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For the block to remain in circular motion without slipping, the centripetal force must be provided by static friction.

The maximum force of static friction is:
\(F_f = \mu m g\)

The required centripetal force for circular motion at angular velocity \(\omega\) is:
\(F_c = m \omega^2 r\)

Setting these equal at the point of slipping:
\(m \omega^2 r = \mu m g\)

Solving for \(\omega\):
\(\omega = \sqrt{\frac{\mu g}{r}}\)

A is the correct answer. (1 mark for equating the maximum frictional force to the centripetal force formula and correctly solving for angular velocity).

Let the point of zero potential be at a distance \(x\) from the charge \(+Q\). Since the point is between the two charges, its distance from the charge \(-4Q\) is \(d - x\).

The electric potential \(V\) due to a point charge is given by \(V = \frac{kQ}{r}\).

The total potential at this point is the sum of the potentials from both charges:
\(V = \frac{kQ}{x} + \frac{k(-4Q)}{d - x} = 0\)

Dividing by \(kQ\):
\(\frac{1}{x} - \frac{4}{d - x} = 0\)

\(\frac{1}{x} = \frac{4}{d - x}\)

\(d - x = 4x\)

\(5x = d\)

\(x = \frac{d}{5}\)

A is the correct answer. (1 mark for setting up the potential sum equal to zero and correctly solving the algebraic expression).

According to Faraday's law of electromagnetic induction, the magnitude of the average induced e.m.f. \(\mathcal{E}\) is equal to the rate of change of magnetic flux linkage:
\(\mathcal{E} = \frac{\Delta \Phi}{\Delta t}\)

Initially, the plane of the coil is perpendicular to the field, so the flux linkage is:
\(\Phi_{\text{initial}} = N B A\)

After rotating by \(90^\circ\), the plane of the coil is parallel to the magnetic field, so the magnetic flux linkage is:
\(\Phi_{\text{final}} = 0\)

The change in flux linkage is:
\(\Delta \Phi = |0 - N B A| = N B A\)

Thus, the average induced e.m.f. is:
\(\mathcal{E} = \frac{N B A}{\Delta t}\)

C is the correct answer. (1 mark for applying Faraday's law using the change in flux linkage from maximum to zero).

The kinetic energy \(E_k\) gained by an electron of charge \(e\) accelerated through a potential difference \(V\) is:
\(E_k = e V\)

Since \(E_k = \frac{p^2}{2m_e}\), the momentum \(p\) of the electron is:
\(p = \sqrt{2 m_e e V}\)

The de Broglie wavelength \(\lambda\) is given by:
\(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m_e e V}}\)

This shows that \(\lambda \propto \frac{1}{\sqrt{V}}\).

If the potential difference is increased by a factor of 4, the new wavelength \(\lambda'\) is:
\(\lambda' = \frac{\lambda}{\sqrt{4}} = \frac{\lambda}{2}\)

C is the correct answer. (1 mark for identifying that wavelength is inversely proportional to the square root of voltage and calculating the resulting factor of 0.5).

Total initial momentum vector:
\(\vec{p}_i = (2mv)\hat{i}\)

Total final momentum vector must equal initial momentum vector due to conservation of momentum:
\(\vec{p}_{1} + \vec{p}_{2} = \vec{p}_i\)

where \(\vec{p}_{1}\) is the momentum of the mass \(2m\) and \(\vec{p}_{2}\) is the momentum of the mass \(m\) after the collision.

Since mass \(2m\) moves with velocity \(\frac{v}{2}\) in the \(y\)-direction:
\(\vec{p}_{1} = (2m)\left(\frac{v}{2}\right)\hat{j} = (mv)\hat{j}\)

Therefore, the momentum of the mass \(m\) is:
\(\vec{p}_{2} = \vec{p}_i - \vec{p}_{1} = (2mv)\hat{i} - (mv)\hat{j}\)

The magnitude of this momentum vector is:
\(|\vec{p}_{2}| = \sqrt{(2mv)^2 + (-mv)^2} = \sqrt{4m^2v^2 + m^2v^2} = \sqrt{5}mv\)

C is the correct answer. (1 mark for using components of momentum and resolving via Pythagoras' theorem).

The activity of a radioactive sample as a function of time is given by:
\(A = A_0 e^{-\lambda t}\)

where \(A_0\) is the original activity.

