imone/ARB huggingface dataset:
Question: When a $5000 \mathrm{lb}$ car driven at $60 \mathrm{mph}$ on a level road is suddenly put into neutral gear (i.e. allowed to coast), the velocity decreases in the following manner:

$$
V=\frac{60}{1+\left(\frac{t}{60}\right)} \mathrm{mph},
$$

where $t$ is the time in sec. Find the horsepower required to drive this car at $30 \mathrm{mph}$ on the same road.

Useful constants: $g=22 \mathrm{mph} / \mathrm{sec}, 1 \mathrm{H} . \mathrm{P} .=550 \mathrm{ft} . l \mathrm{~b} / \mathrm{sec}, 60 \mathrm{mph}=$ $88 \mathrm{ft} / \mathrm{sec}$.
Answer: Let $V_{0}=60 \mathrm{mph}$, then

$$
\frac{t}{60}=\frac{V_{0}}{V}-1 \text {. }
$$

Hence

$$
\frac{d V}{d t}=\frac{-V^{2}}{60 V_{0}},
$$

and the resistance acting on the car is $F=m V^{2} /\left(60 V_{0}\right)$, where $m$ is the mass of the car. The propulsive force must be equal to the resistance $F^{\prime}$ at the speed of $V^{\prime}=30 \mathrm{mph}$ in order to maintain this speed on the same road. It follows that the horsepower required is

$$
\begin{aligned}
P^{\prime} &=F^{\prime} V^{\prime}=\frac{m V^{\prime 3}}{60 V_{0}}=37500 \frac{\mathrm{mph}^{2} . l \mathrm{~b} .}{\mathrm{s}} \\
&=\frac{37500}{g} \frac{\mathrm{mph}^{2} . \mathrm{lb} \text { wt }}{\mathrm{s}}=\frac{37500}{22} \mathrm{mph} . \mathrm{lb} \text { wt } \\
&=\frac{37500}{22} \cdot \frac{88}{60} \frac{\mathrm{ft} . \mathrm{lb} \text { wt }}{\mathrm{s}} \\
&=2500 \frac{\mathrm{ft} . l \mathrm{~b} \text { wt }}{\mathrm{s}}=4.5 \text { H.P. }
\end{aligned}
$$

Note that pound weight (lb wt) is a unit of force and $1 \mathrm{lb} \mathrm{wt}=g \mathrm{ft} . \mathrm{lb} / \mathrm{s}^{2}$. The horsepower is defined as $550 \mathrm{ft} . l \mathrm{~b} \mathrm{wt} / \mathrm{s}$.
Question: One gram each of ice, water, and water vapor are in equilibrium together in a closed container. The pressure is $4.58 \mathrm{~mm}$ of $\mathrm{Hg}$, the temperature is $0.01^{\circ} \mathrm{C}$. Sixty calories of heat are added to the system. The total volume is kept constant. Calculate the masses of water vapor now present in the container. 

(Hint: For water at $0.01^{\circ} \mathrm{C}$, the latent heat of fusion is $80 \mathrm{cal} / \mathrm{g}$, the latent heat of vaporization is $596 \mathrm{cal} / \mathrm{g}$, and the latent heat of sublimation is $676 \mathrm{cal} / \mathrm{g}$. Also note that the volume of the vapor is much larger than the volume of the water or the volume of the ice.)
Answer: It is assumed that the original volume of water vapor is $V$, it volume is also $V$ after heating, and the masses of ice, water, and water vapor are respectively $x, y$ and $z$ at the new equilibrium. We have

$$
\begin{gathered}
x+y+z=3, \\
(1-x) L_{\mathrm{sub}}+(1-y) L_{\mathrm{vap}}=Q=60, \\
\frac{1-x}{\rho_{\mathrm{ice}}}+\frac{(1-y)}{\rho_{\text {water }}}+V_{0}=V . \\
V_{0}=\frac{R T}{\mu p} . \\
V=\frac{z}{\mu p} R T
\end{gathered}
$$

