Hong Kong Certificate of Education Examination (HKCEE)

February 7, 2024February 7, 2024

202402071059 Solution to 1980-CE-PHY-II-16

A light ray passes through a spherical air bubble in water. Which of the following represents the path of the emergent ray?

A. $P$
B. $Q$
C. $R$
D. $S$

Roughwork.

Refractive (/refraction) index of a pure substance, with some exceptions, is proportional to its density; and of a mixture of substances, as a rule, to its mass concentration.

Thus water, higher in density than air, should have a larger refractive index than air:

$n_{\textrm{water}}>n_{\textrm{air}}$ ;

alternatively, as light travels in air faster than in water, from

$\displaystyle{v=\frac{c}{n}}$ ,

it can be seen also.

Let $\theta_{1}$ and $\theta_{2}$ be the angle of incidence and of refraction when light is refracted through the first interface; and $\theta_{2}$ and $\theta_{3}$ , when through the second interface.

Assuming WLOG $n_1\geqslant n_2$ and noting $\sin \theta \geqslant 0$ (non-negative) for $\theta\in \big[ 0,\frac{\pi}{2}\big]$ . By Snell’s Law, write

$\begin{aligned} n_1\sin\theta_1 & = n_2\sin\theta_2 \\ \frac{\sin\theta_1}{\sin\theta_2} & = \frac{n_2}{n_1}\leqslant 1 \\ \sin\theta_1 & \leqslant \sin\theta_2 \\ \theta_1 & \leqslant \theta_2 \\ \end{aligned}$

$\therefore \quad n_1\geqslant n_2\Longrightarrow \theta_2\geqslant \theta_1$

$\therefore\quad n_{\textrm{water}}>n_{\textrm{air}}\Longrightarrow \theta_{\textrm{air}}>\theta_{\textrm{water}}$

Answer. Path $P$ .

Reverse the direction of incidence and emergence:

Light ray along path: $P$ bends away from and then towards the normal; $Q$ bends towards the normal once and again; $R$ bends away from the normal once and again; $S$ bends towards and then away from the normal. Wits have it that $Q$ and $R$ are unlikely. So we turn our attention to light rays $P$ and $S$ .

Fussy enough, path $P$ is symmetric with respect to straight line $L_P$ ; but there across path $S$ is no line of symmetry were it to be $L_S$ .

This problem is not to be attempted.

December 20, 2023December 20, 2023

202312201537 Solution to 1980-CE-PHY-II-13

$6\times 10^{-3}\,\mathrm{m^3}$ of a gas is contained in a vessel at $91^\circ\mathrm{C}$ and a pressure of $4\times 10^5\,\mathrm{Pa}$ . If the density of the gas at s.t.p. ( $0^\circ\mathrm{C}$ and $10^5\,\mathrm{Pa}$ ) is $1.2\,\mathrm{kg\, m^{-3}}$ , what is the mass of the gas?

A. $7.2\,\mathrm{g}$
B. $14.4\,\mathrm{g}$
C. $21.6\,\mathrm{g}$
D. $28.8\,\mathrm{g}$

Official answer: C

Roughwork.

$\textrm{\textbf{\scriptsize{CASE I}}}$ (closed vessel)

Assumed a closed vessel, its volume $V=\textrm{Const.}$ a constant. Then, that $6\times 10^{-3}\,\mathrm{m^3}$ of gas fully occupied the closed vessel, implies that the volume of the vessel is also

$V=6\times 10^{-3}\,\mathrm{m^3}$ .

Hence may we write

$\begin{aligned} \textrm{density }(\rho ) & = \frac{\textrm{mass }(m)}{\textrm{volume }(V)} \\ m & = \rho V \\ & = (1.2\,\mathrm{kg\,m^{-3}})(6\times 10^{-3}\,\mathrm{m^3}) \\ & = 7.2\,\mathrm{g} \\ \end{aligned}$

And we get option A, which is different from the official answer C. Hence, we need to look into

$\textrm{\textbf{\scriptsize{CASE II}}}$ (open vessel)

We set up the background below.

