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Physics Resonance: Problem set 28 -->

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Saturday, 19 November 2016

Problem set 28

  1. The solution of the differential equation \frac{dx}{dt}=2\sqrt{1-x^2}, with initial condition x=0 at t=0 is
    1. \begin{align*} x=\begin{cases} \sin{2t},\quad 0\leq t<\frac{\pi}{4}\\ \sinh{2t},\quad t\geq\frac{\pi}{4} \end{cases} \end{align*}
    2. \begin{align*} x=\begin{cases} \sin{2t},\quad 0\leq t<\frac{\pi}{2}\\ 1,\quad t\geq\frac{\pi}{2} \end{cases} \end{align*}
    3. \begin{align*} x=\begin{cases} \sin{2t},\quad 0\leq t<\frac{\pi}{4}\\ 1,\quad t\geq\frac{\pi}{4} \end{cases} \end{align*}
    4. x=1-\cos{2t},\quad t\geq 0
  2. Given a uniform magnetic field \vec B=B_0\hat k (where B_0 is a constant), a possible choice for the magnetic vector potential \vec A is
    1. B_0y\hat i
    2. -B_0y\hat i
    3. B_0(x\hat j+y\hat i)
    4. B_0(x\hat i-y\hat j)
  3. Consider a charge Q at the origin of 3-dimensional coordinate system. The flux of the electric field through the curved surface of a cone that has a height h and a circular base of radius R is
    1. \frac{Q}{\epsilon_0}
    2. \frac{Q}{2\epsilon_0}
    3. \frac{hQ}{R\epsilon_0}
    4. \frac{QR}{2h\epsilon_0}
  4. A Hermitian operator \hat O has two normalised eigenstates |1 > and |2 > with eigenvalues 1 and 2, respectively. The two states |u > =\cos\theta|1 > +\sin\theta|2 > and |v > =\cos\phi|1 > +\sin\phi|2 > are such that < v|\hat O|v > =7/4 and < u|v > =0. Which of the following are possible values of \theta and \phi?
    1. \theta=-\frac{\pi}{6} and \phi=\frac{\pi}{3}
    2. \theta=\frac{\pi}{6} and \phi=\frac{\pi}{3}
    3. \theta=-\frac{\pi}{4} and \phi=\frac{\pi}{4}
    4. \theta=\frac{\pi}{3} and \phi=-\frac{\pi}{6}
  5. The ground state energy of a particle of mass m in the potential V(x)=V_0\cosh{\left(\frac{x}{L}\right)}, where L and V_0 are constants (and V_0>>\frac{\hbar^2}{2mL^2}) is approximately
    1. V_0+\frac{\hbar}{L}\sqrt{\frac{2V_0}{m}}
    2. V_0+\frac{\hbar}{L}\sqrt{\frac{V_0}{m}}
    3. V_0+\frac{\hbar}{4L}\sqrt{\frac{V_0}{m}}
    4. V_0+\frac{\hbar}{2L}\sqrt{\frac{V_0}{m}}

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