NSEP 2015 Problem 12 | Periodic Motion due to Electrostatic Force

Try this problem on periodic motion due to electrostatic force from NSEP 2015 Problem 12.

NSEP 2015-16 Problem 12

A particle of mass m and charge $-q$ moves along a diameter of a uniform spherical distribution of radius R with total charge $+Q$. The angular frequency of the periodic motion performed by the particle is

a) $\sqrt{\frac{2\pi qQ}{\epsilon_0 mR^3}}$ b) $\sqrt{\frac{qQ}{2\pi \epsilon_0 mR^3}}$ c) $\sqrt{\frac{qQ}{\epsilon_0 mR^3}}$ d) $\sqrt{\frac{qQ}{4\pi \epsilon_0 mR^3}}$

Key Concepts

Periodic Motion

Gauss's Law of Electrostatic

Suggested Book | Source | Answer

Concept of Physics H.C. Verma

University Physics by H. D. Young and R.A. Freedman

Fundamental of Physics D. Halliday, J. Walker and R. Resnick

National Standard Examination in Physics(NSEP) 2015-2016

Option-(d) $\sqrt{\frac{qQ}{4\pi \epsilon_0 mR^3}}$

Try with Hints

The problem describe a structure something like that,

The whole blue region is charged and the black particle oscillates between A and B. We have to find the angular velocity of this oscillation.

To solve this problem we have to find what will be the Electric force inside the sphere. Using the gauss's law we can solve that.

Using the gauss's law, if we consider a Gaussian surface (sphere) of radius $r$, then charge inside that region is $Q_{in} = \frac{Q}{\frac{4}{3}\pi R^3}\frac{4}{3}\pi r^3 = Q\cdot \frac{r^3}{R^3}$

$$E\cdot 4\pi r^2 = \frac{Q_{in}}{\epsilon_0}$$

This means the particle will get a force of $F = -q E = -\frac{qQ}{4\pi \epsilon}\frac{r}{R^3}$

Arranging this in a equation,

$$ F = m\frac{d^2r}{dt^2} = -\frac{qQr}{4\pi \epsilon R^3}$$

This reduces to ,

$$ \frac{d^2r}{dt^2} + \frac{qQ}{4\pi \epsilon mR^3}r =0$$

We know that $w^2 = \frac{qQ}{4\pi \epsilon mR^3}$. This gives $$ w = \sqrt{\frac{qQ}{4\pi \epsilon_0 mR^3}} $$

NSEP 2015 Problem 9 | Pulley problem

Try this problem on pulley problem on inclined plane from NSEP 2015 Problem 9.

NSEP 2015-16 ~ Problem 9

Maases $m_1$ and $m_2$ are connected to a string passing over a pulley as shown. Mass $m_2$ starts from rest and falls through a distance $d$ in time t. Now, by interchanging the masses the time required for $m_1$ to fall through the same distance is $2t$. Therefore, the ratio of masses $m_2 : m_1$

a) $\frac{2}{3}$ b) $\frac{3}{2}$ c) $\frac{5}{2}$ d) $\frac{4}{3}$

$m_1$ and $m_2$ are interchanged from real problem

Key Concepts

Newton's Laws of Motion

Idea of accelerations, velocity and displacement

Suggested Book | Source | Answer

Concept of Physics H.C. Verma

University Physics by H. D. Young and R.A. Freedman

Fundamental of Physics D. Halliday, J. Walker and R. Resnick

National Standard Examination in Physics(NSEP) 2015-2016

Option-(b) $\frac{3}{2}$

Try with Hints

We know at the beginning the blocks have zero velocities. Using the relation $s= ut+\frac{1}{2}at^2$, we can find the relation between the accelerations for two cases (i.e., when they are interchanged).

Knowing the accelerations we can now use the second law of newton to find the ratio of masses.

