datavorous/entrance-exam-dataset huggingface dataset:
\({ }_{98} \mathrm{Cf}^{246}\) was formed along with a neutron when an unknown radioactive substance was bombarded with \({ }_{6} \mathrm{C}^{12}\). The unknown substance was :

Options:
A \({ }_{91} \mathrm{~Pa}^{234}\)
B \({ }_{90} \mathrm{Th}^{234}\)
C \({ }_{92} \mathrm{U}^{235}\)
D \({ }_{92} \mathrm{U}^{238}\)

The correct answer is:
\({ }_{92} \mathrm{U}^{235}\)

Solution:
Let, the unknown radioactive substance be ${ }_{x} A^{y}$.
$x A^{y}+{ }_{6} C^{12} \longrightarrow{ }_{98} C f^{246}+{ }_{0} n^{1}$
On comparing atomic numbers, we get
$$
\begin{array}{r}
x+6=98+0 \\
x=98-6=92
\end{array}
$$
On comparing mass numbers, we get
$$
y+12=246+1
$$
$$
\begin{aligned}
y &=247-12 \\
&=235
\end{aligned}
$$
Since, uranium has the atomic number 92 , thus, the given unknown substance was ${ }_{92} \mathrm{U}^{235}$.

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2014
$0.126 \mathrm{g}$ of an acid is needed to completely neutralise $20 \mathrm{mL} 0.1$ (N) NaOH solution. The equivalent weight of the acid is

Options:
A 53
B 40
C 45
D 63

The correct answer is:
63

Solution:
$\because$ Number of milli equivalents of NaOH
$\left(n^{\prime}\right)=N V=0.1 \times 20=2$
$\therefore$ Number of equivalent $=0.002$
For Neutralisation Number of equivalent of base (NaOH) Number of equivalent of acid
$0.002=\frac{W}{E}=\frac{0.126}{E}$
$\therefore \quad E=\frac{0.126}{0.002}=63$

Topic: Chemistry, Redox Reactions, WBJEE, WBJEE 2017
\(1.56 \times 10^5 \mathrm{~J}\) of heat is conducted through a \(2 \mathrm{~m}^2\) wall of \(12 \mathrm{~cm}\) thick in one hour. Temperature difference between the two sides of the wall is \(20^{\circ} \mathrm{C}\). The thermal conductivity of the material of the wall is (in \(\mathrm{W} \mathrm{m}^{-1} \mathrm{~K}^{-1}\) )

Options:
A 0.11
B 0.13
C 0.15
D 1.2

The correct answer is:
0.13

Solution:
Hints: \(\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{KA} \Delta \mathrm{T}}{\mathrm{x}}\)
\(\begin{aligned}
& \frac{1.56 \times 10^5}{3600}=\frac{\mathrm{K} \times 2 \times 20}{12 \times 10^{-2}} \\
& \mathrm{~K}=\frac{1.56 \times 10^5 \times 12 \times 10^{-2}}{3600 \times 2 \times 20} \\
& =\frac{1.56}{12}=0.13
\end{aligned}\)

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2011
$5.75 \mathrm{mg}$ of sodium vapour is converted to sodium ion. If the ionisation energy of sodium is $490 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and atomic weight is 23 units, the amount of energy needed for this conversion will be

Options:
A $1.96 \mathrm{~kJ}$
B $1960 \mathrm{~kJ}$
C $122.5 \mathrm{~kJ}$
D $0.1225 \mathrm{~kJ}$

The correct answer is:
$0.1225 \mathrm{~kJ}$

Solution:
$\mathrm{Na}(\mathrm{g}) \longrightarrow \mathrm{Na}^{+}(\mathrm{g})+\mathrm{e}^{-}$
$1 \mathrm{~mol} \mathrm{Na}(\mathrm{g})$
requires $490 \mathrm{~kJ} /$ mol for ionization.
$\therefore \quad \frac{5.75 \times 10^{-3}}{23}$ mol $\quad$ requires $490 \mathrm{~kJ} \times \frac{5.75 \times 10^{-3}}{23}=0.1225 \mathrm{~kJ}=122.5 \mathrm{~J}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, WBJEE, WBJEE 2021
$58.4 \mathrm{g}$ of $\mathrm{NaCl}$ and $180 \mathrm{g}$ of glucose were separately dissolved in $1000 \mathrm{mL}$ of water. Identify the correct statement regarding the elevation of boiling point (b.p) of the resulting solutions.

