• 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 / (4πϵ_oR)

(ii)

Define Capacitance and show expression:

Capacitance is the ratio of charge and potential or q/V

C = q/V = 4πϵ_oR

(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:

N_p / N_s = V_p / V_s

Vp = (2)^1/2 x V_rms = (2)^1/2 x 240 V = 339.4V

V_s = 9V

Ratio = V_p / V_s = 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}