Substituting \(t = \frac{2}{\lambda}\):
\(A = A_0 e^{-\lambda \left(\frac{2}{\lambda}\right)} = A_0 e^{-2}\)

The fraction of the original activity remaining is:
\(\frac{A}{A_0} = e^{-2}\)

A is the correct answer. (1 mark for substituting the time expression into the exponential decay formula).

1. According to the kinetic theory of gases, the average kinetic energy of the molecules is directly proportional to the absolute temperature:
\(\frac{1}{2}m\langle c^2 \rangle = \frac{3}{2}kT \Rightarrow \langle c^2 \rangle \propto T\)
Therefore, doubling the temperature doubles the mean square speed \(\langle c^2 \rangle\).

2. From the ideal gas equation, \(pV = nRT\). Since volume \(V\) is constant:
\(p \propto T\)
Therefore, doubling the temperature also doubles the pressure \(p\).

A is the correct answer. (1 mark for linking absolute temperature proportionally to both mean square speed and pressure).

A baryon is composed of three quarks.

1. The strangeness of \(-1\) indicates the presence of exactly one strange (\(s\)) quark, which has a charge of \(-\frac{1}{3}e\).

2. The remaining two quarks must be combinations of up (\(u\), charge \(+\frac{2}{3}e\)) and/or down (\(d\), charge \(-\frac{1}{3}e\)) quarks.

3. Let the total charge of the other two quarks be \(Q_{\text{other}}\).
\(Q_{\text{total}} = Q_{\text{other}} + Q(s) \Rightarrow +1e = Q_{\text{other}} - \frac{1}{3}e \Rightarrow Q_{\text{other}} = +\frac{4}{3}e\).

4. This charge is achieved by having two up quarks: \(+\frac{2}{3}e + \frac{2}{3}e = +\frac{4}{3}e\).

Thus, the quark composition is \(uus\).

A is the correct answer. (1 mark for identifying that three quarks are needed, and using conservation of strangeness and charge to find the correct combination).

For a charged particle moving perpendicular to a magnetic field \(B\), the magnetic force acts as the centripetal force:

\(B q v = \frac{m v^2}{r}\)

Rearranging for radius \(r\):

\(r = \frac{m v}{q B}\)

Since momentum \(p = m v\) can be written in terms of kinetic energy \(E_k\) as \(p = \sqrt{2mE_k}\), the expression for the radius is:

\(r = \frac{\sqrt{2mE_k}}{q B}\)

For the second particle, let its radius be \(R_{\text{second}}\):

\(R_{\text{second}} = \frac{\sqrt{2(4m)E_k}}{(2q) B} = \frac{2\sqrt{2mE_k}}{2q B} = \frac{\sqrt{2mE_k}}{q B} = R\)

Thus, the radius of the path of the second particle is also \(R\).

1 mark for selecting B (R).

The relationship between pressure, volume, and the mean square speed of the molecules of an ideal gas is given by:

\(p V = \frac{1}{3} N m \langle c^2 \rangle\)

where \(N\) is the number of molecules and \(m\) is the mass of a single molecule.

When the volume is doubled to \(2V\) and the pressure is halved to \(\frac{1}{2}p\), the new product of pressure and volume is:

\(p_{\text{new}} V_{\text{new}} = \left(\frac{1}{2}p\right)(2V) = pV\)

Since the product of pressure and volume remains unchanged, and the mass of the gas is fixed (so \(N\) and \(m\) are constant), the mean square speed \(\langle c^2 \rangle\) must remain unchanged. Therefore, the root-mean-square speed \(c_{\text{rms}} = \sqrt{\langle c^2 \rangle}\) also remains unchanged.

1 mark for selecting C (It remains unchanged).

Using \(r = \frac{mv}{Bq}\), we can express the velocity as \(v = \frac{Bqr}{m}\). Substituting the given values: \(v = \frac{0.35\text{ T} \times 1.60 \times 10^{-19}\text{ C} \times 1.2\text{ m}}{1.67 \times 10^{-27}\text{ kg}} = 4.02 \times 10^7\text{ m s}^{-1}\). The kinetic energy of the proton is \(E_k = \frac{1}{2}mv^2 = 0.5 \times 1.67 \times 10^{-27}\text{ kg} \times (4.02 \times 10^7\text{ m s}^{-1})^2 = 1.35 \times 10^{-12}\text{ J}\). Converting this energy to MeV: \(E_k = \frac{1.35 \times 10^{-12}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} = 8.44 \times 10^6\text{ eV} = 8.44\text{ MeV}\). To two significant figures, this is \(8.4\text{ MeV}\).

