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Question 639: [Current of Electricity + Measurements]

(a) Define electrical resistance.

(b) A circuit is set up to measure resistance R of a metal wire. Potential difference (p.d.) V across the wire and the current І in the wire are to be measured.

(i) Draw a circuit diagram of apparatus that could be used to make these measurements.

(ii) Readings for p.d. V and the corresponding current І are obtained. These are shown in Fig.1.

Explain how Fig.1 indicates that the readings are subject to

1. a systematic uncertainty,

2. random uncertainties.

(iii) Use data from Fig.1 to determine R. Explain your working.

(c) In another experiment, a value of R is determined from the following data:

Current І = 0.64 ± 0.01 A and p.d. V = 6.8 ± 0.1 V.

Calculate value of R, together with its uncertainty. Give your answer to an appropriate number of significant figures.

Reference: Past Exam Paper – November 2012 Paper 21 Q2

Solution 639:

(a) Electric resistance of a component is defined as the ratio of potential difference across it to the current flowing through it.

(b).

(i)

The circuit should consist of

metal wire in series with power supply and ammeter

voltmeter in parallel with metal wire

EITHER rheostat in series with power supply or potential divider arrangement

OR variable power supply

(ii)

1. The (y-) intercept on the graph (is not at I = 0)

2. The scatter of the readings about the best fit line.

(iii)

There should be a correction for zero error for the value of current taken since the y-intercept is not at zero. There is a zero error of +0.05A. So, all values of current from the graph should be reduced by 0.05A.

Use of V and corrected І values from graph:

Consider V = 4.0V

From graph, I = 0.23A

Correct value of current to use = 0.23 – 0.05 = 0.18A

Resistance = V / І = 22.(2) Ω              [e.g. 4.0 / 0.18]

(c)

Resistance R = 6.8 / 0.64 = 10.625 Ω            

(Percentage uncertainties) %R = %V + %І

%R = (0.1/6.8)×100 + (0.01/0.64)×100 = 1.47% + 1.56% (= 3.03%)

ΔR (= %R × R) = 0.0303 × 10.625 = 0.32 Ω

Resistance R = 10.6 ± 0.3 Ω

Question 640: [Waves]

Period of an electromagnetic wave is 1.0 ns.

What are frequency and wavelength of the wave?

frequency / Hz            wavelength / m

A         1.0                               3.0 × 10⁸

B         1.0 × 10⁶                      300

C         1.0 × 10⁹                      0.30

D         1.0 × 10¹²                    3.0 × 10^–4

Reference: Past Exam Paper – June 2012 Paper 12 Q27

Solution 640:

Answer: C.

Frequency f = 1/T       where T is the period

Frequency f = 1 / (1×10^-9) = 1×10⁹ Hz

There is no need to waste time calculating the wavelength.

Question 641: [Nuclear Physics]

(a) Nuclear reaction occurs when uranium-235 nucleus absorbs a neutron. The reaction may be represented by equation:

²³⁵₉₂U   +          ^W_X n     - - - >  ⁹³₃₇Rb +         ¹⁴¹_ZCs +          Y^W_X n

State the number represented by letter

W

X

Y

Z

(b) The sum of masses on left-hand side of equation in (a) is not the same as the sum of the masses on right-hand side.

Explain why mass seems not to be conserved.

Reference: Past Exam Paper – June 2012 Paper 22 Q7

Solution 641:

Go to
A nuclear reaction occurs when a uranium-235 nucleus absorbs a neutron. The reaction may be represented by the equation

Question 642: [Waves > Intensity]

A small source emits spherical waves.

Wave intensity I at any point P, a distance r from source, is inversely proportional to r².

What is the relationship between wave amplitude a and distance r?

A a² α 1/r                     B a α 1/r                      C a α 1/r²                     D a α 1/r⁴

Reference: Past Exam Paper – June 2014 Paper 13 Q27

Solution 642:

Answer: B.

Intensity I is directly proportional to (amplitude a)².

Intensity I is inversely proportional to r².

So, a² is proportional to 1 / r².

Amplitude a is proportional to 1 / r.