(Original post by mashmammad)
these are teachercol's answers.
Usual disclaimer. These are just my answers and are in no sense official.
They may contain erros and typos.
Q1 a) Power = rate of doing work or Work done per unit time (1)
b) i) P=VI so I = 40/230 = 0.17(4) A (2)
ii) R-V/I (or use P = I^2R) R = 230/0.174 = 1320 ohm (2)
c) R = rho L/A so L = 1320 x 3.0E-8/7.0E-5 = 0.567m (3) ecf
d) If same PD bigger power must have more current (P=VI)
so R must be smaller so A must be bigger so must be thicker (3)
e) i) Q=It = 0.174 x 8 x 3600 = 5011C (2) ecf
ii) 1) kW.h = ENERGY converted by a device of power 1kW operating
for a time of 1h (1)
2) Energy = 40/1000 x 8 = 0.32kWh
so cost = 0.32 x 22 = 7(.04)p (2)
Q2 a) i) 1) R is infinite below 1.4V (1)
2) When V=1.8v, I = 10mA so R=1.8/10E-3 = 180ohm (2)
ii) R gets smaller. Current is increasing faster than PD.
or just calculate R at different points. (3)
NB Ohm's law doesnt apply here- graph not a striaght line through the origin
R is NOT the gradient. R=V/I not dV/dI
b) Draw LED (right way round ) with ammeter/ voltmeter and resistor in right places.(4)
c) LED used for torch bulbs / car headlights.
More efficient becasue they produce less heat. Lower running costs. Longer lifetime. (2)
Q3 a) When temp rises, Rtherm falls, Total R falls, cureent increases. (3)
b) Either use potential divider or use Total R and work out current then PD.
Rtherm = 40ohm so ratio of R is 200:40 so PD must split in same ratio 5:1
so PD=5v (4)
c) i) Draw LDR symbol (1)
ii) When light intensity increases, R falls so ratio Rldr:R200 must drop
so PD across 200ohm resistor must increase.
or total R drops so current increases so PD across 200 increases. (2)
Q4 a) PD across 6ohm = IR = 0.30A x 6.0ohm = 1.2v
In parallel so same PD so I = V/R = 1.2/6.0 = 0.20A
b) i) Kirchoff's 1st law (1)
ii) I = I1 + I2 = 0.30 + 0.20 = 0.50A (1)
c) 1/Rp = 1/R1 + 1/R2 = 1/6 + 1/4 = 5/12
Rp = 12/5 = 2.4ohm
Total R = 2.4+5.6 = 8.0 ohm (3)
d) i) Emf = energy converted per unit charge
from one form of energy to electrical energy in a source (2)
ii) Terminal PD = PD across combination = 0.50 x 8.0 = 4.0v (1)
(or add up Pds)
iii) V = E - Ir
4.0 = 5.0 - 0.50r so r = 2.0 ohm (2)
Q5 a) e has charge / photon doesnt
e has mass / photon doesnt
photn travels at c / e cant (2)
b) i) Gain in energy = QV = 1.6E-19 v 5000 = 8.0E-16J (2)
ii) Energy = 1/2 m v^2
8.0E-16 = 1/2 x 9,1E-31 x v^2 so v = 4.2E7 ms-1 (3) ecf
c) i) Wavelength of probability wave associated with electron (1)
(not sure what they expect here. )
d) E = hf
8.0E-16 = 6.6E-34 x f so f = 1.21E18 Hz
lambda = c/f = 3.0E8/1.21E18 = 2.48E-10m (3)
e) i) Photoelectric effect (1)
ii) max KE = hf - phi
= 9.0E-19 - 7.2E-19 = 1.8E-19
iii) phi is min energy needed to release an electron from surface.
deeper electrons need more energy
so have less KE. (2)
Q6 a) i) Displacment = distance moved from equilibrium in a specified direction
Amplitude = max displacment (2)
ii) Frequency = no of complete oscillation per unit time
Phase difference = measure of how much out of step two
oscialltions are measured in fractions of a cycle
or as ana angle 0-360 or 0 - 2pi (2)
b) Diagram is essentially mirror imaged. Wave travels 0.25m every 0.5s (4)
Q7 a) i) Photon is emitted when atom moves from higher to lower energy level
Energy of photon = hf = hc/lambda= difference in energy of two levels.
so One wavelength corresponds to one transition between higher / lower energy level
three possible transitions (3-2, 3-1 and 2-1) so three different wavelengths (3)
ii) 1) f = v/lambda = 3.0E8/6.56E-7 = 4.57E14Hz
E = hf = 6.6E-34 x 4.57E14 = 3.0E-19J (2)
2) Photon energy Must be difference btween energy of levels
so must be 3-2 (1)
b) i) 1) lambda = d sin theta so d = 6.56E-7 / sin 11.4 = 3.32E-6
2) N = 1/d = 301306 lines per m
= 301 lines per mm (1)
ii) Blue has smaller lambda
sin theta = lambda / d d is same so theta is smaller (1)
Q8 a) same speed / travel through vacuum / E and B fields oscillate (2)
b) uwave = 1E-4 / UV = 1E-8 / gamma = 1E-12 / IR = 1E-6 (2)
c) i) Receiver reflects incident wave back at 90 but some is absorbed (freq stays same)
so reflected wave has lower amplitude and same freq than incident wave (2)
ii) Waves superpose. R detects resultant amplitude
At some points waves are in phase / reinforce and give maxima
At other points waves are 180 out / cancel and we get minima (3)
iii) max - min = 7.5mm
so lambda = 4 x 7.5 = 30 mm (1)
iV) Max amp = 0.8a + 0.6a = 1.4a
Min amp = 0.8a - 0.6a = 0.2a
I is prop to square of amp
so ratio = (1.4a/).2a) ^2 = 49 (3)
I thought this was quite an easy paper but the feedback I've had was mixed.
Some people really liked it, others hated it.
There wasnt much in the way of descriptive stuff, a lot of calcualtion.
It probably suits mathematically inclined students.
I still think its easy so grade boundaries may not be as low as they have been.
No - I cant tell you how many UMS you'll get. Dont ask.