2023 OCR AS Level Physics A - H156 模擬試題連答案詳解

First, calculate the energy \(E\) of the incident photons:

\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{310 \times 10^{-9}} \approx 6.42 \times 10^{-19}\text{ J}\)

Convert this photon energy to electronvolts (\(eV\)):

\(E = \frac{6.42 \times 10^{-19}}{1.60 \times 10^{-19}} \approx 4.01\text{ eV}\)

Now, use the photoelectric equation to find the maximum kinetic energy \(E_{k\text{,max}}\):

\(E_{k\text{,max}} = E - \Phi = 4.01\text{ eV} - 2.4\text{ eV} = 1.61\text{ eV} \approx 1.6\text{ eV}\)

1 mark for the correct option A.

First, find the final velocity \(v\) at the end of the uniform acceleration period using the equations of motion:

\(s_1 = \frac{u + v}{2} t_1\)

Since the car starts from rest, \(u = 0\):

\(120 = \frac{0 + v}{2} \times 8.0 \implies 120 = 4.0 v \implies v = 30\text{ m s}^{-1}\)

Next, calculate the distance \(s_2\) travelled during the \(4.0\text{ s}\) constant-velocity phase:

\(s_2 = v \times t_2 = 30 \times 4.0 = 120\text{ m}\)

Therefore, the total distance travelled is:

\(s_{\text{total}} = s_1 + s_2 = 120 + 120 = 240\text{ m}\)

1 mark for the correct option C.

Determine the percentage uncertainty for each variable:

* For \(V\): \(\frac{0.12}{6.00} \times 100\% = 2.0\%\)
* For \(d\): \(\frac{0.01}{0.50} \times 100\% = 2.0\%\). Since the formula contains \(d^2\), this contributes \(2 \times 2.0\% = 4.0\%\) to the total percentage uncertainty.
* For \(I\): \(\frac{0.05}{2.00} \times 100\% = 2.5\%\)
* For \(L\): \(\frac{0.002}{1.000} \times 100\% = 0.2\%\)

Sum the percentage uncertainties to find the overall percentage uncertainty in \(\rho\):

\(\% \text{ uncertainty in } \rho = 2.0\% + 4.0\% + 2.5\% + 0.2\% = 8.7\%\)

1 mark for the correct option C.

First, calculate the total resistance of the circuit:

\(R_{\text{total}} = R + r = 6.5 + 1.5 = 8.0\ \Omega\)

Next, calculate the current \(I\) flowing through the circuit:

\(I = \frac{E}{R_{\text{total}}} = \frac{12.0}{8.0} = 1.5\text{ A}\)

Now, calculate the terminal potential difference \(V\):

\(V = E - Ir = 12.0 - (1.5 \times 1.5) = 12.0 - 2.25 = 9.75\text{ V}\)

Finally, calculate the power \(P_r\) dissipated in the internal resistance \(r\):

\(P_r = I^2 r = (1.5)^2 \times 1.5 = 2.25 \times 1.5 = 3.375\text{ W} \approx 3.38\text{ W}\)

1 mark for the correct option B.

First, determine the extension \(\Delta L\) of the wire using the formula for Young modulus \(E = \frac{FL}{A\Delta L}\):

\(\Delta L = \frac{FL}{AE} = \frac{120 \times 2.0}{1.2 \times 10^{-6} \times 2.0 \times 10^{11}} = \frac{240}{2.4 \times 10^5} = 1.0 \times 10^{-3}\text{ m}\)

Now, compute the elastic potential energy \(E_p\) stored:

\(E_p = \frac{1}{2} F \Delta L = \frac{1}{2} \times 120 \times 1.0 \times 10^{-3} = 0.060\text{ J}\)

1 mark for the correct option B.

First, calculate the wavelength \(\lambda\) of the electromagnetic wave:

\(\lambda = \frac{c}{f} = \frac{3.00 \times 10^8}{6.0 \times 10^{14}} = 5.0 \times 10^{-7}\text{ m}\)

Now, find the phase difference \(\Delta \phi\) in radians:

\(\Delta \phi = \frac{2\pi \Delta x}{\lambda} = \frac{2\pi \times 1.5 \times 10^{-7}}{5.0 \times 10^{-7}} = 0.6\pi\text{ rad}\)

1 mark for the correct option B.

First, calculate the fringe spacing \(w\) using the double-slit formula:

\(w = \frac{\lambda D}{s} = \frac{633 \times 10^{-9} \times 1.8}{0.45 \times 10^{-3}} = 2.532 \times 10^{-3}\text{ m} = 2.532\text{ mm}\)

The third-order bright fringe is at a distance of \(3w\) from the central bright fringe:

\(x = 3 w = 3 \times 2.532 = 7.596\text{ mm} \approx 7.6\text{ mm}\)

1 mark for the correct option C.

The pressure \(P\) exerted on the table is given by:

\(P = \frac{F}{A} = \frac{mg}{A} = \frac{\rho V g}{A}\)

Since \(V = A \times h\), the equation simplifies to:

\(P = \rho h g\)

Substitute the given values:

\(P = 7800\text{ kg m}^{-3} \times 0.40\text{ m} \times 9.81\text{ m s}^{-2} = 30607.2\text{ Pa} \approx 31\text{ kPa}\)

1 mark for the correct option C.

