Sunday, October 19, 2014

9702 June 2014 Paper 42 Worked Solutions | A-Level Physics

  • 9702 June 2014 Paper 42 Worked Solutions | A-Level Physics


Paper 42


SECTION A

Question 1
{Detailed explanations for this question is available as Solution 860 at Physics 9702 Doubts | Help Page 172 - http://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-172.html}




Question 2
{Detailed explanations for this question is available as Solution 868 at Physics 9702 Doubts | Help Page 174 - http://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-174.html}




Question 3
Microwave cooker uses electromagnetic waves of frequency 2450 MHz. Microwaves warm food in cooker by causing water molecules in food to oscillate with a large amplitude at frequency of the microwaves.
(a)
Name given to this phenomenon:
Resonance

(b)
Effective microwave power of cooker is 750 W. Temperature of mass of 280 g of water rises from 25 °C to 98 °C in a time of 2.0 minutes. Value for specific heat capacity of water:
Pt = mcΔθ
750 (2 x 60) = 0.28 c (98 – 25)
Specific capacity of water, c = 4400Jkg-1K-1

(c)
Value of specific heat capacity determined from data in (b) is greater than accepted value. Student gives as reason for this difference: ‘heat lost to the surroundings’. In more detail than that given by student, possible reason for difference:
Example:
There is some microwave leakage from the cooker
The container for the water is also heater



Question 4
{Detailed explanations for this question is available as Solution 697 at Physics 9702 Doubts | Help Page 126 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-126.html}




Question 5
{Detailed explanations for this question is available as Solution 461 at Physics 9702 Doubts | Help Page 89 - http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-89.html}



Question 6
(a)
Use of uniform electric field and uniform magnetic field for selection of velocity of charged particle:
The electric and magnetic fields are normal to each other. EITHER The charged particle enters the region normal to both fields OR Correct B direction with respect to E for zero deflection. For no deflection, the velocity of the charged particle, v = E / B.

(b)
Ions, all of same isotope, are travelling in vacuum with speed of 9.6 × 104 m s−1. Ions are incident normally on uniform magnetic field of flux density 640 mT. Ions follow semicircular paths A and B before reaching a detector, as shown.
Data for diameter of paths shown
Path                 Diameter / cm
A                     6.2
B                     12.4
Ions in path B each have charge +1.6x10-19C.
(i)
Mass, in u, of ions in path B:
m = Bqr / v = (640x10-3) (1.6x10-19) (12.4x10-2) / (9.6x104) = 6.61x10-26kg
Mass, m (in u) = (6.61x10-26) / (1.66x10-27) = 40u

(ii)
Reason for difference in radii of paths A and B of ions:
The ratio q/m is proportional to 1/r OR Mass m is constant and charge q is proportional to 1/r. The ratio q/m for path A is twice that for path B. So, the ions in path A have (the same mass but) twice the charge (of the ions in path B)



Question 7
{Detailed explanations for this question is available as Solution 544 at Physics 9702 Doubts | Help Page 106 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-106.html}



Question 8
Light of wavelength 590 nm incident normally on surface, as illustrated. Power of light is 3.2 mW. Light is completely absorbed by surface.
(a)
Number of photons incident on surface in 1.0s:
Photon energy of 1 photon = hc / λ = (6.63x10-34) (3.0x108) / (590x10-9) = 3.37x10-19J 
{Power = energy / time = energy in 1.0s. So, in 1.0s, number of photons = total energy in 1.0s [this is the power = 3.2mW] / energy of 1 photon}
Number of photons = (3.2x10-3) / (3.37x10-19) = 9.5x1015

(b)
Use answer in (a) to determine
(i)
Total momentum of photons arriving at surface in 1.0s:
Momentum of 1 photon, p = h / λ = (6.63x10-34) / (590x10-9) = 1.12x10-27 kgms-1
Total momentum = (9.5x1015) (1.12x10-27) = 1.06x10-11kgms-1

(ii)
Force exerted on surface by light:
Force = 1.06x10-11N



Question 9
Some water contaminated with radioactive iodine-131 (13153I). Activity of iodine-131 in 1.0 kg of this water is 460 Bq. Half-life of iodine-131 is 8.1 days.
(a)
Radioactive half-life is defined as the time for the number of atoms / nuclei / activity (of the isotope) to be reduced to one half (of its initial value)

(b)
(i)
Number of iodine-131 atoms in 1.0 kg of this water:
A = λN
460 = N (ln2 / [8.1 x 24 x 60 x 60])
Number of iodine-131 atoms, N = 4.6x108

(ii)
Amount of 1.0 mol of water has mass of 18 g. Ratio of number of molecules of water in 1.0 kg of water to number of atoms of iodine-131 in 1.0 kg of contaminated water:
Number of water molecules in 1.0kg = (6.02x1023) / (18x10-3) = 3.3x1025


Ratio = (3.3x1025) / (4.6x108) = 7.2 (7.3) x1016


(c)
Acceptable limit for activity of iodine-131 in water has been set as 170 Bq kg−1. Time, in days, for activity of contaminated water to be reduced to this acceptable level:
Activity, A = Aoexp(- λt)        and      λt½ = ln 2
170 = 460exp(-{t ln2} / 8.1)
Time, t = 11.6 days




SECTION B


Question 10
(a)
Function of comparator circuit incorporating operational amplifier (op-amp):
The circuit compares the potentials / voltages at the (inverting and non-inverting) inputs. EITHER The output (potential) is dependent on which input is the larger OR V+ > V-, the VOUT is positive. + States the other condition

(b)
Ideal op-amp is incorporated into circuit.
(i)
On Fig, draw circle around part of circuit that is being used as an output device:
A circle is drawn around both the LEDs (and the series resistors)

