datavorous/entrance-exam-dataset huggingface dataset:
A student performed the experiment to measure the speed of sound in air using resonance air-column method. Two resonances in the air-column were obtained by lowering the water level. The resonance with the shorter air-column is the first resonance and that with the longer air column is the second resonance. Then,

Options:
A
the intensity of the sound heard at the first resonance was more than that at the second resonance

B
the prongs of the tuning fork were kept in a horizontal place above the resonance tube

C
the amplitude of vibration of the ends of the prongs is typically around $1 \mathrm{~cm}$

D
the length of the air-column at the first resonance was somewhat shorter than $1 / 4$ th of the wavelength of the sound in air

The correct answers are:
the intensity of the sound heard at the first resonance was more than that at the second resonance
,
the length of the air-column at the first resonance was somewhat shorter than $1 / 4$ th of the wavelength of the sound in air

Solution:
$1 < \frac{\lambda}{4}$
Further, larger the length of air column, feebler is the intensity.

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2009 (Paper 2)
A student performs a titration with different burettes and finds titre values of $25.2 \mathrm{~mL}, 25.25 \mathrm{~mL}$, and $25.0 \mathrm{~mL}$.
The number of significant figures in the average titre value is

Options:

The correct answer is:
3

Solution:
Average
$$
\begin{aligned}
& =\frac{25.2+25.25+25.0}{3}=\frac{75.45}{3} \\
& =25.15=25.2 \mathrm{~mL}
\end{aligned}
$$
Number of significant figure is 3 .
Basics in chemistry
Rule application
III (Difficulty level is high because of shear ignorance of such topics)

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Advanced, JEE Advanced 2010 (Paper 1)
A student performs an experiment for determination of $g\left(=\frac{4 \pi^2 l}{T^2}\right), l \approx 1 \mathrm{~m}$, and he commits an error of $\Delta l$. For $T$ he takes the time of $n$ oscillations with the stop watch of least count $\Delta T$ and he commits a human error of $0.1 \mathrm{~s}$. For which of the following data, the measurement of $g$ will be most accurate?

Options:
A
$\Delta L=0.5, \Delta T=0.1, n=20$

B
$\Delta L=0.5, \Delta T=0.1, n=50$

C
$\Delta L=0.5, \Delta T=0.01, n=20$

D
$\Delta L=0.1, \Delta T=0.05, n=50$

The correct answer is:
$\Delta L=0.1, \Delta T=0.05, n=50$

Solution:
$\frac{\Delta g}{g}=\frac{\Delta l}{l}+2 \frac{\Delta T}{n}$
In option (d) error in $\Delta g$ is minimum and number of observations made are maximum. Hence, in this case error in $g$ will be minimum.

Topic: Physics, Mathematics in Physics, JEE Advanced, JEE Advanced 2006
A student performs an experiment to determine the Young's modulus of a wire, exactly $2 \mathrm{~m}$ long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be $0.8 \mathrm{~mm}$ with an uncertainty of $\pm 0.05 \mathrm{~mm}$ at a load of exactly $1.0 \mathrm{~kg}$. The student also measures the diameter of the wire to be $0.4 \mathrm{~mm}$ with an uncertainty of $\pm 0.01 \mathrm{~mm}$. Take $g=9.8 \mathrm{~m} / \mathrm{s}^2$ (exact). The Young's modulus obtained from the reading is

Options:
A
$(2.0 \pm 0.3) 10^{11} \mathrm{~N} / \mathrm{m}^2$

B
$(2.0 \pm 0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

C
$(2.0 \pm 0.1) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

D
$(2.0 \pm 0.05) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

The correct answer is:
$(2.0 \pm 0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

Solution:
$\begin{aligned} Y & =\frac{F L}{A l}=\frac{4 F L}{\pi d^2 l} \\ & =\frac{(4)(1.0 \times 9.8)(2)}{\pi\left(0.4 \times 10^{-3}\right)^2\left(0.8 \times 10^{-3}\right)}=20 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\end{aligned}$
Further $\quad \frac{\Delta Y}{Y}=2\left(\frac{\Delta d}{d}\right)+\left(\frac{\Delta l}{l}\right)$
$$
\begin{aligned}
\therefore \quad \Delta Y & =\left\{2\left(\frac{\Delta d}{d}\right)+\left(\frac{\Delta l}{l}\right)\right\} Y=\left\{2 \times \frac{0.01}{0.4}+\frac{0.05}{0.8}\right\} \times 2.0 \times 10^{11} \\
& =0.2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \text { or }(Y+\Delta Y)=(2+0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^2
\end{aligned}
$$
The correct option is (b).

