2025 HKDSE Physics Practice Paper with Answers

Total electrical energy supplied by the heater:
\[ E_{\text{total}} = P \times t = 150 \times (6 \times 60) = 54\,000\text{ J} \]
Since \(20\%\) of energy is lost, the useful heat absorbed by the liquid is:
\[ Q = E_{\text{total}} \times (1 - 0.20) = 54\,000 \times 0.80 = 43\,200\text{ J} \]
Using \(Q = mc\Delta T\):
\[ 43\,200 = 0.50 \times c \times (50 - 20) = 15 \times c \]
\[ c = \frac{43\,200}{15} = 2880\text{ J kg}^{-1}\text{ }^\circ\text{C}^{-1} \]

B (1 mark)

Initial kinetic energy of the car:
\[ E_k = \frac{1}{2} m u^2 = \frac{1}{2} (0.80)(4.0)^2 = 6.4\text{ J} \]
Total work done against the resistive force is \(\Delta E_k = 6.4\text{ J}\).
For uniform deceleration to rest:
\[ s = \frac{u + v}{2} t \implies 3.2 = \frac{4.0 + 0}{2} t \implies t = 1.6\text{ s} \]
Average power dissipated:
\[ P_{\text{avg}} = \frac{\Delta E_k}{t} = \frac{6.4\text{ J}}{1.6\text{ s}} = 4.0\text{ W} \]

B (1 mark)

(1) By conservation of linear momentum (taking right as positive):
\[ m_A u_A + m_B u_B = m_A v_A + m_B v_B \]
\[ (2.0)(3.0) + 0 = (2.0)(1.0) + (1.0)(v_B) \implies 6.0 = 2.0 + v_B \implies v_B = 4.0\text{ m s}^{-1} \]
Statement (1) is correct.

(2) Total initial kinetic energy:
\[ E_{k1} = \frac{1}{2}(2.0)(3.0)^2 = 9.0\text{ J} \]
Total final kinetic energy:
\[ E_{k2} = \frac{1}{2}(2.0)(1.0)^2 + \frac{1}{2}(1.0)(4.0)^2 = 1.0 + 8.0 = 9.0\text{ J} \]
Since total kinetic energy is conserved, the collision is elastic. Statement (2) is correct.

(3) By Newton's third law, the interaction forces are equal in magnitude and act for the same time, so the impulses received have equal magnitudes (\(|\Delta p_A| = |2.0(1.0 - 3.0)| = 4.0\text{ N s}\) and \(|\Delta p_B| = |1.0(4.0 - 0)| = 4.0\text{ N s}\)). Statement (3) is correct.

D (1 mark)

Using the interference fringe separation formula:
\[ \Delta y = \frac{\lambda D}{a} \]
where \(\Delta y = 0.85\text{ m}\), \(D = 4.0\text{ m}\), and \(a = 1.6\text{ m}\).
\[ \lambda = \frac{a \Delta y}{D} = \frac{1.6 \times 0.85}{4.0} = 0.34\text{ m} \]
Using the wave equation \(v = f \lambda\):
\[ f = \frac{v}{\lambda} = \frac{340\text{ m s}^{-1}}{0.34\text{ m}} = 1000\text{ Hz} \]

C (1 mark)

For the initial setup:
\[ u = 15\text{ cm}, \quad m = \frac{v}{u} = 2 \implies v = 30\text{ cm} \]
Using the lens formula:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{15} + \frac{1}{30} = \frac{3}{30} = \frac{1}{10} \implies f = 10\text{ cm} \]
For the new setup with \(u' = 30\text{ cm}\):
\[ \frac{1}{v'} = \frac{1}{f} - \frac{1}{u'} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15} \implies v' = 15\text{ cm} \]
Since \(v' > 0\) and \(u' > 2f\), the new image is real, inverted, diminished, and formed at \(15\text{ cm}\) on the opposite side of the lens.

A (1 mark)

Terminal potential difference is given by:
\[ V = \mathcal{E} \left(\frac{R}{R + r}\right) \]
For the first case:
\[ 8.0 = \mathcal{E} \left(\frac{4.0}{4.0 + r}\right) \implies \mathcal{E} = 2.0(4.0 + r) = 8.0 + 2.0r \quad \text{--- (1)} \]
For the second case:
\[ 9.0 = \mathcal{E} \left(\frac{9.0}{9.0 + r}\right) \implies \mathcal{E} = 9.0 + r \quad \text{--- (2)} \]
Equating (1) and (2):
\[ 8.0 + 2.0r = 9.0 + r \implies r = 1.0\ \Omega \]
Substituting \(r = 1.0\ \Omega\) into (2):
\[ \mathcal{E} = 9.0 + 1.0 = 10.0\text{ V} \]

A (1 mark)

According to Faraday's law of electromagnetic induction:
\[ |\varepsilon| = N \frac{\Delta \Phi}{\Delta t} = N A \frac{\Delta B}{\Delta t} \]
Given:
\[ N = 200, \quad A = 5.0 \times 10^{-3}\text{ m}^2, \quad \Delta B = 0.40\text{ T} - 0 = 0.40\text{ T}, \quad \Delta t = 0.050\text{ s} \]
\[ |\varepsilon| = 200 \times (5.0 \times 10^{-3}) \times \frac{0.40}{0.050} = 1.0 \times 8.0 = 8.0\text{ V} \]

D (1 mark)

Let \(N_0\) be the initial number of \(X\) nuclei.
At time \(t\), the number of remaining \(X\) nuclei is \(N_X\) and the number of formed \(Y\) nuclei is \(N_Y = N_0 - N_X\).
Given:
\[ \frac{N_Y}{N_X} = \frac{N_0 - N_X}{N_X} = 7 \implies \frac{N_0}{N_X} - 1 = 7 \implies \frac{N_X}{N_0} = \frac{1}{8} \]
Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), exactly \(3\) half-lives have passed.
\[ t = 3 \times T_{1/2} = 3 \times 4.0\text{ hours} = 12\text{ hours} \]

B (1 mark)

The rate of heat supplied is given by $P = m c \frac{\Delta T}{\Delta t}$, which rearranges to $\frac{\Delta T}{\Delta t} = \frac{P}{m c}$. Since the power $P$ and mass $m$ are identical for both blocks, the rate of temperature rise is inversely proportional to specific heat capacity $c$. Therefore, $\frac{c_X}{c_Y} = \frac{(\Delta T / \Delta t)_Y}{(\Delta T / \Delta t)_X} = \frac{1}{1.5} = \frac{2}{3}$.

1 mark for option A. [1A for recognizing inverse proportionality between rate of temperature rise and specific heat capacity: $c_X / c_Y = 1 / 1.5 = 2 : 3$].

During the powered ascent ($0 \le t \le 4.0\text{ s}$):
Velocity at burnout: $v_1 = u + a t = 0 + (2.5)(4.0) = 10.0\text{ m s}^{-1}$.
Height attained at burnout: $s_1 = \frac{1}{2} a t^2 = \frac{1}{2}(2.5)(4.0)^2 = 20.0\text{ m}$.
During free flight under gravity until top of trajectory ($v = 0$):
$v^2 = v_1^2 - 2 g s_2 \implies 0 = 10.0^2 - 2(9.81) s_2 \implies s_2 = \frac{100}{19.62} \approx 5.10\text{ m}$.
Total maximum height $H = s_1 + s_2 = 20.0 + 5.10 = 25.1\text{ m}$.

1 mark for option B. [1A for correctly summing powered ascent distance $20.0\text{ m}$ and free-flight distance $5.1\text{ m}$ to obtain $25.1\text{ m}$].

Let $M = 4.0\text{ kg}$ and $m = 1.0\text{ kg}$.
Friction acting on $M$: $f_k = \mu_k M g = 0.15 \times 4.0 \times 9.81 = 5.886\text{ N}$.
Equation of motion for the system: $(M + m)a = m g - f_k \implies 5.0 a = (1.0)(9.81) - 5.886 = 3.924\text{ N} \implies a = 0.7848\text{ m s}^{-2}$.
Tension in the string: $T = m(g - a) = 1.0 \times (9.81 - 0.7848) \approx 9.03\text{ N} \approx 9.0\text{ N}$.

1 mark for option C. [1A for determining system acceleration $a = 0.785\text{ m s}^{-2}$ and computing $T = m(g - a) \approx 9.0\text{ N}$].

Wavelength $\lambda = \frac{v}{f} = \frac{4.0}{2.0} = 2.0\text{ m}$.
(1) Phase difference $\Delta \phi = \frac{2\pi}{\lambda}\Delta x = \frac{2\pi}{2.0} \times 0.50 = \frac{\pi}{2}\text{ rad}$. Statement (1) is correct.
(2) In one period $T$, the wave travels one wavelength $\lambda = 2.0\text{ m}$. The particle vibrates through a total distance of $4A$ (where $A$ is amplitude), which is generally not equal to $\lambda$. Statement (2) is incorrect.
(3) By definition of a transverse wave, the particle oscillation is perpendicular to the direction of propagation. Statement (3) is correct.
Therefore, only (1) and (3) are correct.

