datavorous/entrance-exam-dataset huggingface dataset:
Two charges $+q$ and $-q$ are placed at a distance $a$ in a uniform electric fleld. The dipole moment of the combination is $2 q a(\cos \theta i+\sin \theta)$. where, $\theta$ is the angle between the direction of the field and the line joining the two charges. Which of the following statement(s) is/ase correct?

Options:
A The torque exerted by the field on the dipole vanishes
B The net force on the dipole vanishes
C The torque is independent of the choice of coordinates
D The net force is independent of a

The correct answers are:
The net force on the dipole vanishes, The torque is independent of the choice of coordinates, The net force is independent of a

Solution:
The situation can be diagrammatically as,

The dipole moment,
$$
\mathbf{P}=2 q a(\cos \theta \hat{\mathbf{i}}+\sin \theta \hat{\mathbf{j}})
$$
The net force on the dipole is always zero while net torque on the dipole is not zero.

Topic: Physics, Electrostatics, WBJEE, WBJEE 2015
Two circle $S_1=p x^2+p y^2+2 g^{\prime} x+2 f^{\prime} y+d=0$ and $S_2=x^2+y^2+2 g x+2 f y+d^{\prime}=0$ have a common chord $P Q$. The equation of $\mathrm{PQ}$ is

Options:
A $S_1-S_2=0$
B $S_1+S_2=0$
C $\mathrm{S}_1-\mathrm{pS}_2=0$
D $\mathrm{S}_1+\mathrm{pS}_2=0$

The correct answer is:
$\mathrm{S}_1-\mathrm{pS}_2=0$

Solution:
$\frac{\mathrm{S}_1}{\mathrm{p}}-\mathrm{S}_2=0 \Rightarrow \mathrm{S}_1-\mathrm{pS}_2=0$

Topic: Mathematics, Circle, WBJEE, WBJEE 2022
Two coherent monochromatic beams of intensities $I$ and $4 I$ respectively are superposed. The maximum and minimum intensitien in the reaulting pattern are

Options:
A 5I and 3I
B 9I and 3I
C 4I and I
D 9I and I

The correct answer is:
9I and I

Solution:
We know that The maximum intensities
$$
I_{\max }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}
$$
The minimum intensities
$$
I_{\min }=\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}
$$
So, the ratio of the maximum and minimum of intensities is
$$
\begin{aligned}
\frac{I_{\max }}{I_{\min }} &=\frac{\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}}{\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}} \\
\frac{I_{\max }}{I_{\min }} &=\frac{(\sqrt{4 I}+\sqrt{I})^{2}}{(\sqrt{4 I}-\sqrt{I})^{2}} \\
&=\left(\frac{3 \sqrt{I}}{\sqrt{I}}\right)^{2} \\
=& \frac{9}{1}=9: 1
\end{aligned}
$$

Topic: Physics, Wave Optics, WBJEE, WBJEE 2014
Two coils of self-inductances $6 \mathrm{mH}$ and $8 \mathrm{mH}$ are connected in series and are adjusted for highest coefficient of coupling. Equivalent self-inductance $L$ for the assembly is approximately

Options:
A $50 \mathrm{mH}$
B $36 \mathrm{mH}$
C $28 \mathrm{mH}$
D $18 \mathrm{mH}$

The correct answer is:
$28 \mathrm{mH}$

Solution:
Given, $L_{1}=6 \mathrm{mH}=6 \times 10^{-3} \mathrm{H}$
$$
L_{2}=8 \mathrm{mH}=8 \times 10^{-3} \mathrm{H}
$$
If two coills $L_{1}$ and $L_{2}$ are connected in series, the equivalent self-inductance $L$ for assembly
$$
L=L_{1}+L_{2}+2 \sqrt{L_{1} L_{2}}
$$
$$
\begin{array}{l}
=6 \times 10^{-3}+8 \times 10^{-3}+2 \sqrt{6 \times 8 \times 10^{-6}} \\
=14 \times 10^{-3}+2 \sqrt{48} \times 10^{-3} \\
=14 \times 10^{-3}+2 \times 7 \times 10^{-3} \\
=28 \times 10^{-3} \mathrm{H}=28 \mathrm{mH}
\end{array} \quad(\because \sqrt{48}=6.92=7)
$$