1. Rearranging for velocity \(v = \frac{Bqr}{m}\) or equating centripetal and magnetic force: 1 mark.
2. Calculating the velocity of the proton \(v = 4.0 \times 10^7\text{ m s}^{-1}\): 1 mark.
3. Finding kinetic energy in Joules \(E_k = 1.35 \times 10^{-12}\text{ J}\): 1 mark.
4. Converting energy to MeV to give \(8.4\text{ MeV}\) (accept \(8.4\text{ MeV}\) to \(8.5\text{ MeV}\)): 1 mark.

At the top of the vertical loop, the forces acting towards the center of the circular motion are the weight of the car (\(mg\)) and the normal contact force (\(N\)). The centripetal force \(F_c\) is the sum of these forces: \(F_c = N + mg = \frac{mv^2}{r}\). The weight is \(mg = 450\text{ kg} \times 9.81\text{ m s}^{-2} = 4414.5\text{ N}\). The total centripetal force is \(F_c = 1500\text{ N} + 4414.5\text{ N} = 5914.5\text{ N}\). Using \(\frac{mv^2}{r} = 5914.5\text{ N}\), we get \(v = \sqrt{\frac{5914.5\text{ N} \times 8.0\text{ m}}{450\text{ kg}}} = 10.26\text{ m s}^{-1}\). This is \(10.3\text{ m s}^{-1}\) to three significant figures.

1. Recognizing that \(F_c = N + mg\): 1 mark.
2. Correct calculation of weight as \(4410\text{ N}\) or \(4415\text{ N}\): 1 mark.
3. Correct substitution into the centripetal force equation: 1 mark.
4. Correct final speed of \(10.3\text{ m s}^{-1}\) (accept \(10.2\text{ m s}^{-1}\) to \(10.3\text{ m s}^{-1}\)): 1 mark.

The decay constant is \(\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{12.7\text{ h}} = 0.05458\text{ h}^{-1}\). The decay equation is \(A = A_0 e^{-\lambda t}\). Substituting \(t = 48.0\text{ h}\) and \(A_0 = 3.5 \times 10^7\text{ Bq}\), we obtain: \(A = 3.5 \times 10^7\text{ Bq} \times e^{-0.05458\text{ h}^{-1} \times 48.0\text{ h}} = 3.5 \times 10^7\text{ Bq} \times e^{-2.62} = 2.56 \times 10^6\text{ Bq}\). To two significant figures, this is \(2.6 \times 10^6\text{ Bq}\).

1. Calculation of decay constant \(\lambda\) as \(0.0546\text{ h}^{-1}\) or \(1.52 \times 10^{-5}\text{ s}^{-1}\): 1 mark.
2. Use of \(A = A_0 e^{-\lambda t}\): 1 mark.
3. Correct substitution of time in appropriate units: 1 mark.
4. Correct final activity in the range \(2.5 \times 10^6\text{ Bq}\) to \(2.6 \times 10^6\text{ Bq}\): 1 mark.

Convert the initial temperature to Kelvin: \(T_1 = 20 + 273.15 = 293.15\text{ K}\). Using the ideal gas equation \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\), we rearrange to find \(T_2\): \(T_2 = \frac{P_2 V_2 T_1}{P_1 V_1} = \frac{1.8 \times 10^5\text{ Pa} \times 0.030\text{ m}^3 \times 293.15\text{ K}}{1.2 \times 10^5\text{ Pa} \times 0.025\text{ m}^3} = 527.67\text{ K}\). Convert the final temperature back to Celsius: \(\theta_2 = 527.67 - 273.15 = 254.52\text{ }^\circ\text{C}\). Expressed to three significant figures, this is \(255\text{ }^\circ\text{C}\).

1. Conversion of initial temperature to Kelvin (\(293\text{ K}\)): 1 mark.
2. Use of \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\): 1 mark.
3. Correct determination of the final Kelvin temperature (\(528\text{ K}\)): 1 mark.
4. Correct final Celsius temperature in the range \(254\text{ }^\circ\text{C}\) to \(255\text{ }^\circ\text{C}\): 1 mark.

