DODOMA REGION FORM SIX MID TERM EXAMINATION
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PHYSICS 1 - Examination Answers
SECTION A (70 Marks)
Answer all questions from this section
Question 1 (10 Marks)
(a) (i) The dimension of magnetic flux is [ML²T⁻²I⁻¹].
(a) (ii) Avogadro's number is a dimensionless quantity.
(a) (iii)
Let V ∝ T^a L^b m^c
Dimensions: [V] = [LT⁻¹], [T] = [MLT⁻²], [L] = [L], [m] = [M]
Equating dimensions: [LT⁻¹] = [MLT⁻²]^a [L]^b [M]^c
For M: 0 = a + c ⇒ c = -a
For L: 1 = a + b ⇒ b = 1 - a
For T: -1 = -2a ⇒ a = 1/2
Thus c = -1/2, b = 1/2
So V ∝ T^(1/2) L^(1/2) m^(-1/2) = √(TL/m) = √(T/(m/L))
Therefore V ∝ √(Tension/Mass per unit length)
(b)
Given: L = 325 cm, ΔL = 0.1 cm
Extension, e = 0.227 cm, Δe = 0.001 cm
Diameter, d = 0.043 cm, Δd = 0.001 cm
Young's modulus Y = (FL)/(Ae) = (4FL)/(πd²e)
Maximum permissible error in Y:
ΔY/Y = ΔL/L + 2(Δd/d) + Δe/e
= (0.1/325) + 2(0.001/0.043) + (0.001/0.227)
= 0.000308 + 0.0465 + 0.00441 ≈ 0.0512 or 5.12%
Question 2 (10 Marks)
(i) The letter is called a projectile, and the path followed is a parabola.
(ii)
The boy should throw the letter with an initial velocity that has both horizontal and vertical components.
Let u be initial velocity, θ be angle with horizontal.
Horizontal distance: 40 = u cosθ × t
Vertical distance: 10 = u sinθ × t - ½gt²
Solving these equations gives the required velocity and angle.
(iii) Assumptions:
1. Air resistance is negligible
2. Acceleration due to gravity is constant
3. The stone is thrown from ground level (or window height is accounted for)
Question 3 (10 Marks)
(a) (i) In an accelerating lift, the apparent weight changes. When ascending and accelerating upward, apparent weight increases. When descending and accelerating downward, apparent weight decreases.
(a) (ii) Mud flies off tangentially due to inertia - the mud tends to continue moving in a straight line tangent to the wheel's circular path.
(b)
Using conservation of momentum and energy:
Let m be mass of each ball, u be initial velocity of first ball.
After collision: first ball at 30° with speed v₁, second ball at angle φ with speed v₂.
Conservation of momentum in x-direction: mu = mv₁cos30° + mv₂cosφ
Conservation of momentum in y-direction: 0 = mv₁sin30° - mv₂sinφ
Conservation of energy: ½mu² = ½mv₁² + ½mv₂²
Solving these equations gives v₂ and φ.
Question 4 (9 Marks)
(a) (i) The motion of a simple pendulum is not strictly simple harmonic because the restoring force is proportional to sinθ rather than θ, and this approximation is only valid for small angles.
(a) (ii) In SHM, velocity is maximum when displacement is zero, and acceleration is maximum when displacement is maximum. They are 90° out of phase.
(b) (i) Maximum speed v_max = Aω = A√(k/m) = 0.03 × √(20/0.5) = 0.03 × √40 ≈ 0.19 m/s
(b) (ii) Kinetic energy = ½k(A² - x²) = ½ × 20 × (0.03² - 0.02²) = 10 × (0.0009 - 0.0004) = 10 × 0.0005 = 0.005 J
(c)
When mass falls from height h, velocity before impact v = √(2gh)
After sticking to pan, maximum compression x is given by:
Energy conservation: mg(h+x) = ½kx²
Solving gives x, which is the amplitude.
Energy of oscillation = ½kA²
Question 5 (10 Marks)
(a) (i) A person in an artificial satellite feels weightless because both the person and the satellite are in free fall towards Earth, with the same acceleration.
(a) (ii) The satellite remains in orbit because its horizontal velocity is sufficient to keep it falling around the Earth rather than into it, balancing gravitational pull with centripetal force.
(b)
g = GM/R²
For Earth: g_e = GM_e/R_e²
For planet: g_p = GM_p/R_p²
Since density is same: M_p/M_e = (R_p/R_e)³ = (1/3)³ = 1/27
So g_p = G(M_e/27)/(R_e/3)² = (G M_e/R_e²) × (1/27)/(1/9) = g_e × (1/3) = 9.8/3 ≈ 3.27 m/s²
Question 6 (10 Marks)
(a) (i) The pupil appears black because it absorbs most of the light entering the eye, with very little reflected back.
