id
int64 0
467
| images
images listlengths 0
10
| problem
stringlengths 221
32.8k
| answer
stringlengths 1
679
| reason
stringlengths 18
12.1k
| error range
stringclasses 2
values | source
stringclasses 26
values | answer_type
stringclasses 5
values |
|---|---|---|---|---|---|---|---|
0
|
Consider a neutron star with a mass of $\quad 2.0 .10^{30} \mathrm{~kg}$, an average radius of $\quad 1.0 .10^{4} \mathrm{~m}$, and a rotation period of $2.0 .10^{-2} \mathrm{~s}$. a - Calculate the flattening factor, given that the gravitational constant is $6.67 .10^{-11}$ N. m ${ }^{2} . \mathrm{kg}^{-2}$.
|
$3,7.10^{-4}$
|
For equilibrium we have $\mathrm{F}_{\mathrm{c}}=\mathrm{F}_{\mathrm{g}}+\mathrm{N}$ where N is normal to the surface. Resolving into horizontal and vertical components, we find: $$ \begin{gathered} F_{g} \cdot \cos (\phi)=F_{c}+N \cdot \sin (\alpha) \\ \quad F_{g} \cdot \sin (\phi)=N \cdot \cos (\alpha) \end{gathered} \rightarrow F_{g} \cdot \cos (\phi)=F_{c}+F_{g} \cdot \sin (\phi) \cdot \operatorname{tg}(\alpha) $$ From: $$ F_{g}=\frac{G \cdot M}{r^{2}}, \quad F_{c}=\omega^{2} \cdot r, \quad x=r \cdot \cos (\phi), \quad y=r \cdot \sin (\phi) \text { en } \operatorname{tg}(\alpha)=\frac{d y}{d x} $$ we find: $$ y \cdot d y+\left(1-\frac{\omega^{2} \cdot r^{3}}{G \cdot M}\right) \cdot x \cdot d x=0 $$ where: $$ \frac{\omega^{2} . r^{3}}{G . M} \approx 7.10^{-4} $$ This means that, although r depends on x and y , the change in the factor in front of xdx is so slight that we can take it to be constant. The solution of Eq. (1) is then an ellipse: $$ \frac{x^{2}}{r_{e}^{2}}+\frac{y^{2}}{r_{p}^{2}}=1 \rightarrow \frac{r_{p}}{r_{e}}=\sqrt{1-\frac{\omega^{2} \cdot r^{3}}{G \cdot M}} \approx 1-\frac{\omega^{2} \cdot r^{3}}{2 \cdot G \cdot M} $$ and from this it follows that: $$ \epsilon=\frac{r_{e}-r_{p}}{r_{e}}=\frac{\omega^{2} \cdot r^{3}}{2 . G . M} \approx 3,7.10^{-4} $$
|
IPHO 1990
|
value
|
||
1
|
In the long run (over many years) the rotation of the star slows down, due to energy loss, and this leads to a decrease in the flattening. The star has however a solid crust that floats on a liquid interior. The solid crust resists a continuous adjustment to equilibrium shape. Instead, starquakes occur with sudden changes in the shape of the crust towards equilibrium. During and after such a star-quake the angular velocity is observed to change according to figure 1. <image> Figure 1 time (days) --> A sudden change in the shape of the crust of a neutron star results in a sudden change of the angular velocity. b - Calculate the average radius of the liquid interior, using the data of Fig. 1. Make the approximation that the densities of the crust and the interior are the same. (Ignore the change in shape of the interior).
|
0.95
|
For a point mass of 1 kg on the surface, $$ U_{p o t}=-\frac{G \cdot M}{r} \quad U_{k i n}=\frac{1}{2} \cdot \omega^{2} \cdot r^{2} \cdot \cos ^{2}(\phi) $$ The form of the surface is such that $\mathrm{U}_{\text {pot }}-\mathrm{U}_{\text {kin }}=$ constant. For the equator ( $\Phi=0$, $r=r_{e}$ ) and for the pole ( $\Phi=\pi / 2, r=r_{p}$ ) we have: $$ \frac{G . M}{r_{p}}=\frac{G . M}{r_{e}}+\frac{1}{2} \cdot \omega^{2} \cdot r_{e}^{2} \rightarrow \frac{r_{e}}{r_{p}}=1+\frac{\omega^{2} \cdot r_{e}^{3}}{2 . G . M} $$ Thus: $$ \epsilon=\frac{r_{e}-r_{p}}{r_{e}}=\frac{1+\frac{\omega^{2} \cdot r_{e}^{3}}{2 \cdot G \cdot M}-1}{1+\frac{\omega^{2} \cdot r_{e}^{3}}{2 \cdot G \cdot M}} \approx \frac{\omega^{2} \cdot r_{e}^{3}}{2 \cdot G \cdot M} \approx 3,7 \cdot 10^{-4} $$ b - As a consequence of the star-quake, the moment of inertia of the crust $I_{m}$ decreases by $\Delta I_{m}$. From the conservation of angular momentum, we have: $$ I_{m} \cdot \omega_{0}=\left(I_{m}-\Delta I_{m}\right) \cdot \omega_{1} \quad \rightarrow \quad \Delta I_{m}=I_{m} \cdot \frac{\omega_{1}-\omega_{0}}{\omega_{1}} $$ After the internal friction has equalized the angular velocities of the crust and the core, we have: $$ \begin{gathered} \left(I_{m}+I_{c}\right) \cdot \omega_{0}=\left(I_{m}+I_{c}-\Delta I_{m}\right) \cdot \omega_{2} \rightarrow \Delta I_{m}=\left(I_{m}+I_{c}\right) \cdot \frac{\omega_{2}-\omega_{0}}{\omega_{2}} \\ \frac{I_{m}}{I_{m}+I_{c}}=\frac{\left(\omega_{2}-\omega_{0}\right) \cdot \omega_{1}}{\left(\omega_{1}-\omega_{0}\right) \cdot \omega_{2}} \rightarrow 1-\frac{I_{c}}{I_{m}+I_{c}}=\frac{\left(\omega_{2}-\omega_{0}\right) \cdot \omega_{1}}{\left(\omega_{1}-\omega_{0}\right) \cdot \omega_{2}} \\ I(:) R^{2} \\ \rightarrow \frac{I_{c}}{I_{m}+I_{c}}=\frac{r_{c}^{2}}{r^{2}} \rightarrow \frac{r_{c}}{r}=\sqrt{1-\frac{\left(\omega_{2}-\omega_{0}\right) \cdot \omega_{1}}{\left(\omega_{1}-\omega_{0}\right) \cdot \omega_{2}}} \approx 0.95 \end{gathered} $$
|
IPHO 1990
|
value
|
||
2
|
An electron after leaving a device, which accelerated it with the potential difference $U$, falls into a region with an inhomogeneous field $\boldsymbol{B}$ generated with a system of stationary coils $L_{1}, L_{2}, \ldots, L_{n}$. The known currents in the coils are $i_{1}, i_{2}, \ldots, i_{n}$, respectively. What should the currents $i_{1}{ }^{\prime}, i_{2}{ }^{\prime}, \ldots, i_{n}$ ' in the coils $L_{1}, L_{2}, \ldots, L_{n}$ be, in order to guide the proton (initially accelerated with the potential difference $-U$ ) along the same trajectory (and in the same direction) as that of the electron?
|
$$ i^{\prime}{ }_{n}=-35.0 i_{n} . $$
|
At the beginning one should notice that the kinetic energy of the electron accelerated with the potential difference $U=511 \mathrm{kV}$ equals to its rest energy $E_{0}$. Therefore, at least in the case of the electron, the laws of the classical physics cannot be applied. It is necessary to use relativistic laws. The relativistic equation of motion of a particle with the charge $e$ in the magnetic field B has the following form: $$ \frac{d}{d t} \mathbf{p}=\mathbf{F}_{L} $$ where $\mathbf{p}=m_{0} \gamma \mathbf{v}$ denotes the momentum of the particle (vector) and $$ \mathbf{F}_{L}=e \mathbf{v} \times \mathbf{B} $$ is the Lorentz force (its value is $e v B$ and its direction is determined with the right hand rule). $m_{0}$ denotes the (rest) mass of the particle and $v$ denotes the velocity of the particle. The quantity $\gamma$ is given by the formula: $$ \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} $$ The Lorentz force $\mathbf{F}_{L}$ is perpendicular to the velocity $\mathbf{v}$ of the particle and to its momentum $\mathbf{p}=m_{0} \gamma \mathbf{v}$. Hence, $$ \mathbf{F}_{L} \cdot \mathbf{v}=\mathbf{F}_{L} \cdot \mathbf{p}=0 . $$ Multiplying the equation of motion by $\mathbf{p}$ and making use of the hint given in the text of the problem, we get: $$ \frac{1}{2} \frac{d}{d t} p^{2}=0 . $$ It means that the value of the particle momentum (and the value of the velocity) is constant during the motion: $$ p=m_{0} v \gamma=\text { const; } \quad v=\text { const. } $$ The same result can be obtained without any formulae in the following way: The Lorentz force $\mathbf{F}_{L}$ is perpendicular to the velocity $\mathbf{v}$ (and to the momentum $p$ as $\mathbf{p}=m_{0} \gamma \mathbf{v}$ ) and, as a consequence, to the trajectory of the particle. Therefore, there is no force that could change the component of the momentum tangent to the trajectory. Thus, this component, whose value is equal to the length of $\mathbf{p}$, should be constant: $\mathbf{p}=$ const. (The same refers to the component of the velocity tangent to the trajectory as $\mathbf{p}=m_{0} \gamma \mathbf{v}$ ). Let $s$ denotes the path passed by the particle along the trajectory. From the definition of the velocity, we have: $$ \frac{d s}{d t}=v . $$ Using this formula, we can rewrite the equation of motion as follows: $$ \begin{gathered} v \frac{d}{d s} \mathbf{p}=\frac{d s}{d t} \frac{d}{d s} \mathbf{p}=\frac{d}{d t} \mathbf{p}=\mathbf{F}_{L} \\ \frac{d}{d s} \mathbf{p}=\frac{\mathbf{F}_{L}}{v} \end{gathered} $$ Dividing this equation by $p$ and making use of the fact that $p=$ const, we obtain: $$ v \frac{d}{d s} \frac{\mathbf{p}}{p}=\frac{\mathbf{F}_{L}}{v p} $$ and hence $$ \frac{d}{d s} \mathbf{t}=\frac{\mathbf{F}_{L}}{v p} $$ where $\mathbf{t}=\mathbf{p} / p=\mathbf{v} / v$ is the versor (unit vector) tangent to the trajectory. The above equation is exactly the same for both electrons and protons if and only if the vector quantity: $$ \frac{\mathbf{F}_{L}}{v p} $$ is the same in both cases. Denoting corresponding quantities for protons with the same symbols as for the electrons, but with primes, one gets that the condition, under which both electrons and protons can move along the same trajectory, is equivalent to the equality: $$ \frac{\mathbf{F}_{L}}{v p}=\frac{\mathbf{F}_{L}^{\prime}}{v^{\prime} p^{\prime}} . $$ However, the Lorentz force is proportional to the value of the velocity of the particle, and the directions of any two vectors of the following three: $\mathbf{t}$ (or $\mathbf{v}$ ), $\mathbf{F}_{\mathrm{L}}, \mathbf{B}$ determine the direction of the third of them (right hand rule). Therefore, the above condition can be written in the following form: $$ \frac{e \mathbf{B}}{p}=\frac{e^{\prime} \mathbf{B}^{\prime}}{p^{\prime}} $$ Hence, $$ \mathbf{B}^{\prime}=\frac{e}{e^{\prime}} \frac{p^{\prime}}{p} \mathbf{B}=\frac{p^{\prime}}{p} \mathbf{B} $$ This means that at any point the direction of the field $\mathbf{B}$ should be conserved, its orientation should be changed into the opposite one, and the value of the field should be multiplied by the same factor $p^{\prime} / p$. The magnetic field $\mathbf{B}$ is a vector sum of the magnetic fields of the coils that are arbitrarily distributed in the space. Therefore, each of this fields should be scaled with the same factor $-p^{\prime} / p$. However, the magnetic field of any coil is proportional to the current flowing in it. This means that the required scaling of the fields can only be achieved by the scaling of all the currents with the same factor $-p^{\prime} / p$ : $$ i_{n}^{\prime}=-\frac{p^{\prime}}{p} i_{n} $$ Now we shall determine the ratio $p^{\prime} / p$. The kinetic energies of the particles in both cases are the same; they are equal to $E_{k}=e|U|=511 \mathrm{keV}$. The general relativistic relation between the total energy $E$ of the particle with the rest energy $E_{0}$ and its momentum $p$ has the following form: $$ E^{2}=E_{0}^{2}+p^{2} c^{2} $$ where c denotes the velocity of light. The total energy of considered particles is equal to the sum of their rest and kinetic energies: $$ E=E_{0}+E_{k} . $$ Using these formulae and knowing that in our case $E_{k}=e|U|=E_{e}$, we determine the momenta of the electrons ( $p$ ) and the protons ( $p^{\prime}$ ). We get: a) electrons: $$ \begin{gathered} \left(E_{e}+E_{e}\right)^{2}=E_{e}^{2}+p^{2} c^{2} \\ p=\frac{E_{e}}{c} \sqrt{3} \end{gathered} $$ b) protons $$ \begin{gathered} \left(E_{p}+E_{e}\right)^{2}=E_{p}^{2}+p^{\prime 2} c^{2} \\ p^{\prime}=\frac{E_{e}}{c} \sqrt{\left(\frac{E_{p}}{E_{e}}+1\right)^{2}-\left(\frac{E_{p}}{E_{e}}\right)^{2}} \end{gathered} $$ Hence, $$ \frac{p^{\prime}}{p}=\frac{1}{\sqrt{3}} \sqrt{\left(\frac{E_{p}}{E}+1\right)^{2}-\left(\frac{E_{p}}{E_{e}}\right)^{2}} \approx 35.0 $$ and $$ i^{\prime}{ }_{n}=-35.0 i_{n} . $$ It is worthwhile to notice that our protons are 'almost classical', because their kinetic energy $E_{k}\left(=E_{e}\right)$ is small compared to the proton rest energy $E_{p}$. Thus, one can expect that the momentum of the proton can be determined, with a good accuracy, from the classical considerations. We have: $$ \begin{gathered} E_{e}=E_{k}=\frac{p^{\prime 2}}{2 m_{p}}=\frac{p^{\prime 2} c^{2}}{2 m_{p} c^{2}}=\frac{p^{\prime 2} c^{2}}{2 E_{p}}, \\ p^{\prime}=\frac{1}{c} \sqrt{2 E_{e} E_{p}} . \end{gathered} $$ On the other hand, the momentum of the proton determined from the relativistic formulae can be written in a simpler form since $E_{\mathrm{p}} / E_{\mathrm{e}} » 1$. We get: $$ p^{\prime}=\frac{E_{e}}{c} \sqrt{\left(\frac{E_{p}}{E_{e}}+1\right)^{2}-\left(\frac{E_{p}}{E_{e}}\right)^{2}}=\frac{E_{e}}{c} \sqrt{2 \frac{E_{p}}{E_{e}}+1} \approx \frac{E_{e}}{c} \sqrt{2 \frac{E_{p}}{E_{e}}}=\frac{1}{c} \sqrt{2 E_{e} E_{p}} . $$ In accordance with our expectations, we have obtained the same result as above.
|
IPHO 1989
|
equation
|
||
3
|
How many times would the resolving power of the above microscope increase or decrease if the electron beam were replaced with the proton beam? Assume that the resolving power of the microscope (i.e. the smallest distance between two point objects whose circular images can be just separated) depends only on the wave properties of the particles.
|
35
|
The resolving power of the microscope (in the meaning mentioned in the text of the problem) is proportional to the wavelength, in our case to the length of the de Broglie wave: $$ \lambda=\frac{h}{p} $$ where $h$ denotes the Planck constant and $p$ is the momentum of the particle. We see that $\lambda$ is inversely proportional to the momentum of the particle. Therefore, after replacing the electron beam with the proton beam the resolving power will be changed by the factor $p / p^{\prime} \approx 1 / 35$. It means that our proton microscope would allow observation of the objects about 35 times smaller than the electron microscope.
|
IPHO 1989
|
value
|
||
4
|
Consider two liquids A and B insoluble in each other. The pressures $p_{i}(i=\mathrm{A}$ or B$)$ of their saturated vapors obey, to a good approximation, the formula: $$ \ln \left(p_{i} / p_{o}\right)=\frac{\alpha_{i}}{T}+\beta_{i} ; \quad i=\mathrm{A} \text { or } \mathrm{B}, $$ where $p_{o}$ denotes the normal atmospheric pressure, $T-$ the absolute temperature of the vapor, and $\alpha_{i}$ and $\beta_{i}(i=\mathrm{A}$ or B$)-$ certain constants depending on the liquid. (The symbol ln denotes the natural logarithm, i.e. logarithm with base $e=2.7182818 \ldots$ ) The values of the ratio $p_{i} / p_{0}$ for the liquids A and B at the temperature $40^{\circ} \mathrm{C}$ and $90^{\circ} \mathrm{C}$ are given in Tab. 1.1. Table 1.1 | $t\left[{ }^{\circ} \mathrm{C}\right]$ | $p_{i} / p_{0}$ | | | :---: | :---: | :---: | | | $i=\mathrm{A}$ | $i=\mathrm{B}$ | | 40 | 0.284 | 0.07278 | | 90 | 1.476 | 0.6918 | The errors of these values are negligible. A. Determine the boiling temperatures of the liquids A under the pressure $p_{0}$. ( unit: $\circ \mathrm{C}$ )
|
77
|
The liquid boils when the pressure of its saturated vapor is equal to the external pressure. Thus, in order to find the boiling temperature of the liquid $i(i-\mathrm{A}$ or B$)$, one should determine such a temperature $T_{b i}$ (or $t_{b i}$ ) for which $p_{i} / p_{0}=1$. Then $\ln \left(p_{i} / p_{0}\right)=0$, and we have: $$ T_{b i}=-\frac{\alpha_{i}}{\beta_{i}} . $$ The coefficients $\alpha_{i}$ and $\beta_{i}$ are not given explicitly. However, they can be calculated from the formula given in the text of the problem. For this purpose one should make use of the numerical data given in the Tab. 1.1. For the liquid A , we have: $$ \begin{aligned} & \ln 0.284=\frac{\alpha_{A}}{(40+273.15) \mathrm{K}}+\beta_{A}, \\ & \ln 1.476=\frac{\alpha_{A}}{(90+273.15) \mathrm{K}}+\beta_{A} . \end{aligned} $$ After subtraction of these equations, we get: $$ \begin{gathered} \ln 0.284-\ln 1.476=\alpha_{A}\left(\frac{1}{40+273.15}-\frac{1}{90+273.15}\right) \mathrm{K}^{-1} \\ \alpha_{A}=\frac{\ln \frac{0.284}{1.476}}{\frac{1}{40+273.15}-\frac{1}{90+273.15}} \mathrm{~K} \approx-3748.49 \mathrm{~K} \end{gathered} $$ Hence, $$ \beta_{A}=\ln 0.284-\frac{\alpha_{A}}{(40+273.15) \mathrm{K}} \approx 10.711 . $$ Thus, the boiling temperature of the liquid A is equal to $$ T_{b A}=3748.49 \mathrm{~K} / 10.711 \approx 349.95 \mathrm{~K} . $$ In the Celsius scale the boiling temperature of the liquid A is $$ t_{b A}=(349.95-273.15)^{\circ} \mathrm{C}=76.80^{\circ} \mathrm{C} \approx 77^{\circ} \mathrm{C} . $$
|
IPHO 1989
|
value
|
||
5
|
A cylindrical wheel of uniform density, having the mass $\mathrm{M}=0,40 \mathrm{~kg}$, the radius $\mathrm{R}=0,060 \mathrm{~m}$ and the thickness $\mathrm{d}=0,010 \mathrm{~m}$ is suspended by means of two light strings of the same length from the ceiling. Each string is wound around the axle of the wheel. Like the strings, the mass of the axle is negligible. When the wheel is turned manually, the strings are wound up until the centre of mass is raised $1,0 \mathrm{~m}$ above the floor. If the wheel is allowed to move downward vertically under the pulling force of the gravity, the strings are unwound to the full length of the strings and the wheel reaches the lowest point. The strings then begin to wound in the opposite sense resulting in the wheel being raised upwards. Analyze and answer the following questions, assuming that the strings are in vertical position and the points where the strings touch the axle are directly below their respective suspending points (see fig. 19.5). <image> Fig. 19.5Determine the angular speed of the wheel when the centre of mass of the wheel covers the vertical distance s. ( unit: $\frac{\mathrm{rad}}{\mathrm{~s}$ )
|
72.4
|
conservation of energy: $$ \mathrm{M} \cdot \mathrm{~g} \cdot \mathrm{~s}=\frac{1}{2} \cdot \mathrm{I}_{\mathrm{A}} \cdot \omega^{2} $$ where $\omega$ is the angular speed of the wheel and $\mathrm{I}_{\mathrm{A}}$ is the moment of inertia about the axis through A. Note: If we would take the moment of inertia about S instead of A we would have $\mathrm{M} \cdot \mathrm{g} \cdot \mathrm{s}=\frac{1}{2} \cdot \mathrm{l}_{\mathrm{s}} \cdot \omega^{2}+\frac{1}{2} \cdot \mathrm{~m} \cdot \mathrm{v}^{2}$ where v is the speed of the centre of mass along the vertical. This equation is the same as the above one in meanings since $\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{S}}+\mathrm{M} \cdot \mathrm{r}^{2}$ and $\mathrm{I}_{\mathrm{S}}=\mathrm{M} \cdot \mathrm{R}^{2}$ From (1) we get $$ \omega=\sqrt{\frac{2 \cdot \mathrm{M} \cdot \mathrm{~g} \cdot \mathrm{~S}}{\mathrm{I}_{\mathrm{A}}}} $$ substitute $$ I_{A}=\frac{1}{2} \cdot M \cdot r^{2}+M \cdot R^{2} $$ $\omega=\sqrt{\frac{2 \cdot \mathrm{~g} \cdot \mathrm{~S}}{\mathrm{r}^{2}+\frac{\mathrm{R}^{2}}{2}}}$ Putting in numbers we get $$ \omega=\sqrt{\frac{2 \cdot 9,81 \cdot 0,50}{9 \cdot 10^{-6}+\frac{1}{2} \cdot 36 \cdot 10^{-4}}} \approx 72,4 \frac{\mathrm{rad}}{\mathrm{~s}} $$
|
IPHO 1988
|
value
|
||
6
|
A cylindrical wheel of uniform density, having the mass $\mathrm{M}=0,40 \mathrm{~kg}$, the radius $\mathrm{R}=0,060 \mathrm{~m}$ and the thickness $\mathrm{d}=0,010 \mathrm{~m}$ is suspended by means of two light strings of the same length from the ceiling. Each string is wound around the axle of the wheel. Like the strings, the mass of the axle is negligible. When the wheel is turned manually, the strings are wound up until the centre of mass is raised $1,0 \mathrm{~m}$ above the floor. If the wheel is allowed to move downward vertically under the pulling force of the gravity, the strings are unwound to the full length of the strings and the wheel reaches the lowest point. The strings then begin to wound in the opposite sense resulting in the wheel being raised upwards. Analyze and answer the following questions, assuming that the strings are in vertical position and the points where the strings touch the axle are directly below their respective suspending points (see fig. 19.5). <image> Fig. 19.5Determine the kinetic energy of the linear motion of the centre of mass $\mathrm{E}_{\mathrm{r}}$ after the wheel travels a distance $\mathrm{s}=0,50 \mathrm{~m}$, and calculate the ratio between $\mathrm{E}_{\mathrm{r}}$ and the energy in any other form in this problem up to this point. Radius of the axle $=0,0030 \mathrm{~m}$
|
$5,13 \cdot 10^{-3}$
|
Kinetic energy of linear motion of the centre of mass of the wheel is $\mathrm{E}_{\mathrm{T}}=\frac{1}{2} \cdot \mathrm{M} \cdot \mathrm{v}^{2}=\frac{1}{2} \cdot \mathrm{M} \cdot \omega^{2} \cdot \mathrm{r}^{2}=\frac{1}{2} \cdot 0,40 \cdot 72,4^{2} \cdot 9 \cdot 10^{-6}=9,76 \cdot 10^{-3} \mathrm{~J}$ Potential energy of the wheel $\mathrm{E}_{\mathrm{P}}=\mathrm{M} \cdot \mathrm{g} \cdot \mathrm{S}=0,40 \cdot 9,81 \cdot 0,50=1,962 \mathrm{~J}$ Rotational kinetic energy of the wheel $E_{R}=\frac{1}{2} \cdot I_{S} \cdot \omega^{2}=\frac{1}{2} \cdot 0,40 \cdot 1,81 \cdot 10^{-3} \cdot 72,4^{2}=1,899 \mathrm{~J}$ $$ \frac{E_{T}}{E_{R}}=\frac{9,76 \cdot 10^{-3}}{1,899}=5,13 \cdot 10^{-3} $$
|
IPHO 1988
|
value
|
||
7
|
A cylindrical wheel of uniform density, having the mass $\mathrm{M}=0,40 \mathrm{~kg}$, the radius $\mathrm{R}=0,060 \mathrm{~m}$ and the thickness $\mathrm{d}=0,010 \mathrm{~m}$ is suspended by means of two light strings of the same length from the ceiling. Each string is wound around the axle of the wheel. Like the strings, the mass of the axle is negligible. When the wheel is turned manually, the strings are wound up until the centre of mass is raised $1,0 \mathrm{~m}$ above the floor. If the wheel is allowed to move downward vertically under the pulling force of the gravity, the strings are unwound to the full length of the strings and the wheel reaches the lowest point. The strings then begin to wound in the opposite sense resulting in the wheel being raised upwards. Analyze and answer the following questions, assuming that the strings are in vertical position and the points where the strings touch the axle are directly below their respective suspending points (see fig. 19.5). <image> Fig. 19.5Determine the tension in the string while the wheel is moving downward. ( unit: $\mathrm{~N}$ )
|
1.96
|
Let $\frac{\mathrm{T}}{2}$ be the tension in each string. Torque $\tau$ which causes the rotation is given by $\tau=\mathrm{M} \cdot \mathrm{g} \cdot \mathrm{r}=\mathrm{I}_{\mathrm{A}} \cdot \alpha$ where $\alpha$ is the angular acceleration $\quad \alpha=\frac{\mathrm{M} \cdot \mathrm{g} \cdot \mathrm{r}}{\mathrm{I}_{\mathrm{A}}}$ The equation of the motion of the wheel is M.g-T = M.a Substituting $\mathrm{a}=\alpha \cdot \mathrm{r}$ and $\mathrm{I}_{\mathrm{A}}=\frac{1}{2} \cdot \mathrm{M} \cdot \mathrm{r}^{2}+\mathrm{M} \cdot \mathrm{R}^{2}$ we get $T=M \cdot g+\frac{M \cdot g \cdot r^{2}}{\frac{1}{2} \cdot M \cdot R^{2}+M \cdot r^{2}}=M \cdot g \cdot\left(1+\frac{2 \cdot r^{2}}{R^{2}+2 \cdot r^{2}}\right)$ Thus for the tension $\frac{T}{2}$ in each string we get $\frac{\mathrm{T}}{2}=\frac{\mathrm{M} \cdot \mathrm{g}}{2} \cdot\left(1+\frac{2 \cdot \mathrm{r}^{2}}{\mathrm{R}^{2}+2 \cdot \mathrm{r}^{2}}\right)=\frac{0,40 \cdot 9,81}{2} \cdot\left(1+\frac{2 \cdot 9 \cdot 10^{-6}}{3,6 \cdot 10^{-3}+2 \cdot 9 \cdot 10^{-6}}\right)=1,96 \mathrm{~N}$ $$ \frac{\mathrm{T}}{2}=1,96 \mathrm{~N} $$
|
IPHO 1988
|
value
|
||
8
|
A gas consists of positive ions of some element (at high temperature) and electrons. The positive ion belongs to an atom of unknown mass number Z. It is known that this ion has only one electron in the shell (orbit). Let this ion be represented by the symbol $A^{(Z-1)+}$ Constants: electric field constant $$ \varepsilon_{\mathrm{O}}=8,85 \cdot 10^{-12} \frac{\mathrm{~A} \cdot \mathrm{~s}}{\mathrm{~V} \cdot \mathrm{~m}} $$ elementary charge $$ \begin{aligned} & \mathrm{e}= \pm 1,602 \cdot 10^{-19} \mathrm{~A} \cdot \mathrm{~s} \\ & \mathrm{q}^{2}=\frac{\mathrm{e}^{2}}{4 \cdot \pi \cdot \varepsilon_{0}}=2,037 \cdot 10^{-28} \mathrm{~J} \cdot \mathrm{~m} \end{aligned} $$ Planck's constant $$ \hbar=1,054 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{~s} $$ (rest) mass of an electron $$ \mathrm{m}_{\mathrm{e}}=9,108 \cdot 10^{-31} \mathrm{~kg} $$ Bohr's atomic radius $$ \mathrm{r}_{\mathrm{B}}=\frac{\hbar}{\mathrm{m} \cdot \mathrm{q}^{2}}=5,92 \cdot 10^{-11} \mathrm{~m} $$ Rydberg's energy $$ E_{R}=\frac{q^{2}}{2 \cdot r_{B}}=2,180 \cdot 10^{-18} \mathrm{~J} $$ (rest) mass of a proton $$ m_{p} \cdot c^{2}=1,503 \cdot 10^{-10} \mathrm{~J} $$Assume that the ion which has just one electron left the shell. $\mathrm{A}^{(\mathrm{Z}-1)+}$ is in the ground state. In the lowest energy state, the square of the average distance of the electron from the nucleus or $\mathrm{r}^{2}$ with components along $\mathrm{x}-, \mathrm{y}$ - and z -axis being $(\Delta \mathrm{x})^{2},(\Delta \mathrm{y})^{2}$ and $(\Delta \mathrm{z})^{2}$ respectively and $\mathrm{r}_{\mathrm{O}}^{2}=(\Delta \mathrm{x})^{2}+(\Delta \mathrm{y})^{2}+(\Delta \mathrm{z})^{2} \quad$ and also the square of the average momentum by $\mathrm{p}_{\mathrm{o}}^{2}=\left(\Delta \mathrm{p}_{\mathrm{x}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{y}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{z}}\right)^{2}$, whereas $\Delta \mathrm{p}_{\mathrm{x}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{x}}, \Delta \mathrm{p}_{\mathrm{y}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{y}}$ and $\Delta \mathrm{p}_{\mathrm{z}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{z}}$. Write inequality involving $\left(\mathrm{p}_{\mathrm{O}}\right)^{2} \cdot\left(\mathrm{r}_{\mathrm{O}}\right)^{2}$ in a complete form.
