# Physics Paper 1 Questions and Answers - Maranda Pre-Mock Examinations seccccc022

QUESTIONS
SECTION A: 25 MARK

1. Figure 1 shows a ball bearing of mass 0.0025 kg is held between the anvil and spindle of a micrometer screw gauge. The reading on the gauge when the jaws are closed without anything in between is 0.011cm. Use this information and the position of the scale in the figure below to answer the questions (a) and (b) below:

1. What is the diameter of the ball bearing? (1 mark)
2. Find the density of the ball bearing correct to 3 three significant figures (2 marks)
2. Explain why solids are good conductors of heat (2 marks)
3. The barometrid height at sea level is 76 cm of mercury while at a point on a highland it is 72 cm of mercury. What is the altitude of the point? (3 marks)
4. Figure 2 shows an arrangement of identical springs A, B, C and D.

Each spring extends by 2 cm when a force of 5N it. Determine the extension of the system. (3 marks)
5. A bathroom shower has 200 holes each 2.5 mm2 in area. Water flows from a pipe of cross-section area of 15 cm2 5m/s to the shower. Determine the speed of the spray. (2 marks)
6. Explain how a ball and ring apparatus can be used to demonstrate contraction in solids (2 marks)
7. Smoke particles are observed through the eye piece of a microscope. They are seen to move randomly. Explain what causes this motion. (2 marks)
8. Explain how a cyclist maintains the stability of a moving bicycle. (2 marks)
9. Figure 3 shows a U-tube manometer open at one end and the other end connected to the gas súpply.

Given that the atmospheric pressure is 1.0 x 105 pa , determine the pressure of the gas(3 marks)
10. State one advantage of alcohol-in-glass thermometer over mercury-in glass thermometer. (1 mark)
11. In the crushing can experiment, it is observed that the can crushes on cooling. Explain this observation. (2 marks)

SECTION B (55 MARKS)

1.
1. Explain why it is important to wipe yourself with a towel after bathing. (2 marks)
2. 200g of a solid was uniformly heated by a 0.2kW heater for some time. The graph in figure 4 shows how the temperature of the solid changed with time.

1. Explain what is happening between BC and CD. (2 marks)
2. Calculate the specific heat capacity of the solid. (3 marks)
3. Calculate the specific latent heat of fusion Ly of the solid. (3 marks)
2.
1. Figure 5 shows a simple set up for pressure law apparatus

1. State the measurements to be taken (2 marks)
2. State the physical quantities that are kept constant (2 marks)
3. Explain how the measurements taken in (a) above can be used to verify the law. (3 mark)
2. The graph in figure 6 shows the relationship between the pressure and temperature for a fixed mass of an ideal gas at constant volume.

1. Given that the relationship between pressure. P and temperature, T in Kelvin is of the form P=kT+C where k and C are constants, determine from the graph, the value of:
1. k (2 marks)
2. c (1 mark)
2. Explain why it would be impossible for the pressure of the gas to be reduced to zero in practice. (1 mark)
3. A gas is put into a container of fixed volume at a pressure of 2.1 x 105 Nm-2 and temperature 27°C. The gas is then heated to a temperature of 327°C. Determine the new pressure. (2 marks)
3.
1. Figure 7(a) and (b) shows a set up used by a student to determine upthrust in paraffin

Given that:
• Weight of solid Win air that balances with solid of weight W1N when in equilibrium in air = WN
• Perpendicular distance of solid W from the pivot = d1cm
• Perpendicular distance of solid from the pivot = d cm
• Apparent weight of solid S in water = W2N
• Apparent weight of solid S in paraffin = W3
• Perpendicular distance between solid W and pivot when solid S is immersed in water = d2cm
• Perpendicular distance between solid Wand pivot when solid S is immersed in paraffin =d2cm
Show that:
1. Relative density of solid S is given by    d 1 (3 marks)
d- d2
2. Relative density of paraffin is given by d- d3
d- d(3 marks)
2. Figure 8 shows a block of wood of dimensions 14 cm x 7 cm x 2 cm floating with 1/3 of its size submerged in a liquid.

During an experiment with the set-up, the following results were obtained:
• Initial reading of the top pan balance with empty beaker = 32 g
• Final reading of the top pan balance = 186 g
Use the above results to determine:
1. the density of the block(3 marks)
2. the density of the liquid.
3. Explain the purpose of the wide bulb of a hydrometer. (1 mark)
4.
1. Explain why a body moving in a uniform circular path with constant speed accelerates. (1 mark)
2. A wooden block of mass 150 g is placed at various distances from the centre of a turntable which is rotating at constant angular velocity. It is found that at a distance of 8.0 cm from the centre, the block just starts to slide off the table. If the force of friction between the block and the table is 0.5 N Calculate
1. The angular velocity of the table (2 marks)
2. The force required to hold the block at a distance of 14.0 cm from the centre of the table (3 marks)
3. A similar block of mass 300 g is now placed at distance of 8.0 cm from the centre of the turntable in (i) above and the turntable rotated at the same angular velocity, State with a reason whether or not the ball will slide off. (2 marks)
3. A funfair ride of diameter 12 m makes 0.5 revolutions per second.
1. Determine the angular velocity of the funfair. (2 marks)
2. If the mass of the child is 30 kg, find the centripetal force that keeps the child in the motion. (3 marks)
5. A student performed an experiment using a pulley as shown in figure 9

