• 9702 June 2012 Paper 42 Worked Solutions | A-Level Physics

SECTION A

Question 1

{Detailed explanations for this question is available as Solution 711 at Physics 9702 Doubts | Help Page 144 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-144.html}

Question 2

{Detailed explanations for this question is available as Solution 510 at Physics 9702 Doubts | Help Page 99 - http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-99.html}

Question 3

(a)

The internal energy of a system is the sum of the potential energy and the kinetic energy of the atoms / molecules / particles which are randomly distributed in the system.

(b)

Explain change in internal energy of following systems:

(i)

Lump of ice at 0^oC melts to form liquid water at 0^oC:

As the lattice structure is ‘broken’ / the bonds are broken / the forces between the molecules are reduced, the potential energy of the molecules increases while no change occur in the kinetic energy. So, the internal energy of the system increases.

(ii)

Cylinder containing gas at constant volume in sunlight so that its temperature rises from 25^oC to 35^oC:

The molecules / atoms / particles move faster / the mean squared speed, <c²> is increasing  /  Kinetic energy of the gas molecules increases with an increase in temperature. So, no change occurs to the potential energy of the molecules while their kinetic energy increases. Thus, internal energy increases.

Question 4

{Detailed explanations for this question is available as Solution 994 at Physics 9702 Doubts | Help Page 206 - http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-206.html}

Question 5

A horseshoe magnet is placed on a balance. A stiff metal wire is clamped horizontally between the poles, as illustrated in Fig. 5.1.

Question 6

{Detailed explanations for this question is available as Solution 727 at Physics 9702 Doubts | Help Page 147 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-147.html}
 




Question 7

(a)

Work function energy is the minimum energy required to remove an electron from the metal/surface.

(b)

Variation with frequency f of maximum kinetic energy Ek of photoelectrons emitted from surface of sodium metal.

Use gradient to determine value of Planck constant h:

Gradient = 4.17x10^-15 (allow 4.1 – 4.3)

h = (4.15x10^-15)(1.6x10^-19)                  or h = (4.1 to 4.3) x10^-15 eVs

h = 6.6x10^-34Js

(c)

Sodium metal (Work function = 2.4eV) replaced by Calcium (Work function = 2.9eV).

Draw line to show variation with f of max Ek of photoelectrons emitted from surface of calcium:

Graph: 

a straight line parallel to the given line with intercept at frequency between 6.9x10¹⁴Hz and 7.1x10¹⁴Hz

Question 8

Strontium has at least 16 isotopes. One is strontium-89, with half-life = 52days

(a)

Isotopes of an element are nuclei of that element having the same number of protons(atomic number) but different number of neutrons

(b)

Probability per second of decay of a nucleus of strontium-89:

The probability of decay per unit time is the decay constant.

λ = ln2 / t_½ = 0.693 / (52x24x3600) = 1.54x10^-7s^-1    

(c)

Activity of strontium-89 source measured 21 days after preparation of source and found to be = 7.4x10⁶Bq.

At time it was prepared,

(i)

Activity:

A = A_oexp(-λt)

7.4x10⁶ = A_oexp(-1.54x10^-7(21x24x3600))

A_o = 9.8x10⁶Bq

(ii)

Mass of strontium-89:

A­_o = λN                       and      mass = N x 89/N­_A

mass = (9.8x10⁶ x 89) / (1.54x10^-7 x 6.02x10²³) = 9.4x10^-9g  

SECTION B

Question 9

(a)

4 properties of an ideal operational amplifier (op-amp):

Any 4:

e.g Infinite input impedance/resistance

Zero output impedance/resistance

Infinite (open loop) gain

Infinite bandwidth

Infinite slew rate

(b)

Comparator: V_OUT = A_o (V⁺ - V^-). Here V⁺ = 0V and V^- = V_IN

Graph: square wave with a 180^o phase change and amplitude 5.0V

(c)

Connections of the 2 LEDs to output of op-amp – red when V_out +ve and green when V_out -ve:

The correct symbol of LED

Diodes are connected correctly between V_out and earth

Diodes are identified

Question 10

(a)

Aluminum block placed near small source of X-ray radiation. Reasons why intensity of X-ray beam at point B not as great as the intensity at point A:

Any 2 sensible suggestions:

The beam is divergent / obeys an inverse square law

Due to absorption (in the block)

Due to scattering (of the beam in the block)

Reflection (occurs at the boundaries)

(b)

Thickness of model = 2.4cm. Thickness of bone in model = 1.1cm

Linear attenuation (absorption) coefficient for bone = 3.0cm^-1, for soft tissue = 0.27cm^-1

Ratio of intensity of X-ray beam incident on model to intensity of X-ray beam emergent from model:

(i)

For beam AB:

I = I_oexp(-μx)

Ratio = I_o/I = exp(0.27x2.4)  = 1.9

(ii)

For beam CD:

Ratio = I_o/I = exp(0.27x1.3) x exp(3.0x1.1)

Ratio = 1.42 x 27.1 = 38.5

(A common error was to calculate the ratios for the bone and the soft tissue and then to add these two quantities, rather than find the product.)

(c)

Why, for this model, X-ray image with good image contrast may be obtained:

Either There is much greater absorption in the bone than in the soft tissue

Or I_o/I is much greater for the bone than for the soft tissue

Question 11

Signal transmitted over long distance will be attenuated and it will pick up noise.

(a)

(i)

Attenuation is the loss of (signal) power.

(ii)

Noise is the random unwanted power that gets picked up on the signal.

(b)

Why regenerator amplifiers do not amplify noise that has been picked up on digital signals:

For a digital signal, only the ‘high’ and the ‘low’ \ 1 and 0 are necessary. So, any variation between ‘highs’ and ‘lows’ which are caused by noise is not required.

(c)

Transmitter on Earth produces signal of power 2.4kW. Signal, when received by satellite, is attenuated by 195dB.

Signal power received by satellite:

Attenuation = 10log(P₂/P₁)

Either 195 = 10log({2.4x10³}/P₁)

Or –195 = 10log(P₁/{2.4x10³})

P₁ = 7.6x10^-17W  

Question 12

Simplified block diagram of circuitry for a mobile-phone handset.

(a)

(i)

X = modulator

(ii)

Y = serial-to-parallel converter

(b)

Explain purpose of

(i)

Switch:

The switch enables one aerial to be used for transmission and receipt of signals.

(ii)

Parallel-to-serial converter:

All the bits for one number arrive at the same time. The parallel-to-serial converter allows the bits to be sent out one after another.