|
$\mathrm{p}_{0}^{2} \cdot \mathrm{r}_{0}^{2} \geq \frac{9}{4} \cdot \hbar^{2}$
|
$\mathrm{r}_{0}^{2}=(\Delta \mathrm{x})^{2}+(\Delta \mathrm{y})^{2}+(\Delta \mathrm{z})^{2}$ $\mathrm{p}_{0}^{2}=\left(\Delta \mathrm{p}_{\mathrm{x}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{y}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{z}}\right)^{2}$ since $\Delta \mathrm{p}_{\mathrm{x}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{x}} \quad \Delta \mathrm{p}_{\mathrm{y}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{y}} \quad \Delta \mathrm{p}_{\mathrm{z}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{z}}$ gives $\mathrm{p}_{0}^{2} \geq \frac{\hbar^{2}}{4} \cdot\left[\frac{1}{(\Delta \mathrm{x})^{2}}+\frac{1}{(\Delta \mathrm{y})^{2}}+\frac{1}{(\Delta \mathrm{z})^{2}}\right]$ and $(\Delta x)^{2}=(\Delta y)^{2}=(\Delta z)^{2}=\frac{r_{0}^{2}}{3}$ thus $\quad \mathrm{p}_{0}^{2} \cdot \mathrm{r}_{0}^{2} \geq \frac{9}{4} \cdot \hbar^{2}$
|
IPHO 1988
|
expression
|
||
9
|
A gas consists of positive ions of some element (at high temperature) and electrons. The positive ion belongs to an atom of unknown mass number Z. It is known that this ion has only one electron in the shell (orbit). Let this ion be represented by the symbol $A^{(Z-1)+}$ Constants: electric field constant $$ \varepsilon_{\mathrm{O}}=8,85 \cdot 10^{-12} \frac{\mathrm{~A} \cdot \mathrm{~s}}{\mathrm{~V} \cdot \mathrm{~m}} $$ elementary charge $$ \begin{aligned} & \mathrm{e}= \pm 1,602 \cdot 10^{-19} \mathrm{~A} \cdot \mathrm{~s} \\ & \mathrm{q}^{2}=\frac{\mathrm{e}^{2}}{4 \cdot \pi \cdot \varepsilon_{0}}=2,037 \cdot 10^{-28} \mathrm{~J} \cdot \mathrm{~m} \end{aligned} $$ Planck's constant $$ \hbar=1,054 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{~s} $$ (rest) mass of an electron $$ \mathrm{m}_{\mathrm{e}}=9,108 \cdot 10^{-31} \mathrm{~kg} $$ Bohr's atomic radius $$ \mathrm{r}_{\mathrm{B}}=\frac{\hbar}{\mathrm{m} \cdot \mathrm{q}^{2}}=5,92 \cdot 10^{-11} \mathrm{~m} $$ Rydberg's energy $$ E_{R}=\frac{q^{2}}{2 \cdot r_{B}}=2,180 \cdot 10^{-18} \mathrm{~J} $$ (rest) mass of a proton $$ m_{p} \cdot c^{2}=1,503 \cdot 10^{-10} \mathrm{~J} $$The ion represented by $\mathrm{A}^{(\mathrm{Z}-1)+}$ may capture an additional electron and consequently emits a photon. Write down an equation which is to be used for calculation the frequency of an emitted photon.
|
$m_{e} \cdot \vec{v}_{e}+(M+m) \cdot \vec{v}_{i}=\left(M+2 \cdot m_{e}\right) \cdot \vec{v}_{f}+\frac{h \cdot v}{c} \cdot \overrightarrow{1}$
|
$\left|\overrightarrow{\mathrm{v}}_{\mathrm{e}}\right| \ldots .$. speed of the external electron before the capture $\left|\vec{V}_{i}\right| \ldots \ldots$. speed of $\mathrm{A}^{(\mathrm{Z}-1)+}$ before capturing $\left|\vec{V}_{f}\right|$...... speed of $A^{(\mathrm{Z}-1)+}$ after capturing $\mathrm{E}_{\mathrm{n}}=\mathrm{h} . v \ldots .$. energy of the emitted photon conservation of energy: $\frac{1}{2} \cdot m_{e} \cdot v_{e}^{2}+\frac{1}{2} \cdot\left(M+m_{e}\right) \cdot V_{i}^{2}+E\left[A^{(z-1)+}\right]=\frac{1}{2} \cdot\left(M+2 \cdot m_{e}\right) \cdot V_{f}^{2}+E\left[A^{(z-2)+}\right]$ where $\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-1)+}\right)$ and $\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-2)+}\right]$ denotes the energy of the electron in the outermost shell of ions $\mathrm{A}^{(\mathrm{Z}-1)+}$ and $\mathrm{A}^{(\mathrm{Z}-2)+}$ respectively. conservation of momentum: $m_{e} \cdot \vec{v}_{e}+(M+m) \cdot \vec{v}_{i}=\left(M+2 \cdot m_{e}\right) \cdot \vec{v}_{f}+\frac{h \cdot v}{c} \cdot \overrightarrow{1}$ where $\overrightarrow{1}$ is the unit vector pointing in the direction of the motion of the emitted photon.
|
IPHO 1988
|
equation
|
||
10
|
A gas consists of positive ions of some element (at high temperature) and electrons. The positive ion belongs to an atom of unknown mass number Z. It is known that this ion has only one electron in the shell (orbit). Let this ion be represented by the symbol $A^{(Z-1)+}$ Constants: electric field constant $$ \varepsilon_{\mathrm{O}}=8,85 \cdot 10^{-12} \frac{\mathrm{~A} \cdot \mathrm{~s}}{\mathrm{~V} \cdot \mathrm{~m}} $$ elementary charge $$ \begin{aligned} & \mathrm{e}= \pm 1,602 \cdot 10^{-19} \mathrm{~A} \cdot \mathrm{~s} \\ & \mathrm{q}^{2}=\frac{\mathrm{e}^{2}}{4 \cdot \pi \cdot \varepsilon_{0}}=2,037 \cdot 10^{-28} \mathrm{~J} \cdot \mathrm{~m} \end{aligned} $$ Planck's constant $$ \hbar=1,054 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{~s} $$ (rest) mass of an electron $$ \mathrm{m}_{\mathrm{e}}=9,108 \cdot 10^{-31} \mathrm{~kg} $$ Bohr's atomic radius $$ \mathrm{r}_{\mathrm{B}}=\frac{\hbar}{\mathrm{m} \cdot \mathrm{q}^{2}}=5,92 \cdot 10^{-11} \mathrm{~m} $$ Rydberg's energy $$ E_{R}=\frac{q^{2}}{2 \cdot r_{B}}=2,180 \cdot 10^{-18} \mathrm{~J} $$ (rest) mass of a proton $$ m_{p} \cdot c^{2}=1,503 \cdot 10^{-10} \mathrm{~J} $$Calculate the energy of the ion $\mathrm{A}^{(\mathrm{Z}-1)+}$ using the value of the lowest energy. The calculation should be approximated based on the following principles: ### 3.3.A The potential energy of the ion should be expressed in terms of the average value of $\frac{1}{r}$. (ie. $\frac{1}{r_{o}} ; r_{0}$ is given in the problem). ### 3.3.B In calculating the kinetic energy of the ion, use the average value of the square of the momentum given in 3.1 after being simplified by $\left(\mathrm{p}_{\mathrm{O}}\right)^{2} \cdot\left(\mathrm{r}_{\mathrm{O}}\right)^{2} \approx(\hbar)^{2}$
|
$-E_{R} \cdot Z^{2}$
|
Determination of the energy of $\mathrm{A}^{(\mathrm{Z}-1)+}$ : potential energy $=-\frac{Z \cdot e^{2}}{4 \cdot \pi \cdot \varepsilon_{0} \cdot r_{0}}=-\frac{Z \cdot q^{2}}{r_{0}}$ kinetic energy $=\frac{p^{2}}{2 \cdot m}$ If the motion of the electrons is confined within the $x-y$-plane, principles of uncertainty in 3.1 can be written as $\mathrm{r}_{0}^{2}=(\Delta \mathrm{x})^{2}+(\Delta \mathrm{y})^{2}$ $\mathrm{p}_{0}^{2}=\left(\Delta \mathrm{p}_{\mathrm{x}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{y}}\right)^{2}$ $\mathrm{p}_{0}^{2}=\frac{\hbar^{2}}{4} \cdot\left[\frac{1}{(\Delta \mathrm{x})^{2}}+\frac{1}{(\Delta \mathrm{y})^{2}}\right]=\frac{\hbar^{2}}{4} \cdot\left[\frac{2}{\mathrm{r}_{0}^{2}}+\frac{2}{\mathrm{r}_{0}^{2}}\right]=\frac{\hbar^{2}}{4} \cdot \frac{4}{\mathrm{r}_{0}^{2}}$ thus $\mathrm{p}_{0}^{2} \cdot \mathrm{r}_{0}^{2}=\hbar^{2}$ $E\left[A^{(Z-1)+}\right]=\frac{p_{0}^{2}}{2 \cdot m_{e}}-\frac{Z \cdot q^{2}}{r_{0}}=\frac{\hbar^{2}}{2 \cdot m_{e} \cdot r_{e}}-\frac{Z \cdot q^{2}}{r_{0}}$ Energy minimum exists, when $\frac{d E}{d r_{0}}=0$. Hence $$ \frac{\mathrm{dE}}{\mathrm{dr} r_{0}}=-\frac{\hbar^{2}}{\mathrm{~m}_{\mathrm{e}} \cdot \mathrm{r}_{\mathrm{e}}^{3}}+\frac{\mathrm{Z} \cdot \mathrm{q}^{2}}{\mathrm{r}_{0}^{2}}=0 $$ this gives $\quad \frac{1}{r_{0}}=\frac{Z \cdot q^{2} \cdot m_{e}}{\hbar^{2}}$ hence $$ \begin{gathered} E\left[A^{(Z-1)+}\right]=\frac{\hbar^{2}}{2 \cdot m_{e}} \cdot\left(\frac{Z \cdot q^{2} \cdot m_{e}}{\hbar}\right)^{2}-Z \cdot q^{2} \cdot \frac{Z \cdot q^{2} \cdot m_{e}}{\hbar^{2}}=-\frac{m_{e}}{2} \cdot\left(\frac{Z \cdot q^{2}}{\hbar}\right)^{2}=-\frac{q^{2} \cdot Z^{2}}{2 \cdot r_{B}}=-E_{R} \cdot Z^{2} \\ E\left[A^{(Z-1)+}\right]=-E_{R} \cdot Z^{2} \end{gathered} $$
|
IPHO 1988
|
value
|
||
11
|
A gas consists of positive ions of some element (at high temperature) and electrons. The positive ion belongs to an atom of unknown mass number Z. It is known that this ion has only one electron in the shell (orbit). Let this ion be represented by the symbol $A^{(Z-1)+}$ Constants: electric field constant $$ \varepsilon_{\mathrm{O}}=8,85 \cdot 10^{-12} \frac{\mathrm{~A} \cdot \mathrm{~s}}{\mathrm{~V} \cdot \mathrm{~m}} $$ elementary charge $$ \begin{aligned} & \mathrm{e}= \pm 1,602 \cdot 10^{-19} \mathrm{~A} \cdot \mathrm{~s} \\ & \mathrm{q}^{2}=\frac{\mathrm{e}^{2}}{4 \cdot \pi \cdot \varepsilon_{0}}=2,037 \cdot 10^{-28} \mathrm{~J} \cdot \mathrm{~m} \end{aligned} $$ Planck's constant $$ \hbar=1,054 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{~s} $$ (rest) mass of an electron $$ \mathrm{m}_{\mathrm{e}}=9,108 \cdot 10^{-31} \mathrm{~kg} $$ Bohr's atomic radius $$ \mathrm{r}_{\mathrm{B}}=\frac{\hbar}{\mathrm{m} \cdot \mathrm{q}^{2}}=5,92 \cdot 10^{-11} \mathrm{~m} $$ Rydberg's energy $$ E_{R}=\frac{q^{2}}{2 \cdot r_{B}}=2,180 \cdot 10^{-18} \mathrm{~J} $$ (rest) mass of a proton $$ m_{p} \cdot c^{2}=1,503 \cdot 10^{-10} \mathrm{~J} $$Calculate the energy of the ion $\mathrm{A}^{(\mathrm{Z}-2)+}$ taken to be in the ground state, using the same principle as the calculation of the energy of $\mathrm{A}^{(\mathrm{Z}-1)+}$. Given the average distance of each of the two electrons in the outermost shell (same as $r_{0}$ given in 3.3) denoted by $r_{1}$ and $r_{2}$, assume the average distance between the two electrons is given by $r_{1}+r_{2}$ and the average value of the square of the momentum of each electron obeys the principle of uncertainty ie. $\mathrm{p}_{1}^{2} \cdot \mathrm{r}_{1}^{2} \approx \hbar^{2}$ and $\mathrm{p}_{2}^{2} \cdot \mathrm{r}_{2}^{2} \approx \hbar^{2}$ hint: Make use of the information that in the ground state $r_{1}=r_{2}$
|
$-2 \cdot E_{R} \cdot\left(Z-\frac{1}{4}\right)^{2}$
|
In the case of $\mathrm{A}^{(\mathrm{Z}-1)+}$ ion captures a second electron potential energy of both electrons $=-2 \cdot \frac{Z \cdot q^{2}}{r_{0}}$ kinetic energy of the two electrons $=2 \cdot \frac{\mathrm{p}^{2}}{2 \cdot \mathrm{~m}}=\frac{\hbar^{2}}{\mathrm{~m}_{\mathrm{e}} \cdot \mathrm{r}_{0}^{2}}$ potential energy due to interaction between the two electrons $=\frac{q^{2}}{\left|\vec{r}_{1}-\vec{r}_{2}\right|}=\frac{q^{2}}{2 \cdot r_{0}}$ $E\left[A^{(Z-2)+}\right]=\frac{\hbar^{2}}{m_{e} \cdot r_{0}^{2}}-\frac{2 \cdot Z \cdot q^{2}}{r_{0}^{2}}+\frac{q^{2}}{2 \cdot r_{0}}$ total energy is lowest when $\frac{d E}{d r_{0}}=0$ hence $0=-\frac{2 \cdot \hbar^{2}}{\mathrm{~m}_{\mathrm{e}} \cdot \mathrm{r}_{0}^{3}}+\frac{2 \cdot \mathrm{Z} \cdot \mathrm{q}^{2}}{\mathrm{r}_{0}^{3}}-\frac{\mathrm{q}^{2}}{2 \cdot \mathrm{r}_{0}^{2}}$ hence $\frac{1}{r_{0}}=\frac{q^{2} \cdot m_{e}}{2 \cdot \hbar^{2}} \cdot\left(2 \cdot z-\frac{1}{2}\right)=\frac{1}{r_{B}} \cdot\left(z-\frac{1}{4}\right)$ $E\left[A^{(Z-2)+}\right]=\frac{\hbar^{2}}{m_{e}} \cdot\left(\frac{q^{2} \cdot m_{e}}{2 \cdot \hbar^{2}}\right)^{2}-\frac{q^{2} \cdot\left(2 \cdot Z-\frac{1}{2}\right)}{\hbar} \cdot \frac{q^{2} \cdot m_{e} \cdot\left(2 \cdot Z-\frac{1}{2}\right)}{2 \cdot \hbar}$ $E\left[A^{(z-2)+}\right]=-\frac{m_{e}}{4} \cdot\left[\frac{q^{2} \cdot\left(2 \cdot Z-\frac{1}{2}\right)}{\hbar}\right]^{2}=-\frac{m_{e} \cdot\left[q^{2} \cdot\left(Z-\frac{1}{4}\right)\right]^{2}}{\hbar^{2}}=-\frac{q^{2} \cdot\left(Z-\frac{1}{4}\right)^{2}}{\hbar^{2}}$ this gives $$ E\left[A^{(Z-2)+}\right]=-2 \cdot E_{R} \cdot\left(Z-\frac{1}{4}\right)^{2} $$
|
IPHO 1988
|
value
|
||
12
|
A gas consists of positive ions of some element (at high temperature) and electrons. The positive ion belongs to an atom of unknown mass number Z. It is known that this ion has only one electron in the shell (orbit). Let this ion be represented by the symbol $A^{(Z-1)+}$ Constants: electric field constant $$ \varepsilon_{\mathrm{O}}=8,85 \cdot 10^{-12} \frac{\mathrm{~A} \cdot \mathrm{~s}}{\mathrm{~V} \cdot \mathrm{~m}} $$ elementary charge $$ \begin{aligned} & \mathrm{e}= \pm 1,602 \cdot 10^{-19} \mathrm{~A} \cdot \mathrm{~s} \\ & \mathrm{q}^{2}=\frac{\mathrm{e}^{2}}{4 \cdot \pi \cdot \varepsilon_{0}}=2,037 \cdot 10^{-28} \mathrm{~J} \cdot \mathrm{~m} \end{aligned} $$ Planck's constant $$ \hbar=1,054 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{~s} $$ (rest) mass of an electron $$Consider in particular the ion $\mathrm{A}^{(\mathrm{Z}-2)+}$ is at rest in the ground state when capturing an additional electron and the captured electron is also at rest prior to the capturing. Determine the numerical value of Z , if the frequency of the emitted photon accompanying electron capturing is $2,057.10^{17} \mathrm{rad} / \mathrm{s}$. Identify the element which gives rise to the ion.
|
4.1
|
The ion $\mathrm{A}^{(\mathrm{Z}-1)+}$ is at rest when it captures the second electron also at rest before capturing. From the information provided in the problem, the frequency of the photon emitted is given by $v=\frac{\omega}{2 \cdot \pi}=\frac{2,057 \cdot 10^{17}}{2 \cdot \pi} \mathrm{~Hz}$ The energy equation can be simplified to $\quad E\left[A^{(Z-1)+}\right]-E\left[A^{(Z-2)+}\right]=\hbar \cdot \omega=h \cdot v$ that is $-E_{R} \cdot Z^{2}-\left[-2 \cdot E_{R} \cdot\left(Z-\frac{1}{4}\right)^{2}\right]=\hbar \cdot \omega$ putting in known numbers follows $2,180 \cdot 10^{-18} \cdot\left[-Z^{2}+2 \cdot\left(Z-\frac{1}{4}\right)^{2}\right]=1,05 \cdot 10^{-34} \cdot 2,607 \cdot 10^{17}$ this gives $Z^{2}-Z-12,7=0$ with the physical sensuous result $\quad Z=\frac{1+\sqrt{1+51}}{2}=4,1$ $$ \text { This implies } Z=4 \text {, and that means Beryllium } $$
|
IPHO 1988
|
value
|
||
13
|
Moist air is streaming adiabatically across a mountain range as indicated in the figure. Equal atmospheric pressures of 100 kPa are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of 70 kPa at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ} \mathrm{C}$. As the air is ascending, cloud formation sets in at 84.5 kPa . Consider a quantity of moist air ascending the mountain with a mass of 2000 kg over each square meter. This moist air reaches the mountain ridge (station $\mathrm{M}_{2}$ ) after 1500 seconds. During that rise an amount of 2.45 g of water per kilogram of air is precipitated as rain. <image>1. Determine temperature $\mathrm{T}_{1}$ at $\mathrm{M}_{1}$ where the cloud ceiling forms. ( unit: $\mathrm{~K}$ )
|
279
|
Temperature $\mathrm{T}_{1}$ where the cloud ceiling forms $$ \mathrm{T}_{1}=\mathrm{T}_{0} \cdot\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{0}}\right)^{1-\frac{1}{\mathrm{x}}}=279 \mathrm{~K} $$
|
IPHO 1987
|
value
|
||
14
|
Moist air is streaming adiabatically across a mountain range as indicated in the figure. Equal atmospheric pressures of 100 kPa are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of 70 kPa at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ} \mathrm{C}$. As the air is ascending, cloud formation sets in at 84.5 kPa . Consider a quantity of moist air ascending the mountain with a mass of 2000 kg over each square meter. This moist air reaches the mountain ridge (station $\mathrm{M}_{2}$ ) after 1500 seconds. During that rise an amount of 2.45 g of water per kilogram of air is precipitated as rain. <image>2. What is the height $\mathrm{h}_{1}$ (at $\mathrm{M}_{1}$ ) above station $\mathrm{M}_{0}$ of the cloud ceiling assuming a linear decrease of atmospheric density? ( unit: m )
|
1410
|
Height $\mathrm{h}_{1}$ of the cloud ceiling: $$ \begin{aligned} & \mathrm{p}_{0}-\mathrm{p}_{1}=\frac{\rho_{0}+\rho_{1}}{2} \cdot \mathrm{~g} \cdot \mathrm{~h}_{1}, \text { with } \rho_{1}=\rho_{0} \cdot \frac{\mathrm{p}_{1}}{\mathrm{p}_{0}} \cdot \frac{\mathrm{~T}_{0}}{\mathrm{~T}_{1}} \\ & \mathrm{~h}_{1}=1410 \mathrm{~m} \end{aligned} $$
|
IPHO 1987
|
value
|
||
15
|
Moist air is streaming adiabatically across a mountain range as indicated in the figure. Equal atmospheric pressures of 100 kPa are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of 70 kPa at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ} \mathrm{C}$. As the air is ascending, cloud formation sets in at 84.5 kPa . Consider a quantity of moist air ascending the mountain with a mass of 2000 kg over each square meter. This moist air reaches the mountain ridge (station $\mathrm{M}_{2}$ ) after 1500 seconds. During that rise an amount of 2.45 g of water per kilogram of air is precipitated as rain. <image>What temperature $\mathrm{T}_{2}$ is measured at the ridge of the mountain range? ( unit: $\mathrm{~K}$ )
|
271
|
Temperature $T_{2}$ at the ridge of the mountain. The temperature difference when the air is ascending from the cloud ceiling to the mountain ridge is caused by two processes: - adiabatic cooling to temperature $\mathrm{T}_{\mathrm{x}}$, - heating by $\Delta \mathrm{T}$ by condensation. $$ \begin{aligned} & \mathrm{T}_{2}=\mathrm{T}_{\mathrm{x}}+\Delta \mathrm{T} \\ & \mathrm{~T}_{\mathrm{x}}=\mathrm{T}_{1} \cdot\left(\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right)^{1-\frac{1}{x}}=265 \mathrm{~K} \end{aligned} $$ For each kg of air the heat produced by condensation is $\mathrm{L}_{\mathrm{v}} \cdot 2.45 \mathrm{~g}=6.125 \mathrm{~kJ}$. $$ \begin{aligned} & \Delta \mathrm{T}=\frac{6.125}{\mathrm{c}_{\mathrm{p}}} \cdot \frac{\mathrm{~kJ}}{\mathrm{~kg}}=6.1 \mathrm{~K} \\ & \mathrm{~T}_{2}=271 \mathrm{~K} \end{aligned} $$
|
IPHO 1987
|
value
|
||
16
|
Moist air is streaming adiabatically across a mountain range as indicated in the figure. Equal atmospheric pressures of 100 kPa are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of 70 kPa at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ} \mathrm{C}$. As the air is ascending, cloud formation sets in at 84.5 kPa . Consider a quantity of moist air ascending the mountain with a mass of 2000 kg over each square meter. This moist air reaches the mountain ridge (station $\mathrm{M}_{2}$ ) after 1500 seconds. During that rise an amount of 2.45 g of water per kilogram of air is precipitated as rain. <image> Determine the height of the water column (precipitation level) precipitated by the air stream in 3 hours, assuming a homogeneous rainfall between points $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$. ( unit: $\mathrm{~mm}$ )
|
35
|
Height of precipitated water column $$ \mathrm{h}=35 \mathrm{~mm} $$
|
IPHO 1987
|
value
|
||
17
|
Moist air is streaming adiabatically across a mountain range as indicated in the figure. Equal atmospheric pressures of 100 kPa are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of 70 kPa at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ} \mathrm{C}$. As the air is ascending, cloud formation sets in at 84.5 kPa . Consider a quantity of moist air ascending the mountain with a mass of 2000 kg over each square meter. This moist air reaches the mountain ridge (station $\mathrm{M}_{2}$ ) after 1500 seconds. During that rise an amount of 2.45 g of water per kilogram of air is precipitated as rain. <image>What temperature $\mathrm{T}_{3}$ is measured in the back of the mountain range at station $\mathrm{M}_{3}$ ? Discuss the state of the atmosphere at station $\mathrm{M}_{3}$ in comparison with that at station $\mathrm{M}_{0}$. ( unit: $\mathrm{~K}$ )
|
300
|
Temperature $\mathrm{T}_{3}$ behind the mountain $$ \mathrm{T}_{3}=\mathrm{T}_{2} \cdot\left(\frac{\mathrm{p}_{3}}{\mathrm{p}_{2}}\right)^{1-\frac{1}{\mathrm{x}}}=300 \mathrm{~K} $$ The air has become warmer and dryer. The temperature gain is caused by condensation of vapour.