1. State the V.R. of the system (1 mark)
2. Determine the M.A. of the system (2 marks)
3. Calculate the efficieney of the system. (3 marks)

## MARKING SCHEME

1.
1. 4.50 + (26 x 0.01) = 4.76mm - 0.11mm = 4.65
2. ℑ = m/v =               0.0025kg
4/3 x 22/7 x (2.325 x 10-3)3mv
4.747 x 104kg/m3
2. the particles are closely packed making the force between them strong therefore heat transfer occurs by collision of the adjacent vibrating particles
3. hair x 1.25 x 10 = (76 - 73)m x 13600 x 10
100
hair = 435.2m
4. K = 5N/2cm = 2.5Ncm-1
A&B extension = 30/2.5 = 12cm
C extension = 60/2.5 = 24cm
D extension = 60/2.5 = 24cm
Total = 60cm
5. A1V1 = A2V2
V1 x 200 x (2.5 x 10-6)m2 = (15 x 10-4)m2 x 5m/s
V1 = 15m/s
6. At room temperature, the ball goes through the ring. when the ring is cooled below room temperature, its diameter reduces because of contraction making the ball not to go through the ring
7. because of uneven bombardment of smoke particles by the invisible air particles that are always in a continuous random motion
8. by bending his/her body down
this lowers the c.o.g hence increased stability
9. pg = pa + hℑg
1.0 x 105 + (0.20m x 1000 x 10)
102,000pa
10. can record very low temperatures
11. on cooling, steam condenses creating a vaccuum in the container hence decrease in pressure below the atmosphere pressure
A resultant force due to the pressure difference crushes the can
12.
1. as water evaporates from the body, it absorbs latent heat of vaporization from the body which is a lot. The towel minizes loss of this heat by absorbing the water
2.
1. BC - Heat energy absorbed is used to break the bonds hence changing from solid to liquid
CD - Heat energy absorbed is used to raise the temperature of the liquid
2. pt = m.c. T
200W X 100s =  200   x c x (340 - 200)
1000
c = 714.29Jkg-1k-1
3. pt = mLf
200W x (250 - 100) = 200/1000 x Lf
Lf = 150000Jkg-1
13.
1.
1. temperature of the as
pressure of the gas
2. mass of the gas
volume of the gas
3. initial temperature and pressure readings are taken and recorded
a graph of pressure against absolute temperature is plotted. the graph is a straight line, positive gradient, through origins thus verifying the law
2.
1.
1. k = slope of graph =  ΔP (Nm-2)
ΔT   (K)
= (12-8) x 104 Nm-2
(300-200)K
= 400Nm-2K-1
2. c = y - intercept = 0
2. because at absolute temperature (OK) the particle has low kinetic energy and hence do not collide with each other
3. P1/T1 = P2/T2
2.1 x 105 = P2
300        600
P2 = 4.2 x 105Nm-2
14.
1.
1. Taking moment about O. Wd1 - Wd ⇒ W1 = Wd1/d
Wd2 = W2d ⇒ W2 = Wd2/d
R.d =    weight in air     =  W1  =        Wd1/d       =   w/d(d1)
upthrust in water   w1-w2   Wd1/d - Wd2/d     w/d(d1-d2)
d1    (shown)
d1 - d2
2. upthrust in water = w/d(d1 - d2)
upthrust in liquid = w/d(d1 - d2)
r.d of the liquid = upthrust in paraffin w/d(d1 - d3)= d1 - d3
upthrust in water      w/d(d1 - d2)    d1 -d2
2.
1. Mass of block  - mass of water displaced = 186 - 32 = 154g
volume of block = 14 x 7 x 2 = 196cm3
v.block =   154g
196cm3
1 = 0.7857g/cm3
2. upthrust = vℑg
154   x 10)N = 1/3 x 196 x 10-6m3 x ℑ x 10N/kg
1000
= 2357kg/m3
3. to displace large volume of liquid to provide higher upthrust sufficient to keep the hydrometer upright
15.
1. becasue there is change in direction of velocity at every instant as the body moves in a circular path
2.
1. F = mw2r
0.5 = 150/1000 x w2 x 8/100
2. F = mw2r
150/1000 x 6.4552 x 14/100
0.875N
3. the ball will slide off
because the centipedal force is greater than the friction between the block and the table
3.
1. w = 2πf
2 x 22/7 x 0.5
2. Fc = mw2r
30 x 3.1432 x 12
3556N
16.
1. 5
2. M.A = L/E = 3500N
1000N
= 3.5
3. M.A/V.R x 100%
3.5/5 x 100%
70%

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