Options:
A NaCl solution will show higher elevation of boiling point
B Glucose solution will show higher elevation of boiling point
C Both the solutions will show equal elevation of boiling point
D The boiling point elevation will be shown by neither of the solutions

The correct answer is:
NaCl solution will show higher elevation of boiling point

Solution:
Elevation in boiling point, $\Delta T_{b}=i \times K_{b} \times m$
Molality of NaCl solution $=\frac{n}{W} \times 1000$
$=\frac{58.5}{W_{\mathrm{H}_{2} \mathrm{O}}} \times 1000=\frac{1000}{\mathrm{W}_{\mathrm{M}_{2}} \mathrm{O}}$
Molality of $f \mathrm{C}_{6} \mathrm{H}_{1} \not \mathrm{O}_{6}$ solution = $\frac{\frac{180}{180} \times 1000}{w_{\mathrm{H}_{2}} \mathrm{O}}$
$\frac{1000}{W_{M_{2}} O}$. Both the solutions have same molarity but the values for $\mathrm{NaCl}$ and glucose are ${2}$ and 1respectively. $\left.\therefore \quad \Delta T_{b(\mathrm{NaG})}=2 \times \Delta T_{b C C_{6} H_{1} 2} \mathrm{O}_{6}\right)$

Topic: Chemistry, Solutions, WBJEE, WBJEE 2012
1000 droplets of water having $2 \mathrm{mm}$ diameter each coalesce to form a single drop. Given the surface tension of water is $0.072 \mathrm{Nm}^{-1}$. The energy loss in the process is

Options:
A $8.146 \times 10^{-4} \mathrm{J}$
B $4.4 \times 10^{-4} \mathrm{J}$
C $2108 \times 10^{-5} \mathrm{J}$
D $4.7 \times 10^{-1} \mathrm{J}$

The correct answer is:
$8.146 \times 10^{-4} \mathrm{J}$

Solution:
Let the radius of single drop = $r$
Radius of small drop $=\frac{2 m m}{2}$
$$
\begin{array}{l}
=1 \mathrm{mm} \\
=1 \times 10^{-3} \mathrm{m}
\end{array}
$$
Surface tension of water $=0.072 \mathrm{N} / \mathrm{m}$
The volume of large drop must be equal volume of all small drops.
$$
\frac{4}{3} \pi R^{3}=1000\left(\frac{4}{3} \pi r^{3}\right)
$$
$\therefore$
$$
A=(1000)^{\frac{1}{3}} \cdot r=10 r
$$
The initial potential energy of drops system $U_{i}=s \times n \times 4 \pi r^{2}$
$$
\begin{array}{l}
=s \times 1000 \times 4 \pi^{2} \\
=1000 \mathrm{s} \cdot 4 \pi^{2}
\end{array}
$$
When the droplets coalesce, the final potential energy of the system $U_{1}=s \cdot 4 \pi R^{2}$
$$
=s \cdot 4 \pi \times 100 r^{2}
$$
Energy loss in the formation of large drop
$$
\begin{aligned}
\Delta E &=U_{t}-U_{i} \\
&=s \cdot 4 \pi 100 r^{2}-1000 \mathrm{s} \cdot 4 \pi r^{2} \\
&=s \cdot \pi r^{2}(400-4000) \\
&=-3600 \mathrm{s} \cdot \pi^{2} \\
&=-3600 \times 0.72 \times 3.14 \times\left(1 \times 10^{-3}\right)^{2}
\end{aligned}
$$
From catenation
$$
\begin{array}{l}
=813.888 \times 10^{-6} \\
=8.1388 \times 10^{-4} \mathrm{J} \\
=8.146 \times 10^{-4} \mathrm{J}
\end{array}
$$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2016
22320 cal of heat is supplied to 100 g of ice at $0^{\circ} \mathrm{C}$. If the latent heat of fusion of ice is 80 cal $\mathrm{g}^{-1}$ and latent heat of vaporization of water is 540 cal $\mathrm{g}^{-1},$ the final amount of water thus obtained and its temperature respectively
are

