datavorous/entrance-exam-dataset huggingface dataset:
A point mass of $1 \mathrm{~kg}$ collides elastically with a stationary point mass of $5 \mathrm{~kg}$. After their collision, the $1 \mathrm{~kg}$ mass reverses its direction and moves with a speed of $2 \mathrm{~ms}^{-1}$. Which of the following statement(s) is/are correct for the system of these two masses?

Options:
A
Total momentum of the system is $3 \mathrm{~kg}-\mathrm{ms}^{-1}$

B
Momentum of $5 \mathrm{~kg}$ mass after collision is $4 \mathrm{~kg}-\mathrm{ms}^{-1}$

C
Kinetic energy of the centre of mass is $0.75 \mathrm{~J}$

D
Total kinetic energy of the system is $4 \mathrm{~J}$

The correct answers are:
Total momentum of the system is $3 \mathrm{~kg}-\mathrm{ms}^{-1}$
,
Kinetic energy of the centre of mass is $0.75 \mathrm{~J}$

Solution:
$$
\begin{aligned}
& v_1^{\prime}=\left(\frac{m_1-m_5}{m_1+m_5}\right) v_1+\left(\frac{2 m_5}{m_1+m_5}\right) v_5 \\
& -2=\left(\frac{1-5}{1+5}\right) v_1+0 \quad\left(\text { as } v_5=0\right) \\
& \therefore \quad v_1=3 \mathrm{~ms}^{-1} \\
& v_5^{\prime}=\left(\frac{m_5-m_1}{m_1+m_5}\right) v_5+\left(\frac{2 m_1}{m_1+m_2}\right) v_1 \\
& =0+\left(\frac{2 \times 1}{6}\right)(3)=1 \mathrm{~ms}^{-1} \\
& P_{\mathrm{CM}}=P_i=(1)(3) \\
& =3 \frac{\mathrm{kg}-\mathrm{m}}{\mathrm{s}} \\
& P_5{ }^{\prime}=(5)(1)=5 \frac{\mathrm{kg}-\mathrm{m}}{5} \\
& K_{\mathrm{CM}}=\frac{P_{\mathrm{CM}}^2}{2 M_{\mathrm{CM}}} \\
& =\frac{9}{2 \times 6}=0.75 \mathrm{~J} \\
& K_{\text {total }}=\frac{1}{2} \times 1 \times(3)^2 \\
& =4.5 \mathrm{~J} \\
&
\end{aligned}
$$
$\therefore$ correct options are (a) and (c).

Topic: Physics, Center of Mass Momentum and Collision, JEE Advanced, JEE Advanced 2010 (Paper 1)
A point object is placed at distance of $20 \mathrm{~cm}$ from a thin planoconvex lens of focal length $15 \mathrm{~cm}$. The plane surface of the lens is now silvered. The image created by the system is at

Options:
A
$60 \mathrm{~cm}$ to the left of the system

B
$60 \mathrm{~cm}$ to the right of the system

C
$12 \mathrm{~cm}$ to the left of the system

D
$12 \mathrm{~cm}$ to the right of the system

The correct answer is:
$12 \mathrm{~cm}$ to the left of the system

Solution:
Refraction from lens
$$
\begin{aligned}
\frac{1}{v_1}-\frac{1}{-20} & =\frac{1}{15} \\
v & =60 \mathrm{~cm}
\end{aligned}
$$
i.e., first image is formed at $60 \mathrm{~cm}$ to the right of lens system.
Reflection from mirror
After reflection from the mirror, the second image will be formed at a distance of $60 \mathrm{~cm}$ to the left of lens system.
Refraction from lens
$$
\begin{aligned}
\frac{1}{v_3}-\frac{1}{60} & =\frac{1}{15} \\
v_3 & =12 \mathrm{~cm}
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2006
A police car with a siren of frequency $8 \mathrm{kHz}$ is moving with uniform velocity $36 \mathrm{~km} / \mathrm{h}$ towards a tall building which reflects the sound waves. The speed of sound in air is $320 \mathrm{~m} / \mathrm{s}$. The frequency of the siren heard by the car driver is

