datavorous/entrance-exam-dataset huggingface dataset:
If $|z|=1$ and $z \neq \pm 1$, then all the points representing $\frac{z}{1-z^{2}}$ lie on;

Options:
A a line not passing through the origin
B the line $y=x$
C the $x$-axis
D the $y$-axis

The correct answer is:
the $y$-axis

Solution:
Let $z=e^{i \theta}, \theta \neq \pm n \pi ~ n \in I$
Let $w=\frac{e^{i \theta}}{1-e^{i 2 \theta}}=\frac{1}{e^{-i \theta}-e^{i \theta}}=\frac{1}{-2 \cdot i \sin \theta}$

$\therefore$ Locus is y-axis

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2021
If $z_{1}$ and $z_{2}$ be two non-zero complex numbers such that $\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{1}}=1,$ then the origin and the points represented by $z_{1}$ and $z_{2}$

Options:
A lie on a straight line
B form a right angled triangle
C form an equilateral triangle
D form an is asceles triangle

The correct answer is:
form an equilateral triangle

Solution:
We know that, if $z_{1}, z_{2}$ and $z_{3}$ are the vertices of an equilateral triangle. Then,
$$
z_{1}^{2}+z_{2}^{2}+z_{3}^{2}-z_{1} z_{2}-z_{2} z_{3}-z_{3} z_{1}=0
$$
Now, but we have
$$
\begin{aligned}
\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{1}} &=1 \\
\Rightarrow \quad z_{1}^{2}+z_{2}^{2} &=z_{1} z_{2} \\
\Rightarrow \quad z_{1}^{2}+z_{2}^{2}-z_{1} z_{2} &=0 \\
z_{3} &=0
\end{aligned}
$$
Hence, given points form an equilateral triangle.

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2018
If $z=\frac{4}{1-i}$, then $\bar{z}$ is (where $\bar{z}$ is complex conjugate of $z$ )

Options:
A $2(1+i)$
B $(1+i)$
C $\frac{2}{1-\mathrm{i}}$
D $\frac{4}{1+\mathrm{i}}$

The correct answer is:
$\frac{4}{1+\mathrm{i}}$

Solution:
Hints: $z=\frac{4}{1-i}$
$$
\overline{\mathrm{z}}=\frac{4}{1+\mathrm{i}}
$$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2010
If $\frac{z-1}{z+1}$ is purely imaginary, then

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

The correct answer is:
$|z|=1$

Solution:
Let $\omega=\frac{z-1}{z+1}$
Then, using componendo and dividendo, we get
$$
\begin{array}{l}
\quad z=\frac{1+\omega}{1-\omega} \\
\Rightarrow \quad|z|=\left|\frac{\omega+1}{\omega-1}\right| \\
\Rightarrow \quad|z|=1 \\
\Rightarrow \quad|\omega+1|=|\omega-1|
\end{array}
$$
Let $6=u+i v$
Then, $|\omega+1|=|u+i v+1|$
$$
=|(u+1)+i v|=\sqrt{(u+1)^{2}+v^{2}}
$$
and $|\omega-1|=\sqrt{(t u-1)^{2}+v^{2}}$
$\therefore$ From Eq. (i)
$$
\begin{array}{rlr}
& \sqrt{(u+1)^{2}+v^{2}}=\sqrt{(u-1)^{2}+v^{2}} \\
\Rightarrow & (u+1)^{2}+v^{2} & =(v-1)^{2}+v^{2} \\
\Rightarrow & u=0
\end{array}
$$
$\therefore \omega=\frac{z-1}{z+1}$ is a pure imaginary number.

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2012
If $z_{1}, z_{2}, z_{3}$ are imaginary numbers such that $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}\right|=1,$ then
$\left|z_{1}+z_{2}+z_{3}\right|$ is

