# 9702 June 2014 Paper 41 & 43 Worked Solutions | A-Level Physics

## Paper 41 & 43

**SECTION A**

__Question 1__**(a)**

Gravitational potential at a point
is defined as the work done bringing unit mass from infinity (to the point)

**(b)**

Stone of mass m has gravitational
potential energy E

_{P}at point X in gravitational field. Magnitude of gravitational potential at X is φ. Relation between m, E_{P}and φ:
E

_{p}= - mφ**(c)**

Isolated spherical planet of radius
R assumed to have all its mass concentrated at its centre. Gravitational
potential at surface of planet is − 6.30 × 10

^{7}J kg^{−1}. Stone of mass 1.30 kg travelling towards planet such that its distance from centre of planet changes from 6R to 5R. Change in gravitational potential energy of stone:
Gravitational potential , φ α 1/x

EITHER

At a distance of 6R from the centre,
potential = (6.3x10

^{7}) / 6 [= 1.05x10^{7}Jkg^{-1}]__and__at 5R from the centre, potential = (6.3x10

^{7}) / 5 [= 1.26x10

^{7}Jkg

^{-1}]

So, change in energy in energy =
(1.26 – 1.05)x10

^{7}x 1.3 = 2.7x10^{6}J
OR

Change in gravitational potential =
(1/5 – 1/6) x 6.3x10

^{7}
Change in energy = (1/5 – 1/6) x (6.3x10

^{7}) x 1.3 = 2.7x10^{6}J

__Question 2__**(a)**

The Avogadro constant is the number
of atoms in 12g of carbon-12

**(b)**

Argon-40 (

^{40}_{18}Ar) may be assumed to be ideal gas. Mass of 3.2 g of argon-40 has volume of 210 cm^{3}at temperature of 37 °C. For mass of argon-40 gas,
(i)

Amount, in mol:

Amount, in mol = 3.2 / 40 = 0.080
mol

(ii)

Pressure:

pV = nRT

(T = 273 + 37 = 310K)

p (210x10

^{-6}) = 0.080 x 8.831 x 310
Pressure, p = 9.8x10

^{5}Pa
(iii)

Root-mean-square (r.m.s.) speed of
argon atom:

EITHER

pV = (1/3) Nm<c

^{2}>
Number of atoms, N = 0.080 (6.02x10

^{23}) [= 4.82x10^{22}]__and__mass of 1 Ar atom, m = 40 (1.66x10

^{-27}) [= 6.64x10

^{-26}]

(9.8x10

^{5}) (310x10^{-6}) = (1/3) (4.82x10^{22}) (6.64x10^{-26}) <c^{2}>
<c

^{2}> = 1.93x10^{5}
r.m.s. speed, c

_{RMS}= √<c^{2}> = 440ms^{-1}
OR

(Total mass of gas) Nm = 3.2x10

^{-3}
(9.8x10

^{5}) (310x10^{-6}) = (1/3) (3.2x10^{-3}) <c^{2}>
<c

^{2}> = 1.93x10^{5}
r.m.s. speed, c

_{RMS}= √<c^{2}> = 440ms^{-1}
OR

(1/2) m<c

^{2}> = (3/2) kT
0.5 (40 x 1.66x10

^{-27}) <c^{2}> = (3/2) (1.38x10^{-23}) (310)
<c

^{2}> = 1.93x10^{5}
r.m.s. speed, c

_{RMS}= √<c^{2}> = 440ms^{-1}

__Question 3__
Volume of 1.00 kg of water in liquid
state at 100 °C is 1.00 × 10

^{−3}m^{3}. Volume of 1.00 kg of water vapour at 100 °C and atmospheric pressure 1.01 × 10^{5}Pa is 1.69 m^{3}.**(a)**

Show that the work done against the
atmosphere when 1.00 kg of liquid water becomes water vapour is 1.71 × 10

^{5}J:
EITHER change in the volume = 1.69 –
(1.00x10

^{-3})
OR the liquid volume << volume
of the vapour

So, work done (= pΔV) =
(1.01x10

^{5}) x 1.69 = 1.71x10^{5}J**(b)**

(i)

First law of thermodynamics may be
given by expression ΔU = + q + w where ΔU is increase in internal energy of
system. What is meant by

1.

+q: heating of the system / thermal
energy supplied to the system

2.