Topic: Physics, Mathematics in Physics, JEE Advanced, JEE Advanced 2007 (Paper 2)
A student uses a simple pendulum of exactly $1 \mathrm{~m}$ length to determine $g$, the acceleration due to gravity. He uses a stop watch with the least count of $1 \mathrm{~s}$ for this and records $40 \mathrm{~s}$ for 20 oscillations. For this observation, which of the following statement(s) is/are true?

Options:
A
Error $\Delta T$ in measuring $T$, the time period, is $0.05 \mathrm{~s}$

B
Error $\Delta T$ in measuring $T$, the time period, is $1 \mathrm{~s}$

C
Percentage error in the determination of $g$ is $5 \%$

D
Percentage error in the determination of $g$ is $2.5 \%$

The correct answers are:
Error $\Delta T$ in measuring $T$, the time period, is $0.05 \mathrm{~s}$
,
Percentage error in the determination of $g$ is $5 \%$

Solution:
$T=\frac{40 \mathrm{~s}}{20}=2 \mathrm{~s}$.
Further, $t=n T=20 T$ or $\Delta t=20 \Delta T$
$\therefore \quad \frac{\Delta t}{t}=\frac{\Delta T}{T}$
or $\Delta T=\frac{T}{t} \cdot \Delta t=\left(\frac{2}{40}\right)(1)=0.05 \mathrm{~s}$
Further, $T=2 \pi \sqrt{\frac{l}{g}}$ or $\quad T \propto g^{-1 / 2}$
$$
\therefore \frac{\Delta T}{T} \times 100=-\frac{1}{2} \times \frac{\Delta g}{g} \times 100
$$
or $\%$ error in determination of $g$ is
$$
\begin{aligned}
\frac{\Delta g}{g} \times 100 & =-200 \times \frac{\Delta T}{T} \\
& =-\frac{200 \times 0.05}{2}=-5 \%
\end{aligned}
$$
$\therefore$ correct options are (a) and (c).

Topic: Physics, Mathematics in Physics, JEE Advanced, JEE Advanced 2010 (Paper 1)
A tangent $P T$ is drawn to the circle $x^{2}+y^{2}=4$ at the point $P(\sqrt{3}, 1)$. A straight line $L$, perpendicular to $P T$ is a tangent to the circle $(x-3)^{2}+y^{2}-1$.

Question: A possible equation of $L$ is

Options:
A $x-\sqrt{3} y=1$
B $x+\sqrt{3} y=1$
C $x-\sqrt{3} y=-1$
D $x+\sqrt{3} y=5$

The correct answer is:
$x-\sqrt{3} y=1$

Solution:
Equation of tangent $P T$ to the circle $x^{2}+y^{2}=4$ at the point $P(\sqrt{3}, 1)$ is $x \sqrt{3}+y=4$ Let the line $L$, perpendicular to tangent $P T$ be $x-y \sqrt{3}+\lambda=0$

As it is tangent to the circle $(x-3)^{2}+y^{2}=1$ $\therefore \quad$ Length of perpendicular from centre of circle to the Tangent $=$ radius of circle.

$\Rightarrow\left|\frac{3+\lambda}{2}\right|=1 \Rightarrow \lambda=-1 \text { or }-5$

$\therefore$ Equation of $L$ can be $x-\sqrt{3} y=1$ or $x-\sqrt{3} y=5$

Topic: Mathematics, Circle, JEE Advanced, JEE Advanced 2012 (Paper 2)
A tangent $P T$ is drawn to the circle $x^{2}+y^{2}=4$ at the point $P(\sqrt{3}, 1)$. A straight line $L$, perpendicular to $P T$ is a tangent to the circle $(x-3)^{2}+y^{2}-1$.