The kinetic energy gained by the electron is \(E_k = eV = 1.60 \times 10^{-19}\text{ C} \times 2500\text{ V} = 4.0 \times 10^{-16}\text{ J}\). The momentum \(p\) of the electron is related to its kinetic energy by \(E_k = \frac{p^2}{2m_e}\). Therefore, \(p = \sqrt{2 m_e E_k} = \sqrt{2 \times 9.11 \times 10^{-31}\text{ kg} \times 4.0 \times 10^{-16}\text{ J}} = 2.70 \times 10^{-23}\text{ kg m s}^{-1}\). The de Broglie wavelength is \(\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}\text{ J s}}{2.70 \times 10^{-23}\text{ kg m s}^{-1}} = 2.46 \times 10^{-11}\text{ m}\). To two significant figures, this is \(2.5 \times 10^{-11}\text{ m}\).

1. Calculate kinetic energy in Joules (\(4.0 \times 10^{-16}\text{ J}\)): 1 mark.
2. Use of \(E_k = \frac{p^2}{2m}\) or \(v = \sqrt{\frac{2E_k}{m}}\): 1 mark.
3. Correct momentum calculation (\(2.7 \times 10^{-23}\text{ kg m s}^{-1}\)): 1 mark.
4. Correct final wavelength in the range \(2.4 \times 10^{-11}\text{ m}\) to \(2.5 \times 10^{-11}\text{ m}\): 1 mark.

The capacitance of a parallel-plate capacitor with a dielectric is given by \(C = \frac{\varepsilon_r \varepsilon_0 A}{d}\). Substituting the given values: \(C = \frac{2.1 \times 8.85 \times 10^{-12}\text{ F m}^{-1} \times 0.085\text{ m}^2}{0.12 \times 10^{-3}\text{ m}} = 1.316 \times 10^{-8}\text{ F}\). The stored charge is \(Q = CV = 1.316 \times 10^{-8}\text{ F} \times 12\text{ V} = 1.58 \times 10^{-7}\text{ C}\). Expressed to two significant figures, this is \(1.6 \times 10^{-7}\text{ C}\).

1. Use of \(C = \frac{\varepsilon_r \varepsilon_0 A}{d}\): 1 mark.
2. Correct value for capacitance (\(1.3 \times 10^{-8}\text{ F}\)): 1 mark.
3. Use of \(Q = CV\): 1 mark.
4. Correct final charge in the range \(1.5 \times 10^{-7}\text{ C}\) to \(1.6 \times 10^{-7}\text{ C}\): 1 mark.

The centripetal force is provided by gravity: \(\frac{G M m}{r^2} = \frac{m v^2}{r}\), which gives \(v = \sqrt{\frac{G M}{r}}\). Substituting values: \(v = \sqrt{\frac{6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2} \times 6.4 \times 10^{23}\text{ kg}}{9.5 \times 10^6\text{ m}}} = 2119.8\text{ m s}^{-1}\). The orbital period is \(T = \frac{2 \pi r}{v} = \frac{2 \pi \times 9.5 \times 10^6\text{ m}}{2119.8\text{ m s}^{-1}} = 2.816 \times 10^4\text{ s}\). Converting to hours: \(T = \frac{2.816 \times 10^4\text{ s}}{3600\text{ s h}^{-1}} = 7.82\text{ hours}\). This rounds to \(7.8\text{ hours}\).

1. State \(\frac{G M m}{r^2} = \frac{m v^2}{r}\) or \(T^2 = \frac{4 \pi^2 r^3}{G M}\): 1 mark.
2. Calculate orbital velocity \(v \approx 2.1 \times 10^3\text{ m s}^{-1}\) or correct algebraic substitution: 1 mark.
3. Calculate orbital period in seconds (\(2.8 \times 10^4\text{ s}\)): 1 mark.
4. Convert the period to hours to obtain \(7.8\text{ hours}\) (accept \(7.8\text{ hours}\) to \(7.9\text{ hours}\)): 1 mark.

