datavorous/entrance-exam-dataset huggingface dataset:
A solution is saturated with $\mathrm{SrCO}_{3}$ and $\mathrm{Sr} \mathrm{F}_{2}$. The $\left[\mathrm{CO}_{3}^{2-}\right]$ is found to be $1.2 \times 10^{-3} \mathrm{M}$. The concentration of $\mathrm{F}$ - in the solution would be

Options:
A $3.7 \times 10^{-6} \mathrm{M}$
B $3.2 \times 10^{-3} \mathrm{M}$
C $5.1 \times 10^{-7} \mathrm{M}$
D $3.7 \times 10^{-2} \mathrm{M}$
Given : $\mathrm{K}_{\mathrm{sp}}\left(\mathrm{SrCO}_{3}\right)=7.0 \times 10^{-10}, \mathrm{~K}_{\mathrm{sp}}\left(\mathrm{SrF}_{2}\right)=7.9 \times 10^{-10}$

The correct answer is:
$3.7 \times 10^{-2} \mathrm{M}$
Given : $\mathrm{K}_{\mathrm{sp}}\left(\mathrm{SrCO}_{3}\right)=7.0 \times 10^{-10}, \mathrm{~K}_{\mathrm{sp}}\left(\mathrm{SrF}_{2}\right)=7.9 \times 10^{-10}$

Solution:
Hint:
$\begin{aligned}
\mathrm{Sr} \mathrm{CO}_{3}(\mathrm{~s}) \stackrel{\mathrm{S}}{\rightleftarrows} \mathrm{Sr}^{2+}(\mathrm{aq})+\mathrm{CO}_{3}^{2-}(\mathrm{aq}) \\
\mathrm{S}_{1}+\mathrm{S}_{2} \quad \mathrm{~S}_{1} \\
\mathrm{SrF}_{2}(\mathrm{~s}) \rightleftarrows \mathrm{Sr}^{2+}(\mathrm{aq})+2 \mathrm{~F}^{-}(\mathrm{aq}) \\
\mathrm{S}_{1}+\mathrm{S}_{2} \quad 2 \mathrm{~S}_{2}
\end{aligned}$
Where $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ are solubilities of $\mathrm{SrCO}_{3}(\mathrm{~s})$ and $\mathrm{Sr} \mathrm{F}_{2}(\mathrm{~s})$ respectively.
$\begin{array}{l}
\mathrm{S}_{1}=1.2 \times 10^{-3} \mathrm{M}(\text { Given }) \\
\mathrm{K}_{\mathrm{sp}}\left(\mathrm{SrCO}_{3}\right)=\left[\mathrm{Sr}^{2+}\right]\left[\mathrm{CO}_{3}^{-2}\right] \\
7 \times 10^{-10}=\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right) \mathrm{S}_{1}-\ldots-\text { (i) } \\
\mathrm{K}_{\text {sp }}\left(\mathrm{SrF}_{2}\right)=\left[\mathrm{Sr}^{2+}\right]\left[\mathrm{F}^{-}\right]^{2} \\
7.9 \times 10^{-10}=\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)\left(2 \mathrm{~S}_{2}\right)^{2}
\end{array}$
Dividing equation (ii) by equation (i)
$\begin{array}{l}
\frac{7.9 \times 10^{-10}}{7 \times 10^{-10}}=\frac{4 \mathrm{~S}_{2}^{2}}{\mathrm{~S}_{1}} \\
\begin{array}{l}
\therefore \mathrm{S}_{2}=\sqrt{\frac{7.9}{7} \times \frac{1.2 \times 10^{-3}}{4}} \\
\quad=1.84 \times 10^{-2} \\
\therefore\left[\mathrm{F}-\mathrm{]}=2 \mathrm{~S}_{2}=2 \times 1.84 \times 10^{-2}=3.68 \times 10^{-2} \mathrm{M}\right.
\end{array}
\end{array}$

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2020
A solution of $\mathrm{NaNO}_{3}$, when treated with a mixture of $\mathrm{Zn}$ dust and ' $\mathrm{A}$ ' yields ammonia. ' $\mathrm{A}$ ' can be

