Feb Assesment y13

1.       (a)     appropriate shape; lines perpendicular to and touching plate and sphere; (2)
arrows towards negative sphere (1)                                                                  3

  (b)   (i)      By moments, e.g F cos 20 = W sin 20 / by triangle of forces /
by resolution of forces / other suitable method; i.e. justification needed (1)
F = 1.0 × 10–5 tan 20; = 1.0 × 10–5 × 0.364; (= 3.64 × 10–6 N) (2)
triangle of forces gives W/F = tan 70, etc (1)                                           3

  (ii)   E = F/Q; = 3.64 × 10–6 / 1.2 × 10–9 = 3.0 × 103;N C–1 / V m–1                3

  (c)   E = (1/4πεo)Q/r2; 3.0 × 103 = 9 × 109 × 1.2 × 10–9/r2; (2)
or use F = (1/4πεo)Q2/r2; r2 = 3.6 × 10–3 giving r = 6 × 10–2 (m) (1)                 3

  (d)   field line sketch minimum of 5 lines symmetrical about line joining
centres with arrows; (1)
Fig 1 sketch matches RHS of Fig 2/plate analogous to mirror/AW
relating to symmetry (1)                                                                                   2

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  2.     (a)     (i)      Q = VC; W = ½ VC.V ( = ½ CV2) (2)

(ii)     parabolic shape passing through origin (1)
plotted accurately as W = 1.1 V2 (1)                                                       4

  (b)   (i)      T = RC; = 6.8 × 103 × 2.2 = 1.5 × 104 s = 4.16 h                                     2

(ii)     ΔW = ½ C(V12 –V22) = 1.1(25 – 16) ; = 9.9 (J)                                       2

  (iii)  4 = 5 exp(–t/1.5 × 104) ; giving t = 1.5 × 104 × ln 1.25 = 3.3 × 103 (s)    2

(iv)    P = ΔW/Δt = 9.9/3.3 × 103 = 3.0 mW            ecf b(ii) and (iii)                 1
allow P = Vav2 /R = 4.52/6.8 × 103 = 2.98 mW

[11]

 

  3.     nature and features:
α-particle is 2p + 2n/ mass 4 u (1)
charge of +2e (1)
very short range/heavily ionizing/absorbed by paper (1)
spontaneous; and random nature of radioactive decay (2)
energetically more favourable to eject four particles together than a single
one/other comment about energy minimisation/mainly occurs from higher A
 nuclei/AW (1)
small mass decrease/loss provides kinetic energy of α-particle (1)
particle energy of a few MeV; particular decay is monoenergetic (2)
α-particle scattering:
suitable diagram and/or description to illustrate experiment up to 2 marks (2)
most particles have little if any deflection (1)
large deflection of very few shows nucleus is small; and very massive (2)
(Coulomb’s law enables closest approach to) estimate nuclear size
(in case of α-particle back scattering with conservation of energy argument)     max 7
Quality of Written Communication                                                                            2

[9]

 

4.       (a)     (i)      5.0 (V) (1)                                                                                               1

(ii)     10.0 (V) (1)                                                                                             1

  (b)   (i)      Q = CV;= 1.0 × 10–3 (C) (2)                                                                    2

(ii)     The total capacitance of each circuit is the same (namely 100 μF); (1)
because capacitors in series add as reciprocals/ in parallel add/
supply voltage is the same and Q = VC, etc. (1) max 2 marks                 2

  (c)    (i)      A1 will give the same reading as A2; because the two ammeters are (1)
connected in series /AW (1)                                                                    2
answer only in terms of exponential decrease for a maximum of 1 mark

  (ii)   A4 will show the same reading as A2 at all times; (1)
A3 will show half the reading of A2 initially;
and at all subsequent times (2)                                                                3

[11]

 

  5.     (a)     Positive as E-field is downwards/top plate is positive/like charges repel/AW (1)  1

  (b)   (i)      k.e. = QV; = 300 × 1.6 × 10–19 = (4.8 × 10–17 J) (2)                                2

  (ii)   1/2mv2 = 4.8 × 10–17; = 0.5 × 2.3 × 10–26 × v2 so v2 = 4.17 × 109;
(giving v = 6.46 × 104 m s–1) (2)                                                             2

  (c)   E = V/d; so d = V/E = 600/4 × 104 = 0.015 m (2)                                             2

  (d)   (i)      semicircle to right of hole (1) ecf(a); (a) and d(i) to be consistent           1

(ii)     mv2/r; = BQv; (2)
giving r = mv/BQ = 2.3 × 10–26 × 6.5 × 104/(0.17 × 1.6 × 10–19); (1)
r = 55 mm;so distance = 2r = 0.11 m (2)                                                 5

[13]

 

  6.     (i)      leptons;                                                                                                             1

(ii)     neutrino / muon / tau(on);                                                                                 1

[2]

 

 

Y12 Homework on Divided Circuits

Answers are below

Divided Circuits

Answers

1 3 ohm 1.4A
7 ohm 0.6A

2 2 ohm 2A
4 ohm 1A

3 4 ohm 0.5A
8 ohm 0.25A
total 1.75A

4 4 ohm 0.5A
5 ohm 0.4A
20 ohm 0.1A

5 A 4A
B 1A
C 1A
D 4A

6. 3.5V