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
The emission spectrum of atomic hydrogen consists of a number of discrete wavelengths.
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}