Options:
A caustic soda
B dilute sulphuric acid
C concentrated sulphuric acid
D sodium carbonate

The correct answer is:
caustic soda

Solution:
$\mathrm{NaNO}_{3}+4 \mathrm{Zn}+\frac{7 \mathrm{NaOH}}{(\mathrm{A})} \rightarrow 4 \mathrm{Na}_{2} \mathrm{ZnO}_{2}+\mathrm{NH}_{3}+2 \mathrm{H}_{2} \mathrm{O}$
A is caustic soda

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2021
A sphere of radius $R$ has a volume density of charge $\rho=k r,$ where $r$ is the distance from the centre of the sphere and $k$ is constant. The magnitude of the electric field which exists at the surface of the sphere is given by $\left(\varepsilon_{0}=\right.$ permittivity of free space)

Options:
A $\frac{4 \pi k R^{4}}{3 \varepsilon_{0}}$
B $\frac{k R}{3 e_{0}}$
C $\frac{4 \pi k R}{\varepsilon_{0}}$
D $\frac{k R^{2}}{4 \varepsilon_{0}}$

The correct answer is:
$\frac{k R^{2}}{4 \varepsilon_{0}}$

Solution:
By Gauss's theorem
\(\begin{aligned} \mathrm{E}\left(4 \pi \mathrm{r}^{2}\right) &=\frac{\int \mathrm{p} \times 4 \pi \mathrm{r}^{2} \mathrm{dr}}{\varepsilon_{0}} \\ &=\frac{\int \mathrm{kr} \times 4 \pi \mathrm{r}^{2} \mathrm{dr}}{\varepsilon_{0}} \\ &=\mathrm{E}=\frac{\mathrm{Kr}^{2}}{4 \varepsilon_{0}} \end{aligned}\)

Topic: Physics, Electrostatics, WBJEE, WBJEE 2013
A spherical convex surface of power 5 dioptre separates object and image space of refractive indices $1.0$ and $\frac{4}{3}$ respectively. The radius of curvature of the surface is

Options:
A $20 \mathrm{~cm}$
B $1 \mathrm{~cm}$
C $4 \mathrm{~cm}$
D $5 \mathrm{~cm}$

The correct answer is:
$5 \mathrm{~cm}$

Solution:
$P=5 D$
$u=\infty$ then
$\frac{1}{\infty}-\frac{4 / 3}{f}=\frac{1-4 / 3}{R}$
$-\frac{4}{3 f}=-\frac{1}{3 R}$
$\frac{f}{4}=R \quad \quad f=4 R$
$\therefore \frac{100}{5}=4 R \Rightarrow R=\frac{25}{5}=5 \mathrm{~cm}$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2021
A spherical liquid drop is placed on a horizontal plane. A small disturbance causes the volume of the drop to oscillate. The time period of oscillation ( $T$ ) of the liquid drop depends on radius $(n$ of the drop, density $(\rho)$ and surface tension (5) of the liquid. Which among the following will be a possible expression for $T$ (where, $k$ is a dimensionless constant)?

Options:
A $k \sqrt{\frac{\rho r}{S}}$
B $k \sqrt{\frac{\rho^{2} r}{S}}$
C $k \sqrt{\frac{\rho r^{3}}{S}}$
D $k \sqrt{\frac{\rho r^{3}}{s^{2}}}$

The correct answer is:
$k \sqrt{\frac{\rho r^{3}}{S}}$

Solution:
According to the question. time period. $T \alpha r^{a} p^{b} s^{c}$
$r=k r^{a} p^{b} s^{c}$
Thus, putting dimension, we get
Equating the dimensions of both sides, we get
$[T]=[\mathrm{L}]^{a}\left[\mathrm{ML}^{-3}]^ \mathrm{b}\left[\mathrm{MT}^{-2} ]^c\right.\right.$
$[T]=[M]^{b+c} \cdot[L]^{2a-3b} [T]^{-2c}$
$b+c=0, a-3 b=0$
and $-2 x=$
$\begin{array}{l}
c=-\frac{1}{2} \\
b=\frac{1}{2} \\
a=3 b=\frac{3}{2}
\end{array}$
Putting these value of $a, b$ and $c$ into Eq. (i), we get
$T=k \cdot r^{\frac{3}{2}} p^{\frac{1}{2}} \cdot s^{-\frac{1}{2}}$
$=k \sqrt{\frac{\rho}{s} r^{3}}$