First, calculate the energy of the incident photons: \(E = hf = (6.63 \times 10^{-34}\text{ J s}) \times (1.20 \times 10^{15}\text{ Hz}) = 7.96 \times 10^{-19}\text{ J}\). Next, convert the work function to joules: \(\Phi = 3.20\text{ eV} \times (1.60 \times 10^{-19}\text{ J eV}^{-1}) = 5.12 \times 10^{-19}\text{ J}\). The maximum kinetic energy is given by the photoelectric equation: \(E_{k\text{,max}} = hf - \Phi = 7.96 \times 10^{-19}\text{ J} - 5.12 \times 10^{-19}\text{ J} = 2.84 \times 10^{-19}\text{ J}\). This rounds to \(2.8 \times 10^{-19}\text{ J}\).

1 mark for the correct answer A.

Using the equation of motion for constant acceleration: \(s = ut + \frac{1}{2}at^2\). Let upwards be the positive direction. Here, \(u = +12.0\text{ m s}^{-1}\), \(a = -9.81\text{ m s}^{-2}\), and \(t = 3.50\text{ s}\). The displacement \(s\) is: \(s = (12.0 \times 3.50) + \frac{1}{2}(-9.81)(3.50)^2 = 42.0 - 60.086 = -18.086\text{ m}\). The negative sign indicates that the final position is below the starting point, so the height of the cliff is \(18.1\text{ m}\).

1 mark for the correct answer A.

The formula for the density of a cylinder is \(\rho = \frac{m}{V} = \frac{4m}{\pi d^2 L}\). When combining uncertainties, the fractional uncertainties (or percentage uncertainties) add. For variables raised to a power, the percentage uncertainty is multiplied by that power. Therefore, the percentage uncertainty in density is: \(\frac{\Delta \rho}{\rho} \times 100\% = \left( \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L} \right) \times 100\% = 1.5\% + 2(1.2\%) + 2.0\% = 5.9\%\).

1 mark for the correct answer B.

Using the relationship \(E = V + Ir\) and Ohm's law \(I = \frac{V}{R}\). For the first case: \(I_1 = \frac{1.6\text{ V}}{4.0\ \Omega} = 0.40\text{ A}\), leading to \(E = 1.6 + 0.40r\). For the second case: \(I_2 = \frac{1.8\text{ V}}{9.0\ \Omega} = 0.20\text{ A}\), leading to \(E = 1.8 + 0.20r\). Equating the two expressions for \(E\): \(1.6 + 0.40r = 1.8 + 0.20r \Rightarrow 0.20r = 0.2 \Rightarrow r = 1.0\ \Omega\).

1 mark for the correct answer B.

The Young modulus \(E\) is given by \(E = \frac{FL}{A\Delta L}\), so the extension is \(\Delta L = \frac{FL}{AE}\). Since the cross-sectional area \(A = \frac{\pi d^2}{4}\), we have \(\Delta L \propto \frac{L}{d^2}\) because \(F\) and \(E\) are constant for both wires. For wire X, \(\Delta L_X \propto \frac{L}{d^2}\). For wire Y, \(\Delta L_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{1}{2}\frac{L}{d^2}\). Therefore, the ratio \(\frac{\Delta L_X}{\Delta L_Y} = \frac{1}{1/2} = 2\).

1 mark for the correct answer C.

For a point source, the intensity \(I\) decreases with distance \(r\) according to the inverse-square law: \(I \propto \frac{1}{r^2}\). Additionally, intensity is directly proportional to the square of the amplitude: \(I \propto A^2\). Thus, \(A^2 \propto \frac{1}{r^2}\), which gives \(A \propto \frac{1}{r}\). Therefore, at a distance of \(3r\), the amplitude of the wave is \(\frac{1}{3} A_0\).

1 mark for the correct answer B.

The fringe separation \(x\) is given by \(x = \frac{\lambda D}{a}\), where \(\lambda = 600 \times 10^{-9}\text{ m}\), \(D = 1.8\text{ m}\), and \(a = 0.15 \times 10^{-3}\text{ m}\). Calculating \(x\): \(x = \frac{(600 \times 10^{-9}) \times 1.8}{0.15 \times 10^{-3}} = 7.2 \times 10^{-3}\text{ m} = 7.2\text{ mm}\). The distance from the central bright fringe to the third-order bright fringe is \(3x = 3 \times 7.2\text{ mm} = 21.6\text{ mm}\).

1 mark for the correct answer C.

First, find the weight \(W\) of the sphere: \(W = m g = \rho_{\text{brass}} V g = (8.5 \times 10^3\text{ kg m}^{-3}) \times (2.0 \times 10^{-4}\text{ m}^3) \times 9.81\text{ m s}^{-2} = 16.68\text{ N}\). Next, find the upthrust \(U\) acting on the sphere using Archimedes' principle: \(U = \rho_{\text{water}} V g = (1.0 \times 10^3\text{ kg m}^{-3}) \times (2.0 \times 10^{-4}\text{ m}^3) \times 9.81\text{ m s}^{-2} = 1.96\text{ N}\). In equilibrium, the tension \(T\) in the string plus the upthrust equals the weight of the sphere: \(T + U = W \Rightarrow T = W - U = 16.68\text{ N} - 1.96\text{ N} = 14.72\text{ N}\), which rounds to \(14.7\text{ N}\).

1 mark for the correct answer B.

From Einstein's photoelectric equation:

\(hf = \Phi + E_k\)

where \(h\) is the Planck constant and \(f\) is the frequency of the incident light.

When the frequency is doubled to \(2f\), the new maximum kinetic energy \(E_k'\) of the photoelectrons is given by:

\(E_k' = h(2f) - \Phi = 2(hf) - \Phi\)

Substituting \(hf = E_k + \Phi\) into this equation:

\(E_k' = 2(E_k + \Phi) - \Phi = 2E_k + 2\Phi - \Phi = 2E_k + \Phi\)

Therefore, the correct response is B.