where $\mu=18 \mathrm{~g} / \mathrm{mole}, p=4.58 \mathrm{mmHg}, T=273.16 \mathrm{~K}, R=8.2 \times 10^{8}$ $\mathrm{m}^{3} \cdot \mathrm{atm} / \mathrm{mol} \cdot \mathrm{K}, \rho_{\text {ice }}=\rho_{\text {water }}=1 \mathrm{~g} / \mathrm{cm}^{3}, L_{\text {sub }}=676 \mathrm{cal} / \mathrm{g}$, and $L_{\text {vap }}=$ $596 \mathrm{cal} / \mathrm{g}$. Solving the equations we find

$$
x=0.25 \mathrm{~g}, \quad y=1.75 \mathrm{~g}, \quad z=1.00 \mathrm{~g} .
$$

That is, the heat of $60 \mathrm{cal}$ is nearly all used to melt the ice.
Question: The entropy of water at atmospheric pressure and $100^{\circ} \mathrm{C}$ is $0.31$ $\mathrm{cal} / \mathrm{g} \cdot \mathrm{deg}$, and the entropy of steam at the same temperature and pressure is $1.76 \mathrm{cal} / \mathrm{g} \cdot \mathrm{deg}$.

What is the heat of vaporization at this temperature?
Answer: Heat of vaporization is

$$
L=T \Delta S=540 \mathrm{cal} / \mathrm{g} .
$$
Question: For a particle of mass $m$ in a one-dimensional box of length $l$, the eigenfunctions and energies are

$$
\begin{aligned}
\psi_{n}(x) &=\sqrt{\frac{2}{l}} \sin \frac{n \pi x}{l}, \quad 0 \leq x \leq l \\
E_{n} &=\frac{1}{2 m}\left(\frac{n \pi \hbar}{l}\right)^{2}, \quad n=\pm 1, \pm 2, \ldots
\end{aligned}
$$

Suppose the particle is originally in a state $|n\rangle$ and the box length is increased to a length of $2 l(0 \leq x \leq 2 l)$ in a time $t \ll h / E_{n}$. Afterwards what is the probability that the particle will be found in an energy eigenstate with energy $E_{n}$ ?
Answer: First consider the process in which the box length is increased from $l$ to $2 l$. As $t \ll \frac{h}{E_{n}}$, it is reasonable to assume that the state of the particle in the box is unable to respond to the change during such a short time. Therefore the wave function of the particle after the change is completed is

$$
\psi(x)= \begin{cases}\sqrt{\frac{2}{l}} \sin \frac{n \pi x}{l}, & 0 \leq x \leq l \\ 0, & l \leq x \leq 2 l\end{cases}
$$

On the other hand, the eigenstates and eigenvalues of the same particle in the one-dimensional box of length $2 l$ would be, respectively,

$$
\begin{aligned}
\phi_{n^{\prime}}(x) &=\sqrt{\frac{1}{l}} \sin \frac{n^{\prime} \pi x}{2 l}, \quad(0 \leq x \leq 2 l), \\
E_{n^{\prime}} &=\frac{1}{2 m}\left(\frac{n^{\prime} \pi \hbar}{2 l}\right)^{2}, \quad\left(n^{\prime}=\pm 1, \pm 2, \ldots\right) .
\end{aligned}
$$

The energy $E_{n}$ of the particle corresponds to the energy level $E_{n^{\prime}}$ in the $2 l$ box, where $\frac{n}{l}=\frac{n^{\prime}}{2 l}$, i.e., $n^{\prime}=2 n$. The corresponding eigenstate is then $\phi_{2 n}$. Thus the probability amplitude is

$$
A=\int_{-\infty}^{\infty} \phi_{2 n}(x) \psi(x) d x=\frac{\sqrt{2}}{l} \int_{0}^{l} \sin ^{2} \frac{n \pi x}{l} d x, \frac{1}{\sqrt{2}},
$$

and the probability of finding the particle in an eigenstate with energy $E_{n}$ is

$$
P=|A|^{2}=\frac{1}{2} .
$$
Question: Given a classical model of the tritium atom with a nucleus of charge $+1$ and a single electron in a circular orbit of radius $r_{0}$, suddenly the nucleus emits a negatron and changes to charge $+2$.
(The emitted negatron escapes rapidly and we can forget about it.) The electron in orbit suddenly has a new situation.