Options:
A $8 \mathrm{g}, 100^{\circ} \mathrm{C}$
B $100 \mathrm{g} ,90^{\circ} \mathrm{C}$
C $92 \mathrm{g}, 100^{\circ} \mathrm{C}$
D $82 \mathrm{g}, 100^{\circ} \mathrm{C}$

The correct answer is:
$8 \mathrm{g}, 100^{\circ} \mathrm{C}$

Solution:
Heat required to convert ice to water at $100^{\circ} \mathrm{C}$
$$
Q=m \times L+m s \Delta T=18000 \mathrm{cal}
$$
Amount of heat left $=4320 \mathrm{cal}$
$$
\begin{array}{l}
m \times L=4320 \\
m=8 g \text { steam }
\end{array}
$$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2012
$\mathrm{A}=\{1,2,3,4\}, \mathrm{B}=\{1,2,3,4,5,6\}$ are two sets, and function $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$ is defined by $\mathrm{f}(\mathrm{x})=\mathrm{x}+2 \forall \mathrm{x} \in \mathrm{A}$, then the function $\mathrm{f}$ is

Options:
A bijective
B onto
C one-one
D many-one

The correct answer is:
one-one

Solution:
Hints: $\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{y}) \Rightarrow \mathrm{x}+2=\mathrm{y}+2 \Rightarrow \mathrm{x}=\mathrm{y} \quad \therefore$ one-one

Topic: Mathematics, Functions, WBJEE, WBJEE 2010
A $1 \mu \mathrm{F}$ capacitor $C$ is connected to a battery of $10 \mathrm{V}$ through a resistance 1 Mega Ohm. The voltage across $C$ after 1 s is approximately

Options:
A $5.6 \mathrm{V}$
B $7.8 \mathrm{V}$
C $6.3 \mathrm{V}$
D $10 \mathrm{V}$

The correct answer is:
$6.3 \mathrm{V}$

Solution:
The time constant of the circuit
Here,
$$
\begin{aligned}
\tau &=C \cdot R \\
C &=1 \mu F=1 \times 10^{-6}
\end{aligned}
$$
$$
\begin{array}{l}
R=1 \mathrm{M\Omega}=1 \times 10^{6} \Omega \\
\tau=1 \times 10^{-6} \times 1 \times 10^{6}=1 \mathrm{s}
\end{array}
$$
We know that in one (1) time constant $63 \%$ charging is done. $\therefore \quad \frac{63}{100} \times q_{\max }$
$=\frac{63}{100} \times 1 \times 10^{-6} \times 10$
$=\frac{63}{100} \times 1 \mu \mathrm{F} \times 10=6.3 \mu \mathrm{C}$
$V=\frac{\mathrm{q}}{\mathrm{C}}=\frac{6.3 \times 10^{-6} \mathrm{C}}{1 \times 10^{-6} \mathrm{F}}=6.3 \mathrm{V}$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2016
A $5 \mu F$ capacitor is connected in series with $10 \mathrm{\mu F}$ capacitor. When a $300 \mathrm{v}$ potentid dillerence is applied across this combination, the total energy stored in the capacitors is

Options:
A $18 \mathrm{J}$
B $1.5 \mathrm{J}$
C $0.15 \mathrm{J}$
D $0.10 \mathrm{J}$

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

Solution:
According to question the figure can be drawn as below

The equivalent capacitance,
$$
\begin{aligned}
\frac{1}{C_{e q}} &=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{5}+\frac{1}{10} \\
&=\frac{2+1}{10}=\frac{3}{10} \\
\Rightarrow \quad C_{e q} &=\frac{10}{3} \mu \mathrm{F}
\end{aligned}
$$
Now, the energy stored in the capacitor is
$$
\begin{aligned}
U &=\frac{1}{2} C V^{2} \\
&=\frac{1}{2} \times \frac{10}{3} \times 10^{-6} \times 300 \times 300 \\
&=\frac{3}{10 \times 2}=\frac{3}{20}=0.15 \mathrm{J}
\end{aligned}
$$