1 mark for option C. [1A for correctly evaluating statements (1) and (3) as true and (2) as false].

Applying Snell's law at the liquid-glass boundary:
$n_{\text{liquid}} \sin 45^\circ = n_{\text{glass}} \sin 30^\circ$
$n_{\text{liquid}} \left(\frac{1}{\sqrt{2}}\right) = 1.60 \times 0.50 = 0.80$
$n_{\text{liquid}} = 0.80 \sqrt{2} \approx 1.1314$.
At the liquid-air interface, the critical angle $C$ satisfies:
$\sin C = \frac{n_{\text{air}}}{n_{\text{liquid}}} = \frac{1.00}{0.80\sqrt{2}} \approx 0.8839$
$C = \sin^{-1}(0.8839) \approx 62.1^\circ$.

1 mark for option C. [1A for computing $n_{\text{liquid}} = 0.80\sqrt{2} \approx 1.131$ and finding $C = \sin^{-1}(1/n) = 62.1^\circ$].

Terminal potential difference $V = E - I r = E \frac{R}{R + r}$.
For $R_1 = 1.0\,\Omega$ and $V_1 = 3.0\text{ V}$:
$3.0 = \frac{E (1.0)}{1.0 + r} \implies E = 3.0(1.0 + r) = 3.0 + 3.0r$ --- (1)
For $R_2 = 4.0\,\Omega$ and $V_2 = 4.8\text{ V}$:
$4.8 = \frac{E (4.0)}{4.0 + r} \implies 4.8(4.0 + r) = 4.0E \implies 19.2 + 4.8r = 4.0E$ --- (2)
Substituting (1) into (2):
$19.2 + 4.8r = 4.0(3.0 + 3.0r) = 12.0 + 12.0r$
$7.2 = 7.2r \implies r = 1.0\,\Omega$.
Then $E = 3.0(1.0 + 1.0) = 6.0\text{ V}$.

1 mark for option B. [1A for setting up simultaneous equations in $E$ and $r$ and solving for $E = 6.0\text{ V}$].

As the loop leaves the field, the rate of change of magnetic flux is $\frac{\Delta \Phi}{\Delta t} = B w v$.
The induced electromotive force is $\varepsilon = B w v$, resulting in an induced current $I = \frac{\varepsilon}{R} = \frac{B w v}{R}$.
The magnetic braking force on the leading edge within the field is $F_B = I w B = \left(\frac{B w v}{R}\right) w B = \frac{B^2 w^2 v}{R}$.
To move at constant velocity, the external applied force must balance this magnetic force, so $F_{\text{ext}} = \frac{B^2 w^2 v}{R}$.

1 mark for option C. [1A for calculating induced current $I = B w v / R$ and force $F = I w B = B^2 w^2 v / R$].

The activities as functions of time $t$ (in hours) are:
$A_P(t) = 640 \times \left(\frac{1}{2}\right)^{t / 3.0}$
$A_Q(t) = 80 \times \left(\frac{1}{2}\right)^{t / 6.0}$
Setting $A_P(t) = A_Q(t)$:
$640 \times 2^{-t / 3} = 80 \times 2^{-t / 6}$
$\frac{640}{80} = 2^{t/3 - t/6}$
$8 = 2^{t/6}$
$2^3 = 2^{t/6} \implies \frac{t}{6} = 3 \implies t = 18\text{ hours}$.

1 mark for option C. [1A for equating activity expressions $640 \times 2^{-t/3} = 80 \times 2^{-t/6}$ and solving for $t = 18\text{ hours}$].

Heat gained by water:
$$Q_{\text{gain}} = m_w c_w \Delta T_w = (0.25)(4200)(36 - 20) = 16800\text{ J}$$
Assuming no heat lost to surroundings:
$$Q_{\text{lost}} = m_m c_m \Delta T_m = (0.50) c_m (100 - 36) = 32 c_m$$
Equating heat lost and heat gained:
$$32 c_m = 16800 \implies c_m = 525\text{ J kg}^{-1}\ ^\circ\text{C}^{-1}$$
Hence, the correct option is C.

1 mark for correct calculation of heat exchange leading to C.

(1) By the principle of conservation of linear momentum, in the absence of net external force, total momentum is conserved. Thus, (1) is correct.
(2) The collision is completely inelastic because the two trolleys stick together, so mechanical/kinetic energy is not conserved (converted into internal energy/sound). Thus, (2) is incorrect.
(3) By conservation of momentum:
$$m u = (m + 2m) v \implies v = \frac{u}{3}$$
Initial kinetic energy:
$$E_{k1} = \frac{1}{2} m u^2$$
Final kinetic energy:
$$E_{k2} = \frac{1}{2}(3m) v^2 = \frac{1}{2}(3m)\left(\frac{u}{3}\right)^2 = \frac{1}{6} m u^2 = \frac{1}{3} E_{k1}$$
Thus, the ratio of total kinetic energy after to before is $1:3$. So, (3) is correct.
Therefore, (1) and (3) only are correct.

1 mark for identifying statements (1) and (3) as correct.

(1) By Kepler's third law, $T^2 \propto R^3$. Therefore:
$$T' = T \left(\frac{4R}{R}\right)^{3/2} = T (4)^{3/2} = 8T$$
So (1) is correct.

(2) By Newton's law of universal gravitation, $F = \frac{GMm}{R^2}$. When the radius becomes $4R$:
$$F' = \frac{GMm}{(4R)^2} = \frac{1}{16} \frac{GMm}{R^2} = \frac{1}{16}F$$
So (2) is correct.

(3) For a circular orbit, $\frac{mv^2}{R} = \frac{GMm}{R^2} \implies E_k = \frac{1}{2}mv^2 = \frac{GMm}{2R}$. When radius is $4R$:
$$E_k' = \frac{GMm}{2(4R)} = \frac{1}{4} E_k$$
So (3) is correct.

Therefore, (1), (2) and (3) are all correct.

1 mark for identifying that all three statements (1), (2), and (3) are correct.

Applying Snell's law at the interface:
$$n_1 \sin 45^\circ = n_2 \sin 30^\circ$$
$$\frac{n_1}{n_2} = \frac{\sin 30^\circ}{\sin 45^\circ} = \frac{0.5}{1/\sqrt{2}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$$
Since $n_2 > n_1$, total internal reflection can occur when light travels from medium 2 to medium 1.
The critical angle $C$ is given by:
$$\sin C = \frac{n_1}{n_2} = \frac{1}{\sqrt{2}} \implies C = 45^\circ$$
Thus, the critical angle is $45^\circ$.

1 mark for applying Snell's law and critical angle formula to obtain 45 degrees.

The fringe separation is given by $\Delta y = \frac{\lambda D}{a}$, where $\lambda$ is the wavelength, $D$ is the distance to the screen, and $a$ is the slit separation.
(1) Doubling $a$ gives $\Delta y' = \frac{\lambda D}{2a} = \frac{1}{2}\Delta y$ (halved). Thus (1) is correct.
(2) Doubling $D$ gives $\Delta y' = \frac{\lambda (2D)}{a} = 2\Delta y$ (doubled). Thus (2) is incorrect.
(3) Since $c = f\lambda$, doubling the frequency $f$ halves the wavelength $\lambda' = \frac{\lambda}{2}$. Then $\Delta y' = \frac{(\lambda/2) D}{a} = \frac{1}{2}\Delta y$ (halved). Thus (3) is correct.
Therefore, (1) and (3) only will halve the fringe separation.

1 mark for identifying (1) and (3) as correct.

Using the equation $\mathcal{E} = I(R + r)$ for the two cases:
Case 1: $\mathcal{E} = 2.0(3.0 + r) = 6.0 + 2.0r$
Case 2: $\mathcal{E} = 1.0(8.0 + r) = 8.0 + 1.0r$
Equating the two expressions for $\mathcal{E}$:
$$6.0 + 2.0r = 8.0 + 1.0r$$
$$1.0r = 2.0 \implies r = 2.0\ \Omega$$
Hence, the correct option is B.

1 mark for setting up simultaneous equations and solving for r = 2.0 ohms.

