• 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

A microwave cooker uses electromagnetic waves of frequency 2450 MHz. The microwaves warm the food in the cooker by causing water molecules in the food to oscillate with a large amplitude at the frequency of the microwaves.

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

Ions, all of the same isotope, are travelling in a vacuum with a speed of 9.6 × 104 m s-1.

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 = A_oexp(- λt)        and      λt_½ = 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 V_OUT 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 V_IN at inverting input of op-amp shown.

1.

On axes of Fig, draw variation with time t of output potential of op-amp:

V_OUT switches at +1.0V. The maximum value of V_OUT is 5.0V. When the curve is above +1.0V, V_OUT is negative (or vice versa)

{Since V_IN is at the inverting input, when V_IN is greater than the non-inverting input (1.0V), the output has the opposite polarity than V_IN 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 V_IN, 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

Distinguish between an X-ray image of a body structure and a CT scan.

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.