Options:
A
$8.50 \mathrm{kHz}$

B
$8.25 \mathrm{kHz}$

C
$7.75 \mathrm{kHz}$

D
$7.50 \mathrm{kHz}$

The correct answer is:
$8.50 \mathrm{kHz}$

Solution:
Apparent frequency of sound heard by car driver (observer) reflected from the building will be
$$
\begin{aligned}
f^{\prime} & =f\left(\frac{v+v_0}{v-v_s}\right) \\
& =8\left(\frac{320+10}{320-10}\right) \\
& =8.5 \mathrm{kHz}
\end{aligned}
$$
$\therefore$ Correct option is (a).
Analysis of Question
(i) Question is simple.
(ii) Driver will listen two sounds, direct and reflected. Direct sound will be of $8 \mathrm{kHz}$ as driver has no relative motion with the car. But reflected sound is of increased frequency because driver and image of car both are approaching towards each other.

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2011 (Paper 1)
A positron is emitted from ${ }_{11}^{23} \mathrm{Na}$. The ratio of the atomic mass and atomic number of the resulting nuclide is

Options:
A
22/10

B
22/11

C
23/10

D
$23 / 12$

The correct answer is:
23/10

Solution:
$$
{ }_{11}^{23} \mathrm{Na} \longrightarrow{ }_{10}^{23} \mathrm{Na}+{ }_{-1}^0 e
$$

Topic: Chemistry, Chemical Kinetics, JEE Advanced, JEE Advanced 2007 (Paper 2)
A proton is fired from very far away towards a nucleus with charge $Q=120 e$, where $e$ is the electronic charge. It makes a closest approach of $10 \mathrm{fm}$ to the nucleus. The de Broglie wavelength (in units of $\mathrm{fm}$ ) of the proton at its start is: (take the proton mass, $m_{p}=(5 / 3) \times 10^{-27} \mathrm{~kg} ; h / e=4.2 \times$ $10^{-15} \mathrm{~J} . \mathrm{S} / \mathrm{C}$;

$\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{~m} / \mathrm{F} ; 1 \mathrm{fm}=10^{-15} \mathrm{~m}$

Options:

The correct answer is:
7

Solution:
From energy conservation Loss in K.E. of proton = gain in potential energy of the proton - nucleus system

$\begin{array}{c}

\frac{1}{2} m v^{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r} \quad \therefore \frac{p^{2}}{2 m}=\frac{1}{4 \pi \in_{0}} \frac{q_{1} q_{2}}{r} \\

\therefore \frac{1}{2 m}\left(\frac{h^{2}}{\lambda^{2}}\right)=\frac{1}{4 \pi \in_{0}} \frac{q_{1} q_{2}}{r} \quad \therefore \lambda=\sqrt{\frac{4 \pi \epsilon_{0} r \cdot h^{2}}{q_{1} q_{2}(2 m)}}

\end{array}$

Putting the values of $4 \pi \varepsilon_{0}, r, h, q_{1}, q_{2}$ and $m$ we get, deBroglie wavelength of proton, $\lambda=7 \mathrm{fm}$

Topic: Physics, Dual Nature of Matter, JEE Advanced, JEE Advanced 2012 (Paper 1)
A radioactive sample $S_1$ having an activity of $5 \mu \mathrm{Ci}$ has twice the number of nuclei as another sample $S_2$ which has an activity of $10 \mu \mathrm{Ci}$. The half lives of $S_1$ and $S_2$ can be

Options:
A
20 yr and 5 yr, respectively

B
20 yr and 10 yr, respectively

C
10 yr each

D
5 yr each

The correct answer is:
20 yr and 5 yr, respectively

Solution:
Activity of $S_1=\frac{1}{2}$ (activity of $S_2$ )
or
$$
\lambda_2 N_1=\frac{1}{2}\left(\lambda_2 N_2\right) \text { or } \frac{\lambda_1}{\lambda_2}=\frac{N_2}{2 N_1}
$$
or
$$
\begin{aligned}
& \frac{T_1}{T_2}=\frac{2 N_1}{N_2}\left(T=\text { half-life }=\frac{\ln 2}{\lambda}\right) \\
& N_1=2 N_2 \\
& \frac{T_1}{T_2}=4
\end{aligned}
$$
Given
$\therefore$ correct option is (a).

