# 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 × 10

^{6}J B 3.4 × 10^{6}J C 1.7 × 10^{9}J D 3.4 × 10^{9}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 mv

^{2}= ½ (1200 000) (75)^{2}= 3.4 × 10^{9}J
So, the useful work that must be
done on the train is 3.4 × 10

^{9}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}m^{2}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 × 10^{8}) / (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 × 10^{15}
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 × 10^{15}) = 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 / (πD

^{2}/4) = 4ρL / (πD^{2})
For wire X, R = 4ρL / (πD

^{2})
For wire Y, R = 4ρ[2L] / (π[2D]

^{2}) = 2ρL / (πD^{2}).
So, ratio of resistance of X to
resistance of Y = 2 / 1

__Question 565: [Nuclear Physics]__**(a)**Two isotopes of uranium are uranium-235 (

^{235}

_{92}U) and uranium-238 (

^{238}

_{92}U).

(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:

^{235}

_{92}U +

^{W}

_{X}n - - - >

^{148}

_{57}La +

^{Z}

_{Y}Q + 3

^{W}

_{X}n

(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

^{1}_{0}n. 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 = mc^{2}.
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?

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