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YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

## Wednesday, June 16, 2021

### Which statement about electromagnetic radiation is correct?

Question 26

Which statement about electromagnetic radiation is correct?

A Waves of wavelength 5 × 10-9 m are high-energy gamma rays.

B Waves of wavelength 3 × 10-8 m are ultra-violet waves.

C Waves of wavelength 5 × 10-7 m are infra-red waves.

D Waves of wavelength 9 × 10-7 m are light waves.

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q22

Solution:

Answer: B.

Estimates for the different components of the electromagnetic spectrum should be known at A-Level.

All EM waves travel at the same speed of 3.0×108 m s-1.

Gamma rays have the highest frequencies, followed by X-rays and UV.

v = fλ

For the given wavelength of UV waves,

f = v / λ = 3.0×108 / 3.0×10-8 = 1×1016 Hz

This is a correct estimate for the frequency of UV waves.

## Saturday, June 5, 2021

### A monochromatic plane wave of speed c and wavelength λ is diffracted at a small aperture.

Question 36

A monochromatic plane wave of speed c and wavelength λ is diffracted at a small aperture.

The diagram illustrates successive wavefronts.

After what time will some portion of the wavefront GH reach point P?

A 3λ / 2c                      B 2λ / c                                    C 3λ / c                                    D 4λ / c

Reference: Past Exam Paper – March 2017 Paper 12 Q28

Solution:

Answer: C.

Speed of wave = c

The distance between two successive wavefronts is equal to a wavelength.

So, the distance between wavefront GH and the wavefront containing point P is equal to 3 wavelengths (= 3λ).

Speed = distance / time

Time = distance / speed

Time = 3λ / c

## Tuesday, May 25, 2021

### A particle is following a circular path and is observed to have an angular displacement of 10.3°.

Question 3

A particle is following a circular path and is observed to have an angular displacement of 10.3°.

(a) Express this angle in radians (rad). Show your working and give your answer to three significant figures. 

(b) (i) Determine tan10.3° to three significant figures.

(ii) Hence calculate the percentage error that is made when the angle 10.3°, as measured in radians, is assumed to be equal to tan10.3°.



Reference: Past Exam Paper – November 2004 Paper 4 Q1

Solution:

(a)

θ (rad) = 2π × (10.3/360)

θ (rad) = 0.180 rad

(b) (i) tan θ = 0.182

(ii)

{Difference between θ and tanθ = 0.182 – 0.180 = 0.002}

percentage error = (0.002/0.180) x 100

percentage error = 1.1 (%)

## Monday, May 17, 2021

### A steel wire of cross-sectional area 15 mm2 has an ultimate tensile stress of 4.5 × 108 N m-2.

Question 21

A steel wire of cross-sectional area 15 mm2 has an ultimate tensile stress of 4.5 × 108 N m-2.

(a) Calculate the maximum tension that can be applied to the wire. 

(b) The steel of the wire has density 7800 kg m-3. The wire is hung vertically.

Calculate the maximum length of the steel wire that could be hung vertically before the wire breaks under its own weight. 

Reference: Past Exam Paper – November 2015 Paper 23 Q7

Solution:

(a)

{Stress = Force / Area

Note that the tension is a force.}

stress σ = F / A

{Convert area into m2 by multiplying by 10-6.

Tension = Stress × Area}

max. tension = UTS × A = 4.5×108 × 15×10-6 = 6800 (6750) N

(b)

{We first need to relate the length with data available.}

Density ρ = m / V

weight = mg = ρVg = ρALg                 {since Volume V = Area × Length = AL}

{When the wire break under its own weight, the weight should be equal to the maximum tension.

Weight = ρALg

Tension = ρALg}

6750 = 7.8×103 × 15×10-6 × L × 9.81

L = 5.9 (5.88) × 103 m

## Saturday, April 10, 2021

### The analogue signal from a microphone is to be transmitted in digital form.

Question 10

The analogue signal from a microphone is to be transmitted in digital form.

The variation with time t of part of the signal from the microphone is shown in Fig. 5.1.

Fig. 5.1

The microphone output is sampled at a frequency of 5.0 kHz by an analogue-to-digital converter (ADC).

The output from the ADC is a series of 4-bit numbers. The smallest bit represents 1.0 mV.

The first sample is taken at time t = 0.

(a) Use Fig. 5.1 to complete Fig. 5.2.

Fig. 5.2                                                            

(b) After transmission of the digital signal, it is converted back to an analogue signal using a digital-to-analogue converter (DAC).

Using data from Fig. 5.1, draw, on the axes of Fig. 5.3, the output level from the DAC for the transmitted signal from time t = 0 to time t = 1.2 ms.

Fig. 5.3



(c) It is usual in modern telecommunication systems for the ADC and the DAC to have more than four bits in each sample.

State and explain the effect on the transmitted analogue signal of such an increase. 

[Total: 8]

Reference: Past Exam Paper – November 2017 Paper 42 Q5

Solution:

(a)

(0.2 ms)           8.0 (mV)          1000                B1

(0.8 ms)           5.8 (mV)          0101                B1

{4-bit:               23   22   21   20 }

8     4    2     1

8.0 mV            1     0    0     0              = 1(23) + 0(22) + 0(21) + 0(20) = 8

5.8 mV (consider 5.0 mV, NOT 6.0 mV – NOT the nearest whole number)

5.8 mV            0     1   0     1              = 0(23) + 1(22) + 0(21) + 1(20) = 5

(b)

level                 0              8                  10                    15                    5                   8

time / ms         0–0.2     0.2–0.4          0.4–0.6            0.6–0.8            0.8–1.0        1.0–1.2

{Consider the initial time for the intervals.

e.g.      for time: 0–0.2ms, consider voltage at t = 0 ms (= 0 mV = level 0)

for time: 0.2–0.4ms, consider voltage at t = 0.2 ms (= 8 mV = level 8)

…}

(c)

The heights of the step is smaller.

It allows more accurate reproduction (of the input signal). {That is, the output signal represents the input signal more closely (smaller changes between the output and the input).}

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