Topic: Physics, Capacitance, WBJEE, WBJEE 2015
$A$ 10 W electric heater is used to heat a container filled with $0.5 \mathrm{kg}$ of water. It is found that the temperature of water and the container rises by $3^{\circ} \mathrm{K}$ in $15 \mathrm{min}$. The container is then emptied, dried and filled with $2 \mathrm{kg}$ of oil. The same heater now raises the temperature of container-oil system by $2^{\circ} \mathrm{K}$ in $20 \mathrm{min}$. Assuming that there is no heat loss in the process and the specific heat of water is $4200 \mathrm{d} \mathrm{kg}^{-1} \mathrm{K}^{-1}$, the specific heat of oil in the same unit is equal to

Options:
A $1.50 \times 10^{3}$
B $2.55 \times 10^{3}$
C $3.00 \times 10^{3}$
D $5.10 \times 10^{3}$

The correct answer is:
$2.55 \times 10^{3}$

Solution:
$m_{1} s_{1} \Delta t+m_{2} s_{2} \Delta t=$ Work done

$m_{1} s_{1} \Delta t+m_{2} s_{2} \Delta t=P_{1} t_{1}$
where $m_{1}=0.5 \mathrm{kg}$
Specific heat $s_{1}=4200 \mathrm{J} / \mathrm{kg}-\mathrm{K}$
$$
\begin{array}{l}
\Delta t=\Delta t_{1}=\Delta t_{2}=3 \mathrm{K} \\
P_{1}=P_{2}=10 \mathrm{W} \\
t_{1}=15 \times 60=900 \mathrm{s}
\end{array}
$$
$s_{2}=$ Specific heat capacity of container
$$
0.5 \times 4200 \times(3-0)+m_{2} \mathrm{s}_{2} \times(3-0)
$$
$$
\begin{aligned}
&=10 \times 15 \times 60 \\
2100 \times 3+m_{2} s_{2} \times 3 &=9000 \\
m_{2} s_{2} &=\frac{9000-6300}{3} \\
m_{2} s_{2} &=900
\end{aligned}
$$
Similarly, in case of oil
$$
m_{1} s_{0} \Delta t+m_{2} s_{2} \Delta t=P_{2} t_{2}
$$
where $s_{0}=$ specific heat capacity of oil
$$
\begin{aligned}
P_{1} &=P_{2}=10 \mathrm{W} \\
2 \times s_{0} \times 2+900 \times 2 &=10 \times 20 \times 60 \\
4 s_{0}+1800 &=12000 \\
4 s_{0} &=12000-1800 \\
s_{0} &=\frac{10200}{4}=2550 \\
&=2.55 \times 10^{3} \mathrm{J} \mathrm{kg}^{-1} \mathrm{k}^{-1}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2014
A $20 \mathrm{cm}$ long capillary tube is dipped vertically in water and the liquid rises upto $10 \mathrm{cm} .$ If the entire system is kept is a freely falling platform, the length of the water column in the tube will be

Options:
A $5 \mathrm{cm}$
B $10 \mathrm{cm}$
C $15 \mathrm{cm}$
D $20 \mathrm{cm}$

The correct answer is:
$20 \mathrm{cm}$

Solution:
The height raised by liquid in capillary tube
$$
h=\frac{2 I \cos \theta}{\rho g h}
$$
As in freely falling platform, a body experience weightlessness.

So, the liquid will rise upto to length of the capillary.
i.e. height raised by the liquid will be $20 \mathrm{cm}$.

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2015
A $2.0 \mathrm{~cm}$ object is placed $15 \mathrm{~cm}$ in front of a concave mirror of focal length $10 \mathrm{~cm}$. What is the size and nature of the image?