Topic: Physics, Nuclear Physics, JEE Advanced, JEE Advanced 2008 (Paper 2)
A ray of light travelling in water is incident on its surface open to air. The angle of incidence is $\theta$, which is less than the critical angle. Then there will be

Options:
A
only a reflected ray and no refracted ray

B
only a refracted ray and no reflected ray

C
a reflected ray and refracted ray and the angle between them would be less than $180^{\circ}-2 \theta$

D
a reflected ray and a refracted ray and the angle between them would be greater than $180^{\circ}-2 \theta$

The correct answer is:
a reflected ray and refracted ray and the angle between them would be less than $180^{\circ}-2 \theta$

Solution:
Since $\theta < \theta_c$, both reflection and refraction will take place. From the figure we can see that angle between reflected and refracted rays $\alpha$ is less than $180^{\circ}-2 \theta$.
$\therefore$ Option (c) is correct.

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2007 (Paper 1)
A ray $O P$ of monochromatic light is incident on the face $A B$ of prism $A B C D$ near vertex $B$ at an incident angle of $60^{\circ}$ (see figure). If the refractive index of the material of the prism is $\sqrt{3}$, which of the following is (are) correct?

Options:
A
The ray gets totally internally reflected at face $C D$

B
The ray comes out through face $A D$

C
The angle between the incident ray and the emergent ray is $90^{\circ}$

D
The angle between the incident ray and the emergent ray is $120^{\circ}$

The correct answers are:
The ray gets totally internally reflected at face $C D$
,
The ray comes out through face $A D$
,
The angle between the incident ray and the emergent ray is $90^{\circ}$

Solution:
$$
\begin{aligned}
& \sqrt{3}=\frac{\sin 60^{\circ}}{\sin r} \\
& \therefore r=30^{\circ}
\end{aligned}
$$
$$
\theta_C=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)
$$
or $\sin \theta_C=\frac{1}{\sqrt{3}}=0.577$
At point $Q$, angle of incidence inside the prism is $i=45^{\circ}$.
Since $\sin i=\frac{1}{\sqrt{2}}$ is greater than $\sin \theta_C=\frac{1}{\sqrt{2}}$, ray gets totally internally reflected at face $C D$. Path of ray of light after point $Q$ is shown in figure.
From the figure, we can see that angle between incident ray $O P$ and emergent ray $R S$ is $90^{\circ}$.
Therefore, correct options are (a), (b) and (c).

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2010 (Paper 1)
A real gas behaves like an ideal gas if its

Options:
A
pressure and temperature are both high

B
pressure and temperature are both low

C
pressure is high and temperature is low

D
pressure is low and temperature is high

The correct answer is:
pressure is low and temperature is high

Solution:
A real gas behaves like an ideal gas at low pressure and high temperature. $\therefore$ correct option is (d).

Topic: Physics, Kinetic Theory of Gases, JEE Advanced, JEE Advanced 2010 (Paper 1)
A resistance of $2 \Omega$ is connected across one gap of a meter-bridge (the length of the wire is $100 \mathrm{~cm}$ ) and an unknown resistance, greater than $2 \Omega$, is connected across the other gap. When these resistances are interchanged, the balance point shifts by $20 \mathrm{~cm}$. Neglecting any corrections, the unknown resistance is

Options:
A
$3 \Omega$

B
$4 \Omega$

C
$5 \Omega$

D
$6 \Omega$

The correct answer is:
$3 \Omega$

Solution:
$$
\begin{aligned}
& R & >2 \Omega \\
\therefore & 100-x & >x \\
\text { Applying } & \frac{P}{Q} & =\frac{R}{S}
\end{aligned}
$$
We have
$$
\begin{aligned}
& \frac{2}{R}=\frac{x}{100-x} \\
& \frac{R}{2}=\frac{x+20}{80-x}
\end{aligned}
$$
Solving Eqs. (i) and (ii), we get
$$
R=3 \Omega
$$
$\therefore$ Answer is (a).

Topic: Physics, Current Electricity, JEE Main
A resistance of $2 \Omega$ is connected across one gap of a meter-bridge (the length of the wire is $100 \mathrm{~cm}$ ) and an unknown resistance, greater than $2 \Omega$, is connected across the other gap. When these resistances are interchanged, the balance point shifts by $20 \mathrm{~cm}$. Neglecting any corrections, the unknown resistance is

Options:
A
$3 \Omega$

B
$4 \Omega$

C
$5 \Omega$

D
$6 \Omega$

The correct answer is:
$3 \Omega$

Solution:
$$
\begin{aligned}
& R & >2 \Omega \\
\therefore & 100-x & >x \\
\text { Applying } & \frac{P}{Q} & =\frac{R}{S}
\end{aligned}
$$
We have
$$
\begin{aligned}
& \frac{2}{R}=\frac{x}{100-x} \\
& \frac{R}{2}=\frac{x+20}{80-x}
\end{aligned}
$$
Solving Eqs. (i) and (ii), we get
$$
R=3 \Omega
$$
$\therefore$ Answer is (a).