Topic: Physics, Electromagnetic Induction, WBJEE, WBJEE 2016
Two coins are available, one fair and the other two-headed. Choose a coin and toss it once; assume that the unbiased coin is chosen with probability $\frac{3}{4}$, Given that the outcome is head, the probability that the two-headed coin was chosen, is

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

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

Solution:
Let $F$ denotes fair coin $T$ denotes two headed $H$ denotes head occurs
$$
\begin{array}{lc}
\therefore \quad P(F)=\frac{3}{4}, P(T)=1-\frac{3}{4}=\frac{1}{4} \\
\therefore \quad P\left(\frac{T}{H}\right)=\frac{P\left(\frac{H}{T}\right) \cdot P(T)}{P\left(\frac{H}{T}\right) \cdot P(T)+P\left(\frac{H}{F}\right) \cdot P(F)}
\end{array}
$$
By Baye's theorem
$$
=\frac{1 \cdot \frac{1}{4}}{1 \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{3}{4}}=\frac{2}{5}
$$

Topic: Mathematics, Probability, WBJEE, WBJEE 2012
Two decks of playing cards are well shuffled and 26 cards are randomly distributed to a player. Then, the probability that the player gets all distinct cards is

Options:
A ${ }^{52} C_{26} /{ }^{104} C_{26}$
B $2 \times{ }^{52} \mathrm{C}_{26} /{ }^{104} \mathrm{C}_{26}$
C $2^{3} \times{ }^{52} C_{26} /{ }^{104} C_{26}$
D $2^{26} \times{ }^{52} \mathrm{C}_{26} /{ }^{104} \mathrm{C}_{26}$

The correct answer is:
$2^{26} \times{ }^{52} \mathrm{C}_{26} /{ }^{104} \mathrm{C}_{26}$

Solution:
Since, these are 52 distinct cards in decks and each distinct card is 2 in number. Therefore, 2 decks will also contain only 52 distinet cards two each.
$\therefore$ Probability that the player gets all distinct cards
$$
=\frac{{ }^{52} C_{26} \times 2^{26}}{104_{C_{26}}}
$$

Topic: Mathematics, Probability, WBJEE, WBJEE 2012
Two dice are tossed once. The probability of getting an even number at the first die or a total of 8 is

Options:
A $\frac{1}{36}$
B $\frac{3}{36}$
C $\frac{11}{36}$
D None of these

The correct answer is:
None of these

Solution:
Hints: $A=$ getting even no on 1st dice
$$
\mathrm{B}=\text { getting sum } 8
$$
So $|\mathrm{A}|=18 \quad|\mathrm{~B}|=5 \quad|\mathrm{~A} \cap \mathrm{B}|=3$
$$
\text { So } P(A \cup B)=\frac{18+5-3}{36}=\frac{20}{36} \text {}
$$

Topic: Mathematics, Probability, WBJEE, WBJEE 2010
Two elements $A$ and $B$ with atomic numbers $Z_{A}$ and $Z_{B}$ are used to produce characteristic X-rays with frequencies $v_{A}$ and $v_{B}$ respectively. If $Z_{A}: Z_{B}=1: 2,$ then $v_{A}: v_{B}$ will be

Options:
A $1: \sqrt{2}$
B $1: 8$
C $4: 1$
D $1: 4$

The correct answer is:
$1: 4$

Solution:
Given $Z_{A}: Z_{B}=1: 2$ We know
$v \propto Z^{2}$
Hence,
$\begin{array}{l}
\frac{v_{A}}{v_{B}}=\frac{(1)^{2}}{(2)^{2}} \\
v_{A}: v_{B}=1: 4
\end{array}$