Options:
A $4 \mathrm{~cm}$. real
B $4 \mathrm{~cm}$, virtual
C $1.0 \mathrm{~cm}$, real
D None

The correct answer is:
$4 \mathrm{~cm}$. real

Solution:
$\frac{1}{\mathrm{v}}-\frac{1}{15}=\frac{1}{-10} \Rightarrow \mathrm{v}=-30 \mathrm{~cm}$
$m=\frac{-30}{-15}=2$, image size $=4 \mathrm{~cm}$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2009
A $100$ mL, $0.1=\mathrm{M}$ solution of ammonium acetate is diluted by adding $100 \mathrm{mL}$ of water. The pH of the resulting solution will be $\left(p K_{a}\right.$ of acetic acid is nearly equal to $\mathrm{pK}_{b}$ of $\mathrm{NH}_{4} \mathrm{OH}$ )

Options:
A 4.9
B 5
C 7
D 10

The correct answer is:
7

Solution:
Ammonium acetate is a salt of weak acid and weak base. When solutions of such salts diluted resulting pH of solution is calculated from following relationship
$$
\mathrm{pH}=7+\frac{1}{2} \mathrm{p} \mathrm{K}_{a}-\frac{1}{2} \mathrm{p} K_{\mathrm{b}}
$$
$\because p K_{a}=p K_{b}$
$\therefore \quad \mathrm{pH}$ of diluted solution $=7$

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2012
A $12.5$ ev electron beam is used to bombard gaseous hydrogen at ground state. The energy level upto which the hydrogen atoms would be excited is

Options:
A 2
B 3
C 4
D 1

The correct answer is:
3

Solution:
$12.5 < 12.75$

$\therefore$ 3rd state

Topic: Physics, Atomic Physics, WBJEE, WBJEE 2021
A $400 \Omega$ resistor, a $250 \mathrm{mH}$ inductor and a $2.5 \mu \mathrm{F}$ capacitor are connected in series with an AC source of peak voltage $5 \mathrm{~V}$ and angular frequency $2 \mathrm{rad/sec}$. What is the peak value of the electrostatic energy of the capacitor?

Options:
A $2 \mu \mathrm{J}$
B $2.5 \mu \mathrm{J}$
C $3.33 \mu \mathrm{J}$
D $5 \mu \mathrm{J}$

The correct answers are:
$5 \mu \mathrm{J}$

Solution:
Hint:
The angular frequency is $2 \mathrm{KHz}$. The unit given is incorrect Assuming it to be in radian / sec $X_{L}=2 \times 10^{3} \times 250 \times 10^{-3}=500 \Omega$
$\mathrm{X}_{\mathrm{C}}=\frac{1}{2.5 \times 10^{-6} \times 2 \times 10^{3}}=200 \Omega$
$Z=\sqrt{R^{2}+\left(X_{L}-X_{c}\right)^{2}}=500 \Omega$
$\left(\mathrm{V}_{\mathrm{C}}\right)_{\text {peak }}=\mathrm{i}_{\text {peak }} \mathrm{X}_{\mathrm{c}}=\frac{\left(\mathrm{V}_{\mathrm{s}}\right)_{\text {peak }}}{\mathrm{Z}} \mathrm{X}_{\mathrm{C}}=\frac{5}{500} \times 200=2 \mathrm{~V}$
$\left(\mathrm{U}_{\mathrm{C}}\right)_{\max }=\frac{1}{2} \times 2.5 \times 10^{-6} \times(2)^{2}=5 \mu \mathrm{J}$

Topic: Physics, Capacitance, WBJEE, WBJEE 2020
A and $\mathrm{B}$ are independent events. The probability that both A and $\mathrm{B}$ occur is $\frac{1}{20}$ and the probability that neither of them occurs is $\frac{3}{5}$. The probability of occurrence of $\mathrm{A}$ is