Let the direction out of the page be positive.
Initial magnetic flux density: $B_1 = -0.20\text{ T}$
Final magnetic flux density: $B_2 = +0.60\text{ T}$
Change in magnetic field:
$$\Delta B = B_2 - B_1 = 0.60 - (-0.20) = 0.80\text{ T}$$
Change in magnetic flux linkage per turn:
$$\Delta \Phi = A \Delta B = (0.040\text{ m}^2)(0.80\text{ T}) = 0.032\text{ Wb}$$
By Faraday's law of electromagnetic induction, the magnitude of the induced e.m.f. is:
$$\varepsilon = N\frac{\Delta \Phi}{\Delta t} = 50 \times \frac{0.032\text{ Wb}}{0.40\text{ s}} = 4.0\text{ V}$$
Hence, the magnitude of the induced e.m.f. is $4.0\text{ V}$.

1 mark for applying Faraday's law with total delta B = 0.80 T to obtain 4.0 V.

The elapsed time is $t = 18.0\text{ hours}$, which corresponds to:
$$n = \frac{18.0\text{ h}}{6.0\text{ h}} = 3\text{ half-lives}$$
Let the initial number of nuclei of $X$ be $N_0$.
The remaining number of nuclei of $X$ after 3 half-lives is:
$$N_X = N_0 \left(\frac{1}{2}\right)^3 = \frac{1}{8} N_0$$
Since each decayed nucleus of $X$ forms one stable nucleus of $Y$, the number of nuclei of $Y$ formed is:
$$N_Y = N_0 - N_X = N_0 - \frac{1}{8} N_0 = \frac{7}{8} N_0$$
Therefore, the ratio of the number of nuclei of $Y$ to $X$ is:
$$\frac{N_Y}{N_X} = \frac{7/8 N_0}{1/8 N_0} = 7 : 1$$
Hence, the correct option is B.

1 mark for determining the remaining and decayed fractions after 3 half-lives to find the ratio 7 : 1.

By conservation of energy, heat lost by metal = heat gained by liquid:
\[ m_m c_m (100 - T) = m_l c_l (T - 20) \]
\[ (0.50)(400)(100 - T) = (0.20)(2500)(T - 20) \]
\[ 200(100 - T) = 500(T - 20) \]
\[ 2(100 - T) = 5(T - 20) \]
\[ 200 - 2T = 5T - 100 \]
\[ 7T = 300 \implies T \approx 42.9^\circ\text{C} \]

Correct Answer: B (1 mark)

Using the equation of motion \(v^2 = u^2 - 2gs\):
At height \(h\):
\[ \left(\frac{u}{3}\right)^2 = u^2 - 2gh \implies \frac{u^2}{9} = u^2 - 2gh \implies 2gh = \frac{8}{9}u^2 \implies u^2 = \frac{9}{4}gh \]
At the maximum height \(H\), the speed is \(0\):
\[ 0 = u^2 - 2gH \implies H = \frac{u^2}{2g} = \frac{\frac{9}{4}gh}{2g} = \frac{9}{8}h \]

Correct Answer: B (1 mark)

Using the grating equation \(d \sin\theta = n\lambda\):
\[ d \sin(30.0^\circ) = 3 \times (600 \times 10^{-9}\text{ m}) \]
\[ d(0.50) = 1.80 \times 10^{-6}\text{ m} \implies d = 3.60 \times 10^{-6}\text{ m} \]
The number of lines per millimetre is:
\[ N = \frac{1\text{ mm}}{d} = \frac{10^{-3}\text{ m}}{3.60 \times 10^{-6}\text{ m}} \approx 278\text{ lines mm}^{-1} \]

Correct Answer: B (1 mark)

Across parallel boundaries, Snell's law gives:
\[ n_P \sin\theta_P = n_Q \sin\theta_Q = n_R \sin\theta_R \]
For total internal reflection to just occur at the \(Q-R\) boundary, the angle of refraction in medium \(R\) is \(\theta_R = 90^\circ\):
\[ n_P \sin\theta_P = n_R \sin(90^\circ) \]
\[ 1.80 \sin(60^\circ) = n_R \times 1 \]
\[ n_R = 1.80 \times \frac{\sqrt{3}}{2} \approx 1.56 \]

Correct Answer: C (1 mark)

As the resistance of the variable resistor increases:
1. Total resistance of the circuit increases, so the circuit current \(I\) decreases.
2. Terminal voltage \(V = \mathcal{E} - Ir\). As \(I\) decreases, the lost volts \(Ir\) decrease, so \(V\) increases.
3. Power dissipated in the fixed resistor is \(P = I^2 R\). Since \(I\) decreases and \(R\) is fixed, \(P\) decreases.

Correct Answer: B (1 mark)

(1) When entering, the magnetic flux into the page increases. By Lenz's law, the induced current produces a magnetic field directed out of the page to oppose the change, which corresponds to an anticlockwise current. (Correct)
(2) When the loop is entirely inside the uniform magnetic field, the magnetic flux through the loop remains constant (\(\Delta \Phi / \Delta t = 0\)), so the induced e.m.f. and induced current are zero. (Correct)
(3) According to Lenz's law, the induced effects always oppose the motion producing them. Therefore, as the loop leaves the field, the magnetic force acts upwards to oppose the downward fall. (Correct)
Thus, (1), (2), and (3) are all correct.

Correct Answer: D (1 mark)

For a satellite of mass \(m\) in a circular orbit of radius \(r\) around Earth of mass \(M\):
\[ \frac{G M m}{r^2} = \frac{m v^2}{r} \implies E_k = \frac{1}{2} m v^2 = \frac{G M m}{2r} \]
Therefore, the ratio of kinetic energies is:
\[ \frac{E_{k,A}}{E_{k,B}} = \frac{m_A / r_A}{m_B / r_B} = \left(\frac{m_A}{m_B}\right) \left(\frac{r_B}{r_A}\right) = \left(\frac{1}{2}\right) (4) = 2 = 2 : 1 \]

Correct Answer: C (1 mark)

Let \(N_0\) be the initial number of \(X\) nuclei.
Since each decayed nucleus of \(X\) becomes a nucleus of \(Y\):
\[ N_X + N_Y = N_0 \]
Given \(N_Y / N_X = 7\), we have \(N_Y = 7 N_X\):
\[ N_X + 7 N_X = N_0 \implies 8 N_X = N_0 \implies N_X = \frac{1}{8} N_0 = \left(\frac{1}{2}\right)^3 N_0 \]
This corresponds to 3 half-lives.
Thus, \(t = 3 \times 4.0\text{ hours} = 12.0\text{ hours}\).

Correct Answer: B (1 mark)

Let \(t\) be the time of flight of the ball.

For the vertical motion under gravity from rest:
\[ h = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2h}{g}} \]
The vertical component of velocity just before landing is:
\[ v_y = gt = g\sqrt{\frac{2h}{g}} = \sqrt{2gh} \]
Since the ball moves at an angle of \(45^\circ\) to the horizontal upon landing, the magnitude of the horizontal velocity \(v_x\) is equal to the magnitude of the vertical velocity \(v_y\):
\[ \tan 45^\circ = \frac{v_y}{v_x} = 1 \implies v_x = v_y = \sqrt{2gh} \]
Since there is no horizontal acceleration, the initial projection speed is \(u = v_x = \sqrt{2gh}\).

The horizontal distance \(s_x\) travelled is:
\[ s_x = u \times t = \sqrt{2gh} \times \sqrt{\frac{2h}{g}} = \sqrt{4h^2} = 2h \]
Therefore, the correct option is D.

Correct option: D (1 mark)

(a) Energy absorbed by the water:
\[ Q = mc\Delta T = (0.80)(4200)(100 - 24) = 2.5536 \times 10^5\text{ J} \approx 2.55 \times 10^5\text{ J} \]

(b) Total electrical energy supplied:
\[ E_{\text{in}} = P t = (1800)(150) = 2.70 \times 10^5\text{ J} \]
Total heat lost:
\[ Q_{\text{loss}} = E_{\text{in}} - Q = 2.70 \times 10^5 - 2.5536 \times 10^5 = 14640\text{ J} \]
Average rate of heat loss:
\[ \text{Rate} = \frac{Q_{\text{loss}}}{t} = \frac{14640}{150} = 97.6\text{ W} \]
(Using unrounded values gives \(97.6\text{ W}\); accept \(97.3\text{ W}\) to \(98.0\text{ W}\)).

(c) Using a double-walled / vacuum-insulated body (or having a polished/shiny exterior surface to reduce radiation).

