# PHYSICS Paper 1 Questions and Answers - KCSE 2022 Past Papers

SECTION A (25 marks)

Answer all the questions in this section in the spaces provided.

1. State what is meant by "Area". (1 mark)
2. State the SI unit of area. (1 mark)
1. Explain why water in a glass tube forms a concave meniscus. (2 marks)
2. Figure 1 shows how water is drawn from a large tank into a low lying container using a rubber tube.

Explain how the process takes place. (2 marks)
3. State how a piece of paper can be used to demonstrate that matter is made of tiny particles. (1 mark)
4. Figure 2 shows Six's maximum and minimum thermometer.

Explain how increase in temperature causes index A to move upwards. (2 marks)
5. State the difference between heat and temperature. (2 marks)
6. State two factors that affect the stability of a cylindrical container. (2 marks)
7. Figure 3 shows a set up in which a spring with a pointer is attached to a wooden strip that has a hanging hook. A graph paper is fixed along the strip to be used to calibrate the spring.

A mass of 100 g is provided. Explain how the spring balance can be calibrated. (3 marks)
8. Water enters a pipe at a velocity V, at a point where the cross-sectional area is A,. It leaves the pipe at a velocity V2 at a point where the cross sectional area is A2. Show that A, V1 = A2V2. (3 marks)
9. Sketch the displacement-time graph for a body moving with decreasing velocity. (1 mark)
10. Figure 4 shows a graph of force against time when a tennis ball is hit.

Determine the mass of the tennis ball whose velocity is 60 ms-1. (Assume the ball is stationary before it is hit). (3 marks)
11. State the energy transformations that take place as a pendulum bob swings. (1 mark)
12. When determining the specific latent heat of fusion of ice by electrical method, other than mass, voltage and current, state one other measurement that should be taken. (1 mark)

SECTION B (55 marks)

Answer all the questions in this section in the spaces provided.

1.
1. State Boyle's law. (1 mark)
2. Figure 5 (a) shows a column of air of length 6cm trapped by a mercury thread in a tube. Figure 5 (b) shows the same tube in a horizontal position.
1. Draw the mercury thread in Figure 5 (b). (2 marks)
2. Explain why the thread appears as in 14(b)(i). (2 marks)
3.
1. State what is meant by "absolute zero temperature". (1 mark)
2. A balloon contains hydrogen gas at a temperature of 2 °C and a pressure of 6 mmHg. Determine the pressure in the balloon when the temperature is raised to 80°C. (3 marks)
2.
1. State two ways in which the centripetal force acting on a body of mass M can be reduced. (2 marks)
2. A stone of mass 0.5 kg tied to a string is whirled in a vertical plane along a circular path of radius 2m and that its frequency is 2 cycles per second. (π-3.142)
1. Determine the:
1. velocity of the stone  (3 marks)
2. tension in the string when the stone is at the top most part of the circular path (3 marks)
2. State with a reason how the tension in the string changes as the stone gets to the bottom of the circular path. (2 marks)
3.
1. Figure 6 shows a cube of mass 2 kg and sides 5 cm fully immersed in a liquid of density 0.8 gem3. The cube is balanced by a stone of mass M

Given that the gravitational field strength, g, is 10Nm 2, determine the:
1. upthrust acting on the cube (3 marks)
2. apparent weight of the cube (3 marks)
3. weight of the stone (3 marks)
2. A block of mass 500 g floats in water. Determine the volume of the block under the water. (density of water is 1 gem). (3 marks)
4.
1. State two factors that affect the boiling point of a substance. (2 marks)
2. A well lagged calorimeter contains a liquid of mass 200g at a temperature of 10°C. An electric heater rated 80 W is used to heat the liquid. Figure 7 shows a graph of temperature against time for the liquid.