Find the ratio of the electron's energy after to before the emission of the negatron (taking the zero of energy, as usual, to be for zero kinetic energy at infinite distance).
Answer: As the negatron leaves the system rapidly, we can assume that its leaving has no effect on the position and kinetic energy of the orbiting electron.

From the force relation for the electron,

$$
\frac{m v_{0}^{2}}{r_{0}}=\frac{e^{2}}{4 \pi \varepsilon_{0} r_{0}^{2}},
$$

we find its kinetic energy

$$
\frac{m v_{0}^{2}}{2}=\frac{e^{2}}{8 \pi \varepsilon_{0} r_{0}}
$$

and its total mechanical energy

$$
E_{1}=\frac{m v_{0}^{2}}{2}-\frac{e^{2}}{4 \pi \varepsilon_{0} r_{0}}=\frac{-e^{2}}{8 \pi \varepsilon_{0} r_{0}}
$$

before the emission of the negatron. After the emission the kinetic energy of the electron is still $\frac{e^{2}}{8 \pi \varepsilon_{0} r_{0}}$, while its potential energy suddenly changes to

$$
\frac{-2 e^{2}}{4 \pi \varepsilon_{0} r_{0}}=-\frac{-e^{2}}{2 \pi \varepsilon_{0} r_{0}} .
$$

Thus after the emission the total mechanical energy of the orbiting electron is

$$
E_{2}=\frac{m v_{0}^{2}}{2}-\frac{2 e^{2}}{4 \pi \varepsilon_{0} r_{0}}=\frac{-3 e^{2}}{8 \pi \varepsilon_{0} r_{0}}
$$

giving

$$
\frac{E_{2}}{E_{1}}=3 .
$$

In other words, the total energy of the orbiting electron after the emission is three times as large as that before the emission.
Question: An air-spaced coaxial cable has an inner conductor $0.5 \mathrm{~cm}$ in diameter and an outer conductor $1.5 \mathrm{~cm}$ in diameter. When the inner conductor is at a potential of $+8000 \mathrm{~V}$ with respect to the grounded outer conductor, what is the charge per meter on the inner conductor?
Answer: Let the linear charge density for the inner conductor be $\lambda$. By symmetry we see that the field intensity at a point distance $r$ from the axis in the cable between the conductors is radial and its magnitude is given by Gauss' theorem as

$$
E=\frac{\lambda}{2 \pi \varepsilon_{0} r} .
$$

Then the potential difference between the inner and outer conductors is

$$
V=\int_{a}^{b} E d r=\frac{\lambda}{2 \pi \varepsilon_{0}} \ln (b / a)
$$

with $a=1.5 \mathrm{~cm}, b=0.5 \mathrm{~cm}$, which gives

$$
\begin{aligned}
\lambda &=\frac{2 \pi \varepsilon_{0} V}{\ln (b / a)}=\frac{2 \pi \times 8.9 \times 10^{-12} \times 8000}{\ln (1.5 / 0.5)} \\
&=4.05 \times 10^{-7} \mathrm{C} / \mathrm{m} .
\end{aligned}
$$
Question: A small celestial object, held together only by its self-gravitation, can be disrupted by the tidal forces produced by another massive body, if it comes near enough to that body. For an object of diameter $1 \mathrm{~km}$ and density $2 \mathrm{~g} / \mathrm{cm}^{3}$, find the critical distance from the earth (Roche limit).
Answer: Suppose the earth is fixed in space and the small celestial object orbits around it at a distance $l$ away as shown in Fig. 1.73. Let $M$ be the mass of the earth, $m$ the mass and $\rho$ the density of the small celestial object. Consider a unit mass of the object on the line $O C$ at distance $x$ from $C$. We have from the balance of forces on it

$$
(l-x) \omega^{2}=\frac{G M}{(l-x)^{2}}-\frac{G\left(\frac{4}{3}\right) \pi x^{3} \rho}{x^{2}} .
$$