+w: work done on the system

(ii)

Specific latent heat of vaporisation
of water at 100 °C is 2.26 × 10

^{6}J kg^{−1}. Mass of 1.00 kg of liquid water becomes water vapour at 100 °C. Using answer in (a), increase in internal energy of mass of water during vaporization:
ΔU = (2.26x10

^{6}) – (1.71x10^{5}) = 2.09x10^{6}J

__Question 4__
Student investigates energy changes
of mass oscillating on vertical spring, as shown. Student draws graph of
variation with displacement x of energy E of oscillation, as shown.

**(a)**

State whether energy E represents
total energy, potential energy or kinetic energy of the oscillations:

Kinetic energy / KE / E

_{K}**(b)**

Student repeats investigation but
with smaller amplitude. Maximum value of E now found to be 1.8 mJ. Use Fig to
determine change in amplitude:

EITHER Change in energy = 0.60mJ

OR The

__maximum__value of E is proportional (amplitude)^{2}/ equivalent numerical working
The new amplitude is 1.3cm

Change in amplitude = 0.2cm

__Question 5__
Isolated solid metal sphere of
radius r is given positive charge. Distance from centre of sphere is x.

**(a)**

Electric potential at surface of
sphere is V

_{0}. On axes of Fig, sketch graph to show variation with distance x of electric potential due to charged sphere, for values of x from x = 0 to x = 4r:
The graph is a straight line at
constant potential = V

_{0}from x = 0 to x = r. Then, it is a curve with a decreasing gradient passing through (2r, 0.50V_{0}) and (4r, 0.25V_{0})**(b)**

Electric field strength at surface
of sphere is E

_{0}. On axes of Fig, sketch graph to show variation with distance x of electric field strength due to the charged sphere, for values of x from x = 0 to x = 4r:
The graph is a straight line at E =
0 from x = 0 to x = r. Then, it is a curve with a decreasing gradient from (r,
E

_{0}) passing through (2r, ¼E_{0})

__Question 6__
Uncharged capacitor is connected in
series with battery, switch and resistor, as shown. Battery has e.m.f. 9.0 V
and negligible internal resistance. Capacitance of capacitor is 4700 μF. Switch
is closed at time t = 0. During time interval t = 0 to t = 4.0 s, charge
passing through resistor is 22 mC.

**(a)**

(i)

Energy transfer in battery during time
interval t = 0 to t = 4.0 s:

Energy = EQ = 9.0 x (22x10

^{-3}) = 0.20J
(ii)

For capacitor at time t = 4.0s,

1.

Potential difference V across
capacitor:

Capacitance, C = Q / V

Potential difference, V = (22x10

^{-3}) / (4700x10^{-6}) = 4.7V
2.

Energy stored in capacitor:

EITHER Energy, E = ½CV

^{2}= ½(4700x10^{-6})(4.7^{2}) = 5.1x10^{-2}J
OR E = ½QV = ½(22x10

^{-3})(4.7) = 5.1x10^{-2}J
OR E = ½Q

^{2}/C = ½(22x10^{-3})^{2}/ (4700x10^{-6}) = 5.1x10^{-2}J**(b)**

Why answers in (a)(i) and (a)(ii)
part 2 are different:

There is energy lost (as thermal
energy) in the resistance / wires / battery / resistor

__Question 7__**{Detailed explanations for this question is available as Solution 1008 at Physics 9702 Doubts | Help Page 210 -**

*http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-210.html*}

__Question 8__**(a)**

Quantisation of charge is the
discrete and equal amounts (of charge)

**(b)**

Student carries out experiment to
determine elementary charge. Charged oil drop is positioned between two
horizontal metal plates, as shown. Plates are separated by distance of 7.0 mm.
Lower plate is earthed. Potential of upper plate is gradually increased until
drop is held stationary. Potential for drop to be stationary is 680 V. Weight
of oil drop, allowing for upthrust of air, is 4.8 × 10

^{−14}N. Value for charge on oil drop:
(Electric force = qV/d)

Weight of oil drop = qV / d

4.8x10

^{-14}= q (680) / (7x10^{-3})
Charge, q = 4.9x10

^{-19}C**(c)**

Student repeats experiment and
determines the following values for charge on oil drops. 3.3 × 10

^{−19}C 4.9 × 10^{−19}C 9.7 × 10^{−19}C 3.4 × 10^{−19}C
Use values to suggest value for elementary
charge:

The elementary charge = 1.6x10

^{-19}C
EITHER The values are
(approximately) multiples of this value OR It is a common factor. It is the highest
common factor

__Question 9__
For particular metal surface, it is
observed that there is a minimum frequency of light below which photoelectric
emission does not occur. This observation provides evidence for particulate nature
of electromagnetic radiation.

