## Thursday, 28 November 2013

### SPM 2013 Physics Paper 1 Answers

2013 SPM Physics Paper 1 Answers

Here are my answers - hopefully they are the same as the examiners':

 2 A 4 B 7 C 8 B 9 D 11 A 12 D 14 A 17 C 19 A 21 A 22 C 23 A 29 D 30 D 35 B 36 B 37 A 38 C 39 C 40 A 41 D 42 B 43 D 44 A 45 A 46 D 47 B 49 C 50 B

(My answers for SPM 2013 Chemistry Paper 1 can be viewed by clicking here.

My answers for SPM 2013 Modern Maths Paper 1 (1449/1) are here.)

My Comments on Some Physics P1 Questions:

Q 30: The answer is D. More than twice before, 'telescope at normal adjustment' was tested in SPM:
* In 2005 P2 Q11(a)(iii) - diagram @ pg. 387 of Past-Yr Papers 2005-2012 - D;
* In 2006 P2 Q9(c)(i) - diagram @ pg. 393 of Past-Yr Papers 2005-2012 - D; and,

Q32: The answer is D. Reason:

"As water waves enter shallower region (at headland first, then only the bay), their speed slows down with the shortening of wavelength (again at headland first, then only the bay) without any change in wave frequency. This refraction of water waves results in the wavefronts  taking the shape of the shoreline with wavelengths (distance between successive wavefronts) at the headland shorter than their corresponding wavelengths at the bay as shown by the diagram. At the bay, wave energy is spread-out over longer wavefronts resulting in calmer sea due to dispersal of wave energy. Answer: D

For many other questions, I have inserted some hotlinks which you can click on to direct you to the relevant pages of this blog where the answers can be found/deduced. For examples, if you click on the hotlink of:

* Q 33 D, you will be directed to the page on "Diffraction" where its Para 6 has the same diagram as in the answer D (but in different orientation);

* Q 32 D, you will be directed to the page on "Refraction of Waves" where its Para 7 shows similar diagram and the answer should be D;

* Q 31 C, you will be directed to the page on "Diffraction" where its Para 2 shows C should be the answer;

* Q 28 A, you will be directed to the page on "Refraction" where its Para 6, pt 1 shows you the answer;

* Q 27 A, you will be directed to the page on "Light" where its Para 2(d)(v)(2) shows you the answer;

* Q 26 A, you will be directed to the page on "Refraction" where its Para 6, pt 1 shows you the answer;

* Q 25 C, you will be directed to the page on "Refraction" where its 2nd Para 6 shows you the answer;
and, so on......

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## Tuesday, 26 November 2013

### Interesting SPM 2013 Question on Half-Life

1) SPM 2013 Physics Paper 1 has an interesting question on half-life calculation.

"25% of Gallium-65 atoms have decayed after 15 minutes. What is the half-life of Gallium-65?

A 30.0 minutes

B 23.4 minutes

C 15.0 minutes

D 7.5 minutes

(Shouldn't there be another choice, say:

E 36+ minutes?)

My Comments (based on info given in the Q48):

Half-life of Gallium-65 based on info given in the Q48:
25% have decayed means 75% have not decayed. And that means, its half-life > 15 minutes. So was the half-life A, B or none of the 4 given choices? Do you not agree that the half life should be calculated as follows:

The 1/4-life Method:

At the end of 15 minutes (1st 1/4-life):
a) amount decayed = 25% = ¼ of original amount
b) amount undecayed = 75% = ¾ of original amount

At the end of 30 minutes (2nd 1/4-life):
a) amount decayed   ¼ x ¾ of original amount = 3/16 of original amount
b) total amount decayed = (¼ + 3/16) of original amount = 7/16 of the original amount
c) total amount undecayed = (1 – 7/16) of original amount = 9/16 of the original amount

To decay to ½ (or 8/16) of original amount:
a)      A further 1/16 is to be decayed
b)      1/16 = [(1/4) x (9/16)] x (4/9)
c)      Time needed = [15 minutes] x (4/9) =  6.67 minutes

Therefore, half-life = (15 + 15 + 6.67) minutes
= 36.67minutes

Answer (30 minutes) cannot be the acceptable answer because in 30 minutes only 7/16 and not 1/2 (or 8/16) of the radioactive substance has decayed. Conceptually, A is a wrong answer. Q48, to me, has no acceptable answers. This is also confirmed by logarithm method of calculating half-life as follows:

Logarithm Method to Calculate Number of Half-Lives:

Let:      n = number of half-lives
Ao = Original amount of radioactive substance (or original level of radioactivity)
Ac = Current amount of radioactive substance (or current level of radioactivity)

Then,    (1/2)n x Ao = Ac
(1/2)n = Ac/Ao
(1/2)n = 0.75Ao/Ao = 0.75

log both sides:      log (1/2)n = log (0.75)
Hence,       n  = log (0.75) ÷ log (1/2) = 0.415037499 half-life

To Calculate the Half-Life, T1/2

Let       T1/2 = Half-life time
To = Time as at original amount of radioactive substance
Tc = Time as at current amount of radioactive substance
n = number of half-lives

Then,    T1/2 = (Tc - To) ÷ n = 15 minutes ÷ n = 15 min ÷ [log (0.75) ÷ log (1/2)]

T1/2 = 36.14131259

T1/2 = 36.1 minutes

Based on facts given in Q48, logarithm method shows a half-life 36.1 minutes.

But online searches (please click here or here) show that isotopes Gallium-65  actually have a half-life of 15.2 minutes

So, which is the examiners' preferred answer and why?

Isn't this question both interesting and bewildering?! :)

In conclusion: Overall, except for Q48, Paper 1 is still a good paper - of respectable standard! :)

(10.4.2014: The examiners should have replaced the words "Gallium-65" in Q 48 with the words "isotopes-X". Otherwise, it's like a biology question on man that talks about the man having ovaries, uterus, etc. that makes the candidates wonder whether it is a question about man or something else. The more well-read the students, the more disadvantaged he/she would be - that is ridiculous!)

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