Topic: Physics, Current Electricity, JEE Advanced, JEE Advanced 2007 (Paper 1)
A satellite is moving with a constant speed $v$ in a circular orbit about the earth. An object of mass $m$ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is

Options:
A
$\frac{1}{2} m v^2$

B
$m v^2$

C
$\frac{3}{2} m v^2$

D
$2 m v^2$

The correct answer is:
$m v^2$

Solution:
In circular orbit of a satellite, potential energy
$$
\begin{aligned}
& =-2 \times(\text { kinetic energy) } \\
& =-2 \times \frac{1}{2} m v^2=-m v^2
\end{aligned}
$$
Just to escape from the gravitational pull, its total mechanical energy should be zero. Therefore, its kinetic energy should be $+m v^2$.
$\therefore$ Correct answer is (b).
Analysis of Question
(i) Question is moderately difficult.
(ii) In circular orbit of satellite $U=-\frac{G M m}{r}$ and $K=\frac{G M m}{2 r}$ or $U=-2 K$
Here, $U$ is potential energy and $K$ is kinetic energy.

Topic: Physics, Gravitation, JEE Advanced, JEE Advanced 2011 (Paper 2)
A series $R$ - $C$ circuit is connected to $A C$ voltage source. Consider two cases; $(A)$ when $C$ is without a dielectric medium and $(B)$ when $C$ is filled with dielectric of constant 4. The current $I_R$ through the resistor and voltage $V_C$ across the capacitor are compared in the two cases. Which of the following is/are true?

Options:
A
$I_R^A>I_R^B$

B
$I_R^A < I_R^B$
C
$V_C^A>V_C^B$

D
$V_C^A < V_C^B$

The correct answers are:
$I_R^A < I_R^B$,
$V_C^A>V_C^B$

Solution:
$Z=\sqrt{R^2+X_C^2}=\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}$
In case (b) capacitance $C$ will be more.
Therefore, impedance $Z$ will be less.
Hence, current will be more.
$\therefore$ Option (b) is correct.
Further, $\quad \begin{aligned} V_C & =\sqrt{V^2-V_R^2} \\ & =\sqrt{V^2-(I R)^2}\end{aligned}$
In case (b), since current $I$ is more.
Therefore, $V_C$ will be less.
$\therefore$ Option (c) is correct.
$\therefore$ Correct options are (b) and (c).
Analysis of Question
(i) Question is moderately difficult.
(ii) In my opinion problems of alternating currents are not very difficult.
(iii) Topic of $A C$ is small. One can feel comfortable in this topic by putting less efforts.

Topic: Physics, Alternating Current, JEE Advanced, JEE Advanced 2011 (Paper 2)
A series $R-C$ combination is connected to an $\mathrm{AC}$ voltage of angular frequency $\omega=500 \mathrm{rad} / \mathrm{s}$. If the impedance of the $R-C$ circuit is $R \sqrt{1.25}$, the time constant (in millisecond) of the circuit is

Options:

The correct answer is:
4

Solution:
$Z=\sqrt{R^2+X_C^2}=R \sqrt{1.25}$
$$
\therefore \quad R^2+X_C^2=1.25 R^2
$$
or $\quad X_C=\frac{R}{2}$
or $\quad \frac{1}{\omega C}=\frac{R}{2}$
$\begin{aligned} \therefore \text { Time constant } & =C R=\frac{2}{\omega} \\ & =\frac{2}{500} \mathrm{~s}=4 \mathrm{~ms}\end{aligned}$
$\therefore$ Answer is 4 .
Analysis of Question
(i) Question is very simple.
(ii) I think this is one of the simplest formula based question of this paper.

