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 × 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 + WX n - - - > 9337Rb +
141ZCs + YWX 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 toA 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.
Here are the remaining questions for now:
ReplyDelete21/O/N/10 Q.3(c)(ii)
22/O/N/10 Q.4(b)(ii)
21/O/N/11 Q.5(b)
23/O/N/11 Q.6(c)
23/M/J/12 Q.2(c)
21/O/N/12 Q.6(c)
22/O/N/12 Q.3(b)
23/O/N/12 Q.1(e)
For 21/O/N/10 Q.3(c)(ii), see question 16 at
Deletehttp://physics-ref.blogspot.com/2014/10/physics-9702-doubts-help-page-3.html
Hi,
ReplyDeleteThank you so much, your site is extremely valuable but I have a doubt in Qs 641 how do you know that the products/reactants have energy in a nuclear reaction?
This is quite obvious. As soon as the products are formed, they are not likely to be stationary. They would be moving, and thus have kinetic energy.
DeleteFrom the other chapter, energy of a photon / gamma = hf
+ From Einstein's equation, the mass of an element in a nuclear reaction has an equivalent energy given by E = mc^2
I hope that I answered the question
What about 2=y? Why is it 2?
Deletethe explanation has been updated.
Delete