Question: A common tangent of the two circles is

Options:
A $x=4$
B $y=2$
C $x+\sqrt{3} y=4$
D $x+2 \sqrt{2} y=6$

The correct answer is:
$x+2 \sqrt{2} y=6$

Solution:
From the figure it is clear that the intersection point of two direct common tangents lies on $x$-axis.

Also $\Delta P T_{1} C_{1} \sim \Delta P T_{2} C_{2}$

$\therefore \quad P C_{1}: P C_{2}=2: 1$

or $P$ divides $C_{1} C_{2}$ in the ratio $2: 1$ externally

$\therefore \quad$ Coordinates of $P$ are $(6,0)$.

Let the equation of tangent through $P$ be

$y=m(x-6)$

Asit touches $x^{2}+y^{2}=4$

$\therefore\left|\frac{6 m}{\sqrt{m^{2}+1}}\right|=2 \Rightarrow 36 m^{2}=4\left(m^{2}+1\right)$

$\therefore \quad m=\pm \frac{1}{2 \sqrt{2}}$

$\therefore \quad$ Equations of common tangents are

$y=\pm \frac{1}{2 \sqrt{2}}(x-6)$

Also $x=2$ is the common tangent to the two circles.

Topic: Mathematics, Circle, JEE Advanced, JEE Advanced 2012 (Paper 2)
A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting the surface. The force on the ball during the collision is proportional to the length of compression of the ball. Which one of the following sketches describes the variation of its kinetic energy K with t time most appropriately? The figures are only illustrative and not to the scale.

Options:
A

B

C

D

The correct answer is:

Solution:
When ball of mass 'm' dropped from height and strikes to ground.It's kinetic energy is converting into elastic potential energy due to compression.

So after time 't' its final velocity,

v=u+at

u is initial velocity of ball, a is acceleration of ball and t is time.

⇒v=0+-gt

v=-gt

Negative sign shows downward direction.

Now kinetic energy of ball,

K.E.=12mv2=12m-gt2

⇒K.E.=12mg2t2

⇒K.E.∝t2

So, it is upward parabola. But after complete compression, K.E. becomes zero, then ball will reshape again. Kinetic energy will not be constant.

Topic: Physics, Work Power Energy, JEE Main
A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting the surface. The force on the ball during the collision is proportional to the length of compression of the ball. Which one of the following sketches describes the variation of its kinetic energy K with t time most appropriately? The figures are only illustrative and not to the scale.

Options:
A

B

C

D

The correct answer is:

Solution:
When ball of mass 'm' dropped from height and strikes to ground.It's kinetic energy is converting into elastic potential energy due to compression.

So after time 't' its final velocity,

v=u+at

u is initial velocity of ball, a is acceleration of ball and t is time.

⇒v=0+-gt

v=-gt

Negative sign shows downward direction.

Now kinetic energy of ball,

K.E.=12mv2=12m-gt2

⇒K.E.=12mg2t2

⇒K.E.∝t2

So, it is upward parabola. But after complete compression, K.E. becomes zero, then ball will reshape again. Kinetic energy will not be constant.

Topic: Physics, Work Power Energy, JEE Advanced, JEE Advanced 2014 (Paper 2)
A thin flexible wire of length $L$ is connected to two adjacent fixed points and carries a current $I$ in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength $B$ going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is

Options:
A
$I B L$

B
$\frac{I B L}{\pi}$

C
$\frac{I B L}{2 \pi}$

D
$\frac{I B L}{4 \pi}$

The correct answer is:
$\frac{I B L}{2 \pi}$

Solution:
$L=2 \pi R$
$R=\frac{L}{2 \pi}$
$2 T \sin (d \theta)=F_m$
for small angles, $\sin (d \theta) \approx d \theta$
$$
\begin{aligned}
\therefore \quad 2 T(d \theta) & =I(d L) B \sin 90^{\circ} \\
& =I(2 R \cdot d \theta) \cdot B \\
\therefore \quad T & =I R B=\frac{I L B}{2 \pi}
\end{aligned}
$$
$\therefore$ correct option is (c)

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2010 (Paper 1)
A thin flexible wire of length $L$ is connected to two adjacent fixed points and carries a current $I$ in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength $B$ going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is

