UMOJA WA WAZAZI TANZANIA.
WARI SECONDARY SCHOOL
PRE-NATIONAL EXAMINATION SERIES
PHYSICS 1 - SERIES 6
131/01
TIME: 2:30 HRS
JANUARY-MAY, 2023
INSTRUCTIONS
- This paper consists of section A and B with a total of ten (10) questions.
- Answer all questions in section A and two (2) questions from section B.
- Section A carries seventy (70) marks and section B carries thirty (30) marks.
- Marks for each question or part thereof are indicated.
- Mathematical tables and non-programmable calculators may be used.
- Cellular phones and any unauthorised materials are not allowed in the examination room.
- Write your Examination Number on every page of your answer booklet(s).
The following information may be useful:
- Acceleration due to gravity \( g = 9.8 m/s^2 \)
- Density of air \( = 1.3 kg/m^3 \)
- Density of oil \( = 9.2 \times 10^2 kg/m^3 \)
- Stefan – Boltzmann constant, \( \sigma = 5.7 \times 10^{-8} Wm^{-2} K^{-4} \)
- Coefficient of viscosity of oil \( = 8.4 \times 10^{-2} N s/m^{-2} \)
- Electronic charge, \( e = -16 \times 10^{-19} C \)
- \( 1 eV = 1.6 \times 10^{-19} J \)
- Thermal conductivity of ice \( = 2.3 Wm^{-1} K^{-1} \)
- Density of water \( = 1000 kg m^3 \)
- Specific latent heat of fusion of water \( = 3.25 \times 10^{-5} J kg^{-1} \)
- Molar gas constant is \( 8.31 J Mol^{-1} K^{-1} \)
- Pic, \( \pi = 3.14 \)
Answer all questions in this section.
Answer two (2) questions from this section.
Figure 1
Identify the device labelled \(M_1\) and \(M_2\) and state its functions. (03 marks)
UMOJA WA WAZAZI TANZANIA.
WARI SECONDARY SCHOOL
PRE-NATIONAL EXAMINATION SERIES
PHYSICS 1 - SERIES 6
131/01
TIME: 2:30 HRS
JANUARY-MAY, 2023
INSTRUCTIONS
- This paper consists of section A and B with a total of ten (10) questions.
- Answer all questions in section A and two (2) questions from section B.
- Section A carries seventy (70) marks and section B carries thirty (30) marks.
- Marks for each question or part thereof are indicated.
- Mathematical tables and non-programmable calculators may be used.
- Cellular phones and any unauthorised materials are not allowed in the examination room.
- Write your Examination Number on every page of your answer booklet(s).
The following information may be useful:
- Acceleration due to gravity \( g = 9.8 m/s^2 \)
- Density of air \( = 1.3 kg/m^3 \)
- Density of oil \( = 9.2 \times 10^2 kg/m^3 \)
- Stefan – Boltzmann constant, \( \sigma = 5.7 \times 10^{-8} Wm^{-2} K^{-4} \)
- Coefficient of viscosity of oil \( = 8.4 \times 10^{-2} N s/m^{-2} \)
- Electronic charge, \( e = -1.6 \times 10^{-19} C \)
- \( 1 eV = 1.6 \times 10^{-19} J \)
- Thermal conductivity of ice \( = 2.3 Wm^{-1} K^{-1} \)
- Density of water \( = 1000 kg m^3 \)
- Specific latent heat of fusion of water \( = 3.34 \times 10^5 J kg^{-1} \)
- Molar gas constant is \( 8.31 J Mol^{-1} K^{-1} \)
- Pic, \( \pi = 3.14 \)
Answer all questions in this section.
For uniform circular motion, acceleration depends on:
- Velocity (v) - dimensions [LT⁻¹]
- Radius (r) - dimensions [L]
Let acceleration a = k vx ry
Dimensions: [LT⁻²] = [LT⁻¹]x [L]y = [Lx+y T-x]
Equating dimensions:
For L: 1 = x + y
For T: -2 = -x ⇒ x = 2
Then y = 1 - x = 1 - 2 = -1
Therefore, a = k v²/r
From physics, k = 1, so a = v²/r
The given equation is dimensionally incorrect.
