Question 1
(a)
(i) Define pressure. (1 mark)
________________________________________
(ii) Use the answer to (a)(i) to show that the SI base units of pressure are \(\mathrm{kg\,m^{-1}\,s^{-2}}\). (1 mark)
________________________________________
(b) A horizontal pipe has length \(L\) and a circular cross-section of radius \(R\). A liquid of density \(\rho\) flows through the pipe. The mass \(m\) of liquid flowing through the pipe in time \(t\) is given by
\(m=\frac{\pi(p_2-p_1)R^4\rho t}{8kL}\)
where \(p_1\) and \(p_2\) are the pressures at the ends of the pipe and \(k\) is a constant.
Determine the SI base units of \(k\). (3 marks)
SI base units = …………………………………………….
(c) An experiment is performed to determine the value of \(k\) by measuring the values of the other quantities in the equation in (b).
The values of \(L\) and \(R\) each have a percentage uncertainty of \(2\%\).
State and explain, quantitatively, which of these two quantities contributes more to the percentage uncertainty in the calculated value of \(k\). (1 mark)
________________________________________
________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 1.2: SI units — parts (a)(ii) and (b)
• 1.3: Errors and uncertainties — part (c)
▶️ Answer/Explanation
(a)(i) Definition of pressure [1 mark]
Pressure is the force acting per unit area normal to the force.
\(p=\frac{F}{A}\)
Answer: \(\boxed{\text{force per unit area}}\)
(a)(ii) SI base units of pressure [1 mark]
From
\(p=\frac{F}{A}\)
The SI base units of force are
\(\mathrm{N}=\mathrm{kg\,m\,s^{-2}}\)
Therefore,
\(\mathrm{unit\ of\ pressure}=\frac{\mathrm{kg\,m\,s^{-2}}}{\mathrm{m^2}}\)
\(=\mathrm{kg\,m^{-1}\,s^{-2}}\)
Answer: \(\boxed{\mathrm{kg\,m^{-1}\,s^{-2}}}\)
(b) SI base units of \(k\) [3 marks]
The equation is
\(m=\frac{\pi(p_2-p_1)R^4\rho t}{8kL}\)
The relevant SI base units are:
\(m:\ \mathrm{kg}\)
\(p:\ \mathrm{kg\,m^{-1}\,s^{-2}}\)
\(R:\ \mathrm{m}\)
\(\rho:\ \mathrm{kg\,m^{-3}}\)
\(t:\ \mathrm{s}\)
\(L:\ \mathrm{m}\)
Rearranging for \(k\), the base units are
\(\mathrm{unit\ of\ }k=\frac{(\mathrm{kg\,m^{-1}\,s^{-2}})(\mathrm{m^4})(\mathrm{kg\,m^{-3}})(\mathrm{s})}{(\mathrm{kg})(\mathrm{m})}\)
\(=\mathrm{kg\,m^{-1}\,s^{-1}}\)
Answer: \(\boxed{\mathrm{kg\,m^{-1}\,s^{-1}}}\)
(c) Percentage uncertainty [1 mark]
From the equation in (b),
\(k\propto\frac{R^4}{L}\)
Therefore, the percentage uncertainty contributed by \(R\) is multiplied by 4:
\(\text{uncertainty due to }R=4\times2\%=8\%\)
The contribution from \(L\) is only \(2\%\).
Hence, \(R\) contributes more to the percentage uncertainty in \(k\).
Answer: \(\boxed{R}\), because it contributes \(8\%\), compared with \(2\%\) from \(L\).
Question 2
(a) State what is meant by the centre of gravity of an object. (1 mark)
________________________________________
(b) Two blocks are on a horizontal beam that is pivoted at its centre of gravity, as shown in Fig. 2.1.
A large block of weight \(54\,\mathrm{N}\) is a distance of \(0.45\,\mathrm{m}\) from the pivot. A small block of weight \(2.4\,\mathrm{N}\) is a distance of \(0.95\,\mathrm{m}\) from the pivot and a distance of \(0.35\,\mathrm{m}\) from the right-hand end of the beam.
The right-hand end of the beam is connected to the ground by a string that is at an angle of \(30^\circ\) to the horizontal. The beam is in equilibrium.