We also have for the celestial body

$$
m l \omega^{2}=\frac{G M m}{l^{2}},
$$

which gives $\omega^{2}$ to be used in the above. Then as $\frac{x}{l} \ll 1$, retaining only the lowest order in $\frac{x}{l}$, we have

$$
l=\left(\frac{9 M}{4 \pi \rho}\right)^{\frac{1}{3}}
$$

With $M=6 \times 10^{27} \mathrm{~g}, \rho=2 \mathrm{~g} / \mathrm{cm}^{3}$, we find

$$
l=1.29 \times 10^{9} \mathrm{~cm}=1.29 \times 10^{4} \mathrm{~km} .
$$

If $l$ is less than this value, the earth's attraction becomes too large for the unit mass to be held by the celestial body and disruption of the latter occurs.

If the unit mass is located to the right of $C$ on the extended line of $O C$, $x$ is negative but the above conclusion still holds true. We may also consider a unit mass located off the line $O C$ such as the point $P$ in Fig. 1.74. We now have

$$
\sqrt{(l-x)^{2}+y^{2}} \omega^{2} \cos \theta=\frac{G M}{(l-x)^{2}+y^{2}} \cos \theta-\frac{4}{3} \pi \rho G \sqrt{x^{2}+y^{2}} \cos \varphi,
$$


with

$$
\cos \theta=\frac{l-x}{\sqrt{(l-x)^{2}+y^{2}}}, \quad \cos \varphi=\frac{x}{\sqrt{x^{2}+y^{2}}} \text {. }
$$

As $x / l \ll 1, y / l \ll 1$, and retaining only the first-order terms we would obtain the same result.
Question: A self-luminous object of height $h$ is $40 \mathrm{~cm}$ to the left of a converging lens with a focal length of $10 \mathrm{~cm}$. A second converging lens with a focal length of $20 \mathrm{~cm}$ is $30 \mathrm{~cm}$ to the right of the first lens.

Calculate the distance between the final image and the second lens.
Answer: From $\frac{1}{f_{1}}=\frac{1}{u_{1}}+\frac{1}{v_{1}}$, where $f_{1}=10 \mathrm{~cm}, u_{1}=40 \mathrm{~cm}$, we obtain $v_{1}=13 \frac{1}{3} \mathrm{~cm}$.

From $\frac{1}{f_{2}}=\frac{1}{u_{2}}+\frac{1}{v_{2}}$, where $f_{2}=20 \mathrm{~cm}, u_{2}=\left(30-\frac{10}{3}\right) \mathrm{cm}$, we obtain $v_{2}=-100 \mathrm{~cm}$, i.e., the final image is $100 \mathrm{~cm}$ to the left of the second lens.
Question: At room temperature, $k_{\mathrm{B}} T / e=26 \mathrm{mV}$. A sample of cadmium sulfide displays a mobile carrier density of $10^{16} \mathrm{~cm}^{-3}$ and a mobility coefficient $\mu=10^{2} \mathrm{~cm}^{2} /$ volt sec.

Calculate the electrical conductivity of this sample.
Answer: The electrical conductivity is given by $\sigma=n e \mu$. With $n=10^{22} \mathrm{~m}^{-3}$, $e=1.6 \times 10^{-19} \mathrm{C}, \mu=10^{-2} \mathrm{~m}^{2} \mathrm{~V}^{-1} \mathrm{~s}^{-1}$, we have for the material $\sigma=$ $16 \Omega^{-1} \mathrm{~m}^{-1}$.
Question: Consider the situation which arises when a negative muon is captured by an aluminum atom (atomic number $Z=13$ ). After the muon gets inside the "electron cloud" it forms a hydrogen-like muonic atom with the aluminum nucleus. The mass of the muon is $105.7 \mathrm{MeV}$.

