datavorous/entrance-exam-dataset huggingface dataset:
$A$ hollow sphere of external radius $R$ and thickness $t(< < R)$ is made of a metal of density $\rho$. The sphere will float in water, if

Options:
A $t \leq \frac{R}{\rho}$
B $t \leq \frac{R}{3 \rho}$
C $t \leq \frac{R}{2 \rho}$
D $t \geq \frac{R}{3 \rho}$

The correct answer is:
$t \leq \frac{R}{3 \rho}$

Solution:
The density of material is $\rho$

The hollow sphere will float if its weight is less than the weight of the water displaced by the volume of the sphere This implies mass of the sphere is less than that for the same volume of water. Now, mass of spherical cell
$$
m_{1}=4 \pi R^{2} \times t \times \rho
$$
While the mass of water having same volume
$$
m_{2}=\frac{4}{3} \pi R^{3} \times \rho_{g}=\frac{4}{3} \pi R^{3}
$$
where, $\rho_{g}=$ density of water $=\frac{1 \mathrm{kg}}{\mathrm{m}^{3}}$
For the floatation of sphere,
$$
m_{1} \leq m_{2}
$$
$4 \pi R^{2} \times t \times \rho \leq \frac{4}{3} \pi R^{3}$
$\Rightarrow \quad t \rho \leq \frac{R}{3} \Rightarrow t \leq \frac{R}{3 \rho}$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2015
A homonuclear diatomic gas molecule shows 2-electron magnetic moment. The one-electron and two-electron reduced species obtained from above gas molecule can act as both oxidizing and reducing agents. When the gas molecule is one-electron oxidized the bond length decreases compared to the neutral molecule. The gas molecule is

Options:
A $\mathrm{N}_{2}$
B $\mathrm{Cl}_{2}$
C $\mathrm{O}_{2}$
D $\mathrm{B}_{2}$

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

Solution:
Hint:

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2020
A horizontal fire hose with a nozzle of cross-sectional arca $\frac{5}{\sqrt{21}} \times 10^{-3} \mathrm{m}^{2}$ delivers a cubic metre of water in 10 s. What will be the maximum possible increase in the temperature of water while it hits a rigid wall (neglecting the effect of gravity)?

Options:
A $1^{\circ} \mathrm{C}$
B $0.1^{\circ} \mathrm{C}$
C $10^{\circ} \mathrm{C}$
D $0.01^{\circ} \mathrm{C}$

The correct answer is:
$1^{\circ} \mathrm{C}$

Solution:
Given, cross sectional area of nozzle
$$
A=\frac{5}{\sqrt{21}} \times 10^{-3} \mathrm{m}^{2}
$$
and rate of heat transfer $Q=\frac{1}{10}=10^{-1} \mathrm{m}^{3} / \mathrm{s}$
Heat transfer $Q=A V$ $\therefore V=\frac{Q}{A}=\frac{10^{-1} \mathrm{m}^{3} / \mathrm{s}}{\frac{5}{\sqrt{21}} \times 10^{-3} \mathrm{m}^{2}}=2 \sqrt{21} \times 10 \mathrm{m} / \mathrm{s}$
When it hits a rigid wall then maximum possible increase in temperature of water can be expressed as
$$
\frac{1}{2} m v^{2}=m s \Delta T
$$
where, $m=$ mass, $s=$ specific heat of water
$$
=1 \mathrm{cal} / \mathrm{g}=4.2 \times 10^{3} \mathrm{J} / \mathrm{kg}
$$
and $\Delta T=$ increase in temperature
$$
\begin{aligned}
\text { From Eq. }(i), \Delta T=& \frac{1}{2} \frac{v^{2}}{s} \\
&=\frac{\left(2 \sqrt{21} \times 10^{1}\right)^{2}}{2 \times 4.2 \times 10^{3}} \\
&=\frac{84 \times 10^{2}}{8.4 \times 10^{3}} \\
&=1^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2019
A horizontal semi-circular wire of radius $\mathrm{r}$ is connected to a battery through two similar springs $\mathrm{X}$ and $\mathrm{Y}$ to an electric cell, which sends current I through it. A vertically downward uniform magnetic field B is applied on the wire, as shown in the figure. What is the force acting on each spring?