Topic: Physics, Units and Dimensions, WBJEE, WBJEE 2018
A spring of force constant $k$ is cut into three equal parts. The force constant of each part would be

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

The correct answer is:
$3 k$

Solution:
Hints $: \mathrm{K} \propto \frac{1}{l}$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2010
A spring of force constant $\mathrm{k}$ is cut into two equal halves. The force constant of each half is

Options:
A $\frac{\mathrm{k}}{\sqrt{2}}$
B $\mathrm{k}$
C $\frac{\mathrm{k}}{2}$
D $2 \mathrm{k}$

The correct answer is:
$2 \mathrm{k}$

Solution:
As $\quad \mathrm{K} \ell=$ constant
$K^{\prime}=2 K$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2009
A square conducting loop is placed near an infinitely long current carrying wire with one edge parallel to the wire as shown in the figure. If the current in the straight wire is suddenly halved. which of the following statements will be true? "The loop will .......".

Options:
A stay stationary
B move towards the wire
C move away from the wire
D move parallel to the wire

The correct answer is:
move away from the wire

Solution:
As we know that, magnetic field due to a long wire at a distance $x$ from it is given by $B=\frac{\mu_{0} I}{2 \pi x},$ where, $I$ is the current flowing through the wire $\beta \propto I$
Magnetic flux associated with the square loop,
$$
\phi \propto \beta \propto I
$$
Now, if the current increase, then $\phi$ also increases. Direction of long wire will be $\otimes$.
This means, magnetic field due to induced current will be opposite to the existing magnetic field, i.e. according to Lenz's law, The induced current in the loop will be in the anti-clockwise direction. Now,

since, wires attract each other, if current flowing through them is in same direction and repel each other, if currents are in opposite direction.
$\therefore$ Part $C D$ of the loop will experience a force of repulsion, whereas part $A B$ will experience attraction. Parts $B C$ and $A D$ will not experience any force. Thus, the overall force will be a force of repulsion because $A B$ is closer to straight. The force between two current carrying conductors is inversely proportional to the distance between them
$\because$
$$
\begin{array}{l}
F \propto \frac{1}{r} \\
r_{1} < r_{2}
\end{array}
$$
So,
$$
F_{C D}>F_{A B}
$$
$$
F_{\text {net }}=F_{C D}-F_{A B}
$$
Hence, the loop will moves away from the wire.

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2019
A steel and a brass wire, each of length $50 \mathrm{~cm}$ and cross-sectional area $0.005 \mathrm{~cm}^{2}$ hang from a ceiling and are $15 \mathrm{~cm}$ apart. Lower ends of the wires are attached to a light horizontal bar. A suitable downward load is applied to the bar so that each of the wires extends in length by $0.1 \mathrm{~cm}$. At what distance from the steel wire the load must be applied? [Young's modulus of steel is $2 \times 10^{12}$ dynes $/ \mathrm{cm}^{2}$ and that of brass is $1 \times 10^{12}$ dynes $\left./ \mathrm{cm}^{2}\right]$

Options:
A $7.5 \mathrm{~cm}$
B $5 \mathrm{~cm}$
C $10 \mathrm{~cm}$
D $3 \mathrm{~cm}$

The correct answer is:
$5 \mathrm{~cm}$

Solution:
Hint:

At equilibrium, Taking torque about point $\mathrm{O}$
$\begin{aligned}
T_{1} x=T_{2}(15-x) & \Rightarrow Y_{1} x=Y_{2}(15-x) \\
\left[Y=\frac{T}{A} \times \frac{L}{\ell} ; T \propto Y\right] & 1 \times 10^{12} x=2 \times 10^{12}(15-x) ; 3 x=15, \quad \text { So, } x=5 \mathrm{~cm}
\end{aligned}$

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2020
A stone of relative density \(\mathrm{K}\) is released from rest on the surface of a lake. If viscous effects are ignored, the stone sinks in water with an acceleration of