(Slide rule accuracy; neglect nuclear motion)

Compute the mean life of the above muonic atom in the $3 d$ state, taking into account the fact that the mean life of a hydrogen atom in the $3 d$ state is $1.6 \times 10^{-8}$ sec.

Answer: For spontaneous transitions from the $3 d$ state, the largest probability is for $3 d \rightarrow 2 p$. In nonrelativistic approximation, the photon energy is given by

$$
\begin{aligned}
h \nu &=\frac{m_{\mu} Z^{2} e^{4}}{2 \hbar^{2}}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right) \\
&=\frac{m_{\mu} c^{2} Z^{2}}{2}\left(\frac{e^{2}}{\hbar c}\right)^{2}\left(\frac{5}{36}\right) \\
&=\frac{105.7 \times 13^{2}}{2}\left(\frac{1}{137}\right)^{2}\left(\frac{5}{36}\right)=6.61 \times 10^{-2} \mathrm{MeV} .
\end{aligned}
$$

The corresponding wavelength is

$$
\lambda=\frac{c}{\nu}=\frac{h c}{h \nu}=\frac{4.135 \times 10^{-15} \times 3 \times 10^{10}}{6.61 \times 10^{4}}=1.88 \times 10^{-9} \mathrm{~cm} .
$$The transition probability per unit time is

$$
A \propto \omega^{3}\left|\mathbf{r}_{k k^{\prime}}\right|^{2} .
$$

For hydrogen-like atoms, as

$$
\left|\mathbf{r}_{k k^{\prime}}\right| \propto \frac{1}{Z}, \quad \omega \propto m Z^{2}, \quad \text { and so } \quad A \propto m^{3} Z^{4},
$$

the mean life of the muonic atom in the $3 d$ state is

$$
\begin{aligned}
T_{\mu} &=\left(\frac{A_{0}}{A}\right) T_{0}=\left(\frac{m_{0}}{m_{\mu}}\right)^{3} \frac{T_{0}}{Z^{4}} \\
&=\left(\frac{0.51}{105.7}\right)^{3} \times \frac{1}{13^{4}} \times 1.6 \times 10^{-8}=6.3 \times 10^{-20} \mathrm{~s} .
\end{aligned}
$$
Question: The $z$-component of the spin of an electron in free space (no electromagnetic fields) is measured and found to be $+\hbar / 2$. What is the probability of this occuring?


Answer: In the $\sigma_{z}$ representation, the spin wave function is $\left(\begin{array}{l}1 \\ 0\end{array}\right)$, the eigenfunctions of $\sigma_{x}$ are $\frac{1}{\sqrt{2}}\left(\begin{array}{l}1 \\ 1\end{array}\right), \frac{1}{\sqrt{2}}\left(\begin{array}{r}1 \\ -1\end{array}\right)$, corresponding to eigenvalues $+1,-1$ respectively. Expanding $\left(\begin{array}{l}1 \\ 0\end{array}\right)$ in these two states, we see that the possible results of a measurement of $s_{x}$ are $\pm \hbar / 2$ since $\hat{s}_{x}=\frac{1}{2} \sigma_{x}$, the mean value being zero:

$$
\left\langle s_{x}\right\rangle=(1,0) \hat{s}_{x}\left(\begin{array}{l}
1 \\
0
\end{array}\right)=0 .
$$

The probabilities of finding the result to be $+\frac{\hbar}{2}$ and $-\frac{\hbar}{2}$ are $P_{+}$and $P_{-}$respectively:

$$
\begin{aligned}
&P_{+}=\left|\frac{1}{\sqrt{2}}(1,1)\left(\begin{array}{l}
1 \\
0
\end{array}\right)\right|^{2}=\frac{1}{2}, \\
&P_{-}=\left|\frac{1}{\sqrt{2}}(1,-1)\left(\begin{array}{l}
1 \\
0
\end{array}\right)\right|^{2}=\frac{1}{2} .
\end{aligned}
$$
Question: While sitting in front of a color TV with a $25 \mathrm{kV}$ picture tube potential, you have an excellent chance of being irradiated with $\mathrm{X}$-rays.