(i) By taking moments about the pivot, calculate the tension \(T\) in the string. (3 marks)
\(T=\) ……………………………………………… \(\mathrm{N}\)
(ii) The string is cut so that the beam is no longer in equilibrium.
Calculate the magnitude of the resultant moment about the pivot acting on the beam immediately after the string is cut. (1 mark)
resultant moment = …………………………………………… \(\mathrm{N\,m}\)
(c) The beam in (b) rotates when the string is cut and the small block of weight \(2.4\,\mathrm{N}\) is projected through the air. Fig. 2.2 shows the last part of the path of the block before it hits the ground at point Y.
At point X on the path, the block has a speed of \(3.4\,\mathrm{m\,s^{-1}}\) and is at a height of \(1.8\,\mathrm{m}\) above the horizontal ground. Air resistance is negligible.
(i) Calculate the decrease in the gravitational potential energy of the block for its movement from X to Y. (2 marks)
decrease in gravitational potential energy = ………………………………………………. \(\mathrm{J}\)
(ii) Use your answer to (c)(i) and conservation of energy to determine the kinetic energy of the block at Y. (3 marks)
kinetic energy = ………………………………………………. \(\mathrm{J}\)
(iii) State the variation, if any, in the direction of the acceleration of the block as it moves from X to Y. (1 mark)
________________________________________
(iv) The block passes point X at time \(t_X\) and arrives at point Y at time \(t_Y\).
On Fig. 2.3, sketch a graph to show the variation of the magnitude of the horizontal component of the velocity of the block with time from \(t_X\) to \(t_Y\). Numerical values are not required.
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 4.1: Turning effects of forces — parts (b)(i) and (b)(ii)
• 5.2: Gravitational potential energy and kinetic energy — parts (c)(i) and (c)(ii)
• 5.1: Energy conservation — part (c)(ii)
• 3.2: Non-uniform motion — parts (c)(iii) and (c)(iv)
▶️ Answer/Explanation
(a) Centre of gravity [1 mark]
The centre of gravity is the point through which the whole weight of an object may be considered to act.
Answer: \(\boxed{\text{the point where the weight of the object is taken to act}}\)
(b)(i) Tension in the string [3 marks]
Since the beam is in equilibrium, the clockwise moments about the pivot equal the anticlockwise moments.
The distance from the pivot to the right-hand end of the beam is
\(0.95+0.35=1.30\,\mathrm{m}\)
Only the vertical component of the tension produces a moment about the pivot.
Vertical component of tension \(=T\sin30^\circ\).
Therefore,
\(54(0.45)=2.4(0.95)+(T\sin30^\circ)(1.30)\)
\(24.3=2.28+0.65T\)
\(T=33.9\,\mathrm{N}\)
Answer: \(\boxed{34\,\mathrm{N}}\)
(b)(ii) Resultant moment after the string is cut [1 mark]
Immediately after the string is cut, the tension no longer acts. The remaining moments are due to the two blocks.
\(\text{resultant moment}=54(0.45)-2.4(0.95)\)
\(\text{resultant moment}=24.3-2.28\)
\(\text{resultant moment}=22.02\,\mathrm{N\,m}\)
Answer: \(\boxed{22\,\mathrm{N\,m}}\)
(c)(i) Decrease in gravitational potential energy [2 marks]
The block falls through a vertical height of \(1.8\,\mathrm{m}\).
The weight of the block is \(2.4\,\mathrm{N}\), so
\(\Delta E_{\mathrm{P}}=W\Delta h\)
\(\Delta E_{\mathrm{P}}=2.4\times1.8\)
\(\Delta E_{\mathrm{P}}=4.32\,\mathrm{J}\)
Answer: \(\boxed{4.3\,\mathrm{J}}\)
(c)(ii) Kinetic energy at Y [3 marks]
First calculate the kinetic energy of the block at X.
The mass is obtained from \(W=mg\):
\(m=\frac{2.4}{9.81}=0.245\,\mathrm{kg}\)
Therefore,
\(E_{\mathrm{K,X}}=\frac{1}{2}mv^2\)
\(E_{\mathrm{K,X}}=\frac{1}{2}\left(\frac{2.4}{9.81}\right)(3.4)^2\)
\(E_{\mathrm{K,X}}\approx1.4\,\mathrm{J}\)
With negligible air resistance, the decrease in gravitational potential energy is converted into kinetic energy.