Options:
A equal to 1
B less than 1
C greater than 1
D equal to 3

The correct answer is:
equal to 1

Solution:
We have, $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=1$
$\Rightarrow \quad\left|z_{1}\right|^{2}=\left|z_{2}\right|^{2}=\left|z_{3}\right|^{2}=1$
$\Rightarrow \quad \quad z_{1} \bar{z}_{1}=z_{2} \bar{z}_{2}=z_{3} \bar{z}_{3}=1$
$\Rightarrow \quad \frac{1}{z_{1}}=\bar{z}_{1} \frac{1}{z_{2}}=\bar{z}_{2} \frac{1}{z_{3}}=\bar{z}_{3}$
Now. $\quad\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}\right|=1$
$\begin{array}{ll}\Rightarrow & \left|\bar{z}_{1}+\bar{z}_{2}+\bar{z}_{3}\right|=1 \\ \Rightarrow & \left|\overline{z_{1}+z_{2}+z_{3}}\right|=1 \\ \therefore & \left|z_{1}+z_{2}+z_{j}\right|=1\end{array}$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2016
If $|z-25 i| \leq 15$, then Maximum $\arg (z)-$ Minimum $\arg (z)$ is equal to

Options:
A $2 \cos ^{-1}\left(\frac{3}{5}\right)$
B $2 \cos ^{-1}\left(\frac{4}{5}\right)$
C $\frac{\pi}{2}+\cos ^{-1}\left(\frac{3}{5}\right)$
D $\sin ^{-1}\left(\frac{3}{5}\right)-\cos ^{-1}\left(\frac{3}{5}\right)$

The correct answer is:
$2 \cos ^{-1}\left(\frac{4}{5}\right)$

Solution:
$$
\because \cos \theta=\frac{15}{25}=\frac{3}{5}
$$
$\therefore \operatorname{Min} \arg (z)=\cos ^{-1}\left(\frac{3}{5}\right)$
$\operatorname{Max} \arg (z)=\pi-\cos ^{-1}\left(\frac{3}{5}\right)=\frac{\pi}{2}+\sin ^{-1}\left(\frac{3}{5}\right)$
$\therefore$ difference $=\frac{\pi}{2}+\sin ^{-1}\left(\frac{3}{5}\right)-\cos ^{-1}\left(\frac{3}{5}\right)=2 \sin ^{-1}\left(\frac{3}{5}\right)=2 \cos ^{-1}\left(\frac{4}{5}\right)$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2022
If $|z+i|-|z-1|=|z|-2=0$ for a complex number $z$, then $z=$

Options:
A $\sqrt{2}(1+i)$
B $\sqrt{2}(1-i)$
C $\sqrt{2}(-1+i)$
D $\sqrt{2}(-1-\mathrm{i})$

The correct answers are:
$\sqrt{2}(1-i)$, $\sqrt{2}(-1+i)$

Solution:
$|z| =2$ represents circle

$|z+i|=|z-1|$ represents

Solving $x^{2}+y^{2}=4$ and $y=-x$

$z=\sqrt{2}(1-i), \quad \sqrt{2}(-1+i)$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2021
If $Z_{r}=\sin \frac{2 \pi}{11}-i \cos \frac{2 \pi}{11},$ then $\sum_{r=0}^{10} Z_{r}$ is equal to

Options:
A -1
B 0
C i
D $-i$

The correct answer is:
0

Solution:
We have, $Z,=\sin \frac{2 \pi r}{11}-i \cos \frac{2 \pi}{11}$
$$
=-i\left(\cos \frac{2 \pi r}{11}+i \sin \frac{2 \pi r}{11}\right)
$$
$\therefore$
$$
\begin{aligned}
&=-i e^{\frac{i 2 \pi}{11}} \\
\sum_{r=0}^{10} Z_{r} &=-i \sum_{r=0}^{10} e^{\frac{i 2 \pi}{11}}
\end{aligned}
$$
$$
=-i \times 0=0
$$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2018
If $z=\sin \theta-i \cos \theta,$ then for any integer $n$

Options:
A $z^{n}+\frac{1}{z^{n}}=2 \cos \left(\frac{n \pi}{2}-n \theta\right)$
B $z^{n}+\frac{1}{z^{n}}=2 \sin \left(\frac{n \pi}{2}-n \theta\right)$
C $z^{n}-\frac{1}{z^{n}}=2 i \sin \left(n \theta-\frac{n \pi}{2}\right)$
D $z^{n}-\frac{1}{z^{n}}=2 i \cos \left(\frac{n \pi}{2}-n \theta\right)$

The correct answers are:
$z^{n}+\frac{1}{z^{n}}=2 \cos \left(\frac{n \pi}{2}-n \theta\right)$, $z^{n}-\frac{1}{z^{n}}=2 i \sin \left(n \theta-\frac{n \pi}{2}\right)$