|
IPHO 1987
|
value
|
||
18
|
$A$ beam of electrons emitted by a point source $P$ enters the magnetic field $\vec{B}$ of a toroidal coil (toroid) in the direction of the lines of force. The angle of the aperture of the beam $2 \cdot \alpha_{0}$ is assumed to be small ( $2 \cdot \alpha_{0} \ll 1$ ). The injection of the electrons occurs on the mean radius R of the toroid with acceleration voltage $\mathrm{V}_{0}$. Neglect any interaction between the electrons. The magnitude of $\vec{B}, B$, is assumed to be constant. <image>1. To guide the electron in the toroidal field a homogeneous magnetic deflection field $\vec{B}_{1}$ is required. Calculate $\vec{B}_{1}$ for an electron moving on a circular orbit of radius $R$ in the torus. ( unit: $\frac{\mathrm{Vs}}{\mathrm{~m}^{2}}$ )
|
$1.48 \cdot 10^{-2}$
|
Determination of B : The vector of the velocity of any electron is divided into components parallel with and perpendicular to the magnetic field $\overrightarrow{\mathrm{B}}$ : $$ \vec{v}=\vec{v}_{\|}+\vec{v}_{\perp} $$ The Lorentz force $\overrightarrow{\mathrm{F}}=-\mathrm{e} \cdot(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})$ influences only the perpendicular component, it acts as a radial force: $$ \mathrm{m} \cdot \frac{\mathrm{v}_{\perp}{ }^{2}}{\mathrm{r}}=\mathrm{e} \cdot \mathrm{v}_{\perp} \cdot \mathrm{B} $$ Hence the radius of the circular path that has been travelled is $$ \mathrm{r}=\frac{\mathrm{m}}{\mathrm{e}} \cdot \frac{\mathrm{v}_{\perp}}{\mathrm{B}} $$ and the period of rotation which is independent of $\mathrm{v}_{\perp}$ is $$ \mathrm{T}=\frac{2 \cdot \pi \cdot \mathrm{r}}{\mathrm{v}_{\perp}}=\frac{2 \cdot \pi \cdot \mathrm{~m}}{\mathrm{~B} \cdot \mathrm{e}} $$ The parallel component of the velocity does not vary. Because of $\alpha_{0} \ll 1$ it is approximately equal for all electrons: $$ \mathrm{V}_{\| 0}=\mathrm{V}_{0} \cdot \cos \alpha_{0} \approx \mathrm{~V}_{0} $$ Hence the distance b between the focusing points, using eq. (5), is $$ \mathrm{b}=\mathrm{v}_{\| 0} \cdot \mathrm{~T} \approx \mathrm{v}_{0} \cdot \mathrm{~T} $$ From the law of conservation of energy follows the relation between the acceleration voltage $\mathrm{V}_{0}$ and the velocity $\mathrm{v}_{0}$ : $$ \frac{m}{2} \cdot v_{0}^{2}=e \cdot V_{0} $$ Using eq. (7) and eq. (4) one obtains from eq. (6) $$ \mathrm{b}=\frac{2 \cdot \pi}{\mathrm{~B}} \cdot \sqrt{2 \cdot \frac{\mathrm{~m}}{\mathrm{e}} \cdot \mathrm{~V}_{0}} $$ and because of $b=\frac{2 \cdot \pi \cdot R}{4}$ one obtains $$ \mathrm{B}=\frac{4}{\mathrm{R}} \cdot \sqrt{2 \cdot \frac{\mathrm{~m}}{\mathrm{e}} \cdot \mathrm{~V}_{0}}=1.48 \cdot 10^{-2} \frac{\mathrm{Vs}}{\mathrm{~m}^{2}} $$
|
IPHO 1987
|
value
|
||
19
|
$A$ beam of electrons emitted by a point source $P$ enters the magnetic field $\vec{B}$ of a toroidal coil (toroid) in the direction of the lines of force. The angle of the aperture of the beam $2 \cdot \alpha_{0}$ is assumed to be small ( $2 \cdot \alpha_{0} \ll 1$ ). The injection of the electrons occurs on the mean radius R of the toroid with acceleration voltage $\mathrm{V}_{0}$. Neglect any interaction between the electrons. The magnitude of $\vec{B}, B$, is assumed to be constant. <image> Determine the value of $\vec{B}$ which gives four focussing points separated by $\pi / 2$ as indicated in the diagram. ( unit: $\frac{\mathrm{Vs}}{\mathrm{~m}^{2}}$ )
|
$0.37 \cdot 10^{-2} $
|
Determination of $\mathrm{B}_{1}$ : Analogous to eq. (2) $$ m \cdot \frac{v_{0}^{2}}{R}=e \cdot v_{0} \cdot B_{1} $$ must hold. From eq. (7) follows $$ \mathrm{B}_{1}=\frac{1}{\mathrm{R}} \cdot \sqrt{2 \cdot \frac{\mathrm{~m}}{\mathrm{e}} \cdot \mathrm{~V}_{0}}=0.37 \cdot 10^{-2} \frac{\mathrm{Vs}}{\mathrm{~m}^{2}} $$
|
IPHO 1987
|
value
|
||
20
|
When sine waves propagate in an infinite LC-grid (see the figure below) the phase of the acvoltage across two successive capacitors differs by $\Phi$. <image>Determine how $\Phi$ depends on $\omega$, L and C ( $\omega$ is the angular frequency of the sine wave).
|
$\varphi=2 \cdot \arcsin \left(\frac{\omega \cdot \sqrt{\mathrm{~L} \cdot \mathrm{C}}}{2}\right) \text { with } 0 \leq \omega \leq \frac{2}{\sqrt{\mathrm{~L} \cdot \mathrm{C}}}$
|
$$ \text { Current law: } \quad \mathrm{I}_{\mathrm{L}_{\mathrm{n}-1}}+\mathrm{I}_{\mathrm{C}_{\mathrm{n}}}-\mathrm{I}_{\mathrm{L}_{\mathrm{n}}}=0 $$ Voltage law: $\quad \mathrm{V}_{\mathrm{C}_{\mathrm{n}-1}}+\mathrm{V}_{\mathrm{L}_{\mathrm{n}-1}}-\mathrm{V}_{\mathrm{C}_{\mathrm{n}}}=0$ Capacitive voltage drop: $\mathrm{V}_{\mathrm{C}_{\mathrm{n}-1}}=\frac{1}{\omega \cdot \mathrm{C}} \cdot \tilde{\mathrm{I}}_{\mathrm{C}_{\mathrm{n}-1}}$ Note: In eq. (3) $\tilde{\mathrm{I}}_{\mathrm{n}-1}$ is used instead of $\mathrm{I}_{\mathrm{C}_{\mathrm{n}-1}}$ because the current leads the voltage by $90^{\circ}$. Inductive voltage drop: $\quad \mathrm{V}_{\mathrm{L}_{\mathrm{n}-1}}=\omega \cdot \mathrm{L} \cdot \tilde{\mathrm{I}}_{\mathrm{L}_{\mathrm{n}-1}}$ Note: In eq. (4) $\tilde{I}_{L_{n-1}}$ is used instead of $I_{L_{n-1}}$ because the current lags behind the voltage by $90^{\circ}$. The voltage $\mathrm{V}_{\mathrm{C}_{\mathrm{n}}}$ is given by: $\mathrm{V}_{\mathrm{C}_{\mathrm{n}}}=\mathrm{V}_{0} \cdot \sin (\omega \cdot \mathrm{t}+\mathrm{n} \cdot \varphi)$ Formula (5) follows from the problem. From eq. (3) and eq. (5): $\mathrm{I}_{\mathrm{C}_{\mathrm{n}}}=\omega \cdot \mathrm{C} \cdot \mathrm{V}_{0} \cdot \cos (\omega \cdot \mathrm{t}+\mathrm{n} \cdot \varphi)$ From eq. (4) and eq. (2) and with eq. (5) $$ \begin{aligned} & I_{L_{n-1}}=\frac{V_{0}}{\omega \cdot L} \cdot\left[2 \cdot \sin \left(\omega \cdot t+\left(n-\frac{1}{2}\right) \cdot \varphi\right) \cdot \sin \frac{\varphi}{2}\right] \\ & I_{L_{n}}=\frac{V_{0}}{\omega \cdot L} \cdot\left[2 \cdot \sin \left(\omega \cdot t+\left(n+\frac{1}{2}\right) \cdot \varphi\right) \cdot \sin \frac{\varphi}{2}\right] \end{aligned} $$ Eqs. (6), (7) and (8) must satisfy the current law. This gives the dependence of $\varphi$ on $\omega, \mathrm{L}$ and C . $$ 0=\mathrm{V}_{0} \cdot \omega \cdot \mathrm{C} \cdot \cos (\omega \cdot \mathrm{t}+\mathrm{n} \cdot \varphi)+2 \cdot \frac{\mathrm{~V}_{0}}{\omega \cdot \mathrm{~L}} \cdot \sin \frac{\varphi}{2} \cdot\left[2 \cdot \cos (\omega \cdot \mathrm{t}+\mathrm{n} \cdot \varphi) \cdot \sin \left(-\frac{\varphi}{2}\right)\right] $$ This condition must be true for any instant of time. Therefore it is possible to divide by $\mathrm{V}_{0} \cdot \cos (\omega \cdot \mathrm{t}+\mathrm{n} \cdot \varphi)$. Hence $\omega^{2} \cdot \mathrm{~L} \cdot \mathrm{C}=4 \cdot \sin ^{2}\left(\frac{\varphi}{2}\right)$. The result is $$ \varphi=2 \cdot \arcsin \left(\frac{\omega \cdot \sqrt{\mathrm{~L} \cdot \mathrm{C}}}{2}\right) \text { with } 0 \leq \omega \leq \frac{2}{\sqrt{\mathrm{~L} \cdot \mathrm{C}}} $$
|
IPHO 1987
|
expression
|
||
21
|
When sine waves propagate in an infinite LC-grid (see the figure below) the phase of the acvoltage across two successive capacitors differs by $\Phi$. <image> Determine the velocity of propagation of the waves if the length of each unit is $\ell$.
|
$\mathrm{v}=\frac{\ell}{\Delta \mathrm{t}}=\frac{\omega \cdot \ell}{\varphi} \quad \text { or } \quad \mathrm{v}=\frac{\omega \cdot \ell}{2 \cdot \arcsin \left(\frac{\omega \cdot \sqrt{\mathrm{~L} \cdot \mathrm{C}}}{2}\right)}$
|
The distance $\ell$ is covered in the time $\Delta \mathrm{t}$ thus the propagation velocity is $$ \mathrm{v}=\frac{\ell}{\Delta \mathrm{t}}=\frac{\omega \cdot \ell}{\varphi} \quad \text { or } \quad \mathrm{v}=\frac{\omega \cdot \ell}{2 \cdot \arcsin \left(\frac{\omega \cdot \sqrt{\mathrm{~L} \cdot \mathrm{C}}}{2}\right)} $$
|
IPHO 1987
|
expression
|
||
22
|
When sine waves propagate in an infinite LC-grid (see the figure below) the phase of the acvoltage across two successive capacitors differs by $\Phi$. <image>State under what conditions the propagation velocity of the waves is almost independent of $\omega$. Determine the velocity in this case.
|
$\frac{\omega \cdot \sqrt{\mathrm{L} \cdot \mathrm{C}}}{2} \ll 1$ and $ \mathrm{v}_{0}=\frac{\ell}{\sqrt{\mathrm{L} \cdot \mathrm{C}}} $
|
Slightly dependent means arc $\sin \left(\frac{\omega \cdot \sqrt{\mathrm{L} \cdot \mathrm{C}}}{2}\right) \sim \omega$, since v is constant in that case. This is true only for small values of $\omega$. That means $\frac{\omega \cdot \sqrt{\mathrm{L} \cdot \mathrm{C}}}{2} \ll 1$ and therefore $$ \mathrm{v}_{0}=\frac{\ell}{\sqrt{\mathrm{L} \cdot \mathrm{C}}} $$
|
IPHO 1987
|
expression
|
||
23
|
Early this century a model of the earth was proposed in which it was assumed to be a sphere of radius $R$ consisting of a homogeneous isotropic solid mantle down to radius $R_{c}$. The core region within radius $R_{c}$ contained a liquid. Figure 2.1 <image> Figure 2.1 The velocities of longitudinal and transverse seismic waves P and S waves respectively, are constant, $V_{P}$, and $V_{\mathrm{S}}$ within the mantle. In the core, longitudinal waves have a constant velocity $V_{C P},<V_{P}$, and transverse waves are not propagated. An earthquake at E on the surface of the Earth produces seismic waves that travel through the Earth and are observed by a surface observer who can set up his seismometer at any point $X$ on the Earth's surface. The angular separation between E and $\mathrm{X}, 2 \theta$ given by $$ 2 \theta=\text { Angle } E O X $$ where O is the centre of the Earth.
|
$ t=\frac{2 R \sin \theta}{v}, \quad \text { for } \theta \leq \cos ^{-1}\left(\frac{R_{C}}{R}\right) $
|
Figure2.1 $$ \mathrm{EX}=2 R \sin \theta \quad \therefore t=\frac{2 R \sin \theta}{v} $$ where $v=v_{P}$ for P waves and $v=v_{S}$ for S waves. This is valid providing $X$ is at an angular separation less than or equal to $X^{\prime}$, the tangential ray to the liquid core. X ' has an angular separation given by, from the diagram, $$ 2 \phi=2 \cos ^{-1}\left(\frac{R_{C}}{R}\right) $$ Thus $$ t=\frac{2 R \sin \theta}{v}, \quad \text { for } \theta \leq \cos ^{-1}\left(\frac{R_{C}}{R}\right) $$ where $v=v_{\mathrm{P}}$ for P waves and $v=v_{\mathrm{S}}$ for shear waves.
|
IPHO 1986
|
equation
|
||
24
|
Early this century a model of the earth was proposed in which it was assumed to be a sphere of radius $R$ consisting of a homogeneous isotropic solid mantle down to radius $R_{c}$. The core region within radius $R_{c}$ contained a liquid. Figure 2.1 <image> Figure 2.1 The velocities of longitudinal and transverse seismic waves P and S waves respectively, are constant, $V_{P}$, and $V_{\mathrm{S}}$ within the mantle. In the core, longitudinal waves have a constant velocity $V_{C P},<V_{P}$, and transverse waves are not propagated. An earthquake at E on the surface of the Earth produces seismic waves that travel through the Earth and are observed by a surface observer who can set up his seismometer at any point $X$ on the Earth's surface. The angular separation between E and $\mathrm{X}, 2 \theta$ given by $$ 2 \theta=\text { Angle } E O X $$ where O is the centre of the Earth.For some of the positions of X such that the seismic P waves arrive at the observer after two refractions at the mantle-core interface. Draw the path of such a seismic P wave. Obtain a relation between $\theta$ and $i$, the angle of incidence of the seismic P wave at the mantle-core interface, for P waves.
|
$\theta=\left[90-\sin ^{-1}\left(\frac{v_{C P}}{v_{P}} \sin i\right)+i-\sin ^{-1}\left(\frac{R_{C}}{R} \sin i\right)\right]$
|
$\frac{R_{C}}{R}=0.5447 \quad$ and $\quad \frac{v_{C P}}{v_{P}}=0.831 .3$ Figure 2.2 From Figure 2.2 $\theta=\hat{A O C}+E \hat{O A} \Rightarrow \theta=(90-r)+(1-\alpha)$ (ii) Continued Snell's Law gives: $\frac{\sin i}{\sin r}=\frac{v_{P}}{v_{C P}}$. From the triangle EAO , sine rule gives $\frac{R_{C}}{\sin x}=\frac{R}{\sin i}$. Substituting (2) and (3) into (1) $\theta=\left[90-\sin ^{-1}\left(\frac{v_{C P}}{v_{P}} \sin i\right)+i-\sin ^{-1}\left(\frac{R_{C}}{R} \sin i\right)\right]$
|
IPHO 1986
|
expression
|
||
25
|
Early this century a model of the earth was proposed in which it was assumed to be a sphere of radius $R$ consisting of a homogeneous isotropic solid mantle down to radius $R_{c}$. The core region within radius $R_{c}$ contained a liquid. Figure 2.1 <image> Figure 2.1 The velocities of longitudinal and transverse seismic waves P and S waves respectively, are constant, $V_{P}$, and $V_{\mathrm{S}}$ within the mantle. In the core, longitudinal waves have a constant velocity $V_{C P},<V_{P}$, and transverse waves are not propagated. An earthquake at E on the surface of the Earth produces seismic waves that travel through the Earth and are observed by a surface observer who can set up his seismometer at any point $X$ on the Earth's surface. The angular separation between E and $\mathrm{X}, 2 \theta$ given by $$ 2 \theta=\text { Angle } E O X $$ where O is the centre of the Earth. After an earthquake an observer measures the time delay between the arrival of the S wave, following the $P$ wave, as 2 minutes 11 seconds. Deduce the angular separation of the earthquake from the observer using the data given in Section (iii).(iii) Using the data $$ \begin{array}{ll} R & =6370 \mathrm{~km} \\ R_{C} & =3470 \mathrm{~km} \\ v_{C P} & =10.85 \mathrm{~km} \mathrm{~s}^{-1} \\ v_{S} & =6.31 \mathrm{~km} \mathrm{~s}^{-1} \\ v_{C P} & =9.02 \mathrm{~km} \mathrm{~s}^{-1} \end{array} $$
|
$114^{\circ}$
|
Using the result $t=\frac{2 r \sin \theta}{v}$ the time delay $\Delta t$ is given by $\Delta t=2 R \sin \theta\left[\frac{1}{v_{S}}-\frac{1}{v_{P}}\right]$ Substituting the given data $131=2(6370)\left[\frac{1}{6.31}-\frac{1}{10.85}\right] \sin \theta$ Therefore the angular separation of E and X is $2 \theta=17.84^{\circ}$ This result is less than $2 \cos ^{-1}\left(\frac{R_{C}}{R}\right)=2 \cos ^{-1}\left(\frac{3470}{6370}\right)=114^{\circ}$ And consequently the seismic wave is not refracted through the core.
|
IPHO 1986
|
value
|
||
26
|
Three particles, each of mass $m$, are in equilibrium and joined by unstretched massless springs, each with Hooke's Law spring constant $k$. They are constrained to move in a circular path as indicated in Figure 3.1. Figure 3.1 <image>If each mass is displaced from equilibrium by small displacements $u_{1}, u_{2}$ and $u_{3}$ respectively, write down the equation of motion for mass two.
|
$m \frac{d^{2} u_{2}}{d t^{2}}=k\left(u_{3}-u_{2}\right)+k\left(u_{1}-u_{2}\right) $
|
$m \frac{d^{2} u_{2}}{d t^{2}}=k\left(u_{3}-u_{2}\right)+k\left(u_{1}-u_{2}\right)$
|
IPHO 1986
|
equation
|
||
27
|
Three particles, each of mass $m$, are in equilibrium and joined by unstretched massless springs, each with Hooke's Law spring constant $k$. They are constrained to move in a circular path as indicated in Figure 3.1. Figure 3.1 <image>If each mass is displaced from equilibrium by small displacements $u_{1}, u_{2}$ and $u_{3}$ respectively, write down the equation of motion for mass one.
|
$m \frac{d^{2} u_{1}}{d t^{2}}=k\left(u_{2}-u_{1}\right)+k\left(u_{3}-u_{1}\right)$
|
$m \frac{d^{2} u_{1}}{d t^{2}}=k\left(u_{2}-u_{1}\right)+k\left(u_{3}-u_{1}\right)$
|
IPHO 1986
|
equation
|
||
28
|
Three particles, each of mass $m$, are in equilibrium and joined by unstretched massless springs, each with Hooke's Law spring constant $k$. They are constrained to move in a circular path as indicated in Figure 3.1. Figure 3.1 <image>If each mass is displaced from equilibrium by small displacements $u_{1}, u_{2}$ and $u_{3}$ respectively, write down the equation of motion for mass three.
|
$m \frac{d^{2} u_{3}}{d t^{2}}=k\left(u_{1}-u_{3}\right)+k\left(u_{2}-u_{3}\right)$
|
$m \frac{d^{2} u_{3}}{d t^{2}}=k\left(u_{1}-u_{3}\right)+k\left(u_{2}-u_{3}\right)$
|
IPHO 1986
|
equation
|
||
29
|
Three particles, each of mass $m$, are in equilibrium and joined by unstretched massless springs, each with Hooke's Law spring constant $k$. They are constrained to move in a circular path as indicated in Figure 3.1. Figure 3.1 <image> in terms of its displacement and those of the adjacent masses when the particles are displaced from equilibrium.
|
$$ \begin{array}{ll} m \frac{d^{2} u_{n}}{d t^{2}}=k\left(u_{1+n}-u_{n}\right)+k\left(u_{n-1}-u_{n}\right) & n=1,2 \ldots \ldots N \\ \frac{d^{2} u_{n}}{d t^{2}}=k\left(u_{1+n}-u_{n}\right)+\omega_{o}^{2}\left(u_{n-1}-u_{n}\right) & \end{array} $$
|
Equation of motion of the n'th particle: $$ \begin{array}{ll} m \frac{d^{2} u_{n}}{d t^{2}}=k\left(u_{1+n}-u_{n}\right)+k\left(u_{n-1}-u_{n}\right) & n=1,2 \ldots \ldots N \\ \frac{d^{2} u_{n}}{d t^{2}}=k\left(u_{1+n}-u_{n}\right)+\omega_{o}^{2}\left(u_{n-1}-u_{n}\right) & \end{array} $$ Substituting $u_{n}(t)=u_{n}(0) \sin \left(2 n s \frac{\pi}{N}\right) \cos \omega_{s} t$ $$ \begin{aligned} & -\omega_{s}^{2}\left(\sin \left(2 n s \frac{\pi}{N}\right)\right)=\omega_{o}^{2}\left[\sin \left(2(n+1) s \frac{\pi}{N}\right)-2 \sin \left(2 n s \frac{\pi}{N}\right)+\sin \left(2(n-1) s \frac{\pi}{N}\right)\right] \\ & -\omega_{s}^{2}\left(\sin \left(2 n s \frac{\pi}{N}\right)\right)=2 \omega_{o}^{2}\left[\frac{1}{2} \sin \left(2(n+1) s \frac{\pi}{N}\right)+\sin \left(2 n s \frac{\pi}{N}\right)-\frac{1}{2} \sin \left(2(n-1) s \frac{\pi}{N}\right)\right] \\ & -\omega_{s}^{2}\left(\sin \left(2 n s \frac{\pi}{N}\right)\right)=2 \omega_{o}^{2}\left[\sin \left(2 n s \frac{\pi}{N}\right) \cos \left(2 s \frac{\pi}{N}\right)-\sin \left(2 n s \frac{\pi}{N}\right)\right] \\ & \therefore \omega_{s}^{2}=2 \omega_{o}^{2}\left[1-\cos \left(2 s \frac{\pi}{N}\right)\right]: \quad(s=1,2, \ldots . . N) \end{aligned} $$ As $2 \sin ^{2} \theta=1-\cos 2 \theta$ This gives $$ \omega_{s}=2 \omega_{o} \sin \left(\frac{s \pi}{N}\right) \quad(s=1,2, \ldots N) $$ $\omega_{s}$ can have values from 0 to $2 \omega_{o}=2 \sqrt{\frac{k}{m}}$ when $N \rightarrow \infty$; corresponding to range $s=1$ to $\frac{N}{2}$.
