5 \%
$$
Hence, (b) is the correct answer.

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2019
The heat of neutralisation of a strong base acid is $13.7 \mathrm{kcal}$. The heat released when 0.6 mole $\mathrm{HCl}$ solution is added to a 0.25 mole of NaOH is

Options:
A 3.425 kcal
B 8.22 kcal
C 11.645 kcal
D $13.7 \mathrm{kcal}$

The correct answer is:
3.425 kcal

Solution:
Given, the heat of neutralisation of a strong base and a strong acid is $13.7 \mathrm{kcal}$.
The reaction of neutralisation is as follows
$$
\begin{array}{l}
\mathrm{HCl}+\mathrm{NaOH} \longrightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} = -137kcal \\
1 \mathrm{mol} \quad 1 \mathrm{mol}
\end{array}
$$
According to question,
$$
\begin{array}{l}
\mathrm{HCl}+\mathrm{NaOH} \longrightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} \\
0.6 \mathrm{mol} \quad 0.25 \mathrm{mol}
\end{array}
$$
In equation (i), NaOH acts as a limiting reagent. For 1 mole of NaOH and 1 mole of HCl, heat of neutralisation $=137 \mathrm{kcal}$
: For 0.25 mole of NaOH and 0.6 mole of $\mathrm{HCl}$, heat of neutralisation $=137 \times 0.25 \Rightarrow 3.425 \mathrm{kcal}$

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2018
The height of a waterfall is $50 \mathrm{~m}$. If $g=9.8 \mathrm{~ms}^{-2}$ the difference between the temperature at the top and the bottom of the waterfall is:

Options:
A $1.17^{\circ} \mathrm{C}$
B $2.17^{\circ} \mathrm{C}$
C $0.117^{\circ} \mathrm{C}$
D $1.43^{\circ} \mathrm{C}$

The correct answer is:
$0.117^{\circ} \mathrm{C}$

Solution:
$\frac{m g h}{J}=m s \Delta t \Rightarrow \Delta t=0.117^{\circ} \mathrm{C}$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2009
The height vertically above the earth's surface at which the acceleration due to gravity becomes $1 \%$ of its value at the surface is ( $\mathrm{R}$ is the radius of the Earth)

Options:
A $8 \mathrm{R}$
B $9 \mathrm{R}$
C $10 \mathrm{R}$
D $20 \mathrm{R}$

The correct answer is:
$9 \mathrm{R}$

Solution:
$$
\begin{aligned}
& \text { Hints : } g^{\prime}=\frac{g}{\left(1+\frac{h}{\mathrm{R}}\right)^2} \Rightarrow \frac{g}{100}=\frac{g}{\left(1+\frac{h}{\mathrm{R}}\right)^2} \\
& 1+\frac{h}{\mathrm{R}}=10 \Rightarrow \frac{h}{\mathrm{R}}=9, h=9 \mathrm{R}
\end{aligned}
$$

Topic: Physics, Gravitation, WBJEE, WBJEE 2010
The human eye has an approximate angular resolution of $\theta=5.8 \times 10^{-4} \mathrm{rad}$ and typical photo printer prints a minimum of 300 dip (dots per inch, 1 inch $=2.54 \mathrm{~cm}$ ). At what minimal distance $d$ should a printed page be held so that one does not see the individual dots?

Options:
A $20.32 \mathrm{~cm}$
B $29.50 \mathrm{~cm}$
C $14.59 \mathrm{~cm}$
D $6.85 \mathrm{~cm}$

The correct answer is:
$14.59 \mathrm{~cm}$

Solution:
$\theta=\frac{x}{d}$
$5.8 \times 10^{-4}=\frac{2.54}{300} \times \frac{1}{d}$
$d=14.59 \mathrm{~cm}$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2022
The hydrides of the first elements in groups $15-17,$ namely $\mathrm{NH}_{3}, \mathrm{H}_{2} \mathrm{O}$ and $\mathrm{HF}$ respectively show abnormally high values for melting and boiling points. This is due to

Options:
A small size of $\mathrm{N}, \mathrm{O}$ and $\mathrm{F}$
B the ablility to form extensive intermolecula H-bonding
C the ability to form extensive intramolecula H-bonding
D effective van der Waals' interaction

The correct answer is:
the ablility to form extensive intermolecula H-bonding

Solution:
Hydrides of $\mathrm{N}, \mathrm{O}$ and $\mathrm{F}$ because of the small size and high electronegativity of elements, have ability to form extensive intermolecular between two molecules) hydrogen bonding. Thus, a large amount of energy is required to break these bonds, le., the melting and boiling points of hydrides of these elements are abnormally high.