\(E_{\mathrm{K,Y}}=E_{\mathrm{K,X}}+\Delta E_{\mathrm{P}}\)
\(E_{\mathrm{K,Y}}=1.4+4.3\)
\(E_{\mathrm{K,Y}}=5.7\,\mathrm{J}\)
Answer: \(\boxed{5.7\,\mathrm{J}}\)
(c)(iii) Direction of acceleration [1 mark]
Once the block is in free flight and air resistance is negligible, the only force acting on it is its weight.
Therefore, its acceleration remains vertically downwards throughout its motion from X to Y.
Answer: \(\boxed{\text{no variation; acceleration is vertically downwards}}\)
(c)(iv) Horizontal component of velocity [1 mark]
The horizontal acceleration is zero because air resistance is negligible and gravity acts vertically downward.
Therefore, the horizontal component of velocity remains constant and non-zero from \(t_X\) to \(t_Y\).
Answer: \(\boxed{\text{a horizontal straight line at a constant non-zero value}}\)
Question 3
A block is pulled by a force \(X\) in a straight line along a rough horizontal surface, as shown in Fig. 3.1.
Assume that the total resistive force opposing the motion of the block is \(0.80\,\mathrm{N}\) at all speeds of the block.
The variation with time \(t\) of the magnitude of the force \(X\) is shown in Fig. 3.2.
(a)
(i) Define force. (1 mark)
________________________________________
(ii) Determine the change in momentum of the block from time \(t=0\) to time \(t=3.0\,\mathrm{s}\). (2 marks)
change in momentum = …………………………………….. \(\mathrm{kg\,m\,s^{-1}}\)
(b)
(i) Describe and explain the motion of the block between time \(t=3.0\,\mathrm{s}\) and time \(t=6.0\,\mathrm{s}\). (2 marks)
________________________________________
________________________________________
(ii) Force \(X\) produces a total power of \(2.0\,\mathrm{W}\) when moving the block between time \(t=3.0\,\mathrm{s}\) and time \(t=6.0\,\mathrm{s}\).
Calculate the distance moved by the block during this time interval. (3 marks)
distance = ……………………………………………… \(\mathrm{m}\)
(c) The block is at rest at time \(t=0\).
On Fig. 3.3, sketch a graph to show the variation of the momentum of the block with time \(t\) from \(t=0\) to \(t=6.0\,\mathrm{s}\). Numerical values of momentum are not required.
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 3.3: Linear momentum and its conservation — parts (a)(ii) and (c)
• 3.2: Non-uniform motion — part (b)(i)
• 5.1: Energy conservation — part (b)(ii)
▶️ Answer/Explanation
(a)(i) Definition of force [1 mark]
Force is the rate of change of momentum.
\(F=\frac{\Delta p}{\Delta t}\)
Answer: \(\boxed{\text{rate of change of momentum}}\)
(a)(ii) Change in momentum [2 marks]
The resultant force acting on the block is the pulling force minus the resistive force.
\(F_{\mathrm{resultant}}=1.4-0.80\)
\(F_{\mathrm{resultant}}=0.60\,\mathrm{N}\)
Using \(F=\frac{\Delta p}{\Delta t}\),
\(\Delta p=F_{\mathrm{resultant}}\Delta t\)
\(\Delta p=(1.4-0.80)(3.0)\)
\(\Delta p=1.8\,\mathrm{kg\,m\,s^{-1}}\)
Answer: \(\boxed{1.8\,\mathrm{kg\,m\,s^{-1}}}\)
(b)(i) Motion of the block from \(3.0\,\mathrm{s}\) to \(6.0\,\mathrm{s}\) [2 marks]
During this interval, the applied force \(X\) is equal to the total resistive force.
Therefore, the resultant force on the block is zero.
With zero resultant force, the acceleration is zero, so the block moves with constant velocity.