**(a)**

3 further observations from
photoelectric emission that provide evidence for particulate nature of electromagnetic
radiation:

Choose any 3:

There is no time delay between
illumination and emission

The maximum (kinetic) energy of the
electron is dependent on frequency

The maximum (kinetic) energy of the
electron is independent of intensity

The rate of emission of electrons is
dependent on / proportional to the intensity

**(b)**

Some data for variation with
frequency f of maximum kinetic energy E

_{MAX}of electrons emitted from metal surface are shown.
(i)

Why emitted electrons may have
kinetic energy less than maximum at any particular frequency:

The (photon) interaction with the
electron may be below the surface. So, energy is required to bring the electron
to the surface

(ii)

Use Fig to determine

1.

Threshold frequency:

(x-intercept) Threshold
frequency = 5.8x10

^{14}Hz
2.

Work function, in eV, of metal
surface:

Work function, ϕ = hf

_{0}= (6.63x10^{-34}) (5.8x10^{14}) = 3.84x10^{-19}J
Work function, ϕ = (3.84x10

^{-19}) / (1.6x10^{-19}) = 2.4eV
OR

Use equation: hf = ϕ + E

_{MAX}
Choose a point on the line and
substitute values of E

_{MAX}, f and h into the equation with the units of hf converted from J to eV
Work function, ϕ = 2.4eV

__Question 10__**(a)**

The binding energy of a nucleus is
the energy required to separate the nucleons (in a nucleus) to infinity

**(b)**

Data for masses of some particles
are given.

Mass/u

Proton 1.00728

Neutron 1.00867

tritium (

^{3}_{1}H) nucleus 3.01551
polonium (

^{210}_{84}Po) nucleus 209.93722
Energy equivalent of 1.0 u is 930
MeV

(i)

Binding energy, in MeV, of tritium (

^{3}_{1}H) nucleus:
Change in mass, Δm = 2(1.00867) + 1.00728 – 3.01551 = 9.11x10

^{-3}u
Binding energy = (9.11x10

^{-3}) x 930 = 8.47MeV
(ii)

Total mass of separate nucleons that
make up polonium-210 (

^{210}_{84}Po) nucleus is 211.70394 u. Binding energy per nucleon of polonium-210:
Change in mass, Δm = 211.70394 – 209.93722 = 1.76672u

Binding energy per nucleon =
(1.76672 x 930) / 210 = 7.82MeV

**(c)**

1 possible fission reaction is

^{235}_{92}U +^{1}_{0}n - - ->^{141}_{56}Ba +^{92}_{36}Kr + 3^{1}_{0}n
Reference to binding energy, why
this fission reaction is energetically possible:

The

__total__binding energy of barium and krypton is greater than the binding energy of uranium**SECTION B**

__Question 11__**(a)**

Circuit incorporating ideal
operational amplifier (op-amp) is shown.

(i)

Name of this circuit:

Inverting amplifier

(ii)

Why point P is referred to as
virtual earth:

The gain is very large / infinite.
Since V

^{+}is earthed / zero, for the amplifier not to saturate, point P must be (almost) earth / zero.**(b)**

Circuit of Fig. is modified, as
shown. Voltmeter has infinite resistance and its full-scale deflection is 1.0
V. Input potential to circuit is V

_{IN}. Switch position may be changed in order to have different values of resistance in circuit.
(i)

Input potential V

_{IN}and switch position are varied. For each switch position, reading of voltmeter is 1.0 V. Complete Fig. for switch positions shown.
Switch position V

_{IN}/mV Resistance
A 10 R

_{A}= 100kΩ
B 100 R

_{B}= 10kΩ
C 1000 R

_{C}= 1.0kΩ
(V

_{OUT}/V_{IN}= - R_{F}/R_{IN}. So, R_{F}= (V_{OUT}/V_{IN}) R_{IN}. V_{OUT}= (-) 1.0V and R_{IN}= 1.0kΩ {V_{IN}needs to be taken in V not mV})
(ii)

Reference to answers in (i), suggest
use for circuit of Fig:

It can be used as a variable range
meter

__Question 12__**(a)**

Outline briefly principles of CT
scanning:

A series of X-ray images (for one
section / slice) is taken from different angles to give an image of the section
/ slice. This is repeated for many slices to build up a three-dimensional image
(of the whole object)

**(b)**

In model for CT scanning, section is
divided into 4 voxels. Pixel numbers P, Q, R and S of voxels are shown in Fig.