Options:
A $2 \pi \mathrm{rBI}$
B $\frac{1}{2} \pi \mathrm{rBI}$
C $\mathrm{BIr}$
D $2 \mathrm{BIr}$

The correct answer is:
$\mathrm{BIr}$

Solution:
$F_B=i \times 2 r \times B$
$2 F=F_B$
$F=i r B$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2022
A hyperbola, having the transverse axis of length $2 \sin \theta$ is confocal with the ellipse $3 x^{2}+4 y^{2}=12$. It equation is

Options:
A $x^{2} \sin ^{2} \theta-y^{2} \cos ^{2} \theta=1$
B $x^{2} \operatorname{cosec}^{2} \theta-y^{2} \sec ^{2} \theta=1$
C $\left(x^{2}+y^{2}\right) \sin ^{2} \theta=1+y^{2}$
D $x^{2} \operatorname{cosec}^{2} \theta=x^{2}+y^{2}+\sin ^{2} \theta$

The correct answers are:
$x^{2} \operatorname{cosec}^{2} \theta-y^{2} \sec ^{2} \theta=1$

Solution:
Given. $2 a_{1}=2 \sin \theta$
$\Rightarrow \quad a_{1}=\sin \theta$
and $3 x^{2}+4 y^{2}=12$
$\Rightarrow \quad \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$
Here, $a^{2}=4$ and $b^{2}=3$
$\therefore$
$b^{2}=a^{2}\left(1-e^{2}\right)$
$\Rightarrow$
$3=4\left(1-e^{2}\right)$
$\Rightarrow$
$e^{2}=1-\frac{3}{4}=\frac{1}{4}$
$e=\frac{1}{2}$
Focus, $F(a e, 0)=F\left(2 \times \frac{1}{2}, 0\right)$
$=F(1,0)$
For hyperbola foci are same
$a_{1} e_{1}=a e=1$
$\therefore$
$(\sin \theta) e_{1}=1$
$\Rightarrow$
$e_{1}=\operatorname{cosec} \theta$
and $b_{1}^{2}=a_{1}^{2}\left(e_{1}^{2}-1\right)=a_{1}^{2} e_{1}^{2}-a_{1}^{2}$
$\Rightarrow \quad b_{1}^{2}=1-\sin ^{2} \theta=\cos ^{2} \theta$
$\frac{x^{2}}{a_{1}^{2}}-\frac{y^{2}}{b_{1}^{2}}=1$
$\Rightarrow \quad \frac{x^{2}}{\sin ^{2} \theta}-\frac{y^{2}}{\cos ^{2} \theta}=1$
$\Rightarrow x^{2} \operatorname{cosec}^{2} \theta-y^{2} \sec ^{2} \theta=1$

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2018
A is a set containing n elements. $P$ and $Q$ are two subsets of $A$. Then the number of ways of choosing $P$ and $Q$ so that $P \cap Q=\varphi$ is

Options:
A $2^{2 n \space 2 n} C_n$
B $2^n$
C $3^n-1$
D 3^n

The correct answer is:
3^n

Solution:
$\sum_{r=0}^n{ }^n C_r \cdot 2^{n-r}=(2+1)^n=3^n$ (Assuming $P$ contains r elements then $Q$ can be formed with $n-r$ elements in $2^{n-r}$ ways)

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2022
A junction diode has a resistance of $25 \Omega$ when forward biased and $2500 \Omega$ when reverse biased. The current in the diode, for the arrangement shown will be

Options:
A $\frac{1}{15} \mathrm{~A}$
B $\frac{1}{7} \mathrm{~A}$
C $\frac{1}{25} \mathrm{~A}$
D $\frac{1}{180} \mathrm{~A}$

The correct answer is:
$\frac{1}{7} \mathrm{~A}$

Solution:
$\mathrm{R}_{\mathrm{eq}}=25+10=35 \Omega$
Because diode is forward biased. So $I=\frac{V}{R_{e q}}=\frac{5}{35}=\frac{1}{7} \mathrm{~A}$

