Enhance a problem solving ability in Physics for various competitive and qualifying examinations like GRE, GATE, CSIR JRF-NET, SET, UPSC etc.
Notice
Sunday, 23 October 2016
Problem set 13
A rod of length l_0 makes an angle \theta_0 with the y-axis in its rest frame, while the rest frame moves to the right along the x-axis with relativistic speed v with respect to lab frame. If \gamma=\frac{1}{\sqrt{1-v^2/c^2}}, the angle \theta in the lab frame is
\theta=\tan^{-1}{(\gamma\tan\theta_0)}
\theta=\tan^{-1}{(\gamma\cot\theta_0)}
\theta=\tan^{-1}{(\frac{1}{\gamma}\tan\theta_0)}
\theta=\tan^{-1}{(\frac{1}{\gamma}\cot\theta_0)}
The x and y components of length l_0 are given by
l_x=l_0\sin\theta_0, l_y=l_0\cos\theta_0
As the frame is moving along x-axis to the right, there is contraction of length in l_x only
l_x'=(l_0\sin\theta_0)\sqrt{1-v^2/c^2}l_y'=l_0\cos\theta_0\begin{align*}
\tan\theta&=\frac{l_y'}{l_x'}\\
&=\frac{l_0\cos\theta_0}{(l_0\sin\theta_0)\sqrt{1-v^2/c^2}}\\
&=\gamma\cot\theta_0
\end{align*}\theta=\tan^{-1}(\gamma\cot\theta_0)
A particle of mass m moves in a potential V(x)=\frac{1}{2}m\omega^2x^2+\frac{1}{2}m\mu v^2, where x is the position coordinate, v is the speed, and \omega and \mu are constants. The canonical (conjugate) momentum of the particle is
p=m(1+\mu)v
p=mv
p=m\mu v
p=m(1-\mu)v
\begin{align*}
L&=T-V(x)\\
&=\frac{1}{2}m\dot x^2-\frac{1}{2}m\omega^2x^2-\frac{1}{2}m\mu \dot v^2\\
&=\frac{1}{2}m\dot x^2-\frac{1}{2}m\omega^2x^2-\frac{1}{2}m\mu \dot x^2
\end{align*}
Conjugate momentum is given by
\begin{align*}
p&=\frac{\partial L}{\partial\dot x}\\
&=m\dot x-m\mu\dot x=m(1-\mu)v
\end{align*}
A solid sphere of radius R carries a uniform volume charge density \rho. The magnitude of electric field inside the sphere at a distance r from center is
\frac{r\rho}{3\epsilon_0}
\frac{R\rho}{3\epsilon_0}
\frac{R^2\rho}{r\epsilon_0}
\frac{R^3\rho}{r^2\epsilon_0}
According to Gauss law , if the surface encloses a charge q within it, then net flux through the surface is q/\epsilon_0. The flux is given by
\phi_E=\int_S\vec E.d\vec A=\frac{q}{\epsilon_0}
If \rho is the volume charge density, the charge enclosed within a distance r is q=\frac{4}{3}\pi r^3\rho and
\int_S\vec E.d\vec A=E(4\pi r^2)E(4\pi r^2)=\frac{4}{3\epsilon_0}\pi r^3\rhoE=\frac{\rho r}{3\epsilon_0}
The electric field \vec E(\vec r,t) for a circularly polarized electromagnetic wave propagating along the positive z direction is
E_0(\hat x+\hat y)exp[i(kz-\omega t)]
E_0(\hat x+i\hat y)exp[i(kz-\omega t)]
E_0(\hat x+i\hat y)exp[i(kz+\omega t)]
E_0(\hat x+\hat y)exp[i(kz+\omega t)]
Plane wave travelling along the positive z-axis is given by \vec E(z,t)=\vec E_0 cos(kz-\omega t)
Plane wave travelling along the negative z-axis is given by \vec E(z,t)=\vec E_0 cos(kz+\omega t)
A beam linearly polarized along the x-axis and traveling in the positive z-direction can be
represented by:\vec E(z,t)= E_0 cos(kz-\omega t)\hat x
Circularly polarized light can be represented by the expressions:
\begin{align*}
\vec E_{RCP}(z,t)&= E_0 \left(cos(kz-\omega t)\hat x\right.\\
&{}\left.+sin(kz-\omega t)\hat y\right)
\end{align*}\begin{align*}
\vec E_{LCP}(z,t)&= E_0 \left(cos(kz-\omega t)\hat x\right.\\
&\left.-sin(kz-\omega t)\hat y\right)
\end{align*}
Consider option (A)
\begin{align*}
\vec E&=E_0(\hat x+\hat y)exp[i(kz-\omega t)]\\
&=E_0(\hat x+\hat y)\left(cos(kz-\omega t)\right.\\
&{}\left.+isin(kz-\omega t)\right)\\
&=E_0\left[cos(kz-\omega t) \hat x\right.\\
&\left.{}+cos(kz-\omega t)\hat y\right]\\
&{}+iE_0\left[sin(kz-\omega t)\hat x\right.\\
&{}\left.+sin(kz-\omega t)\hat y\right]
\end{align*}
Hence real part of \vec E is
\begin{align*}
Re \vec E&=E_0\left[cos(kz-\omega t) \hat x\right.\\
&{}\left.+cos(kz-\omega t)\hat y\right]
\end{align*}
which represents plane polarized light.
Now consider the option (B)
\begin{align*}
\vec E&=E_0(\hat x+i\hat y)exp[i(kz-\omega t)]\\
&=E_0(\hat x+i\hat y)\left(cos(kz-\omega t)\right.\\
&\left. +isin(kz-\omega t)\right)\\
&=E_0\left[cos(kz-\omega t) \hat x\right.\\
&\left.-sin(kz-\omega t)\hat y\right]\\
&+iE_0\left[sin(kz-\omega t)\hat x\right.\\
&\left.+cos(kz-\omega t)\hat y\right]
\end{align*}\begin{align*}
Re \vec E&=E_0\left[cos(kz-\omega t) \hat x\right.\\
&\left.-sin(kz-\omega t)\hat y\right]
\end{align*}
Which represents a circularly polarized light, as
\begin{align*}
E_x^2+E_y^2&=E_0^2cos^2(kz-\omega t)\\
&+E_0^2sin^2(kz-\omega t)\\
&=E_0^2
\end{align*}
is equation of circle. Hence (B) is correct option.
An unbiased coin is tossed n times. The probability that exactly m heads will
come up is
\frac{n}{2^m}
\frac{1}{2^n}\frac{n!}{m!(n-m)!}
\frac{1}{2^m}\frac{n!}{m!(n-m)!}
\frac{m}{2^n}
This problem is equivalent to 1D random walk problem. In a single toss probability getting head is p=\frac{1}{2} and probability of getting tail is q=\frac{1}{2}. In a sequences of n tosses probability of getting m heads and n-m tails is given by
p^mq^{(n-m)}=\left(\frac{1}{2}\right)^m\left(\frac{1}{2}\right)^{n-m}
However, total combinations of m times head up and (n-m) times tail up are given by
\frac{n!}{m!(n-m)!}
Hence, probability of getting m heads is
\begin{align*}
P&=\frac{n!}{m!(n-m)!}\left(\frac{1}{2}\right)^m\left(\frac{1}{2}\right)^{n-m}\\
&=\frac{n!}{m!(n-m)!}\left(\frac{1}{2}\right)^n
\end{align*}
Hence, answer is (B).
No comments :
Post a Comment