Topic: Physics, Alternating Current, JEE Advanced, JEE Advanced 2011 (Paper 2)
A signal which can be green or red with probability $\frac{4}{5}$ and $\frac{1}{5}$ respectively, is received by station $A$ and then transmitted to station B. The probability of each station receiving the signal correctly is $\frac{3}{4}$. If the signal received at station $B$ is green, then the probability that the original signal green is

Options:
A
$\frac{3}{5}$

B
$\frac{6}{7}$

C
$\frac{20}{23}$

D
$\frac{9}{20}$

The correct answer is:
$\frac{20}{23}$

Solution:
From the tree-diagram it follows that
$$
\begin{gathered}
P\left(B_G \mid G\right)=\frac{10}{16}=\frac{5}{8} \\
\therefore \quad P\left(B_G \cap G\right)=\frac{5}{8} \times \frac{4}{5}=\frac{1}{2} \\
P\left(G \mid B_G\right)=\frac{\frac{1}{2}}{P\left(B_G\right)}=\frac{1}{2} \times \frac{80}{46}=\frac{20}{23}
\end{gathered}
$$

Topic: Mathematics, Probability, JEE Advanced, JEE Advanced 2010 (Paper 2)
A silver sphere of radius $1 \mathrm{~cm}$ and work function $4.7 \mathrm{eV}$ is suspended from an insulating thread in free-space. It is under continuous illumination of $200 \mathrm{~nm}$ wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is $A \times 10^Z$ (where $1 < A < 10$ ). The value of $Z$ is

Options:

The correct answer is:
7

Solution:
Photoemission will stop when potential on silver sphere becomes equal to the stopping potential.
$\therefore \quad \frac{h c}{\lambda}-W=e V_0$
Here, $\quad V_0=\frac{1}{4 \pi \varepsilon_0} \frac{n e}{r}$
$$
\begin{aligned}
& \therefore \quad\left(\frac{1240}{1200} \mathrm{eV}\right)-(4.7 \mathrm{eV}) \\
&=\frac{9 \times 10^9 \times n \times 1.6 \times 10^{-19}}{10^{-2}} \\
&(6.2-4.7)=\frac{9 \times 10^9 \times n \times 1.6 \times 10^{-19}}{10^{-2}} \\
& \text { or } \quad n= \frac{1.5 \times 10^{-2}}{9 \times 1.6 \times 10^{-10}}=1.04 \times 10^7
\end{aligned}
$$
$\therefore$ Answer is 7 .
Analysis of Question
(i) Question is moderately difficult.
(ii) In $\mathrm{eV}_0$ if we substitute the value of $e=1.6 \times 10^{-19} \mathrm{C}$, then answer is in J. If we leave it then, answer is in $e \mathrm{~V}$.

Topic: Physics, Dual Nature of Matter, JEE Advanced, JEE Advanced 2011 (Paper 2)
A simple telescope used to view distant objects has eyepiece and objective lens of focal lengths $f_e$ and $f_0$, respectively. Then

Options:
A
(a) p, (b) r, (c) r, (d) p,q,s

B
(a) q, (b) r, (c) s, (d) p,s

C
(a) p, (b) p, r, (c) q, (d) p,q

D
(a) q, (b) s, (c) r, (d) p,q,s

The correct answer is:
(a) p, (b) r, (c) r, (d) p,q,s

Solution:
(a) $\rightarrow p$
(b) $\rightarrow r$
(c) $\rightarrow r$
(d) $\rightarrow$ p, q, s

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2006
A small block is connected to one end of a massless spring of un-stretched length $4.9 \mathrm{~m}$. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by $0.2 \mathrm{~m}$ and released from rest at $t=0$. It then executes simple harmonic motion with angular frequency $\omega=\pi / 3 \mathrm{rad} / \mathrm{s}$. Simultaneously at $t=0$, a small pebble is projected with speed $v$ form point $P$ at an angle of $45^{\circ}$ as shown in the figure. Point $P$ is at a horizontal distance of $10 \mathrm{~m}$ from $O$. If the pebble hits the block at $t=1 \mathrm{~s}$, the value of $v$ is (take $g$ $=10 \mathrm{~m} / \mathrm{s}^{2}$ )

Options:
A $\sqrt{50} \mathrm{~m} / \mathrm{s}$
B $\sqrt{51} \mathrm{~m} / \mathrm{s}$
C $\sqrt{52} \mathrm{~m} / \mathrm{s}$
D $\sqrt{53} \mathrm{~m} / \mathrm{s}$

