PRE NATIONAL EXAMINATION PHYSICS 1 SERIES 8 (With Comprehensive Answers)

UMOJA WA WAZAZI TANZANIA.

WARI SECONDARY SCHOOL

PRE-NATIONAL EXAMINATION SERIES

PHYSICS 1 - SERIES 8

131/01

TIME: 2:30 HRS

JANUARY-MAY, 2023

INSTRUCTIONS

1. This Paper Consists of Section A and B with total of Ten (10) questions.
2. Answer all questions in section A and any two (2) question from section B.
3. Mathematical tables and non-programmable calculator may be used.
4. Cellular phones and unauthorized materials are not allowed in the Examination room.
5. Write your examination number on every page of your answer booklet (s).

The following information may be useful:

• Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)
• Pie \( \pi = 3.14 \)
• Monoatomic gas constant \( \delta = 1.4 \)
• Specific heat capacity of water \( c = 4.2 \times 10^{-3} \, \text{J/kg K} \)
• Density of water \( \rho w = 1000 \, \text{kg/m}^3 \)
• Universal gravitational constant, \( G = 1.67 \times 10^{-11} \, \text{kg}^{-1} \, \text{m}^3 \, \text{s}^{-2} \)
• Radius of the earth, \( R = 6400 \, \text{km} \)
• Stefan-Boltzmann constant, \( \sigma = 5.67 \times 10^{-8} \, \text{W m}^{-2} \, \text{K}^{-4} \)

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SECTION A (70 Marks)

Answer all questions in this section.

1. (10 marks)

(a) (i) Two or more terms of the same dimensions are not always be added or subtracted together. Give two reasons.

Two reasons why terms with the same dimensions cannot always be added or subtracted:

1. Different physical meanings: Even if dimensions match, the physical quantities may represent different concepts (e.g., work and torque both have dimensions [ML²T⁻²] but cannot be added).
2. Different directions: Some quantities are vectors and have directional properties that prevent addition even if magnitudes have same dimensions.

(ii) The velocity V of a particle depends upon time t according to the relation \[ V = \frac{a}{t} + \frac{b}{t+c} \] What are the dimensions of a, b and c?

Using dimensional analysis:

Velocity V has dimensions [LT⁻¹]

First term: a/t must have dimensions [LT⁻¹]

So [a]/[T] = [LT⁻¹] ⇒ [a] = [LT⁻¹] × [T] = [L]

Second term: b/(t+c) must have dimensions [LT⁻¹]

Since t and c are added, they must have same dimensions ⇒ [c] = [T]

Then [b]/[T] = [LT⁻¹] ⇒ [b] = [LT⁻¹] × [T] = [L]

Therefore:

• Dimensions of a = [L]
• Dimensions of b = [L]
• Dimensions of c = [T]

(b) (i) What is a systematic error?

Systematic error is a consistent, reproducible error that occurs due to problems in the measurement system. These errors are not random and typically affect all measurements in the same way. Examples include zero error in instruments, incorrect calibration, or environmental factors.

(ii) State two ways in which systematic error can be reduced.

Two ways to reduce systematic errors:

1. Calibration: Regularly calibrate instruments against known standards.
2. Proper experimental design: Use control experiments and eliminate known sources of bias.

(iii) The period of oscillation of a rod depends on its radius (r) and velocity (V). Determine the fractional error in calculating the acceleration due to gravity (g) if \[ r = (2 \pm 0.1) \, \text{mm} \] and \[ V = (4 \pm 0.1) \, \text{cm/s} \]. The period of oscillation is measured to be 10 sec using stop watch of scale 0.1 sec. Given that \[ T = \sqrt{\frac{3 \, \text{r} \, \text{V}^2}{\text{K} + gV}} \] Where K is constant value.

First, convert to consistent units:

r = 2 ± 0.1 mm = 0.002 ± 0.0001 m

V = 4 ± 0.1 cm/s = 0.04 ± 0.001 m/s

T = 10 ± 0.1 s (least count of stopwatch)

From the equation: \( T = \sqrt{\frac{3rV^2}{K + gV}} \)

Squaring both sides: \( T^2 = \frac{3rV^2}{K + gV} \)

Rearranging for g: \( K + gV = \frac{3rV^2}{T^2} \)

\( g = \frac{3rV}{T^2} - \frac{K}{V} \)

For error analysis, we'll consider the main terms. The fractional error in g can be found using:

\( \frac{\Delta g}{g} \approx \frac{\Delta r}{r} + \frac{\Delta V}{V} + 2\frac{\Delta T}{T} \)

Calculating fractional errors:

\( \frac{\Delta r}{r} = \frac{0.0001}{0.002} = 0.05 \)

\( \frac{\Delta V}{V} = \frac{0.001}{0.04} = 0.025 \)

\( \frac{\Delta T}{T} = \frac{0.1}{10} = 0.01 \)

So \( \frac{\Delta g}{g} \approx 0.05 + 0.025 + 2(0.01) = 0.095 \)

Fractional error in g ≈ 0.095 or 9.5%

2. (10 marks)

(a) (i) Explain why centripetal force does not do any work in a circular orbit?