For the resulting continuous distribution, calculate the shortest wavelength (maximum energy) X-ray.
Answer: When a high voltage is applied to the picture tube, electrons emitted from the negative electrode will be accelerated by the electric field to strike the screen target. If the energy of the electrons exceeds a certain value, they can knock off inner-shell electrons in the target atoms and make holes in the inner shells. Then as the outer-shell electrons fall in to fill up these holes, X-rays are emitted.
The maximum energy of the X-ray photons produced, $h \nu_{\max }$, is equal to the energy $\mathrm{eV}$ of the incident electrons. Hence the minimum wavelength of the X-rays is

$$
\lambda_{\min }=\frac{h c}{e V}=\frac{12000}{V}=\frac{12000}{25 \times 10^{3}}=0.48 \AA .
$$
Question: Twenty grams of ice at $0^{\circ} \mathrm{C}$ are dropped into a beaker containing 120 grams of water initially at $70^{\circ} \mathrm{C}$. Find the final temperature of the mixture neglecting the heat capacity of the beaker. Heat of fusion of ice is $80$ $\mathrm{cal} / \mathrm{g}$.
Answer: We assume the temperature of equilibrium to be $T$ after mixing. Thus

$$
M_{1} L_{\text {fusion }}+M_{1} C_{p, \text { water }} T=M_{2} C_{p, \text { water }}\left(T_{0}-T\right) .
$$

We substitute $M_{1}=20 \mathrm{~g}, M_{2}=120 \mathrm{~g}, T_{0}=70^{\circ} \mathrm{C}, L_{\text {fusion }}=80 \mathrm{cal} / \mathrm{g}$ and $C_{p, \text { water }}=1 \mathrm{cal} / \mathrm{g}$, and obtain the final temperature $T=48.57^{\circ} \mathrm{C}$.
Question: Two conductors are embedded in a material of conductivity $10^{-4} \Omega / \mathrm{m}$ and dielectric constant $\varepsilon=80 \varepsilon_{0}$. The resistance between the two conductors is measured to be $10^{5} \Omega$. Calculate the capacitance between the two conductors.
Answer: Suppose that the two conductors carry free charges $Q$ and $-Q$. Consider a closed surface enclosing the conductor with the charge $Q$ (but not the other conductor). We have, using Ohm's and Gauss' laws,

$$
I=\oint \mathbf{j} \cdot d \mathbf{S}=\oint \sigma \mathbf{E} \cdot d \mathbf{S}=\sigma \oint \mathbf{E} \cdot d \mathbf{S}=\sigma \frac{Q}{\varepsilon} .
$$

If the potential difference between the two conductors is $V$, we have $V=$ $I R=\frac{\sigma Q}{\varepsilon} R$, whence

$$
C=\frac{Q}{V}=\frac{\varepsilon}{\sigma R} .
$$

Numerically the capacitance between the conductors is

$$
C=\frac{80 \times 8.85 \times 10^{-12}}{10^{-4} \times 10^{5}}=7.08 \times 10^{-11} \mathrm{~F} .
$$
Question: Consider a large number of $N$ localized particles in an external magnetic field $\mathbf{H}$. Each particle has spin $1 / 2$. Find the number of states accessible to the system as a function of $M_{8}$, the $z$-component of the total spin of the system. Determine the value of $M_{\mathrm{s}}$ for which the number of states is maximum.
Answer: The spin of a particle has two possible orientations $1 / 2$ and $-1 / 2$. Let the number of particles with spin $1 / 2$ whose direction is along $\mathbf{H}$ be $N_{\uparrow}$ and the number of particles with spin $-1 / 2$ whose direction is opposite to $\mathbf{H}$ be $N_{\downarrow}$; then the component of the total spin in the direction of $\mathbf{H}$ is $M_{\mathrm{s}}=\frac{1}{2}\left(N_{\uparrow}-N_{\downarrow}\right)$. By $N_{\uparrow}+N_{\downarrow}=N$, we can obtain $N_{\uparrow}=\frac{N}{2}+M_{\mathrm{s}}$ and $N_{\downarrow}=\frac{N}{2}-M_{\mathrm{s}}$. The number of states of the system is