Topic: Chemistry, Hydrogen, WBJEE, WBJEE 2014
The $I$ - $V$ characteristics of a metal wire at two different temperatures $\left(T_{1}\right.$ and $T_{2}$ ) are given in the adjoining figure. Here, we can conclude that

Options:
A $T_{1}>T_{2}$
B $T_{1} < T_{2}$
C $T_{1}=T_{2}$
D $T_{1}=2 T_{2}$

The correct answer is:
$T_{1} < T_{2}$

Solution:
$R$ increases with temperature and slope of $V \cdot I$ graph gives resistance so $T_{1} < T_{2}$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2012
The incentre of an equilateral triangle is (1,1) and the equation of ene side is $3 x+4 y+3=0 .$ Then, the equation of the circumeircle of the triangle is

Options:
A $x^{2}+y^{2}-2 x-2 y-2=0$
B $x^{2}+y^{2}-2 x-2 y-14=0$
C $x^{2}+y^{2}-2 x-2 y+2=0$
D $x^{2}+y^{2}-2 x-2 y+14=0$

The correct answer is:
$x^{2}+y^{2}-2 x-2 y-14=0$

Solution:
since, triangle is equilateral the refore incentre
(1,1) lies on the centroid of the $\Delta A B C$. $\therefore G D=$ Length of perpendicular from the point $G(1,1)$ to the line $3 x+4 y+3=0$

$A G=2 G D=4$
$\therefore$ Equation of circumcircle with centre at (1,1)
$$
\begin{array}{l}
\text { and radius = } \begin{aligned}
4 \text { units } \\
(x-1)^{2}+(y-1)^{2} &=4^{2} \\
\Rightarrow \quad x^{2}+y^{2}-2 x-2 y-14 &=0
\end{aligned}
\end{array}
$$

Topic: Mathematics, Properties of Triangles, WBJEE, WBJEE 2012
The increase in rate constant of a chemical reaction with increasing temperature is (are) due to the fact(s) that

Options:
A the number of collisions among the reactant molecules increases with increasing temperature
B the activation energy of the reaction decreases with increasing temperature
C the concentration of the reactant molecules increases with increasing temperature
D the number of reactant molecules acquiring the activation energy increases with increasing temperature

The correct answers are:
the number of collisions among the reactant molecules increases with increasing temperature, the number of reactant molecules acquiring the activation energy increases with increasing temperature

Solution:
The increase in rate constant of a chemical reaction with increasing temperature are due to the following facts. The number of collisions among the reactant molecules increases with increasing temperature.
The number of reactant molecules acquiring the activation energy increases with increasing temperature.

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2015
The increasing order of $\mathrm{O}-\mathrm{N}-\mathrm{O}$ bond angle in the species $\mathrm{NO}_{2}, \mathrm{NO}_{2}^{+}$ and $\mathrm{NO}_{2}{ }^{-}$ is

Options:
A $\mathrm{NO}_{2}{ }^{+} < \mathrm{NO}_{2} < \mathrm{NO}_{2}^{-}$
B $\mathrm{NO}_{2} < \mathrm{NO}_{2}^{-} < \mathrm{NO}_{2}^{+}$
C $\mathrm{NO}_{2}^{+} < \mathrm{NO}_{2}^{-} < \mathrm{NO}_{2}$
D None of these

The correct answer is:
None of these

Solution:
Option D is correct. As the number of lone pair of electrons increases, bond angle decreases.
$\mathrm{NO}_{2}^{+}$ ion is isoelectronic with $\mathrm{CO}_{2}$, molecule. It is a linear ion and its central atom $\left(\mathrm{N}}\right)$ undergoes sp-hybridisation. Hence, its bond angle is 180 In $\mathrm{NO}_{2}^{-}$ ion, $\mathrm{N}$ -atom undergoes $\mathrm{sp}^{2}$ hybridisation The angle between hybrid orbital should be 120 but one lone pair of electrons is lying on N-atom hence bond angle decreases to $115^{\circ}$ In $\mathrm{NO}_{2}$ molecule, N-atom has one unpared electron in $s p^{2}$ -hybrid orbital. The bond angle should be $120^{\circ}$ but actually, it is $132^{\circ} .$ it may be due to one unpaired electron in $s p^{2}$ -hybrid orbtal
Therefore, the increasing order of bond angle is
$$
\begin{array}{rrr}
\mathrm{NO}_{2}^{-} & < & \mathrm{NO}_{2} & < & \mathrm{NO}_{2}^{+} \\
115^{\circ} & & 132^{\circ} & & 180^{\circ}
\end{array}
$$

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2013
The indicated atom is not a nucleophilic site in

Options:
A $\mathrm{BH}_{4}^{-}$ (atom = B)
B $\mathrm{CH}_{3} \mathrm{Mg}$ (atom = C)
C $\mathrm{CH}_{3} \mathrm{OH}$ (atom = O)
D $\mathrm{CH}_{3} \mathrm{NH}_{2}$ (atom = N)

The correct answer is:
$\mathrm{BH}_{4}^{-}$ (atom = B)

Solution:
In BH ${ }_{4}^{-}$ indicated atom is not a nucleophilic
site because there is no lone-pair on $\mathrm{B}^{\prime}$ atom.
Hence, the option (a) is correct answer.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2019