(a) (ii) If a refrigerator door is kept open, the room will become warmer because the refrigerator transfers heat from inside to outside, and the motor generates additional heat.
(b)
Solar constant S = 1400 W/m²
Total power from sun P = S × 4πd² (where d = 1.5×10¹¹ m)
Also P = σAT⁴ = 4πR²σT⁴ (where R = 7×10⁸ m, σ = 6.0×10⁻⁸)
Equating: 4πd²S = 4πR²σT⁴
T⁴ = (d²S)/(R²σ) = ((1.5×10¹¹)² × 1400)/((7×10⁸)² × 6.0×10⁻⁸)
Solving gives T ≈ 5800 K
(c)
v_rms = √(3RT/M)
For hydrogen: M = 2×10⁻³ kg/mol
T = 27°C = 300 K
R = 8.314 J/mol·K
v_rms = √(3×8.314×300/0.002) ≈ 1934 m/s
Question 7 (11 Marks)
(a) (i) Geothermal energy is thermal energy generated and stored in the Earth, originating from the planet's formation and radioactive decay.
(a) (ii) Three types of geothermal power plants:
1. Dry steam plants - use steam directly from underground
2. Flash steam plants - use high-pressure hot water
3. Binary cycle plants - use heat from moderate-temperature water
(b) Six ped types in soil structure:
1. Granular
2. Platy
3. Blocky
4. Prismatic
5. Columnar
6. Single grain
(c)
Wave power P = ½ρgH²
ρ = 1023 kg/m³, g = 9.8 m/s², H = 3 m
P = ½ × 1023 × 9.8 × 3² ≈ 45,200 W/m
SECTION B (30 Marks)
Answer only two (2) questions from this section
Question 8 (15 Marks)
(a) (i) A potentiometer is a device used to measure potential difference accurately by comparing it with a known voltage.
(a) (ii) It's called a potentiometer because it measures electric potential (voltage).
(a) (iii) Uses: measuring EMF, comparing EMFs, measuring internal resistance, calibrating voltmeters.
(b) (i) The balance length is smaller when a resistor is connected because some voltage is dropped across the internal resistance of the cell.
(b) (ii)
Using formula: r = R(L₁/L₂ - 1)
Where R = 5Ω, L₁ = 90 cm, L₂ = 45 cm
r = 5(90/45 - 1) = 5(2 - 1) = 5Ω
(c) (i)
For AC: E = E₀sin(ωt)
Power P = E²/R = E₀²sin²(ωt)/R
Average power P_avg = (E₀²/R) × average of sin²(ωt)
Since average of sin²(ωt) = ½, P_avg = E₀²/(2R)
For DC with E_rms: P = E_rms²/R
Equating: E_rms²/R = E₀²/(2R) ⇒ E_rms = E₀/√2
(c) (ii)
Given E = 30 sin(30t + π/3)
E₀ = 30 V, so E_rms = 30/√2 ≈ 21.21 V
ω = 30 rad/s, so T = 2π/ω ≈ 0.209 s
Question 9 (15 Marks)
(a) (i) Semiconductors are damaged by heavy current due to excessive heat generation which can melt the material or alter its properties.
(a) (ii) Although carbon, germanium, and silicon all have four valence electrons, carbon has a larger band gap (5.5 eV) making it an insulator, while germanium (0.67 eV) and silicon (1.1 eV) have smaller band gaps, making them semiconductors.
(a) (iii) An integrated circuit is a set of electronic circuits on a small chip of semiconductor material.
(b) (i) A NOT gate is called an inverter because it inverts the input signal (1 becomes 0, 0 becomes 1).
(b) (ii) AM is less efficient than FM because AM is more susceptible to noise and interference, and requires more power for the same coverage area.
(c) (i) Advantages of digital signals:
1. Less affected by noise
2. Can be easily processed and stored
(c) (ii)
Carrier frequency f_c = 500 kHz
Modulating frequency f_m = 1 kHz
Sideband frequencies: f_c ± f_m = 500 ± 1 kHz = 499 kHz and 501 kHz
Question 10 (15 Marks)
(a) (i)
Current gain β = I_C/I_B
Voltage gain A_v = β × R_C/R_i
Power gain A_p = A_v × β = β² × R_C/R_i = β² × Resistance gain
(a) (ii) Common emitter configuration is mostly used because it provides both voltage and current gain, resulting in high power gain.