Options:
A $\frac{1}{2}$
B $\frac{1}{10}$
C $\frac{1}{4}$
D $\frac{1}{5}$

The correct answers are:
$\frac{1}{4}$, $\frac{1}{5}$

Solution:
\(P(A \cap B)=P(A) P(B)=\frac{1}{20} \Rightarrow P(B)=\frac{1}{2 O P(A)}\)
\(P(\bar{A} \cap \bar{B})=\frac{3}{5}=1-P(A \cup B)\)
\(\Rightarrow \frac{3}{5}=1-P(A)-P(B)+P(A \cap B)\)
\(\Rightarrow \frac{3}{5}=1-P(A)-\frac{1}{20 P(A)}+\frac{1}{20}\)
\(\Rightarrow \frac{3}{5}=\frac{21}{20}-P(A)-\frac{1}{20 P(A)}\)
\(\Rightarrow \frac{12-21}{20}=-P(A)-\frac{1}{20 P(A)}\)
Let \(P(A)=x\)
\(\Rightarrow \frac{-9}{20}=\frac{-20 x^2-1}{20 x}\)
\(\Rightarrow 20 x^2-9 x+1=0\)
\(\Rightarrow(4 x-1)(5 x-1)=0\)
\(\Rightarrow x=\frac{1}{4} \frac{1}{5}\)
i. e. \(P(A)=\frac{1}{4} P(A)=\frac{1}{5}\)

Topic: Mathematics, Probability, WBJEE, WBJEE 2020
$A$ and $B$ are two independent events such that $P(A \cup B)=0.8$ and $P(A)=0.3$. The $P(B)$ is

Options:
A $\frac{2}{7}$
B $\frac{2}{3}$
C $\frac{3}{8}$
D $\frac{1}{8}$

The correct answer is:
$\frac{2}{7}$

Solution:
Hints : Let $\mathrm{P}(\mathrm{B})=x$
$$
\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.3+(1-x)-0.3(1-x)
$$
or $\quad 0.8=1-x+0.3 x$
or $\quad 1-0.7 x=0.8$
or $\quad 0.7 x=0.2$
or $\quad x=\frac{2}{7}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2009
$\vec{A}$ and $\vec{B}$ are two vectors given by $\vec{A}=2 \hat{i}+3 \hat{j}$ and $\vec{B}=\hat{i}+\hat{j}$. The magnitude of the component of $\vec{A}$ along $\vec{B}$ is

Options:
A $\frac{5}{\sqrt{2}}$
B $\frac{3}{\sqrt{2}}$
C $\frac{7}{\sqrt{2}}$
D $\frac{1}{\sqrt{2}}$

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

Solution:
Magnitude of components of $\vec{A}$ along $\vec{B}=\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}=\frac{(2 \hat{i}+3 \hat{j})(\hat{i}+\hat{j})}{\sqrt{2}}=\frac{5}{\sqrt{2}}$

Topic: Physics, Mathematics in Physics, WBJEE, WBJEE 2009
\(A, B\) are fixed points with coordinates \((0, a)\) and \((0, b)(a > 0, b > 0)\). \(P\) is variable point \((x, 0)\) referred to rectangular axis. If the angle \(\angle \mathrm{APB}\) is maximum, then

Options:
A \(x^2=a b\)
B \(x^2=a+b\)
C \(x=\frac{1}{a b}\)
D \(x=\frac{a+b}{2}\)

The correct answer is:
\(x^2=a b\)

Solution:
Hint : \(\angle A P B=\theta=\cos ^1\left(\frac{x^2+a^2+x^2+b^2-(a-b)^2}{2 \sqrt{x^2+a^2} \sqrt{x^2+b^2}}\right)\)
For Max; \(\frac{\mathrm{d} \theta}{\mathrm{dx}}=0, \mathrm{x}^2=\mathrm{ab}\)

Topic: Mathematics, Properties of Triangles, WBJEE, WBJEE 2023
$A, B, C$ are mutually exclusive events such that $P(A)=\frac{3 x+1}{3}, P(B)=\frac{1-x}{4}$ and $P(C)=\frac{1-2 x}{2}$. Then the set of possible values of $\mathrm{x}$ are in

Options:
A $[0,1]$
B $\left[\frac{1}{3}, \frac{1}{2}\right]$
C $\left[\frac{1}{3}, \frac{2}{3}\right]$
D $\left[\frac{1}{3}, \frac{13}{3}\right]$