Topic: Physics, Atomic Physics, WBJEE, WBJEE 2012
Two equal resistances, $400 \Omega$ each, are connected in series with a $8 \mathrm{V}$ battery. If the resistance of first one increases by $0.5 \%$, the change required in the resistance of the second one in order to keep the potential difference across it unaltered is to

Options:
A increase it by $1 \Omega$
B increase it by $2 \Omega$
C increase it by $4 \Omega$
D decrease it by $4 \Omega$

The correct answer is:
increase it by $2 \Omega$

Solution:
In series combination,
$$
R=R_{1}+R_{2}
$$
Same current will pass in $A_{1}$ and $R_{2}$. According to question, $R_{1}$ is increased by $0.5 \%$
$\therefore$ Increment in the resistance
$$
\begin{array}{l}
=\frac{0.5}{100} \times 400 \\
=2.0 \Omega
\end{array}
$$
So to keep the potential unchanged in second resistance the change required will be $2.0 ~ \Omega$ increment.

Topic: Physics, Current Electricity, WBJEE, WBJEE 2016
Two gases $X$ (molecular weight $M_{X}$ ) and Y (molecular weight $M_{Y} ; M_{Y}>M_{X}$ ) are at the same temperature $T$ in two different containers. Their root mean square velocities are $C_{X}$ and $C_{Y}$ respectively. If the average kinetic energies per molecule of two gases $X$ and $Y$ are $E_{x}$ and $E_{Y}$ respectively then which of the following relation(s) is(are) true?

Options:
A $E_{X}>E_{Y}$
B $C_{X}>C_{Y}$
C $E_{X}=E_{Y}=(3 / 2) R T$
D $E_{X}=E_{Y}=(3 / 2) k_{B} T$

The correct answers are:
$C_{X}>C_{Y}$, $E_{X}=E_{Y}=(3 / 2) k_{B} T$

Solution:
Given, molecular weight of $X=M_{X}$
Mol. wt. of $Y=M_{Y}$ and $M_{Y}>M_{x}$
Root mean square velocitios $=C_{X}$ and $C_{Y}$ respectively
Average $\mathrm{KE}$/molecule $=E_{X}$ and $E_{Y}$ respectively We know that. Root mean square velocity.
$$
C=\sqrt{\frac{3 R T}{M}} \text { or } C \propto \sqrt{\frac{1}{M}}
$$
$\therefore \quad \frac{C_{X}}{C_{Y}}=\frac{M_{Y}}{M_{X}}$
since $\quad M_{Y}>M_{X}$
$\therefore$
$$
C_{X}>C_{Y}
$$
KE/molecules $=\frac{3}{2} n R T$
$\Rightarrow \quad E_{X} \not \subset E_{Y} \not = \frac{3}{2} R T$
Further. $\quad E_{X}=E_{Y}=\frac{3}{2} \boldsymbol{k}_{\boldsymbol{B}} \boldsymbol{T}$

Topic: Chemistry, States of Matter, WBJEE, WBJEE 2014
Two identical blocks of ice move in opposite directions with equal speed and collide with cach other. What will be the minimum speed required to make both the blocks melt completely, if the initial temperatures of the blocks were $-8^{\circ} \mathrm{C}$ each? (Specific heat of ice is $2100 \mathrm{Jkg}^{-1} \mathrm{K}^{-1}$ and latent heat of fusion of ice is $3.36 \times 10^{5} \mathrm{Jkg}^{-1}$ )

Options:
A $840 \mathrm{ms}^{-1}$
B $420 \mathrm{ms}^{-1}$
C $84 m s^{-1}$
D $84 \mathrm{ms}^{-1}$

