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

Paper 41 & 43

SECTION A

Question 1

A stone of mass m has gravitational potential energy EP at a point X in a gravitational field.

Question 2

A mass of 3.2 g of argon-40 has a volume of 210 cm3 at a temperature of 37 °C.

Question 3

Volume of 1.00 kg of water in liquid state at 100 °C is 1.00 × 10⁻³ m³. Volume of 1.00 kg of water vapour at 100 °C and atmospheric pressure 1.01 × 10⁵ Pa is 1.69 m³.

(a)

Show that the work done against the atmosphere when 1.00 kg of liquid water becomes water vapour is 1.71 × 10⁵ J:

EITHER change in the volume = 1.69 – (1.00x10^-3)

OR the liquid volume << volume of the vapour

So, work done (= pΔV) = (1.01x10⁵) x 1.69 = 1.71x10⁵J

(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 × 10⁶ 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 = (2.26x10⁶) – (1.71x10⁵) = 2.09x10⁶J

Question 4

A student investigates the energy changes of a mass oscillating on a vertical spring, as shown in Fig. 4.1.

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 V₀. 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 = V₀ from x = 0 to x = r. Then, it is a curve with a decreasing gradient passing through (2r, 0.50V₀) and (4r, 0.25V₀)

(b)

Electric field strength at surface of sphere is E₀. 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, E₀) passing through (2r, ¼E₀)

Question 6

An uncharged capacitor is connected in series with a battery, a switch and a resistor, as shown in Fig. 6.1.

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^-19C

(c)

Student repeats experiment and determines the following values for charge on oil drops. 3.3 × 10⁻¹⁹C                  4.9 × 10⁻¹⁹C                      9.7 × 10⁻¹⁹C                3.4 × 10⁻¹⁹C

Use values to suggest value for elementary charge:

The elementary charge = 1.6x10^-19C

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 = 5.8x10¹⁴Hz                

2.

Work function, in eV, of metal surface:

Work function, ϕ = hf₀ = (6.63x10^-34) (5.8x10¹⁴) = 3.84x10^-19J

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 (³₁H) nucleus               3.01551

polonium (²¹⁰₈₄Po) nucleus      209.93722

Energy equivalent of 1.0 u is 930 MeV

(i)

Binding energy, in MeV, of tritium (³₁H) nucleus:

Change in mass, Δm = 2(1.00867) + 1.00728 – 3.01551 = 9.11x10^-3u

Binding energy = (9.11x10^-3) x 930 = 8.47MeV

(ii)

Total mass of separate nucleons that make up polonium-210 (²¹⁰₈₄Po) 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 ²³⁵₉₂U + ¹₀n - - -> ¹⁴¹₅₆Ba  +  ⁹²₃₆Kr  +  3¹₀n

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 D₁, D₂, D₃ and D₄. 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}