Left side: Surface tension γ has dimensions [MT⁻²]
Right side: ln(I/g) is dimensionless, 2cosθ is dimensionless
So the right side is dimensionless, but left side has dimensions [MT⁻²]
Therefore, the equation is dimensionally incorrect.
The correct formula for capillary rise is: h = (2γ cosθ)/(ρgr)
Power = Force × Velocity
P = F × v
23,000 W = T × 0.75 m/s
T = 23,000 / 0.75 = 30,666.67 N
Tension in the cable = 30,667 N (approximately)
The outer rail is raised to:
- Provide the necessary centripetal force for the train to move along the curved path
- Prevent excessive wear on the rails
- Ensure passenger comfort by reducing lateral forces
- Prevent derailment by counteracting the centrifugal force
The banking angle is designed so that the horizontal component of the normal force provides the centripetal force required for circular motion.
According to Newton's third law (action-reaction):
- The helicopter rotor blades push air downward (action)
- The air exerts an equal and opposite upward force on the rotor blades (reaction)
- This upward reaction force provides the lifting force for the helicopter
By changing the angle of attack of the rotor blades, the pilot can control the amount of air pushed downward, thus controlling the lift.
Using conservation of momentum:
m₁ = 10 g = 0.01 kg, v₁ = 100 m/s
m₂ = 990 g = 0.99 kg, v₂ = 0 m/s
m₁v₁ + m₂v₂ = (m₁ + m₂)v
0.01 × 100 + 0.99 × 0 = (0.01 + 0.99)v
1 = 1 × v ⇒ v = 1 m/s
Initial kinetic energy = ½ × 0.01 × (100)² = 50 J
Final kinetic energy = ½ × 1 × (1)² = 0.5 J
Internal energy produced = Initial KE - Final KE = 50 - 0.5 = 49.5 J
Gravitational potential V = -GM/R
G = 6.7 × 10⁻¹¹ Nm²/kg²
M = 6.0 × 10²⁴ kg
R = 6.4 × 10⁶ m
V = -(6.7 × 10⁻¹¹ × 6.0 × 10²⁴) / (6.4 × 10⁶)
V = -(4.02 × 10¹⁴) / (6.4 × 10⁶) = -6.28 × 10⁷ J/kg
Gravitational potential = -6.28 × 10⁷ J/kg
A satellite with a 24-hour period is in a geostationary orbit if it's above the equator. This means:
- The satellite appears stationary relative to a point on Earth
- It remains fixed above the same location on Earth
- It's ideal for communication as ground antennas don't need to track the satellite
For a satellite in circular orbit:
Centripetal force = Gravitational force
mv²/R₀ = GMm/R₀²
Where:
- m = mass of satellite
- M = mass of Earth
- G = universal gravitational constant
- R₀ = orbital radius
- v = orbital velocity
Simplifying: v² = GM/R₀
Also, v = 2πR₀/T, where T = orbital period
So: (2πR₀/T)² = GM/R₀
4π²R₀²/T² = GM/R₀
4π²R₀³ = GMT²
R₀³ = GMT²/(4π²)
Therefore: R₀ = ∛[GMT²/(4π²)]
This is the expression for the orbital radius of a satellite.
Let's analyze the motion:
Vertical velocity before impact: v_y = √[2g(H-h)]
After impact, velocity becomes horizontal, so vertical component becomes 0.
Time to fall from height h: t = √(2h/g)
Horizontal distance traveled: x = v_x × t
But we need to find h/H ratio.
Using energy conservation and projectile motion equations:
The ratio h/H = 1/2
This is because when the object hits the inclined plane, half its initial potential energy has been converted to kinetic energy.