From the graph, determine the:
1. boiling point of the liquid (1 mark)
2. quantity of heat given out by the heater between time t = 1 minute and time 1=4.5 minutes (3 marks)
3. Based on (b)(ii), determine the:
1. temperature change between the time t- 1 minute and time t-4.5 minutes (1 mark)
2. specific heat capacity of the liquid (3 marks)
4. 2g of vapour was collected from the liquid between times t=5.4 minutes and t=6.3 minutes. Determine the specific latent heat of vaporisation of the liquid. (3 marks)
5.
1. A weighing balance placed on the floor of a lift is used to measure the weight of a body of mass 80 kg. Determine the reading on the balance when the lift moves upwards: (acceleration due to gravity g is 10ms 2)
1. with uniform velocity  (3marks)
2. with an acceleration of 3 ms-2 (3 marks)
2. Explain why a person standing on a boat is likely to fall into the water when attempting to jump to the shore. (3 marks)
3. A box is moved 30m along a surface whose frictional force is 1000 N with uniform velocity. Determine the work done against friction. (2 marks)

### MARKING SCHEME

SECTION A (25 MARKS)

 1 (a) Area is a measure of the extent of a surface.(b) m2 (metre squared) 1 1 2 The adhesive forces between water molecules and glass are higher √ than cohesive forces √ in water molecules. 2 3 Remove air from the tube to create a partial vacuum/fill the tube with 2 water, (√) the pressure difference √ between the two ends makes the water to flow. 2 4 The piece paper can be cut into tiny pieces. 1 5 Increase in temperature causes the alcohol in P to expand pushing mercury up on the right side of the u tube. The mercury pushes index A up. 2 6 Heat is a form of energy while temperature is the degree of hotness 2 7 Base area of the cylinder. √ The height of the position of the center of gravity above the base.√ 2 8 With no mass on the hook, mark the pointer position on the graph paper as O g or 0 N mark.√ Suspend the 100 g mass on the hook and mark the pointer position on the graph paper as the 100 g or IN mark. √ Count the number of divisions between the 0 g mark and the 100 g mark, divide and label equal divisions accordingly/ appropriately.√ 3 9 Volume inflow = A1V1Volume outflow=A2V2Since the water is incompressible.√Volume inflow = Volume outflow.A1V1= A2V2 10 11 Ft = mVFt = area under curve    = ½ × 0.5 × 500    = 125125 = m × 60   m = 2.0833 kg 12 Potential → Kinetic → Potential 13 Time 14 The pressure of a fixed mass of a gas is inversely proportional to its volume provided the temperature is kept constant.   √ length of mercury thread √ l2 > l1 In 5 (a) the pressure on the air column is greater than the pressure on the air pressure in 5 (b) hence 1, >1, √Since in (a) pressure in air column = P(atmosphere) + phgWhile in (b) pressure on air column =P(atmosphere)   Absolute zero temperature is the temperature at which the volume of an ideal gas is assumed to be zero P1 = P2   √T1    T2 6   =  P2     √275    353P2 = 7.702mmHg   √ 1 2 2 1 3 15 Reducing the speed of rotation.  √ Increasing the radius of the circular path. √   V = ωr    = 2πfr    √    = 2 × 3.142 × 2 × 2     √    = 25.136ms−1   √ T = mv2 − mg √         r    = 0.5 × 25.1362 − 5   √                2    = 152.955N    =153N √ The tension increases  √At the bottom, the tension acts in thve opposite direction as the weight.  √ 16 Upthrust = pgV    √                 = 800 × 10 × 0.053    √                = 1N    √ Apparent weight = W − upthrust     √                           = 20 − 1    √                           = 19N    √ Sum of clockwise moments = Sum of anticlockwise momentsx × 0.5 = 19 × 0.2    √x = 19 × 0.2    √           0.5   = 7.6N    √ Weight of block = Weight of water displaced0.5 × 10 = 1000 × 10 × V      √∴ V =   0.5×10      √          1000×10       = 0.005m3     √ 17 Impurities    √ Atmospheric pressure     √   Boiling point is 104°C Q = Pt     √    = 80 × 3.5 × 60     √   = 16 800J     √   Δθ = 94 − 30     √     = 64° Q = mcΔθ    √ C = 16800J        √        0.2×64    = 1312.5 Jkg−1K−1   v Δt = 6.3 − 5.4    = 0.9 minutes   = 54 seconds    √L = Pt = 54×80      √      m      0.002   = 2.16 × 106Jkg−1     √ 18 At uniform velocity, W = mg      W = 80 x 10          = 800 N The reaction R = w+ma                         = (80 x 10)+(80x 3) √                         = 1040 N On attempting to jump the person exerts a backward force on the boat which √ exerts an equal forward force on the person because of the low friction between the boat and water, the boat moves backwards √ hence reduces the forward force on the person. V W = F x d   √     = 1000 x 30         =30000J   √

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