From the first hint,

$$ \frac{1}{2}a_1t^2 = \frac{1}{2}a_2 (2t)^2 \to a_1 =4 a_2 $$

Now, we find the value of $a_1$ and $a_2$ using $a = \frac{F}{M}$

$$ \frac{m_2 g - m_1g \sin(30)}{m1+m_2} = 4 \frac{m_1 g - m_2g \sin(30)}{m1+m_2} $$

Rearranging this expression and using $\sin(30) = \frac{1}{2}$,

This gives, $\frac{m_2}{m_1} = \frac{3}{2}$

NSEP 2015 Problem 8 | One Dimensional Motion

Try out this problem on one dimensional motion(Kinematics) from National Standard Examination in Physics 2015-2016.

NSEP 2015-16 ~ Problem 8

A car moving along a straight road at a speed of u m/s applies brakes at $t=0$ seconds. The ratio of distances travelled by the car during $3^{rd}$ and $8^{th}$ seconds is $15:13$. The car covers a distance of $0.25m$ in the last second of it's travel. Therefore, the acceleration a (in $m/s^2$) and the speeed u($m/s$) of the car are respectively,

a) $-0.1,16$

b) $-0.2,12$

c) $-0.5,20$

d) $-0.1,16$

Key Concepts

Basic Equation of motions

Idea of accelerations, velocity and displacement

Suggested Book | Source | Answer

Concept of Physics H.C. Verma

University Physics by H. D. Young and R.A. Freedman

Fundamental of Physics D. Halliday, J. Walker and R. Resnick

National Standard Examination in Physics(NSEP) 2015-2016

Option-(c) $-0.5,20$

Try with Hints

We know that displacement in $t$ sec is given by,

$$ s = ut +\frac{1}{2} a t^2$$

where $u$ is the initial velocity and $a$ is the acceleration. This will be negative for deceleration.

Using this idea we can just find the expression for the displacement in the $n^{th}$ second, for which the information is given to us.

Also, the car come to rest and at the second second it travels a distance of $0.25m$. WE can think it in opposite manner that with an acceleration of $a$ and initial velocity $0m/s$, the car moves a distance of $0.25m$ , hence,

$$0.5 =\frac{1}{2} a (1^2)$$

So, the displacement in $n^{th}$ second is,

$$ s_1 = ut+\frac{1}{2} a n^2 $$

and for the $(n-1)^{th}$ second the displacement is,

$$ s_2= ut+\frac{1}{2} a (n-1)^2 $$

Hence, we have

$$ s_n = s_1-s_2 = u + a(n-\frac{1}{2}) $$

we have,

$$ \frac{15}{13} = \frac{s_3}{s_8} = \frac{u-\frac{5a}{2}}{u - \frac{15a}{2}} $$

Also we have $0.5 =\frac{1}{2} a (1^2)$

Solving this two gives us $a = 0.25m/s^2 $. Putting this in first equation gives us $u = 20 m/s $.

NSEP 2015 Problem 7 | Simple Harmonic Motion

Try out this problem on Simple Harmonic Motion (SHM) from National Standard Examination in Physics 2015-2016.

NSEP 2015-16 ~ Problem 7

A particle execute a periodic motion according to the relation $x^2 = 4 \cos^2(50t) \sin^2(500t)$. Therefore, the motion can be considered to be the superposition of $n$ independent simple harmonic motions, where $n$ is

(a) $2$ (b) $3$ (c) $4$ (d) $5$

Key Concepts

Basic Trigonometry

Equations of SHM

Suggested Book | Source | Answer

Concept of Physics H.C. Verma

University Physics by H. D. Young and R.A. Freedman

Fundamental of Physics D. Halliday, J. Walker and R. Resnick

National Standard Examination in Physics(NSEP) 2015-2016

Option-(b) $3$

Try with Hints

For a simple harmonic motion the equation is $X$ = $A \cos(wt + \phi )$ or $X = A \sin(wt + \phi )$, where X is the position of the particle, A is the amplitude and w is the angular frequency and $\phi $ is the initial phase. We can also consider it to be zero.

If there are $n$ harmonic oscillator then the resultant can be written as ,

$X$=$A_{1} \sin \left(w_{1} t+\phi_{1}\right)$+$A_{2} \sin \left(w_{2} t+\phi_{2}\right)$+$\cdots$+$A_{n} \sin \left(w_{n} t+\phi_{n}\right)$

For this problem, we just have to use suitable trigonometric identities to convert them into sum.