|
IPHO 1986
|
equation
|
||
30
|
Three particles, each of mass $m$, are in equilibrium and joined by unstretched massless springs, each with Hooke's Law spring constant $k$. They are constrained to move in a circular path as indicated in Figure 3.1. Figure 3.1 <image> Determine the ratio $$ u_{n} / u_{n+1} $$ for large $N$ in the case of low frequency solutions.
|
1
|
For s'th mode $$ \frac{u_{n}}{u_{n+1}}=\frac{\frac{u_{n}}{u_{n+1}}=\frac{\sin \left(2 n s \frac{\pi}{N}\right)}{\sin \left(2(n+1) s \frac{\pi}{N}\right)}}{\sin \left(2 n s \frac{\pi}{N}\right) \cos \left(2 s \frac{\pi}{N}\right)+\cos \left(2 n s \frac{\pi}{N}\right) \sin \left(2 s \frac{\pi}{N}\right)} $$ (a) For small $\omega,\left(\frac{s}{N}\right) \approx 0$, thus $\cos \left(2 n s \frac{\pi}{N}\right) \cong 1$ and $=\sin \left(2 n s \frac{\pi}{N}\right) \approx 0$, and so $\frac{u_{n}}{u_{n+1}} \cong 1$.
|
IPHO 1986
|
value
|
||
31
|
Three particles, each of mass $m$, are in equilibrium and joined by unstretched massless springs, each with Hooke's Law spring constant $k$. They are constrained to move in a circular path as indicated in Figure 3.1. Figure 3.1 <image> Determine the ratio $$ u_{n} / u_{n+1} $$ for large $N$ in the case of $\omega=\omega_{\text {max }}$, where $\omega_{\text {max }}$ is the maximum frequency solution.
|
-1
|
For s'th mode $$ \frac{u_{n}}{u_{n+1}}=\frac{\frac{u_{n}}{u_{n+1}}=\frac{\sin \left(2 n s \frac{\pi}{N}\right)}{\sin \left(2(n+1) s \frac{\pi}{N}\right)}}{\sin \left(2 n s \frac{\pi}{N}\right) \cos \left(2 s \frac{\pi}{N}\right)+\cos \left(2 n s \frac{\pi}{N}\right) \sin \left(2 s \frac{\pi}{N}\right)} $$ The highest mode, $\omega_{\max }=2 \omega_{o}$, corresponds to $s=N / 2$ $$ \therefore \frac{u_{n}}{u_{n+1}}=-1 \text { as } \frac{\sin (2 n \pi)}{\sin (2(n+1) \pi)}=-1 $$
|
IPHO 1986
|
value
|
||
32
|
A young radio amateur maintains a radio link with two girls living in two towns. He positions an aerial array such that when the girl living in town A receives a maximum signal, the girl living in town B receives no signal and vice versa. The array is built from two vertical rod aerials transmitting with equal intensities uniformly in all directions in the horizontal plane.[^0]a) Find the parameters of the array, i. e. the distance between the rods, its orientation and the phase shift between the electrical signals supplied to the rods, such that the distance between the rods is minimum.
|
$$ a=\frac{\lambda}{4 \sin \frac{1}{2} \varphi}, \quad \mathcal{G}_{A}=\frac{1}{2} \varphi, \quad \mathcal{G}_{B}=-\frac{1}{2} \varphi \quad \text { and } \quad \delta=\frac{1}{2} \pi-2 \pi N $$
|
a) Let the electrical signals supplied to rods 1 and 2 be $E_{1}=E_{0} \cos \omega t$ and $E_{2}=E_{0} \cos (\omega t+\delta)$, respectively. The condition for a maximum signal in direction $\mathfrak{9}_{A}$ (Fig. 4) is: $$ \frac{2 \pi a}{\lambda} \sin 9_{A}-\delta=2 \pi N $$ and the condition for a minimum signal in direction $\vartheta_{B}$ : $$ \frac{2 \pi a}{\lambda} \sin 9_{B}-\delta=2 \pi N^{\prime}+\pi $$ where $N$ and $N^{\prime}$ are arbitrary integers. In addition, $\vartheta_{A}-\vartheta_{B}=\varphi$, where Fig. 4 $\varphi$ is given. The problem can now be formulated as follows: Find the parameters $a, 9_{A}, 9_{B}, \delta, N$, and $N^{\prime}$ satisfying the above equations such, that $a$ is minimum. We first eliminate $\delta$ by subtracting the second equation from the first one: $$ a \sin \vartheta_{A}-a \sin \vartheta_{B}=\lambda\left(N-N^{\prime}-\frac{1}{2}\right) . $$ Using the sine addition theorem and the relation $\vartheta_{B}=\vartheta_{A}-\varphi$ : $$ 2 a \cos \left(9_{A}-\frac{1}{2} \varphi\right) \sin \frac{1}{2} \varphi=\lambda\left(N-N^{\prime}-\frac{1}{2}\right) $$ or $$ a=\frac{\lambda\left(N-N^{\prime}-\frac{1}{2}\right)}{2 \cos \left(\vartheta_{A}-\frac{1}{2} \varphi\right) \sin \frac{1}{2} \varphi} . $$ The minimum of $a$ is obtained for the greatest possible value of the denominator, i. e.: $$ \cos \left(\vartheta_{A}-\frac{1}{2} \varphi\right)=1, \quad \vartheta_{A}=\frac{1}{2} \varphi $$ and the minimum value of the numerator, i. e.: $$ N-N^{\prime}=1 . $$ The solution is therefore: $$ a=\frac{\lambda}{4 \sin \frac{1}{2} \varphi}, \quad \mathcal{G}_{A}=\frac{1}{2} \varphi, \quad \mathcal{G}_{B}=-\frac{1}{2} \varphi \quad \text { and } \quad \delta=\frac{1}{2} \pi-2 \pi N $$ ( $\mathrm{N}=0$ can be assumed throughout without loosing any physically relevant solution.)
|
IPHO 1985
|
equation
|
||
33
|
A young radio amateur maintains a radio link with two girls living in two towns. He positions an aerial array such that when the girl living in town A receives a maximum signal, the girl living in town B receives no signal and vice versa. The array is built from two vertical rod aerials transmitting with equal intensities uniformly in all directions in the horizontal plane.b) Find the numerical solution if the boy has a radio station transmitting at 27 MHz and builds up the aerial array at Portorož. Using the map he has found that the angles between the north and the direction of A (Koper) and of B (small town of Buje in Istria) are $72^{\circ}$ and $157^{\circ}$, respectively. ( unit: $\circ$ )
|
114.5
|
b) The wavelength $\lambda=c / v=11.1 \mathrm{~m}$, and the angle between directions A and $\mathrm{B}, \varphi=157^{\circ}-72^{\circ}=85^{\circ}$. The minimum distance between the rods is $a=4.1 \mathrm{~m}$, while the direction of the symmetry line of the rods is $72^{\circ}+42.5^{\circ}=114.5^{\circ}$ measured from the north.
|
IPHO 1985
|
value
|
||
34
|
In a long bar having the shape of a rectangular parallelepiped with sides $a$, $b$, and $c(a \gg b \gg c)$, made from the semiconductor InSb flows a current $I$ parallel to the edge $a$. The bar is in an external magnetic field $B$ which is parallel to the edge $c$. The magnetic field produced by the current $I$ can be neglected. The current carriers are electrons. The average velocity of electrons in a semiconductor in the presence of an electric field only is $v=\mu E$, where $\mu$ is called mobility. If the magnetic field is also present, the electric field is no longer parallel to the current. This phenomenon is known as the Hall effect. a) Determine what the magnitude and the direction of the electric field in the bar is, to yield the current described above. ( unit: $\mathrm{~V} / \mathrm{m}$ )
|
4.06
|
a) First the electron velocity is calculated from the current I: $$ I=j S=n e_{0} v b c, \quad v=\frac{I}{n e_{0} b c}=25 \mathrm{~m} / \mathrm{s} $$ The components of the electric field are obtained from the electron velocity. The component in the direction of the current is $$ E_{\|}=\frac{v}{\mu}=3.2 \mathrm{~V} / \mathrm{m} $$ The component of the electric field in the direction $b$ is equal to the Lorentz force on the electron divided by its charge: $$ E_{\perp}=v B=2.5 \mathrm{~V} / \mathrm{m} . $$ The magnitude of the electric field is $$ E=\sqrt{E_{\|}^{2}+E_{\perp}^{2}}=4.06 \mathrm{~V} / \mathrm{m} $$ while its direction is shown in Fig. 5 (Note that the electron velocity is in the opposite direction with respect to the current.) Fig. 5
|
IPHO 1985
|
value
|
||
35
|
In a long bar having the shape of a rectangular parallelepiped with sides $a$, $b$, and $c(a \gg b \gg c)$, made from the semiconductor InSb flows a current $I$ parallel to the edge $a$. The bar is in an external magnetic field $B$ which is parallel to the edge $c$. The magnetic field produced by the current $I$ can be neglected. The current carriers are electrons. The average velocity of electrons in a semiconductor in the presence of an electric field only is $v=\mu E$, where $\mu$ is called mobility. If the magnetic field is also present, the electric field is no longer parallel to the current. This phenomenon is known as the Hall effect. b) Calculate the difference of the electric potential between the opposite points on the surfaces of the bar in the direction of the edge $b$. ( unit: $\mathrm{mV}$ )
|
25
|
b) The potential difference is $$ U_{H}=E_{\perp} b=25 \mathrm{mV} $$
|
IPHO 1985
|
value
|
||
36
|
In a long bar having the shape of a rectangular parallelepiped with sides $a$, $b$, and $c(a \gg b \gg c)$, made from the semiconductor InSb flows a current $I$ parallel to the edge $a$. The bar is in an external magnetic field $B$ which is parallel to the edge $c$. The magnetic field produced by the current $I$ can be neglected. The current carriers are electrons. The average velocity of electrons in a semiconductor in the presence of an electric field only is $v=\mu E$, where $\mu$ is called mobility. If the magnetic field is also present, the electric field is no longer parallel to the current. This phenomenon is known as the Hall effect.c) Find the analytic expression for the DC component of the electric potential difference in case b) if the current and the magnetic field are alternating (AC); $I=I_{0} \sin \omega t$ and $B=B_{0} \sin (\omega t+\delta)$.
|
$$ \bar{U}_{H}=\frac{I_{0} B_{0}}{2 n e_{0} c} \cos \delta $$
|
c) The potential difference $U_{H}$ is now time dependent: $$ U_{H}=\frac{I B b}{n e_{0} b c}=\frac{I_{0} B_{0}}{n e_{0} c} \sin \omega t \sin (\omega t+\delta) $$ The DC component of $U_{H}$ is $$ \bar{U}_{H}=\frac{I_{0} B_{0}}{2 n e_{0} c} \cos \delta $$
|
IPHO 1985
|
expression
|
||
37
|
A particle moves along the positive axis $O x$ (one-dimensional situation) under a force having a projection $F_{x}=F_{0}$ on $O x$, as represented, as function of $x$, in the figure 1.1. In the origin of the $O x$ axis is placed a perfectly reflecting wall. A friction force, with a constant modulus $F_{f}=1,00 \mathrm{~N}$, acts everywhere on the particle. The particle starts from the point $x=x_{0}=1,00 \mathrm{~m}$ having the kinetic energy $E_{c}=10,0 \mathrm{~J}$. a. Find the length of the path of the particle until its' final stop ( unit: m ) <image>
|
20
|
a. It is possible to make a model of the situation in the problem, considering the Ox axis vertically oriented having the wall in its' lower part. The conservative force $F_{x}$ could be the weight of the particle. One may present the motion of the particle as the vertical motion of a small elastic ball elastically colliding with the ground and moving with constant friction through the medium. The friction force is smaller than the weight. The potential energy of the particle can be represented in analogy to the gravitational potential energy of the ball, $m \cdot g \cdot h$, considering $m \cdot g=\left|F_{x}\right| ; h=x$. As is very well known, in the field of a conservative force, the variation of the potential energy depends only on the initial and final positions of the particle, being independent of the path between those positions. For the situation in the problem, when the particle moves towards the wall, the force acting on it is directed towards the wall and has the modulus I $$ \begin{aligned} & F_{\leftarrow}=\left|F_{x}\right|-F_{f} \\ & F_{\leftarrow}=9 \mathrm{~N} \end{aligned} $$ As a consequence, the motion of the particle towards the wall is a motion with a constant acceleration having the modulus $$ a_{\leftarrow}=\frac{F_{\leftarrow}}{m}=\frac{\left|F_{x}\right|-F_{f}}{m} $$ During the motion, the speed of the particle increases. Hitting the wall, the particle starts moving in opposite direction with a speed equal in modulus with the one it had before the collision. When the particle moves away from the wall, in the positive direction of the $O x$ axis, the acting force is again directed towards to the wall and has the magnitude $$ \begin{aligned} & F_{\rightarrow}=\left|F_{x}\right|+F_{f} \\ & F_{\rightarrow}=11 \mathrm{~N} \end{aligned} $$ Correspondingly, the motion of the particle from the wall is slowed down and the magnitude of the acceleration is $$ a_{\rightarrow}=\frac{F_{\rightarrow}}{m}=\frac{\left|F_{x}\right|+F_{f}}{m} $$ During this motion, the speed of the particle diminishes to zero. Because during the motion a force acts on the particle, the body cannot have an equilibrium position in any point on axis - the origin making an exception as the potential energy vanishes there. The particle can definitively stop only in this point. The work of a conservative force from the point having the coordinate $x_{0}=0$ to the point $x, L_{0 \rightarrow x}$ is correlated with the variation of the potential energy of the particle $U(x)-U(0)$ as follows $$ \left\{\begin{array}{l} U(x)-U(0)=-L_{0 \rightarrow x} \\ U(x)-U(0)=-\int_{0}^{x} \vec{F}_{x} \cdot \overrightarrow{d x}=\int_{0}^{x}\left|F_{x}\right| \cdot d x=\left|F_{x}\right| \cdot x \end{array}\right. $$ Admitting that the potential energy of the particle vanishes for $x=0$, the initial potential energy of the particle $U\left(x_{0}\right)$ in the field of conservative force $$ F_{x}(x)=F_{0} $$ can be written $$ U\left(x_{0}\right)=\left|F_{0}\right| \cdot x_{0} $$ The initial kinetic energy $E\left(x_{0}\right)$ of the particle is - as given $$ E\left(x_{0}\right)=E_{c} $$ and, consequently the total energy of the particle $W\left(x_{0}\right)$ is $$ W\left(x_{0}\right)=U\left(x_{0}\right)+E_{c} $$ The draw up of the particle occurs when the total energy of the particle is entirely exhausted by the work of the friction force. The distance covered by the particle before it stops, $D$, obeys $$ \left\{\begin{array}{l} W\left(x_{0}\right)=D \cdot F_{f} \\ U\left(x_{0}\right)+E_{c}=D \cdot F_{f} \\ \left|F_{x}\right| \cdot x_{0}+E_{c}=D \cdot F_{f} \end{array}\right. $$ so that , $D=\frac{\left|F_{x}\right| \cdot x_{0}+E_{c}}{F_{f}}$ and D $=20 \mathrm{~m}$ The relations (1.13) and (1.14) represent the answer to the question I.a.
|
IPHO 1984
|
value
|
||
38
|
Two dispersive prisms having apex angles $\hat{A}_{1}=60^{\circ}$ and $\hat{A}_{2}=30^{\circ}$ are glued as in the figure ( $\hat{C}=90^{\circ}$ ). The dependences of refraction indexes of the prisms on the wavelength are given by the relations $n_{1}(\lambda)=a_{1}+\frac{b_{1}}{\lambda^{2}}$; $n_{2}(\lambda)=a_{2}+\frac{b_{2}}{\lambda^{2}}$ were $a_{1}=1,1, \quad b_{1}=1 \cdot 10^{5} n m^{2}, \quad a_{2}=1,3, \quad b_{2}=5 \cdot 10^{4} n m^{2}$. <image>a. Determine the wavelength $\lambda_{0}$ of the incident radiation that pass through the prisms without refraction on $A C$ face at any incident angle; determine the corresponding refraction indexes of the prisms. ( unit: nm )
|
500
|
a. The ray with the wavelength $\lambda_{0}$ pass trough the prisms system without refraction on $A C$ face at any angle of incidence if : $n_{1}\left(\lambda_{0}\right)=n_{2}\left(\lambda_{0}\right)$ Because the dependence of refraction indexes of prisms on wavelength has the form : $n_{1}(\lambda)=a_{1}+\frac{b_{1}}{\lambda^{2}}$ $n_{2}(\lambda)=a_{2}+\frac{b_{2}}{\lambda^{2}}$ The relation (3.1) becomes: $a_{1}+\frac{b_{1}}{\lambda_{0}{ }^{2}}=a_{2}+\frac{b_{2}}{\lambda_{0}{ }^{2}}$ The wavelength $\lambda_{0}$ has correspondingly the form: $\lambda_{0}=\sqrt{\frac{b_{1}-b_{2}}{a_{2}-a_{1}}}$ Substituting the furnished numerical values $\lambda_{0}=500 \mathrm{~nm}$ The corresponding common value of indexes of refraction of prisms for the radiation with the wavelength $\lambda_{0}$ is: $n_{1}\left(\lambda_{0}\right)=n_{2}\left(\lambda_{0}\right)=1,5$ The relations (3.6) and (3.7) represent the answers of question $\mathbf{a}$.
|
IPHO 1983
|
value
|
||
39
|
Two dispersive prisms having apex angles $\hat{A}_{1}=60^{\circ}$ and $\hat{A}_{2}=30^{\circ}$ are glued as in the figure ( $\hat{C}=90^{\circ}$ ). The dependences of refraction indexes of the prisms on the wavelength are given by the relations $n_{1}(\lambda)=a_{1}+\frac{b_{1}}{\lambda^{2}}$; $n_{2}(\lambda)=a_{2}+\frac{b_{2}}{\lambda^{2}}$ were $a_{1}=1,1, \quad b_{1}=1 \cdot 10^{5} n m^{2}, \quad a_{2}=1,3, \quad b_{2}=5 \cdot 10^{4} n m^{2}$. <image>c. Determine the minimum deviation angle in the system for a ray having the wavelength $\lambda_{0}$. ( unit: $\circ$ )
|
30,7
|
c. In the figure 1.2 is presented the path of ray with wavelength $\lambda_{0}$ at minimum deviation (the angle between the direction of incidence of ray and the direction of emerging ray is minimal). Figure 3.2 In this situation $n_{1}\left(\lambda_{0}\right)=n_{2}\left(\lambda_{0}\right)=\frac{\sin \frac{\delta_{\text {min }}+A^{\prime}}{2}}{\sin \frac{A^{\prime}}{2}}$ where $m\left(\hat{A}^{\prime}\right)=30^{\circ}$, as in the figure 1.1 Substituting in (3.8) the values of refraction indexes the result is $\sin \frac{\delta_{\text {min }}+A^{\prime}}{2}=\frac{3}{2} \cdot \sin \frac{A^{\prime}}{2}$ or $\delta_{\text {min }}=2 \arcsin \left(\frac{3}{2} \cdot \sin \frac{A^{\prime}}{2}\right)-\frac{A^{\prime}}{2}$ Numerically $\delta_{\text {min }} \cong 30,7^{\circ}$ The relation (3.11) represents the answer of question $\mathbf{c}$.
|
IPHO 1983
|
value
|
||
40
|
Two dispersive prisms having apex angles $\hat{A}_{1}=60^{\circ}$ and $\hat{A}_{2}=30^{\circ}$ are glued as in the figure ( $\hat{C}=90^{\circ}$ ). The dependences of refraction indexes of the prisms on the wavelength are given by the relations $n_{1}(\lambda)=a_{1}+\frac{b_{1}}{\lambda^{2}}$; $n_{2}(\lambda)=a_{2}+\frac{b_{2}}{\lambda^{2}}$ were $a_{1}=1,1, \quad b_{1}=1 \cdot 10^{5} n m^{2}, \quad a_{2}=1,3, \quad b_{2}=5 \cdot 10^{4} n m^{2}$. <image>d. Calculate the wavelength of the ray that penetrates and exits the system along directions parallel to DC. ( unit: $\mu m$ )
|
1.2
|
d. Using the figure 1.3 the refraction law on the $A D$ face is $\sin i_{1}=n_{1} \cdot \sin r_{1}$ The refraction law on the $A C$ face is $n_{1} \cdot \sin r_{1}{ }^{\prime}=n_{2} \cdot \sin r_{2}$ Figure 3.3 As it can be seen in the figure 1.3 $$ r_{2}=A_{2} $$ and $i_{1}=30^{\circ}$ Also, $r_{1}+r_{1}{ }^{\prime}=A_{1}$ Substituting (3.16) and (3.14) in (3.13) it results $n_{1} \cdot \sin \left(A_{1}-r_{1}\right)=n_{2} \cdot \sin A_{2}$ or $n_{1} \cdot\left(\sin A_{1} \cdot \cos r_{1}-\sin r_{1} \cdot \cos A_{1}\right)=n_{2} \cdot \sin A_{2}$ Because of (3.12) and (3.15) it results that $\sin r_{1}=\frac{1}{2 n_{1}}$ and $\cos r_{1}=\frac{1}{2 n_{1}} \sqrt{4 n_{1}{ }^{2}-1}$ Putting together the last three relations it results $\sqrt{4 n_{1}{ }^{2}-1}=\frac{2 n_{2} \cdot \sin A_{2}+\cos A_{1}}{\sin A_{1}}$ Because $\hat{A}_{1}=60^{\circ}$ and $\hat{A}_{2}=30^{\circ}$ relation (3.21) can be written as $$ \sqrt{4 n_{1}^{2}-1}=\frac{2 n_{2}+1}{\sqrt{3}} $$ or $$ 3 \cdot n_{1}^{2}=1+n_{2}+n_{2}^{2} $$ Considering the relations (3.1), (3.2) and (3.23) and operating all calculus it results: $$ \lambda^{4} \cdot\left(3 a_{1}^{2}-a_{2}^{2}-a_{2}-1\right)+\left(6 a_{1} b_{1}-b_{2}-2 a_{2} b_{2}\right) \cdot \lambda^{2}+3 b_{1}^{2}-b_{2}^{2}=0 $$ Solving the equation (3.24) one determine the wavelength $\lambda$ of the ray that enter the prisms system having the direction parallel with $D C$ and emerges the prism system having the direction again parallel with $D C$. That is $$ \lambda=1194 \mathrm{~nm} $$ or $$ \lambda \cong 1,2 \mu m $$ The relation (3.26) represents the answer of question $\mathbf{d}$.