1 mark for the correct answer of B.

The percentage uncertainty in a quantity formed by multiplication and division is found by adding the percentage uncertainties of each variable.

When a variable is raised to a power, its percentage uncertainty is multiplied by that power. Here, diameter \(d\) is squared, so its percentage uncertainty is multiplied by 2:

\(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\)

Substituting the given percentages:

Percentage uncertainty in \(\rho = 5\% + 2(2\%) + 1\% = 5\% + 4\% + 1\% = 10\%\)

Therefore, the correct response is B.

1 mark for the correct answer of B.

Using Ohm's law, we determine the current \(I\) flowing through the circuit in each case:

For \(R = 3.0\ \Omega\):
\(I_1 = \frac{V_1}{R_1} = \frac{1.5\text{ V}}{3.0\ \Omega} = 0.5\text{ A}\)

For \(R = 8.0\ \Omega\):
\(I_2 = \frac{V_2}{R_2} = \frac{2.0\text{ V}}{8.0\ \Omega} = 0.25\text{ A}\)

Using the e.m.f. equation \(E = V + Ir\):

From Case 1:
\(E = 1.5 + 0.5r\) (Equation 1)

From Case 2:
\(E = 2.0 + 0.25r\) (Equation 2)

Equating both expressions for \(E\):

\(1.5 + 0.5r = 2.0 + 0.25r\)

\(0.25r = 0.5\)

\(r = 2.0\ \Omega\)

Substituting \(r = 2.0\ \Omega\) back into Equation 1:

\(E = 1.5 + 0.5(2.0) = 2.5\text{ V}\)

Therefore, the correct response is B.

1 mark for the correct answer of B.

The Young modulus \(E\) of a material is defined as:

\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta x / L} = \frac{F L}{A \Delta x}\)

Since the cross-sectional area of a wire of diameter \(d\) is \(A = \frac{\pi d^2}{4}\), we can write the extension \(\Delta x\) as:

\(\Delta x = \frac{4 F L}{\pi d^2 E}\)

Since both wires are made of the same material, they have the same Young modulus \(E\). They also experience the same force \(F\). Thus, extension is proportional to the ratio \(\frac{L}{d^2}\):

\(\Delta x \propto \frac{L}{d^2}\)

For wire X:
\(\Delta x_X \propto \frac{L}{d^2}\)

For wire Y:
\(\Delta x_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{1}{2}\frac{L}{d^2}\)

Taking the ratio:

\(\frac{\Delta x_X}{\Delta x_Y} = \frac{1}{0.5} = 2.0\)

Therefore, the correct response is D.

1 mark for the correct answer of D.

a) The work function is the minimum energy required to remove a single electron from the surface of a metal. b) i) The energy of a photon is given by \(E = \frac{hc}{\lambda}\). Substituting the given values: \(E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{2.40 \times 10^{-7}} = 8.29 \times 10^{-19} \text{ J}\). Converting this energy into electron-volts: \(E = \frac{8.29 \times 10^{-19}}{1.60 \times 10^{-19}} = 5.18 \text{ eV}\), which is approximately \(5.2 \text{ eV}\). ii) According to the photoelectric equation: \(\text{KE}_{\text{max}} = h\nu - \Phi\). Converting the work function to Joules: \(\Phi = 4.30 \times 1.60 \times 10^{-19} = 6.88 \times 10^{-19} \text{ J}\). Thus, \(\text{KE}_{\text{max}} = 8.29 \times 10^{-19} - 6.88 \times 10^{-19} = 1.41 \times 10^{-19} \text{ J}\). iii) The maximum kinetic energy is related to the maximum speed by \(\text{KE}_{\text{max}} = \frac{1}{2} m_e v^2\). Rearranging for \(v\): \(v = \sqrt{\frac{2 \text{KE}_{\text{max}}}{m_e}} = \sqrt{\frac{2 \times 1.41 \times 10^{-19}}{9.11 \times 10^{-31}}} = 5.56 \times 10^5 \text{ m s}^{-1}\).

a) Minimum energy required [1 mark] to remove an electron from the surface of a metal [1 mark]. b) i) Use of \(E = \frac{hc}{\lambda}\) [1 mark], calculation of energy in J as \(8.29 \times 10^{-19} \text{ J}\) [1 mark], division by \(1.60 \times 10^{-19}\) to show value is \(5.18 \text{ eV}\) which rounds to \(5.2 \text{ eV}\) [1 mark]. ii) Correct conversion of work function to Joules (\(6.88 \times 10^{-19} \text{ J}\)) [1 mark], correct subtraction to yield maximum kinetic energy (\(1.41 \times 10^{-19} \text{ J}\), accept \(1.4 \times 10^{-19} \text{ J}\)) [1 mark]. iii) Correct equation for kinetic energy used [1 mark], correct substitution of electron mass (\(9.11 \times 10^{-31} \text{ kg}\)) [1 mark], correct final value of speed (\(5.56 \times 10^5 \text{ m s}^{-1}\), accept \(5.6 \times 10^5 \text{ m s}^{-1}\)) [1 mark].