$******************$

1. Ideal gas law:

An ideal gas satisfies the general gas equation

$pV=nRT$

where $p$ , $V$ , $n$ , $R$ , and $T$ are the pressure, the volume, the amount of substance (/number of moles), the ideal (/universal) gas constant, and the temperature of the gas.

2. Conversion of scales of temperature:

$x\,^\circ\mathrm{C} \approx (x+273)\,\mathrm{K}$

3. Relationship between chemical amount (/number of moles) $n$ , total mass $m$ , and molar mass $M$ , of a substance.

$\displaystyle{n=\frac{m}{M}}$

Note that total mass $m$ and number of moles $n$ are extensive (/extrinsic) properties; whereas molar mass $M$ is an intensive (/intrinsic) property.

$******************$

At the start $t=t_0$ ,

$\begin{aligned} p(t_0) & = 4\times 10^5\,\mathrm{Pa} \\ V(t_0) & = 6\times 10^{-3}\,\mathrm{m^3} \\ T(t_0) & = 364\,\mathrm{K} \\ R & = 8.31\mathrm{m^3\,Pa\,K^{-1}\,mol^{-1}} \\ \end{aligned}$

substituting,

$\begin{aligned} p(t_0)V(t_0) & = n(t_0)RT(t_0) \\ (4\times 10^5)(6\times 10^{-3}) & = n(t_0)(8.31)(364) \\ n(t_0) & = 0.7722\,\mathrm{mol}\quad\mathrm{(4\, s.f.)}\\ \end{aligned}$

at some point later $t=t_{1}$ the gas in standard temperature and pressure (s.t.p.),

$\begin{aligned} p(t_1) & = 10^5\,\mathrm{Pa} \\ V(t_1) & = 6\times 10^{-3}\,\mathrm{m^3} \\ T(t_1) & = 273\,\mathrm{K} \\ R & = 8.31\mathrm{m^3\,Pa\,K^{-1}\,mol^{-1}} \\ \end{aligned}$

substituting,

$\begin{aligned} p(t_1)V(t_1) & = n(t_1)RT(t_1) \\ (10^5)(6\times 10^{-3}) & = n(t_1)(8.31)(273) \\ n(t_1) & = 0.2645\,\mathrm{mol}\quad\mathrm{(4\, s.f.)}\\ \end{aligned}$

From $n(t_1)<n(t_0)$ , it can be inferred that some portion of the gas was leaking from the open vessel to the atmosphere throughout the experiment.

$\begin{aligned} \frac{m(t_1)}{n(t_1)} & = \frac{m(t_0)}{n(t_0)} = M \\ m(t_1) & = \bigg(\frac{n(t_1)}{n(t_0)}\bigg) m(t_0) \\ & = \bigg(\frac{0.2645}{0.7722}\bigg) m(t_0) \\ m(t_1) & = 0.3425m(t_0) \\ \end{aligned}$

Provided $\rho (t_1)=1.2\,\mathrm{kg\,m^{-3}}$ , we know

$\begin{aligned} \rho (t_1) & = \frac{m(t_1)}{V(t_1)} \\ \rho (t_1) & = \frac{Mn(t_1)}{V(t_1)} \\ 1.2 & = \frac{0.2645M}{6\times 10^{-3}} \\ M & = 0.0272212\,\mathrm{kg\,mol^{-1}}\\ \end{aligned}$

Hence

$\begin{aligned} m(t_1) & =Mn(t_1) \\ & = (0.0272212)(0.2645) \\ & = 0.0072000074\,\mathrm{kg} \\ & = 7.2\,\mathrm{g} \\ \end{aligned}$

And we get option A again, which is far from the official answer C.

Neither $\textrm{\textbf{\scriptsize{CASE I}}}$ nor $\textrm{\textbf{\scriptsize{CASE II}}}$ gives option C; have I actually made a circular argument?

Circular reasoning is a logical fallacy in which the reasoner begins with what they are trying to end with.