The correct answer is:
$840 \mathrm{ms}^{-1}$

Solution:
Maximum loss in $K . E=$ Total $K . E$ energy before collision of two ice blocks
$$
K . E_{\max } \text { loss }=\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}=m v^{2}
$$
This maximum loss in kinetic energy will be equal to loss in heat to melt two ice block.
$$
\Rightarrow \quad m v^{2}=(m s \Delta \theta+m L) 2
$$
According to question,
$$
\begin{aligned}
L &=336 \times 10^{5} \mathrm{J} \mathrm{kg}^{-1} \\
S=& 2100 \mathrm{J} \mathrm{kg}^{-1} \mathrm{K}^{-1} \\
\Delta \theta=8^{\circ} \mathrm{C}
\end{aligned}
$$
From Eq.
(i) $v=\sqrt{2(s-\Delta \theta+L)}$
$$
\begin{array}{l}
=\sqrt{2\left(2100 \times 8+336 \times 10^{5}\right)} \\
v=840 \mathrm{ms}^{-1}
\end{array}
$$

Topic: Physics, Center of Mass Momentum and Collision, WBJEE, WBJEE 2019
Two identical cells each of emf $\mathrm{E}$ and internal resistance $\mathrm{r}$ are connected in parallel with an external resistance $\mathrm{R}$. To get maximum power developed across $R$, the value of $R$ is

Options:
A $R=r / 2$
B $\mathrm{R}=\mathrm{r}$
C $R=r / 3$
D $\mathrm{R}=2 \mathrm{r}$

The correct answer is:
$R=r / 2$

Solution:
$R_{e q}=\frac{r}{2}$
$I=\frac{2 E}{r+2 R}$
For max. power consumption. I should be max. So denominator should be min. for that
$r+2 R=(\sqrt{r}, \quad)$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2009
Two identical equiconvex lenses, each of focal length $f$ are placed side by side in contact with each other with a layer of water in between them as shown in the figure. If refractive index of the material of the lenses is greater than that of water, how the combined focal length $F$ is related to $f ?$

Options:
A $F>f$
B $\frac{f}{2} < F < f$
C $F < \frac{f}{2}$
D $F=f$

The correct answer is:
$\frac{f}{2} < F < f$

Solution:
The given combination of lenses

Here, $f_{1}=$ focal length of equiconvex lenses of glass. $f_{2}=$ focal length of lens formed by water (concave). The focal length of the combination.
$$
\therefore \quad \frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}}+\frac{1}{f_{3}}=\frac{1}{f_{1}}-\frac{1}{f_{2}}+\frac{1}{f_{1}}=\frac{2}{f_{1}}-\frac{1}{f_{2}}
$$
$$
\begin{aligned}
\frac{1}{F}=\frac{2 f_{2}-f_{1}}{f_{1} f_{2}} \Rightarrow F &=\frac{f_{1} f_{2}}{2 f_{2}-f_{1}} \\
F &=\frac{f_{1}}{2-\frac{f_{1}}{f_{2}}}
\end{aligned}
$$
$\left[\because f_{3}=f_{1}\right]$
Here, $\frac{1}{f_{1}}=\left(\mu_{g}-1\right) \frac{2}{R},$ for $L_{1}$ and $L_{3}$
and $\frac{1}{f_{2}}=\left(\mu_{w}-1\right)\left(-\frac{2}{R}\right),$ for $L_{2}$
$$
\text { Given, } \mu_{w} < \mu_{g}
$$
Thus, $\frac{f_{1}}{f_{2}} < 1$
So.
$$
F>\frac{f_{1}}{2}
$$
From Eqs.
(i) and (ii), we get
$$
\frac{f_{1}}{2} < F < f_{1} \quad \text { or } \frac{f}{2} < F < f
$$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2018
Two identical light waves, propagating in the same direction, have a phase difference $\delta$. After they superpose the intensity of the resulting wave will be proportional to

Options:
A $\cos \delta$
B $\cos (\delta / 2)$
C $\cos ^2(\delta / 2)$
D $\cos ^2 \delta$

The correct answer is:
$\cos ^2(\delta / 2)$

Solution:
$I=4 I_0 \cos ^2\left(\frac{\delta}{2}\right) \Rightarrow I \propto \cos ^2\left(\frac{\delta}{2}\right)$

