Questions
- An oil drop of volume 0.6mm3 was placed on a clean water surface. It spreads to form a monoatomic circular patch of area 3000mm2. Use this data to calculate the diameter of a molecule of oil. (3 marks)
- State two advantages of alkaline accumulator over lead-acid cell (2 marks)
- The diagram below shows a system in equilibrium with the uniform rule supported at Q and resting horizontally.
The rule is 1m long and weighs 1.8N. calculate the weight of the block X (3 marks) - A boat sent ultrasound signal to the bottom of the sea and its echo received after 10 seconds. If the wavelength of the sound is 0.05m and the frequency of the transmitter is 50KHz, calculate the depth of the sea (3 marks)
-
- Differentiate between (2 marks)
- Distance and displacement
- Speed and velocity
- A body moves 3000 meters due east in 40 sec then 4000 meters due north in 60 sec. calculate
- Its average speed (3 marks)
- Average velocity (4 marks)
- Differentiate between (2 marks)
- A bomber flying horizontally at 100 m/s releases a bomb from a height of 3000 m. calculate the
- Time taken for the bomb to hit the ground (3 marks)
- Horizontal distance travelled before hitting the ground ` (2 marks)
-
- Define the term refraction of light (1 mark)
- A container full of water appears to be 8cm deep. If the speed of light in water is 2.25X108 m/s and in a vacuum is 3.0X108 m/s. calculate
- Refractive index of water (3 marks)
- The actual depth of the container (3 marks)
- Vertical displacement (2 marks)
-
- State the newtons second law of motion (1 mark)
- A bullet of mass 0.006 kg is fired from a gun of mass 0.5kg. if the muzzle velocity of the bullet is 300 m/s, calculate the recoil velocity of the gun (3 marks)
-
- The diagram below shows a u tube filled with two liquids X and Y. if the density of the liquid Y is 1.26g/cm3, determine the density of liquid X (3 marks)
- State two properties of a barometric liquid (2 marks)
- The diagram below shows a u tube filled with two liquids X and Y. if the density of the liquid Y is 1.26g/cm3, determine the density of liquid X (3 marks)
-
- Give FOUR properties of a thermometric liquid (4 marks)
- A faulty thermometer reads 100c when dipped into melting ice and 900c when on steam at normal atmospheric pressure. Determine the reading of this thermometer when dipped into a liquid at 200c (3 marks)
Marking Scheme
- Volume of the oil drop = volume of oil pattern
4/3πr3 = πR2h
0.6mm3 = 3000mm2 xh
h= 0.6
3000
= 0.0002mm
0.0000002M
2.0 x 107M -
- It can be stored in discharged state.
- Requires little maintenance/ attention
-
- Clockwise moment= anticlockwise moment
(1.8 x 0.4)=(1.8 x 0.1)+(x x 0.3)
0.72 = 0.18 + 0.3x
0.54 = 0.3x
x = 0.54
0.3
x= 1.8N
- Clockwise moment= anticlockwise moment
- T= 2D/V
V= fx
= 50000Hz x 0.05
=2500m/s
10= 2D/2500
25000 = 2D
2 2
D= 12 500M -
-
- Distance is the total length covered by a body a long a path while displacement is the distance moved by a body in a specific direction.
- Speed is distance covered per unit time while velocity is the change of displacement per unit time.
-
- Total distance= AB + BC
= 3000 + 4000 = 7000M
Total time taken = 40 + 60= 100s
Average speed = 7000
100
= 70m/s - Displacement = AC
AC= √((AB)2+(BC)2)
= √((3000)2 + (4000)2)
=√25 000 000
= 5000
Magnitude of velocity:
5000 = 50m/s
100
Direction of velocity
tan θ = 4000
3000
Tan θ = 1.333
θ = 53.13°
The bearing = 90-53.13 = 36.89
50m/s on a bearing of 036.87°
- Total distance= AB + BC
-
-
- S= 1/2gt2
3000 = 1/2 x 10 x t2
3000= 5t2
t2 = 600
t= 24.49sec - R= ut
= 100 x 24.49
2449M
- S= 1/2gt2
-
- This is the change of direction of light when it passes from the one optical medium to another.
-
- a∩w = Velocity of light in vacuum
Velocity of light in water
= 3.0 x 108
2.25 x 108
= 1.333 - ∩ = 4/3 = real depth
apparent depth
4/3 = real depth
8
32/8 = real depth
= 10.67cm
- Vertical = Real - apparent
= 10.67 - 8
= 2.67cm
- a∩w = Velocity of light in vacuum
-
- The rate of change of momentum of a body is directly proportional to the resultant force applied and takes place in the direction in which the force acts.
- Momentum before firing = momentum after firing
0 = M1V1 + M2V2
0=(0.006 x 300) + (0.5 x V2)
0= 1.8 x 0.5V2
1.8= 0.5V2
V2= 3.6m/s
- The rate of change of momentum of a body is directly proportional to the resultant force applied and takes place in the direction in which the force acts.
-
- hxρxg = hyρyg
0.2 x ρx x 10 = 0.16 x 1260 x 10
ρx = 0.16 x 1260 x 10
0.2 x 10
= 201.6
0.2
= 1008 kg/m3 -
- Opaque
- High density
- hxρxg = hyρyg
-
-
- Expand and contract uniformly
- Visible/ opaque
- Not wet glass
- Wide range of temperature.
-
- 10°C → 90°C
80 ... 10 → 0
90 → 100
80 → 100
? → 20
20 x 80
100
= 16°C
- 10°C → 90°C
-
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