Answer: \(\boxed{\text{The block moves at constant velocity because the resultant force is zero.}}\)
(b)(ii) Distance moved [3 marks]
The power produced by force \(X\) is
\(P=Fv\)
Therefore,
\(v=\frac{P}{F}\)
\(v=\frac{2.0}{0.80}\)
\(v=2.5\,\mathrm{m\,s^{-1}}\)
The time interval is
\(\Delta t=6.0-3.0=3.0\,\mathrm{s}\)
Since the velocity is constant,
\(s=v\Delta t\)
\(s=2.5\times3.0\)
\(s=7.5\,\mathrm{m}\)
Answer: \(\boxed{7.5\,\mathrm{m}}\)
(c) Momentum-time graph [2 marks]
From \(t=0\) to \(t=3.0\,\mathrm{s}\), the resultant force is constant and positive. Since
\(F=\frac{\Delta p}{\Delta t}\)
the momentum increases at a constant rate. Therefore, the graph is an upward-sloping straight line from the origin.
From \(t=3.0\,\mathrm{s}\) to \(t=6.0\,\mathrm{s}\), the resultant force is zero, so the momentum remains constant.
There is no instantaneous change in momentum at \(t=3.0\,\mathrm{s}\).
Answer: \(\boxed{\text{Straight line increasing from }0\text{ to }3.0\,\mathrm{s},\text{ followed by a horizontal line to }6.0\,\mathrm{s}.}\)
Question 4
A spring is suspended from a fixed point at one end. The spring is extended by a vertical force applied to the other end. The variation of the applied force \(F\) with the length \(L\) of the spring is shown in Fig. 4.1.
For the spring:
(a) State the name of the law that gives the relationship between the force and the extension. (1 mark)
________________________________________
(b) Determine the spring constant, in \(\mathrm{N\,m^{-1}}\). (2 marks)
spring constant = ………………………………………… \(\mathrm{N\,m^{-1}}\)
(c) Determine the elastic potential energy when \(F=6.0\,\mathrm{N}\). (2 marks)
elastic potential energy = ………………………………………………. \(\mathrm{J}\)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
▶️ Answer/Explanation
(a) Law relating force and extension [1 mark]
The relationship between force and extension for an elastic spring is described by Hooke’s law.
Answer: \(\boxed{\text{Hooke’s law}}\)
(b) Spring constant [2 marks]
Hooke’s law is
\(F=kx\)
where \(x\) is the extension of the spring.
From the graph, when \(F=12.0\,\mathrm{N}\), the length of the spring is \(0.240\,\mathrm{m}\), and its original length is \(0.080\,\mathrm{m}\).
Therefore,
\(x=0.240-0.080=0.160\,\mathrm{m}\)
Hence,
\(k=\frac{F}{x}\)
\(k=\frac{12.0}{0.160}\)
\(k=75\,\mathrm{N\,m^{-1}}\)
Answer: \(\boxed{75\,\mathrm{N\,m^{-1}}}\)
(c) Elastic potential energy [2 marks]
When \(F=6.0\,\mathrm{N}\), the graph shows an extension of
\(x=0.160\times\frac{6.0}{12.0}=0.080\,\mathrm{m}\)
The elastic potential energy is equal to the area under the force-extension graph:
\(E_{\mathrm{P}}=\frac{1}{2}Fx\)
\(E_{\mathrm{P}}=\frac{1}{2}(6.0)(0.080)\)
\(E_{\mathrm{P}}=0.24\,\mathrm{J}\)
Answer: \(\boxed{0.24\,\mathrm{J}}\)
Question 5
(a) A progressive wave travels through a medium. The wave causes a particle of the medium to vibrate along a line P. The energy of the wave propagates along a line Q.
Compare the directions of lines P and Q if the wave is:
(i) a transverse wave. (1 mark)
________________________________________
(ii) a longitudinal wave. (1 mark)
________________________________________
(b) A tube is closed at one end. A loudspeaker is placed near the other end of the tube, as shown in Fig. 5.1.
The loudspeaker emits sound of frequency \(1.7\,\mathrm{kHz}\). The speed of sound in the air in the tube is \(340\,\mathrm{m\,s^{-1}}\). A stationary wave is formed with an antinode A at the open end of the tube.
