1. a) At A,B and C the path difference (PD) = n~, where ~ is wavelength.
Hence PD (A) = o~, PD(B) = 1~ and PD(C) = 2~. S2C - S1C = PD(C)=2~.
S1S2 is a straight line where the waves are opposite in direction. Hence stationary waves are formed. Hence between maxima is 0.5~. The distance has 6(o.5~) = 3~.The larger possible answer will be 8(0.5~) which did not appear yet hence less than 4~ and greater than 3~.
ii) the Young double slit formula x = ~D/a. Subs AB = ~(8~)/4~ = 2 ~. Only an approx as D must be much larger than a for the formula to hold.
b) dsin(angle) = 1~. Ans: violet = 4.5 x 10-7m. and red = 7.0 x 10-7 m.
c) y is horizontal amplitude of air molecules as the sound waves are longitudinal waves.
2a) E dir up as force and E dir are opp for negative charges.
b) a = - F/m
Use v = u + at , v = 6.62 x 10 6ms-1
ii) s = ut + 0.5 at 2. s = 0.0177 m.
iii) increase in PE= decrease in KE (or W = Fd) = 5.66 x 10-18 J
Graph horizontal line (value E= 2000 NC-1) and potential ( 160V) straight line with negative gradient. The distance marking should be 0.08 m (no mark given)
3.c) 86.4C, 173 J, 1.94V , 8.08ohms, R= 9.26 ohms.
ii) As current increases, temp increase hence resistance increase.
4. a) 0.33A.b) Vxy =k(Im) and 1.5 V = kL. Hence L=0.9m .
c) Vxy =k(Im) and V = k(0.75m) , hence V= 1.25 V.
current = 1.25/10 = 0.125A. 1.5 - 1.25= 0.125r. Hence 0.25/0.125=r. Solve for r = 2 ohm.
d) Balance length increase if Vxy decrease, hence add resistor in series with driver cell.
5. b) The coil experiences increasing flux, to oppose the increase the induced e.m.f produces B out of page . Hence using right hand grip rule the current flows anticw. Hence current flow from Y to X.
ii) Induced E = BLv, Induced I = BLv/R. Force = BIL= B(BLv)L/R . Power = Fv = B2L2v2/R
iii) Since the coil experiences no change in flux, there will be no induced current and hence no work is required by the machine.
6. Peak power = 100 MW
Mean power = 50 MW
ii) peak voltage = 60kV
Peak current = 1700A
b)square of sine graph with peak value of 100 MW
