# Physics 9702 Doubts | Help Page 127

__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 × 10

^{8}
B 1.0
× 10

^{6}300
C 1.0
× 10

^{9}0.30
D 1.0
× 10

^{12}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}) = 1×10^{9}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:

^{235}

_{92}U +

^{W}

_{X }n - - - >

^{93}

_{37}Rb +

^{141}

_{Z}Cs + 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:__**(a)**

W = 1 and X = 0

Y = 2

Z = 55

**(b)**Mass – energy is conserved.

{Some of the mass is converted
to energy, such that energy E = Δmc

^{2}where Δm is the difference in mass. So, the mass is actually conserved, but it is now on the form of energy.}
Energy is released as gamma or
photons or kinetic energy of products or EM radiation.

__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 r

^{2}.
What is the relationship between
wave amplitude a and distance r?

A a

^{2}α 1/r B a α 1/r C a α 1/r^{2}D a α 1/r^{4}**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 r

^{2}.
So, a

^{2}is proportional to 1 / r^{2}.
Amplitude a is proportional to 1 /
r.