Options:
A \(\mathrm{g}(1-\mathrm{K})\)
B \(\mathrm{g}(1+\mathrm{K})\)
C \(\mathrm{g}\left(1-\frac{1}{\mathrm{~K}}\right)\)
D \(\mathrm{g}\left(1+\frac{1}{\mathrm{~K}}\right)\)

The correct answer is:
\(\mathrm{g}\left(1-\frac{1}{\mathrm{~K}}\right)\)

Solution:
Hints:
\(\begin{aligned}
& F=v \sigma g-v \rho g=v \sigma g\left(1-\frac{\rho}{\sigma}\right)=m g\left(1-\frac{1}{\mathrm{k}}\right) \\
& a=g\left(1-\frac{1}{\mathrm{k}}\right)
\end{aligned}\)

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2011
A straight conductor $0.1 \mathrm{m}$ long moves in uniform magnetic field $0.1 \mathrm{T}$. The velocity the conductor is $15 \mathrm{m} / \mathrm{s}$ and is directer perpendicular to the field. The emf inducod between the two ends of the conductor is

Options:
A $0.10 \mathrm{V}$
B $0.15 \mathrm{V}$
C $1.50 \mathrm{V}$
D $15.00 \mathrm{v}$

The correct answer is:
$0.15 \mathrm{V}$

Solution:
Given, length of conductor $l=0.1
\mathrm{m}$
Mangetic field. $B=0.1 \mathrm{T}$
Velocity of conductor, $v=15 \mathrm{m} / \mathrm{s}$
The angle between $\mathbf{v}$ and $\mathbf{B}$ is $90^{\circ}$
When $\mathbf{v}$ and $\mathbf{B}$ are mutually perpendicular, then emf (induced) is given by
$$
\varepsilon=v B l=15 \times 0.1 \times 0.1=\frac{15}{100}=0.15 \mathrm{V}
$$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2015
A straight line joining the points (1,1,1) and (0,0,0) intersects the plane $2 x+2 y+z=10$ at

Options:
A (1,2,5)
B (2,2,2)
C (2,1,5)
D (1,1,6)

The correct answer is:
(2,2,2)

Solution:
Equation of line joining the points (1,1,1) and (0,0,0) is
$$
\begin{aligned}
\frac{x-0}{1-0} &=\frac{y-0}{1-0}=\frac{z-0}{1-0}=\lambda \\
\Rightarrow & x=y=z=\lambda
\end{aligned}
$$
So, the point is $(\lambda, \lambda, \lambda)$ The point intersects the plane $2 x+2 y+z=10$
$\therefore 2(\lambda)+2(\lambda)+\lambda=10$
$\Rightarrow 5 \lambda=10$
$\Rightarrow \lambda=2$
Hence, the point is (2,2,2) .

Topic: Mathematics, Three Dimensional Geometry, WBJEE, WBJEE 2016
A straight line meets the co-ordinate axes at $\mathrm{A}$ and $\mathrm{B}$. A circle is circumscribed about the triangle $\mathrm{OAB}, \mathrm{O}$ being the origin. If $\mathrm{m}$ and $\mathrm{n}$ are the distances of the tangent to the circle at the origin from the points $\mathrm{A}$ and $\mathrm{B}$ respectively, the diameter of the circle is

Options:
A $m(m+n)$
B $m+n$
C $n(m+n)$
D $\frac{1}{2}(m+n)$

The correct answer is:
$m+n$

Solution:
Clearly, AB is one of diameter
$\because \mathrm{AM}$ and $\mathrm{BN}$ are parallel and $\angle \mathrm{BNO}=\angle \mathrm{AMO}=\pi / 2$
$\therefore$ Points $\mathrm{N}, \mathrm{O}$ and $\mathrm{M}$ are collinear.
$\therefore \triangle \mathrm{BNMA}$ is a rectangle
$\Rightarrow \mathrm{AB}=\mathrm{MN}=\mathrm{m}+\mathrm{n}$

