\int\frac{dx}{\sqrt{1-x^2}}=2\int dt
Let us make substitution x=\sin\theta, dx=\cos\theta\int d\theta=2t+c_1\theta=2t+c_1\sin^{-1}x=c_1+2tx=c+\sin{2t}
At t=0, x=0, hence, c=0x=\sin{2t}
However, maximum value of \sin(2t) is 1
\sin{2t}=1\quad \text{when } 2t=\frac{\pi}{2}\quad\text{ or }t=\frac{\pi}{4}
Hence,
\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*}
Hence, answer is (B).
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
B_0y\hat i
-B_0y\hat i
B_0(x\hat j+y\hat i)
B_0(x\hat i-y\hat j)
Using, \vec B=\vec\nabla \times\vec A, for options
(A)\vec B=\vec\nabla \times B_0y\hat i=-B_0\hat k
(B)\vec B=\vec\nabla \times (-B_0y\hat i)=B_0\hat k
(C)\vec B=\vec\nabla \times (B_0(x\hat j+y\hat i))=0
(D)\vec B=\vec\nabla \times (B_0(x\hat i-y\hat j))=2B_0\hat k
Hence, answer is (B)
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
\frac{Q}{\epsilon_0}
\frac{Q}{2\epsilon_0}
\frac{hQ}{R\epsilon_0}
\frac{QR}{2h\epsilon_0}
Let us consider a sphere of radius of h (if h>R) or of radius R (if R>h) with centre at origin. According to Gauss law the flux through entire sphere is \frac{Q}{\epsilon_0}. The flux through curved surfaces of cone is half of the flux through sphere i.e. \frac{Q}{2\epsilon_0}
Hence, answer is (B)
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?
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
V_0+\frac{\hbar}{L}\sqrt{\frac{2V_0}{m}}
V_0+\frac{\hbar}{L}\sqrt{\frac{V_0}{m}}
V_0+\frac{\hbar}{4L}\sqrt{\frac{V_0}{m}}
V_0+\frac{\hbar}{2L}\sqrt{\frac{V_0}{m}}
\begin{align*}
V(x)&=V_0\cosh{\left(\frac{x}{L}\right)}\\
&=\frac{V_0}{2}\left(e^{x/L}+ e^{-x/L}\right)\\
&=\frac{V_0}{2}\left\{\left[1+\frac{x}{L}+\frac{1}{2!}\left(\frac{x}{L}\right)^2+\cdots\right]\right.\\
&\left. +\left[1-\frac{x}{L}+\frac{1}{2!}\left(\frac{x}{L}\right)^2-\cdots\right]\right\}\\
&=\frac{V_0}{2}\left\{\left[2+2\frac{1}{2!}\left(\frac{x}{L}\right)^2+\cdots\right]\right\}\\
&=V_0+\frac{V_0}{2}\left(\frac{x}{L}\right)^2\\
&=V_0+\frac{1}{2}\frac{V_0}{L^2}x^2\\
&=V_0+\frac{1}{2}kx^2
\end{align*}
where k=\frac{V_0}{L^2}. Thus, particle is under the action of constant potential V_0 and simple harmonic potential
\frac{1}{2}kx^2. For free particle motion lowest energy is V_0 and for harmonic potential lowest energy is \frac{1}{2}\hbar\omega.
Hence, ground state energy is
\begin{align*}
E&=V_0+\frac{1}{2}\hbar\omega\\
&=V_0+\frac{1}{2}\hbar\sqrt{\frac{k}{m}}\\
&=V_0+\frac{1}{2}\hbar\sqrt{\frac{V_0}{L^2m}}
\end{align*}
Hence, answer is (D)
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