_{Wikipedia on Circular reasoning}

Have I really?

No, just that you do so in the way than intended.

July 27, 2019July 25, 2022

201907271536 Solution to 1991-CE-PHY-II-4

$T=\displaystyle{\frac{1}{f}}=1/5=0.2\,\mathrm{s}$ when $f=5\,\mathrm{Hz}$ .

Setup.

Given that from the figure.

Doing substitution for $s$ , $u$ , $a$ , and $t$ in

$s=ut+\displaystyle{\frac{1}{2}}at^2$ ,

I have some equations:

$\begin{aligned} 4 &= s_2 = u_1T + \frac{1}{2}aT^2 \\ 12 &= s_3 = u_1(2T) + \frac{1}{2}a(2T)^2 \\ 24 &= s_4 = u_1(3T) + \frac{1}{2}a(3T)^2 \\ \end{aligned}$

Let $a$ be the subject,

$\begin{aligned} a & = \frac{8}{T^2} - \frac{2u}{T} \\ a & = \frac{6}{T^2} - \frac{u}{T} \\ a & = \frac{16}{3T^2} - \frac{2u}{3T} \end{aligned}$

As is known $T=0.2$ ,

$\begin{aligned} a & = 200- 10u \\ a & = 150 - 5u\\ a & = \frac{400}{3} - \frac{10u}{3} \end{aligned}$

Solving,

$u=10\enspace (\mathrm{cm\, s^{-1}})$ and $a=100\enspace (\mathrm{cm})$ .

And the answer is D.

Roughwork.

Unnecessarily tedious,

,

be simple.

July 25, 2019January 19, 2023

201907251758 Solution to 1980-CE-PHY-II-5

The kinetic energy $E_\mathrm{k}$ of an object of mass $m$ and speed $v$ is given by the formula

$E_\mathrm{k}=\displaystyle{\frac{1}{2}}mv^2$ .

In the situation that the object is thrown upwards with initial speed $u$ , and subjected only to gravity $\mathbf{g}$ , it can be expected that after some time of flight $T$ , the object will return to its initial position, its downward speed in which is equal to the initial upward speed $u$ .

Define a piecewise scalar function $v(t)$ of time $t$ :

$v(t) = \begin{cases} u-gt & \quad \textrm{when }0\leq t\leq \displaystyle{\frac{T}{2}} \\ -u+gt & \quad \textrm{when } \displaystyle{\frac{T}{2}}\leq t\leq T \end{cases}$

or simply

$v:[0,T]\subset \mathbb{R} \rightarrow [0,u]\subset \mathbb{R}$ given by $t\mapsto \big|u-g(T-t)\big|$ .

Then

$\begin{aligned} v^2 & =(u-gt)^2\quad \big( =(-u+gt)^2\big) \\ & = u^2-2ugt+g^2t^2 \quad \big(\forall\, t\in [0,T] \big) \end{aligned}$ .

Thus the kinetic energy $E_\mathrm{k}(t)$ is

$\begin{aligned} E_\mathrm{k}(t) & =\displaystyle{\frac{1}{2}}m(u^2-2ugt+g^2t^2) \\ & = \bigg( \displaystyle{\frac{1}{2}}mg^2 \bigg) t^2 + ( -mug ) t + \bigg( \displaystyle{\frac{1}{2}}mu^2 \bigg) \end{aligned}$

where $m$ , $u$ , $g$ are constants.

The kinetic energy $E_\mathrm{k}(t)$ is set to zero at time $t'$ :

$\begin{aligned} t' & = \displaystyle{\frac{-(-mug)\pm\sqrt{\big( -mug\big)^2-4\big(\frac{1}{2}mg^2\big)\big(\frac{1}{2}mu^2\big)}}{2\big(\frac{1}{2}mg^2\big)}} \\ & = \displaystyle{\frac{mug\pm\sqrt{m^2u^2g^2-m^2u^2g^2}}{mg^2}} \\ & =\displaystyle{\frac{u}{g}} \end{aligned}$