Centripetal force does no work in circular motion because:

1. It always acts perpendicular to the direction of motion (tangential direction)
2. Work done = Force × Displacement × cosθ, where θ is the angle between force and displacement
3. For centripetal force, θ = 90°, so cosθ = 0, hence work done = 0
4. While it changes the direction of velocity, it doesn't change the speed, so kinetic energy remains constant

(ii) A curve in a road has radius of 60m. If the angle of bank of the road is \( 47^\circ \). Calculate the maximum speed of the car without skidding if the coefficient of static friction between the types and the road is 0.8.

For banking with friction, maximum velocity is given by:

\( v_{max} = \sqrt{\frac{rg(\mu + \tan\theta)}{1 - \mu \tan\theta}} \)

Where: r = 60m, θ = 47°, μ = 0.8, g = 9.8 m/s²

First calculate tanθ = tan(47°) ≈ 1.072

Numerator: μ + tanθ = 0.8 + 1.072 = 1.872

Denominator: 1 - μ tanθ = 1 - (0.8)(1.072) = 1 - 0.8576 = 0.1424

So \( v_{max} = \sqrt{\frac{60 \times 9.8 \times 1.872}{0.1424}} \)

= \( \sqrt{\frac{1100.736}{0.1424}} = \sqrt{7728.57} \)

≈ 87.9 m/s

(b) (i) Justify the statement that, projectile motion is two dimensional motion.

Projectile motion is two-dimensional because:

1. It occurs in a vertical plane defined by the initial velocity vector and gravity
2. Motion can be analyzed using two independent components:
   • Horizontal motion (constant velocity)
   • Vertical motion (constant acceleration due to gravity)
3. Only two coordinates (x and y) are needed to describe the position at any time
4. The motion is confined to a single plane throughout its trajectory

(ii) Show that the angle of projection \(\theta^0\) for a projectile launched from the origin is given by \[ \theta^0 = \tan^{-1} \left( \frac{4h}{R} \right) \] Where R stand for horizontal range and h is the maximum vertical height.

We know that for a projectile:

Maximum height \( h = \frac{u^2 \sin^2\theta}{2g} \)

Horizontal range \( R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin\theta \cos\theta}{g} \)

Now, \( \frac{h}{R} = \frac{\frac{u^2 \sin^2\theta}{2g}}{\frac{2u^2 \sin\theta \cos\theta}{g}} = \frac{\sin^2\theta}{4\sin\theta \cos\theta} = \frac{\sin\theta}{4\cos\theta} = \frac{\tan\theta}{4} \)

So \( \tan\theta = \frac{4h}{R} \)

Therefore \( \theta = \tan^{-1} \left( \frac{4h}{R} \right) \)

Hence proved.

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3. (10 marks)

(a) (i) Why is Newton's first law of motion called the law of inertia?

Newton's first law is called the law of inertia because:

1. It describes the inherent property of matter to resist changes in its state of motion
2. Inertia is the tendency of an object to maintain its current state (rest or uniform motion)
3. The law states that an object will remain at rest or in uniform motion unless acted upon by an external force
4. This resistance to change in motion is precisely what we call inertia

(ii) When several passengers are standing in a running bus it is said to be dangerous. Why? Give three (3) reasons.

Three reasons why standing passengers in a moving bus is dangerous:

1. Lack of stability: Standing passengers have higher center of gravity and smaller base, making them more prone to falling during sudden stops or turns.
2. Inertia effects: When the bus stops suddenly, standing passengers continue moving forward due to inertia and may collide with objects or other passengers.
3. Limited grip: Standing passengers have fewer points of contact with the vehicle, reducing their ability to maintain balance during acceleration, braking, or turning.

(iii) A car of mass \( 1.2 \times 10^3 \, \text{kg} \) was travelling east ward at \( 30 \, \text{m/s} \) when it collides with a truck of mass \( 3.6 \times 10^3 \, \text{kg} \) travelling at \( 20 \, \text{m/s} \) in a direction \( 60^\circ \) North East. After collision the two vehicles interlock and move off together. Find the common velocity of the vehicles after collision.

Using conservation of momentum:

Let east be x-direction, north be y-direction

Car: m₁ = 1200 kg, v₁x = 30 m/s, v₁y = 0

Truck: m₂ = 3600 kg, v₂x = 20 cos60° = 10 m/s, v₂y = 20 sin60° = 17.32 m/s

Total mass after collision = 1200 + 3600 = 4800 kg

Momentum conservation in x-direction:

1200×30 + 3600×10 = 4800×v_x

36000 + 36000 = 4800 v_x

72000 = 4800 v_x ⇒ v_x = 15 m/s

Momentum conservation in y-direction:

1200×0 + 3600×17.32 = 4800×v_y

62352 = 4800 v_y ⇒ v_y = 12.99 m/s

Resultant velocity = √(15² + 12.99²) = √(225 + 168.74) = √393.74 ≈ 19.84 m/s

Direction: θ = tan⁻¹(12.99/15) = tan⁻¹(0.866) ≈ 40.9° north of east

7. (10 marks)

(a) State the physical meaning of Kirchhoff's law of electrical network.

