CIE 9702 Physics Paper 42 MJ 2016 Mark Scheme Answers

Q1(a)

(i) The centripetal force acting on both star is provided by the gravitational force acting on each other. The gravitational force, according to Newton's 3rd Law, is equal in magnitude. So, the centripetal force will also has the same magnitude.
(ii) w = 2π / T = 2π / (4.0 x 365 x 24 x 60 x 60)
                        = 5.0x10^-8

Q1(b)

(i)                M_A d w² = M_B (2.8x10⁸ - d) w²
                         M_A d = M_B (2.8x10⁸ - d)
               (M_A/M_B) d = 2.8x10⁸ - d
                          3.0 d = 2.8x10⁸ - d
                               d = 7.0x10⁷ km

(ii)                          M_A d w² = G M_A M_B / (2.8x10¹¹)²
             7.0x10⁷ (5.0x10^-8)² = (6.67x10^-11) M_B / (2.8x10¹¹)²
                                       M_B = 2.0x10²⁶ kg

Q2(a)

(i) Avogadro constant is the number of atoms in 12 g of carbon-12.
Note: This is a formal definition of Avogadro constant
(ii) Mole is the amount of substance containing the same number of atoms as there are in 12 g of carbon-12.
Note: Explaining that 'one mole is total mass / molar mass' or 'one mole is no. of atoms divided by Avogadro constant' will not score any marks

Q2(b)

p V = n R T
2.0 x 10⁷ (1.8 x 10⁴ x 10^-6) = n  (8.31) (273+17)
                                          n = 149 mol
Note:

Q2(c)

(i) pV = nRT, since V and T are constant,
pressure is directly proportional to the amount of gas
amount reduced by 5 %, means pressure reduced by 5 %
so, p = 0.95 (2.0 x 10⁷) = 1.9 x 10⁷
Note: For this type of question, finding the proportionality by eliminating other constants will reduce the chance of making mistakes in calculation. 

(ii) 1.5 x 10¹⁹ molecules leak in 1 second
      0.05 x 149 x N_A molecules leak in y seconds
      y = 3.0 x 10⁵ s

Q3(a)

When two bodies are in thermal equilibrium, they will have the same temperature and there will be no net transfer of thermal energy.
Note: It will be wrong to say that 'both bodies will have the same amount of thermal energies'.

Q3(b)

(i) thermocouple, platinum/metal resistance thermometer, pyrometer
(ii) thermistor, thermocouple

Q3(c)

(i) 38.0 - 26.5 = 11.5 K
(ii) 38.0 +273.15 = 311.2 K
Note: Do not just write '+ 273', the correct conversion requires the two d.p behind.

Q4(a)

(i) w = 2π/T = 2π/ 0.6 = 10.5
(ii) Maximum KE = 1/2 m v_max²
                              = 1/2 mw²(x_o²- x²)   ;  where x = 0
                              = 1/2 mw²x_o²
                              = 1/2 (120x10^-3) (10.5)²(2.0x10^-2)²                         
                              = 2.6 x 10^-3
Note:Energy of the vibration: can be calculated using maximum potential energy or maximum kinetic energy. The consistency in the amplitude shows that there is no energy lost. So at any time, the total energy does not change. Let's take maximum kinetic energy, as it can be derived easily.

Q4(b)&(c)

(b) sketch: smooth curve in correct directions peak at f
                  amplitude never zero and line extends from 0.7f to 1.3f
(c) sketch: peaked line always below a peaked line A
                  peak not as sharp and at (or slightly less than) frequency of peak in line A
Note:

Q5(a)

Amplitude modulation is a process where the amplitude of a carier wave is varied in synchrony with the displacement of the audio signal.

Q5(b)

(i) 203 - 193 = 10 kHz
(ii) 10/2 = 5 kHz
Note: the graph of 193 - 198 and 198 - 203 are the two sidebands produced due to amplitude modulation. Therefore the range of frequency of the audio signal must be the difference between the upper and lower limit these values, i.e 198 - 193 = 5 kHz or 203 -198 = 5 kHz, which also mean that the maximum frequency is 5 kHz  

Q5(c)

(i) 24 = 10 log (P_min / 5.0 x 10^-13)
     P_min = 1.26 x 10^-10
Note: This value carries the meaning that: if the output at the end of the transmission line is lower that this value, it will be too noisy, because the minimum acceptable signal-to-noise ratio is just 24 dB.
(ii) The total attenuation across 45 km of transmission = 2.0 x 45 = 90 dB
      Attenuation = 10 log (P_in  / P_out)
               So, 90 = 10 log ( 500 x 10^-3/ P_out)
                   P_out = 5.0 x 10^-10 
     
       Since P_out is larger than P_min , the transmission is possible.

Q6(a)

The electric field lines are drawn radially from the surface of the spherical conductor. If these lines are extended to the interior of the conductor, they will converge at the center, just like how field lines are drawn for a point charge.