Topic: Mathematics, Circle, WBJEE, WBJEE 2022
A straight line through the origin $\mathrm{O}$ meets the parallel lines $4 \mathrm{x}+2 \mathrm{y}=9$ and $2 \mathrm{x}+\mathrm{y}+6=0$ at $\mathrm{P}$ and $\mathrm{Q}$ respectively. The point O divides the segment $\mathrm{PQ}$ in the ratio

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

The correct answer is:
$3: 4$

Solution:
Hint :
$\Rightarrow \frac{\mathrm{OP}}{\mathrm{OQ}}=\frac{\mathrm{OM}}{\mathrm{ON}}=\Rightarrow \frac{9 / 2}{12 / 2}=\frac{3}{4}$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2020
A straight line through the point (3,-2) is ndined at an angle $60^{\circ}$ to the line $\sqrt{3} x+y=1$. If it intersects the $X$ -axis, then its equation vill be

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

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

Solution:
Given line,
$$
\begin{array}{l}
\sqrt{3} x+y=1 \\
y=-\sqrt{3} x+c
\end{array}
$$
We know that, $m=-\sqrt{3}$
$$
\begin{aligned}
\text { that, } m &=-\sqrt{3} \quad[\because y=m x+c] \\
\tan \theta &=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\
\tan 60^{\circ} &=\left|\frac{m_{1}-(-\sqrt{3})}{1+m_{1}(-\sqrt{3})}\right| \quad\left[\because \theta=60^{\circ}\right] \\
\sqrt{3}=&\left|\frac{m_{1}+\sqrt{3}}{1-\sqrt{3}} m_{1}\right| \\
\pm \sqrt{3}=& \frac{m_{1}+\sqrt{3}}{1-\sqrt{3} m_{1}}
\end{aligned}
$$
\begin{array}{l|l}
\hline taking + sign & taking - sign \\
\hline+\sqrt{3}=\frac{m_{1}+\sqrt{3}}{1-\sqrt{3} m_{1}} & -\sqrt{3}=\frac{m_{1}+\sqrt{3}}{1-\sqrt{3} m_{1}} \\
\sqrt{3}\left(1-\sqrt{3} m_{1}\right)=m_{1}+\sqrt{3} & \Rightarrow-\sqrt{3}\left(1-\sqrt{3} m_{1}\right) \\
& =m_{1}+\sqrt{3} \\
\Rightarrow \sqrt{3}-3 m_{1}=m_{1}+\sqrt{3} & \Rightarrow-\sqrt{3}-3 m_{1}=m_{1}+\sqrt{3} \\
\Rightarrow & \Rightarrow 3 m_{1}-m_{1}=2 \sqrt{3} \\
\Rightarrow 4m_{1}=0 & \Rightarrow \quad 2 m_{1}=2 \sqrt{3} \\
\Rightarrow m_{1}=0 & \Rightarrow \quad m_{1}=\sqrt{3}
\end{array}
given point (3,-2)
$y=m x$ + c
$$
\begin{aligned}
-2 &=3 \sqrt{3}+c \\
c &=-2-3 \sqrt{3}
\end{aligned}
$$
Hence, required equation
$$
\therefore \quad y=m x+c
$$
$$
\begin{array}{l}
y=\sqrt{3} x-2-3 \sqrt{3} \\
y-\sqrt{3} x+2+3 \sqrt{3}=0 \\
y-x \sqrt{3}+2+3 \sqrt{3}=0
\end{array}
$$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2019
A straight line through the point of intersection of the lines $x+2 y=4$ and $2 x+y=4$ meets the coordinate axes at $A$ and $B$. The locus of the mid-point of $A B$ is

