Physics Net Set Slet etc.
Tuesday, 4 October 2016
Monday, 3 October 2016
Problem 4
- For a system performing small oscillations, which of the following statement is correct?
- The number of normal modes and the number of normal coordinates is equal
- The number of normal modes is twice the number of normal coordinates
- The number of normal modes is half the number of normal coordinates
- There is no specific relationship between the number of normal modes and the number of normal coordinates
- For any process, the second law of thermodynamics requires that the change of entropy of the universe be
- Positive only
- Positive or zero
- Zero only
- Negative or zero
- A body of mass $M=m_1+m_2$ at rest splits into two parts of masses $m_1$ and $m_2$ by an internal explosion which generates a kinetic energy $E$. The speed of mass $m_2$ relative to mass $m_1$ is
- $\sqrt{\frac{E}{m_1m_2}}$
- $\sqrt{\frac{2E}{m_1m_2}}$
- $\sqrt{\frac{EM}{m_1m_2}}$
- $\sqrt{\frac{2EM}{m_1m_2}}$
- The value of $$x=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\dots} }}$$
- $\sqrt{2}$
- 1.6
- $\sqrt{3}$
- 0.8
- In Young's double slit experiment, if one of the following parameters ($\lambda$, $d$ and $D$) is increased in the same order keeping the other two same, then the fringe width
- decreases, decreases, increases
- decreases, increases, increases
- increases, decreases, increases
- increases, increases, decreases
- Ideal Atwood machine is nothing but an inextensible string of negligible mass going around the fixed pulley with masses $m_1$ and $m_2$ attached to the ends of the string. If $m_1>m_2$, then the magnitude of acceleration of mass $m_1$ is
- $\frac{m_1g}{(m_1+m_2)}$
- $\frac{m_2g}{(m_1+m_2)}$
- $\frac{(m_1-m_2)g}{(m_1+m_2)}$
- $g$
- A particle of mass $m$ is released from a large height. Resistive force is directly proportional to velocity $\bar v$ with $k$ as a constant of proportionality. Asymptotic value of the velocity of particle is
- $\frac{g}{k}$
- $\frac{k}{m}$
- $\frac{mg}{k}$
- $\frac{g}{km}$
- The momentum of an electron (rest mass $m_0$), which has the same kinetic energy as its rest mass energy, is
- $m_0c$
- $\sqrt{2}m_0c$
- $\sqrt{3}m_0c$
- $2m_0c$
- A planet of mass $m$ moves around the in an elliptic orbit. If $L$ denotes the angular momentum of the planet, then the rate at which area is swept by the radial vector is
- $\frac{L}{2m}$
- $\frac{L}{m}$
- $\frac{2L}{2}$
- $\frac{\sqrt{2}L}{m}$
- The matrix $\begin{pmatrix}8&x&0\\4&0&2\\12&6&0\end{pmatrix}$ will become singular if the value of $x$ is
- $4$
- $6$
- $8$
- $12$
Friday, 30 September 2016
Problem 3
- Calculate equivalent temperature of 1eV energy Answer: $E=3/2kT$, $1eV=11604.52500617K$
- What are unit and dimensions of wavefunction $\psi$ Answer: Wavefunction may be in position space or momentum space. For a 1-dimensional position space wavefunction $\psi(x)$ the normalization condition is $\int\psi^*(x)\psi(x)dx=1$, so $\psi^*(x)\psi(x)$ has units of inverse distance and $\psi(x)$ has units of square root of inverse distance, e.g. $m^{−1/2}$. For a 2-dimensional position space wavefunction $\psi(x,y)$ the normalization condition would be $\iint\psi^*(x,y)\psi(x,y)dxdy=1$, so $\psi^*(x,y)\psi(x,y)$ has units of inverse area and $\psi(x,y )$ has units of square root of inverse area, e.g. $m^{−1}$. Similarly, for 3-dimension $m^{-3/2}$.