Topic: Physics, Wave Optics, WBJEE, WBJEE 2009
Two identical springs are connected to mass \(\mathrm{m}\) as shown \((\mathrm{k}=\) spring constant). If the period of the configuration in (a) is \(2 \mathrm{~S}\), the period of the configuration (b) is

Options:
A \(\sqrt{2} \mathrm{~S}\)
B \(1 \mathrm{~S}\)
C \(\frac{1}{\sqrt{2}} \mathrm{~S}\)
D \(2 \sqrt{2} \mathrm{~s}\)

The correct answer is:
\(1 \mathrm{~S}\)

Solution:
\(\begin{aligned} & \text {Hints: } \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\sqrt{\frac{\mathrm{k}_2}{\mathrm{k}_1}} \Rightarrow \frac{2}{\mathrm{~T}}=\sqrt{\frac{2 \mathrm{k}}{\frac{\mathrm{k}}{2}}}=2 \\ & \therefore \mathrm{T}=1 \mathrm{~S}\end{aligned}\)

Topic: Physics, Oscillations, WBJEE, WBJEE 2011
Two infinite line-charges parallel to each other are moving with a constant velocity $v$ in the same direction as shown in the figure. The separation between two line-charges is $d$. The magnetic attraction balances the electric repulsion when, [c = speed of light in free space]

Options:
A $\quad v=\sqrt{2} c$
B $v=\frac{c}{\sqrt{2}}$
C $v=c$
D $v=\frac{c}{2}$

The correct answer is:
$v=c$

Solution:
$\mathrm{F}_{\mathrm{E}}=\mathrm{q}_2 \mathrm{E}_1=\lambda_2 \ell \frac{2 \mathrm{k} \lambda_1}{\mathrm{~d}}$
$\mathrm{~F}_{\mathrm{B}}=\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 \ell}{2 \pi \mathrm{d}}=\frac{\mu_0 \lambda_1 \mathrm{~V} \lambda_2 \mathrm{~V}}{2 \pi \mathrm{d}} \times \ell$
$\mathrm{F}_{\mathrm{E}}=\mathrm{F}_{\mathrm{B}}$
$\Rightarrow \frac{2 \mathrm{k} \lambda_1 \lambda_2 \ell}{\mathrm{d}}=\frac{\mu_0 \lambda_1 \lambda_2 \mathrm{~V}^2 \ell}{2 \pi \mathrm{d}}$
$\Rightarrow \frac{2 \lambda_1 \lambda_2 \ell}{4 \pi \varepsilon_0}=\frac{\mu_0 \lambda_1 \lambda_2 \mathrm{~V}^2 \ell}{2 \pi}$
$\mathrm{V}^2=\frac{1}{\mu_0 \varepsilon_0}=\mathrm{c}^2 \Rightarrow \mathrm{V}=\mathrm{c}$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2022
Two infinite parallel metal planes, contain electric charges with charge densities $+\sigma$ and $-\sigma$ respectively and they are separated by a small distance in air. If the permittivity of air is $\varepsilon_{0}$, then the magnitude of the field between the two planes with its direction will be

Options:
A $\sigma / \varepsilon_{0}$ towards the positively charged plane
B $\sigma / \varepsilon_{0}$ towards the negatively charged plane
C $\sigma /\left(2 \varepsilon_{0}\right)$ towards the positively charged plane
D 0 and towards any direction

The correct answer is:
$\sigma / \varepsilon_{0}$ towards the negatively charged plane

Solution:
$E=\frac{\sigma}{2 \varepsilon_{0}}+\frac{\sigma}{2 \varepsilon_{0}}$
$E=\frac{\sigma}{\varepsilon_{0}}$ towards negative charged plate.