For simple harmonic motion, the maximum acceleration is given by \(a_{\text{max}} = \omega^2 A\). We can solve for \(\omega^2\): \(\omega^2 = \frac{a_{\text{max}}}{A} = \frac{5.2\text{ m s}^{-2}}{0.045\text{ m}} = 115.6\text{ rad}^2\text{ s}^{-2}\). For a mass-spring system, \(\omega^2 = \frac{k}{m}\). Substituting for the spring constant: \(k = m \omega^2 = 0.35\text{ kg} \times 115.6\text{ rad}^2\text{ s}^{-2} = 40.5\text{ N m}^{-1}\). Expressed to two significant figures, this is \(40\text{ N m}^{-1}\).

1. State or use \(a_{\text{max}} = \omega^2 A\): 1 mark.
2. Calculate the value of \(\omega^2 \approx 116\text{ rad}^2\text{ s}^{-2}\): 1 mark.
3. State or use \(\omega^2 = \frac{k}{m}\): 1 mark.
4. Correct final spring constant in the range \(40\text{ N m}^{-1}\) to \(41\text{ N m}^{-1}\): 1 mark.

The equation for the discharge of a capacitor is:
\(Q = Q_0 e^{-\frac{t}{RC}}\)

Given that the charge decreases to \(25\%\) of its initial value:
\(\frac{Q}{Q_0} = 0.25\)

Therefore:
\(0.25 = e^{-\frac{3.5}{R \times 47 \times 10^{-6}}}\)

Taking the natural logarithm of both sides:
\(\ln(0.25) = -\frac{3.5}{R \times 47 \times 10^{-6}}\)

\(-1.3863 = -\frac{3.5}{R \times 47 \times 10^{-6}}\)

Rearranging to solve for \(R\):
\(R = \frac{3.5}{1.3863 \times 47 \times 10^{-6}}\)
\(R = 5.37 \times 10^4\text{ }\Omega\)

To 2 significant figures, \(R = 5.4 \times 10^4\text{ }\Omega\) (or \(54\text{ k}\Omega\)).

**[1 mark]** Use of \(Q = Q_0 e^{-\frac{t}{RC}}\)
**[1 mark]** Correct substitution of ratio \(\frac{Q}{Q_0} = 0.25\)
**[1 mark]** Rearrangement to find the time constant \(RC \approx 2.52\text{ s}\) or directly solving for \(R\)
**[1 mark]** Correct final answer for \(R\) in the range \(5.3 \times 10^4\text{ }\Omega\) to \(5.4 \times 10^4\text{ }\Omega\) (or \(53\text{ k}\Omega\) to \(54\text{ k}\Omega\)) with unit.

The total internal kinetic energy \(E_k\) of \(n\) moles of a monatomic ideal gas is given by:
\(E_k = \frac{3}{2} n R T\)

The change in internal kinetic energy \(\Delta E_k\) due to a change in temperature \(\Delta T\) is:
\(\Delta E_k = \frac{3}{2} n R \Delta T\)

Substitute the given values:
\(450 = \frac{3}{2} \times 0.18 \times 8.31 \times \Delta T\)
\(450 = 2.2437 \times \Delta T\)

Solve for \(\Delta T\):
\(\Delta T = \frac{450}{2.2437} = 200.56\text{ K}\) (or \(^\circ\text{C}\))

Calculate the final temperature \(T_f\):
\(T_f = T_i + \Delta T = 20 + 200.56 = 220.56\text{ }^\circ\text{C}\)

To 2 or 3 significant figures, this is \(220\text{ }^\circ\text{C}\) or \(221\text{ }^\circ\text{C}\).

**[1 mark]** Use of \(\Delta E_k = \frac{3}{2} n R \Delta T\)
**[1 mark]** Correct substitution of values: \(450 = 1.5 \times 0.18 \times 8.31 \times \Delta T\)
**[1 mark]** Calculation of \(\Delta T = 201\text{ K}\) (or \(200.6\text{ K}\))
**[1 mark]** Addition of \(\Delta T\) to \(20\text{ }^\circ\text{C}\) to find final temperature \(220\text{ }^\circ\text{C}\) or \(221\text{ }^\circ\text{C}\).

