Physics 9702 Doubts | Help Page 110
Question 561: [Frequency
modulation]
(a) Describe what is meant by frequency modulation (FM).
(b) A sinusoidal carrier wave has frequency of 600kHz and amplitude of
5.0V.
The carrier wave is frequency
modulated by a sinusoidal wave of frequency 7.0kHz and amplitude 2.0V.
Frequency deviation of the carrier
wave is 20kHzV-1.
Determine, for modulated carrier
wave,
(i) amplitude
(ii) maximum frequency
(iii) minimum frequency
(iv) number of times per second that
frequency changes from maximum to minimum and then back to maximum
Reference: Past Exam Paper – June 2011 Paper 41 Q11
Solution 561:
(a) The frequency of the carrier wave varies (in synchrony) with the
displacement of the information signal.
(b)
(i) Amplitude = 5.0V
(ii)
{Amplitude of sinusoidal
wave = 2.0V. Frequency deviation of carrier wave is 20kHz per volt. So, for 2V,
deviation = 2x20 = 40kHz}
Maximum frequency = (600 + 40 =)
640kHz
(iii) Minimum frequency = (600 – 40 =)
560kHz
(iv)
(This is the frequency of
the sinusoidal wave (not carrier wave) which is 7kHz = 7000Hz. [Frequency =
1/period – that is, number per unit time (second)])
7000
Question 562: [Work,
Energy and Power]
Railway engine accelerates train of
total mass 1200 tonnes (1 tonne = 1000 kg) from rest to a speed of 75 m s–1.
How much useful work must be done on
the train to reach this speed?
A 1.7 × 106 J B 3.4 × 106 J C 1.7 × 109 J D 3.4 × 109 J
Reference: Past Exam Paper – November 2012 Paper 12 Q20
Solution 562:
Answer: D.
Initially, the train is at rest.
1200 tonnes = 1200 000 kg
When the train reaches the speed of 75
m s–1, its kinetic energy is
Kinetic energy of train = ½ mv2
= ½ (1200 000) (75)2 = 3.4 × 109 J
So, the useful work that must be
done on the train is 3.4 × 109 J.
Question 563: [Quantum
Mechanics > Photon]
(a) State what is meant by photon.
(b) A beam of light is incident normally on a metal surface, as illustrated
in Fig.
Beam of light has cross-sectional
area 1.3 × 10−5 m2 and power 2.7 × 10−3 W.
Light has wavelength 570 nm.
Light energy is absorbed by metal
and no light is reflected.
(i) Show that a photon of this light
has energy of 3.5 × 10−19 J.
(ii) Calculate, for time of 1.0 s,
1. number of photons incident on the
surface,
2. change in momentum of the
photons.
(c) Use answer in (b)(ii) to calculate the pressure that the light
exerts on the metal surface.
Reference: Past Exam Paper – November 2014 Paper 43 Q8
Solution 563:
(a) A photon is a discrete amount / packet / quantum of energy
of electromagnetic radiation / EM radiation.
(b)
(i)
Energy E of a photon = hf = hc / λ
Energy = (6.63 × 10–34 ×
3.0 × 108) / (570 × 10–9) = 3.49 × 10–19 J
(ii)
1.
{Power = rate of change of
energy = energy in a time of 1.0s
Number of photons = total
energy / energy of 1 photon}
Number of photons = (2.7 × 10–3)
/ (3.5 × 10–19) = 7.7 × 1015
2.
{Energy of photon = pc or =
hf. Momentum p of photon = E / c = hf/c = h / λ}
Momentum of 1 photon = h / λ = (6.63
× 10–34) / (570 × 10–9) = 1.16 × 10–27 kg m s–1
{For n photons, the change
in momentum = n x momentum of 1 photon}
Change in momentum = (1.16 × 10–27)
× (7.7 × 1015) = 8.96 × 10–12 kg m s–1
{Note that here, we are
dealing with momentum in Quantum Mechanics (physics of very small particles),
not with the usual momentum in Newtonian Mechanics (physics of large objects)
where momentum = mv. Do not confuse these 2.
Also, further study in
physics will let you know that a photon is actually an elementary particle. At
this level, this can be understand from the wave-particle duality – that is, a
wave can behave as a particle and vice versa. So, it’s possible for a photon to
have a momentum. Do not bother yourself too much with this at this level.}
(c)
{Change is momentum in one
second is the change in momentum per second or the rate of change of momentum.
The rate of change of momentum gives the force.
Pressure = Force / Area}
Pressure = (change in momentum per
second) / area
Pressure = (8.96 × 10–12)
/ (1.3 × 10–5) = 6.9 × 10–7 Pa
Question 564: [Current
of Electricity]
Two electrically-conducting cylinders
X and Y are made from same material.
Their dimensions are as shown.
Resistance between the ends of each
cylinder is measured.
What is the ratio of
resistance of X to resistance of Y?
A 2 / 1 B 1 / 1 C
1 / 2 D 1 / 4
Reference: Past Exam Paper – November 2011 Paper 12 Q34 & June 2014 Paper 11 Q30
Solution 564:
Answer: A.
Resistance R of a wire = ρL/A = ρL / (πD2/4) = 4ρL / (πD2)
For wire X, R = 4ρL / (πD2)
For wire Y, R = 4ρ[2L] / (π[2D]2)
= 2ρL / (πD2).
So, ratio of resistance of X to
resistance of Y = 2 / 1
Question 565: [Nuclear Physics]
(a) Two isotopes of uranium are uranium-235 (23592U)
and uranium-238 (23892U).
(i) Describe in detail an atom of
uranium-235.
(ii) With reference to the two forms
of uranium, explain term isotopes.
(b) When uranium-235 nucleus absorbs a neutron, the following reaction
may occur:
23592U + WXn
- - - > 14857 La + ZYQ +
3WXn
(i) Determine values of Y and Z.
(ii) Explain why sum of the masses
of the uranium nucleus and of the neutron does not equal the total mass of
products of the reaction.
Reference: Past Exam Paper – June 2013 Paper 23 Q7
Solution 565:
(a)
(i)
The nucleus contains 92 protons and
143 neutrons.
Outside / around the nucleus there
are 92 electrons.
Most of the atom is empty space / the
mass concentrated in the nucleus.
The total charge is zero.
The diameter of the atom ~ 10–10
m or the size of the nucleus ~ 10–15 m
(ii) The isotopes of the uranium nucleus
has the same number / 92 protons but one nucleus have 143 neutrons and the
other nucleus has146 neutrons
(b)
(i)
{We have these 2
equations:
235 + W = 148 +Z +3W
92 + X = 57 + Y + 3X
We know that a neutron is 10n.
So, W = 1 and X = 0. Replacing in the equations give}
Y = 35
Z = 85
(ii) Mass-energy is conserved in the
reaction. That is, the mass on the right-hand side of reaction is less {even though the mass number A and proton number Z are equal
on both sides of the equation. The atomic mass is usually expressed in terms of
the unified atomic mass constant, u which is about u = 1.66 × 10–27
kg. The atomic mass is slightly different from the mass number A and thus, when
calculating [note that the atomic mass are not given here, only the mass number
A is given] a difference in mass will be seen. This is very small but the energy
released due to it is quite large} and so energy is released in the reaction
since a mass difference of m gives an energy equal to E = mc2.
Hi, can you explain November 2014 paper 43 Q9 b? Thank you :)
ReplyDeleteSee question 653 at
Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-130.html
w 2014 q 8 pls
ReplyDeletewhich variant?
Delete