• 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               +          ^W_Xn      - - - >   ¹⁴⁸₅₇ La             +          ^Z_YQ     +          3^W_Xn

(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.