Public Examinations in Physics

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.

August 31, 2023September 2, 2023

202308311431 Solution to 2007-AL-PHY-IIA-9

A diver at a depth of $d$ below the water surface looks up and finds that the sky appears to be within a circle of radius $r$ . Which of the following correctly gives the expression for the critical angle?

A. $\tan c=\displaystyle{\frac{r}{d}}$
B. $\sin c=\displaystyle{\frac{r}{d}}$
C. $\tan c=\displaystyle{\frac{d}{r}}$
D. $\sin c=\displaystyle{\frac{d}{r}}$

Background.

The critical angle is the smallest angle of incidence that yields total reflection, or equivalently the largest angle for which a refracted ray exists.

_{Wikipedia on Critical angle (optics)}

The light travelling from air to water cannot suffer total internal reflection because for total internal reflection, the essential condition is that light should travel from a denser medium to a rarer medium with incidence angle more than the critical angle.

_{Toppr on Total internal reflection}

Roughwork. (making a short story long)

Subscripts $\text{}_a$ and $\text{}_w$ stand for air and water respectively.

$\begin{aligned} n_w\sin \theta_w & = n_a\sin\theta_a \\ \textrm{Eq. (1):\qquad }\frac{n_w}{n_a} & = \frac{\sin\theta_a}{\sin\theta_w}} \\ \frac{\mathrm{d}}{\mathrm{d}t}\bigg( \frac{n_w}{n_a} \bigg) & = \frac{\mathrm{d}}{\mathrm{d}t}\bigg( \frac{\sin\theta_a}{\sin\theta_w}\bigg) \\ 0 & = \frac{\dot{\theta}_a\sin\theta_w\cos\theta_a-\dot{\theta}_w\sin\theta_a\cos\theta_w}{\sin^2\theta_w} \\ \textrm{Eq. (2):\qquad }\frac{\dot{\theta}_a}{\dot{\theta}_w} & = \frac{\tan\theta_a}{\tan\theta_w}=\textrm{non-constant}\\ \end{aligned}$

At some height $h$ above the water surface, where $\theta_a$ is the angle of refraction:

$\begin{aligned} \textrm{Eq. (3):\qquad }\tan\theta_a & = \frac{r_a}{h} \\ \textrm{where\qquad }\theta_a & \in [0,90^\circ] \\ \textrm{and \qquad }h & = \textrm{Const.}\in [0,\infty ) \\ r_a & = r_a(\theta_a) \in [0,\infty )\\ \end{aligned}$

To some depth $d$ beneath the water surface, where $\theta_w$ is the angle of incidence:

$\begin{aligned} \textrm{Eq. (4):\qquad }\tan\theta_w & = \frac{r_w}{d} \\ \textrm{where\qquad }\theta_w & \in [0,90^\circ] \\ \textrm{and \qquad }d & = \textrm{Const.}\in [0,\infty ) \\ r_w & = r_w(\theta_w ) \in [0,\infty ) \\ \end{aligned}$

Differentiating $\textrm{Eq. (3)}$ and $\textrm{Eq. (4)}$ :

$\begin{aligned} \textrm{Eq. (3)':\qquad }h\dot{\theta}_a\sec^2\theta_a & = \dot{r}_a \\ \dot{\theta}_a & = \frac{\dot{r}_a}{h\sec^2\theta_a} \\ \end{aligned}$

$\begin{aligned} \textrm{Eq. (4)':\qquad }d\dot{\theta}_w\sec^2\theta_w & = \dot{r}_w \\ \dot{\theta}_w & = \frac{\dot{r}_w}{d\sec^2\theta_w} \\ \end{aligned}$

Dividing $\textrm{Eq. (3)'}$ by $\textrm{Eq. (4)'}$ :

$\textrm{Eq. (5):\qquad }\displaystyle{\frac{\dot{\theta}_a}{\dot{\theta}_w} =\bigg(\frac{\dot{r}_a}{\dot{r}_w}\bigg) \bigg(\frac{d}{h}\bigg)\bigg(\frac{\sec^2\theta_w}{\sec^2\theta_a}\bigg)}$

Equating $\textrm{Eq. (2)}$ and $\textrm{Eq. (5)}$ :

