The sinusoidal carrier wave has a frequency of 750 kHz and an amplitude of 5.0 V. The carrier wave is frequency modulated by a sinusoidal signal of frequency 7.5 kHz and amplitude 1.5 V.

 Question 9

A carrier wave is frequency modulated.

(a) Describe what is meant by frequency modulation. [2]

 

 

(b) The sinusoidal carrier wave has a frequency of 750 kHz and an amplitude of 5.0 V.

The carrier wave is frequency modulated by a sinusoidal signal of frequency 7.5 kHz and amplitude 1.5 V.

 

The frequency deviation of the carrier wave is 20 kHz V^-1.

 

Determine, for the frequency-modulated carrier wave,

(i) the amplitude, [1]

(ii) the minimum frequency, [1]

(iii) the maximum frequency, [1]

(iv) the number of times per second that the frequency changes from its minimum value to its maximum value and then back to the minimum value. [1]

 

 

 

Reference: Past Exam Paper – November 2015 Paper 41 & 42 Q11

 

 

 

Solution:

(a)

Frequency modulation means that the frequency of the carrier wave varies in synchrony with the displacement of the signal (information) wave.

 

 

(b)

(i) 5.0 V          

{In FM, the amplitude does not change but the frequency changes.}

 

(ii) the minimum frequency, [1]

720 kHz

 

{The frequency of the unmodulated carrier wave is 750 kHz.

 

The frequency deviation of the carrier wave is 20 kHz V^-1. This means that the (above) frequency can deviate (from its value of 750 kHz) by 20 kHz per volt of the signal wave.

 

1 V - - - > deviation of 20 kHz

The amplitude of the signal wave is 1.5 V.

1 V - - - > deviation of 20 kHz

1.5 V - - - > 1.5 × 20 = 30 kHz

 

So, the frequency of the carrier wave can deviate (can increase or decrease) by up to 30 kHz.

Minimum frequency = 750 – 30 = 720 kHz}

 

(iii) 780 kHz

{Maximum frequency = 750 + 30 = 780 kHz}

 

(iv) 7500

{This is the frequency of the signal wave (= 7.5 kHz = 7500 Hz)}

 

A radiation detector is placed close to a radioactive source. The detector does not surround the source.

Question 22

(a) A radiation detector is placed close to a radioactive source. The detector does not surround the source.

 

Radiation is emitted in all directions and, as a result, the activity of the source and the measured count rate are different.

 

Suggest two other reasons why the activity and the measured count rate may be different. [2]

 

 

(b) The variation with time t of the measured count rate in (a) is shown in Fig. 12.1.

Fig. 12.1

 

(i) State the feature of Fig. 12.1 that indicates the random nature of radioactive decay. [1]

 

(ii) Use Fig. 12.1 to determine the half-life of the radioactive isotope in the source. [4]

 

 

(c) The readings in (b) were obtained at room temperature.

A second sample of this isotope is heated to a temperature of 500 °C.

The initial count rate at time t = 0 is the same as that in (b).

The variation with time t of the measured count rate from the heated source is determined.

 

State, with a reason, the difference, if any, in

1. the half-life,

2. the measured count rate for any specific time.

[3]

 [Total: 10]

 

 

 

Reference: Past Exam Paper – November 2017 Paper 41 & 43 Q12

 

 

 

Solution:

(a)

Any two points

emission from radioactive daughter products

self-absorption in source

absorption in air before reaching detector

detector not sensitive to all radiations

window of detector may absorb some radiation

dead-time of counter

background radiation

 

 

(b)

(i)

The curve is not smooth.

 

(ii)

clear evidence of allowance for background

half-life determined at least twice

half-life = 1.5 hours

 

 

{As the time increases, the curve should tend towards zero. However, it is observed from the graph, that the curve tends towards a count rate of 10. So,

Background radiation = 10 counts / min

This value should be reduced from the count rate at different times.

 

In 1 half-life, the count rate should be halved (after accounting for the background radiation).

 

1^st set of data:

From graph, when count rate = 160, time = 0.2 hour

This corresponds for a count rate of (160 – 10 =) 150 for the radioactive isotope only.

After 1 half-life, this value would be halved (= 150 / 2 = 75). In the graph (after adding the background radiation), the count rate would be (75 + 10 =) 85.

From graph, when count rate = 85, time = 1.7 hour

Half-life = 1.7 – 0.2 = 1.5 hour

 

2^nd set of data:

From graph, when count rate = 100, time = 1.3 hour

This corresponds for a count rate of (100 – 10 =) 90 for the radioactive isotope only.

After 1 half-life, this value would be halved (= 90 / 2 = 45). In the graph (after adding the background radiation), the count rate would be (45 + 10 =) 55.

From graph, when count rate = 55, time = 2.8 hour

Half-life = 2.8 – 1.3 = 1.5 hour

 

Averaging the 2 values give a half-life of 1.5 hour.}

 

 

(c)

1. There is no change in the half-life because the decay is spontaneous/independent of environment

 

2. The count rate (is likely to be or could be) different

A Hall probe is placed near one end of a solenoid that has been wound on a soft-iron core, as shown in Fig. 9.1.

Question 9

(a) A Hall probe is placed near one end of a solenoid that has been wound on a soft-iron core, as shown in Fig. 9.1.

 Fig. 9.1

 

The current in the solenoid is switched on.

The Hall probe is rotated until the reading VH on the voltmeter is maximum.

 

The current in the solenoid is then varied, causing the magnetic flux density to change.

The variation with time t of the magnetic flux density B at the Hall probe is shown in Fig. 9.2.

 

Fig. 9.2

 

At time t = 0, the Hall voltage is V0.

On Fig. 9.3, draw a line to show the variation with time t of the Hall voltage VH for time t = 0 to time t = t4.

Fig. 9.3

[2]

 

 

(b) The Hall probe in (a) is now replaced by a small coil of wire connected to a sensitive voltmeter, as shown in Fig. 9.4.

Fig. 9.4

 

The magnetic flux density, normal to the plane of the small coil, is again varied as shown in Fig. 9.2.

 

On Fig. 9.5, draw a line to show the variation with time t of the e.m.f. E induced in the small coil for time t = 0 to time t = t4.

Fig. 9.5

[3]

 [Total: 5]

 

Reference: Past Exam Paper – November 2018 Paper 42 Q9

 

Solution:

(a)

0 → t1 horizontal straight line at non-zero value of VH

and

t3 → t4 horizontal straight line at different non-zero VH

 

t1 → t3 straight diagonal line with negative gradient

and

graph line starts at (0, V0) and ends at (t4, –2V0)

 

{Hall voltage: V_H = BI / ntq

The Hall voltage is proportional to the magnetic flux density B.

 

From t= 0 to t1, B is constant, so VH is also constant (= V₀).

 

From t1 to t3, B decreases at a constant rate (constant gradient). So, VH also decreases. At t3 the value of B is negative and twice that at t1, so VH will also be negative and twice larger (= - 2V₀).

 

From t3 to t4, B is constant and so, VH is also constant.}

 

 

(b)

E = 0 for 0 → t1 and t3 → t4

E is non-zero at all points between t1 → t3

E has constant magnitude between t1 → t3

 

{In this case, electromagnetic induction occurs.

From Faraday’s law, the induced e.m.f. is proportional to the rate of change of B.

 

From t = 0 to t1, B is not changing, so no e.m.f. is induced.

 

From t1 to t3, B is decreasing at a constant rate (constant gradient). So, the e.m.f. induced is also constant from t1 to t3.

 

From t3 to t4, B in not changing, so no e.m.f. is induced.}