a) Comparing \(s = 24t - 3.0t^2\) with the kinematic equation \(s = ut + \frac{1}{2}at^2\), we find the initial velocity \(u = 24 \text{ m s}^{-1}\). b) Differentiating the displacement equation with respect to time gives velocity: \(v = \frac{ds}{dt} = 24 - 6.0t\). Differentiating again gives acceleration: \(a = \frac{dv}{dt} = -6.0 \text{ m s}^{-2}\). Since this value is independent of time \(t\), the acceleration is constant with a magnitude of \(6.0 \text{ m s}^{-2}\). c) The car comes to rest when \(v = 0\), so \(24 - 6.0t = 0\), which gives \(t = 4.0 \text{ s}\). Substituting \(t = 4.0\) into the displacement equation gives: \(s = 24(4.0) - 3.0(4.0)^2 = 96 - 48 = 48 \text{ m}\). d) Using Newton's second law: \(F = ma = 1200 \text{ kg} \times 6.0 \text{ m s}^{-2} = 7200 \text{ N}\). e) The velocity-time graph is a straight line starting at \((0, 24)\) on the vertical velocity axis and ending at \((4.0, 0)\) on the horizontal time axis.

a) \(24 \text{ m s}^{-1}\) [1 mark]. b) Differentiates once to find \(v = 24 - 6.0t\) or compares with standard equation to identify \(u = 24\) [1 mark], differentiates again or identifies \(\frac{1}{2}a = -3.0\) to show \(a = -6.0 \text{ m s}^{-2}\) [1 mark], states that because acceleration is independent of time, it is constant, with a magnitude of \(6.0 \text{ m s}^{-2}\) [1 mark]. c) Sets velocity to zero to find \(t = 4.0 \text{ s}\) [1 mark], substitutes \(t = 4.0 \text{ s}\) into displacement equation or uses \(v^2 = u^2 + 2as\) [1 mark], calculates stopping distance as \(48 \text{ m}\) [1 mark]. d) Uses \(F = ma\) to obtain \(7200 \text{ N}\) [1 mark]. e) Describes or draws a straight line with a negative gradient [1 mark], with y-intercept at \(24 \text{ m s}^{-1}\) and x-intercept at \(4.0 \text{ s}\) [1 mark].

a) The diagram must show a single loop containing the cell, a variable resistor, an ammeter, and a switch in series. A voltmeter must be connected in parallel across the terminals of the cell. b) i) The graph of \(V\) against \(I\) is a straight line with a negative gradient. This is because \(V = E - Ir\), which can be rewritten as \(V = -rI + E\). The y-intercept of the line is equal to the e.m.f. \(E\) of the cell, and the magnitude of the gradient of the line is equal to the internal resistance \(r\). c) i) Using the equation \(E = V + Ir\): \(E = 1.40 \text{ V} + (0.25 \text{ A} \times 0.80\ \Omega) = 1.40 + 0.20 = 1.60 \text{ V}\). ii) The efficiency \(\eta\) is the useful power output divided by the total power input: \(\eta = \frac{VI}{EI} = \frac{V}{E}\). Substituting the values: \(\eta = \frac{1.40}{1.60} = 0.875\) or \(87.5\%\).

a) Correct circuit symbols with cell, variable resistor, ammeter, and switch in series [1 mark], voltmeter correctly connected in parallel across the cell [1 mark]. b) Graph described or drawn as a straight line with a negative gradient [1 mark], equation \(V = E - Ir\) or \(V = -rI + E\) stated or implied [1 mark], identifies y-intercept as \(E\) [1 mark], identifies gradient as \(-r\) (or magnitude of gradient as \(r\)) [1 mark]. c) i) Correct equation \(E = V + Ir\) [1 mark], correct calculation yielding \(1.60 \text{ V}\) [1 mark]. ii) Correct formula for efficiency \(\frac{V}{E}\) or \(\frac{I^2 R}{I^2(R+r)}\) [1 mark], correct calculation yielding \(0.875\) or \(87.5\%\) [1 mark].

a) The cross-sectional area is given by \(A = \pi r^2\) or \(A = \frac{\pi d^2}{4}\). With \(d = 0.80 \text{ mm} = 8.0 \times 10^{-4} \text{ m}\): \(A = \frac{\pi \times (8.0 \times 10^{-4})^2}{4} = 5.03 \times 10^{-7} \text{ m}^2\), which is approximately \(5.0 \times 10^{-7} \text{ m}^2\). b) The Young modulus is defined as \(E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{x/L} = \frac{FL}{Ax}\). Rearranging for extension \(x\): \(x = \frac{FL}{AE} = \frac{45 \times 2.20}{5.03 \times 10^{-7} \times 1.10 \times 10^{11}} = 1.79 \times 10^{-3} \text{ m} = 1.79 \text{ mm}\) (accept \(1.8 \text{ mm}\) if using \(5.0 \times 10^{-7} \text{ m}^2\)). c) The elastic strain energy \(E_k\) stored is \(E_k = \frac{1}{2} F x = \frac{1}{2} \times 45 \times 1.79 \times 10^{-3} = 4.03 \times 10^{-2} \text{ J}\) (accept \(4.0 \times 10^{-2} \text{ J}\)). d) i) Since stress is \(\sigma = \frac{F}{A}\) and area \(A \propto d^2\), doubling the diameter increases the area by a factor of 4. Since the force \(F\) is identical, the stress in the second wire is \(\frac{1}{4}\) of the stress in the first wire. The ratio is \(0.25\) (or \(1:4\)). ii) Extension \(x = \frac{FL}{AE}\). Since \(F, L,\) and \(E\) are identical, and the area of the second wire is 4 times larger, the extension is divided by 4. The ratio of extensions is \(0.25\) (or \(1:4\)).