Topic: Physics, Electrostatics, WBJEE, WBJEE 2012
Two long parallel wires separated by $0.1 \mathrm{m}$ carry currents of 1 A and 2 A, respectively in opposite directions. A third current-carrying wire parallel to both of them is placed in the same plane such that it feels no net magnetic force. It is placed at a distance of

Options:
A $0.5 \mathrm{m}$ from the 1 st wire, towards the 2nd wire
B $0.2 \mathrm{m}$ from the 1 st wire, towards the 2nd wire
C $0.1 \mathrm{m}$ from the 1 st wire, away from the 2nd wire
D $0.2 \mathrm{m}$ from the 1 st wire, away from the 2nd wire

The correct answers are:
$0.1 \mathrm{m}$ from the 1 st wire, away from the 2nd wire

Solution:
According to question, the third wire should not feel magnetic force due to wire (1) and
(2). It can be only possible in the condition
that fields due to wire (1) and (2) must act opposite to each other and must be equal. The magnetic field due to long straight wire,
$$
B=\frac{\mu_{0} \cdot I}{2 \pi x}
$$
$\therefore \quad B_{1}=B_{2}$
$$
\frac{\mu_{j_{1}}}{2 \pi x}=\frac{\mu_{d_{2}}}{2 \pi(0 \cdot 1+x)}
$$
Here
$$
\begin{aligned}
i_{1} &=1 \mathrm{A} \\
i_{2} &=2 \mathrm{A} \\
\frac{\mu_{0} \times 1}{2 \pi x} &=\frac{\mu_{0} \times 2}{2 \pi(0 \cdot 1+x)} \\
2 x &=(0 \cdot 1+x) \\
x &=0 \cdot 1 \mathrm{m}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2017
Two massless springs of force constants $\mathrm{K}_1$ and $\mathrm{K}_2$ are joined end to end. The resultant force constant $\mathrm{K}$ of the system is

Options:
A $\mathrm{K}=\frac{\mathrm{K}_1+\mathrm{K}_2}{\mathrm{~K}_1 \mathrm{~K}_2}$
B $\mathrm{K}=\frac{\mathrm{K}_1-\mathrm{K}_2}{\mathrm{~K}_1 \mathrm{~K}_2}$
C $\mathrm{K}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_1+\mathrm{K}_2}$
D $\mathrm{K}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_1-\mathrm{K}_2}$

The correct answer is:
$\mathrm{K}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_1+\mathrm{K}_2}$

Solution:
In series $\mathrm{K}_{\text {eff }}=\frac{K_1 K_2}{K_1+K_2}$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2009
Two metal wires of identical dimensions are connected in series. If $\sigma_{1}$ and $\sigma_{2}$ are the electrical conductivities of the metal wires respectively, the effective conductivity of the combination is

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

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

Solution:
$R_{e q}=R_{1}+R_{2}$
$\frac{1}{\sigma_{e q}} \frac{2 l}{A}=\frac{1}{\sigma_{1}} \frac{1}{A}+\frac{1}{\sigma_{2}} \frac{1}{A}\left[\sigma_{e q}=\frac{2 \sigma_{1} \sigma_{2}}{\sigma_{1}+\sigma_{2}}\right]$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2021
Two metallic spheres of equal outer radii are found to have same moment of inertia about their respective diameters. Then which of the following statement(s) is/are true?

Options:
A The two spheres have equal masses
B The ratio of their masses is nearly $1.67: 1$
C The spheres are made of different materials
D Their rotational kinetic energies will be equal when rotated with equal uniform angular speed about their respective diameters

The correct answers are:
Their rotational kinetic energies will be equal when rotated with equal uniform angular speed about their respective diameters

Solution:
Hint:
Inner radius is not given
$\therefore$ only option (D) is correct

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2020
Two mirrors at an angle $\theta^{\circ}$ produce 5 images of a point. The number of images produced when $\theta$ is decreased to $\theta^{\circ}-30^{\circ}$ is

