- A Burette has an initial reading of 22.5cm3. Determine the final reading after a liquid of volume 11.3 cm3 is removed from the burette. (2mks)
2. In a ball and ring experiment, the ball goes through the rings at room temperature. When it is heated it does not go through the ring, but when left on the ring for some time, it goes through. Explain this observation. (2mks) - Distinguish between speed and velocity. (2mks)
- State how the pressure in a moving fluid varies with sped of the fluid. (1mk)
- Fig 1 Shows a section of partially graduated micrometer screw gauge.
Insert divisions on the sleeve to represent a reading 5.70 mm (2mks)
- The fig 2 below shows a beaker containing water placed on a flat bench.
State and explain the changed in stability of the beaker when the water freezes to ice. (2mks) - Two table tennis balls hang at the same level suspended from a thread a short distance apart. A steam of Air is blown between the balls. Explain what happens to the balls. (2mks)
- A uniform half meter rule is supported by force of 3N and 2N as shown below.
Determine the weight of the metre rule. (3mks) - A stone and a feather are dropped from a building 20m tall if they reach the ground at the same time. Find;
- The velocity with which they reach the ground (take g=10m/s2) (2mks)
- The condition under which they fall. (1mk)
- The forces act on a trolley as shown below.
Find the acceleration of the trolley (2mks) - On the axes below, sketch the graph of density of water against temperature (2mks)
- Water flows in a pipe of cross-sectional area of 10cm2 of a velocity of 0.4m/s. If the pipe has a jet at the end with area of cross section 2.5cm2. Find the velocity of water jetting out. (2mks)
SECTION B: 55 MARKS
- The following results were found by a student in a bid to determine the relationship between the extensions of a spring to the load hooked on its end
Mass (kg)
0
0.02
0.04
0.06
0.08
0.10
Reading (mm)
Force (iv)
Extension (m)
-
- Complete the table above for the values of force F (N) and Extension e(cm) (2mks)
- On the grid provided, plot a graph of force against extension (5mks)
- From the graph determine the spring constant (2mks)
- What would be the extension when a mass of 0.045 kg is hanged from two such springs in series? (2mks)
-
- The diagram below shows a glass prism and incident ray striking the surface XY
-
- Indicate on the diagram the path of the emergent ray. (3mks)
- Calculate the refractive index of the glass prism given that the critical angle of glass is 420. (3mks)
- State ONE condition necessary for total internal reflection to occur. (1mk)
- State ONE application of total internal reflection (1mk)
-
- The figure below shows a velocity the graph for the motion of a body of mass 5kg.
- Use the graph to determine the
- Displacement of the body after 12 seconds (3mks)
- Acceleration after point C (3mks)
- Force acting on a body in part a (ii) (3mks)
- Sketch a displacement time graph for the motion from point B to D (2mks)
- Use the graph to determine the
-
- State Newton second law of motion (1mk)
- A body of mass 100kg is accelerated from a velocity of 2m/s to 5 m/s in 6 seconds. Calculate
- Its initial momentum (2mks)
- Its final momentum. (2mks)
- The resultant force acting on the body. (2mks)
- A bullet of mass 0.01kg travelling horizontally at 100m/s. penetrates a fixed block of wood and comes to rest in 0.02 second. Calculate.
- The distance of penetration of the bullet into the wood. (3mks)
- The average retarding force exerted by the wood on the bullet. (2mks)
- A student stands at a distance 400 from a wall and claps two pieces of wood. After the first clap the student claps whenever an echo is heard from the wall. Another student starts a stopwatch at the first clap and stops it after the twentieth clap. The stopwatch records a time of 50 seconds. Find the speed of sound. (3mks)
- The figure below show a displacement time graph for a wave motion.
What is the frequency of the wave? (2mks) - Give a reason why water is not a suitable liquid for use in a barometer. (1mk)
- An oil drop of radius 0.42 mm when placed in water spreads out to form a circular patch of radius 42 cm using this information.
- Calculate
- Volume of oil drop (2mks)
- Area of oil patch. (2mks)
- Diameter of oil molecule (2mks)
- State one assumption made in this experiment (1mk)
- Calculate
MARKING SCHEME
- Initial reading = 22.5 cm3
Final reading = 11.3 cm3
33.8 cm3 - – Ball expands when heated and goes through the ring
- When left on the ring for some time, it heats the ring causing the ring to expand and ball passes through. - - Speed is the distance covered per unit time
- Velocity is the sped in the specified direction. - Pressure is inversely proportional to the speed
-
- Stability reduces. (1mk)
- They move closer to each other. The blown air increases the speed of the air between them. This reduces the pressure between and external pressure (atmospheric) pushes them closer.
- Sum of clockwise moments= sum of anticlockwise moments
35 x3=2x45 +wx20
105 = 90 +20w
w = 0.75 N -
- V2 = 2gs
V2 = 2 x 10 x 20
v2 = 400
v = 20m/s
- V2 = 2gs
- F = ma
2 = 2 x a
a = 1m/s2 -
- A1V1 = A2V2
10 x 0.4 = 2.5 x V2
V2 = 1.6 m/s
SECTION B: (55 MARKS)
-
-
-
Mass (kg)
0
0.02
0.04
0.06
0.08
0.10
Reading (mm)
120
131
139
149
161
171
Force (N)
0
0.2
0.4
0.6
0.8
1.0
Extension (m)
0.0
0.11
0.19
0.29
0.41
0.51
- Axis = 1 mk (correctly labeled)
Scale – simple and uniform – 1mk
Correct plotting – ½ mk each to maximum 2mks
Line – straight line, passing through 4 correctly plotted prints – 1mk - Spring constant = slope of graph
Correct interval – 1mk
Correct evaluation – 1mk
Correct unit 1mk
-
- e = 2 x F/K
-
-
-
-
- n =1/sin c
n = 1.494
-
- Ray passes from optically denser medium to rarer medium
-Critical angle is exceeded - Prism periscope
- Optic fibre
-
-
-
- Displacement = area under the curve
=1/2 × 20 × 8 + 20 × 4
= 160 m - a = (v - m)/4
a = (0 - 20)/4
a = -5m/s2 - F = ma
= 25N
- Displacement = area under the curve
-
-
- The rate of change of momentum of a body is directly proportional to applied force and takes place in the direction of the force
-
- Initial momentum = 100 x2
= 200 kgm/s - Final momentum = 100 x 5
= 500 kgm/s - F = m
=300/6 = 50N
- Initial momentum = 100 x2
-
- v = u + at
a = - 5000 m/s2
v2 = u2 + 2as
s= 1m - F = ma
= -50N
- v = u + at
- Speed = 2d/t
= 320 m/s - f = 2.5/0.02
= 125 Hz - Due to its low density, the tube needed is very long (10m long tube)
-
-
- V = 4/3πr3
= 4/3 × 3.142 × (0.21)3
= 0.31038mm3 - A = πr2
= 3.142 x (420)2
= 554248.8 mm2 - V = Ah
0.31038 = 554248.8 x h
h = 5.6 x 10-7 mm
- V = 4/3πr3
-
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