All solutions here are SUGGESTED. Casey will hold no liability for any errors. Comments are entirely personal opinions.

- B
- C
- B
- D
- D
- C
- D
- D
- B
- C
- A
- D
- D
- C
- C
- A
- D
- A
- B
- D
- Question 21 is a flawed question. When unpolarised light goes through a polarizer, the I is halved while the A is reduced by a factor of root 2. But based on the information Cambridge provides, the answer is C.
- D
- D
- C
- C
- B
- C
- A
- B
- B
- A
- C
- D
- B
- B
- B
- D
- C
- C
- C

Note to all: Casey will not respond to most of the comments as he is busy. You may contact him by **SMS** at +65 9474 5005 if you have a burning question.

Feel free to explain the answers, if you are confident. Many thanks.

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It will be good if the suggested answers is in a table form. Much neater and clearer.

Thanks for your suggestions.

Will the answers be uploaded?

Yup, Casey is working on it.

But expect slight delays as we are overseas already.

Ok thank you very much!

shldnt the answer for qn 21 be A as the the last polariser is perpendicular to the first so it does not matter what the 2nd polariser does to intensity

It would have been if there weren’t the middle one.

– Casey

Sir, mind reviewing question 28 , I think the answer is C as the question wrote resistance decrease when temperature decrease which is not the conventional thermistor. Thank you.

High temp and high brightness means the R is very small, so the voltmeter reading is small

– Casey

But that is only for ntc thermistors. The question specified that the resistance of thermistor decreases with temperature, which means as T decreases, R decreases. So R is min when T is low, ans should be C.

was this mcq considered hard?

My students range from 27 – 29, 80% more than 30.

I found it easy, except 2 or 3 qns that required further thinking. The rest is managable with careful reading

Hi! could i have H1 answers for phy:)would really appreciate it!!

Hi Sorry but we kinda don’t have the time to produce it!

For 28 the qn says resistance of thermistor decreases with temperature so shouldn’t it be C?

If resistance decrease with temperature, it is the normal thermistor isn’t it? So higher temperature, smaller resistance, potential difference will hence be small.

doesn’t resistance decrease with temperature mean when temperature drops, resistance drops correspondingly? It aint the normal thermistor then right?

hello moderator i did not realise that i left my name, will it be possible to delete that comment? thank you:)

High temp and high brightness means the R is very small, so the voltmeter reading is small

– Casey

Qn 21 is not flawed, answer is I/8, the amplitude is not halved after the first polarizer so the amplitude after 2nd polarizer is given correctly. However intensity is halved for each polarizer passed through.

Correction: A not reduced by a factor of root 2 for the first polarizer.

Hi, it is not possible to half the intensity without A/ root 2 the amplitude.

AS it will go against the laws of I proportional to A^2

So the ensure malus law is obeyed, we assume I doesnt have any change when first one is passed through as Amplitude no change.

Did your students say whether this paper was easier or harder? What do you think the average for this paper is?

What’s the national average for mcq every year? :/ i’m worried I can’t even get a C… i only score 27 for this mcq according to your ans

My students range from 27 – 29, 80% more than 30.

Why qn 5 is D not B

Why qn 38 is C not A?

can someone explain qn 38 thx

Ello, that is only true if the polarizer is before it. Meaning, if the 2nd and 3rd polarizer is 90 degrees, then you are right. It is 0. If in this case, because the first and third is seperated by the second, it is no longer 0 as it is only 45 degrees between the second and third.

OMGGGG what’s the usual cut-off mark for an A??

How many marks is required to score A