$\begin{aligned} \textrm{Eq. (6):\qquad }\frac{\tan\theta_a}{\tan\theta_w} & =\bigg(\frac{\dot{r}_a}{\dot{r}_w}\bigg) \bigg(\frac{d}{h}\bigg)\bigg(\frac{\sec^2\theta_w}{\sec^2\theta_a}\bigg) \\ \frac{\tan\theta_a}{\tan\theta_w} & = \frac{(r_a/h)'}{(r_w/d)'}\bigg(\frac{\sec\theta_w}{\sec\theta_a}\bigg)^2\\ \frac{\tan\theta_a}{\tan\theta_w} & = \frac{(\tan\theta_a)'}{(\tan\theta_w)'}\bigg(\frac{\sec\theta_w}{\sec\theta_a}\bigg)^2\\ \frac{\tan\theta_a}{\tan\theta_w} & = \bigg(\frac{\dot{\theta}_a}{\dot{\theta}_w}\bigg)\bigg(\frac{\sec\theta_a}{\sec\theta_w}\bigg)^2\bigg(\frac{\sec\theta_w}{\sec\theta_a}\bigg)^2=\frac{\dot{\theta}_a}{\dot{\theta}_w}\\ \end{aligned}$

This $\textrm{Eq. (6)}$ is redundant as it is the same as $\textrm{Eq. (2)}$ . Never mind. But rewrite $\textrm{Eq. (6)}$ (/ $\textrm{Eq. (2)}$ ) as follows:

$\begin{aligned} \dot{\theta}_w\cot\theta_w & = \dot{\theta}_a\cot\theta_a \\ \frac{\mathrm{d}}{\mathrm{d}t}\big(\ln |\sin \theta_w|\big) & = \frac{\mathrm{d}}{\mathrm{d}t}\big(\ln |\sin \theta_a|\big) \\ \ln|\sin\theta_w| & = \ln|\sin\theta_a| \\ \theta_w & = k\pi -\theta_a \qquad (k\in\mathbb{Z})\\ \textrm{and}\qquad \theta_a & \geq \theta_w \\ \end{aligned}$

Note that refractive index for air is (approximately) $n_a=1$ ,

$\begin{aligned} \textrm{Eq. (7):\qquad }\frac{\sin\theta_a}{\sin\theta_w} & =n_w \\ \sin\theta_a & = n_w \sin\theta_w \\ \sin\theta_a & = n_w\sin (90^\circ-\theta_a) \\ \sin\theta_a & = n_w\cos\theta_a \\ \textrm{Eq. (7)':\qquad }n_w & =\tan\theta_a\\ \sin (90^\circ -\theta_w) & = n_w\sin\theta_w \\ \cos\theta_w & = n_w\sin\theta_w \\ \textrm{Eq. (7)'':\qquad }n_w & = \frac{1}{\tan\theta_w}} \\ \end{aligned}$

Equating $\textrm{Eq. (7)'}$ and $\textrm{Eq. (7)''}$ :

$\begin{aligned} \tan\theta_a & = \frac{1}{\tan\theta_w} \\ \tan\theta_a\tan\theta_w & = 1 \\ \textrm{Eq. (8):\qquad }\tan (90^\circ -\theta_w)\tan\theta_w & = 1 \\ \end{aligned}$

Taking limit on $\textrm{Eq. (8)}$ ,

$\begin{aligned} \Big(\lim_{\theta_w\to c}\tan (90^\circ -\theta_w)\Big)\Big(\lim_{\theta_w\to c}\tan\theta_w\Big) & =1 \\ (\cot c)\bigg(\frac{r}{d}\bigg) & = 1 \\ \tan c & =\frac{r}{d} \\ \end{aligned}$

And the answer is A.

July 24, 2023July 24, 2023

202307241011 Solution to 2023-DSE-PHY-IA-25

Two cells of negligible internal resistance are connected to two resistors as shown.

Roughwork.

Label the nodes $a$ , $b$ , $c$ , and $d$ :

Identify potential differences $\Delta V_{ab}$ , $\Delta V_{cd}$ , $\Delta V_{ac}$ , and $\Delta V_{bd}$ :

Lemma. (Kirchhoff’s 2^nd law (/voltage law/loop rule))

The directed sum of the potential differences (voltages) around any closed loop is zero.

_{Wikipedia on Kirchhoff’s circuit laws}

$\begin{aligned} \Delta V_{ab}+\Delta V_{bd}+\Delta V_{dc}+\Delta V_{ca} & = 0 \\ (3)+((-I)(1))+(-(1.5))+(-(I)(2)) & = 0 \\ I & = +0.5\,\mathrm{A} \\ \end{aligned}$

This problem is not to be attempted.

January 13, 2023January 13, 2023

202301130941 Solution to 2014-IBHL-PHY-I-5

Two blocks of weight $5\,\mathrm{N}$ and $2\,\mathrm{N}$ are attached to two ropes, $\textrm{X}$ and $\textrm{Y}$ .

The blocks hang vertically. The mass of the ropes is negligible. What is the tension in $\textrm{X}$ and the tension in $\textrm{Y}$ ?

Roughwork.

As always start with free-body diagrams. Take upward positive. First the lower block:

Then the higher block:

Or alternatively,

Ans. $\begin{cases} T_\textrm{X}=7\,\mathrm{N} \\ T_\textrm{Y}=2\,\mathrm{N} \\ \end{cases}$

This problem is not to be attempted.