Topic: Physics, Semiconductors, WBJEE, WBJEE 2009
A ladder $20 \mathrm{ft}$ long leans against a vertical wall. The top end slides downwards at the rate of $2 \mathrm{ft}$ per second. The rate at which the lower end moves on a horizontal floor when it is $12 \mathrm{ft}$ from the wall is

Options:
A $\frac{-8}{3}$
B $\frac{6}{5}$
C $\frac{3}{2}$
D $\frac{17}{4}$

The correct answer is:
$\frac{-8}{3}$

Solution:
$$
x^{2}+y^{2}=(20)^{2}=400
$$
We have. $\frac{d y}{d t}=2 \mathrm{ft} / \mathrm{sec}$
When
$$
x=12
$$
then $\left(13^{2}+y^{2}=400\right.$
$\Rightarrow \quad 144+y^{2}=400$
$\Rightarrow \quad y^{2}=400-144=256$
$\Rightarrow \quad y=16$
Now, $2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=0$
$\Rightarrow \quad x \frac{d x}{d t}=-y \frac{d y}{d t}$
$\Rightarrow \quad 12\left(\frac{d x}{d t}\right)=-16(2)$
$\Rightarrow \quad \frac{d x}{d t}=\frac{-8}{3}$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2018
A letter lock consists of three rings with 15 different letters. If \(\mathrm{N}\) denotes the number of ways in which it is possible to make unsuccessful attempts to open the lock, then

Options:
A 482 divides \(\mathrm{N}\)
B N is the product of two distinct prime numbers.
C \(\mathrm{N}\) is the product of three distinct prime numbers.
D 16 divides \(\mathrm{N}\)

The correct answers are:
482 divides \(\mathrm{N}\), \(\mathrm{N}\) is the product of three distinct prime numbers.

Solution:
Hint : Number of unsuccessful attempts \(=15^3-1=3374=2 \times 1687=2 \times 7 \times 241\)
Divisible by 482
\(\mathrm{N}\) is the product of three distinct prime number

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2023
A light charged particle is revolving in a circle of radius $r$ in electrostatic attraction of a static heavy particle with opposite charge How does the magnetic field vary at the centre of the circle due to the moving charge depend on $r ?$

Options:
A $B \propto \frac{1}{r}$
B $B \propto \frac{1}{r^{2}}$
C $B \propto \frac{1}{r^{\frac{3}{2}}}$
D $B \propto \frac{1}{r^{\frac{5}{2}}}$

The correct answer is:
$B \propto \frac{1}{r^{\frac{5}{2}}}$

Solution:
$$
\begin{aligned}
&\begin{array}{l}
F_{\text {centripetal }}=m r \omega^{2}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}} \\
\therefore \quad \omega^{2}=\frac{1}{4 \pi \varepsilon_{0} m} \cdot \frac{q_{1} q_{2}}{r^{3}} \Rightarrow \omega=\sqrt{\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} m r^{3}}}
\end{array}\\
&\omega \propto \frac{1}{r^{3 / 2}}
\end{aligned}
$$
Current produced due to moving $q_{1}$
$$
i=\frac{\omega}{2 \pi} q_{1}
$$
Thus, magnetic field due to motion of $q_{1}$ in circular path at point $\underline{O}$ $B=\frac{\mu_{0}}{2} \cdot \frac{i}{r}=\frac{\mu_{0}}{2} \cdot \frac{\omega q_{1}}{2 \pi r} \quad[$ using $\mathrm{Eq} \cdot(\mathrm{ii})]$
Hence, $B \propto \frac{\omega}{r}$
Magnetic field $B \propto \frac{1}{r^{3 / 2}}$ $[$ using Eq. (i) ]

Topic: Physics, Electrostatics, WBJEE, WBJEE 2018
A line cuts the $X$ -axis at $A(5,0)$ and the $Y$ -axis at $B(0,-3)$. A variable line $P Q$ is drawn perpendicular to $A B$ cutting the $X$ -axis at $P$ and the $Y$ -axis at $Q$. If $A Q$ and $B P$ meet at $R$, then the locus of $R$ is