Options:
A $3(x+y)=2 x$
B $2(x+y)=3 x y$
C $2(x+y)=x$
D $x+y=3 x y$

The correct answer is:
$2(x+y)=3 x y$

Solution:
Given lines are,
$$
x+2 y=4
$$
and $\quad 2 x+y=4$
Solving Eqs. (i) and (ii), we get
$$
x=\frac{4}{3}, y=\frac{4}{3}
$$
Let $\quad A B: \frac{x}{a}+\frac{y}{b}=1$
Meets $x$ -axis at $A$ and $y$ -axis at $B$. If the straight line (iii) passes through the point $\left(\frac{4}{3}, \frac{4}{3}\right),$ then $\frac{4}{3 a}+\frac{4}{3 b}=1$
$\Rightarrow \quad \frac{1}{a}+\frac{1}{b}=\frac{3}{4}$
Let the mid-point of $A B$ be $(h, k)$
$\therefore \quad h=\frac{a+0}{2}, k=\frac{0+b}{2}$
$\Rightarrow \quad a=2 h, b=2 k$
$\therefore$ From Eq. $(\mathrm{iv}), \frac{1}{2 h}+\frac{1}{2 k}=\frac{3}{4}$
$\Rightarrow \quad 2 h+2 k=3 h k$
Replace $x, y$ from $h, k$ respectively, we get required locus
i.e. $2(x+y)=3 x y$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2012
A straight wire is placed in a magnetic field that varies with distance $x$ from origin as $\vec{B}=B_0\left(2-\frac{x}{a}\right) \hat{k}$. Ends of wire are at $(a, 0)$ and $(2 a, 0)$ and it carries a current $\mathrm{I}$. If force on wire is $\vec{F}=\mathrm{IB}_0\left(\frac{k a}{2}\right) \hat{\mathrm{j}}$, then value of $k$ is

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

The correct answer is:
$-1$

Solution:
$\overrightarrow{\mathrm{dF}}=\mathrm{Idx} \times \overrightarrow{\mathrm{B}}=\operatorname{Idx} \hat{\mathrm{i}} \times \mathrm{B} \hat{\mathrm{k}}=\operatorname{IdxB}(-\hat{\mathrm{j}})$
$\int \mathrm{dF}=\int_{\mathrm{a}}^{2 a} \mathrm{Id} x \mathrm{~B}_0\left(2-\frac{\mathrm{x}}{\mathrm{a}}\right)(-\hat{\mathrm{j}})$
$$
\begin{aligned}
&\mathrm{F}=\mathrm{IB}_0 \int_{\mathrm{a}}^{2 \mathrm{a}}\left(2-\frac{\mathrm{x}}{\mathrm{a}}\right) \mathrm{dx}(-\hat{\mathrm{j}}) \\
&=\operatorname{IB}_0\left(2 \mathrm{x}-\frac{\mathrm{x}^2}{2 \mathrm{a}}\right)_{\mathrm{a}}^{2 \mathrm{a}}(-\hat{\mathrm{j}}) \\
&=\mathrm{IB}_0\left[\left(2 \times 2 \mathrm{a}-\frac{(2 \mathrm{a})^2}{2 \mathrm{a}}\right)-\left(2 \mathrm{a}-\frac{\mathrm{a}^2}{2 \mathrm{a}}\right)\right](-\hat{\mathrm{j}}) \\
&=\mathrm{IB}_0\left[(4 \mathrm{a}-2 \mathrm{a})-\left(\frac{3 \mathrm{a}}{2}\right)\right](-\hat{\mathrm{j}}) \\
&=\mathrm{IB}_0\left[2 \mathrm{a}-\frac{3 \mathrm{a}}{2}\right](-\hat{\mathrm{j}}) \\
&=\operatorname{IB}_0\left[\frac{\mathrm{a}}{2}\right](-\hat{\mathrm{j}})
\end{aligned}
$$
Comparing $\mathrm{F}$ with $\mathrm{IB}_0\left[\frac{\mathrm{ka}}{2}\right] \hat{\mathrm{j}}$, we get $\mathrm{k}=-1$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2022
A straight wire of length $2 \mathrm{m}$ carries a current of $10 \mathrm{A}$. If this wire is placed in a uniform magnetic field of 0.15 T making an angle of $45^{\circ}$ with the magnetic field, the applied force on the wire will be

Options:
A $1.5 \mathrm{N}$
B $3 \mathrm{N}$
C $3 \sqrt{2} \mathrm{N}$
D $\frac{3}{\sqrt{2}} \mathrm{N}$