There is only one other antinode A inside the tube, as shown in Fig. 5.1.
Determine:
(i) the wavelength of the sound. (2 marks)
wavelength = ……………………………………………… \(\mathrm{m}\)
(ii) the length \(L\) of the tube. (1 mark)
\(L=\) ……………………………………………… \(\mathrm{m}\)
(iii) the maximum wavelength of the sound from the loudspeaker that can produce a stationary wave in the tube. (1 mark)
maximum wavelength = ……………………………………………… \(\mathrm{m}\)
(c) Two polarising filters are arranged so that their planes are vertical and parallel. The first filter has its transmission axis at an angle of \(35^\circ\) to the vertical and the second filter has its transmission axis at angle \(\alpha\) to the vertical, as shown in Fig. 5.2.
Angle \(\alpha\) is greater than \(35^\circ\) and less than \(90^\circ\). A beam of vertically polarised light of intensity \(8.5\,\mathrm{W\,m^{-2}}\) is incident normally on the first filter.
(i) Show that the intensity of the light transmitted by the first filter is \(5.7\,\mathrm{W\,m^{-2}}\). (1 mark)
________________________________________
(ii) The intensity of the light transmitted by the second filter is \(5.2\,\mathrm{W\,m^{-2}}\).
Calculate angle \(\alpha\). (2 marks)
\(\alpha=\) ………………………………………………..\(^\circ\)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 7.1: Progressive waves — part (b)(i)
• 8.1: Stationary waves — parts (b)(ii) and (b)(iii)
• 7.5: Polarisation — parts (c)(i) and (c)(ii)
▶️ Answer/Explanation
(a)(i) Transverse wave [1 mark]
For a transverse wave, the particles of the medium vibrate perpendicular to the direction in which the energy propagates.
Answer: \(\boxed{\text{P is perpendicular to Q}}\)
(a)(ii) Longitudinal wave [1 mark]
For a longitudinal wave, the particles of the medium vibrate parallel to the direction in which the energy propagates.
Answer: \(\boxed{\text{P is parallel to Q}}\)
(b)(i) Wavelength of the sound [2 marks]
For a progressive wave,
\(v=f\lambda\)
Therefore,
\(\lambda=\frac{v}{f}\)
\(f=1.7\,\mathrm{kHz}=1.7\times10^3\,\mathrm{Hz}\)
\(\lambda=\frac{340}{1.7\times10^3}\)
\(\lambda=0.20\,\mathrm{m}\)
Answer: \(\boxed{0.20\,\mathrm{m}}\)
(b)(ii) Length of the tube [1 mark]
The tube is closed at one end and open at the other. Therefore, there is a node at the closed end and an antinode at the open end.
With one other antinode inside the tube, the length corresponds to \(\frac{3}{4}\lambda\).
\(L=\frac{3}{4}\lambda\)
\(L=\frac{3}{4}(0.20)\)
\(L=0.15\,\mathrm{m}\)
Answer: \(\boxed{0.15\,\mathrm{m}}\)
(b)(iii) Maximum wavelength [1 mark]
For a tube closed at one end, the fundamental stationary wave has
\(L=\frac{\lambda_{\max}}{4}\)
Therefore,
\(\lambda_{\max}=4L\)
\(\lambda_{\max}=4(0.15)\)
\(\lambda_{\max}=0.60\,\mathrm{m}\)
Answer: \(\boxed{0.60\,\mathrm{m}}\)
(c)(i) Intensity after the first polarising filter [1 mark]
Using Malus’ law,
\(I=I_0\cos^2\theta\)
Here, \(I_0=8.5\,\mathrm{W\,m^{-2}}\) and \(\theta=35^\circ\).
\(I=8.5\cos^2 35^\circ\)
\(I=5.7\,\mathrm{W\,m^{-2}}\)
Answer: \(\boxed{5.7\,\mathrm{W\,m^{-2}}}\)
(c)(ii) Angle \(\alpha\) [2 marks]
The angle between the transmission axes of the two filters is
\(\theta=\alpha-35^\circ\)
Using Malus’ law again,
\(5.2=5.7\cos^2\theta\)
\(\cos^2\theta=\frac{5.2}{5.7}\)
\(\theta\approx17^\circ\)
Therefore,
\(\alpha=35^\circ+17^\circ\)
\(\alpha=52^\circ\)
Answer: \(\boxed{52^\circ}\)
Question 6
(a) The current in a filament lamp decreases.
