JEE 2011 :: General Questions :: Physics

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Phase space diagrams are useful tools in analyzing all kinds of dynamic problems. They are espcially useful in studying the changes in motion as initial position and momenturn are
changed. Here we consider some simple dynamical systems in one dimension. For such sysfems, phase space is a plane in which position is plotted along horizontol axis and momentum is plotted along vertical axis. The phase space diagram is x(t) vs p(t) curue in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative 

The phase space diagram for a simple harmonic motion is circle centerecl at the origin. In the figure , the two circles represent the same osciliator but for different initial conditions , and $E_{1}$ and $E_{2}$ are the total mechanical energies respectively , Then 

Answer:

Explanation:

 $E= \frac{1}{2} m \omega^{2} A^{2}$

 or $ E \propto A^{2}$

 $\frac{E_{2}}{E_{1}}= (\frac{A_{2}}{A_{1}})^{2}$

   =   $\left(\frac{a}{2a}\right)^{2}$

 or $E_{1}=4E_{2}$

 $\therefore$ correct option is (c)

A dense collection of an equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let N be the number density of free electrons, each of mass m. When the electrons are subjected to an electric field, they are displaced relatively away from the
heavy positive ions. lf the electric field becomes zero, the electrons being to oscillate about the positive ions with a natural! angular frequency ω_p, which is called the plasma frequency. To
sustain the oscillations, a time-varying electric field needs to be applied that has an angular frequency ω, where a part of the energy is absorbed and a part of it is reflected. As $\omega$  approaches $\omega_{p}$,  all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of the high reflectivity of metals

Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons $N= 4 \times 10^{27}m^{-1}$. Take  $\epsilon_{0}=10^{-11}$ and $m=10^{-30}$ where these quantities are in proper SI unit

Answer:

Explanation:

 $\omega =2 \pi f =\frac{2 \pi c}{\lambda}$

 $\therefore$  $\lambda = \frac{2 \pi c}{ \omega}= \frac{2 \pi c }{\sqrt{Ne^{2}/m\epsilon_{0}}}$

 Substituting  the values , we get

 $\lambda= 600nm$

 $\therefore$ correct option is (b)

A dense collection of an equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let N be the number density of free electrons, each of mass m. When the electrons are subjected to an electric field, they are displaced relatively away from the
heavy positive ions. lf the electric field becomes zero, the electrons being to oscillate about the positive ions with a natural! angular frequency ω_p, which is called the plasma frequency. To
sustain the oscillations, a time-varying electric field needs to be applied that has an angular frequency ω, where a part of the energy is absorbed and a part of it is reflected. As $\omega$  approaches $\omega_{p}$,  all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of the high reflectivity of metals

Taking the electronic change as e and the permittivity as $\epsilon_{0}$ , use dimensional analysis to determine the correct expression for $\omega_{p}$

Answer:

Explanation:

 N= number of electrons per unit volume 

 $\therefore$   $[N]=[L^{-3}],[e]=[q]$

 = [It]=[AT]

 $ [\epsilon_{0}]=[M^{-1}L^{-3}T^{4}A^{2}]$

 Substituting the dimensions, we can see that 

$\left[\sqrt{\frac{Ne^{2}}{m\epsilon_{0}}}\right]=[T^{-1}]$

 Angular frequency has also the dimension $[T^{-1}]$ 

$\therefore$ correct option  is (c)

A metal rod of length L and mass m is pivoted at one end. A thin disc of mass M and radius R (< L) is attached at its centre to the free end of the rod. Consider two ways the disc is attached. Case  A-the disc is not free to rotate about its centre and case B-the disc is free to rotate about its centre. The rod-disc system performs  SHM in a vertical plane after being released from the same displaced position.

Which of the following statement(s) is/are true?

Answer:

Explanation:

$\tau_{A}=\tau_{B}=mg \frac{L}{2} \sin \theta +MgL \sin \theta$

= Restoring torque about Point O.

In case A, a moment of inertia will be more. Hence, angular acceleration ($\alpha= \tau/I$) will be less. Therefore, the angular frequency will be less.

 $\therefore$ Correct options are (a) and (d)

Analysis of Question
The question is difficult from my point of view. Because this type of SHM is rarely taught in the class and questions of this type are not given in standard books

A spherical metal shell A of radius RAand a solid metal sphere B of radius  $R_{B} <(R_{A})$ are kept far apart and each is given charge +Q. Now, they are connected by a thin metal wire. Then

Answer:

Explanation:

Inside a conducting shell, electric field is always zero. Therefore, option (a) is correct. When the two are connected. their potentials become the same.

 $\therefore$    $V_{A}=V_{B}$

 or $ \frac{Q_{A}}{R_{A}}=\frac{Q_{B}}{R_{B}}$   $\left( \therefore V= \frac{1}{4 \pi \epsilon_{0}}.\frac{Q}{R}\right)$

 Since, $R_{A}  >R_{B}$ 

$\therefore$   $Q_{A} > Q_{B}$

 $\therefore$  option (b) is correct 

 Potential is also equal to 

 $V= \frac{\sigma R}{\epsilon_{0}}$

 $V_{A}=V_{B}$

 $\therefore$   $\sigma_{A} R_{A}=\sigma_{B} R_{B}$

 or $ \frac{ \sigma _{A}}{\sigma_{B}}= \frac{ R_{B}}{R_{A}}$

 $\therefore$  Option (c) is correct

 Electric field on surface

 $E= \frac{\sigma}{E_{0}}$ or  $E \propto \sigma$

 or $ \sigma_{A} < \sigma_{B}$

 Since, $ \sigma_{A} < \sigma_{B}$

 $\therefore$   $E_{A} < E_{B}$

 $\therefore$ Option (d) is also correct

 $\therefore$ Correct options are (a),(b) ,(c) and (d) 

 Analysics of Question

 (i) Question is simple

 (ii)  $E= \frac{1}{4 \pi \epsilon_{0}}\frac{Q}{R^{2}}=\frac{\sigma}{\epsilon_{0}}$ ( on surface)

 As   $ \sigma $= surface charge density

 = $\frac{Q}{4 \pi R^{2}}$

 (iii)   $V= \frac{1}{4 \pi \epsilon_{0}}\frac{Q}{R^{}}=\frac{\sigma R}{\epsilon_{0}}$

• General Questions - Physics
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