a) Uses \(A = \pi r^2\) or \(A = \frac{\pi d^2}{4}\) with correct radius \(0.40 \times 10^{-3} \text{ m}\) or diameter [1 mark], calculates \(5.03 \times 10^{-7} \text{ m}^2\) and states it is approximately \(5.0 \times 10^{-7} \text{ m}^2\) [1 mark]. b) Recalls or uses \(E = \frac{FL}{Ax}\) [1 mark], substitutes values correctly [1 mark], calculates extension as \(1.8 \text{ mm}\) or \(1.79 \text{ mm}\) [1 mark]. c) Recalls and uses \(\text{Energy} = \frac{1}{2} F x\) [1 mark], calculates stored energy as \(4.0 \times 10^{-2} \text{ J}\) (or \(4.1 \times 10^{-2} \text{ J}\) if using \(1.8 \text{ mm}\)) [1 mark]. d) i) Explains that doubling diameter increases cross-sectional area by 4 [1 mark], states stress ratio is \(0.25\) (or \(1:4\)) [1 mark]. ii) Correctly states extension ratio is \(0.25\) (or \(1:4\)) [1 mark].

a) First, calculate the volume of the block: \(V = \text{width} \times \text{length} \times \text{height} = 0.050 \times 0.080 \times 0.120 = 4.80 \times 10^{-4} \text{ m}^3\). The mass is \(m = \rho V = 8500 \text{ kg m}^{-3} \times 4.80 \times 10^{-4} \text{ m}^3 = 4.08 \text{ kg}\). b) i) Hydrostatic pressure increases with depth because of the equation \(p = h \rho g\). The bottom face of the submerged block is at a greater depth than the top face, so the pressure on the bottom face is higher. This difference in pressure produces a net upward force, which is the upthrust. ii) Upthrust is equal to the weight of the fluid displaced (Archimedes' principle): \(U = V_{\text{sub}} \rho_{\text{water}} g\). Substituting the values: \(U = 4.80 \times 10^{-4} \text{ m}^3 \times 1000 \text{ kg m}^{-3} \times 9.81 \text{ m s}^{-2} = 4.71 \text{ N}\), which is approximately \(4.7 \text{ N}\). iii) The forces acting on the submerged block are the downward weight \(W\), the upward tension from the spring balance \(T\) (which is the reading), and the upward upthrust \(U\). For equilibrium: \(T = W - U\). Calculate weight: \(W = mg = 4.08 \text{ kg} \times 9.81 \text{ m s}^{-2} = 40.02 \text{ N}\). Thus, the reading on the spring balance is: \(T = 40.02 \text{ N} - 4.71 \text{ N} = 35.31 \text{ N}\) (accept \(35.3 \text{ N}\)).

a) Correct calculation of volume \(4.80 \times 10^{-4} \text{ m}^3\) [1 mark], correct calculation of mass \(4.08 \text{ kg}\) [1 mark]. b) i) States that pressure increases with depth [1 mark], states that the pressure difference between the bottom and top surface results in a net upward force [1 mark]. ii) Recalls upthrust formula \(U = V \rho g\) or calculates pressure difference \(\Delta p = \Delta h \rho g\) [1 mark], substitutes water density and block volume correctly [1 mark], calculates upthrust as \(4.71 \text{ N}\) [1 mark]. iii) Correctly relates tension, weight, and upthrust as \(T = W - U\) [1 mark], calculates weight of block as \(40.0 \text{ N}\) or \(40.02 \text{ N}\) [1 mark], calculates final reading as \(35.3 \text{ N}\) (accept \(35.3 \text{ N}\) to \(35.4 \text{ N}\) depending on rounding) [1 mark].

Part (a): Work function in Joules is calculated by multiplying the work function in eV by the elementary charge: \\Phi = 2.36 \\times 1.60 \\times 10^{-19} = 3.78 \\times 10^{-19}\\text{ J}. Part (b): The energy of each incident photon is E = hc / \\lambda = (6.63 \\times 10^{-34} \\times 3.00 \\times 10^8) / (240 \\times 10^{-9}) = 8.29 \\times 10^{-19}\\text{ J}. The maximum kinetic energy of the photoelectrons is E_k = E - \\Phi = 8.29 \\times 10^{-19} - 3.78 \\times 10^{-19} = 4.51 \\times 10^{-19}\\text{ J}. Part (c): (i) The maximum kinetic energy remains unchanged because the energy of individual photons (E = hf) depends only on frequency/wavelength, which has not changed. (ii) The rate of photoelectron emission doubles. Doubling the intensity of the light source doubles the number of photons incident per second. Since there is a one-to-one interaction between photons and electrons, this results in twice as many photoelectrons being emitted per second.

Part (a) [2 marks]: 1 mark for using W = eV conversion factor (1.60 x 10^-19), 1 mark for correct final value of 3.78 x 10^-19 J (allow 3.776 x 10^-19 J). Part (b) [4 marks]: 1 mark for using photon energy equation E = hc/\\lambda, 1 mark for correct calculation of photon energy as 8.29 x 10^-19 J, 1 mark for using Einstein's photoelectric equation E_k = E - \\Phi, 1 mark for correct subtraction yielding 4.51 x 10^-19 J (allow 4.50 to 4.52 x 10^-19 J). Part (c) [4 marks]: 1 mark for stating that maximum kinetic energy remains unchanged, 1 mark for explaining that photon energy only depends on frequency/wavelength (which is constant), 1 mark for stating that rate of emission doubles, 1 mark for explaining that double intensity means double the number of photons per second, leading to more one-to-one interactions with electrons.