Kirchhoff's Current Law (KCL): The algebraic sum of currents meeting at any junction in an electrical circuit is zero. Physically, this means charge is conserved - no charge accumulates at any point in the circuit.

Kirchhoff's Voltage Law (KVL): The algebraic sum of potential differences around any closed loop in a circuit is zero. Physically, this means energy is conserved - the net energy gained per unit charge around a closed path is zero.

(b) Study the circuit diagram in figure 1 below and answer the questions that follows:

(i) Determine the equivalent resistance of circuit diagram in figure 1 above

From the description: R₁ = 100Ω, R₂ = R₃ = 50Ω, R₄ = 75Ω

Assuming R₂ and R₃ are in parallel: 1/R_parallel = 1/50 + 1/50 = 2/50 = 1/25

So R_parallel = 25Ω

This parallel combination is in series with R₁ and R₄:

R_eq = R₁ + R_parallel + R₄ = 100 + 25 + 75 = 200Ω

(ii) Determine the current that flows in \( R_4 \).

Total current from battery: I_total = V/R_eq = 6/200 = 0.03 A

Since R₄ is in series with the battery, the same current flows through it:

Current through R₄ = 0.03 A or 30 mA

(c) Briefly explain the meaning of Root Mean Square value (RMS) of alternating e.m.f and give three (3) importance of R.M.S value.

RMS value of alternating e.m.f is the square root of the mean of squares of all instantaneous values over one complete cycle. It represents the equivalent DC value that would produce the same heating effect.

Three importance of RMS value:

1. It allows comparison between AC and DC circuits in terms of power dissipation
2. Most AC measuring instruments are calibrated to read RMS values
3. Electrical power calculations for AC circuits use RMS values (P = I_RMS × V_RMS)

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SECTION B (30 Marks)

Answer two (02) questions from this section.

8. (15 marks)

(a) (i) Why is the amount of impurity added to a pure semiconductor closely controlled?

The amount of impurity in semiconductors is closely controlled because:

1. It determines the conductivity of the semiconductor
2. Too little impurity won't significantly change conductivity
3. Too much impurity can create metallic conduction properties, losing semiconductor characteristics
4. The type and concentration of impurities determine whether it becomes n-type or p-type semiconductor

(ii) Why is Silicon preferred to Germanium in the manufacture of semiconductor devices? Give three (3) reasons.

Three reasons Silicon is preferred over Germanium:

1. Higher thermal stability: Silicon has a higher band gap (1.1 eV vs 0.67 eV for Ge), so it can operate at higher temperatures
2. Better oxide formation: Silicon forms a stable native oxide (SiO₂) which is excellent for insulation and device fabrication
3. Abundance: Silicon is more abundant in nature, making it cheaper to produce

(iii) How does the conductance of a semiconductor change with the rise in temperature?

The conductance of a semiconductor increases with rise in temperature because:

1. More electrons gain sufficient energy to jump from valence band to conduction band
2. This creates more electron-hole pairs, increasing the number of charge carriers
3. Unlike metals where resistance increases with temperature, semiconductors show negative temperature coefficient of resistance

10. (15 marks)

(a) (i) Briefly explain the following climatic factors on how they influence plant growth: Temperature, Relative humidity and Wind.

Temperature: Affects enzyme activity, photosynthesis rate, and metabolic processes. Each plant has optimal temperature ranges for growth.

Relative humidity: Influences transpiration rate and water uptake. High humidity reduces transpiration but may increase disease risk.

Wind: Affects transpiration, pollen dispersal, and physical damage. Moderate wind strengthens plants, but strong winds can cause damage.

(ii) Explain three (3) techniques applicable for improving soil environment for the best plant growth.

Three techniques for improving soil environment:

1. Adding organic matter: Improves soil structure, water retention, and nutrient content
2. Proper irrigation: Maintains optimal moisture levels without waterlogging
3. Soil aeration: Prevents compaction and ensures oxygen reaches plant roots

(b) (i) Briefly explain three (3) effects of seismic waves.

Three effects of seismic waves:

1. Ground shaking: Causes buildings and infrastructure to vibrate, potentially leading to collapse
2. Soil liquefaction: Saturated soil temporarily loses strength and behaves like liquid
3. Tsunamis: Underwater earthquakes can generate large ocean waves

(ii) Briefly explain two (2) causes of Earth quake.

Two main causes of earthquakes:

1. Tectonic activity: Movement of tectonic plates causing stress buildup and sudden release at faults
2. Volcanic activity: Movement of magma beneath volcanoes can cause rock fractures and earthquakes

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