- The typical wavelengths emitted by diatomic molecules in purely vibrational and purely rotational transitions are respectively in the region of
- infrared and visible
- visible and infrared
- infrared and microwaves
- microwaaves and infrared
- Solar cell is a type of :
- Photo-conductive device
- Photo-emissive device
- Photo-voltaic device
- Electromagnetic device
- KCL and KBr are alkali halides, both having the NaCl crystal structure. However, in the X-ray diffraction certain reflections are absent for KCl as compared to KBr, for example (111), (311), (331). The difference in the two similar geometrical structures is because of the following:
- Atomic form factors of K and Cl are similar, but of K and Br are very different
- Atomic form factors of K and Cl are different, but of K and Br are similar
- The structure factors of KCl and KBr are different
- Structure factors of KCl and KBr are different and the form factors of K and Br are also similar.
- For any process, the second law of thermodynamics requires that the change of entropy of the universe be
- Positive only
- Positive or zero
- Zero only
- Negative or zero
In momentum space of 1-dimensional case we have $p^{−1/2}$ i.e. $\frac{1}{\sqrt{kg-m}}$
Wednesday, 28 September 2016
Problem 2
- What would be the approximate length of the day if the earth spun so
fast that bodies floated on the equator? Take the radius of the earth $R=6\times 10^6 m$
and $g=9.8 m/s^2$
(A) 12 hours (B) 6 hours (C) 3 hours (D) 1.5 hourAnswer: Body remains on earth's surface because
$gravitational~ pull ~i.e.~ weight ~of~ the~ body\geq centrifugal~ force ~due ~to~ rotation$
Body will start floating when
$centrifugal~force~due~ to ~rotation> gravitational~pull~i.e. ~weight~of~the~ body$
$$m\omega^2r>mg$$ or $$\omega>\sqrt{\frac{g}{r}}$$ But $$\omega=\frac{2\pi}{T}$$ $$T(in~ seconds)>2\pi\sqrt{\frac{r}{g}}$$
$$T(in~ hours)>2\pi\sqrt{\frac{r}{g}}\frac{1}{3600}=\frac{2\times3.14\times1000\times\sqrt{6}}{3600\times\sqrt{9.8}}=1.5~ hours$$ - The real matrix $A=\begin{pmatrix}a&-f&-g\\ f&a&h\\g&-h&a\end{pmatrix}$ is skew symmetric when
(a) $a=0$ (b) $f=0$ (c) $g=h$ (d) $f=g$ $$$$ Answer: Matrix $A$ is skew symmetric when $A_{ij}=-A_{ji}$. Hence $a=0$ - If $A$ and $B$ are matrices such that $AB=B$ and $BA=A$ then $A^2+B^2$ equals Answer: $$A^2+B^2=(BA)^2+(AB)^2=BABA+ABAB$$ $$A^2+B^2=B\underline{AB}A+A\underline{BA}B=B\underline{BA}+A\underline{AB}=BA+AB=A+B$$
- The average value of function $f(x)=4x^3$ in the interval 1 to 3 is
Answer: $<f(x)>=\frac{\int_1^3f(x)dx}{\int_1^3dx}=\frac{\int_1^34^3dx}{\int_1^3dx}=\frac{[x^4]_1^3}{[x]_1^3}=\frac{81-1}{3-1}=40$
Tuesday, 27 September 2016
Problem 1
Wave function of a particle moving in free space is given by, $\psi=e^{ikx}+2e^{-ikx}$. Find the energy of the particle.
Answer: one dimensional Schrödinger equation is $-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V\psi=E\psi$
For a free particle $V=0$
$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi$
$\frac{d^2\psi}{dx^2}=-k^2\left(e^{ikx}+2e^{-ikx}\right)=-k^2\psi$
$E=\frac{\hbar^2k^2}{2m}$
Answer: one dimensional Schrödinger equation is $-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V\psi=E\psi$
For a free particle $V=0$
$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi$
$\frac{d^2\psi}{dx^2}=-k^2\left(e^{ikx}+2e^{-ikx}\right)=-k^2\psi$
$E=\frac{\hbar^2k^2}{2m}$
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