|
IPHO 1983
|
value
|
||
41
|
A photon of wavelength $\lambda_{i}$ is scattered by a moving, free electron. As a result the electron stops and the resulting photon of wavelength $\lambda_{0}$ scattered at an angle $\theta=60^{\circ}$ with respect to the direction of the incident photon, is again scattered by a second free electron at rest. In this second scattering process a photon with wavelength of $\lambda_{f}=1,25 \times 10^{-10} \mathrm{~m}$ emerges at an angle $\theta=60^{\circ}$ with respect to the direction of the photon of wavelength $\lambda_{0}$. Find the de Broglie wavelength for the first electron before the interaction. The following constants are known: $h=6,6 \times 10^{-34} \mathrm{~J} \cdot \mathrm{~s}$ - Planck's constant $m=9,1 \times 10^{-31} \mathrm{~kg}$ - mass oh the electron $c=3,0 \times 10^{8} \mathrm{~m} / \mathrm{s}$ - speed of light in vacuum ( unit: m )
|
$1,24 \times 10^{-10} $
|
The purpose of the problem is to calculate the values of the speed, momentum and wavelength of the first electron. To characterize the photons the following notation are used: Table 4.1 | | initial <br> photon | photon - <br> after the <br> first scattering | final <br> photon | | :--- | :--- | :--- | :--- | | momentum | $\vec{p}_{i}$ | $\vec{p}_{0}$ | $\vec{p}_{f}$ | | energy | $E_{i}$ | $E_{0}$ | $E_{f}$ | | wavelength | $\lambda_{i}$ | $\lambda_{i}$ | $\lambda_{f}$ | To characterize the electrons one uses Table 4.2 | | first electron <br> before collision | first electron <br> after collision | second electron <br> before collision | Second electron <br> after collision | | :--- | :--- | :--- | :--- | :--- | | momentum | $\vec{p}_{1 e}$ | 0 | 0 | $\vec{p}_{2 e}$ | | energy | $E_{1 e}$ | $E_{0 e}$ | $E_{0 e}$ | $E_{2 e}$ | | speed | $\vec{v}_{1 e}$ | 0 | 0 | $\vec{v}_{2 e}$ | The image in figure 4.1 presents the situation before the first scattering of photon. <img class="imgSvg" id = "mdy6e52byqpgywb77uq" src="data:image/svg+xml;base64,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"/> Figure 4.1 Figure 4.3 Figure 4.2 Figure 4.4 To characterize the initial photon we will use his momentum $\vec{p}_{i}$ and his energy $E_{i}$ $$ \left\{\begin{array}{l} \vec{P}_{i}=\frac{h}{\lambda_{i}}=\frac{h \cdot f_{i}}{c} \\ E_{i}=h \cdot f_{i} \end{array}\right. $$ $$ f_{i}=\frac{C}{\lambda_{i}} $$ is the frequency of initial photon. For initial, free electron in motion the momentum $\vec{p}_{o e}$ and the energy $E_{o e}$ are $$ \left\{\begin{array}{l} \vec{P}_{o e}=m \cdot \vec{v}_{1 e}=\frac{m_{0} \cdot \vec{v}_{1 e}}{\sqrt{1-\beta^{2}}} \\ E_{o e}=m \cdot c^{2}=\frac{m_{0} \cdot c^{2}}{\sqrt{1-\beta^{2}}} \end{array}\right. $$ where $m_{0}$ is the rest mass of electron and $m$ is the mass of moving electron. As usual, $\beta=\frac{V_{1 e}}{c}$. De Broglie wavelength of the first electron is $\lambda_{\text {oe }}=\frac{h}{p_{0 e}}=\frac{h \cdot}{m_{0} \cdot v_{1 e}} \sqrt{1-\beta^{2}}$ The situation after the scattering of photon is described in the figure 4.2. To characterize the scattered photon we will use his momentum $\vec{p}_{0}$ and his energy $E_{0}$ $\left\{\begin{array}{l}\vec{P}_{0}=\frac{h}{\lambda_{0}}=\frac{h \cdot f_{0}}{c} \\ E_{0}=h \cdot f_{0}\end{array}\right.$ where $f_{0}=\frac{C}{\lambda_{0}}$ is the frequency of scattered photon. The magnitude of momentum of the electron ( that remains in rest) after the scattering is zero; his energy is $E_{1 e}$. The mass of electron after collision is $m_{0}$ - the rest mass of electron at rest. So, $E_{1 e}=m_{0} \cdot c^{2}$ To determine the moment of the first moving electron, one can write the principles of conservation of moments and energy. That is $\vec{P}_{i}+\vec{p}_{\text {oe }}=\vec{p}_{0}$ and $E_{i}+E_{0 e}=E_{0}+E_{1 e}$ The conservation of moment on $O x$ direction is written as $\frac{h \cdot f_{i}}{c}+m \cdot v_{1 e} \cdot \cos \alpha=\frac{h \cdot f_{0}}{c} \cos \theta$ and the conservation of moment on $O y$ is $m \cdot v_{1 e} \cdot \sin \alpha=\frac{h \cdot f_{0}}{c} \sin \theta$ To eliminate $\alpha$, the last two equation must be written again as $\left\{\begin{array}{l}\left(m \cdot v_{1 e} \cdot \cos \alpha\right)^{2}=\frac{h^{2} \cdot}{c^{2}}\left(f_{0} \cdot \cos \theta-f_{i}\right)^{2} \\ \left(m \cdot v_{1 e} \cdot \sin \alpha\right)^{2}=\left(\frac{h \cdot f_{0}}{c} \sin \theta\right)^{2}\end{array}\right.$ and then added. The result is $m^{2} \cdot v_{1 e}^{2}=\frac{h^{2} \cdot}{c^{2}}\left(f_{0}^{2}+f_{1}^{2}-2 f_{0} \cdot f_{i} \cdot \cos \theta\right)$ or $\frac{m_{0}^{2} \cdot c^{2}}{1-\left(\frac{v_{1 e}}{c}\right)^{2}} \cdot v_{1 e}^{2}=h^{2} \cdot\left(f_{0}^{2}+f_{1}^{2}-2 f_{0} \cdot f_{i} \cdot \cos \theta\right)$ The conservation of energy (4.7) can be written again as $m \cdot c^{2}+h \cdot f_{1}=m_{0} \cdot c^{2}+h \cdot f_{0}$ or $\frac{m_{0} \cdot c^{2}}{\sqrt{1-\left(\frac{v_{1 e}}{c}\right)^{2}}}=m_{0} \cdot c^{2}+h \cdot\left(f_{0}-f_{1}\right)$ Squaring the last relation results $\frac{m_{0}^{2} \cdot c^{4}}{1-\left(\frac{v_{1 e}}{c}\right)^{2}}=m_{0}^{2} \cdot c^{4}+h^{2} \cdot\left(f_{0}-f_{1}\right)^{2}+m_{0} \cdot h \cdot c^{2} \cdot\left(f_{0}-f_{1}\right)$ Subtracting (4.12) from (4.15) the result is $2 m_{0} \cdot c^{2} \cdot h \cdot\left(f_{0}-f_{1}\right)+2 h^{2} \cdot f_{1} \cdot f_{0} \cdot \cos \theta-2 h^{2} \cdot f_{1} \cdot f=0$ or $\frac{h}{m_{0} \cdot c}(1-\cos \theta)=\frac{c}{f_{1}}-\frac{c}{f_{0}}$ Using $\Lambda=\frac{h}{m_{0} \cdot c}$ the relation (4.17) becomes $\Lambda \cdot(1-\cos \theta)=\lambda_{i}-\lambda_{0}$ The wavelength of scattered photon is $\lambda_{0}=\lambda_{i}-\Lambda \cdot(1-\cos \theta)$ shorter than the wavelength of initial photon and consequently the energy of scattered photon is greater that the energy of initial photon. $\left\{\begin{array}{l}\lambda_{i}<\lambda_{0} \\ E_{i}>E_{0}\end{array}\right.$ Let's analyze now the second collision process that occurs in point $N$. To study that, let's consider a new referential having $O x$ direction on the direction of the photon scattered after the first collision. The figure 4.3 presents the situation before the second collision and the figure 4.4 presents the situation after this scattering process. The conservation principle for moment in the scattering process gives $\left\{\begin{array}{l}\frac{h}{\lambda_{0}}=\frac{h}{\lambda_{f}} \cos \theta+m \cdot v_{2 e} \cdot \cos \beta \\ \frac{h}{\lambda_{f}} \sin \theta-m \cdot v_{2 e} \cdot \sin \beta=0\end{array}\right.$ To eliminate the unknown angle $\beta$ must square and then add the equations (4.22) That is $\left\{\begin{array}{l}\left(\frac{h}{\lambda_{0}}-\frac{h}{\lambda_{f}} \cos \theta\right)^{2}=\left(m \cdot v_{2 e} \cdot \cos \beta\right)^{2} \\ \left(\frac{h}{\lambda_{f}} \sin \theta\right)^{2}=\left(m \cdot v_{2 e} \cdot \sin \beta\right)^{2}\end{array}\right.$ or $\left(\frac{h}{\lambda_{f}}\right)^{2}+\left(\frac{h}{\lambda_{0}}\right)^{2}-\frac{2 \cdot h^{2}}{\lambda_{0} \cdot \lambda_{f}} \cos \theta=\left(m \cdot v_{2 e}\right)^{2}$ The conservation principle of energy in the second scattering process gives $\frac{h \cdot c}{\lambda_{0}}+m_{0} \cdot c^{2}=\frac{h \cdot c}{\lambda_{f}}+m \cdot c^{2}$ (4.24) and (4.25) gives $\frac{h^{2} \cdot c^{2}}{\lambda_{f}^{2}}+\frac{h^{2} \cdot c^{2}}{\lambda_{0}^{2}}-\frac{2 \cdot h^{2} \cdot c^{2}}{\lambda_{0} \cdot \lambda_{f}} \cos \theta=m^{2} \cdot c^{2} \cdot v_{2 e}^{2}$ and $h^{2} \cdot c^{2} \cdot\left(\frac{1}{\lambda_{f}}-\frac{1}{\lambda_{0}}\right)^{2}+m_{0}^{2} \cdot c^{4}+2 h \cdot c^{3} \cdot m_{0} \cdot\left(\frac{1}{\lambda_{f}}-\frac{1}{\lambda_{0}}\right)=m^{2} \cdot c^{4}$ Subtracting (4.26) from (1.27), one obtain $\left\{\begin{array}{l}\frac{h}{m_{0} \cdot c} \cdot(1-\cos \theta)=\lambda_{f}-\lambda_{0} \\ \lambda_{f}-\lambda_{0}=\Lambda \cdot(1-\cos \theta)\end{array}\right.$ That is $\left\{\begin{array}{l}\lambda_{f}>\lambda_{0} \\ E_{f}<E_{0}\end{array}\right.$ Because the value of $\lambda_{f}$ is know and $\Lambda$ can be calculate as $$ \left\{\begin{array}{l} \lambda_{f}=1,25 \times 10^{-10} \mathrm{~m} \\ \Lambda=\frac{6,6 \times 10^{-34}}{9,1 \times 10^{-31} \cdot 3 \times 10^{8}} \mathrm{~m}=2,41 \times 10^{-12} \mathrm{~m}=0,02 \times 10^{-10} \mathrm{~m} \end{array}\right. $$ the value of wavelength of photon before the second scattering is $$ \lambda_{0}=1,23 \times 10^{-10} \mathrm{~m} $$ Comparing (4.28) written as: $$ \lambda_{f}=\lambda_{0}+\Lambda \cdot(1-\cos \theta) $$ and (4.20) written as $$ \lambda_{i}=\lambda_{0}+\Lambda \cdot(1-\cos \theta) $$ clearly results $$ \lambda_{i}=\lambda_{f} $$ The energy of the double scattered photon is the same as the energy of initial photon. The direction of "final photon" is the same as the direction of "initial" photon. Concluding, the final photon is identical with the initial photon. The result is expected because of the symmetry of the processes. Extending the symmetry analyze on electrons, the first moving electron that collides the initial photon and after that remains at rest, must have the same momentum and energy as the second electron after the collision - because this second electron is at rest before the collision. That is $$ \left\{\begin{array}{l} \vec{p}_{1 e}=\vec{p}_{2 e} \\ E_{1 e}=E_{2 e} \end{array}\right. $$ Taking into account (4.24), the moment of final electron is $$ p_{2 e}=h \sqrt{\frac{1}{\lambda_{f}^{2}}+\frac{1}{\left(\lambda_{f}-\Lambda(1-\cos \theta)\right)^{2}}-\frac{2 \cdot \cos \theta}{\lambda_{f} \cdot\left(\lambda_{f}-\Lambda(1-\cos \theta)\right)}} $$ The de Broglie wavelength of second electron after scattering (and of first electron before scattering) is $$ \lambda_{1 e}=\lambda_{2 e}=1 /\left(\sqrt{\frac{1}{\lambda_{f}^{2}}+\frac{1}{\left(\lambda_{f}-\Lambda(1-\cos \theta)\right)^{2}}-\frac{2 \cdot \cos \theta}{\lambda_{f} \cdot\left(\lambda_{f}-\Lambda(1-\cos \theta)\right)}}\right) $$ Numerical value of this wavelength is $$ \lambda_{1 e}=\lambda_{2 e}=1,24 \times 10^{-10} \mathrm{~m} $$
|
IPHO 1983
|
value
|
||
42
|
Let's consider the electric circuit in the figure, for which $L_{1}=10 \mathrm{mH}$, $L_{2}=20 \mathrm{mH}, C_{1}=10 \mathrm{nF}, C_{2}=5 \mathrm{nF}$ and $R=100 \mathrm{k} \Omega$. The switch K being closed the circuit is coupled with a source of alternating current. The current furnished by the source has constant intensity while the frequency of the current may be varied. a. Find the ratio of frequency $f_{m}$ for which the active power in circuit has the maximum value $P_{m}$ and the frequency difference $\Delta f=f_{+}-f_{-}$of the frequencies $f_{+}$and $f_{-}$for which the active power in the circuit is half of the maximum power $P_{m}$. <image> <image>
|
150
|
a. As is very well known in the study of AC circuits using the formalism of complex numbers, a complex inductive reactance $\overline{X_{L}}=L \cdot \omega \cdot j,(j=\sqrt{-1})$ is attached to the inductance $L$ - part of a circuit supplied with an alternative current having the pulsation $\omega$. Similar, a complex capacitive reactance $\overline{X_{C}}=-\frac{j}{C \cdot \omega}$ is attached to the capacityC. A parallel circuit will be characterized by his complex admittance $\bar{Y}$. The admittance of the AC circuit represented in the figure is $\left\{\begin{array}{l}\bar{Y}=\frac{1}{R}+\frac{1}{L_{1} \cdot \omega \cdot j}+\frac{1}{L_{2} \cdot \omega \cdot j}-\frac{C_{1} \cdot \omega}{j}-\frac{C_{2} \cdot \omega}{j} \\ \bar{Y}=\frac{1}{R}+j \cdot\left[\left(C_{1}+C_{2}\right)-\left(\frac{1}{L_{1}}+\frac{1}{L_{2}}\right)\right]\end{array}\right.$ The circuit behave as if has a parallel equivalent capacity $C$ $C=C_{1}+C_{2}$ and a parallel equivalent inductance $L$ $\left\{\begin{array}{l}\frac{1}{L}=\frac{1}{L_{1}}+\frac{1}{L_{2}} \\ L=\frac{L_{1} L_{2}}{L_{1}+L_{2}}\end{array}\right.$ The complex admittance of the circuit may be written as $\bar{Y}=\frac{1}{R}+j \cdot\left(C \cdot \omega-\frac{1}{L \cdot \omega}\right)$ and the complex impedance of the circuit will be $$ \left\{\begin{array}{l} \bar{Z}=\frac{1}{\bar{Y}} \\ \bar{Z}=\frac{\frac{1}{R}+j \cdot\left(\frac{1}{L \cdot \omega}-C \cdot \omega\right)}{\sqrt{\left(\frac{1}{R}\right)^{2}+\left(C \cdot \omega-\frac{1}{L \cdot \omega}\right)^{2}}} \end{array}\right. $$ The impedance $Z$ of the circuit, the inverse of the admittance of the circuit $Y$ is the modulus of the complex impedance $\bar{Z}$ $Z=|\bar{Z}|=\frac{1}{\sqrt{\left(\frac{1}{R}\right)^{2}+\left(C \cdot \omega-\frac{1}{L \cdot \omega}\right)^{2}}}=\frac{1}{Y}$ The constant current source supplying the circuit furnish a current having a momentary value $i(t)$ $i(t)=1 \cdot \sqrt{2} \cdot \sin (\omega \cdot t)$, where $I$ is the effective intensity (constant), of the current and $\omega$ is the current pulsation (that can vary) . The potential difference at the jacks of the circuit has the momentary value $u(t)$ $u(t)=U \cdot \sqrt{2} \cdot \sin (\omega \cdot t+\varphi)$ where $U$ is the effective value of the tension and $\varphi$ is the phase difference between tension and current. The effective values of the current and tension obey the relation $U=I \cdot Z$ The active power in the circuit is $P=\frac{U^{2}}{R}=\frac{Z^{2} \cdot I^{2}}{R}$ Because as in the enounce, $\left\{\begin{array}{l}I=\text { constant } \\ R=\text { constant }\end{array}\right.$ the maximal active power is realized for the maximum value of the impedance that is the minimal value of the admittance . The admittance $Y=\sqrt{\left(\frac{1}{R}\right)^{2}+\left(C \cdot \omega-\frac{1}{L \cdot \omega}\right)^{2}}$ has- as function of the pulsation $\omega$ - an „the smallest value" $Y_{\text {min }}=\frac{1}{R}$ for the pulsation $\omega_{m}=\frac{1}{\sqrt{L \cdot C}}$ In this case $\left(C \cdot \omega-\frac{1}{L \cdot \omega}\right)=0$. So, the minimal active power in the circuit has the value $P_{m}=R \cdot I^{2}$ and occurs in the situation of alternative current furnished by the source at the frequency $f_{m}$ $f_{m}=\frac{1}{2 \pi} \omega_{m}=\frac{1}{2 \pi \cdot \sqrt{C \cdot L}}$ To ensure that the active power is half of the maximum power it is necessary that $\left\{\begin{array}{l}P=\frac{1}{2} P_{m} \\ \frac{Z^{2} \cdot I^{2}}{R}=\frac{1}{2} R \cdot I^{2} \\ \frac{2}{R^{2}}=\frac{1}{Z^{2}}=Y^{2}\end{array}\right.$ That is $\left\{\begin{array}{l}\frac{2}{R^{2}}=\frac{1}{R^{2}}+\left(C \cdot \omega-\frac{1}{L \cdot \omega}\right)^{2} \\ \pm \frac{1}{R}=C \cdot \omega-\frac{1}{L \cdot \omega}\end{array}\right.$ The pulsation of the current ensuring an active power at half of the maximum power must satisfy one of the equations $\omega^{2} \pm \frac{1}{R \cdot C} \omega-\frac{1}{L \cdot C}=0$ The two second degree equation may furnish the four solutions $\omega= \pm \frac{1}{2 R \cdot C} \pm \frac{1}{2} \sqrt{\left(\frac{1}{R \cdot C}\right)^{2}+\frac{4}{L \cdot C}}$ Because the pulsation is every time positive, and because $\sqrt{\left(\frac{1}{R \cdot C}\right)^{2}+\frac{4}{L \cdot C}}>\frac{1}{R \cdot C}$ the only two valid solutions are $\omega_{ \pm}=\frac{1}{2} \sqrt{\left(\frac{1}{R \cdot C}\right)^{2}+\frac{4}{L \cdot C}} \pm \frac{1}{2 R \cdot C}$ It exist two frequencies $f_{ \pm}=\frac{1}{2 \pi} \omega_{ \pm}$allowing to obtain in the circuit an active power representing half of the maximum power. $$ \left\{\begin{array}{l} f_{+}=\frac{1}{2 \pi}\left(\frac{1}{2} \sqrt{\left(\frac{1}{R \cdot C}\right)^{2}+\frac{4}{L \cdot C}}+\frac{1}{2 R \cdot C}\right) \\ f_{-}=\frac{1}{2 \pi}\left(\frac{1}{2} \sqrt{\left(\frac{1}{R \cdot C}\right)^{2}+\frac{4}{L \cdot C}}-\frac{1}{2 R \cdot C}\right) \end{array}\right. $$ The difference of these frequencies is $\Delta f=f_{+}-f_{-}=\frac{1}{2 \pi} \frac{1}{R \cdot C}$ the bandwidth of the circuit - the frequency interval around the resonance frequency having at the ends a signal representing $1 / \sqrt{2}$ from the resonance signal. At the ends of the bandwidth the active power reduces at the half of his value at the resonance. The asked ratio is $\left\{\begin{array}{l}\frac{f_{m}}{\Delta f}=\frac{R \cdot C}{\sqrt{L \cdot C}}=R \sqrt{\frac{C}{L}} \\ \frac{f_{m}}{\Delta f}=R \sqrt{\frac{\left(C_{1}+C_{2}\right) \cdot\left(L_{1}+L_{2}\right)}{L_{1} \cdot L_{2}}}\end{array}\right.$ Because $\left\{\begin{array}{l}C=15 n F \\ L=\frac{20}{3} m H\end{array}\right.$ it results that $\omega_{m}=10^{5} \mathrm{rad} \cdot \mathrm{s}^{-1}$ and $\frac{f_{m}}{\Delta f}=R \sqrt{\frac{C}{L}}=100 \times 10^{3} \cdot \sqrt{\frac{3.15 \times 10^{-9}}{20 \times 10^{-3}}}=150$ The (2.26) relation is the answer at the question $\mathbf{a}$.
|
IPHO 1983
|
value
|
||
43
|
Let's consider the electric circuit in the figure, for which $L_{1}=10 \mathrm{mH}$, $L_{2}=20 \mathrm{mH}, C_{1}=10 \mathrm{nF}, C_{2}=5 \mathrm{nF}$ and $R=100 \mathrm{k} \Omega$. The switch K being closed the circuit is coupled with a source of alternating current. The current furnished by the source has constant intensity while the frequency of the current may be varied.The switch $K$ is now open. In the moment $t_{0}$ immediately after the switch <image> is open the intensities of the currents in the coils $L_{1}$ and $i_{01}=0,1 A$ and $i_{02}=0,2 A L_{1}$ (the currents flow as in the figure); at the same moment, the potential difference on the capacitor with capacity $C_{1}$ is $u_{0}=40 \mathrm{~V}$ : b. Calculate the frequency of electromagnetic oscillation in $L_{1} C_{1} C_{2} L_{2}$ circuit; ( unit: $\mathrm{~Hz}$ ) <image>
|
$\frac{10^{5}}{2 \pi}$
|
b. The fact that immediately after the source is detached it is a current in the coils, allow as to admit that currents dependents on time will continue to flow through the coils. Figure 2.1 The capacitors will be charged with charges variable in time. The variation of the charges of the capacitors will results in currents flowing through the conductors linking the capacitors in the circuit. The momentary tension on the jacks of the coils and capacitors - identical for all elements in circuit - is also dependent on time. Let's admit that the electrical potential of the points $C$ and $D$ is $u(t)$ and the $\qquad$ potential of the points $A$ and $B$ is zero. If through the inductance $L_{1}$ passes the variable current having the momentary value $i_{1}(t)$, the relation between the current and potentials is $u(t)-L_{1} \frac{d i_{1}}{d t}=0$ The current passing through the second inductance $i_{2}(t)$ has the expression, $u(t)-L_{2} \frac{d i_{2}}{d t}=0$ If on the positive plate of the capacitor having the capacity $C_{1}$ is stocked the charge $q_{1}(t)$, then at the jacks of the capacitor the electrical tension is $u(t)$ and $q_{1}=C_{1} \cdot u$ Deriving this relation it results $\frac{d q_{1}}{d t}=C_{1} \cdot \frac{d u}{d t}$ But $\frac{d q_{1}}{d t}=-i_{3}$ because the electrical current appears because of the diminishing of the electrical charge on capacitor plate. Consequently $i_{3}=-C_{1} \cdot \frac{d u}{d t}$ Analogous, for the other capacitor, $i_{4}=-C_{4} \cdot \frac{d u}{d t}$ Considering all obtained results $\left\{\begin{array}{l}\frac{d i_{1}}{d t}=\frac{u}{L_{1}} \\ \frac{d i_{2}}{d t}=\frac{u}{L_{2}}\end{array}\right.$ respectively $\left\{\begin{array}{l}\frac{d i_{3}}{d t}=-C_{1} \frac{d^{2} u}{d t^{2}} \\ \frac{d i_{4}}{d t}=C_{2} \frac{d^{2} u}{d t^{2}}\end{array}\right.$ Denoting $i_{5}(t)$ the momentary intensity of the current flowing from point $B$ to the point $A$, then the same momentary intensity has the current through the points $C$ and $D$. For the point $A$ the Kirchhoff rule of the currents gives $i_{1}+i_{5}=i_{3}$ For $B$ point the same rule produces $i_{4}+i_{5}=i_{2}$ Considering (2.37) and (2.38) results $i_{1}-i_{3}=i_{4}-i_{2}$ and deriving $\frac{d i_{1}}{d t}-\frac{d i_{3}}{d t}=\frac{d i_{4}}{d t}-\frac{d i_{2}}{d t}$ that is $\left\{\begin{array}{l}-\frac{u}{L_{1}}-\frac{u}{L_{2}}=C_{1} \frac{d^{2} u}{d t^{2}}+C_{2} \frac{d^{2} u}{d t^{2}} \\ -u \cdot\left(\frac{1}{L_{1}}+\frac{1}{L_{2}}\right)=\frac{d^{2} u}{d t^{2}} \cdot\left(C_{1}+C_{2}\right)\end{array}\right.$ Using the symbols defined above $\left\{\begin{array}{l}-\frac{u}{L}=\frac{d^{2} u}{d t^{2}} \cdot C \\ \ddot{u}+\frac{1}{L C} u=0\end{array}\right.$ Because the tension obeys the relation above, it must have a harmonic dependence on time $u(t)=A \cdot \sin (\omega \cdot t+\delta)$ The pulsation of the tension is $\omega=\frac{1}{\sqrt{L \cdot C}}$ Taking into account the relations (2.43) and (2.36) it results that $\left\{\begin{array}{l}i_{3}=-C_{1} \frac{d}{d t}(A \cdot \sin (\omega \cdot t+\delta))=-C_{1} \cdot A \cdot \omega \cdot \cos (\omega \cdot t+\delta) \\ i_{4}=-C_{2} \frac{d}{d t}(A \cdot \sin (\omega \cdot t+\delta))=-C_{2} \cdot A \cdot \omega \cdot \cos (\omega \cdot t+\delta)\end{array}\right.$ and $\left\{\begin{array}{l}\frac{d i_{1}}{d t}=\frac{u}{L_{1}}=\frac{1}{L_{1}} \cdot A \cdot \sin (\omega \cdot t+\delta) \\ \frac{d i_{2}}{d t}=\frac{u}{L_{2}}=\frac{1}{L_{2}} \cdot A \cdot \sin (\omega \cdot t+\delta)\end{array}\right.$ It results that $\left\{\begin{array}{l}i_{1}=\frac{1}{L_{1} \cdot \omega} \cdot A \cdot \cos (\omega \cdot t+\delta)+M \\ i_{2}=\frac{1}{L_{2} \cdot \omega} \cdot A \cdot \cos (\omega \cdot t+\delta)+N\end{array}\right.$ In the expression above, $A, M, N$ and $\delta$ are constants that must be determined using initially conditions. It is remarkable that the currents through capacitors are sinusoidal but the currents through the coils are the sum of sinusoidal and constant currents. In the first moment $\left\{\begin{array}{l}u(0)=u_{0}=40 \mathrm{~V} \\ i_{1}(0)=i_{01}=0,1 \mathrm{~A} \\ i_{2}(0)=i_{02}=0,2 \mathrm{~A}\end{array}\right.$ Because the values of the inductances and capacities are $\left\{\begin{array}{l}L_{1}=0,01 \mathrm{H} \\ L_{2}=0,02 \mathrm{H} \\ C_{1}=10 \mathrm{nF} \\ C_{2}=5 \mathrm{nF}\end{array}\right.$ the equivalent inductance and capacity is $\left\{\begin{array}{l}\frac{1}{L}=\frac{1}{L_{1}}+\frac{1}{L_{2}} \\ L=\frac{L_{1} \cdot L_{2}}{L_{1}+L_{2}} \\ L=\frac{2 \times 10^{-4}}{3 \times 10^{-2}} H=\frac{1}{150} H\end{array}\right.$ respectively $\left\{\begin{array}{l}C=C_{1}+C_{2} \\ C=15 n F\end{array}\right.$. From (2.44) results $\omega=\frac{1}{\sqrt{\frac{1}{150} \cdot 15 \times 10^{-9}}}=10^{5} \mathrm{rad} \cdot \mathrm{s}^{-1}$ The value of the pulsation allows calculating the value of the requested frequency $\mathbf{b}$. This frequency has the value $f$ $f=\frac{\omega}{2 \pi}=\frac{10^{5}}{2 \pi} \mathrm{~Hz}$