As the magnet falls, it creates a changing magnetic flux linkage through the copper tube. According to Faraday's law, this change in flux induces an electromotive force (e.m.f.) in the tube. Since copper is an electrical conductor, the induced e.m.f. causes eddy currents to flow around the tube. Lenz's law states that the direction of the induced current opposes the change in magnetic flux that created it. Consequently, the magnetic field produced by these eddy currents exerts an upward magnetic force on the falling magnet, opposing its gravitational force. This reduces the net downward force and thus its acceleration, causing it to fall more slowly and eventually reach a lower terminal velocity. In terms of energy conservation, the loss of gravitational potential energy is converted not only into kinetic energy but also into electrical energy, which is subsequently dissipated as thermal energy in the copper tube due to its electrical resistance.

Marking points (Max 8 marks): 1. Change in magnetic flux linkage through the copper tube as magnet falls. 2. Reference to Faraday's law: induced e.m.f. is proportional to the rate of change of magnetic flux linkage. 3. Copper is a conductor, so eddy currents are established. 4. Reference to Lenz's law: the induced current/magnetic field opposes the change/motion of the magnet. 5. This results in an upward magnetic force acting on the magnet. 6. The net downward force (and therefore acceleration) is reduced. 7. The magnet reaches a terminal velocity where magnetic force + air resistance = weight. 8. Conservation of energy: Gravitational potential energy is transferred to kinetic energy, electrical energy, and thermal energy in the tube.

At the top of the vertical loop, the forces acting on the car are its weight \(mg\) and the normal reaction force \(R\) from the track, both acting vertically downwards. The resultant downward force provides the centripetal acceleration, so \(R + mg = \frac{mv^2}{r}\, where \)v\) is the speed at the top of the loop. For the car to maintain contact with the track, the normal reaction force must be greater than or equal to zero (\(R \ge 0\)). The minimum speed \(v_{min}\) occurs at the limit where \(R = 0\), yielding \(mg = \frac{mv_{min}^2}{r}\, which simplifies to \)v_{min}^2 = gr\). By the conservation of mechanical energy, the initial potential energy at height \(h\) equals the total mechanical energy at the top of the loop (at height \(2r\)). Thus, \(mgh = mg(2r) + \frac{1}{2}mv_{min}^2\). Substituting \(v_{min}^2 = gr\) into the energy equation gives \(mgh = 2mgr + \frac{1}{2}mgr\). Dividing both sides by \(mg\), we obtain \(h = 2r + 0.5r = 2.5r\).

Marking points (Max 8 marks): 1. Identify forces at the top of the loop as weight \(mg\) and normal reaction \(R\) both acting downwards. 2. Write centripetal force equation: \(R + mg = \frac{mv^2}{r}\). 3. State that for minimum speed to not lose contact, the normal reaction force \(R = 0\). 4. Derivation of minimum speed at the top: \(v = \sqrt{gr}\). 5. Apply conservation of energy: Initial GPE = GPE at top + KE at top. 6. Express initial energy as \(mgh\) and energy at the top as \(mg(2r) + \frac{1}{2}mv^2\). 7. Substitute \(v^2 = gr\) into the energy equation: \(mgh = 2mgr + 0.5mgr\). 8. Correctly solve for \(h = 2.5r\).

Technetium-99m is highly suitable as an internal diagnostic tracer because it emits gamma radiation, which is weakly ionising and highly penetrating, allowing it to easily pass through the patient's body and be detected by a gamma camera. Its short half-life of 6 hours is long enough to complete the diagnostic imaging procedure but short enough that the isotope quickly decays, minimizing the long-term radiation dose received by the patient. In contrast, Cobalt-60 is used as an external radiation source for cancer therapy. It emits high-energy gamma rays that can penetrate deep into body tissues to target and destroy cancer cells. Its long half-life of 5.27 years ensures that the source activity remains relatively constant over long periods, meaning the medical equipment does not require frequent replacement or recalibration. Storage and handling precautions differ significantly: Technetium-99m can be stored in short-term lead shield containers until it decays to safe levels. Cobalt-60, due to its long-term high activity and high penetrative power, must be permanently housed in thick, heavy lead or concrete shielding, manipulated using remote-control mechanisms, and stored in highly secured facilities with restricted access.