Part (a): Using v = u + at where u = 0, a = 3.5 m s^-2, and t = 4.0 s: v = 0 + 3.5 * 4.0 = 14 m s^-1. Part (b): The graph consists of three linear segments: a straight line from (0,0) to (4.0 s, 14 m s^-1), a horizontal line from (4.0 s, 14 m s^-1) to (10.0 s, 14 m s^-1), and a straight line descending from (10.0 s, 14 m s^-1) to (13.0 s, 0). Part (c): Total distance is the area under the velocity-time graph. Area of trapezium = 1/2 * (sum of parallel sides) * height = 1/2 * (6.0 + 13.0) * 14 = 0.5 * 19.0 * 14 = 133 m. Part (d): Average velocity = total displacement / total time = 133 m / 13.0 s = 10.23 m s^-1 (rounded to 3 significant figures as 10.2 m s^-1).

Part (a) [1 mark]: 1 mark for clear substitution into v = u + at to show v = 14 m s^-1. Part (b) [3 marks]: 1 mark for describing constant acceleration from 0 to 4.0 s up to 14 m s^-1, 1 mark for constant velocity of 14 m s^-1 from 4.0 s to 10.0 s, 1 mark for uniform deceleration to 0 at t = 13.0 s. Part (c) [3 marks]: 1 mark for indicating that distance is the area under the v-t graph, 1 mark for correct calculation of individual areas or trapezium formula setup, 1 mark for final distance of 133 m. Part (d) [3 marks]: 1 mark for definition: average velocity = total displacement / total time, 1 mark for substituting 133 / 13.0, 1 mark for correct final value of 10.2 m s^-1 (accept 10 m s^-1 if 2 s.f. is used).

Part (a): A micrometer screw gauge or digital vernier calipers are suitable instruments because they can measure to a precision of 0.01 mm / 0.01 cm. Part (b): The volume of the cylinder V = (\\pi * d^2 * L) / 4 = (\\pi * 1.54^2 * 8.42) / 4 = 15.684 cm^3. The density \\rho = m / V = 145.2 / 15.684 = 9.258 g cm^-3. Part (c): Percentage uncertainty in mass: (0.1 / 145.2) * 100 = 0.069%. Percentage uncertainty in length: (0.02 / 8.42) * 100 = 0.238%. Percentage uncertainty in diameter: (0.01 / 1.54) * 100 = 0.649%. The formula for percentage uncertainty in density is %\\Delta \\rho = %\\Delta m + %\\Delta L + 2 * %\\Delta d = 0.069% + 0.238% + 2 * 0.649% = 1.605% (or 1.6%). Part (d): Absolute uncertainty in density = 9.258 * (1.605 / 100) = 0.149 g cm^-3. If we round uncertainty to 1 s.f., it is +/- 0.1 g cm^-3 or +/- 0.2 g cm^-3. Thus, density = (9.3 +/- 0.2) g cm^-3 (or keeping 2 s.f. in uncertainty: (9.26 +/- 0.15) g cm^-3).

Part (a) [1 mark]: 1 mark for recommending micrometer (screw gauge) or vernier calipers. Part (b) [4 marks]: 1 mark for correct formula for the volume of a cylinder V = \\pi d^2 L / 4, 1 mark for volume calculation of 15.68 cm^3, 1 mark for density formula \\rho = m/V, 1 mark for calculating density = 9.26 g cm^-3 (or 9.3 g cm^-3). Part (c) [3 marks]: 1 mark for calculating individual percentage uncertainties, 1 mark for adding percentage uncertainties with diameter's uncertainty doubled, 1 mark for final percentage uncertainty of 1.6% (accept 1.61%). Part (d) [2 marks]: 1 mark for calculating absolute uncertainty of approx 0.15 g cm^-3, 1 mark for quoting final value with matching decimal places, e.g., (9.3 +/- 0.2) g cm^-3 or (9.26 +/- 0.15) g cm^-3.

Part (a): The circuit diagram must show a cell/battery with internal resistance represented (typically inside a dashed box) in series with an ammeter and a variable resistor. A voltmeter must be connected in parallel across the variable resistor (or across the terminals of the cell). Part (b): The equation is V = E - Ir. Comparing this to the equation of a straight line y = mx + c: a graph of V on the y-axis against I on the x-axis gives a straight line with a negative gradient. The y-intercept of the line is equal to the e.m.f. E of the cell, and the magnitude of the gradient is equal to the internal resistance r. Part (c): Using E = I(R + r): Case 1: E = 0.60 * (2.2 + r) = 1.32 + 0.60r. Case 2: E = 0.30 * (5.0 + r) = 1.50 + 0.30r. Equating the two expressions for E: 1.32 + 0.60r = 1.50 + 0.30r, which yields 0.30r = 0.18 => r = 0.60 \\Omega. Substituting r back to find E: E = 0.60 * (2.2 + 0.60) = 0.60 * 2.80 = 1.68 V.

Part (a) [2 marks]: 1 mark for correct symbols for cell/battery, ammeter, and variable resistor connected in series, 1 mark for voltmeter connected in parallel across the variable resistor. Part (b) [4 marks]: 1 mark for stating the correct equation V = E - Ir, 1 mark for identifying V as y-axis and I as x-axis, 1 mark for identifying the y-intercept as the e.m.f. E, 1 mark for identifying the gradient as negative internal resistance (-r). Part (c) [4 marks]: 1 mark for setting up two simultaneous equations using E = I(R + r), 1 mark for equating the equations or substituting to eliminate one variable, 1 mark for correct calculation of r = 0.60 \\Omega, 1 mark for correct calculation of E = 1.68 V (accept 1.7 V).