January 13, 2023January 13, 2023

202301130918 Solution to 2011-IBHL-PHY-I-3

A skydiver of mass $80\,\mathrm{kg}$ falls vertically with a constant speed of $50\,\mathrm{m\,s^{-1}}$ . The upward force acting on the skydiver is approximately

A. $0\,\mathrm{N}$ .
B. $80\,\mathrm{N}$ .
C. $800\,\mathrm{N}$ .
D. $4000\,\mathrm{N}$ .

Roughwork.

As always start with a free-body diagram.

Take upward positive. Write the equation of motion by Newton’s 2^nd law:

$\begin{aligned} \mathbf{F}_{\textrm{net}} & = m\mathbf{a} \\ f-mg & = m(0) \\ f & = mg \\ & \approx (80)(10) \\ & = 800 \\ \end{aligned}$

And the answer is C.

This problem is not to be attempted.

January 12, 2023January 17, 2023

202301120908 Solution to 2002-IBHL-PHY-I-9

A pendulum has a bob of mass $m$ and swings in the arc shown below.

As the bob swings through the lowest point of its motion, the tension in the string will be

A. zero.
B. less than $mg$ .
C. equal to $mg$ .
D. greater than $mg$ .

Roughwork.

Assumed the string is $l$ long, taut and inextensible.

Derivation. Read More

$\begin{aligned} T & = \frac{1}{2}ml^2\dot{\theta}^2 \\ V & = mgl(1-\cos\theta ) \\ \mathcal{L} & = T-V \\ \frac{\partial \mathcal{L}}{\partial\theta} & = -mgl\sin\theta \\ \frac{\partial \mathcal{L}}{\partial\dot{\theta}} & = ml^2\dot{\theta} \\ 0 & =\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial\mathcal{L}}{\partial \dot{\theta}}\bigg) - \frac{\partial\mathcal{L}}{\partial \theta} \\ \end{aligned}$

A simple pendulum, under gravity, has

$\textrm{EoM: }\qquad\displaystyle{\ddot{\theta}+\frac{g}{l}\cdot\sin\theta =0}$

the equation of motion, independent of mass $m$ , neglecting friction.

Lemma. (not in use) Read More

$\begin{aligned} \begin{bmatrix} \hat{\mathbf{e}}_r \\ \hat{\mathbf{e}}_\theta \end{bmatrix} & = \begin{bmatrix} \displaystyle{\frac{\partial x}{\partial r}} & \displaystyle{\frac{\partial y}{\partial r}} \\[2mm]\displaystyle{\frac{1}{r}\frac{\partial x}{\partial\theta}} & \displaystyle{\frac{1}{r}\frac{\partial y}{\partial\theta}} \end{bmatrix}\begin{bmatrix} \hat{\mathbf{e}}_x \\ \hat{\mathbf{e}}_y \end{bmatrix} \\ \begin{bmatrix} \hat{\mathbf{e}}_x \\ \hat{\mathbf{e}}_y \end{bmatrix} & = \begin{bmatrix} \displaystyle{\frac{1}{r}\frac{\partial y}{\partial \theta}} & \displaystyle{-\frac{\partial y}{\partial r}} \\[2mm]\displaystyle{-\frac{1}{r}\frac{\partial x}{\partial\theta}} & \displaystyle{\frac{\partial x}{\partial r}} \end{bmatrix} \begin{bmatrix} \hat{\mathbf{e}}_r \\ \hat{\mathbf{e}}_\theta \end{bmatrix} \\ \end{aligned}$

The net force for this non-uniform circular motion, being centripetal force producing radial plus tangential accelerations,

$\mathbf{a} = -l(\dot{\theta}^2\cos\theta +\ddot{\theta}\sin\theta)\,\hat{\mathbf{i}} + l(\ddot{\theta}\cos\theta -\dot{\theta}^2\sin\theta )\,\hat{\mathbf{j}}$

is the resultant of tension

$\begin{aligned} T & = |\mathbf{T}| \\ \mathbf{T} & =T\sin\theta\,\hat{\mathbf{i}}+T\cos\theta\,\hat{\mathbf{j}}} \end{aligned}$

and weight $\mathbf{W}=m\mathbf{g}=-mg\,\hat{\mathbf{j}}$ . Write, with a free-body diagram in mind,

$\begin{aligned} \because\enspace & \mathbf{F}_{\textrm{net}} =\mathbf{T}+\mathbf{W}=m\mathbf{a} \\ \therefore\enspace & \begin{cases} T\sin\theta = -ml(\dot{\theta}^2\cos\theta+\ddot{\theta}\sin\theta ) \\ T\cos\theta - mg = ml(\ddot{\theta}\cos\theta-\dot{\theta}^2\sin\theta ) \\ \end{cases} \\ \end{aligned}$

When the bob is at the lowest point, i.e., $\theta =0$ ,

$\begin{aligned} T & = mg + ml\ddot{\theta} \\ & = mg - mg\sin\theta \qquad (\textrm{by EoM})\\ & \approx mg-mg\theta \qquad (\because\enspace \sin\theta\approx\theta\textrm{ for small }\theta)\\ & \lneq mg \\ \end{aligned}$

The official answer is D but mine B. Could that be E: more or less $mg$ ? If I am wrong, I must have omitted some simple facts of noble truth.

