Q. 1.0
1.1 As the copper tuns in the coil cut the horizontal magnetic lines each turn adds an emf. So the EMF will slowly increase until the movement stops.
1.2 Emf will be high to begin with and slowly decline as the turns come out of the magnetic field. The current will be in the opposite direction.
1.3 The effect will be as in 1.1 but the current will be in the other direction.
1.4 Similar to what happened in 1.2.
Q. 2.0
2.1 clock-wise.
2.2 Counter clock-wise.
Q.3.0
3.1 No current.
3.2 No current
3.3 B to A (Right hand rule)
Q . 4.0
4,1 N to S
4.2 B to A. (Use right hand rule)
4.3 V = B. l. v
V = 5x 0.25 x 5
= 6.25V.
4.4
4.5
Q. 5.0
5.1 Step down Transformer capable getting several voltages.
5.2 Vp /Vs = Np/Ns
230/ Vs = 1000/ 400
Vs= 230x400/1000
= 92V
5.3
230 V 1000
V 800
V = 230x800 / 1000
204
Q. 6.0
6.1 B ( use the left hand rule)
6.2 Current coming into the coil will change as the brushes will contact opposite ends. ( B will get connected to positive and so on)
5x2= 10 marks) Energy Cycle
Q.1.0
1 D, 2 B, 3 A 4. C, 5 A,
6 C, 7. B, 8 .A, 9. D. 10-A
2.0 Give the energy made use of and the main form of energy released.
Device / Action | Starting Energy | Final energy formed or released. |
Green plant exposed to sun. | Light/Radiant | Chemical |
Turning a bicycle dynamo . | Kinetic | Electric |
Flashing a torch light. | Chemical | Electric- radiant/light |
Wood burner. | Chemical | Thermal- radiant/light. |
Stopping a moving car. | Kinetic | Thermal - vibrational/sound. |
2x10 = 20 marks.
Q.3.0
3-1 500 x10 x60 =300,000 J
3-2
Bread…...990x2 ……= 1980 J
Egg ……… = 600 J
Butter…. 5 x31.2… = 156 J
Sugar…...3x16……. = 48 J
Total content……..= 2784 J 25% of this ….= 25 x 2784 ÷ 100
= 696 J.
3-3 Kinetic energy = ½ x mv2
= 1200 x100 x100 J
= 12000000 J…..12,000.kJ.
3.4 Capacity ……………………...= 3x 1000 kW.
No. of houses to serve = 3000x 1000 ÷ 500 …= 6000.Houses
Q.4.0.
4-1 Both bulbs use equal amounts.
4-2 Fluorescent bulb.
F
Forces - Extensions.
1.0 1.1 For 5 N the extension is 0.03m For 10 N the extension is =0.03 x2 = 0.06 Therefore the new length is = 0.10 + 0.06 = 0.16 m 1.2 Spring constant = Force / extension
= 5 / .03 = 166.67 nm-1
2.0 2.1 11 cm. 2.2 12 cm. 2.3 The weight of the lower spring gets added to the load.
3.0 3.1 Downward force = + mg Tension = - kx At equilibrium they are balanced mg = -kx m x 9.8 = - 6 x 0.1 m = - 0.6 / 9.8 = 6.1 x 10-2 kg. 3.2 P.E = 0.5 x kx2 = 0.5 x 6 x 0.1 x0.1 = 0.03 J
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4.0 4.1 This is the length at zero force . = 10 cm. 4.2 This is where the graph changes = 20 N 4.3 K = Force / Extension = 1. 
5.0 F = kx k = F / x = 2450 / 0.08 =
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Forces - Friction
1.1 A –. Fluid friction,
B – Unwanted C - Streamlined to reduce
1.2 A - Fluid friction,
B – Unwanted C - Streamlined to reduce.
1.3 A – Sliding
friction, B – Unwanted, C = Lubricated to reduce.
1.4 – Rolling friction,
B – unwanted C - Axle is greased to reduce.
OR A- Sliding
friction B – Un wanted C Reduced by wheels.
1.5 – Fluid friction.
B – Wanted. C – Large area of
parachute.
2.0
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Material or Method.
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How or where used?
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Effect.
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2.1. Using rubber
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On sliding surfaces.
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Increases friction
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2. 2 Using ball bearings.
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Around axles of rotating wheels.
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Reduces rubbing friction.
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2.3 Treads.
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Shoe soles, tyres.
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Increases depending on material and depth of tread.
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2.4 Magnetic force.
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Trains lifted over rails.