The correct answer is:
$\sqrt{50} \mathrm{~m} / \mathrm{s}$

Solution:
Time of flight of projectile,

$T=\frac{2 V \sin \theta}{g}$

$\therefore \quad 1=\frac{2 V \sin 45^{\circ}}{g} \quad \therefore \quad V=\sqrt{50} \mathrm{~ms}^{-1}$

Hence pebble is projected with a speed $V=\sqrt{50} \mathrm{~ms}^{-1}$

Topic: Physics, Motion In Two Dimensions, JEE Main
A small block is connected to one end of a massless spring of un-stretched length $4.9 \mathrm{~m}$. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by $0.2 \mathrm{~m}$ and released from rest at $t=0$. It then executes simple harmonic motion with angular frequency $\omega=\pi / 3 \mathrm{rad} / \mathrm{s}$. Simultaneously at $t=0$, a small pebble is projected with speed $v$ form point $P$ at an angle of $45^{\circ}$ as shown in the figure. Point $P$ is at a horizontal distance of $10 \mathrm{~m}$ from $O$. If the pebble hits the block at $t=1 \mathrm{~s}$, the value of $v$ is (take $g$ $=10 \mathrm{~m} / \mathrm{s}^{2}$ )

Options:
A $\sqrt{50} \mathrm{~m} / \mathrm{s}$
B $\sqrt{51} \mathrm{~m} / \mathrm{s}$
C $\sqrt{52} \mathrm{~m} / \mathrm{s}$
D $\sqrt{53} \mathrm{~m} / \mathrm{s}$

The correct answer is:
$\sqrt{50} \mathrm{~m} / \mathrm{s}$

Solution:
Time of flight of projectile,

$T=\frac{2 V \sin \theta}{g}$

$\therefore \quad 1=\frac{2 V \sin 45^{\circ}}{g} \quad \therefore \quad V=\sqrt{50} \mathrm{~ms}^{-1}$

Hence pebble is projected with a speed $V=\sqrt{50} \mathrm{~ms}^{-1}$

Topic: Physics, Motion In Two Dimensions, JEE Advanced, JEE Advanced 2012 (Paper 1)
A small block of mass of $0.1 \mathrm{~kg}$ lies on a fixed inclined plane PQ which makes an angle $\theta$ with the horizontal. A horizontal force of $1 \mathrm{~N}$ acts on the block through its centre of mass as shown in the figure.

The block remains stationary if (take $g=10 \mathrm{~m} / \mathrm{s}^{2}$ )

Options:
A $\theta=45^{\circ}$
B $\theta>45^{\circ}$ and a frictional force acts on the block towards P.
C $\theta>45^{\circ}$ and a frictional force acts on the block towards Q.
D $\theta < 45^{\circ}$ and a frictional force acts on the block towards Q.

The correct answers are:
$\theta=45^{\circ}$, $\theta>45^{\circ}$ and a frictional force acts on the block towards Q.

Solution:
The various forces acting on the block are as shown in the figure.

When $\theta=45^{\circ}, \sin \theta=\cos \theta$

The block will remain stationary and the frictional force is zero.

When $\theta>45^{\circ}, \sin \theta>\cos \theta$

Therefore a frictional force acts towards $Q$.

When $\theta < 45^{\circ}, \cos \theta>\sin \theta$

Therefore a frictional force acts towards $P$.

Topic: Physics, Laws of Motion, JEE Advanced, JEE Advanced 2012 (Paper 1)
A small circular loop of area A and resistance R is fixed on a horizontal xy-plane with the center of the loop always on the axis n^ of a long solenoid. The solenoid has m turns per unit length and carries current I counter clockwise as shown in the figure. The magnetic field due to the solenoid is in n^ direction. List-I gives time dependences of n^ in terms of a constant angular frequency ω. List-II gives the torques experienced by the circular loop at time t=π6ω, Let α=A2μ02m2I2ω2R.

List-I	List-II

(i)	12sinωtj^+cosωtk^	(p)	0

(ii)	12sinωti^+cosωtj^	(q)	-α4i^

(iii)	12sinωti^+cosωtk^	(r)	3α4i^

(iv)	12cosωtj^+sinωtk^	(s)	α4j^

(t)	-3α4i^

Which one of the following options is correct?

Options:
A i→q,ii→p,iii→s,iv→t
B i→s,ii→t,iii→q,iv→p
C i→q,ii→p,iii→s,iv→r
D i→t,ii→q,iii→p,iv→r

The correct answer is:
i→q,ii→p,iii→s,iv→r

Solution:
Note: This question was given bonus by JEE council.