Options:
A 9
B 10
C 11
D 12

The correct answer is:
11

Solution:
Hints: No. of images $=5$
$\therefore \theta=60^{\circ}$
New angle $=\theta-30^{\circ}=30^{\circ}$. No of images $=\frac{360^{\circ}}{30^{\circ}}-1=11$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2010
Two monochromatic coherent light beams $A$ and $B$ have intensities $L$ and $\frac{L}{4},$ respectively. If these beams are superposed, the maximum and minimum intensities will be

Options:
A $\frac{9 L}{4}, \frac{L}{4}$
B $\frac{5 L}{4}, 0$
C $\frac{5 L}{2}, 0$
D $2 L, \frac{L}{2}$

The correct answer is:
$\frac{9 L}{4}, \frac{L}{4}$

Solution:
For maximum intensity,
$$
I_{\max }=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}} \cos \delta
$$
$$
\begin{aligned}
\delta &=0 \\
I_{\max } &=L+\frac{L}{4}+2 \sqrt{L \times \frac{L}{4}} \cdot \cos 0 \\
&=\frac{5 L}{4}+\frac{2 L}{2} \\
&=\frac{5 L+4 L}{4} \\
&=\frac{9 L}{4} \\
I_{\min } &=I_{1}+I_{2}-2 \sqrt{I_{1} I_{2}} \cos \theta \\
\theta &=0^{\circ} \\
&=L+\frac{L}{4}-2 \sqrt{L \times \frac{L}{4}} \times 1 \\
&=\frac{5 L}{4}-L=\frac{L}{4}
\end{aligned}
$$
Thus, maximum and minimum intensities will be $\left(\frac{9 L}{4}, \frac{L}{4}\right)$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2017
"Two particles $A$ and $B$ are moving as shown. in the figure.

Their total angular momentum about the point $O$ is

Options:
A $9.8 \mathrm{kg} \mathrm{m}^{2} / \mathrm{s}$
B zero
C $52.7 \mathrm{kg} \mathrm{m}^{2} / \mathrm{s}$
D $37.9 \mathrm{kg} \mathrm{m}^{2} / \mathrm{s}$

The correct answer is:
$9.8 \mathrm{kg} \mathrm{m}^{2} / \mathrm{s}$

Solution:
Total angular momentum about $O$ is given as,
$$
\begin{aligned}
\mathbf{L} &=\mathbf{L}_{1}+\mathbf{L}_{2}=m_{1} V_{1} I_{1}+m_{2} V_{2} I_{2} \\
&=-6.5 \times 2.2 \times 1.5+3.1 \times 3.6 \times 2.8 \\
&=-21.45+31.248=9.8 \mathrm{kgm}^{2} / \mathrm{s}
\end{aligned}
$$

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2015
Two particles $A$ and $B$ (both initially at rest) start moving towards each other under a mutual force of attraction. At the instant, when the speed of $A$ is $v$ and the speed of $B$ is $2 v$, the speed of the centre of mass is

Options:
A zero
B $v$
C $\frac{3 v}{2}$
D $-\frac{3 v}{2}$

The correct answer is:
zero

Solution:
In the absence of external force the speed
$$
u_{\mathrm{co}}=0
$$
$F_{ext}=0$ then
$$
v_{\mathrm{cm} .}=0 \text { always }
$$

Topic: Physics, Center of Mass Momentum and Collision, WBJEE, WBJEE 2017
Two particles $A$ and $B$ having different masses are projected from a tower with same speed. $A$ is projected vertically upward and $B$ vertically downward. On reaching the ground

Options:
A velocity of $A$ is greator than that of $B$
B velocity of $B$ is greater than that of $A$
C both $A$ and $B$ attain the same velocity
D the particle with the larger mass attains higher velocily

The correct answer is:
both $A$ and $B$ attain the same velocity

Solution:
The situation is diagrammatically as below

When $A$ is projected with a vertical speed after sometime it comes on the same level with same
speed as it was projected. Now, the downward speeds of $A$ and $B$ at the level- $X$ is same. So, on reaching the ground, velocity of $A$ and $B$ are same.