Options:
A $x^{2}+y^{2}-5 x+3 y=0$
B $x^{2}+y^{2}+5 x+3 y=0$
C $x^{2}+y^{2}+5 x-3 y=0$
D $x^{2}+y^{2}-5 x-3 y=0$

The correct answer is:
$x^{2}+y^{2}-5 x+3 y=0$

Solution:
Equation of line $A B$ is
$$
\begin{array}{l}
\frac{x}{5}+\frac{y}{-3}=1 \\
3 x-5 y=15
\end{array}
$$
Perpendicular line to $A B$ is
$$
5 x+3 y=\lambda
$$
Coordinate of $P$ is $\left(\frac{\lambda}{5}, 0\right)$ and coordinate of $Q$ is $(0, \lambda / 3)$
Now, equation of line $A Q$ is
$x / 5+\frac{y}{\lambda / 3}=1$
$\Rightarrow \quad \frac{x}{5}+\frac{3 y}{\lambda}=1$
$\Rightarrow \quad \frac{3 y}{\lambda}=1-\frac{x}{5}$
$\Rightarrow \quad \frac{1}{\lambda}=\frac{1}{3 y}\left(1-\frac{x}{5}\right)$
and equation of line $B P$ is
$\frac{x}{\lambda / 5}+\frac{y}{-3}=1$
$\Rightarrow \quad \frac{5 x}{\lambda}-\frac{y}{3}=1$
$\Rightarrow \quad \frac{1}{\lambda}=\frac{1}{5 x}\left(\frac{y}{3}+1\right)$
From Eqs. (i) and (ii),
$\frac{1}{3 y}\left(1-\frac{x}{5}\right)=\frac{1}{5 x}\left(\frac{y}{3}+1\right)$
$\Rightarrow \quad 5 x\left(1-\frac{x}{5}\right)=3 y\left(\frac{y}{3}+1\right)$
$\Rightarrow \quad 5 x-x^{2}=y^{2}+3 y$
$\Rightarrow x^{2}+y^{2}-5 x+3 y=0$
which is a circle.

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2018
A line passes through the point $(-1,1)$ and makes an angle $\sin ^{-1}\left(\frac{3}{5}\right)$ in the positive direction of $x$-axis. If this line meets the curve $x^2=4 y-9$ at $A$ and $B$, then $|A B|$ is equal to

Options:
A $\frac{4}{5}$ unit
B $\frac{5}{4}$ unit
C $\frac{3}{5}$ unit
D $\frac{5}{3}$ unit

The correct answer is:
$\frac{5}{4}$ unit

Solution:
$4 y-3 x=7$ and $x^2=4 y-9$
On solving, $x^2-3 x+2=0 \Rightarrow x=1,2$ and $y=\frac{5}{2}, \frac{13}{4}$
$A\left(1, \frac{5}{2}\right)$ and $B\left(2, \frac{13}{4}\right)$
$$
|A B|=\frac{5}{4}
$$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2022
A line passing through the point of intersection of $x+y=4$ and $x-y=2$ makes an angle $\tan ^{-1}\left(\frac{3}{4}\right)$ with the $X$ -axis. It intersects the parabola $y^{2}=4(x-3)$ at points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right),$ respectively. Then $\left|x_{1}-x_{2}\right|$ is equal to