$X$ =$ x^2$ = $4 \cos (50t) \sin (500t) \cos (50t)$, Now, we will use $2\cos(\theta)\sin(\phi)$= $\sin(\theta +\phi)$ +$ \sin(\theta -\phi)$, hence,

$X$=$2(\sin (550 t)$+$\sin (450 t)) \cos (50 t)$=$2 \sin (550 t) \cos (50 t)$+$2 \sin (450 t) \cos (50 t)$

Again applying the same identity,

$X$=$\sin (600 t)$+$2 \sin (500 t)$+$\sin (400 t)$

Hence, there are $3$ SHM.

NSEP 2015 Problem 6 | Surface Tension and Pressure

Try out this problem on the Surface Tension and Pressure from National Standard Examination in Physics 2015-2016.

NSEP 2015-16 ~ Problem 6

Two air bubble with radius $r_1$ and $r_2$ $(r_2>r_1)$ formed of the same liquid stick to each other to form a common interface. Therefore, the radius of curvature of the common surface is

(a) $\sqrt{r_1 r_2}$ (b) Infinity (c) $\frac{r_2}{r_1}\sqrt{{r_2}^2-{r_2}^2}$ (d) $\frac{r_1 r_2}{r_2 - r_1}$

Key Concepts

Basic surface Tension

Relation between surface Tension and Radius of curvature

Suggested Book | Source | Answer

Concept of Physics H.C. Verma

University Physics by H. D. Young and R.A. Freedman

Fundamental of Physics D. Halliday, J. Walker and R. Resnick

National Standard Examination in Physics(NSEP) 2015-2016

Option-(d) $\frac{r_1 r_2}{r_2 - r_1}$

Try with Hints

For a bubble of radius $r$ with double surface and whose inside pressure is $p_{in}$ and outside pressure is $p_{out}$ and also the surface tension is $T$, the relation between pressure and radius is,

$$p_{in} - p_{out} = \Delta p = \frac{4T}{r}$$

When 2 bubble merge on their common interface the pressure difference is Just the difference between the pressure(inside) of both two bubbles. Also, the surface tension remains same.

We know $p_1 - p_0 = \frac{4T}{r_1}$ and $p_2 - p_0 = \frac{4T}{r_2}$, Then,

$$ p_1 - p_2 = \frac{4T}{R} $$

here $R$ is the radius of curvature of the interface.

Hence,

$p_{1}-p_{2}$=$\left(p_{1}-p_{0}\right)-\left(p_{2}-p_{0}\right)$=$4 T\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)$

This gives us,

$$ \frac{1}{R} = \frac{1}{r_1}-\frac{1}{r_2} \to R= \frac{r_1 r_2}{r_2 - r_1} $$

NSEP 2015 Problem 5 | Rotational Mechanics & Small Oscillation

Try out this problem on the one rotational mechanics and small oscillation from National Standard Examination in Physics 2015-2016.

NSEP 2015-16 ~ Problem 5

A thin rod of length l in the shape of a semicircle is pivoted at one of its ends such that it is free to oscillate in its own place. The frequency f of small oscillation of the semicircular rod is

(a) $\frac{1}{2\pi} \sqrt{\frac{g\pi}{2l}}$ (b) $\frac{1}{2\pi} \sqrt{\frac{g\sqrt{\pi^2+4}}{2l}}$ (c) $\frac{1}{2\pi} \sqrt{\frac{g(\pi+4)}{l}}$ (d) $\frac{1}{2\pi} \sqrt{\frac{g(\pi^2+1)}{2\pi l}}$

Key Concepts

Rotational Mechanics and Moment of Inertia

Small Oscillations

Suggested Book | Source | Answer

Concept of Physics H.C. Verma

University Physics by H. D. Young and R.A. Freedman

Fundamental of Physics D. Halliday, J. Walker and R. Resnick

National Standard Examination in Physics(NSEP) 2015-2016

Option-(b) $\frac{1}{2\pi} \sqrt{\frac{g\sqrt{\pi^2+4}}{2l}}$

Try with Hints

Suppose, we have $\Gamma = -k\theta $, where $\theta $ is the angular displacement and $\Gamma $ is the torque.