Options:
A
$I B L$

B
$\frac{I B L}{\pi}$

C
$\frac{I B L}{2 \pi}$

D
$\frac{I B L}{4 \pi}$

The correct answer is:
$\frac{I B L}{2 \pi}$

Solution:
$L=2 \pi R$
$R=\frac{L}{2 \pi}$
$2 T \sin (d \theta)=F_m$
for small angles, $\sin (d \theta) \approx d \theta$
$$
\begin{aligned}
\therefore \quad 2 T(d \theta) & =I(d L) B \sin 90^{\circ} \\
& =I(2 R \cdot d \theta) \cdot B \\
\therefore \quad T & =I R B=\frac{I L B}{2 \pi}
\end{aligned}
$$
$\therefore$ correct option is (c)

Topic: Physics, Magnetic Effects of Current, JEE Main
A thin ring of mass $2 \mathrm{~kg}$ and radius $0.5 \mathrm{~m}$ is rolling without slipping on a horizontal plane with velocity 1 $\mathrm{m} / \mathrm{s}$. A small ball of mass $0.1 \mathrm{~kg}$, moving with velocity $20 \mathrm{~m} / \mathrm{s}$ in the opposite directions, hits the ring at a height of $0.75 \mathrm{~m}$ and goes vertically up with velocity 10 $\mathrm{m} / \mathrm{s}$. Immediately after the collision

Options:
A
the ring has pure rotation about its stationary $\mathrm{CM}$

B
the ring comes to a complete stop

C
friction between the ring and the ground is to the left

D
there is no friction between the ring and the ground

The correct answers are:
the ring has pure rotation about its stationary $\mathrm{CM}$
,
friction between the ring and the ground is to the left

Solution:
The data is incomplete. Let us assume that friction from ground on ring is not impulsive during impact.
From linear momentum conservation in horizontal direction, we have
$$
\begin{aligned}
& (-2 \times 1)+(0.1 \times 20) \\
& =(0.1 \times 0)+(2 \times v) \stackrel{\text {-ve }}{\leftarrow} \stackrel{+v e}{\rightarrow}
\end{aligned}
$$
Here, $v$ is the velocity of $\mathrm{CM}$ of ring after impact.
Solving the above equation, we have
$$
v=0
$$
Thus, CM becomes stationary.
$\therefore$ Correct answer is (a).
Linear impulse during impact
(i) In horizontal direction $J_1=\Delta P=0.1 \times 20=2 \mathrm{Ns}$
(ii) In vertical direction
$$
J_2=\Delta P=0.1 \times 10=1 \mathrm{Ns}
$$
Writing the equation (about $\mathrm{CM}$ )
Angular impulse
$=$ Change in angular momentum
$$
\begin{aligned}
& 1 \times\left(\frac{\sqrt{3}}{2} \times \frac{1}{2}\right)-2 \times 0.5 \times \frac{1}{2} \\
& =2 \times(0.5)^2\left[\omega-\frac{1}{0.5}\right]
\end{aligned}
$$
Solving this equation $\omega$ comes out to be positive or $\omega$ anti-clockwise. So just after collision rightwards slipping is taking place.
Hence, friction is leftwards.
Therefore, option (c) is also correct.
$\therefore$ Correct options are (a) and (c).
Note In JEE 2011 official answer key, correct option were given $\mathrm{a}, \mathrm{ac}$.
Analysis of Question
(i) Question is moderately difficult.
(ii) In such type of problems impulse due to friction during collision is ignored.

Topic: Physics, Center of Mass Momentum and Collision, JEE Advanced, JEE Advanced 2011 (Paper 2)
A thin uniform annular disc (see figure) of mass $M$ has outer radius $4 R$ and inner radius $3 R$. The work required to take a unit mass from point $P$ on its axis to infinity is

