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Sunday, 6 November 2016
Problem set 20
Let v, p and E denote the speed, the magnitude of the momentum, and the energy of a free particle of rest mass m. Then
dE/dp=constant
p=mv
v=cp/\sqrt{p^2+m^2c^2}
E=mc^2
As particle is moving with velocity v and having rest mass m, its relativistic mass is given by
m_{rel}=\frac{m}{\sqrt{1-v^2/c^2}}
Its momentum is given by
p=m_{rel}v=\frac{mv}{\sqrt{1-v^2/c^2}}
Squaring and rearranging this equation, we get,
p^2\left(1-v^2/c^2\right)=m^2v^2
On solving this equation, we get, v=cp/\sqrt{p^2+m^2c^2}. Hence, option (C) is correct.
The rest mass energy and kinetic energy of particle are mc^2 and K=\frac{mc^2}{\sqrt{1-v^2/c^2}}-mc^2, respectively.
Here, option (D) is wrong, because, total energy E, is sum of kinetic energy plus rest mass energy i.e. E=\frac{mc^2}{\sqrt{1-v^2/c^2}}.
Option (A) is wrong, because, velocity is involved in the expression for E.
A binary star system consists of two stars S_1 and S_2, with masses m and 2m, respectively, separated by a distance r. If both S_1 and S_2 individually follow circular orbits around the centre of mass with instantaneous speeds V_1 and V_2 respectively, the speeds ratio V_1/V_2 is
\sqrt{2}
1
1/2
2
Let the stars are at distances r_1 and r_2 from centre of mass, hence, we have
m_1r_1=m_2r_2\quad\text{and}\quad r_1+r_2=r
For the motion of first star in circular orbit, we have
\frac{Gm_1m_2}{r}=\frac{m_1v_1^2}{r_1}
For the motion of second star in circular orbit, we have
\frac{Gm_1m_2}{r}=\frac{m_2v_2^2}{r_2}
Taking ratio of these two equations, we get,
\frac{v_1}{v_2}=\sqrt{\frac{r_1m_2}{r_2m_1}}
Substituting \frac{r_1}{r_2}=\frac{m_2}{m_1}\frac{v_1}{v_2}=\frac{m_2}{m_1}=2
Hence, answer is (D)
Three charges are located on the circumference of a circle of radius R as shown in the figure below. The two charges Q subtend an angle 90^o at the centre of the circle. The charge q is symmetrically placed with respect to the charges Q. If the electric field at the centre of the circle is zero, what is the magnitude of Q?
q/\sqrt{2}
\sqrt{2}q
2q
2q
From figure, for balancing, we must have
E_q=2E_Q\cos{(45)}
i.e. \frac{1}{4\pi\epsilon_0}\frac{q}{R^2}=2\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}\frac{1}{\sqrt{2}}. Hence, Q=q/\sqrt{2}.
Consider a hollow charged shell of inner radius a and outer radius b. The volume charge density is \rho(r) =\frac{k}{r^2} (k is a constant) in the region a < r < b. The magnitude of the electric field produced at distance r > a is
\frac{k(b-a)}{\epsilon_0r^2} for all r > a
\frac{k(b-a)}{\epsilon_0r^2} for a < r < b and \frac{kb}{\epsilon_0r^2} for r > b
\frac{k(r-a)}{\epsilon_0r^2} for a < r < b and \frac{k(b-a)}{\epsilon_0r^2} for r > b
\frac{k(r-a)}{\epsilon_0a^2} for a < r < b and \frac{k(b-a)}{\epsilon_0a^2} for r > b
Let us divide the region r > a into two regions viz. a\leq r\leq b and r > b.
For region a\leq r\leq b, using Gauss law, we have,
\int\vec E\cdot d\vec s=\frac{q_{enclosed}}{\epsilon_0}\begin{align*}
E\;4\pi r^2&=\frac{1}{\epsilon_0}\int_a^r\rho\:dV\\
&=\frac{1}{\epsilon_0}\int_a^r\frac{k}{r^2}\:4\pi r^2\;dr\\
&=\frac{4\pi k}{\epsilon_0}(r-a)
\end{align*}E=\frac{k(r-a)}{\epsilon_0 r^2}
For region r > b, using Gauss law, we have,
\int\vec E\cdot d\vec s=\frac{q_{enclosed}}{\epsilon_0}\begin{align*}
E\;4\pi r^2&=\frac{1}{\epsilon_0}\int_a^b\rho\:dV\\
&=\frac{1}{\epsilon_0}\int_a^b\frac{k}{r^2}\:4\pi r^2\;dr\\
&=\frac{4\pi k}{\epsilon_0}(b-a)
\end{align*}E=\frac{k(b-a)}{\epsilon_0 r^2}
Hence, answer is (C)
Consider the interference of two coherent electromagnetic waves whose electric field vectors are given by \vec E_1 = \hat i E_0 \cos{(\omega t+\phi)} and \vec E_2 = \hat j E_0 \cos{(\omega t+\phi)} where \phi is the phase difference. The intensity of the resulting wave is given by \frac{\epsilon_0}{2}\left < E^2\right >, where \left < E^2\right > is the time average of E^2. The total intensity is
0
\epsilon_0E_0^2
\epsilon_0E_0^2\sin^2{\phi}
\epsilon_0E_0^2\cos^2{\phi}
Resultant electric field is given by
\begin{align*}
\vec E &= \hat i E_0 \cos{(\omega t+\phi)}\\
&+\hat j E_0 \cos{(\omega t+\phi)}
\end{align*} E^2 = 2 E_0^2 \cos^2{(\omega t+\phi)}\begin{align*}
\left < E^2\right > &=\frac{\int_0^TE^2\,dt}{\int_0^T\,dt}\\
&=\frac{2E_0^2}{T}\int_0^T\cos^2{(\omega t+\phi)}\;dt\\
&={\scriptstyle\frac{2E_0^2}{T}\frac{1}{2}\int_0^T\left[1-\cos2{(\omega t+\phi)}\right]\;dt}\\
&=\frac{E_0^2}{T}T=E_0^2
\end{align*}I=\frac{\epsilon_0}{2}\left < E^2\right > =\frac{\epsilon_0E_0^2}{2}
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