Solution:
Given, $z=\sin \theta-i \cos \theta$
$=\cos \left(\theta-\frac{\pi}{2}\right)+i \sin \left(\theta-\frac{\pi}{2}\right)=e^{i(\theta-\frac{\pi}{2}).}$
Now. $z^{n}=e^{i(n\theta-\frac{n\pi}{2}).}=\cos \left(n \theta-\frac{n \pi}{2}\right)+i \sin \left(n \theta-\frac{n \pi}{2}\right)$
$\Rightarrow \frac{1}{z^{n}}=e^{i\left(\frac{n \pi}{2}-n\theta\right)}=\cos \left(n \theta-\frac{n \pi}{2}\right)-i \sin \left(n \theta-\frac{n \pi}{2}\right)$
$\therefore \quad z^{n}+\frac{1}{z^{n}}=2 \cos \left(n \theta-\frac{n \pi}{2}\right)=2 \cos \left(\frac{n \pi}{2}-n \theta\right)$
and $z^{n}-\frac{1}{z^{n}}=2 i \sin \left(n \theta-\frac{n \pi}{2}\right)$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2016
If $z=x+i y,$ where $x$ and $y$ are real numbers and $i=\sqrt{-1}$, then the points $(x, y)$ for which $\frac{z-1}{z-i}$ is real, lie on

Options:
A an ellipse
B a circle
C a parabola
D a straight line

The correct answer is:
a straight line

Solution:
Given, $z=x+i y$
Now, $\frac{z-1}{z-i}=\frac{(x+y)-1}{(x+i y)-i}$
$=\frac{(x-1)+i y}{x+i(y-1)} \times \frac{x-i(y-1)}{x-i(y-1)}$
$=\frac{x(x-1)+i x y-i(x-1)(y-1)+y(y-1)}{x^{2}+(y-1)^{2}}$
$=\frac{\left(x^{2}+y^{2}-x-y\right)+i(x y-x y+y+x-1)}{\left(x^{2}+y^{2}+1^{-2 y}\right)}$
$=\left(\frac{x^{2}+y^{2}-x-y}{x^{2}+y^{2}-2 y+1}\right)+i\left(\frac{x+y-1}{x^{2}+y^{2}-2 y+1}\right) \ldots(0)$
Given, $\frac{z-1}{z-i}$ is real.
So, its imaginary part should be zero. i.e.. $\quad \frac{x+y-1}{x^{2}+y^{2}-2 y+1}=0$
$\Rightarrow \quad x+y=1$
$\left(\because x^{2}+y^{2}-2 y+1 \neq 0\right)$
Which represent a straight line.

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2013
If $z=x$ - iy and $z^{1 / 3}=p+$ iq $(x, y, p, q \in \mathbb{R})$, then $\frac{\left(\frac{x}{p}+\frac{y}{q}\right)}{\left(p^2+q^2\right)}$ is equal to

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

The correct answer is:
$-2$

Solution:
$z=p^3-3 p q^2+i\left(3 p^2 q-q^3\right)=x-i y$
$\begin{aligned}
&x=p\left(p^2-3 q^2\right) \\
&y=-q\left(3 p^2-q^2\right) \\
&\therefore \frac{x}{p}+\frac{y}{q}=p^2-3 q^2-3 p^2+q^2=-2\left(p^2+q^2\right)
\end{aligned}$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2022
If \(z_1\) and \(z_2\) are two complex numbers satisfying the equation \(\left|\frac{z_1+z_2}{z_1-z_2}\right|=1\), then \(\frac{z_1}{z_2}\) may be

Options:
A real positive
B real negative
C zero
D purely imaginary

The correct answers are:
zero, purely imaginary

Solution:
Hint : \(\left|\frac{z_1+z_2}{z_1-z_2}\right|=1\)
\(\left|z_1 / z_2+1\right|=\left|z_1 / z_2-1\right|\)
Distance of \(\frac{z_1}{z_2}\) from -1 and 1 are equal
So, locus of \(\frac{z_1}{z_2}\) is perpendicular bisector of line joining \((-1,0) ~\&~(1,0)\)
\(\therefore \frac{\boldsymbol{z}_1}{\mathbf{z}_2} \Rightarrow\) purely imaginary or 0

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2023
In 2 -butene, which one of the following statements is true?