Options:
A
$\frac{2 G M}{7 R}(4 \sqrt{2}-5)$

B
$-\frac{2 G M}{7 R}(4 \sqrt{2}-5)$

C
$\frac{G M}{4 R}$

D
$\frac{2 G M}{5 R}(\sqrt{2}-1)$

The correct answer is:
$\frac{2 G M}{7 R}(4 \sqrt{2}-5)$

Solution:
$$
\text { } \begin{aligned}
W=\Delta U & =U_f-U_i=U_{\infty}-U_p \\
& =-U_P \\
& =-m V_P \\
& =-V_P \quad(\text { as } m=1)
\end{aligned}
$$
Potential at point $P$ will be obtained by integration as given below
Let $d M$ be the mass of small ring as shown
$$
\begin{aligned}
d M & =\frac{M}{\pi(4 R)^2-\pi(3 R)^2}(2 \pi r) d r \\
& =\frac{2 M r d r}{7 R^2} \\
d V_P & =-\frac{G \cdot d M}{\sqrt{16 R^2+r^2}} \\
& =-\frac{2 G M}{7 R^2} \int_{3 R}^{4 R} \frac{r}{\sqrt{16 R^2+r^2}} \cdot d r \\
& =-\frac{2 G M}{7 R}(4 \sqrt{2}-5) \\
\therefore \quad W & =+\frac{2 G M}{7 R}(4 \sqrt{2}-5)
\end{aligned}
$$
$\therefore$ correct option is (a).

Topic: Physics, Gravitation, JEE Main
A thin uniform annular disc (see figure) of mass $M$ has outer radius $4 R$ and inner radius $3 R$. The work required to take a unit mass from point $P$ on its axis to infinity is

Options:
A
$\frac{2 G M}{7 R}(4 \sqrt{2}-5)$

B
$-\frac{2 G M}{7 R}(4 \sqrt{2}-5)$

C
$\frac{G M}{4 R}$

D
$\frac{2 G M}{5 R}(\sqrt{2}-1)$

The correct answer is:
$\frac{2 G M}{7 R}(4 \sqrt{2}-5)$

Solution:
$$
\text { } \begin{aligned}
W=\Delta U & =U_f-U_i=U_{\infty}-U_p \\
& =-U_P \\
& =-m V_P \\
& =-V_P \quad(\text { as } m=1)
\end{aligned}
$$
Potential at point $P$ will be obtained by integration as given below
Let $d M$ be the mass of small ring as shown
$$
\begin{aligned}
d M & =\frac{M}{\pi(4 R)^2-\pi(3 R)^2}(2 \pi r) d r \\
& =\frac{2 M r d r}{7 R^2} \\
d V_P & =-\frac{G \cdot d M}{\sqrt{16 R^2+r^2}} \\
& =-\frac{2 G M}{7 R^2} \int_{3 R}^{4 R} \frac{r}{\sqrt{16 R^2+r^2}} \cdot d r \\
& =-\frac{2 G M}{7 R}(4 \sqrt{2}-5) \\
\therefore \quad W & =+\frac{2 G M}{7 R}(4 \sqrt{2}-5)
\end{aligned}
$$
$\therefore$ correct option is (a).

Topic: Physics, Gravitation, JEE Advanced, JEE Advanced 2010 (Paper 1)
A thin uniform cylindrical shell, closed at both ends, is partially filled with water. It is floating vertically in water in half-submerged state. If $\rho_{c}$ is the relative density of the material of the shell with respect to water, then the correct statement is that the shell is

Options:
A more than half-filled if $\rho_{c}$ is less than $0.5$.
B more than half-filled if $\rho_{c}$ is more than 1.0.
C half-filled if $\rho_{c}$ is more than $0.5$.
D less than half-filled if $\rho_{c}$ is less than $0.5$.

The correct answer is:
more than half-filled if $\rho_{c}$ is less than $0.5$.

Solution:
Let $V_{1}=$ total material volume of cylindrical shell $V_{2}=$ total inside volume of cylindrical shell and $y=$ fraction of $V_{2}$ volume filled with water.

Total weight $=$ Upthrust (Condition of floatation)

$\therefore \quad V_{1} \rho_{c} g+\left(y V_{2}\right)(1) g=\left(\frac{V_{1}+V_{2}}{2}\right)(1) g$

or $y=0.5+\left(0.5-\rho_{c}\right) \frac{V_{1}}{V_{2}}$

So, if $\rho_{c} < 0.5$ then $y>0.5$

Topic: Physics, Mechanical Properties of Fluids, JEE Advanced, JEE Advanced 2012 (Paper 2)
A thin uniform rod, pivoted at $O$, is rotating in the horizontal plane with constant angular speed $\omega$, as shown in the figure. At time $t=0$, a small insect starts from $O$ and moves with constant speed $v$, with respect to the rod towards the other end. It reaches the end of the rod at $t=T$ and stops. The angular speed of the system remains $\omega$ throughout. The magnitude of the torque $(|\vec{\tau}|)$ about $O$, as a function of time is best represented by which plot?

