Thursday, October 29, 2015

Physics 9702 Doubts | Help Page 220

  • Physics 9702 Doubts | Help Page 220



Question 1048: [Applications > Communicating information]
(a) State two reasons why frequencies in the gigahertz (GHz) range are used in satellite communication.

(b) In one particular satellite communication system, the frequency of the signal transmitted from Earth to the satellite (the up-link) is 6 GHz. The frequency of the signal transmitted back to Earth from the satellite (the down-link) is 4 GHz.
Explain why the two signals are transmitted at different frequencies.

(c) A signal transmitted from Earth has a power of 3.1 kW.
This signal, received by a satellite, has been attenuated by 185 dB.
Calculate the power of the signal received by the satellite.

Reference: Past Exam Paper – November 2013 Paper 43 Q12



Solution 1048:
(a) e.g.
no / little ionospheric reflection
large information carrying capacity

(b)
This prevents the (very) low power signal received at satellite from being swamped by the high-power transmitted signal       

(c)
Attenuation / dB = 10 lg(P2/P1)
{P2 = power of signal transmitted from Earth and P1 = power of (attenuated) signal received by satellite}
185 = 10 lg({3.1×103} / P)
P = 9.8 × 10–16 W










Question 1049: [Matter > Thermal properties of materials > Specific latent heat]
(a) Define specific latent heat of fusion.

(b) Some crushed ice at 0 °C is placed in a funnel together with an electric heater, as shown in Fig. 2.1.


The mass of water collected in the beaker in a measured interval of time is determined with the heater switched off. The mass is then found with the heater switched on. The energy supplied to the heater is also measured.
For both measurements of the mass, water is not collected until melting occurs at a constant rate.
The data shown in Fig. 2.2 are obtained.
(i) State why the mass of water is determined with the heater switched off.
(ii) Suggest how it can be determined that the ice is melting at a constant rate.
(iii) Calculate a value for the specific latent heat of fusion of ice.

Reference: Past Exam Paper – November 2008 Paper 4 Q2



Solution 1049:
(a) Specific latent heat of fusion is defined as (thermal) energy / heat required to convert unit mass of solid to liquid at its normal melting point / without any change in temperature.

(b)
(i) To make allowance for heat gains from the atmosphere

(ii) Choose any 1:
By the constant rate of production of droplets from the funnel
When a constant mass of water is collected per minute in the beaker

(iii)
{Even with the heater off, some of the ice would still melts, forming water. In 10.0min 16.6g of water has been formed. So, in 5.0min half that amount of water is formed.
To calculate the specific latent heat of ice, we need to consider the amount of ice melted only due to the heater being switched on. So, the amount of water formed without the heater being on should be subtracted from the total mass of water formed in 5.0min with the heater being on.}
Mass melted by heater in 5 minutes = 64.7 – (½ × 16.6) = 56.4g
{Mass × Specific latent heat of fusion = Amount of heat energy supplied to the heater}
56.4×10-3 × L = 18 kJ
Specific latent heat of fusion, L = 320 kJ kg-1










Question 1050 [Electric field > Electric potential]
Two point charges A and B each have a charge of + 6.4 × 10–19 C. They are separated in a vacuum by a distance of 12.0 μm, as shown in Fig. 4.1.

Points P and Q are situated on the line AB. Point P is 3.0 μm from charge A and point Q is 3.0 μm from charge B.
(a) Calculate the force of repulsion between the charges A and B.

(b) Explain why, without any calculation, when a small test charge is moved from point P to point Q, the net work done is zero.

(c) Calculate the work done by an electron in moving from the midpoint of line AB to point P.

Reference: Past Exam Paper – June 2010 Paper 42 & 43 Q4



Solution 1050:
(a)
Force = q1q2 / 4πϵox2  
Force = (6.4×10-19)2 / [4π × (8.85×10-12) × (12×10-6)2]
Force = 2.56×10-17N 

(b)
{Electric potential, V = q / 4πϵor
The electric potential V at a point is the work done in bringing unit positive charge from infinity to that point.
Consider a charge at point P. It is affected by the field of 2 charges: a charge of + 6.4 × 10–19 C at a distance of 3.0μm (located at A) and a charge of + 6.4 × 10–19 C at a distance of 9.0μm (located at B).
Now consider a charge at point Q. It will be affected in the field in a similar way, except that the charge at A is at a distance of 9.0μm and the charge at B is at a distance of 3.0μm.
Thus, P and Q are at the same electric potential.}
The potential at P is the same as the potential at Q. The work done = qΔV. Since ΔV = 0, the net work done is zero.

(c)
{V = q / 4πϵor. At midpoint, potential = VA + VB. The distance r is 6.0μm. Since both potentials are the same, we can take twice the value of one of the potentials.}
At the midpoint, potential = 2 × [(6.4×10-19) / 4πϵo (6×10-6)]
{At point P, the value of r in VA = 3.0μm and the value of r for VB is 9.0μm.}
At point P, potential = [(6.4×10-19) / 4πϵo (3×10-6)] + [(6.4×10-19) / 4πϵo×(9×10-6)]
{At the midpoint, potential = 2×[(6.4×10-19) / 4πϵo (6×10-6)] = (6.4×10-19) / 4πϵo (3×10-6)
When taking the difference in potentials at P and at the midpoint, this terms ‘(6.4×10-19) / 4πϵo (3×10-6)’ would get eliminated.}
Change in potential (= potential at P – potential at midpoint) = (6.4×10-19) / 4πϵo×(9×10-6)
Energy (= qV) = (1.6×10-19) × [(6.4×10-19) / 4πϵo×(9×10-6)]  
{q is the charge of the electron.}
Energy = 1.0×10-22J











Question 1051: [Quantum Physics > Line Spectra]
White light is incident on a cloud of cool hydrogen gas, as illustrated in Fig. 8.1.

The spectrum of the light emerging from the gas cloud is found to contain a number of dark lines.
(a) Explain why these dark lines occur.

(b) Some electron energy levels in a hydrogen atom are illustrated in Fig. 8.2.

One dark line is observed at a wavelength of 435 nm.
(i) Calculate the energy, in eV, of a photon of light of wavelength 435 nm.
(ii) On Fig. 8.2, draw an arrow to indicate the energy change that gives rise to this dark line.

Reference: Past Exam Paper – November 2014 Paper 41 & 42 Q8



Solution 1051:
(a) The photon is ‘absorbed’ by the electrons. The photon has energy equal to the difference in energy of two energy levels. The electron de-excites emitting photon (of same energy) in any direction.

(b)
(i)
Energy E = hc / λ
E = (6.63×10–34 × 3×108) / (435×10–9)
E = 4.57 × 10–19 J
{To convert energy in J to eV, we divide by the charge of an electron.}
E = (4.57 × 10–19) / (1.6 × 10–19) (eV)
E = 2.86 eV

(ii) arrow pointing in either direction between –3.41 eV and –0.55 eV
{We need to consider 2 energy levels that have a difference in energy of 2.86eV. These are the energy levels –3.41 eV and –0.55 eV.}



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