P Q

S R

Section is viewed from 4 directions
D

_{1}, D_{2}, D_{3}and D_{4}. Detector readings for each direction are noted. Detector readings are summed as shown.
49 61

73 55

Background reading is 34. Determine
pixel numbers P, Q, R and S as shown in Fig:

The background must be deducted from
the readings

(15 27)

(39 21)

Divide the values by 3

5 9

13 7

__Question 13__
Signal from microphone is to be transmitted
in digital form. Block diagram of part of transmission system is shown.

**(a)**

2 advantages of transmission of
signal in digital form rather than in analogue form:

Choose any 2:

Noise can be eliminated / the
waveform can be regenerated

Some extra bits of data can be added
to check for errors

It is cheaper / more reliable

It has a greater

__rate__of transfer of data**(b)**

Function of parallel-to-serial
converter:

A parallel-to-serial converter receives
the bits all at one time and transmits the bits one after another.

**(c)**

In particular telephone system, sampling
frequency is 8 kHz. In manufacture of compact disc, sampling frequency is
approximately 44 kHz. Explain why sampling frequency is much higher for compact
disc:

The sampling frequency must be
higher than / (at least) twice the frequency to be sampled.

EITHER Higher (range of) frequencies
are reproduced on the disc OR Lower (range of) frequencies on the phone.

Additionally, EITHER a higher
quality (of sound) may be needed on the disc OR high quality (of sound) is not
required for the phone

__Question 14__**{Detailed explanations for this question is available as Solution 611 at Physics 9702 Doubts | Help Page 121 -**

*http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-121.html*}
can you please graph 2c from w14 qp 43?

ReplyDelete2b (same paper) is it a straight line passing through origin because f=bqv where v is proportional to f?

Check question 624 at

Deletehttp://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-123.html

and 4 c for the same paper.

ReplyDeleteFor question 4, check solution 632 at

Deletehttp://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-125.html

ReplyDeletein 10 b i) can you explain"op-amp has very large / infinite gain and so saturates" this statement?whats saturation and infinite gain ?

Firstly, I would like to point out that if you read the application booklet carefully, these are already explained well.

DeleteBut for a quick explanation:

What saturation means?

The supply line usually provide a potential of about 9.0V, 6.0V, ... (like these values) to the op-amp. If there is a significant difference between the input potentials (inverting and non-inverting or V- and V+ here), the output voltage would be very much larger than the voltages supplied by the supply line. (see application booklet for examples)

From the conservation of energy, this is not possible - a supply voltage of 9.0V cannot give an output voltage of, say, 10 000V. SO, then the output would be equal to the supply voltage, 9.0V here. THat is what we call saturation. Theoretically, the output should be 10 000V, but in reality this is not the case. We say that the op-amp has been saturated.

Infinite gain is that very large value of output obtained theoretically, when compared to the input potential.

However, saturation does not occur if the inputs are approximately the same. V+ is connected to the ground, so its potential is 0.0V. For the op-amp not to saturate, V- should be approximately equal to V+. Since V+ is zero, V- should be approximately equal to zero. That's why it's called virtual earth

Can you explain to me the answers in 14)c) in detail? Thanks :)

DeleteCan you explain to me the answer for 14c) in detail?

DeleteDo you mind explaining to me answer for O/N 2014/41 question 5b)ii) as well?

Your help is much appreciated :)

Details have been updated

DeleteFor Nov 2014 P41 Q5, see question 619 at

Deletehttp://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-122.html

for Nov 2014 P41 Q5 b i, how is electric potential constant? as you go to the right,the potential decreases. Moreover, even if the potential is constant. why would that prevent the radius from being any bigger?

DeleteEven though this not directly shown in the graph, it is known that the electric potential starts to decrease from the surface of the sphere from further away and that the electric potential is constant inside the sphere. Since the graph starts at x = 1cm (and the part before x = 1cm is not shown [it is not be zero]), we can infer that before x = 1cm, we are inside the sphere.

DeleteIf the radius was greater than 1.0cm, the electric potential should be constant for some values of x after x = 1cm.

11 ciii) why is it multiplied instead of dividing?and 12 c it says greater number of voltage and then smaller voltage which doesnt make sense for w 14 qp 43. Thank you very much.

ReplyDeleteFor the first one, see solution 615 at

Deletehttp://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-121.html

Hye can you explain question 7 June 2014

ReplyDeleteQ7 has been explained

Delete