Options:
A
B
C
D

The correct answer is:

Solution:
Angular momentum, $|\mathbf{L}|$ or $L=I \omega$ (about axis of rod) Moment of inertia of the rod-insect system. $I=I_{\text {rod }}+m x^{2}=I_{\text {rod }}+m v^{2} t^{2}$ Here, $m=$ mass of insect $\therefore \quad L=\left(I_{\mathrm{rod}}+m v^{2} t^{2}\right) \omega$

Now $|\tau|=\frac{d L}{d t}=\left(2 m v^{2} t \omega\right)$ or $|\tau| \propto t$

i.e. the graph is straight line passing through origin.

After time $T, L=$ constant

$\therefore \quad|\tau| \text { or } \frac{d L}{d t}=0$

i.e., when the insect stops moving, $L$ does not change and therefore $T$ becomes constant.

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2012 (Paper 1)
A tiny spherical oil drop carrying a net charge $q$ is balanced in still air with a vertical uniform electric field of strength $\frac{81 \pi}{7} \times 10^5 \mathrm{Vm}^{-1}$. When the field is switched off, the drop is observed to fall with terminal velocity $2 \times 10^{-3} \mathrm{~ms}^{-1}$. Given $g=9.8 \mathrm{~ms}^{-2}$, viscosity of the air $=1.8 \times 10^{-5} \mathrm{Ns} \mathrm{m}^{-2}$ and the density of oil $=900 \mathrm{~kg} \mathrm{~m}^{-3}$, the magnitude of $q$ is

Options:
A
$1.6 \times 10^{-19} \mathrm{C}$

B
$3.2 \times 10^{-19} \mathrm{C}$

C
$4.8 \times 10^{-19} \mathrm{C}$

D
$8.0 \times 10^{-19} \mathrm{C}$

The correct answer is:
$8.0 \times 10^{-19} \mathrm{C}$

Solution:
$$
\begin{aligned}
& q E=m g \\
& \quad 6 \pi \eta r v=m g \\
& \frac{4}{3} \pi r^3 \rho g=m g \\
& \therefore \quad r=\left(\frac{3 m g}{4 \pi \rho g}\right)^{1 / 3}
\end{aligned}
$$
Substituting the value of $r$ in Eq. (ii) we get,
$$
6 \pi \eta v\left(\frac{3 m g}{4 \pi \rho g}\right)^{1 / 3}=m g
$$
or $\quad(6 \pi \eta v)^3\left(\frac{3 m g}{4 \pi \rho g}\right)=(m g)^3$
Again substituting $m g=q E$ we get,
$$
(q E)^2=\left(\frac{3}{4 \pi \rho g}\right)(6 \pi \eta v)^3
$$
or $\quad q E=\left(\frac{3}{4 \pi \rho g}\right)^{1 / 2}(6 \pi \eta v)^{3 / 2}$
$$
\therefore \quad q=\frac{1}{E}\left(\frac{3}{4 \pi \rho g}\right)^{1 / 2}(6 \pi \eta v)^{3 / 2}
$$
Substituting the values we get ,
$$
\begin{aligned}
q= & \frac{7}{81 \pi \times 10^5} \sqrt{\frac{3}{4 \pi \times 900 \times 9.8} \times 216 \pi^3} \\
& \times \sqrt{\left(1.8 \times 10^{-5} \times 2 \times 10^{-3}\right)^3} \\
& =8.0 \times 10^{-19} \mathrm{C}
\end{aligned}
$$
$\therefore$ correct option is $(\mathrm{d})$.