Then, we can write the angular acceleration $\alpha $ as,

$$ \alpha = \frac{\Gamma}{I}= -\frac{k}{I}\theta = w^2 \theta$$

where I is the moment of inertia, i.e., mass's counterpart in rotational mechanics.

Finally we have, $w = 2\pi f= \sqrt{\frac{\alpha }{\theta}} = \sqrt{\frac{\Gamma }{I\theta}} $, where f is the frequency of the oscillation.

Now for a force $\vec{F}$, $\Gamma = F r \sin(\theta)$. Now using small angle approximation $\sin(\theta) = \theta $. Hence,

$$ f = \frac{1}{2\pi }\sqrt{\frac{Fr}{I}} $$

Now, we can consider the semi-circle of mass m as a point of mass m concentrated into it's centre of mass. This will give us as shown in the figure.

Now, for a semi-circle the centre of mass is at a height of $\frac{2R}{\pi}$ distance above the radius, i.e., $AO = \frac{2R}{\pi}$, where $AP=R=\frac{l}{\pi}$.

Now, $F r = mg l_{c} = mg \sqrt{R^2 + (\frac{2R}{\pi})^2}$

and $I = 2mR^2$

Putting these values gives us,

$f=\frac{1}{2 \pi} \sqrt{\frac{F r}{I}}$

=$\frac{1}{2 \pi} \sqrt{\frac{g \sqrt{\pi^{2}+4}}{2 \pi R}}$

=$\frac{1}{2 \pi} \sqrt{\frac{g \sqrt{\pi^{2}+4}}{2 l}}$

NSEP 2015 Problem 4 | Rotational Motion

Try out this problem on the one dimensional motion from National Standard Examination in Physics 2015-2016.

NSEP 2015-16 ~ Problem 4

A body released from a height H hits elastically an inclined plane at a point P. After the impact the body starts moving horizontally and hits the ground. The height at which point P should be situated so as to have the total time of travel maximum is

(a) H (b) 2H (c) $\frac{H}{4} (d) $\frac{H}{2}$

Key Concepts

Freely falling body

Time needed for a free body

Suggested Book | Source | Answer

Concept of Physics H.C. Verma

University Physics by H. D. Young and R.A. Freedman

Fundamental of Physics D. Halliday, J. Walker and R. Resnick

National Standard Examination in Physics(NSEP) 2015-2016

Option-(d) $\frac{H}{2}$

Try with Hints

From the height H up to the point P, the ball will have a free fall. Hence, the time is fixed.

We just have to think how to get the path of the maximum time in the next part of the motion.

Hint: It can also be considered as free fall by only considering y direction.

B is the point at height H. From P the ball goes horizontally. Then due to gravity it has some projectile motion.

The path is randomly drawn.

From P to ground, the height is h. Then, as the height is h, the time needed to fall is,

$$ t_{PA} = \sqrt{\frac{2x}{g}}$$

From B to P, the time needed is,

$$ t_{BP} = \sqrt{\frac{2(H-x)}{g}}$$

The total time is,

$$t = t_{BP}+t_{PA} = \sqrt{\frac{2(H-x)}{g}} +\sqrt{\frac{2x}{g}}$$

For maximum,

$$ \frac{dt}{dx}=0 $$

Calculating this,

$$ \frac{1}{2}\sqrt{\frac{2}{g}}\frac{-1}{\sqrt{H-x}} + \frac{1}{2}\sqrt{\frac{2}{g}}\frac{1}{\sqrt{x}} $$

solving this,

$$x=\frac{H}{2}$$

NSEP 2015 Problem 3 | Rotational Motion

Try out this problem on the concept of Angular momentum from National Standard Examination in Physics 2015-2016.