Options:
A $\mathrm{C}_{1}-\mathrm{C}_{2}$ bond is a $\mathrm{sp}^{3}-\mathrm{sp}^{3}-\mathrm{\sigma}$ -bond
B $\mathrm{C}_{2}-\mathrm{C}_{3}$ bond is $s \mathrm{p}^{3}-\mathrm{sp}^{2}-\mathrm{\sigma}$ -bond
C $\mathrm{C}_{1}-\mathrm{C}_{2}$ bond is a $s \mathrm{p}^{3}-\mathrm{sp}^{2}-\mathrm{\sigma}$ -bond
D $\mathrm{C}_{1}-\mathrm{C}_{2}$ bond is a $s \mathrm{p}^{2}-\mathrm{sp}^{2}-\mathrm{\sigma}$ -bond

The correct answer is:
$\mathrm{C}_{1}-\mathrm{C}_{2}$ bond is a $s \mathrm{p}^{3}-\mathrm{sp}^{2}-\mathrm{\sigma}$ -bond

Solution:
The structure of 2 -butene is as

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2012
In a 12 storied building, 3 persons enter a lift cabin. It is known that they will leave the lift at different floors. In how many ways can they do so if the lift does not stop at the second floor?

Options:
A 36
B 120
C 240
D 720

The correct answer is:
720

Solution:
Hint:
Total no. of ways $=10_{p_{3}}=720$ (except the floor they enter and second floor)

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2020
In a $\triangle A B C, a, b, c$ are the sides of the triangle opposite to the angles $A, B, C,$ respectively. Then, the value of $a^{3} \sin (B-C)$ $+b^{3} \sin (C-A)+c^{3} \sin (A-B)$ is equal to

Options:
A 0
B 1
C 3
D 2

The correct answer is:
0

Solution:
$\sum a^{3} \sin (B-C)$
$=\sum k^{3} \sin ^{3} A \sin (B-C)$
$$
\begin{aligned}\left[\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\right] \\=\sum k^{3}\left[\sin ^{2} A \sin (B+C) \sin (B-C)\right] \\=k^{3}\left[\left\{\sin ^{2} A \times \frac{1}{2}(\cos 2 C-\cos 2 B)\right\}\right.\\+\sin ^{2} B \times \frac{1}{2}(\cos 2 A-\cos 2 C) \\\left.+\sin ^{2} C \times \frac{1}{2}(\cos 2 B-\cos 2 A)\right] \end{aligned}
$$
$=\frac{k^{3}}{2}\left[\sin ^{2} A\left(1-2 \sin ^{2} C-1+2 \sin ^{2} B\right)\right.$
$+\sin ^{2} B\left(1-2 \sin ^{2} A-1+2 \sin ^{2} C\right)$
$\left.+\sin ^{2} C\left(1-2 \sin ^{2} B-1+2 \sin ^{2} A\right)\right]$
$=\frac{k^{3}}{2}\left[-2 \sin ^{2} A \sin ^{2} C+2 \sin ^{2} A \sin ^{2} B\right.$
$-2 \sin ^{2} B \sin ^{2} A+2 \sin ^{2} B \sin ^{2} C$
$\left.-2 \sin ^{2} C \sin ^{2} B+2 \sin ^{2} C \sin ^{2} A\right]$
$=\frac{k^{3}}{2}[0]=0$

Topic: Mathematics, Properties of Triangles, WBJEE, WBJEE 2014
In a $\triangle A B C$, tan $A$ and $\tan B$ are the roota $p q\left(x^{2}+1\right)=r^{2} x$ Then, $\Delta A B C$ is

Options:
A a right angled triangle
B an acute angled triangle
C an obtuse angled triangle
D an equilateral triangle

The correct answer is:
a right angled triangle

Solution:
Given equation can be rewritten as $x^{2}-\frac{r^{2}}{p q} x+1=0$
$\therefore \quad \tan A+\tan B=\frac{r^{2}}{p q}$
and $\tan A \tan B=1$
We know, $A+B+C=180^{\circ}$
$\Rightarrow \quad(A+B)=180^{\circ}-C$
$\Rightarrow \quad \tan (A+B)=\tan \left(180^{\circ}-C\right)$
$\Rightarrow \quad \frac{\tan A+\tan B}{1-\tan A \tan B}=-\tan C$
$\Rightarrow \quad \frac{r^{2} / p q}{1-1}=-\tan C$
$\Rightarrow \quad \infty=-\tan C \Rightarrow C=90^{\circ}$
Hence, $\triangle A B C$ is a right angled triangle.