Marking points (Max 8 marks): 1. Identify Tc-99m emits gamma radiation which is highly penetrating and escapes the body. 2. Explain Tc-99m's short half-life (6 hours) reduces patient radiation exposure. 3. State Tc-99m has low ionising power, reducing damage to healthy tissue. 4. Identify Co-60 emits high-energy gamma which penetrates deep into the body to kill cancer cells. 5. Explain Co-60's long half-life (5.27 years) provides a stable, long-lived source activity requiring less replacement. 6. Safe handling of Tc-99m requires short-term storage and local shielding (e.g. syringe shields). 7. Safe handling of Co-60 requires thick permanent shielding (lead/concrete) and remote operation. 8. Logical organization showing comparative structure of both diagnostic and therapeutic applications.

According to the kinetic theory, the temperature of an ideal gas is directly proportional to the average kinetic energy of its molecules, as given by \(E_k = \frac{3}{2}kT\). Increasing the temperature increases the mean square speed of the gas molecules. The pressure exerted by the gas is due to the collisions of these molecules with the container walls. When a molecule collides with a wall, it undergoes a change in momentum because its velocity changes direction. According to Newton's Second Law, the force exerted on the wall during each collision is equal to the rate of change of momentum of the molecule. Since the molecules are moving faster at a higher temperature, they experience a greater change in momentum during each collision. Additionally, because they are moving faster in a container of fixed volume, they travel between walls more quickly, leading to an increased frequency of collisions with the walls. Therefore, both the greater momentum change per collision and the higher frequency of collisions increase the total rate of change of momentum, resulting in a larger average force exerted on the walls. Since pressure is defined as force per unit area (\(P = F/A\)) and the surface area of the rigid container is constant, this increased force results in an increased pressure.

Marking points (Max 8 marks): 1. State temperature is proportional to the average kinetic energy of the molecules. 2. Conclude that higher temperature means higher mean square speed of molecules. 3. State that gas pressure is caused by collisions of molecules with container walls. 4. Explain that each collision results in a change of momentum of the molecules. 5. State that force is equal to the rate of change of momentum (Newton's Second Law). 6. Explain that faster molecules have a greater change in momentum per collision. 7. Explain that faster molecules collide more frequently with the walls. 8. Combine these to state that total average force increases, and since pressure is force/area (with constant area), the pressure increases.

In both accelerators, charged particles are accelerated across gaps by an alternating electric field/potential difference. A LINAC consists of a series of collinear, hollow metal drift tubes. Acceleration occurs only in the gaps between these tubes where the electric field exists; inside the tubes, the electric field is zero (shielding effect). Because the particles accelerate and speed up, they cover greater distances in the same time. To keep the particles in phase with the alternating electric voltage of constant frequency, the drift tubes must be made progressively longer, so that the time spent in each tube remains constant at \(T/2\). A Cyclotron consists of two hollow D-shaped electrodes (Dees) placed inside a vacuum chamber, with a uniform magnetic field applied perpendicularly to the plane of the Dees. The magnetic field provides a centripetal force (\(Bqv = \frac{mv^2}{r}\)), making the particles move in circular arcs, but does not change their speed since the force is always perpendicular to velocity. Acceleration occurs only when particles cross the gap between the Dees where an alternating electric field is applied. As the particles gain speed in the gaps, the radius of their path increases (\(r = \frac{mv}{Bq}\)). The time to complete a semi-circle (\(T/2 = \frac{\pi m}{Bq}\)) is independent of the particle's speed. Consequently, a constant frequency alternating voltage source is used without needing to alter the geometry of the Dees.

Marking points (Max 8 marks): 1. State that both devices use alternating electric fields to accelerate charged particles across gaps. 2. Describe the LINAC structure as collinear, hollow metal drift tubes of increasing length. 3. Explain that the electric field inside the LINAC drift tubes is zero, shielding the particles. 4. Explain why LINAC drift tubes increase in length (to maintain constant time \(T/2\) in each tube as speed increases, keeping them in phase with constant AC frequency). 5. Describe the Cyclotron structure as two hollow D-shaped electrodes (Dees) under a perpendicular magnetic field. 6. Explain the role of the magnetic field in a Cyclotron (provides centripetal force to curve path, does no work / does not change speed). 7. Explain that the radius of the path in the Cyclotron increases as speed increases (\(r = \frac{mv}{Bq}\)). 8. Explain that the time spent in each Dee is constant (\(t = \frac{\pi m}{Bq}\)), which is independent of speed, allowing a constant frequency AC supply.