NSEP 2015-16 ~ Problem 3

A particle of mass 10g starts from rest at $t=0$s from point (0m, 4m) and gets accelerated at $0.5m/ s^2$ along $x-\sqrt{3}y + 4\sqrt{3}=0$ in XY plane. The angular momentum of the particle about the origin (in SI units) at $t=2$s is
(a) $-0.01\sqrt{3} \hat{k}$
(b) $-0.02\sqrt{3}\hat{k}$
(c) Zero
(d) $-20\sqrt{3}\hat{k}$

Key Concepts

Circular Motion

Center of Curvature for generalised motion

Vector and their Components

Suggested Book | Source | Answer

Concept of Physics H.C. Verma

University Physics by H. D. Young and R.A. Freedman

Fundamental of Physics D. Halliday, J. Walker and R. Resnick

National Standard Examination in Physics(NSEP) 2015-2016

Option-(b) $-0.02\sqrt{3}\hat{k}$

Try with Hints

The first thing to remember is that Even if a particle is not moving in a circle, then also it can have circular motion, i.e., Angular Momentum, Torque, Centripetal forces can exist.

This is the case because any random curves can be approximated bu circles. Even straight lines are circles with radius $\to \infty$.

You can use the circle's centre as the centre of rotation or you can also use the origin. Using the origin is often easier as the positions can be directly given using position vectors.

In our problem, The motion is happening on the line $y = \frac{x}{\sqrt{3}}+4$.

If at any moment of time it's velocity along the curve is $\vec{v}=v_x \hat{i}+v_y \hat{j}$ and it's position vector is $\vec{r}=x \hat{i}+y \hat{j}$ then, It's angular momentum ($\vec{L}$) with respect to origin can be given by,

$$\vec{L} = m(\vec{v}\times \vec{r}) $$

where m is the mass of the particle.

From the previous image, if $(x,y)$ is then they must satisfy $y = \frac{x}{\sqrt{3}}+4$. Differentiating with respect to time,

$$\frac{dy}{dt} = \frac{1}{\sqrt{3}}\frac{dx}{dt} \to v_y = \frac{v_x}{\sqrt{3}}$$

Again differentiating w.r.t time,

$$a_y = \frac{a_x}{\sqrt{3}}$$ where $\vec{a} = a_x\hat{i}+a_y\hat{j}$ is the acceleration (we will not need this but I think it should be written here as we can also use this in solution).

Now at $t=0$, $v_x=0$. Hence, at $t=2s$, $v_x $=$ a_xt $= $a\cos(30^{\circ})t $= $0.5\frac{\sqrt{3}}{2} 2 $= $\frac{\sqrt{3}}{2}$

Now, $x$ = $v_x(t=0 )t+\frac{1}{2}a_x t^2$ = $0 + \frac{1}{2}a\cos(30^{\circ})2^2$ = $\frac{\sqrt{3}}{2} $

Then, $v_y = \frac{v_x}{\sqrt{3}}= \frac{1}{2}$ and $y = \frac{x}{\sqrt{3}} + 4 = \frac{1}{2} + 4 = \frac{9}{2}$.

Hence, $\vec{r}$ = $\frac{\sqrt{3}}{2}\hat{i} + \frac{9}{2}\hat{j}$ and $\vec{v}$ = $\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}$.

$\vec{L}$=$m(\vec{r} \times \vec{v})$=$-m 2 \sqrt{3} \hat{k}$=$-\left(\frac{10}{1000}\right) 2 \sqrt{3} \hat{k}$=$-0.02 \sqrt{3} \hat{k}$

Note: The answer is asked in S.I unit. So, m = 10g = $\frac{10}{1000}$kg = 0.01.

NSEP 2015 Problem 2 | Rotational Motion

Try out this problem on Rotational Motion from National Standard Examination in Physics 2015-2016.

NSEP 2015-16 ~ Problem 2

A body of mass m and radius R rolling horizontally without slipping at a speed v climbs a ramp to a height $\frac{3v^2}{4g}$. The rolling body can be
(a) A Sphere
(b) A circular Ring
(c) A Spherical Shell
(d) A Circular Disc

Key Concepts

Rotational Mechanics

Rolling without Slipping

Energy Conservation

Suggested Book | Source | Answer

Concept of Physics H.C. Verma

University Physics by H. D. Young and R.A. Freedman

Fundamental of Physics D. Halliday, J. Walker and R. Resnick

National Standard Examination in Physics(NSEP) 2015-2016

Option-(d) A Circular Disc

Try with Hints

Notice that It's a motion without slipping. This implies that the point of contact between the body and the ground has zero relative velocity.