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2014
In $\mathrm{a} \triangle \mathrm{ABC}, 2 \operatorname{acsin} \frac{\mathrm{A}-\mathrm{B}+\mathrm{C}}{2}$ is equal to

Options:
A $\mathrm{a}^2+\mathrm{b}^2-\mathrm{c}^2$
B $c^2+a^2-b^2$
C $\mathrm{b}^2-\mathrm{a}^2-\mathrm{c}^2$
D $\mathrm{c}^2-\mathrm{a}^2-\mathrm{b}^2$

The correct answer is:
$c^2+a^2-b^2$

Solution:
Hints : $2 \mathrm{ac} \sin \left(\frac{\mathrm{A}+\mathrm{C}-\mathrm{B}}{2}\right) \quad\left[\frac{\mathrm{A}+\mathrm{C}}{2}=\frac{\pi}{2}-\frac{\mathrm{B}}{2}\right],=2 \mathrm{ac} \sin \left(\frac{\pi}{2}-\mathrm{B}\right)=2 \mathrm{ac} \cos \mathrm{B} \quad=\mathrm{a}^2+\mathrm{c}^2-\mathrm{b}^2$

Topic: Mathematics, Properties of Triangles, WBJEE, WBJEE 2010
In $\Delta A B C$, if $a^{2} \cos ^{2} A-b^{2}-c^{2}=0,$ then

Options:
A $\frac{\pi}{4} < A < \frac{\pi}{2}$
B $\frac{\pi}{2} < A < \pi$
C $A=\frac{\pi}{2}$
D $A < \frac{\pi}{4}$

The correct answer is:
$\frac{\pi}{2} < A < \pi$

Solution:
We have, $a^{2} \cos ^{2} A-b^{2}-c^{2}=0$
$\Rightarrow \quad a^{2} \cos ^{2} A=b^{2}+c^{2}$
Also in $\triangle A B C$, we have $\begin{aligned} \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c} &=\frac{a^{2} \cos ^{2} A-a^{2}}{2 b c} \\=& \frac{-a^{2}\left(1-\cos ^{2} A\right)}{2 b c}=\frac{-a^{2} \sin ^{2} A}{2 b c} < 0 \\ & \quad \because 0 < A < \pi, a, b, c>0 \end{aligned}$
$\Rightarrow \cos A < 0 \Rightarrow A$ lies in IInd quadrant.
$\Rightarrow \quad \frac{\pi}{2} < A < \pi$

Topic: Mathematics, Properties of Triangles, WBJEE, WBJEE 2015
In a certain test, there are n questions. In this test $2^{n-i}$ students gave wrong answers to at least i questions, where i $=1,2, \ldots \ldots \mathrm{n}$. If the total number of wrong answers given is 2047 , then $\mathrm{n}$ is equal to

Options:
A 10
B 11
C 12
D 13

The correct answers are:
11

Solution:
Hint:
Total students was gave wrong answer to exactly i - questions $=2^{n-i}-2^{n-(i+1)}$
Total wrong answer given \(\sum_{i=0}^{n-1} i \times \left(2^{n-i}-2^{n-(i+1)}\right)+n \times 1\)
$\Rightarrow 2^{n-1}+\ldots \ldots 1=2047$
$\Rightarrow 2^{n}=2048$
$\Rightarrow \mathrm{n}=11$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2020
In a certain town, $60 \%$ of the families own a car, $30 \%$ own a house and $20 \%$ own both car and house. If a family is randomly chosen. then what is the probability that this family owns a car or a house but not both?