Using projectile motion equations:
Maximum height formula: h = (v² sin²θ)/(2g)
2.2 = (8² × sin²θ)/(2 × 9.8)
2.2 = (64 × sin²θ)/19.6
sin²θ = (2.2 × 19.6)/64 = 43.12/64 = 0.674
sinθ = √0.674 = 0.821
Horizontal velocity v_x = v cosθ = 8 × cosθ
cosθ = √(1 - sin²θ) = √(1 - 0.674) = √0.326 = 0.571
Horizontal velocity = 8 × 0.571 = 4.57 m/s
Time of flight T = (2v sinθ)/g
v = 8 m/s, sinθ = 0.821, g = 9.8 m/s²
T = (2 × 8 × 0.821)/9.8 = 13.136/9.8 = 1.34 seconds
Evidence for the first law of thermodynamics (conservation of energy):
- Joule's experiment showing mechanical equivalent of heat
- Conservation of energy in all observed physical processes
- Impossibility of perpetual motion machines of the first kind
- Consistent energy accounting in all thermodynamic processes
According to Wien's displacement law: λ_max T = constant
As a black body is constantly heated:
- Its temperature increases
- The peak wavelength of emitted radiation decreases (shifts to shorter wavelengths)
- The body glows with different colors: red → orange → yellow → white → blue
- Total radiated power increases according to Stefan-Boltzmann law (P ∝ T⁴)
For monatomic gas at constant volume: Q = nC_vΔT
Helium is monatomic: C_v = (3/2)R
R = 8.31 J/mol·K
C_v = 1.5 × 8.31 = 12.465 J/mol·K
Q = nC_vΔT
60 = 2 × 12.465 × ΔT
ΔT = 60 / (2 × 12.465) = 60 / 24.93 = 2.41 K
The temperature rises by approximately 2.41 K.
Answer two (2) questions from this section.
Step-up transformer vs Step-down transformer:
- Step-up transformer: Increases voltage from primary to secondary coil (has more turns in secondary coil)
- Step-down transformer: Decreases voltage from primary to secondary coil (has fewer turns in secondary coil)
Both work on the principle of electromagnetic induction and conserve power (ignoring losses).
Electricity is transmitted at high voltage to:
- Reduce energy losses due to resistance (P_loss = I²R)
- For the same power (P = VI), higher voltage means lower current
- Lower current means less energy lost as heat in transmission lines
- More efficient power transmission over long distances
Charging circuit equation: V_source = E + I(R + r)
Where: V_source = 100 V, E = 50 V, I = 2 A, r = 2 Ω
100 = 50 + 2(R + 2)
50 = 2(R + 2)
25 = R + 2
R = 25 - 2 = 23 Ω
The required resistance is 23 Ω.
Figure 1
Identify the device labelled \(M_1\) and \(M_2\) and state its functions. (03 marks)
In a circuit for measuring resistance:
- M₁: Likely an ammeter - measures current flowing through the wire
- M₂: Likely a voltmeter - measures potential difference across the wire
Using Ohm's law (R = V/I), the resistance can be calculated from these measurements.
Alloys are used for standard resistance coils because:
- They have high resistivity
- They have low temperature coefficient of resistance (resistance doesn't change much with temperature)
- They are stable and don't oxidize easily
- Common examples: Constantan, Manganin, Nichrome
Using the formula: R_t = R_0[1 + α(t - t_0)]
At 20°C: 10.8 = R_0[1 + 20α] ...(1)
At 100°C: 14.1 = R_0[1 + 100α] ...(2)
Dividing (2) by (1): 14.1/10.8 = [1 + 100α]/[1 + 20α]
1.3056 = [1 + 100α]/[1 + 20α]
1.3056(1 + 20α) = 1 + 100α
1.3056 + 26.112α = 1 + 100α
0.3056 = 73.888α
α = 0.3056/73.888 = 0.00414 °C⁻¹
Now find R_0 using equation (1):
10.8 = R_0[1 + 20×0.00414] = R_0[1 + 0.0828] = R_0×1.0828
R_0 = 10.8/1.0828 = 9.97 Ω
Temperature coefficient = 0.00414 °C⁻¹, Resistance at 0°C = 9.97 Ω
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