If the relative velocity between the ground and object is 0, then we can use the relation $v = \omega R$, where $\omega $ is the angular velocity.

We know the kinetic energy for the rotational motion is,

$$K_{rot} = \frac{1}{2}I^2 \omega = \frac{Iv^2}{2R^2}$$

Hence, the total kinetic energy is,

$K_{\text {tot }}=K_{\text {linear }}+K_{\text {rot }}$=$\frac{1}{2} m v^{2}+\frac{I v^{2}}{2 R^{2}}$=$\frac{m v^{2}}{2}\left(1+\frac{I}{m R^{2}}\right)$

The object can only go upto the height $\frac{3v^2}{4g}$. At that height it's kinetic energy will be zero and all the energy will be potential energy. So, the potential energy at that height is,

$$ V = mgh = mg\frac{3v^2}{4g} = \frac{3}{4}mv^2 $$

Using the conservation of energy,

$\frac{m v^{2}}{2}\left(1+\frac{I}{m R^{2}}\right)=\frac{3}{4} m v^{2} \rightarrow I$

$=\frac{1}{2} m R^{2}$

We know this is the Moment of Inertia of a Circular disc. Hence, the object is a circular disc.

There is a nice way to remember those moment of inertia values of some standard bodies. , Maybe we will see that in some later blog.

Installing Julia in Ubuntu

Author: Kazi Abu Rousan

C is hard but fast
But you need to be on guard to last.
Python is easy but slow
But you can use it to glow.
But if you have julia
Beautiful rhythms will flow.
---Me

Julia is a high-level, high-performance, dynamic programming language.

Most of you guys have heard or learnt languages like C, C++, Python, Ruby, Java and many more. All of these are great and have helped us in many things. Each type of people love different languages.

But when it comes to science community, most guys use Matlab, Mathematica, Ruby or Python.

This all are great but this all have changed now.

It's because of julia. It is a language which is fast as C (almost close to that with pure julia code), have functions with structure like maths( in pen and paper) and doesn't have Two-Language problem.

Due to this, we people of cheenta is going to make a tutorial on Basic Julia and then we will use Julia in different topics of Math , Physics and also Statistics.

This blog is written just to share few commands to install Julia in Ubuntu 20.04.

List of Commands

First open a terminal inside the folder where you want to save julia.

Then write this command:

wget https://julialang-s3.julialang.org/bin/linux/x64/1.7/julia-1.7.2-linux-x86_64.tar.gz

This will download the latest julia which is 1.7.2 in my case. You can also download the file from here.

Download

Now extract the .tar.gz file using the command

tar -xvzf julia-1.7.2-linux-x86_64.tar.gz

in folder_name/opt folder,i.e., we will create a folder inside the folder julia-1.7.2 with a name opt.

sudo cp -r julia-1.7.2 /opt/

Now we will create a symbolic link to Julia using the command

sudo ln -s /opt/julia-1.7.2/bin/julia /usr/local/bin/julia

It's done!!

Now, enjoy the Julia -- The elegant and fast language.

NSEP 2015-16 Problem 12

Key Concepts

Suggested Book | Source | Answer

Try with Hints

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NSEP 2015-16 ~ Problem 9

Key Concepts

Suggested Book | Source | Answer

Try with Hints

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NSEP 2015-16 ~ Problem 8

Key Concepts

Suggested Book | Source | Answer

Try with Hints

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NSEP 2015-16 ~ Problem 7

Key Concepts

Suggested Book | Source | Answer

Try with Hints

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NSEP 2015-16 ~ Problem 6

Key Concepts

Suggested Book | Source | Answer

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NSEP 2015-16 ~ Problem 5

Key Concepts

Suggested Book | Source | Answer

Try with Hints

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NSEP 2015-16 ~ Problem 4

Key Concepts

Suggested Book | Source | Answer

Try with Hints

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NSEP 2015-16 ~ Problem 3

Key Concepts

Suggested Book | Source | Answer

Try with Hints

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NSEP 2015-16 ~ Problem 2

Key Concepts

Suggested Book | Source | Answer

Try with Hints

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List of Commands