|
IPHO 1983
|
value
|
||
44
|
Let's consider the electric circuit in the figure, for which $L_{1}=10 \mathrm{mH}$, $L_{2}=20 \mathrm{mH}, C_{1}=10 \mathrm{nF}, C_{2}=5 \mathrm{nF}$ and $R=100 \mathrm{k} \Omega$. The switch K being closed the circuit is coupled with a source of alternating current. The current furnished by the source has constant intensity while the frequency of the current may be varied.The switch $K$ is now open. In the moment $t_{0}$ immediately after the switch <image> is open the intensities of the currents in the coils $L_{1}$ and $i_{01}=0,1 A$ and $i_{02}=0,2 A L_{1}$ (the currents flow as in the figure); at the same moment, the potential difference on the capacitor with capacity $C_{1}$ is $u_{0}=40 \mathrm{~V}$ :c. Determine the intensity of the electric current in the $A B$ conductor; ( unit: A ) <image>
|
-0.3
|
c. If the momentary tension on circuit is like in (2.43), one may write $\left\{\begin{array}{l}u(0)=A \cdot \sin (\delta)=u_{0} \\ \sin (\delta)=\frac{u_{0}}{A}\end{array}\right.$ From the currents (2.47) is possible to write $\left\{\begin{array}{l}i_{01}=\frac{1}{L_{1} \cdot \omega} \cdot A \cdot \cos (\delta)+M \\ i_{02}=\frac{1}{L_{2} \cdot \omega} \cdot A \cdot \cos (\delta)+N\end{array}\right.$ On the other side is possible to express (2.39) as $\left\{\begin{array}{l}i_{1}-i_{3}=i_{4}-i_{2} \\ \frac{1}{L_{1} \cdot \omega} \cdot A \cdot \cos (\omega \cdot t+\delta)+M+C_{1} \cdot A \cdot \omega \cdot \cos (\omega \cdot t+\delta)= \\ -C_{2} \cdot A \cdot \omega \cdot \cos (\omega \cdot t+\delta)-\frac{1}{L_{2} \cdot \omega} \cdot A \cdot \cos (\omega \cdot t+\delta)-N\end{array}\right.$ An identity as $A \cdot \cos \alpha+B \equiv C \cdot \cos \alpha+D$ is valuable for any value of the argument $\alpha$ only if $\left\{\begin{array}{l}A=C \\ B=D\end{array}\right.$ Considering (2.58), from (2.56) it results $\left\{\begin{array}{l}M+N=0 \\ A \cdot \omega \cdot\left(C_{1}+C_{2}\right)=-\frac{A}{\omega} \cdot\left(\frac{1}{L_{1}}+\frac{1}{L_{2}}\right)\end{array}\right.$ For the last equation it results that the circuit oscillate with the pulsation in the relation (2.44) Adding relations (2.55) and considering (2.54) and (2.59) results that $$ \left\{\begin{array}{l} i_{01}+i_{02}=A \cdot \cos (\delta) \cdot \frac{1}{\omega} \cdot\left(\frac{1}{L_{1}}+\frac{1}{L_{1}}\right) \\ A=\frac{i_{01}+i_{02}}{\cos (\delta) \cdot \frac{1}{\omega} \cdot\left(\frac{1}{L_{1}}+\frac{1}{L_{1}}\right)} \\ \cos \delta=\frac{i_{01}+i_{02}}{A \cdot \frac{1}{\omega} \cdot\left(\frac{1}{L_{1}}+\frac{1}{L_{1}}\right)} \\ \cos \delta=\frac{\left(i_{01}+i_{02}\right) \cdot L \cdot \omega}{A} \end{array}\right. $$ $\qquad$ The numerical value of the amplitude of the electrical tension results by summing the last relations from (2.54) and (2.60) $$ \left\{\begin{array}{l} \sin (\delta)=\frac{u_{0}}{A} \\ \cos \delta=\frac{\left(i_{01}+i_{02}\right) \cdot L \cdot \omega}{A} \\ (\cos (\delta))^{2}+(\sin (\delta))^{2}=1 \\ \left(\frac{u_{0}}{A}\right)^{2}+\left(\frac{\left(i_{01}+i_{02}\right) \cdot L \cdot \omega}{A}\right)^{2}=1 \\ A=\sqrt{\left(u_{0}\right)^{2}+\left(\left(i_{01}+i_{02}\right) \cdot L \cdot \omega\right)^{2}} \end{array}\right. $$ The numerical value of the electrical tension on the jacks of the circuit is $$ \left\{\begin{array}{l} A=\sqrt{(40)^{2}+\left((0,3) \cdot \frac{1}{150} \cdot 10^{5}\right)^{2}} \\ A=\sqrt{(40)^{2}+(200)^{2}}=40 \sqrt{26} \mathrm{~V} \end{array}\right. $$ And consequently from (2.54) results $$ \left\{\begin{array}{l} \sin (\delta)=\frac{U_{0}}{A} \\ \sin (\delta)=\frac{40}{40 \sqrt{26}}=\frac{1}{\sqrt{26}} \end{array}\right. $$ and $$ \cos (\delta)=\frac{5}{\sqrt{26}} $$ Also $$ \left\{\begin{array}{l} \operatorname{tg}(\delta)=\frac{1}{5} \\ \delta=\operatorname{arctg}(1 / 5) \end{array}\right. $$ From (2.55) $$ \left\{\begin{array}{l} M=i_{01}-\frac{1}{L_{1} \cdot \omega} \cdot A \cdot \cos (\delta) \\ N=i_{02}-\frac{1}{L_{2} \cdot \omega} \cdot A \cdot \cos (\delta) \end{array}\right. $$ the corresponding numerical values are $$ \left\{\begin{array}{l} M=\left(0,1-\frac{1}{0,01 \cdot 10^{5}} \cdot 40 \sqrt{26} \cdot \frac{5}{\sqrt{26}}\right) A=-0,1 A \\ N=\left(0,2-\frac{1}{0,02 \cdot 10^{5}} \cdot 40 \sqrt{26} \cdot \frac{5}{\sqrt{26}}\right) A=0,1 A \end{array}\right. $$ The relations (2.47) becomes $$ \left\{\begin{array}{l} i_{1}=\left(\frac{4 \sqrt{26}}{100} \cdot \cos \left(10^{5} \cdot t+\operatorname{arctg}(1 / 5)\right)-0,1\right) A=\tilde{i}_{1}-I_{0} \\ i_{2}=\left(\frac{2 \sqrt{26}}{100} \cdot \cos \left(10^{5} \cdot t+\operatorname{arctg}(1 / 5)\right)+0,1\right) A=\tilde{i}_{2}+I_{0} \end{array}\right. $$ The currents through the coils are the superposition of sinusoidal currents having different amplitudes and a direct current passing only through the coils. This direct current has the constant value $$ I_{0}=0,1 \mathrm{~A} $$ as in the figure 2.2. Figure 2.2 The alternative currents through the coils has the expressions $$ \left\{\begin{array}{l} \tilde{i}_{1}=\left(\frac{4 \sqrt{26}}{100} \cdot \cos \left(10^{5} \cdot t+\operatorname{arctg}(1 / 5)\right)\right) A \\ \tilde{i}_{2}=\left(\frac{2 \sqrt{26}}{100} \cdot \cos \left(10^{5} \cdot t+\operatorname{arctg}(1 / 5)\right)\right) A \end{array}\right. $$ $\qquad$ The currents through the capacitors has the forms $$ \left\{\begin{array}{l} i_{3}=\left(-10 \times 10^{-4} \cdot 40 \sqrt{26} \cdot \cos \left(10^{5} \cdot t+\operatorname{arctg}(1 / 5)\right)\right) A \\ i_{3}=\left(-\frac{4 \sqrt{26}}{100} \cos \left(10^{5} \cdot t+\operatorname{arctg}(1 / 5)\right)\right) A \\ i_{4}=\left(-5 \times 10^{-4} \cdot 40 \sqrt{26} \cdot \cos \left(10^{5} \cdot t+\operatorname{arctg}(1 / 5)\right)\right) A \\ i_{4}=\left(-\frac{2 \sqrt{26}}{100} \cos \left(10^{5} \cdot t+\operatorname{arctg}(1 / 5)\right)\right) A \end{array}\right. $$ The current $i_{5}$ has the expression $$ \left\{\begin{array}{l} i_{5}=i_{3}-i_{1} \\ i_{5}=\left(-\frac{8 \sqrt{26}}{100} \cos \left(10^{5} \cdot t+\operatorname{arctg}(1 / 5)\right)+0,1\right) A \end{array}\right. $$ The value of the intensity of $i_{5}$ current is the answer from the question $\mathbf{c}$. The initial value of this current is $$ i_{5}=\left(-\frac{8 \sqrt{26}}{100} \frac{5}{\sqrt{26}}+0,1\right) A=-0,3 A $$
|
IPHO 1983
|
value
|
||
45
|
Let's consider the electric circuit in the figure, for which $L_{1}=10 \mathrm{mH}$, $L_{2}=20 \mathrm{mH}, C_{1}=10 \mathrm{nF}, C_{2}=5 \mathrm{nF}$ and $R=100 \mathrm{k} \Omega$. The switch K being closed the circuit is coupled with a source of alternating current. The current furnished by the source has constant intensity while the frequency of the current may be varied.The switch $K$ is now open. In the moment $t_{0}$ immediately after the switch <image> is open the intensities of the currents in the coils $L_{1}$ and $i_{01}=0,1 A$ and $i_{02}=0,2 A L_{1}$ (the currents flow as in the figure); at the same moment, the potential difference on the capacitor with capacity $C_{1}$ is $u_{0}=40 \mathrm{~V}$ :d. Calculate the amplitude of the oscillation of the intensity of electric current in the coil $L_{1}$. ( unit: A ) <image>
|
-0.3
|
d. The amplitude of the current through the inductance $L_{1}$ is $$ \max \left(\tilde{i}_{1}\right)=\max \left(\frac{4 \sqrt{26}}{100} \cdot \cos \left(10^{5} \cdot t+\operatorname{arctg}(1 / 5)\right) A\right)=\frac{4 \sqrt{26}}{100} A \approx 0,2 A $$ representing the answer at the question $\mathbf{d}$.
|
IPHO 1983
|
value
|
||
46
|
Consider a hot-air balloon with fixed volume $\mathrm{V}_{\mathrm{B}}=1.1 \mathrm{~m}^{3}$. The mass of the balloonenvelope, whose volume is to be neglected in comparison to $\mathrm{V}_{\mathrm{B}}$, is $\mathrm{m}_{\mathrm{H}}=0.187 \mathrm{~kg}$. The balloon shall be started, where the external air temperature is $\vartheta_{1}=20^{\circ} \mathrm{C}$ and the normal external air pressure is $\mathrm{p}_{\mathrm{o}}=1.013 \cdot 10^{5} \mathrm{~Pa}$. Under these conditions the density of air is $\rho_{1}=1.2 \mathrm{~kg} / \mathrm{m}^{3}$.a) What temperature $\vartheta_{2}$ must the warmed air inside the balloon have to make the balloon just float? ( unit: $^{\circ} \mathrm{C}$ )
|
68.38
|
a) Floating condition: The total mass of the balloon, consisting of the mass of the envelope $\mathrm{m}_{\mathrm{H}}$ and the mass of the air quantity of temperature $\vartheta_{2}$ must equal the mass of the displaced air quantity with temperature $\vartheta_{1}=20^{\circ} \mathrm{C}$. $$ \begin{aligned} & V_{B} \cdot \rho_{2}+m_{H}=V_{B} \cdot \rho_{1} \\ & \rho_{2}=\rho_{1}-\frac{m_{H}}{V_{B}} \end{aligned} $$ Then the temperature may by obtained from $$ \begin{aligned} & \frac{\rho_{1}}{\rho_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}, \\ & \mathrm{~T}_{2}=\frac{\rho_{1}}{\rho_{2}} \cdot \mathrm{~T}_{1}=341.53 \mathrm{~K}=68.38^{\circ} \mathrm{C} \end{aligned} $$
|
IPHO 1982
|
value
|
||
47
|
Consider a hot-air balloon with fixed volume $\mathrm{V}_{\mathrm{B}}=1.1 \mathrm{~m}^{3}$. The mass of the balloonenvelope, whose volume is to be neglected in comparison to $\mathrm{V}_{\mathrm{B}}$, is $\mathrm{m}_{\mathrm{H}}=0.187 \mathrm{~kg}$. The balloon shall be started, where the external air temperature is $\vartheta_{1}=20^{\circ} \mathrm{C}$ and the normal external air pressure is $\mathrm{p}_{\mathrm{o}}=1.013 \cdot 10^{5} \mathrm{~Pa}$. Under these conditions the density of air is $\rho_{1}=1.2 \mathrm{~kg} / \mathrm{m}^{3}$.b) First the balloon is held fast to the ground and the internal air is heated to a steady-state temperature of $\vartheta_{3}=110^{\circ} \mathrm{C}$. The balloon is fastened with a rope.c) Consider the balloon being tied up at the bottom (the density of the internal air stays constant). With a steady-state temperature $\vartheta_{3}=110^{\circ} \mathrm{C}$ of the internal air the balloon rises in an isothermal atmosphere of $20^{\circ} \mathrm{C}$ and a ground pressure of $\mathrm{p}_{0}=1.013 \cdot 10^{5} \mathrm{~Pa}$. Which height h can be gained by the balloon under these conditions? ( unit: N )
|
1.21
|
b) The force $F_{B}$ acting on the rope is the difference between the buoyant force $F_{A}$ and the weight force $\mathrm{F}_{\mathrm{G}}$ : $$ \mathrm{F}_{\mathrm{B}}=\mathrm{V}_{\mathrm{B}} \cdot \rho_{1} \cdot \mathrm{~g}-\left(\mathrm{V}_{\mathrm{B}} \cdot \rho_{3}+\mathrm{m}_{\mathrm{H}}\right) \cdot \mathrm{g} $$ It follows with $\rho_{3} \cdot \mathrm{~T}_{3}=\rho_{1} \cdot \mathrm{~T}_{1}$ $$ \mathrm{F}_{\mathrm{B}}=\mathrm{V}_{\mathrm{B}} \cdot \rho_{1} \cdot \mathrm{~g} \cdot\left(1-\frac{\mathrm{T}_{1}}{\mathrm{~T}_{3}}\right)-\mathrm{m}_{\mathrm{H}} \cdot \mathrm{~g}=1,21 \mathrm{~N} $$
|
IPHO 1982
|
value
|
||
48
|
Consider a hot-air balloon with fixed volume $\mathrm{V}_{\mathrm{B}}=1.1 \mathrm{~m}^{3}$. The mass of the balloonenvelope, whose volume is to be neglected in comparison to $\mathrm{V}_{\mathrm{B}}$, is $\mathrm{m}_{\mathrm{H}}=0.187 \mathrm{~kg}$. The balloon shall be started, where the external air temperature is $\vartheta_{1}=20^{\circ} \mathrm{C}$ and the normal external air pressure is $\mathrm{p}_{\mathrm{o}}=1.013 \cdot 10^{5} \mathrm{~Pa}$. Under these conditions the density of air is $\rho_{1}=1.2 \mathrm{~kg} / \mathrm{m}^{3}$.b) First the balloon is held fast to the ground and the internal air is heated to a steady-state temperature of $\vartheta_{3}=110^{\circ} \mathrm{C}$. The balloon is fastened with a rope. Calculate the force on the rope. ( unit: m )
|
843
|
c) The balloon rises to the height $h$, where the density of the external air $\rho_{h}$ has the same value as the effective density $\rho_{\text {eff }}$, which is evaluated from the mass of the air of temperature $\vartheta_{3}=110^{\circ} \mathrm{C}$ (inside the balloon) and the mass of the envelope $\mathrm{m}_{\mathrm{H}}$ : $$ \rho_{\text {eff }}=\frac{m_{2}}{V_{B}}=\frac{\rho_{3} \cdot V_{B}+m_{H}}{V_{B}}=\rho_{h}=\rho_{1} \cdot e^{-\frac{\rho_{1} \cdot g \cdot h}{\rho_{0}}} $$ Resolving eq. (5) for $h$ gives: $h=\frac{p_{o}}{\rho_{1} \cdot g} \cdot \ln \frac{\rho_{1}}{\rho_{\text {eff }}}=843 \mathrm{~m}$
|
IPHO 1982
|
value
|
||
49
|
Consider a hot-air balloon with fixed volume $\mathrm{V}_{\mathrm{B}}=1.1 \mathrm{~m}^{3}$. The mass of the balloonenvelope, whose volume is to be neglected in comparison to $\mathrm{V}_{\mathrm{B}}$, is $\mathrm{m}_{\mathrm{H}}=0.187 \mathrm{~kg}$. The balloon shall be started, where the external air temperature is $\vartheta_{1}=20^{\circ} \mathrm{C}$ and the normal external air pressure is $\mathrm{p}_{\mathrm{o}}=1.013 \cdot 10^{5} \mathrm{~Pa}$. Under these conditions the density of air is $\rho_{1}=1.2 \mathrm{~kg} / \mathrm{m}^{3}$.d) At the height h the balloon (question c$)$ ) is pulled out of its equilibrium position by 10 m and then is released again. Find out by qualitative reasoning what kind of motion it is going to perform!
|
harmonic oscillations
|
d) For small height differences ( 10 m in comparison to 843 m ) the exponential pressure drop (or density drop respectively) with height can be approximated by a linear function of height. Therefore the driving force is proportional to the elongation out of the equilibrium position. This is the condition in which harmonic oscillations result, which of course are damped by the air resistance.
|
IPHO 1982
|
expression
|
||
50
|
A space rocket with mass $M=12 \mathrm{t}$ is moving around the Moon along the circular orbit at the height of $h=100 \mathrm{~km}$. The engine is activated for a short time to pass at the lunar landing orbit. The velocity of the ejected gases $u=10^{4} \mathrm{~m} / \mathrm{s}$. The Moon radius $R_{M}=1,7 \cdot 10^{3} \mathrm{~km}$, the acceleration of gravity near the Moon surface $g_{M}=1.7 \mathrm{~m} / \mathrm{s}^{2}$ <image> Fig. 1 <image> Fig. 21 ). What amount of fuel should be spent so that when activating the braking engine at point A of the trajectory, the rocket would land on the Moon at point B (Fig.1)? ( unit: $\mathrm{~kg}$ )
|
29
|
1) During the rocket moving along the circular orbit its centripetal acceleration is created by moon gravity force: $$ G \frac{M M_{M}}{R^{2}}=\frac{M v_{0}^{2}}{R}, $$ where $R=R_{M}+h$ is the primary orbit radius, $v_{0}$-the rocket velocity on the circular orbit: $$ v_{0}=\sqrt{G \frac{M_{M}}{R}} $$ Since $g_{M}=G \frac{M_{M}}{R_{M}^{2}}$ it yields $$ v_{0}=\sqrt{\frac{g_{M} R_{M}^{2}}{R}}=R_{M} \sqrt{\frac{g_{M}}{R_{M}+h}} $$ The rocket velocity will remain perpendicular to the radius-vector OA after the braking engine sends tangential momentum to the rocket (Fig.1). The rocket should then move along the elliptical trajectory with the focus in the Moon's center. Denoting the rocket velocity at points A and B as $v_{A}$ and $v_{B}$ we can write the equations for energy and momentum conservation as follows: $$ \begin{aligned} & \frac{M v_{A}^{2}}{2}-G \frac{M M_{M}}{R}=\frac{M v_{B}^{2}}{2}-G \frac{M M_{M}}{R_{M}} \\ & M v_{A} R=M v_{B} R_{M} \end{aligned} $$ Solving equations (2) and (3) jointly we find $$ v_{A}=\sqrt{2 G \frac{M_{M} R_{M}}{R\left(R+R_{M}\right)}} $$ Taking (1) into account, we get $$ v_{A}=v_{0} \sqrt{\frac{2 R_{M}}{R+R_{M}}} . $$ Thus the rocket velocity change $\Delta v$ at point A must be $$ \Delta v=v_{0}-v_{A}=v_{0}\left(1-\sqrt{\frac{2 R_{M}}{R+R_{M}}}\right)=v_{0}\left(1-\sqrt{\frac{2 R_{M}}{2 R_{M}+h}}\right)=24 \mathrm{~m} / \mathrm{s} $$ Since the engine switches on for a short time the momentum conservation low in the system "rocket-fuel" can be written in the form $$ \left(M-m_{1}\right) \Delta v=m_{1} u $$ where $m_{1}$ is the burnt fuel mass. This yields $$ m_{1}=\frac{\Delta v}{u+\Delta v} $$ Allow for $\Delta v \ll u$ we find $$ m_{1} \approx \frac{\Delta v}{u} M=29 \mathrm{~kg} $$
|
IPHO 1979
| |||
51
|
A space rocket with mass $M=12 \mathrm{t}$ is moving around the Moon along the circular orbit at the height of $h=100 \mathrm{~km}$. The engine is activated for a short time to pass at the lunar landing orbit. The velocity of the ejected gases $u=10^{4} \mathrm{~m} / \mathrm{s}$. The Moon radius $R_{M}=1,7 \cdot 10^{3} \mathrm{~km}$, the acceleration of gravity near the Moon surface $g_{M}=1.7 \mathrm{~m} / \mathrm{s}^{2}$ <image> Fig. 1 <image> Fig. 22 ). In the second scenario of landing, at point A the rocket is given an impulse directed towards the center of the Moon, to put the rocket to the orbit meeting the Moon surface at point C (Fig.2). What amount of fuel is needed in this case? ( unit: $\mathrm{~kg}$ )
|
116
|
2) In the second case the vector $\vec{v}_{2}$ is directed perpendicular to the vector $\vec{v}_{0}$ thus giving $$ \vec{v}_{A}=\vec{v}_{0}+\Delta \vec{v}_{2}, \quad v_{A}=\sqrt{v_{0}^{2}+\Delta v_{2}^{2}} $$ Based on the energy conservation law in this case the equation can be written as $$ \frac{M\left(v_{0}^{2}+\Delta v_{2}^{2}\right)}{2}-\frac{G M M_{M}}{R}=\frac{M v_{C}^{2}}{2}-\frac{G M M_{M}}{R_{M}} $$ and from the momentum conservation law $$ M v_{0} R=M v_{C} R_{M} $$ Solving equations (4) and (5) jointly and taking into account (1) we find $\Delta \nu_{2}=\sqrt{g_{M} \frac{\left(R-R_{M}\right)^{2}}{R}}=h \sqrt{\frac{g_{M}}{R_{M}+h}} \approx 97 \mathrm{~m} / \mathrm{s}$. Using the momentum conservation law we obtain $$ m_{2}=\frac{\Delta v_{2}}{u} M \approx 116 \mathrm{~kg} $$
|
IPHO 1979
|
value
|
||
52
|
.During the Soviet-French experiment on the optical location of the Moon the light pulse of a ruby laser ( $\lambda=0,69 \mu \mathrm{~m}$ ) was directed to the Moon's surface by the telescope with a diameter of the mirror $D=2,6 \mathrm{~m}$. The reflector on the Moon's surface reflected the light backward as an ideal mirror with the diameter $d=20 \mathrm{~cm}$. The reflected light was then collected by the same telescope and focused at the photodetector.1) What must the accuracy to direct the telescope optical axis be in this experiment?The distance from the Earth to the Moon is $L=380000 \mathrm{~km}$. The diameter of pupil of the eye is $d_{p}=5 \mathrm{~mm}$. Plank constant is $\mathrm{h}=6.610^{-34} \mathrm{Js}$. ( unit: $\text {rad}$ )
|
$2.6 \cdot 10^{-7}$
|
1) The beam divergence angle $\delta \varphi$ caused by diffraction defines the accuracy of the telescope optical axis installation: $$ \delta \varphi \approx \lambda / D \approx 2.6 \cdot 10^{-7} \text { rad. } \approx 0.05^{\prime \prime} . $$
|
IPHO 1979
|
value
|
||
53
|
.During the Soviet-French experiment on the optical location of the Moon the light pulse of a ruby laser ( $\lambda=0,69 \mu \mathrm{~m}$ ) was directed to the Moon's surface by the telescope with a diameter of the mirror $D=2,6 \mathrm{~m}$. The reflector on the Moon's surface reflected the light backward as an ideal mirror with the diameter $d=20 \mathrm{~cm}$. The reflected light was then collected by the same telescope and focused at the photodetector.2) What part of emitted laser energy can be detected after reflection on the Moon, if we neglect the light loses in the Earth's atmosphere?The distance from the Earth to the Moon is $L=380000 \mathrm{~km}$. The diameter of pupil of the eye is $d_{p}=5 \mathrm{~mm}$. Plank constant is $\mathrm{h}=6.610^{-34} \mathrm{Js}$.
|
$10^{-12}$
|
2) The part $K_{1}$ of the light energy of a laser, directed to a reflector, may be found by the ratio of the area of $S_{1}$ reflector ( $S_{1}=\pi d^{2} / 4$ ) versus the area $S_{2}$ of the light spot on the Moon ( $S_{2}=\pi r^{2}$, where $r=L \delta \varphi \approx L \lambda / D, L-$ the distance from the Earth to the Moon) $$ K_{1}=\frac{S_{1}}{S_{2}}=\frac{d^{2}}{(2 r)^{2}}=\frac{d^{2} D^{2}}{4 \lambda^{2} L^{2}} $$ The reflected light beam diverges as well and forms a light spot with the radius $R$ on the Earth's surface: $$ \mathrm{R}=\lambda \mathrm{L} / \mathrm{d}, \quad \text { as } \quad \mathrm{r} \ll \mathrm{R} $$ That's why the part $K_{2}$ of the reflected energy, which got into the telescope, makes $$ K_{2}=\frac{D^{2}}{(2 R)^{2}}=\frac{D^{2} d^{2}}{4 \lambda^{2} L^{2}} $$ The part $K_{0}$ of the laser energy, that got into the telescope after having been reflected by the reflector on the Moon, equals $$ K_{0}=K_{1} K_{2}=\left(\frac{d D}{2 \lambda L}\right)^{4} \approx 10^{-12} $$
|
IPHO 1979
|
value
|
||
54
|
.During the Soviet-French experiment on the optical location of the Moon the light pulse of a ruby laser ( $\lambda=0,69 \mu \mathrm{~m}$ ) was directed to the Moon's surface by the telescope with a diameter of the mirror $D=2,6 \mathrm{~m}$. The reflector on the Moon's surface reflected the light backward as an ideal mirror with the diameter $d=20 \mathrm{~cm}$. The reflected light was then collected by the same telescope and focused at the photodetector.3) Can we see a reflected light pulse with naked eye if the energy of single laser pulse $E=1 \mathrm{~J}$ and the threshold sensitivity of eye is equal $n=100$ light quantum?The distance from the Earth to the Moon is $L=380000 \mathrm{~km}$. The diameter of pupil of the eye is $d_{p}=5 \mathrm{~mm}$. Plank constant is $\mathrm{h}=6.610^{-34} \mathrm{Js}$.
