Thursday, April 30, 2015

Physics 9702 Doubts | Help Page 127

  • 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:
(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 = Δmc2 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 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.




4 comments:

  1. Here are the remaining questions for now:

    21/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)

    ReplyDelete
    Replies
    1. For 21/O/N/10 Q.3(c)(ii), see question 16 at
      http://physics-ref.blogspot.com/2014/10/physics-9702-doubts-help-page-3.html

      Delete
  2. Hi,
    Thank 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?

    ReplyDelete
    Replies
    1. 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.

      From 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

      Delete

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