$$
Q=\frac{N !}{N_{\uparrow} ! N_{\downarrow} !} \frac{N !}{\left[\frac{N}{2}+M_{\mathrm{s}}\right] !\left[\frac{N}{2}-M_{\mathrm{s}}\right] !}
$$

Using Stirling's formula, one obtains

$$
\begin{aligned}
\ln Q &=\ln \frac{N !}{N_{\uparrow} ! N_{\downarrow} !} \\
&=N \ln N-N_{\uparrow} \ln N_{\uparrow}-N_{\downarrow} \ln N_{\downarrow} \\
&=N \ln N-N_{\uparrow} \ln N_{\uparrow}-\left(N-N_{\uparrow}\right) \ln \left(N-N_{\uparrow}\right) .
\end{aligned}
$$

By

$$
\frac{\partial \ln Q}{\partial N_{\uparrow}}=-\ln N_{\uparrow}+\ln \left(N-N_{\uparrow}\right)=0,
$$

we get $N_{\uparrow}=\frac{N}{2}$, i.e., $M_{\mathrm{s}}=0$ when the number of states of the system is maximum.
Question: Assume a visible photon of $3 \mathrm{eV}$ energy is absorbed in one of the cones (light sensors) in your eye and stimulates an action potential that produces a $0.07$ volt potential on an optic nerve of $10^{-9} \mathrm{~F}$ capacitance.

Calculate the energy of the action potential.
Answer: $Q=V C=0.07 \times 10^{-9}=7 \times 10^{-11}$ Coulomb.$E=\frac{Q V}{2}=\frac{1}{2} \times 7 \times 10^{-11} \times 0.07=2.5 \times 10^{-12}$ joule.
Question: A spherical dust particle falls through a water mist cloud of uniform density such that the rate of accretion onto the droplet is proportional to the volume of the mist cloud swept out by the droplet per unit time. If the droplet starts from rest in the cloud, find the value of the acceleration of the drop for large times.
Answer: Suppose the spherical dust particle initially has mass $M_{0}$ and radius $R_{0}$. Take the initial position of the dust particle as the origin and the $x$-axis along the downward vertical. Let $M(t)$ and $R(t)$ be the mass and radius of the droplet at time $t$ respectively. Then

$$
M(t)=M_{0}+\frac{4}{3} \pi\left(R^{3}-R_{0}^{3}\right) \rho,
$$

where $\rho$ is the density of the water mist, giving

$$
\frac{d M}{d t}=\rho 4 \pi R^{2} \frac{d R}{d t} .
$$

The droplet has a cross section $\pi R^{2}$ and sweeps out a cylinder of volume $\pi R^{2} \dot{x}$ in unit time, where $\dot{x}$ is its velocity. As the rate of accretion is proportional to this volume, we have

$$
\frac{d M}{d t}=\alpha \pi R^{2} \dot{x},
$$

$\alpha$ being a positive constant. Hence

$$
\dot{x}=\frac{4 \rho}{\alpha} \dot{R} .
$$

The momentum theorem gives

$$
M(t+d t) \dot{x}(t+d t)-M(t) \dot{x}(t)=M g d t .
$$

Using Taylor's theorem to expand $M(t+d t)$ and $\dot{x}(t+d t)$ and retaining only the lowest-order terms, we obtain

$$
\dot{x} \frac{d M}{d t}+M \ddot{x}=M g .
$$

For large $t, M(t) \approx \frac{4}{3} \pi R^{3} \rho, d M / d t \approx 3 M \dot{R} / R$, and the above becomes