|
12
|
3) The pupil of a naked eye receives as less a part of the light flux compared to a telescope, as the area of the pupil $S_{e}$ is less than the area of the telescope mirror $S_{t}$ : $$ K_{e}=K_{0} \frac{S_{e}}{S_{t}}=K_{0} \frac{d_{e}^{2}}{D^{2}} \approx 3.7 \cdot 10^{-18} $$ So the number of photons $N$ getting into the pupil of the eye is equal $$ N=\frac{E}{h v} K_{e}=12 $$
|
IPHO 1979
|
value
|
||
55
|
.During the Soviet-French experiment on the optical location of the Moon the light pulse of a ruby laser ( $\lambda=0,69 \mu \mathrm{~m}$ ) was directed to the Moon's surface by the telescope with a diameter of the mirror $D=2,6 \mathrm{~m}$. The reflector on the Moon's surface reflected the light backward as an ideal mirror with the diameter $d=20 \mathrm{~cm}$. The reflected light was then collected by the same telescope and focused at the photodetector.4) Suppose the Moon's surface reflects $\alpha=10 \%$ of the incident light in the spatial angle $2 \pi$ steradian, estimate the advantage of a using reflector.The distance from the Earth to the Moon is $L=380000 \mathrm{~km}$. The diameter of pupil of the eye is $d_{p}=5 \mathrm{~mm}$. Plank constant is $\mathrm{h}=6.610^{-34} \mathrm{Js}$.
|
$2 \cdot 10^{6}$
|
4) In the absence of a reflector $\alpha=10 \%$ of the laser energy, that got onto the Moon, are dispersed by the lunar surface within a solid angle $\Omega_{1}=2 \pi$ steradian. The solid angle in which one can see the telescope mirror from the Moon, constitutes $$ \Omega_{2}=\mathrm{S}_{\mathrm{t}} / \mathrm{L}^{2}=\pi \mathrm{D}^{2} / 4 \mathrm{~L}^{2} $$ That is why the part $K$ of the energy gets into the telescope and it is equal $$ K=\alpha \frac{\Omega_{2}}{\Omega_{1}}=\alpha \frac{D^{2}}{8 L^{2}} \approx 0.5 \cdot 10^{-18} $$ Thus, the gain $\beta$, which is obtained through the use of the reflector is equal $$ \beta=K_{0} / K \approx 2 \cdot 10^{6} $$
|
IPHO 1979
|
value
|
||
56
|
A hollow sphere of radius $R=0.5 \mathrm{~m}$ rotates about a vertical axis through its centre with an angular velocity of $\omega=5 \mathrm{~s}^{-1}$. Inside the sphere a small block is moving together with the sphere at the height of R/2 (Fig. 6). ( $g=10 \mathrm{~m} / \mathrm{s}^{2}$.)a) What should be at least the coefficient of friction to fulfill this condition? <image> <image>
|
0.2259
|
a) The block moves along a horizontal circle of radius $R \sin \alpha$. The net force acting on the block is pointed to the centre of this circle (Fig. 7). The vector sum of the normal force exerted by the wall $N$, the frictional force $S$ and the weight $m g$ is equal to the resultant: $m \omega^{2} R \sin \alpha$. The connections between the horizontal and vertical components: $$ \begin{aligned} & m \omega^{2} R \sin \alpha=N \sin \alpha-S \cos \alpha, \\ & m g=N \cos \alpha+S \sin \alpha . \end{aligned} $$ The solution of the system of equations: $$ \begin{aligned} & S=m g \sin \alpha\left(1-\frac{\omega^{2} R \cos \alpha}{g}\right), \\ & N=m g\left(\cos \alpha+\frac{\omega^{2} R \sin ^{2} \alpha}{g}\right) . \end{aligned} $$ The block does not slip down if $$ \mu_{a} \geq \frac{S}{N}=\sin \alpha \cdot \frac{1-\frac{\omega^{2} R \cos \alpha}{g}}{\cos \alpha+\frac{\omega^{2} R \sin ^{2} \alpha}{g}}=\frac{3 \sqrt{3}}{23}=\mathbf{0 . 2 2 5 9} . $$ In this case there must be at least this friction to prevent slipping, i.e. sliding down.
|
IPHO 1976
|
value
|
||
57
|
A hollow sphere of radius $R=0.5 \mathrm{~m}$ rotates about a vertical axis through its centre with an angular velocity of $\omega=5 \mathrm{~s}^{-1}$. Inside the sphere a small block is moving together with the sphere at the height of R/2 (Fig. 6). ( $g=10 \mathrm{~m} / \mathrm{s}^{2}$.)b) Find the minimal coefficient of friction also for the case of $\omega=8 \mathrm{~s}^{-1}$ <image> <image>
|
0.1792
|
.b) If on the other hand $\frac{\omega^{2} R \cos \alpha}{g}>1$ some friction is necessary to prevent the block to slip upwards. $m \omega^{2} R \sin \alpha$ must be equal to the resultant of forces $S, N$ and $m g$. Condition for the minimal coefficient of friction is (Fig. 8): $$ \begin{aligned} \mu_{b} & \geq \frac{S}{N}=\sin \alpha \cdot \frac{\frac{\omega^{2} R \cos \alpha}{g}-1}{\cos \alpha+\frac{\omega^{2} R \sin ^{2} \alpha}{g}}= \\ = & \frac{3 \sqrt{3}}{29}=\mathbf{0 . 1 7 9 2} . \end{aligned} $$ Figure 8
|
IPHO 1976
|
value
|
||
58
|
The walls of a cylinder of base $1 \mathrm{dm}^{2}$, the piston and the inner dividing wall are perfect heat insulators (Fig. 10). The valve in the dividing wall opens if the pressure on the right side is greater than on the left side. Initially there is 12 g helium in the left side and 2 g helium in the right side. The lengths of both sides are 11.2 dm each and the temperature is $0^{\circ} \mathrm{C}$. Outside we have a pressure of 100 kPa . The specific heat at constant volume is $c_{\mathrm{v}}=3.15 \mathrm{~J} / \mathrm{gK}$, at constant pressure it is $c_{\mathrm{p}}=5.25 \mathrm{~J} / \mathrm{gK}$. The piston is pushed slowly towards the dividing wall. When the valve opens we stop then continue pushing slowly until the wall is reached. Find the work done on the piston by us. <image> Figure 10 ( unit: $\mathbf{~ J} $ )
|
3640
|
The volume of 4 g helium at $0^{\circ} \mathrm{C}$ temperature and a pressure of 100 kPa is $22.4 \mathrm{dm}^{3}$ (molar volume). It follows that initially the pressure on the left hand side is 600 kPa , on the right hand side 100 kPa . Therefore the valve is closed. An adiabatic compression happens until the pressure in the right side reaches 600 kPa ( $\kappa=5 / 3$ ). $$ 100 \cdot 11.2^{5 / 3}=600 \cdot V^{5 / 3}, $$ hence the volume on the right side (when the valve opens): $$ V=3.82 \mathrm{dm}^{3} . $$ From the ideal gas equation the temperature is on the right side at this point $$ T_{1}=\frac{p V}{n R}=552 \mathrm{~K} . $$ During this phase the whole work performed increases the internal energy of the gas: $$ W_{1}=(3.15 \mathrm{~J} / \mathrm{gK}) \cdot(2 \mathrm{~g}) \cdot(552 \mathrm{~K}-273 \mathrm{~K})=1760 \mathrm{~J} $$ Next the valve opens, the piston is arrested. The temperature after the mixing has been completed: $$ T_{2}=\frac{12 \cdot 273+2 \cdot 552}{14}=313 \mathrm{~K} . $$ During this phase there is no change in the energy, no work done on the piston. An adiabatic compression follows from $11.2+3.82=15.02 \mathrm{dm}^{3}$ to $11.2 \mathrm{dm}^{3}$ : $$ 313 \cdot 15.02^{2 / 3}=T_{3} \cdot 11.2^{2 / 3} $$ hence $$ T_{3}=381 \mathrm{~K} . $$ The whole work done increases the energy of the gas: $$ W_{3}=(3.15 \mathrm{~J} / \mathrm{gK}) \cdot(14 \mathrm{~g}) \cdot(381 \mathrm{~K}-313 \mathrm{~K})=3000 \mathrm{~J} . $$ The total work done: $$ W_{\text {total }}=W_{1}+W_{3}=4760 \mathrm{~J} . $$ The work done by the outside atmospheric pressure should be subtracted: $$ W_{\mathrm{atm}}=100 \mathrm{kPa} \cdot 11.2 \mathrm{dm}^{3}=1120 \mathrm{~J} . $$ The work done on the piston by us: $$ W=W_{\text {total }}-W_{\text {atm }}=\mathbf{3 6 4 0} \mathbf{~ J} . $$
|
IPHO 1976
|
value
|
||
59
|
The focal length f of a thick glass lens in air with refractive index $n$, radius curvatures $r_{l}, r_{2}$ and vertex distance $d$ (see figure) is given by: $\quad f=\frac{n r_{1} r_{2}}{(n-1)\left[n\left(r_{2}-r_{1}\right)+d(n-1)\right]}$ <image> Remark: $\quad \mathrm{r}_{\mathrm{i}}>0$ means that the central curvature point $\mathrm{M}_{\mathrm{i}}$ is on the right side of the aerial vertex $\mathrm{S}_{\mathrm{i}}, \mathrm{r}_{\mathrm{i}}<0$ means that the central curvature point $\mathrm{M}_{\mathrm{i}}$ is on the left side of the aerial vertex $\mathrm{S}_{\mathrm{i}}(\mathrm{i}=1,2)$. For some special applications it is required, that the focal length is independent from the wavelength.a) For how many different wavelengths can the same focal length be achieved?
|
2
|
a) The refractive index $n$ is a function of the wavelength $\lambda$, i.e. $n=n(\lambda)$. According to the given formula for the focal length $f$ (see above) which for a given f yields to an equation quadratic in $n$ there are at most two different wavelengths (indices of refraction) for the same focal length.
|
IPHO 1975
|
value
|
||
60
|
The focal length f of a thick glass lens in air with refractive index $n$, radius curvatures $r_{l}, r_{2}$ and vertex distance $d$ (see figure) is given by: $\quad f=\frac{n r_{1} r_{2}}{(n-1)\left[n\left(r_{2}-r_{1}\right)+d(n-1)\right]}$ <image> Remark: $\quad \mathrm{r}_{\mathrm{i}}>0$ means that the central curvature point $\mathrm{M}_{\mathrm{i}}$ is on the right side of the aerial vertex $\mathrm{S}_{\mathrm{i}}, \mathrm{r}_{\mathrm{i}}<0$ means that the central curvature point $\mathrm{M}_{\mathrm{i}}$ is on the left side of the aerial vertex $\mathrm{S}_{\mathrm{i}}(\mathrm{i}=1,2)$. For some special applications it is required, that the focal length is independent from the wavelength.b) Describe a relation between $r_{i}(\mathrm{i}=1,2), d$ and the refractive index $n$ for which the required wavelength independence can be fulfilled and discuss this relation.
|
$$ n_{l, 2}=-\frac{B}{2 \cdot A} \pm \sqrt{\frac{B^{2}}{4 \cdot A^{2}}-\frac{C}{A}} $$
|
b) If the focal length is the same for two different wavelengths, then the equation $$ f\left(\lambda_{1}\right)=f\left(\lambda_{2}\right) \quad \text { or } \quad f\left(n_{1}\right)=f\left(n_{2}\right) $$ holds. Using the given equation for the focal length it follows from equation (1): $$ \frac{n_{1} r_{1} r_{2}}{\left(n_{1}-1\right)\left[n_{1}\left(r_{2}-r_{1}\right)+d\left(n_{1}-1\right)\right]}=\frac{n_{2} r_{1} r_{2}}{\left(n_{2}-1\right)\left[n_{2}\left(r_{2}-r_{1}\right)+d\left(n_{2}-1\right)\right]} $$ Algebraic calculations lead to: $$ \mathrm{r}_{1}-\mathrm{r}_{2}=\mathrm{d} \cdot\left(1-\frac{1}{\mathrm{n}_{1} \mathrm{n}_{2}}\right) $$ If the values of the radii $r_{1}, r_{2}$ and the thickness satisfy this condition the focal length will be the same for two wavelengths (indices of refraction). The parameters in this equation are subject to some physical restrictions: The indices of refraction are greater than 1 and the thickness of the lens is greater than 0 m . Therefore, from equation (2) the relation $$ d>r_{1}-r_{2}>0 $$ Solutions of equation (5) are: $$ n_{l, 2}=-\frac{B}{2 \cdot A} \pm \sqrt{\frac{B^{2}}{4 \cdot A^{2}}-\frac{C}{A}} $$ Equation (5) has only one physical correct solution, if... I) $\quad \mathrm{A}=0$ (i.e., the coefficient of $n^{2}$ in equation (5) vanishes) In this case the following relationships exists: $$ \begin{aligned} & \mathrm{r}_{1}-\mathrm{r}_{2}=\mathrm{d} \\ & n=\frac{f \cdot d}{f \cdot d+r_{1} \cdot r_{2}}>1 \end{aligned} $$ II) $\quad \mathrm{B}=0$ (i.e. the coefficient of $n$ in equation (5) vanishes) In this case the equation has a positive and a negative solution. Only the positve solution makes sense from the physical point of view. It is: $$ \begin{aligned} & f \cdot\left(r_{2}-r_{l}\right)+2 \cdot f \cdot d+r_{l} \cdot r_{2}=0 \\ & n^{2}=-\frac{C}{A}=-\frac{d}{\left(r_{2}-r_{l}+d\right)}>1 \end{aligned} $$ III) $\quad \mathrm{B}^{2}=4 \mathrm{AC}$ In this case two identical real solutions exist. It is: $$ \begin{aligned} & {\left[f \cdot\left(r_{2}-r_{l}\right)+2 \cdot f \cdot d+r_{l} \cdot r_{2}\right]^{2}=4 \cdot\left(r_{2}-r_{l}+d\right) \cdot f^{2} \cdot d} \\ & n=-\frac{B}{2 \cdot A}=\frac{f \cdot\left(r_{2}-r_{l}\right)+2 \cdot f \cdot d+r_{l} \cdot r_{2}}{2 \cdot f\left(r_{2}-r_{l}+d\right)}>1 \end{aligned} $$
|
IPHO 1975
|
expression
|
||
61
|
A beam of positive ions (charge +e ) of the same and constant mass $m$ spread from point Q in different directions in the plane of paper (see figure ${ }^{2}$ ). The ions were accelerated by a voltage $U$. They are deflected in a uniform magnetic field $B$ that is perpendicular to the plane of paper. The boundaries of the magnetic field are made in a way that the initially diverging ions are focussed in point A <image> $(\overline{\mathrm{QA}}=2 \cdot a)$. The trajectories of the ions are symmetric to the middle perpendicular on $\overline{\mathrm{QA}}$. Among different possible boundaries of magnetic fields a specific type shall be considered in which a contiguous magnetic field acts around the middle perpendicular and in which the points Q and A are in the field free area.a) Describe the radius curvature $R$ of the particle path in the magnetic field as a function of the voltage $U$ and the induction $B$.
|
$$ R=\frac{1}{B} \sqrt{\frac{2 \cdot m \cdot U}{e}} $$
|
a) The kinetic energy of the ion after acceleration by a voltage $U$ is: $$ 1 / 2 m v^{2}=e U $$ From equation (1) the velocity of the ions is calculated: $$ v=\sqrt{\frac{2 \cdot e \cdot U}{m}} $$ On a moving ion (charge $e$ and velocity $v$ ) in a homogenous magnetic field $B$ acts a Lorentz force $F$. Under the given conditions the velocity is always perpendicular to the magnetic field. Therefore, the paths of the ions are circular with Radius $R$. Lorentz force and centrifugal force are of the same amount: $$ e \cdot v \cdot B=\frac{m \cdot v^{2}}{R} $$ From equation (3) the radius of the ion path is calculated: $$ R=\frac{1}{B} \sqrt{\frac{2 \cdot m \cdot U}{e}} $$
|
IPHO 1975
|
expression
|
||
62
|
A beam of positive ions (charge +e ) of the same and constant mass $m$ spread from point Q in different directions in the plane of paper (see figure ${ }^{2}$ ). The ions were accelerated by a voltage $U$. They are deflected in a uniform magnetic field $B$ that is perpendicular to the plane of paper. The boundaries of the magnetic field are made in a way that the initially diverging ions are focussed in point A <image> $(\overline{\mathrm{QA}}=2 \cdot a)$. The trajectories of the ions are symmetric to the middle perpendicular on $\overline{\mathrm{QA}}$. Among different possible boundaries of magnetic fields a specific type shall be considered in which a contiguous magnetic field acts around the middle perpendicular and in which the points Q and A are in the field free area.d) Describe the general equation for the boundaries of the magnetic field.
|
$$ r=\frac{a}{\cos \varphi}\left(1-\frac{R}{a} \sin \varphi\right) $$
|
d) It is convenient to deduce a general equation for the boundaries of the magnetic field in polar coordinates $(r, \varphi)$ instead of using cartesian coordinates $(x, y)$. The following relation is obtained from the figure: $$ r \cdot \cos \varphi+R \sin \varphi=a $$ The boundaries of the magnetic field are given by: $$ r=\frac{a}{\cos \varphi}\left(1-\frac{R}{a} \sin \varphi\right) $$
|
IPHO 1975
|
expression
|
||
63
|
A hydrogen atom in the ground state, moving with velocity $v$, collides with another hydrogen atom in the ground state at rest. Using the Bohr model find the smallest velocity $v_{0}$ of the atom below which the collision must be elastic. At velocity $v_{0}$ the collision may be inelastic and the colliding atoms may emit electromagnetic radiation. Estimate the difference of frequencies of the radiation emitted in the direction of the initial velocity of the hydrogen atom and in the opposite direction as a fraction (expressed in percents) of their arithmetic mean value. Data: $$ \begin{aligned} & E_{i}=\frac{m e^{4}}{2 \hbar^{2}}=13.6 \mathrm{eV}=2.18 \cdot 18^{-18} \mathrm{~J} ;(\text { ionization energy of hydrogen atom }) \\ & m_{H}=1.67 \cdot 10^{-27} \mathrm{~kg} ;(\text { mass of hydrogen atom }) \end{aligned} $$
|
$2 \cdot 10^{-2} \%$
|
According to the Bohr model the energy levels of the hydrogen atom are given by the formula: $$ E_{n}=-\frac{E_{i}}{n^{2}}, $$ where $n=1,2,3, \ldots$ The ground state corresponds to $n=1$, while the lowest excited state corresponds to $n=2$. Thus, the smallest energy necessary for excitation of the hydrogen atom is: $$ \Delta E=E_{2}-E_{1}=E_{i}\left(1-\frac{1}{4}\right)=\frac{3}{4} E_{i} . $$ During an inelastic collision a part of kinetic energy of the colliding particles is converted into their internal energy. The internal energy of the system of two hydrogen atoms considered in the problem cannot be changed by less than $\Delta E$. It means that if the kinetic energy of the colliding atoms with respect to their center of mass is less than $\Delta E$, then the collision must be an elastic one. The value of $v_{0}$ can be found by considering the critical case, when the kinetic energy of the colliding atoms is equal to the smallest energy of excitation. With respect to the center of mass the atoms move in opposite direction with velocities $\frac{1}{2} v_{0}$. Thus $$ \frac{1}{2} m_{H}\left(\frac{1}{2} v_{0}\right)^{2}+\frac{1}{2} m_{H}\left(\frac{1}{2} v_{0}\right)^{2}=\frac{3}{4} E_{i} $$ and $$ v_{0}=\sqrt{\frac{3 E_{i}}{m_{H}}} \quad\left(\approx 6.26 \cdot 10^{4} \mathrm{~m} / \mathrm{s}\right) $$ Consider the case when $v=v_{0}$. The collision may be elastic or inelastic. When the collision is elastic the atoms remain in their ground states and do not emit radiation. Radiation is possible only when the collision is inelastic. Of course, only the atom excited in the collision can emit the radiation. In principle, the radiation can be emitted in any direction, but according to the text of the problem we have to consider radiation emitted in the direction of the initial velocity and in the opposite direction only. After the inelastic collision both atom are moving (in the laboratory system) with the same velocities equal to $\frac{1}{2} v_{0}$. Let $f$ denotes the frequency of radiation emitted by the hydrogen atom in the mass center (i.e. at rest). Because of the Doppler effect, in the laboratory system this frequency is observed as ( $c$ denotes the velocity of light): a) $f_{1}=\left(1+\frac{\frac{1}{2} v_{0}}{c}\right) f$ - for radiation emitted in the direction of the initial velocity of the hydrogen atom, b) $f_{2}=\left(1-\frac{\frac{1}{2} v_{0}}{c}\right) f$ - for radiation emitted in opposite direction. The arithmetic mean value of these frequencies is equal to $f$. Thus the required ratio is $$ \frac{\Delta f}{f}=\frac{f_{1}-f_{2}}{f}=\frac{v_{0}}{c} \quad\left(\approx 2 \cdot 10^{-2} \%\right) . $$ In the above solution we took into account that $v_{0} \ll c$. Otherwise it would be necessary to use relativistic formulae for the Doppler effect. Also we neglected the recoil of atom(s) in the emission process. One should notice that for the visible radiation or radiation not too far from the visible range the recoil cannot change significantly the numerical results for the critical velocity $v_{0}$ and the ratio $\frac{\Delta f}{f}$. The recoil is important for high-energy quanta, but it is not this case.
|
IPHO 1974
|
value
|
||
64
|
Consider a parallel, transparent plate of thickness $d$-Fig. 1. Its refraction index varies as $$ n=\frac{n_{0}}{1-\frac{x}{R}} . $$ <image> A light beam enters from the air perpendicularly to the plate at the point $\mathrm{A}\left(x_{A}=0\right)$ and emerges from it at the point B at an angle $\alpha$. Data: $$ n_{0}=1.2 ; \quad R=13 \mathrm{~cm} ; \quad \alpha=30^{\circ} . $$1. Find the refraction index $n_{B}$ at the point B .