The magnetic field in a solenoid is given by, B→=μ0mIn^. Therefore, the flux through the loop will be,

ϕ=Ak^·μ0mIn^

For case I:

ϕ=BA2cosωt

The emf will be,

ε=BAω2sinωt

Therefore, the current in the loop will be,

i=BAω2Rsinωt

The magnetic moment of the loop will be,

m→=iAk^=BA2ω2Rsinωtk^

Therefore, the torque

τ→=m→×B→=B2A2ω2Rsinωtk^×n^

⇒τ→=-B2A2ω2Ri^sin2ωt

⇒τ→=-B2A2ω2Rsin2π6=-α4i^

Hence, I→q

For case II

ϕ=0

Therefore, we can directly write τ=0

Hence II→p

Following the same process as in case I for case III

ϕ=BA2cosωt

i=BAω2Rsinωt

m→=BA2ω2Rsinωtk^

τ→=m→×B→=B2A2ω2×2Rsinωtk^×sinωti^+cosωtk^

τ→=B2A2ωsinωt2Rsinωtj^

τ→=B2A2ω2Rsin2ωtj^

τ→=α4j^

Hence, III→s

Following the same process as in case I for case IV

ϕ=BA2sinωt

i=-BAω2Rcosωt

m→=-BA2ω2Rcosωtk^

τ→=m→×B→=-B2A2ω2Rk^×j^cos2ωt

τ→=+B2A2ω2Ri^·cos2π6

τ→=+34αi^

Hence, IV→r

Topic: Physics, Magnetic Effects of Current, JEE Main
A small circular loop of area A and resistance R is fixed on a horizontal xy-plane with the center of the loop always on the axis n^ of a long solenoid. The solenoid has m turns per unit length and carries current I counter clockwise as shown in the figure. The magnetic field due to the solenoid is in n^ direction. List-I gives time dependences of n^ in terms of a constant angular frequency ω. List-II gives the torques experienced by the circular loop at time t=π6ω, Let α=A2μ02m2I2ω2R.

List-I	List-II

(i)	12sinωtj^+cosωtk^	(p)	0

(ii)	12sinωti^+cosωtj^	(q)	-α4i^

(iii)	12sinωti^+cosωtk^	(r)	3α4i^

(iv)	12cosωtj^+sinωtk^	(s)	α4j^

(t)	-3α4i^

Which one of the following options is correct?

Options:
A i→q,ii→p,iii→s,iv→t
B i→s,ii→t,iii→q,iv→p
C i→q,ii→p,iii→s,iv→r
D i→t,ii→q,iii→p,iv→r

The correct answer is:
i→q,ii→p,iii→s,iv→r

Solution:
Note: This question was given bonus by JEE council.

The magnetic field in a solenoid is given by, B→=μ0mIn^. Therefore, the flux through the loop will be,

ϕ=Ak^·μ0mIn^

For case I:

ϕ=BA2cosωt

The emf will be,

ε=BAω2sinωt

Therefore, the current in the loop will be,

i=BAω2Rsinωt

The magnetic moment of the loop will be,

m→=iAk^=BA2ω2Rsinωtk^

Therefore, the torque

τ→=m→×B→=B2A2ω2Rsinωtk^×n^

⇒τ→=-B2A2ω2Ri^sin2ωt

⇒τ→=-B2A2ω2Rsin2π6=-α4i^

Hence, I→q

For case II

ϕ=0

Therefore, we can directly write τ=0

Hence II→p

Following the same process as in case I for case III

ϕ=BA2cosωt

i=BAω2Rsinωt

m→=BA2ω2Rsinωtk^

τ→=m→×B→=B2A2ω2×2Rsinωtk^×sinωti^+cosωtk^

τ→=B2A2ωsinωt2Rsinωtj^

τ→=B2A2ω2Rsin2ωtj^

τ→=α4j^

Hence, III→s

Following the same process as in case I for case IV

ϕ=BA2sinωt

i=-BAω2Rcosωt

m→=-BA2ω2Rcosωtk^

τ→=m→×B→=-B2A2ω2Rk^×j^cos2ωt

τ→=+B2A2ω2Ri^·cos2π6

τ→=+34αi^

Hence, IV→r

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2022 (Paper 1)