Options:
A 0.5
B 0.7
C 0.1
D 0.9

The correct answer is:
0.5

Solution:
Let $A$ be the set of families who own a car and $B$ be the set of families who own a house.
Then, $\quad P(A)=60 \%, P(B)=30 \%$
and $P(A \cap B)=20 \%$
Now, $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$=60+30-20=70 \%$
Now, families who owns a car or a house but not both $=A \Delta B$
$$
\begin{aligned}
\Rightarrow \quad P(A \Delta B) &=P(A \cup B)-P(A \cap B) \\
&=70-20=50 \%=0.5
\end{aligned}
$$

Topic: Mathematics, Probability, WBJEE, WBJEE 2015
In a closed circuit there is only a coil of inductance $L$ and resistance $100 \Omega$. The coil is situated in a uniform magnetic field. All on a sudden, the magnetic flux linked with the circuit changes by $5 \mathrm{Weber}$. What amount of charge will flow in the circuit as a result?

Options:
A $500 \mathrm{C}$
B $0.05 \mathrm{C}$
C $20 \mathrm{C}$
D Value of $L$ is to be known to find the charge flown

The correct answer is:
$0.05 \mathrm{C}$

Solution:
Emf $=-\frac{\mathrm{d} \phi}{\mathrm{dt}}($ Faraday's Law $)$
$\mathrm{i}_{\text {induced }}=\frac{\mathrm{Emf}}{\mathrm{R}}=-\frac{\mathrm{d} \phi}{\mathrm{dt}} \times \frac{1}{\mathrm{R}}$
$\Rightarrow \frac{\mathrm{dq}}{\mathrm{dt}}=-\frac{1}{\mathrm{R}} \frac{\mathrm{d} \phi}{\mathrm{dt}} \Rightarrow \mathrm{dq}=-\frac{\mathrm{d} \phi}{\mathrm{R}} \Rightarrow \int \mathrm{dq}=-\int \frac{\mathrm{d} \phi}{\mathrm{R}}$
$\Rightarrow \mathrm{q}_{\text {flown }}=\frac{\phi_{\mathrm{i}}-\phi_{\mathrm{f}}}{\mathrm{R}}=\frac{5 \text { weber }}{100}=0.05 \mathrm{C} .$

Topic: Physics, Alternating Current, WBJEE, WBJEE 2022
In a close-packed body-centred cubic lattice of potassium, the correct relation between the atomic radius $(r)$ of potassium and the edge-length of the cube is

Options:
A $r=\frac{a}{\sqrt{2}}$
B $r=\frac{a}{\sqrt{3}}$
C $r=\frac{\sqrt{3}}{2} a$
D $r=\frac{\sqrt{3}}{4} a$

The correct answer is:
$r=\frac{\sqrt{3}}{4} a$

Solution:
$\because$ For be packing, $\sqrt{3} a=4 r \therefore r=\frac{\sqrt{3}}{4} \cdot a$
where $a=$ edge length $r=$ radius of lattice sphere

Topic: Chemistry, Solid State, WBJEE, WBJEE 2017
In a common emitter configuration, a transistor has $\beta=50$ and input resistance $1 \mathrm{k} \Omega$. If the peak value of a.c. input is $0.01 \mathrm{~V}$ then the peak value of collector current is

Options:
A $0.01 \mu \mathrm{A}$
B $0.25 \mu \mathrm{A}$
C $100 \mu \mathrm{A}$
D $500 \mu \mathrm{A}$

The correct answer is:
$500 \mu \mathrm{A}$

Solution:
$$
\text { Hints: } \beta=50 \Rightarrow \beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}} \Rightarrow \Delta \mathrm{I}_{\mathrm{C}}=\beta \times \Delta \mathrm{I}_{\mathrm{B}}
$$
$$
\begin{aligned}
& \Delta \mathrm{I}_{\mathrm{B}}=\frac{0.01}{10^3}=10^{-2} \times 10^{-3}=10^{-5} \\
& \Delta \mathrm{I}_{\mathrm{C}}=50 \times 10^{-5}=500 \times 10^{-6}=500 \mu \mathrm{A}
\end{aligned}
$$

Topic: Physics, Semiconductors, WBJEE, WBJEE 2010
In a flask, the weight ratio of $\mathrm{CH}_{4}(g)$ and $\mathrm{SO}_{2}(g)$ at $298 \mathrm{K}$ and 1 bar is $1: 2 .$ The ratio of the number of molecules of $\mathrm{SO}_{2}(\mathrm{g})$ and $\mathrm{CH}_{4}(g)$ is