Part (a): Tensile stress \\sigma = Force / Area = 90 / (4.5 \\times 10^{-7}) = 2.0 \\times 10^8\\text{ Pa}. Part (b): Young modulus E = stress / strain = \\sigma / \\epsilon. Therefore, strain \\epsilon = \\sigma / E = (2.0 \\times 10^8) / (2.0 \\times 10^{11}) = 1.0 \\times 10^{-3}. Extension x = \\epsilon * L = 1.0 \\times 10^{-3} * 2.4 = 2.4 \\times 10^{-3}\\text{ m} (or 2.4 mm). Part (c): Elastic strain energy E_s = 1/2 * Force * extension = 0.5 * 90 * 2.4 \\times 10^{-3} = 0.108 J (which rounds to 0.11 J). Part (d): In steel, strong metallic bonds must be stretched, requiring very high forces to produce small deformations. In contrast, polymeric materials involve weak intermolecular forces (such as van der Waals forces) between polymer chains that can easily slide or uncoil, resulting in a much lower Young modulus.

Part (a) [2 marks]: 1 mark for using stress = F / A, 1 mark for correct final value of 2.0 x 10^8 Pa. Part (b) [4 marks]: 1 mark for relating Young modulus, stress, and strain: E = stress / strain, 1 mark for calculating strain = 1.0 x 10^-3, 1 mark for relating extension to strain and original length: x = strain * L, 1 mark for correct extension of 2.4 x 10^-3 m (or 2.4 mm). Part (c) [3 marks]: 1 mark for using work done / strain energy formula E_s = 1/2 F x, 1 mark for substituting values: 0.5 * 90 * 2.4 x 10^-3, 1 mark for correct energy of 0.11 J (or 0.108 J). Part (d) [1 mark]: 1 mark for mentioning strong metallic bonds in steel versus weak intermolecular forces / uncoiling of chains in polymers.

Part (a): Phase difference is defined as the fraction of a cycle by which two points or waves are out of step with each other, expressed as an angle in degrees or radians. Part (b): A phase difference of \\frac{\\pi}{3}\\text{ rad} corresponds to a fraction of a full cycle given by \\frac{\\pi/3}{2\\pi} = \\frac{1}{6}. Since there are no other peaks or troughs between P and Q, the physical distance between them is equal to \\frac{1}{6} of a wavelength: \\Delta x = \\frac{\\lambda}{6}. Therefore, \\lambda = 6 \\times 0.15\\text{ m} = 0.90\\text{ m}. Part (c): Using the wave equation: v = f \\lambda = 50\\text{ Hz} \\times 0.90\\text{ m} = 45\\text{ m s}^{-1}. Part (d): The vertical displacement of P and Q can be written as y_P = A \\sin(\\omega t) and y_Q = A \\sin(\\omega t - \\pi/3). The difference in displacement is \\Delta y = y_P - y_Q = A \\sin(\\omega t) - A \\sin(\\omega t - \\pi/3). Using the trigonometric identity \\sin A - \\sin B = 2 \\cos(\\frac{A+B}{2}) \\sin(\\frac{A-B}{2}), we get \\Delta y = 2 A \\cos(\\omega t - \\frac{\\pi}{6}) \\sin(\\frac{\\pi}{6}). Since \\sin(\\frac{\\pi}{6}) = 0.5, the equation simplifies to \\Delta y = A \\cos(\\omega t - \\frac{\\pi}{6}). The maximum value of the cosine term is 1, so the maximum vertical separation is exactly equal to the amplitude, which is 4.0 mm. (Alternatively, this occurs when \\omega t = \\frac{\\pi}{6}, giving y_P = A \\sin(\\frac{\\pi}{6}) = 0.5 A = 2.0 mm, and y_Q = A \\sin(-\\frac{\\pi}{6}) = -0.5 A = -2.0 mm, yielding a total separation of 4.0 mm).

Part (a) [2 marks]: 1 mark for describing 'out of step' or relative position of two waves/points on a wave, 1 mark for expressing this as an angle in degrees/radians or fraction of a cycle. Part (b) [3 marks]: 1 mark for stating that a phase difference of \\pi/3 rad corresponds to 1/6 of a wavelength, 1 mark for writing the relation \\Delta x = \\lambda / 6, 1 mark for substituting 0.15 m to show \\lambda = 0.90 m. Part (c) [2 marks]: 1 mark for using v = f \\lambda, 1 mark for correct calculation: 45 m s^-1. Part (d) [3 marks]: 1 mark for stating that maximum vertical separation is 4.0 mm, 1 mark for explaining that the separation is maximized when the displacements are equal in magnitude but opposite in sign (e.g., +2.0 mm and -2.0 mm), 1 mark for demonstrating via trigonometry or phasor representation that \\Delta y_max = A = 4.0 mm.

To find the Young modulus \( E \) of a metal wire, we use the formula \( E = \frac{\text{stress}}{\text{strain}} = \frac{FL}{Ae} \), where \( F \) is the applied force, \( L \) is the original length, \( A \) is the cross-sectional area, and \( e \) is the extension.

**Experimental Setup & Procedure:**
1. Suspend a long, thin wire from a rigid clamp, passing it over a pulley at the end of the bench, and attach a mass hanger to the free end. Alternatively, use a double-wire system (one test wire, one reference wire) to eliminate temperature expansion errors.
2. Measure the original length \( L \) of the test wire from the clamp to a reference marker on the wire using a metre rule.
3. Measure the diameter \( d \) of the wire at several different positions and orientations along its length using a micrometer screw gauge. Calculate the average diameter and then the cross-sectional area: \( A = \frac{\pi d^2}{4} \).
4. Add masses to the hanger (increasing the force \( F = mg \)) and measure the new position of the marker using a vernier scale or a ruler with a travelling microscope. Calculate the extension \( e \) for each mass added.
5. Repeat the process as masses are unloaded to check for elastic behaviour (the wire should return to its original length).