(ii)
Show that potential at non-inverting input of op-amp is 1.0V:
V+ = [2.4 / (1.2 + 2.4)] x 1.5 = 1.0V

(iii)
Variation with time t of potential VIN at inverting input of op-amp shown.
1.
On axes of Fig, draw variation with time t of output potential of op-amp:
VOUT switches at +1.0V. The maximum value of VOUT is 5.0V. When the curve is above +1.0V, VOUT is negative (or vice versa)


{Since VIN is at the inverting input, when VIN is greater than the non-inverting input (1.0V), the output has the opposite polarity than VIN and its magnitude is equal to the power supply voltage, 5.0V (it cannot be greater – from energy considerations). When the non-inverting input (1.0V) is greater than VIN, the output polarity will be the same as the non-inverting input (that is, positive) and its magnitude also equal to power supply voltage, 5.0V}
 
2.
Whether each diode is emitting light or not emitting light at time t1 and at time t2:
At time t1, diode R will emit light and diode G will not emit
At time t2, diode R will not emit and diode G will emit



Question 11
(a)
Distinguish between X-ray image of body structure and CT scan:
An X-ray image is a flat / shadow / 2D image regardless of the depth of the object / depth not indicated.
The image from a CT scan is built up from (many) images at different angles. The image is three-dimensional and the image can be rotated / viewed at different angles.

(b)
Data for linear absorption (attenuation) coefficient μ of X-ray radiation of energy 80 keV given.
Metal                           μ / mm-1
Aluminium                  0.46
Copper                      0.69
Parallel X-ray beam incident on copper filter, as shown. Intensity of incident beam is I0.
(i)
Thickness of copper required to reduce intensity of emergent beam to 0.25I0:
I = I0exp(-μx)
0.25 = exp(-0.69x)
Thickness, x = 2.0mm

(ii)
Aluminium filter of thickness 2.4 mm now placed in X-ray beam, together with copper filter in (i). Fraction of incident intensity that emerges after passing through the 2 filters:
For aluminium, I / I0 = exp(-0.46(2.4)) = 0.33
So, fraction of incident intensity = 0.33 x 0.25 = 0.083

(iii)
Answer in (ii) as gain in decibels (dB):
Gain / dB = 10 lg(I/I0) = 10 lg(0.083) = -10.8dB



Question 12
2 people, living in different regions of Earth, communicate either using link provided by geostationary satellite or using optic fibres.
(a)
(i)
A geostationary satellite is a satellite which is in equatorial orbit, travelling from west to east with a period of 24 hours / 1 day

(ii)
Uplink frequency for communication with satellite is 6 GHz and downlink has frequency of 4 GHz. Why frequencies are different:
EITHER The uplink signal is highly attenuated OR The signal is highly amplified (before transmission) as the downlink signal. This prevents the downlink signal from swamping the uplink signal.

(b)
Comment on time delays experienced by the 2 people when communicating either using geostationary satellites or using optic fibres:
The speed of the signal is of the same order of magnitude in both systems. The optic fibre link is (much) shorter than via satellite. So, the time delay using optic fibre is less.



24 comments:

  1. pls explain qn 10. b) iii)

    ReplyDelete
    Replies
    1. the graph and some explanations have been added. See if it helps

      Delete
  2. Please explain q8(a)

    ReplyDelete
    Replies
    1. details for this part if the question has been added

      Delete
  3. And also for q4(b)(ii), what does it mean by short range?

    ReplyDelete
    Replies
    1. The short range of the force is explained above

      Delete
    2. how can there be repulsion outside of nucleus between protons when proton are in the nucleus to begin with? shouldnt it be repulsion between electrons?

      Delete
    3. I have updated the explanations. See if it helps

      Delete
  4. Could you please add the graph for 5(a) ? Thanks a lot.

    ReplyDelete
  5. Excuse me, isn't the temperature for 3b suppose to be in kelvins instead of celcius?

    ReplyDelete
    Replies
    1. Yes, but since this is a change in temperature, even if we converted both temperatures in kelvin, the final answer would be the same.

      Delete
  6. Explain 7 b i please, dont understand why it is calculated that way

    ReplyDelete
    Replies
    1. A diagram along with some notes have been included.

      Delete
  7. can you please explain why there is no potential energy in 2 c?is it beacuse it's an ideal gas?

    and give an example where there is potential energy in a gas?

    ReplyDelete
    Replies
    1. Yes, it's because it's an ideal gas. One of the assumptions is that there is no (intermolecular) forces between the molecules.

      In a real gas, there will be potential energy

      Delete
  8. for 12 a ii) is saying "downlink is highly attenuated" wrong ,if yes then why?
    I don't get "signal is highly amplified (before transmission) as downlink signal"

    ReplyDelete
    Replies
    1. It is the uplink that is highly attenuated.

      Before the satellite sends the received (uplink) signal down, it amplifies it (increases its amplitude). So, even if the downlink signal is attenuated by an amount, since it has already been highly amplified before being sent down, the amount of attenuation does not affect it. So, even if it may be attenuated by an amount, this does not affect. Thus, it's better not to say that the downlink signal is attenuated.

      Delete
  9. 11 b iii) what does it mean if the gain is negative?

    ReplyDelete
    Replies
    1. It means that there is an attenuation (reduction in amplitude)

      Delete
  10. Q2 (a). Why P constant when V/T constant? Can't I use PV=nRT and since nR is constant I compare values of T/V? Urgent help needed. Thank you.

    ReplyDelete
    Replies
    1. Some details have been added to the solution

      Delete
  11. In Question 1 how do we know x=r ?

    ReplyDelete

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