(a)
1M: Use of \(Q = mc\Delta T\)
1A: \(2.55 \times 10^5\text{ J}\) (or \(255\text{ kJ}\))

(b)
1M: Finding total energy supplied \(Pt\) and computing \(Q_{\text{loss}} / t\) or \(P - mc\Delta T/t\)
1A: \(97.6\text{ W}\) (accept \(97.3\text{ W} - 98.0\text{ W}\))

(c)
1A: Double-walled vacuum structure / plastic casing / shiny reflective exterior body (any ONE valid reason)

(a) Considering vertical motion from rest:
\[ s_y = \frac{1}{2}gt^2 \implies 1.25 = \frac{1}{2}(9.81)t^2 \]
\[ t = \sqrt{\frac{2 \times 1.25}{9.81}} = 0.5048\text{ s} \approx 0.505\text{ s} \]

(b) Horizontal speed:
\[ u_x = \frac{s_x}{t} = \frac{3.0}{0.5048} = 5.943\text{ m s}^{-1} \approx 5.94\text{ m s}^{-1} \]

(c) Vertical velocity just before hitting the ground:
\[ v_y = gt = (9.81)(0.5048) = 4.952\text{ m s}^{-1} \text{ (downwards)} \]
Magnitude of velocity:
\[ v = \sqrt{u_x^2 + v_y^2} = \sqrt{(5.943)^2 + (4.952)^2} = 7.736\text{ m s}^{-1} \approx 7.74\text{ m s}^{-1} \]
Direction:
\[ \theta = \tan^{-1}\left(\frac{v_y}{u_x}\right) = \tan^{-1}\left(\frac{4.952}{5.943}\right) = 39.8^\circ \text{ below the horizontal} \]

(a)
1M: Use of \(h = \frac{1}{2}gt^2\)
1A: \(t = 0.505\text{ s}\) (accept \(0.50\text{ s} - 0.51\text{ s}\))

(b)
1M: Use of \(u_x = s_x / t\)
1A: \(5.94\text{ m s}^{-1}\) (or e.c.f. from (a))

(c)
1M: Finding \(v_y = gt\) and combining components \(v = \sqrt{v_x^2 + v_y^2}\)
1A: \(7.74\text{ m s}^{-1}\) at \(39.8^\circ\) below the horizontal (accept \(39.7^\circ - 40.0^\circ\))

(a) The forces are:
1. Weight \(W = mg\) acting vertically downwards.
2. Normal reaction \(N\) acting vertically upwards.
Equation of motion:
\[ mg - N = \frac{mv^2}{R} \]

(b) Normal reaction \(N\):
\[ N = m\left(g - \frac{v^2}{R}\right) = 1200\left(9.81 - \frac{14.0^2}{45}\right) \]
\[ N = 1200(9.81 - 4.3556) = 1200(5.4544) = 6545.3\text{ N} \approx 6550\text{ N} \text{ (or } 6.55\text{ kN)} \]

(c) At the verge of losing contact, \(N = 0\):
\[ mg = \frac{mv_{\max}^2}{R} \implies v_{\max} = \sqrt{gR} \]
\[ v_{\max} = \sqrt{(9.81)(45)} = 21.01\text{ m s}^{-1} \approx 21.0\text{ m s}^{-1} \]

(a)
1A: Identifying weight downwards and normal reaction upwards
1A: Correct net force equation \(mg - N = \frac{mv^2}{R}\)

(b)
1M: Correct substitution into the equation for \(N\)
1A: \(N = 6550\text{ N}\) (accept \(6540\text{ N} - 6550\text{ N}\))

(c)
1M: Setting \(N = 0\) to get \(v = \sqrt{gR}\)
1A: \(v_{\max} = 21.0\text{ m s}^{-1}\)

(a) Wavelength:
\[ \lambda = \frac{v}{f} = \frac{340}{680} = 0.500\text{ m} \]

(b) Using the fringe separation formula for double-source interference:
\[ \Delta y = \frac{\lambda D}{a} \]
where \(D = 4.00\text{ m}\), \(a = 1.50\text{ m}\), and \(\lambda = 0.500\text{ m}\):
\[ \Delta y = \frac{(0.500)(4.00)}{1.50} = 1.333\text{ m} \approx 1.33\text{ m} \]

(c) By \(v = f\lambda\), when frequency \(f\) increases, the wavelength \(\lambda\) decreases since the wave speed \(v\) remains constant.
Since fringe spacing \(\Delta y = \frac{\lambda D}{a} \propto \lambda\), the separation between consecutive maxima will decrease.

(a)
1M: Use of \(v = f\lambda\)
1A: \(\lambda = 0.500\text{ m}\)

(b)
1M: Use of \(\Delta y = \frac{\lambda D}{a}\)
1A: \(\Delta y = 1.33\text{ m}\) (accept \(\frac{4}{3}\text{ m}\))

(c)
1A: Stating that wavelength decreases as frequency increases
1A: Explaining that fringe spacing decreases because \(\Delta y \propto \lambda\)

(a) By Snell's Law:
\[ n_{\text{air}} \sin i = n_{\text{glass}} \sin r \]
\[ (1.00)\sin(50.0^\circ) = 1.52 \sin r \]
\[ \sin r = \frac{\sin 50.0^\circ}{1.52} = \frac{0.7660}{1.52} = 0.5040 \]
\[ r = \sin^{-1}(0.5040) = 30.26^\circ \approx 30.3^\circ \]

(b) Critical angle \(C\):
\[ \sin C = \frac{1}{n} = \frac{1}{1.52} = 0.6579 \]
\[ C = \sin^{-1}(0.6579) = 41.14^\circ \approx 41.1^\circ \]

(c) No, total internal reflection will not occur.
The ray emerges from the center of the flat side and travels along a radius of the semicircle. Therefore, it strikes the curved surface normally (at an angle of incidence of \(0^\circ\)), which is less than the critical angle \(C\), so it passes straight out without deviation.

(a)
1M: Use of \(n_1 \sin \theta_1 = n_2 \sin \theta_2\)
1A: \(r = 30.3^\circ\) (accept \(30.2^\circ - 30.3^\circ\))

(b)
1M: Use of \(\sin C = \frac{1}{n}\)
1A: \(C = 41.1^\circ\)

(c)
1A: Stating that the light ray travels along the radius / hits normally (angle of incidence \(= 0^\circ\))
1A: Concluding no total internal reflection occurs because the incident angle is less than the critical angle

(a) By the potential divider formula:
\[ V_1 = V_{\text{total}} \times \frac{R_1}{R_1 + R_{\text{LDR}}} = 12.0 \times \frac{4.0}{4.0 + 1.0} = 12.0 \times \frac{4}{5} = 9.60\text{ V} \]

(b) When \(V_1 = 3.0\text{ V}\), the voltage across the LDR is:
\[ V_{\text{LDR}} = 12.0 - 3.0 = 9.0\text{ V} \]
Since current is the same throughout the series circuit:
\[ \frac{R_{\text{LDR}}}{R_1} = \frac{V_{\text{LDR}}}{V_1} \implies R_{\text{LDR}} = 4.0\text{ k}\Omega \times \frac{9.0}{3.0} = 12.0\text{ k}\Omega \]

(c) When the surroundings become darker, the resistance of the LDR increases.
As the total circuit resistance increases, the current decreases (or the LDR takes a greater fraction of the total voltage), so the potential difference across the fixed resistor \(R_1\) decreases, causing the voltmeter reading to decrease.

(a)
1M: Use of potential divider formula \(V_1 = V \frac{R_1}{R_1 + R_{\text{LDR}}}\) or finding current \(I = \frac{12}{5000}\text{ A}\)
1A: \(V_1 = 9.60\text{ V}\)

(b)
1M: Expressing ratio \(V_1 / V_{\text{LDR}} = R_1 / R_{\text{LDR}}\) or solving \(3.0 = 12\frac{4}{4 + R_{\text{LDR}}}\)
1A: \(R_{\text{LDR}} = 12.0\text{ k}\Omega\) (or \(1.20 \times 10^4\ \Omega\))

(c)
1A: Stating that LDR resistance increases in darker conditions
1A: Explaining that smaller proportion of voltage drops across \(R_1\) / circuit current decreases, hence voltmeter reading decreases

(a) Induced e.m.f.:
\[ \varepsilon = B L v = (0.80)(0.25)(4.0) = 0.80\text{ V} \]

(b) Induced current:
\[ I = \frac{\varepsilon}{R} = \frac{0.80}{0.50} = 1.60\text{ A} \]

(c) (i) Magnetic force on the rod:
\[ F_B = B I L = (0.80)(1.60)(0.25) = 0.32\text{ N} \]

(c) (ii) By Lenz's law, the magnetic force opposes the motion of the rod, so its direction is opposite to the velocity / direction of motion of the rod.

(a)
1M: Use of \(\varepsilon = B L v\)
1A: \(\varepsilon = 0.80\text{ V}\)

(b)
1A: \(I = 1.60\text{ A}\)

(c)(i)
1M: Use of \(F = B I L\)
1A: \(F = 0.32\text{ N}\)

(c)(ii)
1A: Opposite to the direction of motion (or opposing the motion)

(a) Corrected initial count rate \(C_0 = 520 - 40 = 480\text{ cpm}\).
Corrected count rate after \(18.0\text{ hours}\): \(C(t) = 100 - 40 = 60\text{ cpm}\).