State and explain how the resistance of the lamp changes. (1 mark)
________________________________________
(b) A cylindrical wire has length \(L\) and resistance \(R\). The total number of free electrons (charge carriers) contained in the volume of the wire is \(N\). Each free electron has charge \(e\). The potential difference between the ends of the wire is \(V\).
Determine expressions, in terms of some or all of the symbols \(e\), \(L\), \(N\), \(R\) and \(V\) for:
(i) the current in the wire. (1 mark)
current = …………………………………………………
(ii) the average drift speed of the free electrons. (2 marks)
average drift speed = …………………………………………………
(iii) the average time taken for a free electron to move along the full length of the wire. (1 mark)
time taken = …………………………………………………
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 9.1: Electric current — parts (b)(i), (b)(ii) and (b)(iii)
▶️ Answer/Explanation
(a) Change in resistance of the filament lamp [1 mark]
The current decreases, so the temperature of the filament decreases. A decrease in temperature causes the resistance of the filament lamp to decrease.
Answer: \(\boxed{\text{Resistance decreases}}\)
(b)(i) Current in the wire [1 mark]
Using Ohm’s law,
\(V=IR\)
Therefore,
\(I=\frac{V}{R}\)
Answer: \(\boxed{\frac{V}{R}}\)
(b)(ii) Average drift speed of the free electrons [2 marks]
For charge carriers moving through a conductor,
\(I=Anve\)
The volume of the cylindrical wire is
\(V_{\mathrm{wire}}=AL\)
Hence, the number density of free electrons is
\(n=\frac{N}{AL}\)
Substituting into \(I=Anve\),
\(I=A\left(\frac{N}{AL}\right)ve\)
\(I=\frac{Nve}{L}\)
Therefore,
\(v=\frac{IL}{Ne}\)
Using \(I=\frac{V}{R}\),
\(v=\frac{VL}{RNe}\)
Answer: \(\boxed{\frac{VL}{RNe}}\)
(b)(iii) Average time taken [1 mark]
Using
\(v=\frac{L}{t}\)
Therefore,
\(t=\frac{L}{v}\)
Substituting \(v=\frac{VL}{RNe}\),
\(t=\frac{L}{VL/(RNe)}\)
\(t=\frac{RNe}{V}\)
Answer: \(\boxed{\frac{RNe}{V}}\)
Question 7
(a) A battery of electromotive force (e.m.f.) \(9.0\,\mathrm{V}\) and negligible internal resistance is connected to a light-dependent resistor (LDR) and a fixed resistor, as shown in Fig. 7.1.
The LDR and fixed resistor have resistances of \(1800\,\Omega\) and \(1200\,\Omega\) respectively.
Calculate the potential difference across the LDR. (2 marks)
potential difference = ……………………………………………… \(\mathrm{V}\)
(b) The circuit in (a) is now modified by adding a uniform resistance wire XY and a galvanometer, as shown in Fig. 7.2.
The length of the wire XY is \(1.2\,\mathrm{m}\). The movable connection Z is positioned on the wire XY so that the galvanometer reading is zero.
(i) Calculate the length XZ along the resistance wire. (2 marks)
length XZ = ……………………………………………… \(\mathrm{m}\)
(ii) The environmental conditions change causing a decrease in the resistance of the LDR.
The temperature of the LDR remains constant.
State whether there is a decrease, increase or no change to:
• the intensity of the light illuminating the LDR
________________________________________
• the total power produced by the battery
________________________________________
• the length XZ so that the galvanometer reads zero. (3 marks)
________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 10.3: Potential dividers — parts (a), (b)(i) and (b)(ii), including the potentiometer, galvanometer null method and LDR potential-divider application
▶️ Answer/Explanation
(a) Potential difference across the LDR [2 marks]
The LDR and fixed resistor are connected in series, so the same current flows through both.