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

The correct answer is:
$1 : 2$

Solution:
Given, weight ratio : $W_{\mathrm{C H_{4}},}: W_{\mathrm{SO}_{2}}=1: 2$ $\because \quad n=\frac{w}{m} \quad\left[\begin{array}{l}n=\text { no. of mole } \\ w=\text { mass } \\ M=\text { molar mas: }\end{array}\right.$
$\therefore \quad \frac{n_{1}}{n_{2}}=\frac{W_{1}}{M_{1}} \times \frac{M_{2}}{W_{2}}\left\{\begin{array}{l}n_{1}=n_{\mathrm{so}_{1}} \\ n_{2}=n_{\mathrm{Oi}_{2}}\end{array}\right.$
$=\frac{2}{64} \times \frac{16}{1}$
$\frac{n_{1}}{n_{2}}=\frac{1}{2},$ i.e. 1: 2
Also $n \propto N$
$\therefore$ Ratio of number of molecules is 1: 2.

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2017
In a Frounhofer diffraction experiment, a single slit of width $0.5 \mathrm{~mm}$ is illuminated by a monochromatic light of wavelength $600 \mathrm{~nm}$. The diffraction pattern is observed on a screen at a distance of $50 \mathrm{~cm}$ from the slit. What will be the linear separation of the first order minima?

Options:
A $1.0 \mathrm{~mm}$
B $1.1 \mathrm{~mm}$
C $0.6 \mathrm{~mm}$
D $1.2 \mathrm{~mm}$

The correct answer is:
$1.2 \mathrm{~mm}$

Solution:
Hint:
$\mathrm{d}=0.5 \mathrm{~mm} .$
$\lambda=600 \mathrm{~nm} .$ Width of central maxima $=\frac{2 \lambda \mathrm{D}}{\mathrm{d}}$
$\mathrm{D}=50 \mathrm{~cm}$
$\frac{2 \lambda \mathrm{D}}{\mathrm{d}}=\frac{2 \times 600 \times 50 \times 10^{-2} \times 10^{-9}}{0.5 \times 10^{-3}}=1.2 \mathrm{~mm}$

Topic: Physics, Wave Optics, WBJEE, WBJEE 2020
In a GP series consisting of positive terms, each term is equal to the sum of next two terms. Then, the common ratio of this GP series is

Options:
A $\sqrt{5}$
B $\frac{\sqrt{5}-1}{2}$
C $\frac{\sqrt{5}}{2}$
D $\frac{\sqrt{5}+1}{2}$

The correct answer is:
$\frac{\sqrt{5}-1}{2}$

Solution:
Let $a_{n}$ be the general term of a GP whose first term is $a$ and common ratio is $r$. Now according to the question,
$$
\begin{aligned}
a_{n} &=a_{n+1}+a_{n+2} \\
a r^{n-1} &=a r^{n}+a r^{n+1}
\end{aligned}
$$
$\Rightarrow \quad r^{n-1}=r^{n}+r^{n+1}$
$\Rightarrow \quad 1=\frac{r^{n}}{r^{n-1}}+\frac{r^{n+1}}{r^{n-1}}$
$\Rightarrow \quad 1=r+r^{2}$
$\Rightarrow r^{2}+r-1=0$
$\Rightarrow \begin{aligned}r=& \frac{-1 \pm \sqrt{(1)^{2}-4(1)(-1)}}{2(1)} \\ &=\frac{-1 \pm \sqrt{1+4}}{2}=\frac{-1 \pm \sqrt{5}}{2} \end{aligned}$
since, GP consists only positive terms
$\therefore$
$r=\frac{\sqrt{5}-1}{2}$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2017
In a group of 14 males and 6 females. 8 and 3 of the males and females, respectively are aged above 40 yr. The probability that a selected at random from the group is person aged above $40 \mathrm{yr}$ given that the selected person is a female, is

Options:
A $\frac{2}{7}$
B $\frac{1}{2}$
C $\frac{1}{4}$
D $\frac{5}{6}$

The correct answer is:
$\frac{1}{2}$

Solution:
Number of males = 14
and females $=6$
Total =14+6=20
Males above $40 \mathrm{yr}=8$
and females above $40 \mathrm{y}=3$
Now, P (selected person is female and above 40 yn) $=\frac{3}{6}=\frac{1}{2}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2016
In a mercury thermometer, the ice point (lower fixed point) is marked as $10^{\circ}$ and the steam point (upper fixed point) is marked as $130^{\circ}$. At $40^{\circ} \mathrm{C}$ temperature, what will this thermometer read?