$$
\ddot{R}+\frac{3 \dot{R}^{2}}{R}=\frac{\alpha g}{4 \rho} .
$$

For a particular solution valid for large $t$, setting

$$
R(t)=a t^{2},
$$

where $a$ is a constant, in the above we obtain

$$
a=\frac{\alpha g}{56 \rho} .
$$

Thus for large $t$,

$$
\dot{x}=\frac{4 \rho}{\alpha} \cdot 2 a t=\frac{g t}{7} .
$$

Hence the acceleration for large times is $g / 7$.
Question: A solenoid is designed to generate a magnetic field over a large volume. Its dimensions are as follows: length $=2$ meters, radius $=0.1$ meter, number of turns $=1000$.
(Edge effects should be neglected.)

Calculate the self-inductance of the solenoid in Henrys.
Answer: inside it is

Suppose the solenoid carries a current $I$. The magnetic induction

$$
B=\mu_{0} n I=\mu_{0} N I / l,
$$

and the magnetic flux linkage is

$$
\psi=N B S=N \frac{\mu_{0} N I}{l} \cdot \pi r^{2}=\frac{I \mu_{0} N^{2} \pi r^{2}}{l} .
$$

Hence the self-inductance is

$$
\begin{aligned}
L=\frac{\psi}{I} &=\frac{\mu_{0} N^{2} \pi r^{2}}{l}=\frac{4 \pi \times 10^{-7} \times 1000^{2} \times \pi \times 0.1^{2}}{2} \\
&=1.97 \times 10^{-2} \mathrm{H} .
\end{aligned}
$$
Question: Suppose two compact stars, each of one solar mass, are in a circular orbit around each other with a radius of one solar radius. What is the time scale for decay of this orbit? Take

$$
\begin{aligned}
\text { solar mass } &=2 \times 10^{33} \mathrm{gm}, \\
\text { solar radius } &=7 \times 10^{10} \mathrm{~cm}, \\
G &=6.7 \times 10^{-8} \mathrm{~cm}^{3} \mathrm{gm}^{-1} \mathrm{sec}^{-2}, \\
c &=3 \times 10^{10} \mathrm{~cm} \mathrm{sec}^{-1} .
\end{aligned}
$$
Answer: The total mass-energy of an isolated system is conserved, so it is not possible to radiate monopole gravitational radiation. Also as the total momentum of an isolated system is conserved, which means that the second time derivative of the mass dipole moment $\sum m r$ is zero, it is not possible to radiate dipole gravitational radiation. The lowest multipole gravitational radiation is quadrupole.
For a system of two stars in a circular orbit around their common center of mass, the rate of energy loss by radiation of gravitational waves is

$$
-\frac{d E}{d t}=\frac{32 G^{4}}{5 c^{5} r^{5}}\left(m_{1} m_{2}\right)^{2}\left(m_{1}+m_{2}\right)
$$

where $r$ is their mutual distance, which is constant for motion in a circular orbit. With $m_{1}=m_{2}=m$ and the data given we have

$$
-\frac{d E}{d t}=\frac{64 G^{4}}{5 c^{5} r^{5}} m^{5}=1.57 \times 10^{31} \mathrm{erg} \mathrm{s}^{-1} .
$$

With

$$
\begin{gathered}
E=-\frac{G m^{2}}{2 r}, \\
\frac{d r}{d t}=\frac{2 r^{2}}{G m^{2}} \frac{d E}{d t}=-\frac{128 G m^{3}}{5 c^{5} r^{3}} .
\end{gathered}
$$

As $\frac{d r}{d t}$ is the rate at which the two stars approach each other, the time taken for the complete collapse of the orbit is

$$
\begin{aligned}
\tau=\int_{r}^{0} \frac{d r}{\left(\frac{d r}{d t}\right)} &=-\frac{5 c^{5}}{128 G^{3} m^{3}} \int_{r}^{0} r^{3} d r \\
&=\frac{5 c^{5}}{512 G^{3} m^{3}} r^{4}=2.4 \times 10^{15} \mathrm{~s} .
\end{aligned}
$$