The correct answer is:
$\frac{3}{\sqrt{2}} \mathrm{N}$

Solution:
Given $i=10 \mathrm{A}, B=0.15 \mathrm{T}, \theta=45^{\circ}$ and $l=2 \mathrm{m}$
Here,
$$
\begin{aligned}
F &=i L B \sin \theta \\
&=10 \times 2 \times 0.15 ~ x ~ \sin 45^{\circ} \\
&=\frac{3}{\sqrt{2}} \mathrm{N}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2012
A stream of electrons and protons are directed towards a narrow slit in a screen (see figure). The intervening region has a uniform electric field (vertically downwards) and a uniform magnetic field $\mathbf{B}$ (out of the plane of the figure) as shown. Then

Options:
A electrons and protons with speed $\frac{|\mathrm{E}|}{|\mathbf{B}|}$ will pass through the slit
B protons with speed $|\mathrm{E}| /|\mathrm{B}|$ will pass through the slit, electrons of the same speed will not
C neither electrons nor protons will go through the sit irrespective of their speed
D electrons will always be deflected upwards irrespective of their speed

The correct answers are:
neither electrons nor protons will go through the sit irrespective of their speed, electrons will always be deflected upwards irrespective of their speed

Solution:
of their speed.

And electrons will always be deflected upwards irrespective of their speed.
So, (c) and (d) are correct options.

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2014
A student answers a multiple choice question with 5 alternatives, of which exactly one is correct. The probability that he knows the correct answer is $p, 0 < p < 1 .$ If he does not know the correct answer, he randomly ticks one answer. Given that he has answered the question correctly, the probability that he did not tick the answer randomly, is

Options:
A $\frac{3 p}{4 p+3}$
B $\frac{5 p}{3 p+2}$
C $\frac{5 p}{4 p+1}$
D $\frac{4 p}{3 p+1}$

The correct answer is:
$\frac{5 p}{4 p+1}$

Solution:
Let, $\quad E_{1}=$ Student does not know the answer
$E_{2}=$ Student knows the answer
and $\quad E=$ Student answer correctly.
$\therefore$
$P\left(E_{1}\right)=1-p$
$P\left(E_{2}\right)=p$
$P\left(\frac{E}{E_{2}}\right)=1$ and $P\left(\frac{E}{E_{1}}\right)=\frac{1}{5}$
$\therefore$ The probability that student did not tick the answer randomly
= The probability that student tick the answer correctly $=\frac{P\left(E_{2}\right) P\left(\frac{E}{E_{2}}\right)}{P\left(E_{1}\right) P\left(\frac{E}{E_{1}}\right)+P\left(E_{2}\right) P\left(\frac{E}{E_{2}}\right)}$
$=\frac{p(1)}{(1-p) \frac{1}{5}+p(1)}$
$=\frac{p}{\frac{1-p+5 p}{5}}=\frac{5 p}{1+4 p}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2014
$\mathrm{A}$ student appears for tests 1 , 11 and $\mathrm{III}$. The student is successful if he passes in tests I,II or I, III. The probabilities of the student passing in tests I. II and III are respectively,
p,q and $1 / 2 .$ If the probability of the student to be successful is $1 / 2 .$ Then

Options:
A $p(1+q)=1$
B $q(1+p)=1$
C $p q=1$
D $\frac{1}{p}+\frac{1}{q}=1$

The correct answer is:
$p(1+q)=1$

Solution:
Let $X$ be the event that student will be successful. $X_{1}$ be the event that student will be pass in test-I. $X_{2}$ be the event that student will be pass in test-II. $X_{3}$ be the event that student will be pass in test-III. $\therefore P(X)=P\left(X_{1} \cap X_{2} \cap X_{3}^{\prime}\right)+P\left(X_{1} \cap X_{2}^{\prime} \cap X_{3}\right)$
$P\left(X_{1}^{1} \cap X_{2} \cap X_{3}\right).$
$\Rightarrow P(X)=P\left(X_{1}\right) \cdot P\left(X_{2}\right) \cdot P\left(X_{3}^{1}\right)+P\left(X_{1}\right) \cdot P\left(X_{2}^{\prime}\right) \cdot P\left(X_{3}\right)$
$+P\left(X_{1}^{1}\right) \cdot P\left(X_{2}\right) \cdot P\left(X_{3}\right)$
$\Rightarrow \quad \frac{1}{2}=p \cdot q \cdot \frac{1}{2}+p(1-q) \frac{1}{2}+p q \frac{1}{2}$
$\Rightarrow \quad \frac{1}{2}=p \cdot q \cdot \frac{1}{2}+p \cdot \frac{1}{2}-p \cdot q \cdot \frac{1}{2}+p \cdot q \cdot \frac{1}{2}$
$\Rightarrow \quad \frac{1}{2}=p \cdot \frac{1}{2}+p \cdot q \cdot \frac{1}{2}$
$\Rightarrow \quad(p+p q)=1$
$\Rightarrow \quad p(1+q)=1$