The correct answer is:
$\left[\frac{1}{3}, \frac{1}{2}\right]$

Solution:
$0 \leq \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C}) \leq 1$ and $0 \leq \mathrm{P}(\mathrm{A}) \leq 1,0 \leq \mathrm{P}(\mathrm{B}) \leq 1,0 \leq \mathrm{P}(\mathrm{C}) \leq 1 \Rightarrow \frac{1}{3} \leq \mathrm{x} \leq \frac{1}{2}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2022
\(A B C\) is an isosceles triangle with an inscribed circle with centre \(O\). Let \(P\) be the midpoint of \(B C\). If \(A B=A C=15\) and \(B C=10\), then \(\mathrm{OP}\) equals

Options:
A \(\frac{\sqrt{5}}{\sqrt{2}}\) unit
B \(\frac{5}{\sqrt{2}}\) unit
C \(2 \sqrt{5}\) unit
D \(5 \sqrt{2}\) unit

The correct answer is:
\(\frac{5}{\sqrt{2}}\) unit

Solution:
Hint :
\(\begin{aligned} & A P=\sqrt{15^2-5^2}=10 \sqrt{2} \\ & \Delta=\frac{1}{2} \cdot B C \cdot A P=50 \sqrt{2} \\ & s=\frac{15+15+10}{2}=20 \\ & \because r=\frac{\Delta}{s}=\frac{50 \sqrt{2}}{20}=\frac{5}{\sqrt{2}}\end{aligned}\)

Topic: Mathematics, Properties of Triangles, WBJEE, WBJEE 2023
$A B$ is a variable chord of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. If $A B$ subtends a right angle at the origin $O$, then $\frac{1}{OA^2}+\frac{1}{OB^2}$ equals to

Options:
A $\frac{1}{a^2}+\frac{1}{b^2}$
B $\frac{1}{a^2}-\frac{1}{b^2}$
C $a^2+b^2$
D $a^2-b^2$

The correct answer is:
$\frac{1}{a^2}+\frac{1}{b^2}$

Solution:

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2022
A ball is projected horizontally with a velocity of $5 \mathrm{~m} / \mathrm{s}$ from the top of a building $19.6 \mathrm{~m}$ high. How long will the ball take of hit the ground?

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

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

Solution:
Hints: $\mathrm{T}=\sqrt{\frac{2 \mathrm{H}}{g}}=\sqrt{\frac{2 \times 19.6}{9.8}}=2 \mathrm{sec}$

Topic: Physics, Motion In Two Dimensions, WBJEE, WBJEE 2010
A balloon starting from rest is ascending from ground with uniform acceleration of \(4 \mathrm{ft} / \mathrm{sec}^2\). At the end of \(5 \mathrm{sec}\), a stone is dropped from it. If \(\mathrm{T}\) be the time to reach the stone to the ground and \(\mathrm{H}\) be the height of the balloon when the stone reaches the ground, then

Options:
A \(\mathrm{T}=6 \mathrm{sec}\)
B \(\mathrm{H}=112.5 \mathrm{ft}\)
C \(T=5 / 2 \mathrm{sec}\)
D \(225 \mathrm{ft}\)

The correct answers are:
\(T=5 / 2 \mathrm{sec}\)

Solution:
Hint :
Let after \(5 \mathrm{sec}\) ballon is at \(B\)
Given: \(a=4 \mathrm{ft} / \sec ^2 u(\) Initial velocity \()=0\)
\(\mathrm{t}=5 \mathrm{sec}\)
Distance covered by ballon in \(5 \mathrm{sec}\);
\(S=u t+\frac{1}{2} \mathrm{at}^2=\frac{1}{2} \times 4 \times 5^2=50 \mathrm{ft}\)
Let Speed of balloon at \(B\) be \(\mathrm{V}\)
\(V=u+\) at \(=0+4 \times 5=20 \mathrm{ft} / \mathrm{sec}\)
So, speed of ball when it's dropped
at point \(B\) is \(v_1=20 \mathrm{ft} / \mathrm{sec}\).
\(S=v_1 t+\frac{1}{2} a_1 t^2\)
\(\Rightarrow-50=20 t+\frac{1}{2}(-32) t^2 \Rightarrow t=\frac{5}{2} \mathrm{sec}\)

Topic: Mathematics, Heights and Distances, WBJEE, WBJEE 2023
A bar magnet falls from rest under gravity through the centre of a horizontal ring of conducting wire as shown in figure. Which of the following graph best represents the speed \((v) v s\). time (t) graph of the bar magnet?