(b) Ratio of corrected count rates:
\[ \frac{C(t)}{C_0} = \frac{60}{480} = \frac{1}{8} = \left(\frac{1}{2}\right)^3 \]
This corresponds to 3 half-lives:
\[ 3 t_{1/2} = 18.0\text{ hours} \implies t_{1/2} = \frac{18.0}{3} = 6.0\text{ hours} \]

(c) Any TWO differences:
1. Beta particles have higher penetrating power (can pass through paper, stopped by a few mm of aluminum) than alpha particles.
2. Beta particles have lower ionizing power than alpha particles.
3. Beta particles are negatively charged fast-moving electrons, whereas alpha particles are positively charged helium-4 nuclei.
4. Beta particles have much smaller mass than alpha particles.

(a)
1A: Both corrected values correct: \(480\text{ cpm}\) and \(60\text{ cpm}\)

(b)
1M: Recognizing \(60/480 = (1/2)^n\) with \(n = 3\) or using \(N = N_0 e^{-kt}\)
1A: \(t_{1/2} = 6.0\text{ hours}\) (or \(2.16 \times 10^4\text{ s}\))

(c)
2A: Any TWO correct contrasting properties (1 mark each): higher penetrating power, lower ionizing power, negatively charged / fast-moving electrons vs helium nuclei, smaller deflection mass/charge in magnetic field.

(a) Vertical height reached by the block:
\[ h = s \sin 30^\circ = 2.4 \times \sin 30^\circ = 1.20\text{ m} \]
Gain in gravitational potential energy:
\[ \Delta E_p = mgh = 0.50 \times 9.81 \times 1.20 = 5.886\text{ J} \approx 5.89\text{ J} \]

(b) Initial kinetic energy of the block:
\[ E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.50 \times (6.0)^2 = 9.00\text{ J} \]
Work done against friction:
\[ W_f = E_k - \Delta E_p = 9.00 - 5.886 = 3.114\text{ J} \]
Average frictional force \(f\):
\[ f = \frac{W_f}{s} = \frac{3.114}{2.4} = 1.2975\text{ N} \approx 1.30\text{ N} \]

(c) The magnitude of acceleration is less than that when sliding up.
When the block slides up, the component of gravity along the incline and the frictional force act in the same direction (down the plane):
\[ F_{\text{net, up}} = mg\sin\theta + f \implies a_{\text{up}} = g\sin\theta + \frac{f}{m} \]
When the block slides down, the frictional force acts up the plane, opposing gravity:
\[ F_{\text{net, down}} = mg\sin\theta - f \implies a_{\text{down}} = g\sin\theta - \frac{f}{m} \]
Hence, the net downward force and acceleration are smaller when sliding down.

(a) Height \(h = 2.4\sin 30^\circ = 1.2\text{ m}\) and \(\Delta E_p = mgh = (0.50)(9.81)(1.2)\) [1M]
\(\Delta E_p = 5.89\text{ J}\) (or \(5.886\text{ J}\)) [1A]

(b) \(\frac{1}{2}mv^2 - \Delta E_p = f \cdot s\) or \(9.00 - 5.886 = f(2.4)\) [1M]
\(f = 1.30\text{ N}\) (accept \(1.29\text{ N}\) to \(1.30\text{ N}\)) [1A]

(c) State that acceleration is less than when sliding up [1A]
Explain that the friction opposes the motion in both cases, so the net force along the plane is \(mg\sin\theta - f\) down the incline, which is smaller than \(mg\sin\theta + f\) up the incline [1A]

(a) Area swept by the rod per unit time projected onto the horizontal plane is \(\frac{\Delta A_h}{\Delta t} = L \cdot (v \Delta t \cos\theta) / \Delta t = L v \cos\theta\).
By Faraday's law, \(\mathcal{E} = \frac{\Delta \Phi}{\Delta t} = B \frac{\Delta A_h}{\Delta t} = B L v \cos\theta\).

(b) As the rod slides down, the upward magnetic flux linked with the circuit increases.
According to Lenz's law, the induced current produces an opposing effect, resulting in a horizontal magnetic force directed towards the left (opposing the horizontal component of the motion).

(c) (i) Induced current \(I = \frac{\mathcal{E}}{R} = \frac{B L v \cos\theta}{R}\).
The total horizontal magnetic force is \(F_B = B I L = \frac{B^2 L^2 v \cos\theta}{R}\).
The component along the inclined plane opposing downward motion is:
\(F_\text{incline} = F_B \cos\theta = \frac{B^2 L^2 v \cos^2\theta}{R}\).

(ii) At terminal speed \(v_t\), the net force along the plane is zero:
\(m g \sin\theta = \frac{B^2 L^2 v_t \cos^2\theta}{R}\)
\(v_t = \frac{m g R \sin\theta}{B^2 L^2 \cos^2\theta}\)
\(v_t = \frac{(0.04)(9.81)(1.5) \sin 30^\circ}{(0.60)^2 (0.25)^2 (\cos 30^\circ)^2} = \frac{0.2943}{0.36 \times 0.0625 \times 0.75} = \frac{0.2943}{0.016875} \approx 17.44\text{ m s}^{-1} \approx 17.4\text{ m s}^{-1}\).

(d) Rate of loss of GPE \(= m g v_t \sin 30^\circ = (0.04)(9.81)(17.44)(0.50) = 3.422\text{ W} \approx 3.42\text{ W}\).
Electrical power dissipated \(P = \frac{\mathcal{E}^2}{R} = \frac{(B L v_t \cos 30^\circ)^2}{R} = \frac{[(0.60)(0.25)(17.44)(\cos 30^\circ)]^2}{1.5} = \frac{(2.2655)^2}{1.5} = 3.422\text{ W} \approx 3.42\text{ W}\).
Thus, the rate of loss of GPE equals the electrical power dissipated in \(R\).

(a) Component of velocity perpendicular to B is \(v \cos\theta\) (or magnetic field perpendicular to plane is \(B \cos\theta\)) [1M]
Correct expression \(\mathcal{E} = B L v \cos\theta\) derived clearly [1A]

(b) Magnetic flux increases / induced current creates downward opposing flux [1M]
Magnetic force is horizontal directed to the left (opposing motion) [1A]

(c)(i) State \(I = \frac{B L v \cos\theta}{R}\) and \(F_B = B I L\) [1M]
Correct resolution along incline: \(F_\text{incline} = \frac{B^2 L^2 v \cos^2\theta}{R}\) [1A]

(c)(ii) Equating \(m g \sin\theta = F_\text{incline}\) [1M]
Correct substitution of values [1M]
\(v_t = 17.4\text{ m s}^{-1}\) (accept \(17.4\text{ m s}^{-1}\) to \(17.5\text{ m s}^{-1}\)) [1A]

(d) Calculate rate of GPE loss \(= 3.42\text{ W}\) [1M]
Calculate electrical power \(P = 3.42\text{ W}\) and state they are equal [1A]

(a) Fringe separation \(\Delta y = \frac{\lambda D}{a}\)
\(\Delta y = \frac{(632.8 \times 10^{-9})(1.80)}{0.20 \times 10^{-3}} = 5.6952 \times 10^{-3}\text{ m} \approx 5.70\text{ mm}\).

(b) (i) Since \(\Delta y \propto \lambda\) and the wavelength decreases (from \(632.8\text{ nm}\) to \(450\text{ nm}\)), the fringe separation \(\Delta y\) decreases (fringes become narrower / more closely spaced), and the bright fringes appear blue.
(ii) The two-slit interference pattern disappears; instead, a broader single-slit diffraction pattern with a wide central maximum and much dimmer secondary maxima is observed.

(c) (i) The extra optical path length is given by:
\(\Delta L = (n - n_\text{air}) t = (1.50 - 1.00)(1.20 \times 10^{-6}\text{ m}) = 6.00 \times 10^{-7}\text{ m}\).

(ii) The condition for the central maximum (zero net path difference) requires:
\(\frac{a \Delta Y}{D} = \Delta L\)
\(\Delta Y = \frac{\Delta L \cdot D}{a} = \frac{(6.00 \times 10^{-7})(1.80)}{0.20 \times 10^{-3}} = 5.40 \times 10^{-3}\text{ m} = 5.40\text{ mm}\)
(Alternatively: shift in fringe units \(N = \frac{\Delta L}{\lambda} = \frac{6.00 \times 10^{-7}}{6.328 \times 10^{-7}} = 0.9481\); shift \(= N \times \Delta y = 0.9481 \times 5.695\text{ mm} = 5.40\text{ mm}\)).