Options:
A $\frac{16}{9}$
B $\frac{32}{9}$
C $\frac{40}{9}$
D $\frac{80}{9}$

The correct answer is:
$\frac{32}{9}$

Solution:
Given equations are
$x+y=4 ...(i)$
and $\quad x-y=2$ ...(ii)
From Eqs. (i) and $(ii),$ we get
$x=3$ and $y=1$
The line through this point making an angle tan $^{-1} \frac{3}{4}$ with the $X$ -axis is
$(y-1)=\frac{3}{4}(x-3) \quad\left[\because m=\frac{3}{4}\right]$
$\Rightarrow \quad y=\frac{3 x}{4}-\frac{5}{4}=\frac{3 x-5}{4}$
Since, this line intersects the parabola
$y^{2}=4(x-3)$ at points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$, respectively.
$\therefore$ Putting $y=\frac{3 x-5}{4}$ in equation of parabola, we get
$$
\left(\frac{3 x-5}{4}\right)^{2}=4(x-3)
$$
$\Rightarrow \quad 9 x^{2}-94 x+217=0$
$\Rightarrow \quad x_{1}+x_{2}=\frac{94}{9}$ and $x_{1} x_{2}=\frac{217}{9}$
$\begin{aligned} \therefore \quad\left|x_{1}-x_{2}\right| &=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\ &=\sqrt{\left(\frac{94}{9}\right)^{2}-4 \times \frac{217}{9}} \\ &=\frac{32}{9} \end{aligned}$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2016
A line passing through the point of intersection of $x+y=4$ and $x-y=2$ makes an angle $\tan ^{-1}\left(\frac{3}{4}\right)$ with the $x$ -axis. It intersects the parabola $y^{2}=4(x-3)$ at points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ respectively. Then, $\left|x_{1}-x_{2}\right|$ is equal to

Options:
A $\frac{16}{9}$
B $\frac{32}{9}$
C $\frac{40}{9}$
D $\frac{80}{9}$

The correct answer is:
$\frac{32}{9}$

Solution:
Given lines are
$$
x+y=4 \text { and } x-y=2
$$
On solving these lines, we get
$x=3$ and $y=1$
Now, the equation of line which passes through the intersection point (3, 1) having slope
$\theta=\tan ^{-1}\left(\frac{3}{4}\right)$ is $(y-1)=\frac{3}{4}(x-3)$
$\Rightarrow \quad 4 y-4=3 x-9$
$\Rightarrow \quad 3 x-4 y=5$
Now for the intersection point of the line (i) with parabola $y^{2}=4(x-3)$. Put $y=\left(\frac{3 x-5}{4}\right)$
then we get $\quad \frac{(3 x-5)^{2}}{16}=4(x-3)$
$\Rightarrow \quad 9 x^{2}+25-30 x=64 x-192$
$\Rightarrow \quad 9 x^{2}-94 x+217=0$
$\Rightarrow \quad x=\frac{94 \pm \sqrt{8836-7812}}{18}=\frac{94 \pm \sqrt{1024}}{18}$
$\Rightarrow \quad x=\frac{94 \pm 32}{18}=\frac{126}{18}$ or $\frac{62}{18}$
$\Rightarrow \quad x_{1}=\frac{21}{3}=7$
and $\quad x_{2}=\frac{31}{9}$
$\therefore \quad\left|x_{1}-x_{2}\right|=\left|7-\frac{31}{9}\right|=\left|\frac{32}{9}\right|=\frac{32}{9}$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2013
A line through the point $\mathrm{A}(2,0)$ which makes an angle of $30^{\circ}$ with the positive direction of $x$-axis is rotated about A in clockwise direction through an angle $15^{\circ}$. Then the equation of the straight line in the new position is

Options:
A $(2-\sqrt{3}) x+y-4+2 \sqrt{3}=0$
B $(2-\sqrt{3}) x-y-4+2 \sqrt{3}=0$
C $(2-\sqrt{3}) x-y+4+2 \sqrt{3}=0$
D $(2-\sqrt{3})_x+y+4+2 \sqrt{3}=0$

The correct answer is:
$(2-\sqrt{3}) x-y-4+2 \sqrt{3}=0$

Solution:
Hints : Equation of line in new position :
$$
\begin{aligned}
& y-0=\tan 15^{\circ}(x-2) \\
& \Rightarrow y=\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)(x-2) \\
& \Rightarrow y=\frac{(\sqrt{3}-1)^2}{2}(x-2)
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow 2 y=(4-2 \sqrt{3})(x-2) \\
& \Rightarrow y=(2-\sqrt{3})(x-2) \\
& \Rightarrow(2-\sqrt{3}) x-y-4+2 \sqrt{3}=0
\end{aligned}
$$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2009
A line with positive direction cosines passes through the point $\mathrm{P}(2,-1,2)$ and makes equal angle with co-ordinate axes. The line meets the plane $2 x+y+z=9$ at point $Q$. The length of the line segment $P Q$ equals.