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2010 (Paper 2)
A train is moving along a straight line with a constant acceleration a. A boy standing in the train throws a ball forward with a speed of $10 \mathrm{~m} / \mathrm{s}$, at an angle of $60^{\circ}$ to the horizontal. The boy has to move forward by $1.15 \mathrm{~m}$ inside the train to catch the ball back at the initial height. The acceleration of the train in $\mathrm{m} / \mathrm{s}^2$, is

Options:

The correct answer is:
5

Solution:
$\begin{aligned} t & =T=\frac{2 u \sin \theta}{g} \\ & =\frac{2 \times 10 \times \sin 60^{\circ}}{10}=\sqrt{3} \mathrm{~s}\end{aligned}$
Displacement of train in time $t=\frac{1}{2} a t^2$
Displacement of boy with respect to train
$$
=1.15 \mathrm{~m}
$$
$\therefore$ Displacement of boy with respect to ground $=\left(1.15+\frac{1}{2} a t^2\right)$
Displacement of ball with respect to ground $=\left(u \cos 60^{\circ}\right) t$
To catch the ball back at initial height,
$$
\begin{aligned}
1.15+\frac{1}{2} a t^2 & =\left(u \cos 60^{\circ}\right) t \\
\therefore 1.15+\frac{1}{2} a(\sqrt{3})^2 & =10 \times \frac{1}{2} \times \sqrt{3}
\end{aligned}
$$
Solving this equation, we get
$$
a=5 \mathrm{~ms}^{-2}
$$
$\therefore$ Answer is 5 .
Analysis of Question
(i) Question is moderately tough.
(ii) Velocity of ball given in the question is with respect to ground.

Topic: Physics, Motion In Two Dimensions, JEE Advanced, JEE Advanced 2011 (Paper 2)
A transverse sinusoidal wave moves along a string in the positive $x$-direction at a speed of $10 \mathrm{~cm} / \mathrm{s}$. The wavelength of the wave is $0.5 \mathrm{~m}$ and its amplitude is $10 \mathrm{~cm}$. At a particular time $t$, the snap-shot of the wave is shown in figure. The velocity of point $P$ when its displacement is $5 \mathrm{~cm}$ is

Options:
A
$\frac{\sqrt{3} \pi}{50} \hat{\mathbf{j}} \mathrm{m} / \mathrm{s}$

B
$-\frac{\sqrt{3} \pi}{50} \hat{j} \mathrm{~m} / \mathrm{s}$

C
$\frac{\sqrt{3} \pi}{50} \hat{\mathbf{i}} \mathrm{m} / \mathrm{s}$

D
$-\frac{\sqrt{3} \pi}{50} \hat{\mathbf{i}} \mathrm{m} / \mathrm{s}$

The correct answer is:
$\frac{\sqrt{3} \pi}{50} \hat{\mathbf{j}} \mathrm{m} / \mathrm{s}$

Solution:
Particle velocity $v_p=-v$ (slope of $y-x$ graph)
Here,$v=+v e$, as the wave is travelling in positive $x$-direction.
Slope at $P$ is negative.
$\therefore$ Velocity of particle is in positive $y(+\overrightarrow{\mathbf{j}})$ direction.
$\therefore$ correct option is (a).

Topic: Physics, Waves and Sound, JEE Main
A transverse sinusoidal wave moves along a string in the positive $x$-direction at a speed of $10 \mathrm{~cm} / \mathrm{s}$. The wavelength of the wave is $0.5 \mathrm{~m}$ and its amplitude is $10 \mathrm{~cm}$. At a particular time $t$, the snap-shot of the wave is shown in figure. The velocity of point $P$ when its displacement is $5 \mathrm{~cm}$ is

Options:
A
$\frac{\sqrt{3} \pi}{50} \hat{\mathbf{j}} \mathrm{m} / \mathrm{s}$

B
$-\frac{\sqrt{3} \pi}{50} \hat{j} \mathrm{~m} / \mathrm{s}$

C
$\frac{\sqrt{3} \pi}{50} \hat{\mathbf{i}} \mathrm{m} / \mathrm{s}$

D
$-\frac{\sqrt{3} \pi}{50} \hat{\mathbf{i}} \mathrm{m} / \mathrm{s}$

The correct answer is:
$\frac{\sqrt{3} \pi}{50} \hat{\mathbf{j}} \mathrm{m} / \mathrm{s}$

Solution:
Particle velocity $v_p=-v$ (slope of $y-x$ graph)
Here,$v=+v e$, as the wave is travelling in positive $x$-direction.
Slope at $P$ is negative.
$\therefore$ Velocity of particle is in positive $y(+\overrightarrow{\mathbf{j}})$ direction.
$\therefore$ correct option is (a).

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2008 (Paper 2)