Thursday, July 3, 2014

9702 June 2013 Paper 42 Worked Solutions | A-Level Physics

  • 9702 June 2013 Paper 42 Worked Solutions | A-Level Physics


SECTION A


Question 1
(a)
A geostationary orbit is an equatorial orbit (above the equator) in which an object / satellite moves in the same direction as Earth spins (from west to east), with the same period(24 hours) as the spinning of Earth.

(b)
Satellite mass = m : in circular orbit about planet of mass M, concentrated at its centre.
Radius, R of orbit of satellite:
Gravitational force provides/is the centripetal force for circular motion.
GMm / R2 = mRω2      or         GMm / R2 = mv2/R
ω = 2π/T                      or v = 2πR/T
Work out to obtain R3 = (GMT2) / (4π2)

(c)
Mass of Earth = 6.0 x 1024kg. Determine radius of geostationary orbit about Earth.
R3 = 6.67x10-11 x 6.0x1024 x (24x3600)2 / (4π2) = 7.57x1022m
R = 4.2x107m






Question 2
{Detailed explanations for this question is available as Solution 580 at Physics 9702 Doubts | Help Page 114 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-114.html}






Question 3
{Detailed explanations for this question is available as Solution 586 at Physics 9702 Doubts | Help Page 115 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-115.html}





Question 4
(a)
Insulated metal sphere of radius R in vacuum. Charge q of sphere considered to be point charge at centre of sphere
(i)
Formula for potential V on surface of sphere:
V = q / (4πϵoR)


(ii)
Define Capacitance and show expression:
Capacitance is the ratio of charge and potential or q/V
C = q/V = 4πϵoR

(b)
Radius 45cm
(i)
Capacitance of sphere in picofarad: (1pico = 1x10-12)
C = 4π (8.85x10-12) x 0.45 = 50pF

(ii)
Sphere charged to potential of 9.0x105V. Spark occurs, partially discharging sphere so that its potential is reduced to 3.6x105V. Determine energy of spark:
Either energy = ½ CV2           or energy = ½ QV and Q = CV
Energy of spark = ½ x (50x10-12) {(9.0x105)2 – (3.6x105)2} = 17J





Question 5
{Detailed explanations for this question is available as Solution 593 at Physics 9702 Doubts | Help Page 117 - http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-117.html}





Question 6
(a)
Faraday’s law of electromagnetic induction states that the (induced) e.m.f is proportional to the rate of change of (magnetic) flux (linkage).

(b)
(i)
Positive terminal identified (upper connection to load)

(ii)
Ratio of number of turns on primary coil to number of turns on secondary coil:
Np / Ns = Vp / Vs
Vp = (2)1/2 x Vrms = (2)1/2 x 240 V = 339.4V
Vs = 9V
Ratio = Vp / Vs = 339.4 / 9 = 37.7 ≈ 38

(c)
(i)
Due to smoothing, (output) p.d / voltage / current does not fall to zero
(the range of (output) p.d / voltage / current is reduced)

(ii)
Variation with time t of the smoothed output potential difference
Sketch:
same peak value at start of discharge
correct shape between one peak and the next



Question 7
(a)
Emission spectrum of atomic hydrogen consists of a number of discrete wavelengths. How this observation leads to understanding that there are discrete electron energy levels in atoms:
Each wavelength is associated with a discrete change in energy. The discrete energy change / difference implies discrete levels.

(b)
Longest wavelength: 4.0 x 10-6m
(i)
1.
Mark with letter L: transition giving rise to wavelength of 4.0 x 10-6m:
Arrow from – 0.54eV to – 0.85eV, labeled L

2.
Mark with letter S: transition giving rise to shortest wavelength:
Arrow from – 0.54eV to – 3.4eV, labeled S

(ii)
Wavelength of (i) part 2:
ΔE = hc / λ  
(3.4 – 0.54) x 1.6x10-19 = (6.63x10-34 x 3.0x108) / λ
λ = 4.35x10-7m

(c)
Photon energies in visible spectrum vary approximately between 3.66eV and 1.83eV.  Determine energies of photons in visible spectrum that are produced by transitions between energy levels shown in Fig 7.1:
-1.50 to -3.4 = 1.9eV
-0.85 to -3.4 = 2.55eV
-0.54 to -3.4 = 2.86eV


Question 8
(a)
The mass of an α – particle is less than the total mass of two individual protons and two individual neutrons:
Energy is given out / released on formation of the α – particle.
Either E = mc2, so the mass is less
Or reference to mass-energy equivalence