(b) (i)
Given I_C = 100I_B, V_CC = 10V, R_C = 1kΩ
From circuit: V_CC = I_CR_C + V_CE
Also I_B = (V_CC - V_BE)/R_B
With V_BE ≈ 0.7V for silicon
Solving gives I_B and V_CE
(b) (ii)
Voltage gain A_v = β × R_C/R_i
With R_i = 2000Ω, R_C = 1000Ω, β = 100
A_v = 100 × 1000/2000 = 50
PHYSICS 2 - Examination Answers
Question 1 (20 Marks)
(a) (i) Account for a suction effect phenomenon based on Bernoulli's Theorem
According to Bernoulli's theorem, when the speed of a fluid increases, its pressure decreases. The suction effect occurs when a fluid (air) flows over a surface, creating a region of low pressure. This pressure difference causes the fluid to be "sucked" toward the low-pressure region.
(a) (ii) Work done by gravitational force on raindrop
Radius of raindrop = 2 mm = 0.002 m
Volume = (4/3)πr³ = (4/3)π(0.002)³ ≈ 3.35 × 10⁻⁸ m³
Mass = density × volume = 1000 × 3.35 × 10⁻⁸ = 3.35 × 10⁻⁵ kg
Height = 500 m
Work done in first half (with acceleration) = mgh/2 = 3.35 × 10⁻⁵ × 9.8 × 250 ≈ 0.082 J
Work done in second half (terminal velocity) = mgh/2 ≈ 0.082 J
Total work done = 0.164 J
(b) Water flow from cylindrical tank
i. Initial speed:
Using Torricelli's theorem: v = √(2gh) = √(2 × 9.8 × 5) ≈ 9.9 m/s
ii. Time to empty tank:
Using formula for time to empty cylindrical tank: t = (2A√H)/(a√(2g))
Where A = πr² = π(1)² ≈ 3.14 m², a = 10⁻² m², H = 5 m
t = (2 × 3.14 × √5)/(0.01 × √(2 × 9.8)) ≈ (14.04)/(0.0443) ≈ 317 seconds
(c) (i) Velocity-time graph for ball falling in viscous fluid
(c) (ii) Terminal velocity of combined water drops
For spherical objects in viscous medium: v ∝ r²
Two identical drops combine: volume doubles, so radius increases by ∛2 ≈ 1.26
New terminal velocity = (1.26)² × 10 cm/s ≈ 1.587 × 10 ≈ 15.87 cm/s
(d) (i) Pressure at reduced cross-section
Using continuity equation: A₁v₁ = A₂v₂
0.02 × 2 = 0.01 × v₂ ⇒ v₂ = 4 m/s
Using Bernoulli's equation: P₁ + ½ρv₁² = P₂ + ½ρv₂²
4000 + ½×1000×4 = P₂ + ½×1000×16
4000 + 2000 = P₂ + 8000 ⇒ P₂ = -2000 N/m² (gauge pressure)
(d) (ii) Rate of flow of glycerin
Using Poiseuille's equation: Q = πr⁴ΔP/(8ηL)
Need more information (viscosity η and length L) to calculate.
Question 2 (20 Marks)
(a) (i) Beats heard by observer
Frequency of source = 500 Hz
Speed of source = 1.5 m/s
Speed of sound = 336 m/s
Apparent frequency from wall: f' = f × v/(v - v_s) = 500 × 336/(336 - 1.5) ≈ 502.24 Hz
Beats per second = |f' - f| ≈ 2.24 beats/s
(a) (ii) Width of central maximum
Slit width a = 0.1 mm = 10⁻⁴ m
Wavelength λ = 5000 Å = 5 × 10⁻⁷ m
Distance to screen D = 1 m
Width of central maximum = 2λD/a = 2 × 5 × 10⁻⁷ × 1 / 10⁻⁴ = 0.01 m = 1 cm
(b) (i) Frequency of stretched wire
Length L = 140 cm = 1.4 m
Mass m = 0.52 × 10⁻³ kg
Tension T = 16 × 9.8 = 156.8 N
Mass per unit length μ = m/L = 0.52 × 10⁻³ / 1.4 ≈ 3.71 × 10⁻⁴ kg/m
Frequency f = 1/(2L) × √(T/μ) = 1/(2×1.4) × √(156.8 / 3.71×10⁻⁴) ≈ 0.357 × 650 ≈ 232 Hz
(b) (ii) Positions of bridges on sonometer wire
Fundamental frequencies in ratio 1:2:3
Since f ∝ 1/L, lengths will be in ratio 1:1/2:1/3 = 6:3:2
Total parts = 6+3+2 = 11
First bridge at (6/11)×110 = 60 cm from one end
Second bridge at (9/11)×110 = 90 cm from the same end
(b) (iii) Radii of Newton's rings
Radius of curvature R = 50 cm = 0.5 m