Options:
A $78^{\circ}$
B $66^{\circ}$
C $62^{\circ}$
D $58^{\circ}$

The correct answer is:
$58^{\circ}$

Solution:
$\frac{x-10}{130-10}=\frac{40}{100}$
$$
\begin{array}{r}
100 x-1000=4800 \\
100 x=5800 \\
x=58^{\circ}
\end{array}
$$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2012
In a mixture, two enantioners are found to be present in 85\% and $15 \%$ respectively. The enantiomeric. excess (ee) is

Options:
A $85 \%$
B $15 \%$
C $70 \%$
D $60 \%$

The correct answer is:
$70 \%$

Solution:
$15 \%$ will form racemic mixture with another $15 \%$. Hence. the enantiomeric excess is $=(85-15)=70 \%$

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2015
In a periodic table the basic character of oxides

Options:
A increases from left to right and decreases from top to bottom
B decreases from right to left and increases from top to bottom
C decreases from left to right and increases from top to bottom
D decreases from left to right and increases from bottom to top

The correct answer is:
decreases from left to right and increases from top to bottom

Solution:
No Solution Available

Topic: Chemistry, Classification of Elements and Periodicity in Properties, WBJEE, WBJEE 2010
In a region, the intensity of an electric field is given by $\mathbf{E}=2 \mathbf{i}+3 \mathbf{j}+\mathbf{k}$ in $\mathrm{NC}^{-1}$. The electric flux through a surface $\mathbf{S}=10 \mathbf{i} \mathrm{m}^{2}$ in the region is

Options:
A $5 \mathrm{Nm}^{2} \mathrm{C}^{-1}$
B $10 \mathrm{Nm}^{2} \mathrm{C}^{-1}$
C $15 \mathrm{Nm}^{2} \mathrm{C}^{-1}$
D $20 \mathrm{Nm}^{2} \mathrm{C}^{-1}$

The correct answer is:
$20 \mathrm{Nm}^{2} \mathrm{C}^{-1}$

Solution:
Given $\mathbf{E}=(2 \mathbf{i}+3 \mathbf{j}+\mathbf{k}) \mathrm{NC}^{-1}$
and $\quad \mathrm{S}=10 \mathrm{i} \mathrm{m}^{2}$
We know that
$$
\begin{array}{l}
\phi=\mathbf{E} \cdot \mathbf{S} \\
\phi=(2 \mathbf{i}+3 \mathbf{j}+\mathbf{k}) \cdot(10 \mathbf{i}) \\
\phi=20 \mathrm{Nm}^{2} \mathrm{C}^{-1}
\end{array}
$$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2012
In a right-angled triangle, the sides are $a, b$ and $c$, with $c$ as hypotenuse, and $c-b \neq 1, c+b \neq 1$. Then the value of $\left(\log _{c+b} a+\log _{c-b} a\right) /\left(2 \log _{c+b} a \times \log _{c-b} a\right)$ will be

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

The correct answer is:
1

Solution:
Hints : $c^2=a^2+b^2$
$\Rightarrow \mathrm{c}^2-\mathrm{b}^2=\mathrm{a}^2$
$\frac{\frac{\log a}{\log (c+b)}+\frac{\log a}{\log (c-b)}}{\frac{2 \log a \times \log a}{\log (c+b) \log (c-b)}}=\frac{\log a\left(\log \left(c^2-b^2\right)\right)}{2 \log a \log a}=\frac{\log a^2}{\log a^2}=1$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2010
In a simple harmonic motion, let \(\mathrm{f}\) be the acceleration and \(\mathrm{T}\) be the time period. If \(x\) denotes the displacement, then |fT| vs. X graph will look like,

Options:
A

B

C

D

The correct answer is:

Solution:
\(\begin{gathered}\text {Hint: } F \cdot T=\omega^2 x \cdot \frac{2 \pi}{\omega}=2 \pi \omega x \\ \because \text { F.T }=2 \pi \omega x\end{gathered}\)

Topic: Physics, Oscillations, WBJEE, WBJEE 2023