Topic: Mathematics, Probability, WBJEE, WBJEE 2018
A survey of people in a given region showed that $20 \%$ were smokers. The probability of death due to lung cancer given that a person smoked was 10 times the probability of death due to lung cancer, given that a person did not smoke. If the probability of death due to lung cancer in the region is 0.006 . What is the probability of death due to lung cancer given that a person is a smoker?

Options:
A $1 / 140$
B $1 / 70$
C $3 / 140$
D $1 / 10$

The correct answer is:
$3 / 140$

Solution:
Let $S=$ Event that person is smoker
$\mathrm{NS}=$ Event that person is non-smoker
$D=$ Event that death is due to lung cancer.
Now, probability of death due to cancer that a person is a smoker, $P(D)=P(S) \cdot P\left(\frac{D}{S}\right)+P(N S) \cdot P\left(\frac{D}{N S}\right)$
$\Rightarrow 0.006=\frac{20}{100} \times P\left(\frac{D}{S}\right)+\frac{80}{100} \times \frac{1}{10} \times P\left(\frac{D}{S}\right)$
$\Rightarrow \frac{6}{1000}=\frac{2}{10} P\left(\frac{D}{S}\right)+\frac{8}{100} P\left(\frac{D}{S}\right)$
$\Rightarrow \quad P\left(\frac{D}{S}\right)\left[\frac{2}{10}+\frac{8}{100}\right]=\frac{6}{1000}$
$\Rightarrow \quad P\left(\frac{D}{S}\right)\left[\frac{20+8}{100}\right]=\frac{6}{1000}$
$\therefore \quad P\left(\frac{D}{S}\right)=\frac{6}{1000} \times \frac{100}{28}$
$=\frac{3}{140}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2015
A thin charged rod is bent into the shape of a small circle of radius $\mathrm{R}$ the charge per unit length of the rod being $\lambda$. The circle is rotated about its axis with a time period $T$ and it is found that the magnetic field at a distance ' $d$ ' away $(d>>R)$ from the center and on the axis, varies as $\frac{\mathrm{R}^{\mathrm{m}}}{\mathrm{d}^{\mathrm{n}}} .$ The values of $\mathrm{m}$ and $\mathrm{n}$ respectively are

Options:
A $m=2, n=2$
B $m=2, n=3$
C $m=3, n=2$
D $m=3, n=3$

The correct answer is:
$m=3, n=3$

Solution:
$\mathrm{B}=\frac{\mu_{0}}{4 \pi} \frac{2 \pi \mathrm{R}^{2} \mathrm{I}}{\left(\mathrm{R}^{2}+\mathrm{d}^{2}\right)^{3 / 2}} \quad \mathrm{I}=\frac{\mathrm{q}}{\mathrm{T}}$
$=\frac{2 \pi \mathrm{R} \lambda}{\mathrm{T}}$
$=\frac{\mu_{0}}{4 \pi} \frac{2 \pi \mathrm{R}^{2} \mathrm{I}}{\mathrm{d}^{3}}(\mathrm{~d}>>\mathrm{R})$
$\frac{\mu_{0}}{2} \frac{\mathrm{R}^{2}}{\mathrm{~d}^{3}} \times \frac{2 \pi \mathrm{R} \lambda}{\mathrm{T}}$
$\mathrm{B} \propto \frac{\mathrm{R}^{3}}{\mathrm{~d}^{3}}$
$\therefore \mathrm{m}=\mathrm{n}=3$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2021