What Are the Answers to the Physics Olympiad Past Papers?

[Vincit]

So my teacher entered me for the Physics Challenge for GCSE students(test from the makers of the Physics Olympiad, for gifted students), which is intended for students of 16 years of age; year 11. I'm 15, and I'm in year 10, so i don't know why my teacher entered me..

Anyway, I have to self-learn this before january, so I'm here to ask for help after going over the past papers and being completely befuddled..

2. A student performs an experiment to measure the half life of a
radioactive isotope. First they use a suitable detector and counter and
measure the average background radiation to be 120 counts per minute
(cpm). Next they measure the activity with the radioactive isotope in place
and the record a reading of 1080 cpm. Finally they repeat the experiment
12 hours later with the radioactive isotope still in place and record a count
rate of 240 cpm.
The half life of the sample is approximately:
A. 6 hours
B. 4 hours
C. 3 hours
D. 2 hours
E. cannot be determined from the given information

The answers is B.4 hours, and I don't know why. 240 doesn't go into 1080, and that is why I'm so confused.

All of the questions are found here:

http://www.physics.ox.ac.uk/olympiad/Downloads/PastPapers/BPhO_PC_2008_QP.pdf

I need help with ALL OF THEM!(Multiple choice questions) If someone could please explain each answer for me, then I would be so grateful.

The answers are :

1.A
2.B
3.D
4.B
5.C
6.E
7.A
8.A
9.B
10.D

I don't know how to get any of them though, except for question 1. I'm only 15 so don't ridicule me :/

 

[espen180]

For 2, you have to correct for the average background radiation.

 

[Vincit]

For 2, you have to correct for the average background radiation.

how would i do that?

 

[espen180]

By subtracting it from the measurement. Remember that you want to measure the counts that are due to your sample and want to filter out any other sources.

 

[Vincit]

By subtracting it from the measurement. Remember that you want to measure the counts that are due to your sample and want to filter out any other sources.

so 1080-120=960

960/2=480
/2=240

2 x halflife =12 hours

halflife= 12/2

halflife= 6 hours.. so how is it 4?
EDIT: forgot to subtract from 240!

960/2=480
/2=240
/2=120

so 3 x halflife= 12 hours

halflife= 12/3

halflife= 4 hours!

Thank you, now the rest!

 

[novop]

Subtract it from both measurements. You have 960 cpm initially and 120 cpm after 12 hours.

960 /2 = 480
480 /2 = 240
240 /2 = 120

That's 3 "halfings" in 12 hours, so the average halflife is 12 hours / 3 = 4 hours.

 

[Vincit]

By subtracting it from the measurement. Remember that you want to measure the counts that are due to your sample and want to filter out any other sources.

Subtract it from both measurements. You have 960 cpm initially and 120 cpm after 12 hours.

960 /2 = 480
480 /2 = 240
240 /2 = 120

That's 3 "halfings" in 12 hours, so the average halflife is 12 hours / 3 = 4 hours.

yup, I just worked it out in the previous post. Help me with the rest :)

 

[espen180]

Try it yourself first. Post your attempts at a solution and we'll help you on the way. That's how you learn. If you learn how to do a problem by doing it, you can do it on the exam. If we give you the answers, chances are you'll get stuck.

 

[Vincit]

Try it yourself first. Post your attempts at a solution and we'll help you on the way. That's how you learn. If you learn how to do a problem by doing it, you can do it on the exam. If we give you the answers, chances are you'll get stuck.

I've tried but cannot find any solutions at all.

 

[espen180]

Then you should consult a textbook, not a forum. Check your school library for physics books and read on Wikipedia.

 

[Vincit]

Then you should consult a textbook, not a forum. Check your school library for physics books and read on Wikipedia.

It's the summer holidays and I have no textbooks including any of this stuff! Wikipedia is too convoluted for me..

Why shouldn't I consult a forum? If I'm correct, this is a physics forums, here to offer help and answer questions. If that's not what it is for, then sorry.

 

[espen180]

You are right, but people here generally don't have the time to explain every one of those problems to you unless you do most of the work yourself.

If I were you, I'd consult my local university and use their library.