Q6(b)

(i) EF = Q₁Q₂ / 4πε_or² = (1.6 x 10^-19)² / 4πε_ox²

    GF = GM₁M₂ / r² = G(1.67 x 10^-27)² / x²

EF / GF =(1.6 x 10^-19)² (8.99 x 10⁹) / [ (6.67 x 10^-11) (1.67 x 10^-27)² ]
              = 1.24 x 10³⁶

(ii) The magnitude of electrical force is 10³⁶ of gravitational force, which means gravitational force is negligible comparatively.
Note: Comment must involve the ratio found in (b)(i).

Q7(a)

Acceptable answers include: storing energy , blocking d.c. , in oscillator circuits , in tuning circuits , in timing circuits.

Q7(b)

(i) 1/4μ = 1/C + 1/C + 1/6μ      C = 24 μF

(ii) C = Q / V
     4μ = Q / 12
     Q = 48 μC

Note: Use total capacitance and total p.d. to obtain the total charge stored.

(iii) 1. 48 μC
Note: Since C is in series, it contains the amount of charge as much as for the whole combination of capacitors. 
       2. 24 μC
Note: Since 3.0 μF is parallel with another 3.0 μF, they will share the same portion of charge out of 48 μC.

Q8(a)

(i) voltage gain = voltage output / voltage input
(ii) infinite slew rate means the output responses/changes immediately with the input
Note:

Q8(b)

For non-inverting amplifier, gain = 1 + [ R / 1.5k ]
                                                      12 = 1 + [ R / 1.5k ]
                                                       R = 16.5 k

Q8(c)

12 = 9.0 / Vin
Vin = 0.75 V
This means output voltage saturated at 9.0 V when input voltage increases to 0.75 V

Note: Output graph overlaid on input graph for clearer comparison

Q9(a)

(i) The symbol n represent the number of free electrons (or charge carriers) per unit volume
(ii) PX, QY or RZ
Note: t must the thickness of the slice where the magnetic field passes through.

Q9(b)

(i) According to the equation given, VH is inversly proportional to n, with the same dimensions, I and B. Semiconductor has a much smaller value of n compared to metal, therefore producing much larger VH.
(ii) Using Fleming's Left Hand rule, the direction of force acting on the charge carriers is towards PQXY. This means, if charge carriers are electrons, the negative charge carrirers, PQXY will be negative; and if charge carriers are holes, the positive charge carriers, PQXY will be positive. So, the Hall voltage will be opposite.
Note:

Q10(a)

When iron rod is being inserted into coil P, magnetic field strength around coil P increases. This changing magnetic flux is cut by coil Q, therefore inducing an emf in coil Q.

Q10(b)

Change in current represent the change in magnetic flux density. According to Faraday's law, induced emf is directly proportional to the rate of change of flux linkage, and in this case, to the rate of change of flux density. So, the gradient of I-t graph is the value of V-t graph.

Note: V-t graph overlaid on I-t graph for clearer comparison

Q11(a)

Point P is at the lower end of L

Q11(b)

      V_rms = Vo / √2 = 6.0 / √2 = 4.24 V
      V_rms = I_rms R
       4.24 = I_rms (2.4k)
        I_rms = 1.8 x 10^-3 A

Q11(c)

(i) Capacitor connected parallel with the load
(ii)

Note: A full wave rectification should be shown. Lowest point must be at 3.0 V. Gradient of curve decreases towards lowest point.

Q12(a)

(i) When the high-energy electrons are stopped or accelerated suddenly in the metal target, X-ray photons are produced. There is a range of acceleration of the electrons, resulted in a range of energy of the X-ray photons, indicated by the distribution of the wavelengths in the graph
Note: Those high-energy electrons have the same amount of kinetic energy when they arrived at the target because they are supplied with equal amount of electrical potential energy.

(ii) The sharp cut-off indicates that there are rare cases where X-ray photons have the highest energy, which will only happen when an electron gives all its energy to one photon. This photon will be stopped in a single collision.

(iii) The peaks represent photons that are produced when orbital electrons in the metal target de-excite. A higher orbital electron drop to lower orbital when there is a space is left empty by an orbital electron that got kick out by a high-energy electron. Since orbital electrons occupy only a certain place, the de-excitation will produce photon with only a certain energy. Therefore the peaks appear only at certain wavelength.
Note:

Q12(a)

(i) The photons can be filtered out by an aluminium filter fitted at the window of the X-ray tube.
(ii) To avoid absorption of low energy photons that can not pass through the body.
Note:

Q13(a)

Gamma radiation is an electromagnetic wave emitted from a decaying nuclei

Q13(b)

Rearrange the given equation to give:
ln C = ln C_o – μx
– μ is the gradient of the best-fit line drawn on the graph.
μ = 0.061 (within ±0.004)

Q13(c)

Aluminium is less absorbing compared to lead. Gradient of the graph will be smaller, hence a smaller μ.