The total resistance is
\(R_{\mathrm{total}}=1800+1200=3000\,\Omega\)
The current is
\(I=\frac{9.0}{3000}=3.0\times10^{-3}\,\mathrm{A}\)
Therefore, the potential difference across the LDR is
\(V_{\mathrm{LDR}}=IR\)
\(V_{\mathrm{LDR}}=(3.0\times10^{-3})(1800)\)
\(V_{\mathrm{LDR}}=5.4\,\mathrm{V}\)
Answer: \(\boxed{5.4\,\mathrm{V}}\)
(b)(i) Length XZ [2 marks]
When the galvanometer reads zero, the potential difference across XZ equals the potential difference across the LDR.
For a uniform resistance wire, the potential difference is proportional to length.
Therefore,
\(\frac{XZ}{XY}=\frac{V_{\mathrm{LDR}}}{V_{\mathrm{battery}}}\)
\(\frac{XZ}{1.2}=\frac{5.4}{9.0}\)
\(XZ=1.2\times\frac{5.4}{9.0}\)
\(XZ=0.72\,\mathrm{m}\)
Answer: \(\boxed{0.72\,\mathrm{m}}\)
(b)(ii) Changes in the circuit [3 marks]
Intensity of light illuminating the LDR:
For an LDR, its resistance decreases when the light intensity increases.
Therefore, the intensity of the light has increased.
Answer: \(\boxed{\text{increase}}\)
Total power produced by the battery:
The decrease in LDR resistance causes the total resistance of the circuit to decrease.
Since the e.m.f. is constant, the current increases.
The power produced by the battery is
\(P=\mathcal{E}I\)
Therefore, the total power produced by the battery increases.
Answer: \(\boxed{\text{increase}}\)
Length XZ for zero galvanometer reading:
When the LDR resistance decreases, the potential difference across the LDR decreases.
For the galvanometer to remain at zero, the potential difference across XZ must also decrease.
Since the resistance wire is uniform, its potential difference is proportional to its length.
Therefore, the length XZ must decrease.
Answer: \(\boxed{\text{decrease}}\)
Question 8
(a) Nucleus P and nucleus Q are isotopes of the same element.
Nucleus Q is unstable and emits a \( \beta^- \) particle to form nucleus R.
(i) For nuclei P and Q, compare:
• the number of protons
• the number of neutrons (2 marks)
(ii) When nucleus Q decays to form nucleus R, the quark composition of a nucleon changes.
State the change to the quark composition of the nucleon. (1 mark)
(iii) State the name of another particle that must be emitted from nucleus Q in addition to the \( \beta^- \) particle. (1 mark)
(b) A hadron consists of two charm quarks and one bottom quark.
Determine, in terms of the elementary charge \( e \), the charge of the hadron. (2 marks)
charge = ………………………………………………. \(e\)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 11.2: Fundamental particles — parts (a)(ii) and (b)
▶️ Answer/Explanation
(a)(i) Comparing nuclei P and Q [2 marks]
P and Q are isotopes of the same element, so they have the same proton number.
Number of protons: equal / the same.
Isotopes have different numbers of neutrons.
Number of neutrons: unequal / different.
Answer: \(\boxed{\text{same number of protons; different number of neutrons}}\)
(a)(ii) Change in quark composition [1 mark]
In \( \beta^- \) decay, a down quark changes into an up quark.
Therefore, the quark composition changes from \(udd\) to \(uud\).
Answer: \(\boxed{\text{down quark changes to an up quark}}\)
(a)(iii) Additional particle emitted [1 mark]
A \( \beta^- \) decay also produces an electron antineutrino.
Answer: \(\boxed{\text{electron antineutrino}}\)
(b) Charge of the hadron [2 marks]
The charge of a charm quark is
\(q_{\mathrm{c}}=+\frac{2}{3}e\)
The charge of a bottom quark is
\(q_{\mathrm{b}}=-\frac{1}{3}e\)
The hadron contains two charm quarks and one bottom quark, so
\(q=2\left(+\frac{2}{3}e\right)-\frac{1}{3}e\)
\(q=\frac{4}{3}e-\frac{1}{3}e\)
\(q=+e\)
Answer: \(\boxed{+1e}\)