(a) Use of \(\Delta y = \frac{\lambda D}{a}\) with proper substitution [1M]
\(\Delta y = 5.70\text{ mm}\) (or \(5.695 \times 10^{-3}\text{ m}\)) [1A]

(b)(i) Fringe separation decreases / fringes become closer [1A]
Fringe colour becomes blue [1A]

(b)(ii) Double-slit interference fringes disappear [1A]
Single-slit diffraction pattern formed [1A]

(c)(i) Use of \(\Delta L = (n - 1)t\) [1M]
\(\Delta L = 6.00 \times 10^{-7}\text{ m}\) [1A]

(c)(ii) State \(\frac{a \Delta Y}{D} = \Delta L\) or \(N = \frac{\Delta L}{\lambda}\) [1M]
Correct substitution [1M]
\(\Delta Y = 5.40\text{ mm}\) (or \(5.40 \times 10^{-3}\text{ m}\)) [1A]

(a) By conservation of energy for the spring-block system:
\(\frac{1}{2} k x^2 = \frac{1}{2} m_A v_A^2\)
\(\frac{1}{2} (600)(0.20)^2 = \frac{1}{2} (0.60) v_A^2\)
\(12.0 = 0.30 v_A^2 \implies v_A^2 = 40.0 \implies v_A = \sqrt{40} \approx 6.325\text{ m s}^{-1} \approx 6.32\text{ m s}^{-1}\).

(b) (i) By conservation of linear momentum:
\(m_A v_A = (m_A + m_B) v'\)
\((0.60)(\sqrt{40}) = (0.60 + 1.20) v'\)
\(v' = \frac{0.60 \times 6.3246}{1.80} = \frac{\sqrt{40}}{3} \approx 2.108\text{ m s}^{-1} \approx 2.11\text{ m s}^{-1}\).

(ii) Kinetic energy before collision \(E_{k1} = 12.0\text{ J}\).
Kinetic energy after collision \(E_{k2} = \frac{1}{2} (m_A + m_B) (v')^2 = \frac{1}{2} (1.80) \left(\frac{40}{9}\right) = 4.00\text{ J}\).
Loss in mechanical energy \(= 12.0 - 4.00 = 8.00\text{ J}\).

(c) (i) Frictional force on the combined blocks \(f_k = \mu (m_A + m_B) g = (0.20)(1.80)(9.81) = 3.5316\text{ N}\).
Work done against friction \(W_f = f_k s = (3.5316)(0.70) = 2.4721\text{ J}\).
By work-energy theorem:
\(E_{k3} = E_{k2} - W_f = 4.00 - 2.4721 = 1.5279\text{ J}\).
\(\frac{1}{2} (m_A + m_B) v^2 = 1.5279\)
\(\frac{1}{2} (1.80) v^2 = 1.5279 \implies v = \sqrt{\frac{2 \times 1.5279}{1.80}} = \sqrt{1.6976} \approx 1.303\text{ m s}^{-1} \approx 1.30\text{ m s}^{-1}\).

(ii) By conservation of mechanical energy on the smooth ramp:
\((m_A + m_B) g h = E_{k3}\)
\((1.80)(9.81) h = 1.5279\)
\(h = \frac{1.5279}{17.658} \approx 0.0865\text{ m} = 8.65\text{ cm}\).

(a) Equating elastic PE to KE: \(\frac{1}{2} k x^2 = \frac{1}{2} m_A v_A^2\) [1M]
\(v_A = 6.32\text{ m s}^{-1}\) (or \(\sqrt{40}\text{ m s}^{-1}\)) [1A]

(b)(i) Conservation of momentum: \(m_A v_A = (m_A + m_B) v'\) [1M]
\(v' = 2.11\text{ m s}^{-1}\) [1A]

(b)(ii) Calculate KE after collision \(= 4.00\text{ J}\) [1M]
Loss \(= 8.00\text{ J}\) [1A]

(c)(i) Calculate work done against friction \(W_f = \mu (m_A+m_B) g s = 2.47\text{ J}\) [1M]
Applying energy principle: \(E_k - W_f = \frac{1}{2} (m_A + m_B) v^2\) [1M]
\(v = 1.30\text{ m s}^{-1}\) [1A]

(c)(ii) Conservation of energy on ramp: \(m g h = \frac{1}{2} m v^2\) [1M]
\(h = 0.0865\text{ m}\) (accept \(8.65\text{ cm}\) to \(8.66\text{ cm}\)) [1A]

Galileo's discovery that moons orbited Jupiter showed clearly that not all celestial objects orbit Earth, directly contradicting the fundamental claim of the Ptolemaic geocentric model.

Observations of sunspots and lunar craters showed that heavenly bodies were imperfect and dynamic, challenging Aristotelian concepts of celestial perfection, but they did not directly demonstrate orbits around non-Earth centres.

Therefore, only (1) is correct.

A (1 mark)
- Award 1 mark for selecting option A.
- Deduct 0 marks for incorrect options.

By conservation of angular momentum (or Kepler's second law):
\[ r_p v_p = r_a v_a \implies r_a = r_p \left(\frac{v_p}{v_a}\right) = 1.50 r_p \]

For an ellipse with semi-major axis \(a\):
\[ r_p + r_a = 2a \]
\[ r_p + 1.50 r_p = 2(4.00\text{ AU}) \]
\[ 2.50 r_p = 8.00\text{ AU} \implies r_p = \frac{8.00}{2.50} = 3.20\text{ AU} \]

B (1 mark)
- Award 1 mark for selecting option B.
- Deduct 0 marks for incorrect options.

Stellar parallax angle \(p\) is directly proportional to the baseline (orbital radius \(r\)):
\[ p = \frac{r}{d} \]

Given \(p_{\text{Earth}} = 0.080''\) with \(r_{\text{Earth}} = 1.0\text{ AU}\):
\[ p_{\text{Jupiter}} = \left(\frac{r_{\text{Jupiter}}}{r_{\text{Earth}}}\right) p_{\text{Earth}} = 5.2 \times 0.080'' = 0.416'' \approx 0.42'' \]

C (1 mark)
- Award 1 mark for selecting option C.
- Deduct 0 marks for incorrect options.

Both star \(P\) and star \(Q\) have a low surface temperature (\(3200\text{ K}\)), meaning they are cool, reddish stars.

- Star \(P\) has a very large positive absolute magnitude (\(+12.0\)), indicating a very low luminosity. Since it is cool and intrinsically faint, its radius is small, so it is a red dwarf (main-sequence star).
- Star \(Q\) has a very negative absolute magnitude (\(-5.0\)), indicating an enormous luminosity. To produce high luminosity at such a low temperature, it must have an extremely large surface area, so it is a red supergiant.

B (1 mark)
- Award 1 mark for selecting option B.
- Deduct 0 marks for incorrect options.

Luminosity is given by Stefan-Boltzmann law:
\[ L = 4\pi R^2 \sigma T^4 \]
\[ \frac{L_X}{L_Y} = \left(\frac{R_X}{R_Y}\right)^2 \left(\frac{T_X}{T_Y}\right)^4 = 3^2 \times 2^4 = 9 \times 16 = 144 \]

Apparent brightness (intensity) is:
\[ I = \frac{L}{4\pi d^2} \]
Since \(I_X = I_Y\):
\[ \frac{L_X}{4\pi d_X^2} = \frac{L_Y}{4\pi d_Y^2} \implies \left(\frac{d_X}{d_Y}\right)^2 = \frac{L_X}{L_Y} = 144 \implies \frac{d_X}{d_Y} = \sqrt{144} = 12 \]

B (1 mark)
- Award 1 mark for selecting option B.
- Deduct 0 marks for incorrect options.

(1) Correct. Along the main sequence, stars at the upper-left are more massive than stars at the lower-right.
(2) Correct. More massive stars burn their core hydrogen at a vastly higher rate, resulting in significantly shorter main-sequence lifetimes.
(3) Incorrect. Stars of the same spectral class share the same surface temperature, but their luminosity depends on their size (luminosity class), e.g., an M-type red supergiant is far more luminous than an M-type main-sequence red dwarf.

C (1 mark)
- Award 1 mark for selecting option C.
- Deduct 0 marks for incorrect options.

Calculate the redshift \(z\):
\[ z = \frac{\Delta \lambda}{\lambda_0} = \frac{510.4 - 486.1}{486.1} = \frac{24.3}{486.1} \approx 0.04999 \]

Recession speed \(v\):
\[ v = z c = 0.04999 \times 3.00 \times 10^5\text{ km s}^{-1} \approx 1.500 \times 10^4\text{ km s}^{-1} \]

By Hubble's law \(v = H_0 d\):
\[ d = \frac{v}{H_0} = \frac{1.500 \times 10^4}{70} \approx 214\text{ Mpc} \]

C (1 mark)
- Award 1 mark for selecting option C.
- Deduct 0 marks for incorrect options.