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

The correct answer is:
$\sqrt{3}$ unit

Solution:
Let d.c. of line is $(\mathrm{k}, \mathrm{k}, \mathrm{k})$
$\begin{array}{l}
\Rightarrow 3 \mathrm{k}^{2}=1 \Rightarrow \mathrm{k}=\frac{1}{\sqrt{3}} \\
\Rightarrow \text { d.c. of line is }\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)
\end{array}$
Equation of a line passing through point $p$ is
$\begin{array}{l}
\frac{x-2}{\frac{1}{\sqrt{3}}}=\frac{y+1}{\frac{1}{\sqrt{3}}}=\frac{z-2}{\frac{1}{\sqrt{3}}}=k \\
\Rightarrow x=k+2, y=k-1, z=k+2 \\
\therefore 2(k+2)+(k-1)+(k+2)=9 \\
4 k=4 \Rightarrow k=1
\end{array}$
$\therefore$ Point $\mathrm{Q}$ is $(3,0,3)$
$P Q=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}$ units.

Topic: Mathematics, Three Dimensional Geometry, WBJEE, WBJEE 2021
A liquid of bulk modulus $k$ is compressed by applying an external pressure such that its density increases by $0.01 \% .$ The pressure applied on the liquid is

Options:
A $\frac{k}{10000}$
B $\frac{k}{1000}$
C $1000 \mathrm{k}$
D $0.01 k$

The correct answer is:
$\frac{k}{10000}$

Solution:
The Bulk modulus $k$
$$
k=-\frac{p}{\Delta V / V}
$$
Where, $p=$ Pressure
$$
\begin{aligned}
\Delta V &=\text { Change in volume } \\
V &=\text { Volume of liquid }
\end{aligned}
$$
From (i)
$$
\frac{p}{k}=-\frac{\Delta V}{V}=\frac{\Delta p}{p}
$$
$\Rightarrow \quad p=\frac{k \Delta p}{p}$
$$
\begin{aligned}
\Delta p &=0 \cdot 01 \%=0 \cdot 01 / 100 \\
p &=\frac{k}{10000}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Solids, WBJEE, WBJEE 2017
A long conducting wire carrying a current $I$ is bent at $120^{\circ}$ (see figure). The magnetic field $B$ at a point $P$ on the right bisector of bending angle at a distance $d$ from the bend is $\left(\mu_{0}\right.$ is the permeability of free space)

Options:
A $\frac{3 \mu_{0} I}{2 \pi d}$
B $\frac{\mu_{0} I}{2 \pi d}$
C $\frac{\mu_{0} I}{\sqrt{3} n d}$
D $\frac{\sqrt{3} \mu_{0} I}{2 \pi d}$

The correct answer is:
$\frac{\sqrt{3} \mu_{0} I}{2 \pi d}$

Solution:
We know that

$$
\begin{aligned}
B_{\text {net }} &=2\left[\frac{\mu_{0} i}{4 \pi r}\left(\sin \theta_{1}+\sin \theta_{2}\right)\right] \\
&=2\left[\frac{\mu_{0}}{4 \pi} \times \frac{i}{\frac{d \sqrt{3}}{2}} \times\left(\sin 90^{\circ}+\sin 30^{\circ}\right)\right] \\
&=2\left[\frac{\mu_{0}}{4 \pi} \times \frac{2 i}{d \sqrt{3}} \times\left(1+\frac{1}{2}\right)\right] \\
&=2\left[\frac{\mu_{0}}{4 \pi} \times \frac{2 i}{d \sqrt{3}} \times \frac{3}{2}\right]=\frac{\sqrt{3} \mu_{0} i}{2 \pi d}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2014
A longitudinal wave is represented by $x=x_0 \sin 2 \pi(n t-x / \lambda)$. The maximum particle velocity will be four times the wave velocity if:

Options:
A $\lambda=\frac{\pi x_0}{4}$
B $\lambda=2 \pi x_0$
C $\lambda=\frac{\pi x_0}{2}$
D $\lambda=4 \pi x_0$

