(Original post by **Cosmologist**)
It's 8am in the UK. My answers for the paper:

1. a) Why micrometer?

-The precision (0.001mm) of the micrometer allows to obtain values of the diameter to 3 d.p.

- The scale division is 0.001mm

- It gives small %U

(1 mark)

b) What technique would you use to make you results as accurate as possible?

- Measure the diameter of the wire at multiple orientations and then average the result

- Check for zero error on the micrometer

(1 mark)

c) Find the value for the resistivity.

- I really forgot the answer, but it was smth*10^-7

(3 marks)

d) Percentage uncertainty in resistivity.

- I found it 3.5% (calculated value was 3.45%, so I rounded it up to 3.5%, but I think 3.4% is ok)

(2 marks)

2. a)How would you use a metre rule to make your readings of h as reliable as possible?

- Use set square on the floor to make sure the metre rule is vertically placed

- Align your eyes with a scale to avoid parallax, whilst measuring the height h

- Mark the centre of the ball to measure the height precisely to the centre of the ball

- Repeat taking measurements of h to obtain many values (reduce random error) and then average the result

(3 marks)

b) Describe a technique to measure the period of oscillations T as accurate as possible?

- Record a time for 10-15 complete oscillations (using stopwatch 0.01s) and divide by the number of oscillations to obtain a value for the period of oscillations T. This method would reduce %U.

- Use timing marker in the equilibrium position to see full oscillations (easier to count oscillations)

- Repeat recording the time for 15 oscillations and average to minimise random error (use lap-timer for successive recording of time for 15 oscillations)

(3 marks)

c) Find the gradient of the graph?

- It was (-4pi^2/g)

(1 mark)

d) Show that the ratio intercept/ gradient is equal to H.

- It's weird, since it turns out to be (-H), but I took the modulus of the gradient and obtained H.

(2 marks)

3.a) Find the value of k.

I filled out the table, as following:

k /10^-18Vm^2

1.26

1.14

and then averaged to get 1.20*10^-18 to 3 s.f.

(3 marks)

b) The percentage uncertainty in k

- I came up with 2 ways to do it, but finally realised that %U in k is best found just taking the half-range divided by mean (as we had only 1 mark for that), so (1.26-1.14)/ 1.20 * 100% = 10%

(1 mark)

c) Find the h from h^2=2*m*e*k.

- I computed it to be 5.91*10^-34 to 3 s.f.

(2 marks)

d) Hence, write down the %U in h.

- It's just %U in k halved, so 5%

(1 mark)

e) Comment on the validity of your experimental value of h.

%D = (6.63-5.91)/6.63 *100%=10.8% ~11%

Since %D (between theoretical value for h and our 5.91*10^-34) is significantly greater than %U in experimental value, then h is likely to be invalid.

(2 marks)

4. a)Thermistor is a semiconductor, so as the temperature increases, more electrons get released and the resistance reduces.

(2 marks)

b) Plan the experiment to investigate how the resistance of the thermistor varies with temperature. (i) Describe the apparatus to be used, you may draw a diagram, if you wish. (ii) How would you vary the temperature in the range 0â„ƒ and 100â„ƒ? (iii) Some words about tech precautions to improve the accuracy of results.

- Well, I wont draw it here, but there should be the water bath, Bunsen burner, thermistor and ohmmeter/ multimeter connected by wire, thermometer (both thermometer and the circuit are fixed to the wall, or smth)

- Put ice in the icy water to approach 0â„ƒ and boil the water to reach 100â„ƒ.

- Remove the heat source, while taking the readings.

- Place thermometer and thermistor close together and away from walls and bottom of the container.

- Allow some time for the apparatus to come to thermal equilibrium

- Make sure the thermistor is fully immersed in the water.

(5 marks)

c) Prove that lnR against Éµ would produce the straight graph.

- Just compare to y= mx+c and state that (-a) is the gradient and lnR0 is the intercept

(1 mark)

d) Plot this weird graph.

- This data table for the graph is quite unreasonable as ln0.906<0 and we're provided with positive sector of the axis. I multiplied values in R-column by 10 to get 9.06, and changed R/kâ„¦ to R/10^2â„¦, then plotted a graph with a good enough range and scale. I am sure, many guys just sketched it as it was given with negative sign, but I struggled with that for too long. Doubtful task...

(4 marks)

e) Use your graph to calculate the value for a.

Gradient= -a=-0.03884...So a=3.88*10^-2 (~3.9*10^-2 or 3.8*10^-2)

(3 marks)

Please, do share your answers

)