(1) Correct: The CMBR is the remnant thermal radiation from the early hot and dense universe, predicted by the Big Bang model.
(2) Incorrect: The presence of black holes is an astrophysical phenomenon related to stellar collapse and galactic evolution, not direct primary evidence for the Big Bang.
(3) Correct: The cosmological redshift shows the universe is expanding uniformly, which is fundamental evidence for an expanding universe originating from a compact initial state.

B (1 mark)
- Award 1 mark for selecting option B.
- Deduct 0 marks for incorrect options.

(a) (i) Orbital radius:
\(r = R_M + h = 3.39 \times 10^6\text{ m} + 3.00 \times 10^5\text{ m} = 3.69 \times 10^6\text{ m}\)

The gravitational force provides the centripetal force:
\(\frac{G M_M m}{r^2} = \frac{m v_0^2}{r}\)
\(v_0 = \sqrt{\frac{G M_M}{r}} = \sqrt{\frac{(6.67 \times 10^{-11})(6.42 \times 10^{23})}{3.69 \times 10^6}} = 3.407 \times 10^3\text{ m s}^{-1} \approx 3.41\text{ km s}^{-1}\)

(ii) Orbital period \(T\):
\(T = \frac{2\pi r}{v_0} = \frac{2\pi (3.69 \times 10^6\text{ m})}{3407\text{ m s}^{-1}} = 6805.8\text{ s} = \frac{6805.8}{60}\text{ min} \approx 113\text{ min}\) (or \(1.89\text{ h}\))

(b) (i) By conservation of mechanical energy for escaping to infinity:
\(E_{\text{total}} = \frac{1}{2} m v_1^2 - \frac{G M_M m}{r} = 0\)
\(v_1 = \sqrt{\frac{2 G M_M}{r}} = \sqrt{2} v_0 = \sqrt{2} \times 3407\text{ m s}^{-1} = 4.818 \times 10^3\text{ m s}^{-1} \approx 4.82\text{ km s}^{-1}\)

(ii) Energy supplied by thrusters:
\(\Delta E = E_k(\text{after}) - E_k(\text{before}) = \frac{1}{2} m (v_1^2 - v_0^2) = \frac{1}{2} m v_0^2\)
\(\Delta E = \frac{1}{2} (850\text{ kg}) (3407\text{ m s}^{-1})^2 = 4.93 \times 10^9\text{ J}\) (or \(4.93\text{ GJ}\))

(c) The claim is incorrect. The gravitational force acting on the probe is non-zero and acts as the necessary centripetal force keeping the probe in circular orbit (both the probe and its contents are in free fall).

(a) (i)
- \(\frac{G M_M m}{r^2} = \frac{m v_0^2}{r}\) with \(r = 3.69 \times 10^6\text{ m}\) (1M)
- Correct substitution leading to \(v_0 = 3.41\text{ km s}^{-1}\) (or \(3.407 \times 10^3\text{ m s}^{-1}\)) (1A)

(a) (ii)
- \(T = \frac{2\pi r}{v_0}\) (1M)
- \(T = 113\text{ min}\) (or \(6.81 \times 10^3\text{ s}\), accept \(113 \sim 114\text{ min}\)) (1A)

(b) (i)
- Conservation of energy statement: \(\frac{1}{2} m v_1^2 - \frac{G M_M m}{r} = 0\) (1M)
- Correct substitution of values (1M)
- \(v_1 = 4.82\text{ km s}^{-1}\) (accept \(4.81 \sim 4.83\text{ km s}^{-1}\)) (1A)

(b) (ii)
- \(\Delta E = \frac{1}{2} m v_1^2 - \frac{1}{2} m v_0^2\) OR \(\Delta E = \frac{G M_M m}{2r}\) (1M)
- \(\Delta E = 4.93 \times 10^9\text{ J}\) (accept \(4.90 \times 10^9 \sim 4.95 \times 10^9\text{ J}\)) (1A)

(c)
- States incorrect AND explains that gravitational force provides the centripetal force / the spacecraft is in free fall (1A)

Statement (1) is correct: The electrostatic repulsive force between the α-particle and the nucleus is given by Coulomb's law \(F = \frac{1}{4\pi \varepsilon_0} \frac{q_1 q_2}{r^2}\). The force is inversely proportional to the square of the separation \(r\), so it reaches a maximum at the distance of closest approach where \(r\) is minimum.

Statement (2) is correct: The impact parameter is the perpendicular distance between the initial trajectory of the α-particle and the parallel line passing through the centre of the nucleus. A smaller impact parameter brings the α-particle closer to the concentrated positive charge of the nucleus, resulting in a stronger repulsive force and thus a larger scattering angle.

Statement (3) is incorrect: Because the nucleus occupies an extremely small fraction of the atom's total volume, the vast majority of α-particles pass straight through or suffer only very small deflections; only a tiny fraction (about 1 in 8000) are scattered through angles greater than \(90^\circ\).

Hence, only (1) and (2) are correct.

A (1 mark)

Statement (1) is correct: The standing wave condition on the Bohr orbit is \(2\pi r_n = n\lambda\). For \(n = 3\), the circumference \(2\pi r_3 = 3\lambda\).

Statement (2) is correct: The possible downward electron transitions from \(n = 3\) are \(3 \to 2\), \(2 \to 1\), and \(3 \to 1\). Thus, there are at most 3 distinct emission lines.

Statement (3) is incorrect: Since orbital angular momentum is \(m v_n r_n = \frac{n h}{2\pi}\) and the radius scales as \(r_n \propto n^2\), the orbital speed is \(v_n \propto \frac{1}{n}\). Therefore, the speed in \(n = 3\) is \(\frac{1}{3}\) of that in \(n = 1\), not three times.

Hence, only (1) and (2) are correct.

C (1 mark)

The first excited state corresponds to the principal quantum number \(n = 2\).

The energy of the \(n = 2\) state is:
\[E_2 = -\frac{54.4}{2^2} = -\frac{54.4}{4} = -13.6\text{ eV}\]

Ionization corresponds to removing the electron to \(n = \infty\) where \(E_\infty = 0\text{ eV}\).

The minimum energy required is:
\[\Delta E = E_\infty - E_2 = 0 - (-13.6\text{ eV}) = 13.6\text{ eV}\]

C (1 mark)

According to Einstein's photoelectric equation:
\[e V_\text{s} = h f - \Phi \implies V_\text{s} = \left(\frac{h}{e}\right) f - \frac{\Phi}{e}\]

The slope of the graph is \(\frac{h}{e}\), which depends only on Planck's constant \(h\) and the elementary charge \(e\), both of which are fundamental physical constants.

The threshold frequency is \(f_0 = \frac{\Phi}{h}\), which depends solely on the work function \(\Phi\) of the metal material.

Changing the intensity of light changes the number of incident photons per second (and hence the saturation photocurrent), but does not alter the energy of individual photons, the slope, or the threshold frequency. Both remain unchanged.

A (1 mark)

The kinetic energy gained by a particle of charge \(q\) accelerated through potential difference \(V\) is \(E_\text{k} = qV\).

The momentum is \(p = \sqrt{2m E_\text{k}} = \sqrt{2mqV}\).

The de Broglie wavelength is:
\[\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}\]

Given \(\lambda_\text{p} = \lambda_\alpha\), we have:
\[2 m_\text{p} q_\text{p} V_\text{p} = 2 m_\alpha q_\alpha V_\alpha\]

Substituting \(m_\alpha = 4m_\text{p}\) and \(q_\alpha = 2q_\text{p}\):
\[m_\text{p} q_\text{p} V_\text{p} = (4m_\text{p})(2q_\text{p}) V_\alpha = 8 m_\text{p} q_\text{p} V_\alpha\]
\[\frac{V_\text{p}}{V_\alpha} = 8 = 8 : 1\]

D (1 mark)

The kinetic energy of the electrons is:
\[E_\text{k} = eV = (1.60 \times 10^{-19}\text{ C})(25\times 10^3\text{ V}) = 4.00 \times 10^{-15}\text{ J}\]

The momentum of the electron is:
\[p = \sqrt{2m_\text{e} E_\text{k}} = \sqrt{2 \times (9.11 \times 10^{-31}\text{ kg}) \times (4.00 \times 10^{-15}\text{ J})} = 8.537 \times 10^{-23}\text{ kg m s}^{-1}\]

The de Broglie wavelength is:
\[\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}\text{ J s}}{8.537 \times 10^{-23}\text{ kg m s}^{-1}} \approx 7.77 \times 10^{-12}\text{ m}\]

A (1 mark)

Statement (1) is correct: In an STM, quantum tunnelling allows electrons to pass across a narrow potential barrier (the vacuum/air gap of sub-nanometre width) between the sharp metallic tip and the conductive specimen.

Statement (2) is correct: The tunnelling current \(I\) has an exponential dependence on the separation distance \(d\), namely \(I \propto e^{-2\kappa d}\), providing extremely high vertical sensitivity.