(b)
(i)
Change in mass, in u, associated with reaction
Mass of reactants – mass of products
Mass change = (4.00260u + 14.00307u) – (16.99913u + 1.00728u)
Mass change = 18.00567u – 18.0064u
                      = 7.4 x 10-4u (sign not required)

(ii)
Energy associated with change in mass:
Energy = Δmc2 = (7.4x10-4) x (1.66x10-27) x (3.0x108)2
Energy = 1.1 x 10-13J


(iii)
For reaction to occur, helium-4 nucleus must have a minimum speed:
Either
The mass of the products is greater than the mass of the reactants. By having some velocity, the kinetic energy of the helium-4 nucleus would provide the necessary energy such that the energy of the reactants is now greater than that of the products and the reaction may occur.
Or
Since the nucleus and the alpha-particle are both positively charged, they would repel each other. So, energy is required to overcome the electrostatic repulsion and cause fusion.




SECTION B


Question 9
(a)
Range of volume over which volume has a linear relationship to voltmeter reading:
30 litres to 54 litres 

(b)
(i)
Why: when tank nearly full, voltmeter readings give impression that fuel consumption is low:

There is because at larger fuel volumes, for example, from 70 litres to 80 litres, there is a change of only about 0.1V in the readings. So, the gradient above about 60 litres is small, compared to the gradient at about 40 litres.

(ii)
Why: when voltmeter first indicates that tank is nearly empty, there is more fuel remaining than is expected:

The voltmeter reading is nearly zero even when fuel is still left. The voltmeter reads only about 0.1V when 10 litres of fuel is left in the tank.


Question 10
(a)
Acoustic impedance is the product of the density of the medium and the speed of the ultrasound wave in the medium.

(b)
Importance of difference between Z1 and Z2 for transmission of ultrasound across boundary:

If (Z1 – Z2) is small, there is mostly transmission of the ultrasound .
If (Z1 – Z2) is large, there is mostly reflection of the ultrasound.
Either reflection / transmission also depends on (Z1 + Z2)
Or intensity reflection coefficient = (Z1 – Z2)2 / (Z1 + Z2)2

(c)
Advantage of use of high-frequency ultrasound compared with lower-frequency ultrasound:

Smaller structures can be distinguished because shorter wavelength (higher frequency) gives better resolution.


Question 11
(a)
How hardness of X-ray beam is controlled by accelerating voltage in X-ray tube:
Changing the anode voltage changes the energy / speed of the electrons.
The changing electron energy changes the maximum energy of the X-ray photon.

(b)
(i)
1.
Attenuation of the beam is the loss of its power / energy / intensity.

2.
Why expression applies to a parallel beam:
The intensity of a non-parallel beam changes, even there is no absorption. It decreases when the beam is divergent.

(ii)
Linear attenuation coefficients for X-rays in bone and in soft tissue are 2.9cm-1 and 0.95cm-1 respectively.

Ratio = (exp{-2.9 x 2.5}) / (exp{-0.95 x 6.0})
Ratio = 0.21 (min. 2 sig. fig)






Question 12
{Detailed explanations for this question is available as Solution 667 at Physics 9702 Doubts | Help Page 134 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-134.html}

17 comments:

  1. for question 2 b) in the mark scheme, from where did they get 25j.?
    3aii 2) did they get that from differentiating a i?)
    5b ii) permeability of free space is 8.85 times 10^-12 so why did they use 4pi time 10^7?

    ReplyDelete
    Replies
    1. The details for Q2 has been updated.

      The others will be explained later

      Delete
    2. Details have been added for question 3

      Delete
    3. The value you gave above is for the PERMITTIVITY of free space and the required value is that for the PERMEABILITY of free space. So not confuse these 2. + look at the symbol used, it may help

      Delete
  2. can you please graph 12 b c ?

    ReplyDelete
  3. 12 d) why does increasing number of bits reduce step height ?

    ReplyDelete
  4. in s13 qp 41, 5 c ii) why did they multiply by 2?

    ReplyDelete
    Replies
    1. See solution 628 at
      http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-124.html

      Delete
  5. Replies
    1. Check question 651 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-130.html

      Delete
  6. could u pls give me some op amp notes and how to draw LEDs in them, i really need them.
    thnks....

    ReplyDelete
    Replies
    1. Try to look for some such questions in past exam papers and tell me the reference. I remember explaining such questions, but cannot recall from which papers they were

      Delete
  7. In q7 b part (ii), why aren't we converting eV to joules by multiplying it by 1.6*10^-13?
    Isn't energy supposed to be in joules?

    ReplyDelete
    Replies
    1. 1.6x10-19 has been used. The one you gave is for converting from MeV.

      Delete

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