Options:
A

B

C

D

The correct answer is:

Solution:
Hint :

Topic: Physics, Electromagnetic Induction, WBJEE, WBJEE 2023
A bar magnet has a magnetic moment of $200 \mathrm{Am}^{2}$. The magnet is suspended in a magnetic field of $0.30 \mathrm{NA}^{-1} \mathrm{m}^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^{\circ},$ will be

Options:
A $30 \mathrm{N}$ -m
B $30 \sqrt{3} \mathrm{N}-\mathrm{m}$
C $60 \mathrm{N}$ -m
D $60 \sqrt{3} \mathrm{N} \cdot \mathrm{m}$

The correct answer is:
$30 \mathrm{N}$ -m

Solution:
Given, $\mathrm{M}=200 \mathrm{A} \mathrm{m}^{2}$
$\mathrm{B}=0.30 \mathrm{NA}^{-} \mathrm{M}^{-1}$
and $\theta=30^{\circ}$
We know that the Torque,
$$
\begin{aligned}
\tau &=M \times B \\
\Rightarrow \quad|t| &=M B \sin \theta=200 \times 0.3 \times \frac{1}{2} \\
&=100 \times 0.3=30 \mathrm{N} \mathrm{m}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2013
A bar of length L carrying a small mass $m$ at one of its ends rotates with a uniform angular speed $\omega$ in a vertical plane about the mid point of the bar. During the rotation, at some instant of time when the bar is horizontal, the mass is detached from the bar but the bar continues to rotate with some $\omega .$ The mass moves vertically up, comes back and reaches the bar at the same point. At that place, the acceleration due to gravity is $g$

Options:
A This possible if the quantity $\frac{\omega^{2} I}{2 \pi g}$ is an integer
B The total time of flight of the mass is proportional to $\omega^{2}$
C The total distance travelled by the mass n in air is proportional to $\omega^{2}$
D The total distance travelled by the mass in air and its total time of flight are both independent on its mass.

The correct answers are:
This possible if the quantity $\frac{\omega^{2} I}{2 \pi g}$ is an integer, The total distance travelled by the mass n in air is proportional to $\omega^{2}$, The total distance travelled by the mass in air and its total time of flight are both independent on its mass.

Solution:
We have, $v=\frac{1}{2} \omega l$

$$
T=\frac{2 v}{g}=\frac{\omega l}{g} \text { or } n \frac{2 \pi}{\omega }=\frac{\omega l}{g}
$$
(as completes $n$ rotations with in $I$ ) $\therefore \quad n=\frac{/ \omega^{2}}{2 \pi g}$
Distance travelled $=2 h=\frac{2 v^{2}}{2 g}=\frac{l^{2} \omega^{2}}{4 g}$

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2013
A battery of emf $\mathrm{E}$ and internal resistance $\mathrm{r}$ is connected with an external resistance $\mathrm{R}$ as shown in the figure. The battery will act as a constant voltage source if

Options:
A $r< < R$
B $r>>R$
C $r=R$
D It will never act as a constant voltage source

The correct answer is:
$r< < R$

Solution:
$V_R=\frac{E}{(R+r)} \times R, \quad V_R=\frac{E}{(1+r / R)}$
$r < R, V_R \approx E$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2022
A beam of monochromatic blue light of wavelength $4200 Å$ in air travels in water of refractive index $4 / 3$. Its wavelength in water will be:

Options:
A $4200 Å$
B $5800 Å$
C $4150 Å$
D $3150 Å$

The correct answer is:
$3150 Å$

Solution:
In water $\lambda=\frac{4200}{\frac{4}{3}}=3150 Å$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2009