Statement (3) is incorrect: An STM relies on the flow of a tunnelling electric current; therefore, the sample must be electrically conductive or semiconductive. Insulating materials cannot be imaged using standard STM without conductive coating (an Atomic Force Microscope is used instead for insulators).

Hence, only (1) and (2) are correct.

C (1 mark)

Carbon nanotubes consist of rolled cylindrical sheets of graphene where each carbon atom is bonded to three neighbours via strong in-plane \(sp^2\) covalent bonds. These strong covalent bonds along the wall of the nanotube give it an exceptionally high tensile strength and elastic modulus along its axis.

A (1 mark)

(a) (i) The kinetic energy gained by an electron is:
\[ E_k = e V \]
Since kinetic energy is related to momentum \(p\) by \(E_k = \frac{p^2}{2m_e}\), we have:
\[ p = \sqrt{2 m_e e V} \]
Using the de Broglie relation \(\lambda = \frac{h}{p}\):
\[ \lambda = \frac{h}{\sqrt{2 m_e e V}} \]

(ii) Substituting the values:
\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times (9.11 \times 10^{-31}) \times (1.60 \times 10^{-19}) \times 150}} \]
\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{4.3728 \times 10^{-47}}} = \frac{6.63 \times 10^{-34}}{6.613 \times 10^{-24}} = 1.003 \times 10^{-10}\text{ m} \approx 1.00 \times 10^{-10}\text{ m} \]

(b) (i) Diffraction and interference are characteristic wave behaviors. The electrons are diffracted by the regular atomic lattice planes of the graphite crystals and constructively interfere at specific angles, demonstrating that moving particles exhibit wave properties.

(ii) As \(V\) increases, the kinetic energy and momentum \(p\) of the electrons increase, so the de Broglie wavelength \(\lambda\) decreases (since \(\lambda \propto V^{-1/2}\)). According to the diffraction condition (\(\sin\theta \propto \lambda\)), a shorter wavelength results in a smaller diffraction angle \(\theta\). Therefore, the diameter of the diffraction rings decreases.

(c) The energy levels of hydrogen are:
- Ground state (\(n=1\)): \(E_1 = -13.60\text{ eV}\)
- First excited state (\(n=2\)): \(E_2 = -\frac{13.6}{4} = -3.40\text{ eV}\), excitation energy \(\Delta E_{1 \to 2} = -3.40 - (-13.60) = 10.20\text{ eV}\)
- Second excited state (\(n=3\)): \(E_3 = -\frac{13.6}{9} = -1.51\text{ eV}\), excitation energy \(\Delta E_{1 \to 3} = -1.51 - (-13.60) = 12.09\text{ eV}\)
- Third excited state (\(n=4\)): \(E_4 = -\frac{13.6}{16} = -0.85\text{ eV}\), excitation energy \(\Delta E_{1 \to 4} = -0.85 - (-13.60) = 12.75\text{ eV}\)

Since an colliding electron with kinetic energy \(12.5\text{ eV}\) can transfer part of its kinetic energy to the bound electron via inelastic collision, it can provide \(10.20\text{ eV}\) (retaining \(2.30\text{ eV}\)) or \(12.09\text{ eV}\) (retaining \(0.41\text{ eV}\)). However, \(12.5\text{ eV} < 12.75\text{ eV}\), so excitation to \(n=4\) or higher is impossible.
Thus, the atom can be excited to \(n = 2\) or \(n = 3\).

(a) (i)
- Equating kinetic energy \(E_k = eV = \frac{p^2}{2m_e}\) to find \(p = \sqrt{2m_e eV}\): 1M
- Substituting into de Broglie equation \(\lambda = \frac{h}{p}\) to arrive at the expression: 1A

(a) (ii)
- Correct substitution of values into the formula: 1M
- Correct final answer \(\lambda = 1.00 \times 10^{-10}\text{ m}\) (accept \(1.00 \times 10^{-10} \sim 1.01 \times 10^{-10}\text{ m}\) or \(0.100\text{ nm}\)): 1A

(b) (i)
- Stating that diffraction / interference is a wave property: 1A
- Explaining that circular rings result from constructive interference / diffraction of electron waves by crystal lattice planes in random orientations: 1A

(b) (ii)
- Explaining that higher \(V\) implies smaller wavelength \(\lambda\): 1A
- Concluding that smaller \(\lambda\) leads to smaller diffraction angle \(\theta\), so the diameter of rings decreases: 1A

(c)
- Calculating excitation energies for \(n=2\) (\(10.2\text{ eV}\)), \(n=3\) (\(12.1\text{ eV}\)), and \(n=4\) (\(12.8\text{ eV}\)): 1M
- Correctly identifying both possible states \(n = 2\) and \(n = 3\): 1A

(a) (i) The kinetic energy gained by an electron is:
\[ E_k = e V \]
Since kinetic energy is related to momentum \(p\) by \(E_k = \frac{p^2}{2m_e}\), we have:
\[ p = \sqrt{2 m_e e V} \]
Using the de Broglie relation \(\lambda = \frac{h}{p}\):
\[ \lambda = \frac{h}{\sqrt{2 m_e e V}} \]

(ii) Substituting the values:
\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times (9.11 \times 10^{-31}) \times (1.60 \times 10^{-19}) \times 150}} \]
\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{4.3728 \times 10^{-47}}} = \frac{6.63 \times 10^{-34}}{6.613 \times 10^{-24}} = 1.003 \times 10^{-10}\text{ m} \approx 1.00 \times 10^{-10}\text{ m} \]

(b) (i) Diffraction and interference are characteristic wave behaviors. The electrons are diffracted by the regular atomic lattice planes of the graphite crystals and constructively interfere at specific angles, demonstrating that moving particles exhibit wave properties.

(ii) As \(V\) increases, the kinetic energy and momentum \(p\) of the electrons increase, so the de Broglie wavelength \(\lambda\) decreases (since \(\lambda \propto V^{-1/2}\)). According to the diffraction condition (\(\sin\theta \propto \lambda\)), a shorter wavelength results in a smaller diffraction angle \(\theta\). Therefore, the diameter of the diffraction rings decreases.

(c) The energy levels of hydrogen are:
- Ground state (\(n=1\)): \(E_1 = -13.60\text{ eV}\)
- First excited state (\(n=2\)): \(E_2 = -\frac{13.6}{4} = -3.40\text{ eV}\), excitation energy \(\Delta E_{1 \to 2} = -3.40 - (-13.60) = 10.20\text{ eV}\)
- Second excited state (\(n=3\)): \(E_3 = -\frac{13.6}{9} = -1.51\text{ eV}\), excitation energy \(\Delta E_{1 \to 3} = -1.51 - (-13.60) = 12.09\text{ eV}\)
- Third excited state (\(n=4\)): \(E_4 = -\frac{13.6}{16} = -0.85\text{ eV}\), excitation energy \(\Delta E_{1 \to 4} = -0.85 - (-13.60) = 12.75\text{ eV}\)

Since an colliding electron with kinetic energy \(12.5\text{ eV}\) can transfer part of its kinetic energy to the bound electron via inelastic collision, it can provide \(10.20\text{ eV}\) (retaining \(2.30\text{ eV}\)) or \(12.09\text{ eV}\) (retaining \(0.41\text{ eV}\)). However, \(12.5\text{ eV} < 12.75\text{ eV}\), so excitation to \(n=4\) or higher is impossible.
Thus, the atom can be excited to \(n = 2\) or \(n = 3\).

(a) (i)
- Equating kinetic energy \(E_k = eV = \frac{p^2}{2m_e}\) to find \(p = \sqrt{2m_e eV}\): 1M
- Substituting into de Broglie equation \(\lambda = \frac{h}{p}\) to arrive at the expression: 1A

(a) (ii)
- Correct substitution of values into the formula: 1M
- Correct final answer \(\lambda = 1.00 \times 10^{-10}\text{ m}\) (accept \(1.00 \times 10^{-10} \sim 1.01 \times 10^{-10}\text{ m}\) or \(0.100\text{ nm}\)): 1A

(b) (i)
- Stating that diffraction / interference is a wave property: 1A
- Explaining that circular rings result from constructive interference / diffraction of electron waves by crystal lattice planes in random orientations: 1A

(b) (ii)
- Explaining that higher \(V\) implies smaller wavelength \(\lambda\): 1A
- Concluding that smaller \(\lambda\) leads to smaller diffraction angle \(\theta\), so the diameter of rings decreases: 1A

(c)
- Calculating excitation energies for \(n=2\) (\(10.2\text{ eV}\)), \(n=3\) (\(12.1\text{ eV}\)), and \(n=4\) (\(12.8\text{ eV}\)): 1M
- Correctly identifying both possible states \(n = 2\) and \(n = 3\): 1A