The correct answer is:
$\lambda=\frac{\pi x_0}{2}$

Solution:
Hints: Maximum particle velocity $\left(\mathrm{V}_{\mathrm{p}}\right)=\mathrm{A} \omega=2 \pi n x_0$
Wave velocity $\left(\mathrm{V}_\omega\right)=n \lambda$
Here, $\mathrm{V}_{\mathrm{P}}=4 \mathrm{~V}_\omega$
$$
2 \pi n x_0=4 n \lambda
$$
$$
\lambda=\frac{\pi}{2} x_0
$$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2010
A luminous object is separated from a screen by distance $d$. A convex lens is placed between the object and the screen such that it forms a distinct image on the screen. The maximum possible focal length of this convex lens is

Options:
A $4 d$
B $2 d$
C $\frac{d}{2}$
D $\frac{d}{4}$

The correct answer is:
$\frac{d}{4}$

Solution:
Focal length of a convex lens by displacement method
Focal length of the convex lens
$$
f=\frac{a^{2}-b^{2}}{4 a}
$$
where $a$ = distance between the image and object and $b=$ distance between two positions of lens
Fact The distance bemeen the ho points should be 4 times greater than the focal iength of the convex lens, ie, $d>4 f$.
So, the maximum possible focal length of this convex lens is $\frac{d}{4}$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2014
A magnetic field $B=2 t+4 t^{2}$ (where, $t_{x}$ time) is applied perpendicular to the plany of a circular wire of radius $r$ and resistance
R. If all the units are in SI the electric charge that flows through the circular wire during $t=0 \mathrm{s}$ to $t=2 \mathrm{s}$ is

Options:
A $\frac{6 \pi^{2}}{R}$
B $\frac{20 \pi r^{2}}{R}$
C $\frac{32 \pi^{2}}{R}$
D $\frac{48 \pi^{2}}{R}$

The correct answer is:
$\frac{20 \pi r^{2}}{R}$

Solution:
Given, $B=2 t+4 t^{2}$
at $\quad t=0, B_{1}=0$
and at $t=2, B_{2}=2 \times 2+4(2)^{2}$
$$
=4+16=20 \mathrm{Wb} / \mathrm{m}^{2}
$$
We have, $\Delta Q=\frac{\Delta \phi}{R}=\frac{\pi^{2}\left(B_{2}-B_{1}\right)}{R}$
$$
=\frac{\pi r^{2} \left[20-0\right]}{R}=\frac{20 \pi r^{2}}{R}
$$

Topic: Physics, Electromagnetic Induction, WBJEE, WBJEE 2013
A magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of $60^{\circ}$ and the work done is $W$. The torque on the magnetic needle at this position is

Options:
A $2 \sqrt{3} w$
B $\sqrt{3} w$
C $\frac{\sqrt{3}}{2} w$
D $\frac{\sqrt{3}}{4} \mathrm{W}$

The correct answer is:
$\sqrt{3} w$

Solution:
Given, work done $=W$ and $\theta=60^{\circ}$ We know that
Hence,
$$
\begin{array}{l}
W=M B(1-\cos \theta) \\
W=M B\left(1-\cos 60^{\circ}\right) \\
W=\frac{M B}{2} \\
|\tau|=M B \sin 60^{\circ}=\sqrt{3} W
\end{array}
$$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2012
A Mapping from IN to IN is defined as follows :
$$
\begin{aligned}
& \mathrm{f}: \mathrm{IN} \rightarrow \mathrm{IN} \\
& \mathrm{f}(\mathrm{n})=(\mathrm{n}+5)^2, \mathrm{n} \in \mathrm{IN}
\end{aligned}
$$
(IN is the set of natural numbers). Then

Options:
A $\mathrm{f}$ is not one-to-one
B $\mathrm{f}$ is onto
C $\mathrm{f}$ is both one-to-one and onto
D $\mathrm{f}$ is one-to-one but not onto