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Reduces sliding friction.
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2.5 Ice
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Skating.
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Ice melts due to pressure reducing frictiongreatly.
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3.0
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3.1 R = W cos 35˚.
= 250 x 0.82
=204.8 N.
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3..2 F = W sin 35˚.
= 250 x 0.54
=134
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3.3 Fr = F
= 134 N
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3.4 Coefficient of Fr = F/ R
=
134 / 204.8 = 0.65.
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4.0 Answers.
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4.1
240 up.
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4.2
120 West.
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4.3
C = Fr/R
= 120 / 240
= 0.5
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4.4
New reaction force 240 + 200 =
440 N
Using the formula
C = Fr/ R
We get
0.5 = Fr / 440
Fr =
440 x 0 .5
= 220 N
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G
H
Heat energy
Q.1.0 Answers
1.1 Temparature.
1.2 Conduction, 1.3 Joule
, 1.4 Latent heat,
1.5 Specific heat.
Q.2.0 Answers
2.1 Heat = m.c.t, That is
= 25 x 4.18 x 80.
=
8360. J
2.2. Heat
gained by Al = 150 x0.4 x 80.
= 4800 J
2.3. Yes. When heat is produced outside vessel can
waste a lot by convection and radiation
2.4.
Silver. It reduces neat loss by
radiation.
Q.3.0 Answer.
Let C be the specific heat of the object.
Heat gained by water
= 40x 4.18 x 8 J
Heat lost by object = 20 x c x (100- 28).
That is 40x4.18x8
= 20x xc x72
C= 1337
/ 1440 =0 ..93 J/g per ˚C
Q. 4.0 Answer
Let L be the latent heat of melting.
Then heat gained by ice = 41 x L.
Electric energy given = Vit.
41 x L = 1.5 x 0.03 x300.
L
=
(1.5 x0.03 x300)/ 42
L = 0.33J
Q. 5.0
1. Protects the glass bottle. Prevents heat loss to some extent.
2. Prevents convection and evaporation.
3. Contains the liquid. Double walled flask holds the vacuum and offers a surface foe silver coating.
4. Prevents radiation.
5. Prevents mainly conduction.
I
J
K
Kirchhoff's Law
Q. 1.0 1.1.1 V= ir, V = 1.5 x 4 = 6 V 1. 1. According to Kirchhof first law Incoming current and leaving current should add up to zero.. Lets name Current from A to B as X . Then X + 1-3 = 0 X = 2 A. 2 2. Let V be the voltage of the bulb V – 6 +6 – 12 = 0 V = 12 . 2. 2. Finding V1. V = ir. V = 2 x 6 = 12 V. 3. 3 Finding V2. V = ir. V = 1 x 6 = 6 V. 4 Finding V3. Use Kirchhoff’s voltage Law. Going clockwise in outer loop. + 6 - V 2 + V1 + V3 = 05 + 6 – 6 + 12 + v3 = 0 V3 = - 12 V. 5 Finding V 4 Going clockwise in outer loop + 6 – V4 – V3 = 0 +6 –V4 – 12 = 0 V4 = 6 V. . Q 3.0. Loop 1 1.Find A1. V= ir 12= ix 6 I = 12/6 = 2 A . 3.0. Loop 1 1. Find A1. V= ir 12= ix 6 I = 12/6 =2 At Loop 2 2. Find A 2. V=ir 6= ix6 I = 1 A
| Find A3. Incoming current = leaving current. 2 + 1 = 3 A. 4. Find V1 V1 = 2x 6 = 12 V. 5 . Find V2. V2 = 1x6 = 6 V. Outer loop 6. Find V3. Using Kirchhoff law:- 6 + V-12 = 0 V = 6 V Q. 4.0 W 1. I/r1 + !/r2 = 1?R 1/3 +1/3 =1/ R 1/R = 2/3 R = 3/2 =1.5 Ω . 2. r1 + r2 =R R = 3 + 1.5 Ω. R = 4.5 Ω. 3. Total voltage in the loop = 6 v + 6 v = 12 v As V = ir 12 = i x 4.5 I = 12/4.5 = 2.67 A 4. Current in ‘b’ has to be the same as in ‘a’. 5. C V =ir V =2.7 x 3 V=8,1 V 6. d. As the current gets split into 2 parallel resistors current in d is half of 2.67 = 1.3 A 7 e. Using Kirchhoff law 8.1 – 6 +e -6 = 0 e = 12- 8.1 = 3.9 V
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Q.1.0 1.1 ….At 45°