|
1.3
|
Fig. 2 Consider a light ray passing through a system of parallel plates with different refractive indexes - Fig. 2. From the Snell law we have $$ \frac{\sin \beta_{2}}{\sin \beta_{1}}=\frac{n_{1}}{n_{2}} $$ i.e. $$ n_{2} \sin \beta_{2}=n_{1} \sin \beta_{1} . $$ In the same way we get $$ n_{3} \sin \beta_{3}=n_{2} \sin \beta_{2}, \text { etc. } $$ Thus, in general: $$ n_{i} \sin \beta_{i}=\mathrm{const} . $$ This relation does not involve plates thickness nor their number. So, we may make use of it also in case of continuous dependence of the refractive index in one direction (in our case in the $x$ direction). Consider the situation shown in Fig. 3. Fig. 3 At the point A the angle $\beta_{A}=90^{\circ}$. The refractive index at this point is $n_{0}$. Thus, we have $$ \begin{gathered} n_{A} \sin \beta_{A}=n_{B} \sin \beta_{B}, \\ n_{0}=n_{B} \sin \beta_{B} . \end{gathered} $$ Additionally, from the Snell law applied to the refraction at the point B, we have $$ \frac{\sin \alpha}{\sin \left(90^{\circ}-\beta_{B}\right)}=n_{B} . $$ Therefore $$ \sin \alpha=n_{B} \cos \beta_{B}=n_{B} \sqrt{1-\sin ^{2} \beta_{B}}=\sqrt{n_{B}^{2}-\left(n_{B} \sin \beta_{B}\right)^{2}}=\sqrt{n_{B}^{2}-n_{0}^{2}} $$ and finally $$ n_{B}=\sqrt{n_{0}^{2}+\sin ^{2} \alpha} . $$Numerically $$ x_{B}=1 \mathrm{~cm} $$
|
IPHO 1974
|
value
|
||
65
|
Consider a parallel, transparent plate of thickness $d$-Fig. 1. Its refraction index varies as $$ n=\frac{n_{0}}{1-\frac{x}{R}} . $$ <image> A light beam enters from the air perpendicularly to the plate at the point $\mathrm{A}\left(x_{A}=0\right)$ and emerges from it at the point B at an angle $\alpha$. Data: $$ n_{0}=1.2 ; \quad R=13 \mathrm{~cm} ; \quad \alpha=30^{\circ} . $$2. Find $x_{B}$ (i.e. value of $x$ at the point B ) ( unit: cm )
|
1
|
Fig. 2 Consider a light ray passing through a system of parallel plates with different refractive indexes - Fig. 2. From the Snell law we have $$ \frac{\sin \beta_{2}}{\sin \beta_{1}}=\frac{n_{1}}{n_{2}} $$ i.e. $$ n_{2} \sin \beta_{2}=n_{1} \sin \beta_{1} . $$ In the same way we get $$ n_{3} \sin \beta_{3}=n_{2} \sin \beta_{2}, \text { etc. } $$ Thus, in general: $$ n_{i} \sin \beta_{i}=\mathrm{const} . $$ This relation does not involve plates thickness nor their number. So, we may make use of it also in case of continuous dependence of the refractive index in one direction (in our case in the $x$ direction). Consider the situation shown in Fig. 3. Fig. 3 At the point A the angle $\beta_{A}=90^{\circ}$. The refractive index at this point is $n_{0}$. Thus, we have $$ \begin{gathered} n_{A} \sin \beta_{A}=n_{B} \sin \beta_{B}, \\ n_{0}=n_{B} \sin \beta_{B} . \end{gathered} $$ Additionally, from the Snell law applied to the refraction at the point B, we have $$ \frac{\sin \alpha}{\sin \left(90^{\circ}-\beta_{B}\right)}=n_{B} . $$ Therefore $$ \sin \alpha=n_{B} \cos \beta_{B}=n_{B} \sqrt{1-\sin ^{2} \beta_{B}}=\sqrt{n_{B}^{2}-\left(n_{B} \sin \beta_{B}\right)^{2}}=\sqrt{n_{B}^{2}-n_{0}^{2}} $$ and finally $$ n_{B}=\sqrt{n_{0}^{2}+\sin ^{2} \alpha} . $$ Numerically $$ n_{B}=\sqrt{\left(\frac{12}{10}\right)^{2}+\left(\frac{5}{10}\right)^{2}}=1.3 $$ The value of $x_{B}$ can be found from the dependence $n(x)$ given in the text of the problem. We have $$ \begin{gathered} n_{B}=n\left(x_{B}\right)=\frac{n_{0}}{1-\frac{x_{B}}{R}}, \\ x_{B}=R\left(1-\frac{n_{0}}{n_{B}}\right), \end{gathered} $$ Numerically $$ x_{B}=1 \mathrm{~cm} $$
|
IPHO 1974
|
value
|
||
66
|
Consider a parallel, transparent plate of thickness $d$-Fig. 1. Its refraction index varies as $$ n=\frac{n_{0}}{1-\frac{x}{R}} . $$ <image> A light beam enters from the air perpendicularly to the plate at the point $\mathrm{A}\left(x_{A}=0\right)$ and emerges from it at the point B at an angle $\alpha$. Data: $$ n_{0}=1.2 ; \quad R=13 \mathrm{~cm} ; \quad \alpha=30^{\circ} . $$3. Find the thickness $d$ of the plate. ( unit: cm )
|
5
|
Fig. 2 Consider a light ray passing through a system of parallel plates with different refractive indexes - Fig. 2. From the Snell law we have $$ \frac{\sin \beta_{2}}{\sin \beta_{1}}=\frac{n_{1}}{n_{2}} $$ i.e. $$ n_{2} \sin \beta_{2}=n_{1} \sin \beta_{1} . $$ In the same way we get $$ n_{3} \sin \beta_{3}=n_{2} \sin \beta_{2}, \text { etc. } $$ Thus, in general: $$ n_{i} \sin \beta_{i}=\mathrm{const} . $$ This relation does not involve plates thickness nor their number. So, we may make use of it also in case of continuous dependence of the refractive index in one direction (in our case in the $x$ direction). Consider the situation shown in Fig. 3. Fig. 3 At the point A the angle $\beta_{A}=90^{\circ}$. The refractive index at this point is $n_{0}$. Thus, we have $$ \begin{gathered} n_{A} \sin \beta_{A}=n_{B} \sin \beta_{B}, \\ n_{0}=n_{B} \sin \beta_{B} . \end{gathered} $$ Additionally, from the Snell law applied to the refraction at the point B, we have $$ \frac{\sin \alpha}{\sin \left(90^{\circ}-\beta_{B}\right)}=n_{B} . $$ Therefore $$ \sin \alpha=n_{B} \cos \beta_{B}=n_{B} \sqrt{1-\sin ^{2} \beta_{B}}=\sqrt{n_{B}^{2}-\left(n_{B} \sin \beta_{B}\right)^{2}}=\sqrt{n_{B}^{2}-n_{0}^{2}} $$ and finally $$ n_{B}=\sqrt{n_{0}^{2}+\sin ^{2} \alpha} . $$ Numerically $$ n_{B}=\sqrt{\left(\frac{12}{10}\right)^{2}+\left(\frac{5}{10}\right)^{2}}=1.3 $$ The value of $x_{B}$ can be found from the dependence $n(x)$ given in the text of the problem. We have $$ \begin{gathered} n_{B}=n\left(x_{B}\right)=\frac{n_{0}}{1-\frac{x_{B}}{R}}, \\ x_{B}=R\left(1-\frac{n_{0}}{n_{B}}\right), \end{gathered} $$ Numerically $$ x_{B}=1 \mathrm{~cm} $$ The answer to the third question requires determination of the trajectory of the light ray. According to considerations described at the beginning of the solution we may write (see Fig. 4): $$ n(x) \sin \beta(x)=n_{0} . $$ Thus $$ \sin \beta(x)=\frac{n_{0}}{n(x)}=\frac{R-x}{R} . $$ Fig. 4 Consider the direction of the ray crossing a point C on the circle with radius $R$ and center in point O as shown in Fig. 4. We see that $$ \sin \angle \mathrm{COC}^{\prime}=\frac{R-x}{R}=\sin \beta(x) . $$ Therefore, the angle $\angle \mathrm{COC}^{\prime}$ must be equal to the angle $\beta(x)$ formed at the point C by the light ray and $\mathrm{CC}^{\prime}$. It means that at the point C the ray must be tangent to the circle. Moreover, the ray that is tangent to the circle at some point must be tangent also at farther points. Therefore, the ray cannot leave the circle (as long as it is inside the plate)! But at the beginning the ray (at the point A ) is tangent to the circle. Thus, the ray must propagate along the circle shown in Fig. 4 until reaching point $B$ where it leaves the plate. Already we know that $A^{\prime} B=1 \mathrm{~cm}$. Thus, $B^{\prime} B=12 \mathrm{~cm}$ and from the rectangular triangle $\mathrm{BB}^{\prime} \mathrm{O}$ we get $$ d=\mathrm{B}^{\prime} \mathrm{O}=\sqrt{13^{2}-12^{2}} \mathrm{~cm}=5 \mathrm{~cm} . $$
|
IPHO 1974
|
value
|
||
67
|
Three cylinders with the same mass, the same length and the same external radius are initially resting on an inclined plane. The coefficient of sliding friction on the inclined plane, $\mu$, is known and has the same value for all the cylinders. The first cylinder is empty (tube), the second is homogeneous filled, and the third has a cavity exactly like the first, but closed with two negligible mass lids and filled with a liquid with the same density like the cylinder's walls. The friction between the liquid and the cylinder wall is considered negligible. The density of the material of the first cylinder is $n$ times greater than that of the second or of the third cylinder. Determine:a) The linear acceleration of the cylinders in the non-sliding case. Compare all the accelerations.
|
$$ \mathrm{I}_{1}>\mathrm{I}_{2}>\mathrm{I}_{3} $$
|
The inertia moments of the three cylinders are: $$ I_{1}=\frac{1}{2} \rho_{1} \pi\left(R^{4}-r^{4}\right) h, \quad I_{2}=\frac{1}{2} \rho_{2} \pi R^{4} h=\frac{1}{2} m R^{2} \quad, \quad I 3=\frac{1}{2} \rho_{2} \pi\left(R^{4}-r^{4}\right) h, $$ Because the three cylinders have the same mass : $$ m=\rho_{1} \pi\left(R^{2}-r^{2}\right) h=\rho_{2} \pi R^{2} h $$ it results: $$ r^{2}=R^{2}\left(1-\frac{\rho_{2}}{\rho_{1}}\right)=R^{2}\left(1-\frac{1}{n}\right), n=\frac{\rho_{1}}{\rho_{2}} $$ The inertia moments can be written: $$ I_{1}=I_{2}\left(2-\frac{1}{n}\right)>I_{2}, \quad I_{3}=I_{2}\left(2-\frac{1}{n}\right) \cdot \frac{1}{n}=\frac{I_{1}}{n} $$ In the expression of the inertia momentum $I_{3}$ the sum of the two factors is constant: $$ \left(2-\frac{1}{n}\right)+\frac{1}{n}=2 $$ independent of n , so that their products are maximum when these factors are equal: $2-\frac{1}{n}=\frac{1}{n}$; it results $\mathrm{n}=1$, and the products $\left(2-\frac{1}{n}\right) \cdot \frac{1}{n}=1$. In fact $\mathrm{n}>1$, so that the products is les than 1 . It results: $$ \mathrm{I}_{1}>\mathrm{I}_{2}>\mathrm{I}_{3} $$
|
IPHO 1972
|
equation
|
||
68
|
Three cylinders with the same mass, the same length and the same external radius are initially resting on an inclined plane. The coefficient of sliding friction on the inclined plane, $\mu$, is known and has the same value for all the cylinders. The first cylinder is empty (tube), the second is homogeneous filled, and the third has a cavity exactly like the first, but closed with two negligible mass lids and filled with a liquid with the same density like the cylinder's walls. The friction between the liquid and the cylinder wall is considered negligible. The density of the material of the first cylinder is $n$ times greater than that of the second or of the third cylinder. Determine:b) Condition for angle $\alpha$ of the inclined plane so that no cylinders is sliding.
|
$$ \begin{array}{r} \mathrm{F}_{\mathrm{f}}<\mu \mathrm{N}=\mu \mathrm{mgsin} \alpha \\ \operatorname{tg} \alpha<\mu\left(1+\frac{m R^{2}}{I_{1}}\right) \end{array} $$
|
For a cylinder rolling over freely on the inclined plane (fig. 1.1) we can write the equations: $$ \begin{aligned} & m g \sin \alpha-F_{f}=m a \\ & N-m g \cos \alpha=0 \\ & F_{f} R=I \varepsilon \end{aligned} $$ where $\varepsilon$ is the angular acceleration. If the cylinder doesn't slide we have the condition: $$ a=\varepsilon R $$ Solving the equation system (6-8) we find: $$ a=\frac{g \sin \alpha}{1+\frac{I}{m R^{2}}}, \quad F_{f}=\frac{m g \sin \alpha}{1+\frac{m R^{2}}{I}} $$ The condition of non-sliding is: $$ \begin{array}{r} \mathrm{F}_{\mathrm{f}}<\mu \mathrm{N}=\mu \mathrm{mgsin} \alpha \\ \operatorname{tg} \alpha<\mu\left(1+\frac{m R^{2}}{I_{1}}\right) \end{array} $$ Fig. 1.1 In the case of the cylinders from this problem, the condition necessary so that none of them slides is obtained for maximum I: $$ \operatorname{tg} \alpha\left\langle\mu\left(1+\frac{m R^{2}}{I_{1}}\right)=\mu \frac{4 n-1}{2 n-1}\right. $$ The accelerations of the cylinders are: $$ a_{1}=\frac{2 g \sin \alpha}{3+\left(1-\frac{1}{n}\right)}, a_{2}=\frac{2 g \sin \alpha}{3}, a_{3}=\frac{2 g \sin \alpha}{3-\left(1-\frac{1}{n}\right)^{2}} $$
|
IPHO 1972
|
expression
|
||
69
|
Three cylinders with the same mass, the same length and the same external radius are initially resting on an inclined plane. The coefficient of sliding friction on the inclined plane, $\mu$, is known and has the same value for all the cylinders. The first cylinder is empty (tube), the second is homogeneous filled, and the third has a cavity exactly like the first, but closed with two negligible mass lids and filled with a liquid with the same density like the cylinder's walls. The friction between the liquid and the cylinder wall is considered negligible. The density of the material of the first cylinder is $n$ times greater than that of the second or of the third cylinder. Determine:c) The reciprocal ratios of the angular accelerations in the case of roll over with sliding of all the three cylinders. Make a comparison between these accelerations.d) The interaction force between the liquid and the walls of the cylinder in the case of sliding of this cylinder, knowing that the liquid mass is $\mathrm{m}_{1}$.
|
$$ \mathrm{a}_{1}<\mathrm{a}_{2}<\mathrm{a}_{3} $$
|
Fig. 1.1 In the case of the cylinders from this problem, the condition necessary so that none of them slides is obtained for maximum I: $$ \operatorname{tg} \alpha\left\langle\mu\left(1+\frac{m R^{2}}{I_{1}}\right)=\mu \frac{4 n-1}{2 n-1}\right. $$ The accelerations of the cylinders are: $$ a_{1}=\frac{2 g \sin \alpha}{3+\left(1-\frac{1}{n}\right)}, a_{2}=\frac{2 g \sin \alpha}{3}, a_{3}=\frac{2 g \sin \alpha}{3-\left(1-\frac{1}{n}\right)^{2}} $$ The relation between accelerations: $$ \mathrm{a}_{1}<\mathrm{a}_{2}<\mathrm{a}_{3} $$
|
IPHO 1972
|
equation
|
||
70
|
Three cylinders with the same mass, the same length and the same external radius are initially resting on an inclined plane. The coefficient of sliding friction on the inclined plane, $\mu$, is known and has the same value for all the cylinders. The first cylinder is empty (tube), the second is homogeneous filled, and the third has a cavity exactly like the first, but closed with two negligible mass lids and filled with a liquid with the same density like the cylinder's walls. The friction between the liquid and the cylinder wall is considered negligible. The density of the material of the first cylinder is $n$ times greater than that of the second or of the third cylinder. Determine:
|
$m_{l} g \frac{\cos \alpha}{\cos \phi}$
|
In the case than all the three cylinders slide: $$ F_{f}=\mu N=\mu m g \cos \alpha $$ and from (7) results: $$ \varepsilon=\frac{R}{I} \mu m g \cos \alpha $$ for the cylinders of the problem: $$ \begin{aligned} \varepsilon_{1}: & \varepsilon_{2}: \varepsilon_{3}=\frac{1}{I_{1}}: \frac{1}{I_{2}}: \frac{1}{I_{3}}=1:\left(1-\frac{1}{n}\right): n \\ & \varepsilon_{1}<\varepsilon_{2}<\varepsilon_{3} \end{aligned} $$ In the case that one of the cylinders is sliding: $$ \begin{aligned} & m g \sin \alpha-F_{f}=m a, \quad F_{f}=\mu m g \cos \alpha \\ & a=g(\sin \alpha-\mu \cos \alpha) \end{aligned} $$ Let $\vec{F}$ be the total force acting on the liquid mass $\mathrm{m}_{1}$ inside the cylinder (fig.1.2), we can write: $$ \begin{aligned} & F_{x}+m_{l} g \sin \alpha=m_{l} a=m_{l} g(\sin \alpha-\mu \cos \alpha), \quad F_{y}-m_{l} g \cos \alpha=0 \\ & \quad F=\sqrt{F_{x}^{2}+F_{y}^{2}}=m_{l} g \cos \alpha \cdot \sqrt{1+\mu^{2}}=m_{l} g \frac{\cos \alpha}{\cos \phi} \end{aligned} $$ where $\phi$ is the friction angle $(\operatorname{tg} \phi=\mu)$. Fig. 1.2
|
IPHO 1972
|
value
|
||
71
|
Two cylinders A and B, with equal diameters have inside two pistons with negligible mass connected by a rigid rod. The pistons can move freely. The rod is a short tube with a valve. The valve is initially closed (fig. 2.1). <image> Fig. 2.1 The cylinder A and his piston is adiabatically insulated and the cylinder B is in thermal contact with a thermostat which has the temperature $\theta=27^{\circ} \mathrm{C}$. Initially the piston of the cylinder A is fixed and inside there is a mass $\mathrm{m}=32 \mathrm{~kg}$ of argon at a pressure higher than the atmospheric pressure. Inside the cylinder B there is a mass of oxygen at the normal atmospheric pressure. Liberating the piston of the cylinder A, it moves slowly enough (quasi-static) and at equilibrium the volume of the gas is eight times higher, and in the cylinder B de oxygen's density increased two times. Knowing that the thermostat received the heat $Q^{\prime}=747,9.10^{4} \mathrm{~J}$, determine:a) Establish on the base of the kinetic theory of the gases, studying the elastic collisions of the molecules with the piston, that the thermal equation of the process taking place in the cylinder A is $\mathrm{TV}^{2 / 3}=$ constant.
|
$$ T V^{2 / 3}=\text { const. } $$
|
a) We consider argon an ideal mono-atomic gas and the collisions of the atoms with the piston perfect elastic. In such a collision with a fix wall the speed $\vec{v}$ of the particle changes only the direction so that the speed $\vec{v}$ and the speed $\vec{v}^{\prime}$ after collision there are in the same plane with the normal and the incident and reflection angle are equal. $$ v_{n}^{\prime}=-v_{n}, v_{t}^{\prime}=v_{t} $$ In the problem the wall moves with the speed $\vec{u}$ perpendicular on the wall. The relative speed of the particle with respect the wall is $\vec{v}-\vec{u}$. Choosing the Oz axis perpendicular on the wall in the sense of $\vec{u}$, the conditions of the elastic collision give: $$ \begin{gathered} (\vec{v}-\vec{u})_{z}=-\left(\vec{v}^{\prime}-\vec{u}\right)_{z},(\vec{v}-\vec{u})_{x, y}=\left(\vec{v}^{\prime}-\vec{u}\right)_{x, y} ; \\ v_{z}-u=-\left(v_{z}^{\prime}-u\right), v_{z}^{\prime}=2 u-v_{z}, v_{x, y}^{\prime}=v_{x, y} \end{gathered} $$ The increase of the kinetic energy of the particle with mass $m_{o}$ after collision is: $$ \frac{1}{2} m_{o} v^{\prime 2}-\frac{1}{2} m_{o} v^{2}=\frac{1}{2} m_{o}\left(v_{z}^{\prime 2}-v_{z}^{2}\right)=2 m_{o} u\left(u-v_{z}\right) \cong-2 m_{o} u v_{z} $$ because u is much smaller than $v_{z}$. If $n_{k}$ is the number of molecules from unit volume with the speed component $v_{z k}$, then the number of molecules with this component which collide in the time dt the area dS of the piston is: $$ \frac{1}{2} n_{k} v_{z k} d t d S $$ These molecules will have a change of the kinetic energy: $$ \frac{1}{2} n_{k} v_{z k} d t d S\left(-2 m_{o} u v_{z k}\right)=-m_{o} n_{k} v_{z k}^{2} d V $$ where $d V=u d t d S$ is the increase of the volume of gas. The change of the kinetic energy of the gas corresponding to the increase of volume dV is: $$ d E_{c}=-m_{o} d V \sum_{k} n_{k} v_{z k}^{2}=-\frac{1}{3} n m_{o} \bar{v}^{2} d V $$ and: $$ d U=-\frac{2}{3} N \frac{m_{o} \bar{v}^{2}}{2} \cdot \frac{d V}{V}=-\frac{2}{3} U \frac{d V}{V} $$ Integrating equation (7) results: $$ U V^{2 / 3}=\text { const. } $$ The internal energy of the ideal mono-atomic gas is proportional with the absolute temperature T and the equation (8) can be written: $$ T V^{2 / 3}=\text { const. } $$
|
IPHO 1972
|
equation
|
||
72
|
Two cylinders A and B, with equal diameters have inside two pistons with negligible mass connected by a rigid rod. The pistons can move freely. The rod is a short tube with a valve. The valve is initially closed (fig. 2.1). <image> Fig. 2.1 The cylinder A and his piston is adiabatically insulated and the cylinder B is in thermal contact with a thermostat which has the temperature $\theta=27^{\circ} \mathrm{C}$. Initially the piston of the cylinder A is fixed and inside there is a mass $\mathrm{m}=32 \mathrm{~kg}$ of argon at a pressure higher than the atmospheric pressure. Inside the cylinder B there is a mass of oxygen at the normal atmospheric pressure. Liberating the piston of the cylinder A, it moves slowly enough (quasi-static) and at equilibrium the volume of the gas is eight times higher, and in the cylinder B de oxygen's density increased two times. Knowing that the thermostat received the heat $Q^{\prime}=747,9.10^{4} \mathrm{~J}$, determine:b) Calculate the parameters $\mathrm{p}$, and T of argon in the initial and final states. ( unit: $\mathrm{~N} / \mathrm{m}^{2}$ )
|
$64,9 \cdot 10^{5}$
|
b) The oxygen is in contact with a thermostat and will suffer an isothermal process. The internal energy will be modified only by the adiabatic process suffered by argon gas: $$ \Delta U=v C_{V} \Delta T=m c_{V} \Delta T $$ where $v$ is the number of kilomoles. For argon $C_{V}=\frac{3}{2} R$. For the entire system $\mathrm{L}=0$ and $\Delta U=Q$. We will use indices 1 , respectively 2 , for the measures corresponding to argon from cylinder A , respectively oxygen from the cylinder B : $$ \Delta U=\frac{m_{1}}{\mu_{1}} \cdot \frac{3}{2} \cdot R\left(T_{1}^{\prime}-T_{1}\right)=Q=\frac{m_{1}}{\mu_{1}} \cdot \frac{3}{2} R T_{1}\left[\left(\frac{V_{1}}{V_{1}^{\prime}}\right)^{2 / 3}-1\right] $$ From equation (11) results: $$ \begin{gathered} T_{1}=\frac{2}{3} \cdot \frac{\mu_{1}}{m_{1}} \cdot \frac{Q}{R} \cdot \frac{1}{\left(\frac{V_{1}}{V_{1}^{\prime}}\right)^{2 / 3}-1}=1000 \mathrm{~K} \\ T_{1}^{\prime}=\frac{T_{1}}{4}=250 \mathrm{~K} \end{gathered} $$ For the isothermal process suffered by oxygen: $$ \frac{\rho_{2}^{\prime}}{\rho_{2}}=\frac{p_{2}^{\prime}}{p_{2}} $$ $p_{2}^{\prime}=2,00 \mathrm{~atm}=2,026 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}$ From the equilibrium condition: $$ p_{1}^{\prime}=p_{2}^{\prime}=2 \mathrm{~atm} $$ For argon: $$ \begin{aligned} & p_{1}=p_{1}^{\prime} \cdot \frac{V_{1}^{\prime}}{V_{1}} \cdot \frac{T_{1}}{T_{1}^{\prime}}=64 \mathrm{~atm}=64,9 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}$$
|
IPHO 1972
|
value
|
||
73
|
Two cylinders A and B, with equal diameters have inside two pistons with negligible mass connected by a rigid rod. The pistons can move freely. The rod is a short tube with a valve. The valve is initially closed (fig. 2.1). <image> Fig. 2.1 The cylinder A and his piston is adiabatically insulated and the cylinder B is in thermal contact with a thermostat which has the temperature $\theta=27^{\circ} \mathrm{C}$. Initially the piston of the cylinder A is fixed and inside there is a mass $\mathrm{m}=32 \mathrm{~kg}$ of argon at a pressure higher than the atmospheric pressure. Inside the cylinder B there is a mass of oxygen at the normal atmospheric pressure. Liberating the piston of the cylinder A, it moves slowly enough (quasi-static) and at equilibrium the volume of the gas is eight times higher, and in the cylinder B de oxygen's density increased two times. Knowing that the thermostat received the heat $Q^{\prime}=747,9.10^{4} \mathrm{~J}$, determine:b) Calculate the parameters $\mathrm{V}$, and T of argon in the initial and final states. ( unit: $\mathrm{~m}^{3}$ )
|
8,16
|
b) The oxygen is in contact with a thermostat and will suffer an isothermal process. The internal energy will be modified only by the adiabatic process suffered by argon gas: $$ \Delta U=v C_{V} \Delta T=m c_{V} \Delta T $$ where $v$ is the number of kilomoles. For argon $C_{V}=\frac{3}{2} R$. For the entire system $\mathrm{L}=0$ and $\Delta U=Q$. We will use indices 1 , respectively 2 , for the measures corresponding to argon from cylinder A , respectively oxygen from the cylinder B : $$ \Delta U=\frac{m_{1}}{\mu_{1}} \cdot \frac{3}{2} \cdot R\left(T_{1}^{\prime}-T_{1}\right)=Q=\frac{m_{1}}{\mu_{1}} \cdot \frac{3}{2} R T_{1}\left[\left(\frac{V_{1}}{V_{1}^{\prime}}\right)^{2 / 3}-1\right] $$ From equation (11) results: $$ \begin{gathered} T_{1}=\frac{2}{3} \cdot \frac{\mu_{1}}{m_{1}} \cdot \frac{Q}{R} \cdot \frac{1}{\left(\frac{V_{1}}{V_{1}^{\prime}}\right)^{2 / 3}-1}=1000 \mathrm{~K} \\ T_{1}^{\prime}=\frac{T_{1}}{4}=250 \mathrm{~K} \end{gathered} $$ For the isothermal process suffered by oxygen: $$ \frac{\rho_{2}^{\prime}}{\rho_{2}}=\frac{p_{2}^{\prime}}{p_{2}} $$ $p_{2}^{\prime}=2,00 \mathrm{~atm}=2,026 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}$ From the equilibrium condition: $$ p_{1}^{\prime}=p_{2}^{\prime}=2 \mathrm{~atm} $$ For argon: $$ \begin{aligned} & V_{1}=\frac{m_{1}}{\mu_{1}} \cdot \frac{R T_{1}}{p_{1}}=1,02 \mathrm{~m}^{3}, V_{1}^{\prime}=8 V_{1}=8,16 \mathrm{~m}^{3} \end{aligned} $$
|
IPHO 1972
|
value
|
||
74
|
Two cylinders A and B, with equal diameters have inside two pistons with negligible mass connected by a rigid rod. The pistons can move freely. The rod is a short tube with a valve. The valve is initially closed (fig. 2.1). <image> Fig. 2.1 The cylinder A and his piston is adiabatically insulated and the cylinder B is in thermal contact with a thermostat which has the temperature $\theta=27^{\circ} \mathrm{C}$. Initially the piston of the cylinder A is fixed and inside there is a mass $\mathrm{m}=32 \mathrm{~kg}$ of argon at a pressure higher than the atmospheric pressure. Inside the cylinder B there is a mass of oxygen at the normal atmospheric pressure. Liberating the piston of the cylinder A, it moves slowly enough (quasi-static) and at equilibrium the volume of the gas is eight times higher, and in the cylinder B de oxygen's density increased two times. Knowing that the thermostat received the heat $Q^{\prime}=747,9.10^{4} \mathrm{~J}$, determine: ( unit: $\mathrm{~N} / \mathrm{m}^{2}$ )
|
$2,23 \cdot 10^{5}$
|
c) When the valve is opened the gases intermix and at thermal equilibrium the final pressure will be $p^{\prime}$ and the temperature T . The total number of kilomoles is constant: $$ \begin{aligned} v_{1}+v_{2}=v^{\prime}, \frac{p_{1}^{\prime} V_{1}^{\prime}}{R T_{1}^{\prime}}+\frac{p_{2}^{\prime} V_{2}^{\prime}}{R T}=\frac{p\left(V_{1}^{\prime}+V_{2}^{\prime}\right)}{R T} \\ p_{1}^{\prime}+p_{2}^{\prime}=2 \mathrm{~atm}, T_{2}=T_{2}^{\prime}=T=300 \mathrm{~K} \end{aligned} $$ The total volume of the system is constant: $$ V_{1}+V_{2}=V_{1}^{\prime}+V_{2}^{\prime}, \quad \frac{V_{2}^{\prime}}{V_{2}}=\frac{\rho_{2}}{\rho_{2}^{\prime}}, \quad V_{2}^{\prime}=\frac{V_{2}}{2}=7,14 m^{3} $$ From equation (18) results the final pressure: $$ p=p_{1}^{\prime} \cdot \frac{1}{V_{1}+V_{2}} \cdot\left(V_{1}^{\prime} \cdot \frac{T}{T_{1}^{\prime}}+V_{2}^{\prime}\right)=2,2 \mathrm{~atm}=2,23 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2} $$
|
IPHO 1972
|
value
|
End of preview. Expand
in Data Studio
README.md exists but content is empty.
- Downloads last month
- 22