**Graphical Analysis:**
1. Plot a graph of applied force \( F \) (y-axis) against extension \( e \) (x-axis).
2. Identify the linear region of the graph which represents elastic deformation.
3. Calculate the gradient \( m \) of this linear region, where \( m = \frac{F}{e} \).
4. Since \( E = \frac{FL}{Ae} \), the Young modulus is calculated as: \( E = \text{gradient} \times \frac{L}{A} \).

**Uncertainty and Mitigation:**
- **Uncertainty in extension:** The extension is typically very small, leading to a high percentage uncertainty.
- **Mitigation:** Use a long wire (e.g., > 2 m) and a thin wire to maximise extension for a given load, and measure the extension using a high-precision device such as a travelling microscope or a vernier scale.

**Level 3 (5-6 marks):**
- Comprehensive description of the experimental setup, clear list of measurements (with correct instruments), logical graphical analysis to find the Young modulus, and a valid uncertainty mitigation technique.
- There is a well-developed, clear, and logically structured line of reasoning.

**Level 2 (3-4 marks):**
- Detailed description of the setup and measurements. The analysis explains how to find the Young modulus but may lack some detail (e.g., forgets to explain how to get area from diameter, or uses a simple ratio instead of a gradient of a graph).
- There is a line of reasoning with some structure.

**Level 1 (1-2 marks):**
- Limited description of the setup, measurements, or analysis. One or two correct instruments or equations mentioned.

**Indicative Content:**
- **Apparatus/Measurements:** Metre rule for original length \( L \); Micrometer for diameter \( d \) (multiple positions/orientations); Masses for load \( F \); Vernier scale/microscope/ruler for extension \( e \).
- **Analysis:** Cross-sectional area \( A = \pi d^2 / 4 \); Plot \( F \) vs \( e \); Young modulus \( E = \text{gradient} \times \frac{L}{A} \).
- **Uncertainty:** Small extension is a major source of uncertainty; mitigated by using a long/thin wire or a highly precise measurement tool (travelling microscope).

**Experimental Setup & Circuit:**
1. Connect the chemical cell in a series circuit with a variable resistor (rheostat), an ammeter, and a switch.
2. Connect a voltmeter in parallel directly across the terminals of the chemical cell to measure the terminal potential difference \( V \).

**Procedure:**
1. With the switch open, record the voltmeter reading (this provides an initial estimate of the e.m.f. \( E \)).
2. Close the switch and quickly adjust the variable resistor to a setting that gives a readable current on the ammeter and potential difference on the voltmeter.
3. Record the current \( I \) from the ammeter and the terminal potential difference \( V \) from the voltmeter.
4. Open the switch immediately after taking the readings to prevent continuous current flow.
5. Adjust the variable resistor to a new setting, close the switch, take the new readings for \( V \) and \( I \), and reopen the switch.
6. Repeat this process to obtain at least six sets of values for \( I \) and \( V \) over a wide range.

**Graphical Analysis:**
1. Use the equation relating terminal potential difference, e.m.f., and internal resistance: \( V = E - Ir \).
2. Rearrange this equation into the linear form \( y = mx + c \): \( V = -rI + E \).
3. Plot a graph of terminal potential difference \( V \) (y-axis) against current \( I \) (x-axis).
4. Draw a straight line of best fit through the data points.
5. The magnitude of the gradient of the line of best fit represents the internal resistance \( r \) of the cell (since \( \text{gradient} = -r \)).
6. The y-intercept of the line of best fit represents the e.m.f. \( E \) of the cell.

**Minimising Temperature Errors:**
- As current flows through the cell, electrical energy is dissipated as heat within its internal resistance (\( P = I^2 r \)), which increases the temperature of the cell.
- An increase in temperature can alter the rate of chemical reactions and the resistivity of the electrolyte, thereby changing both the e.m.f. and the internal resistance of the cell during the experiment.
- **Mitigation:** The switch must be opened between readings to minimise the total time current flows, keeping the temperature of the cell constant.

**Level 3 (5-6 marks):**
- Clear and complete circuit description (or diagram details) with correct placement of voltmeter and ammeter. Full experimental procedure detailed with systematic temperature control (opening switch). Comprehensive graphical analysis explaining how to find both \( E \) and \( r \) from the graph's y-intercept and gradient.
- There is a well-developed, clear, and logically structured line of reasoning.

**Level 2 (3-4 marks):**
- Identifies the correct circuit setup and describes how to take measurements of \( V \) and \( I \). Explains how to plot \( V \) against \( I \) and relates gradient to \( r \) or intercept to \( E \), but may lack detail on temperature mitigation or have slight omissions in graphical interpretation.
- There is a line of reasoning with some structure.

**Level 1 (1-2 marks):**
- Fragmented description of the circuit or procedure. Mentions plotting \( V \) and \( I \) or states the equation \( V = E - Ir \) without clear experimental context.

**Indicative Content:**
- **Circuit:** Cell, variable resistor, ammeter in series; voltmeter in parallel across cell.
- **Data Collection:** Measure pairs of \( V \) and \( I \) for multiple variable resistor settings.
- **Analysis:** Plot \( V \) vs \( I \). Gradient \( = -r \), y-intercept \( = E \).
- **Temperature Control:** Continuous current heats the cell, changing internal resistance; open switch between readings to minimise heating.