This problem is not to be attempted.

January 11, 2023January 11, 2023

202301111028 Solution to 2003-IBHL-I-4

Two forces of magnitudes $7\,\mathrm{N}$ and $5\,\mathrm{N}$ act at a point. Which one of the following is not a possible value for the magnitude of the resultant force?

A. $1\,\mathrm{N}$
B. $3\,\mathrm{N}$
C. $5\,\mathrm{N}$
D. $7\,\mathrm{N}$

Roughwork.

Suppose

$\begin{aligned} \mathbf{F}_1 & = 7\sin\alpha\,\hat{\mathbf{i}} + 7\cos\alpha\,\hat{\mathbf{j}} \\ \mathbf{F}_2 & = 5\sin\beta\,\hat{\mathbf{i}} + 5\cos\beta\,\hat{\mathbf{j}} \\ \end{aligned}$

where $\alpha ,\beta\in [0,2\pi )$ .

$\begin{aligned} \mathbf{F}_3 & = \mathbf{F}_1+\mathbf{F}_2 \\ & = (7\sin\alpha + 5\sin\beta )\,\hat{\mathbf{i}} + (7\cos\alpha +5\cos\beta )\,\hat{\mathbf{j}} \\ F_3 & = |\mathbf{F}_3| \\ & = \sqrt{(7\sin\alpha + 5\sin\beta )^2 + (7\cos\alpha +5\cos\beta )^2}\\ & = \sqrt{74+70(\sin\alpha\sin\beta +\cos\alpha\cos\beta )}\\ & = \sqrt{74+70\cos (\alpha-\beta)} \\ \end{aligned}$

$\begin{aligned} \because \quad & \begin{cases} \max (F_3) = \sqrt{74+70(1)} = 12 \\ \min (F_3) = \sqrt{74+70(-1)} = 2 \\ \end{cases} \\ \therefore\quad & F_3 \in [2,12] \\ \end{aligned}$

One can actually do without vector analysis for the sake of a mental exercise.

This problem is not to be attempted.

January 11, 2023January 11, 2023

202301110923 Solution to 1980-AL-PHY-IA-26

A charged ball $X$ is suspended by a string.

When a uniform electric field $E$ is applied horizontally, the ball is displaced a horizontal distance $a$ such that the string makes an angle $\theta$ with the vertical.

Roughwork.

As always start with a free-body diagram. Hence, draw

Resolving components, write

$\begin{cases} \textrm{Horizontal:}\quad & T\sin\theta = qE \\ \textrm{Vertical:}\quad & T\cos\theta = mg \\ \end{cases}$

Squaring and summing, write

$\begin{aligned} (T\sin\theta )^2+ (T\cos\theta)^2 &= (qE)^2 + (mg)^2\\ T & = \sqrt{q^2E^2+m^2g^2} \\ \end{aligned}$

By considering the electric field strength $E$ as an independent variable, and the angle of elevation $\theta$ the dependent variable, make differentiation

$\displaystyle{\mathrm{d}E = \bigg(\frac{T\cos\theta}{q}\bigg)\,\mathrm{d}\theta}\qquad(\theta\in [0,90^\circ])$

This problem is not to be attempted.

January 10, 2023January 10, 2023

202301101150 Solution to 1999-IBHL-PHY-I-7

Two $10\,\mathrm{kg}$ blocks on a smooth horizontal surface are tied together. They are accelerated by a horizontal force of $30\,\mathrm{N}$ which acts as shown below:

If frictional effects are negligible, what is the tension in the connecting rope?

Roughwork.

As always start with free-body diagrams. Take rightward positive. Considering first the bodies as a whole:

so as to write the equation of motion by Newton 2^nd law:

$\begin{aligned} F_{\textrm{net}} & =ma \\ 30 & = 20a\\ a & = 1.5\,\mathrm{m\,s^{-2}} \\ \end{aligned}$

where the tension regarded as internal force does not count. Considering then as individuals,

so as to write:

$\begin{aligned} F_{\textrm{net}} & =ma \\ 30 - T & = (10)(1.5)\\ T & = 15\,\mathrm{N} \\ \end{aligned}$

One may also do with the left block, but an exercise left the reader.