Topic: Physics, Motion In One Dimension, WBJEE, WBJEE 2015
Two particles, $A$ and $B$, having equal charges, after being accelerated through the same potential difference enter into a region of uniform magnetic field and the particles describe circular paths of radii $R_{1}$ and $R_{2}$ resporctively. The ratio of the masses of $A$ and $B$ is

Options:
A $\sqrt{R_{1}} / R_{2}$
B $R_{1} / R_{2}$
C $\left(R_{1} / R_{2}\right)^{2}$
D $\left(R_{2} / R_{1}\right)^{2}$

The correct answer is:
$\left(R_{1} / R_{2}\right)^{2}$

Solution:
Radius of circular path followed by charged particle is given by
$$
R=\frac{m v}{q B}=\frac{\sqrt{2 m K}}{q B} \quad[\because p=m v=\sqrt{2 m K}]
$$
where, $K$ is kinetic energy of particle. Charged particle $q$ is accelerated through some potential difference $V$, such that kinetic energy of particle is
$\therefore$
$$
\begin{aligned}
K &=q V \\
R &=\frac{\sqrt{2 m q V}}{q B}
\end{aligned}
$$
As the two charged particles of same magnitude and being accelerated through same potential, enters into a uniform magnetic field region, then $R \propto \sqrt{m}$
So,
$$
\frac{R_{1}}{R_{2}}=\sqrt{\frac{m_{A}}{m_{B}}} \Rightarrow \frac{m_{A}}{m_{B}}=\left(\frac{R_{1}}{R_{2}}\right)^{2}
$$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2015
Two particles $A$ and $B$ move from rest along a straight line with constant accelerations $f$ and $f$ ' respectively. If $A$ takes $\mathrm{m}$ sec. more than that of $\mathrm{B}$ and describes $\mathrm{n}$ units more than that of $\mathrm{B}$ in acquiring the same velocity, then

Options:
A $\left(f+f^{\prime}\right) m^{2}=$ ff'n
B $\left(f-f f^{\prime}\right) m^{2}=f f^{\prime} n$
C $\left(f^{\prime}-f\right) n=\frac{1}{2} f f^{\prime} m^{2}$
D $\frac{1}{2}\left(\mathrm{f}+\mathrm{f}^{\prime}\right) \mathrm{m}=\mathrm{ff}^{\prime} \mathrm{n}^{2}$

The correct answer is:
$\left(f^{\prime}-f\right) n=\frac{1}{2} f f^{\prime} m^{2}$

Solution:
$\begin{array}{llll}A:\quad u=0 & a_{1}=f & t_{1}=t+m & s_{1}=n+s \\ B: \quad u=0 & a_{2}=f^{\prime} & t_{2}=t & s_{2}=s\end{array}$
$\mathrm{s}+\mathrm{n}=\frac{1}{2} \mathrm{f} .(\mathrm{t}+\mathrm{m})^{2} \ldots \ldots(\mathrm{i}) \quad$ and $\mathrm{s}=\frac{1}{2} \mathrm{f}^{\prime}(\mathrm{t})^{2} \ldots(\mathrm{ii}) \quad \therefore \mathrm{f}^{\prime}(\mathrm{t})^{2}+\mathrm{n}=\frac{1}{2} \mathrm{f}(\mathrm{t}+\mathrm{m})^{2} \ldots (iii)$
$\begin{array}{l}
v_{1}=u_{1}+a_{1} t_{1}=0+f .(t+m) \\
v_{2}=u_{2}+a_{2} t_{2}=0+f^{\prime} . t \\
\therefore f(t+m)=f^{\prime} t \\
t=\frac{f m}{f^{\prime}-f}
\end{array}$
from (iii)
$\frac{1}{2} f^{\prime} \cdot\left(\frac{f m}{f^{\prime}-f}\right)^{2}+n=\frac{1}{2} f\left(\frac{f m}{f^{\prime}-f}+m\right)^{2}$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2021