The correct answer is:
$\mathrm{f}$ is one-to-one but not onto

Solution:
$$
\begin{aligned}
& \text { Hints: } \mathrm{f}: \mathrm{IN} \rightarrow \mathrm{IN} ; \mathrm{f}(\mathrm{n})=(\mathrm{n}+5)^2 \\
& \left(\mathrm{n}_1+5\right)^2=\left(\mathrm{n}_2+5\right)^2 \\
& \Rightarrow\left(\mathrm{n}_1-\mathrm{n}_2\right)\left(\mathrm{n}_1+\mathrm{n}_2+10\right)=0 \\
& \Rightarrow \mathrm{n}_1=\mathrm{n}_2 \rightarrow \text { one-to-one }
\end{aligned}
$$
There does not exist $n \in$ IN such that $(n+5)^2=1$
Hence $\mathrm{f}$ is not onto

Topic: Mathematics, Functions, WBJEE, WBJEE 2009
A mass $M$ at rest is broken into two pieces having masses $m$ and $(M-m) .$ The two masses are then separated by a distance $r$ The gravitational force between them will be the maximum when the ratio of the masses $[m:(M-m)]$ of the two parts is

Options:
A 1: 1
B 1: 2
C 1: 3
D 1: 4

The correct answer is:
1: 1

Solution:
The gravitation force between two masses,
$$
F=\frac{G m_{1} m_{2}}{r}
$$
here, $\quad m_{1}=m$
and
$$
\begin{array}{l}
m_{2}=(M-m) \\
F=\frac{G m(M-m)}{r^{2}}
\end{array}
$$
$\therefore$
For maximum gravitational force, $\frac{d F}{d m}=0$
$\therefore$
$$
\frac{d}{d m}[m(M-m)]=0
$$
By solving, we get $m=\frac{M}{2}$
So, $\quad \frac{m}{(M-m)}=\frac{1}{1}$

Topic: Physics, Gravitation, WBJEE, WBJEE 2013
A mass of $1 \mathrm{kg}$ is suspended by means of a thread. The system is
(i) lifted up with an acceleration of $4.9 \mathrm{ms}^{-2}$
(ii) lowered with an acceleration of $4.9 \mathrm{ms}^{-2}$. The ratio of tension in the first and second case is

Options:
A 3: 1
B 1: 2
C 1: 3
D 2: 1

The correct answer is:
3: 1

Solution:
When the body of mass m lifted up the forces acting on the body

$$
\begin{aligned}
T_{1}-m g &=\frac{m g}{2} \\
T_{1} &=m g+\frac{m g}{2}=\frac{3 m g}{2}
\end{aligned}
$$
(ii) When the body lowered the acting forces

$$
\left(m g-T_{2}\right)=\frac{m g}{2}
$$
$$
T_{2}=m g-\frac{m g}{2}=\frac{m q}{2}
$$
$\therefore$
$$
\begin{aligned}
\frac{T_{1}}{T_{2}} &=\frac{3 m g / 2}{m g / 2}=\frac{3}{1} \\
&=3: 1
\end{aligned}
$$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2016
A material has Poisson's ratio 0.50. If a uniform rod of it suffers a longitudinal strain of \(2 \times 10^{-3}\), then the percentage change in volume is

Options:
A 0.6
B 0.4
C 0.2
D zero

The correct answer is:
zero

Solution:
Hints : Poisson's ratio is 0.5 so there is no change in volume

Topic: Physics, Mechanical Properties of Solids, WBJEE, WBJEE 2011
A metal (M) forms two oxides. The ratio $\mathrm{M}: \mathrm{O}$ (by weight) in the two oxides are $25: 4$ and $25: 6$. The minimum value of atomic mass of $M$ is

Options:
A 50
B 100
C 150
D 200

The correct answer is:
100

Solution:
Let two oxide be
$\mathrm{M}_2 \mathrm{O}_x$ and $\mathrm{M}_2 \mathrm{Oy}$
$\underline{\text{As per question}}$
$\frac{2 a}{16 x}=\frac{25}{4}$ and $\frac{2 a}{16 y}=\frac{25}{6}$
$x=\frac{a}{50}, y=\frac{3 a}{100}$ where $a=$ atomic mass of Metal
As $x$ and $y$ to be an integer,
If we take $a=50$, then $x=1, y=1.5$ (not possible)
If we take $a=100$ then $x=2, y=3$ (possible)
$\therefore$ Minimum Atomic Mass $=100 \mathrm{u}$

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2022