Wavelength λ = 5 × 10⁻⁷ m
For bright rings: r = √((m - ½)λR)
Fifth bright ring (m=5): r₅ = √((5 - ½) × 5×10⁻⁷ × 0.5) = √(4.5 × 2.5×10⁻⁷) ≈ 1.06 × 10⁻³ m = 1.06 mm
Tenth bright ring (m=10): r₁₀ = √((10 - ½) × 5×10⁻⁷ × 0.5) = √(9.5 × 2.5×10⁻⁷) ≈ 1.54 × 10⁻³ m = 1.54 mm
(c) Changes in Young's double slit experiment
i. White light: Colored fringes with white central fringe
ii. One slit covered: No interference pattern, only diffraction
iii. Source moved closer: Fringes become wider
iv. Conditions for interference: Coherent sources, same frequency, constant phase difference
v. Fraunhofer diffraction: Occurs when source and screen are at infinite distance from obstacle
(d) Angle of air wedge
Distance between 1st and 11th bright fringe = 10 fringe separations = 8.1 mm
Fringe separation β = 8.1/10 = 0.81 mm = 8.1 × 10⁻⁴ m
Wavelength λ = 540 nm = 5.4 × 10⁻⁷ m
For air wedge: β = λ/(2θ) ⇒ θ = λ/(2β) = 5.4×10⁻⁷/(2×8.1×10⁻⁴) ≈ 3.33 × 10⁻⁴ rad
Question 3 (20 Marks)
(a) (i) Pressure increase with volume reduction
According to Boyle's law (PV = constant at constant temperature), reducing volume increases the number of collisions per unit time between gas molecules and container walls, thus increasing pressure.
(a) (ii) Avogadro's law in kinetic theory
Avogadro's law states that equal volumes of gases at same temperature and pressure contain equal number of molecules. In kinetic theory, pressure depends on number of molecules per unit volume and their average kinetic energy. At same T and P, the number density must be equal.
(b) (i) RMS speed and speed of gas molecule
Root mean square speed is the square root of the average of the squares of the speeds of all molecules. It's a statistical measure of molecular speeds in a gas.
(b) (ii) RMS speed of hydrogen at 27°C
T = 27°C = 300 K
Mass of H₂ molecule = 2 × 1.674 × 10⁻²⁷ = 3.348 × 10⁻²⁷ kg
v_rms = √(3kT/m) = √(3 × 1.38×10⁻²³ × 300 / 3.348×10⁻²⁷) ≈ √(3.71 × 10⁶) ≈ 1926 m/s
(c) Assumptions of kinetic theory
1. Gas consists of large number of tiny particles
2. Molecules are in continuous random motion
3. Volume of molecules is negligible compared to container volume
4. No intermolecular forces except during collisions
5. Collisions are perfectly elastic
6. Time of collision is negligible compared to time between collisions
Question 4 (20 Marks)
(a) (i) Electron motion in uniform electric field
In uniform electric field E, force on electron F = eE (opposite to field direction)
Acceleration a = F/m = eE/m (constant)
Motion equations: x = vₓt, y = ½at²
Eliminating t: y = ½a(x/vₓ)² = (eE/(2mvₓ²))x²
This is equation of parabola: y ∝ x²
(a) (ii) Time for electron to reach upper plate
Distance h = 2 cm = 0.02 m
Potential difference V = 2400 V
Electric field E = V/h = 2400/0.02 = 1.2 × 10⁵ V/m
Acceleration a = eE/m = (1.6×10⁻¹⁹ × 1.2×10⁵)/(9.11×10⁻³¹) ≈ 2.11 × 10¹⁶ m/s²
Using h = ½at² ⇒ t = √(2h/a) = √(2×0.02/2.11×10¹⁶) ≈ 1.37 × 10⁻⁹ s
(b) (i) Equal charges in vacuum
F = 0.1 N, r = 50 cm = 0.5 m
Using Coulomb's law: F = (1/(4πε₀)) × q²/r²
q² = F × 4πε₀ × r² = 0.1 × (1/(9×10⁹)) × 0.25 ≈ 2.78 × 10⁻¹²
q ≈ 1.67 × 10⁻⁶ C = 1.67 μC
(b) (ii) Charges in insulating liquid
If permittivity is 10 times vacuum: ε = 10ε₀
Force F = (1/(4πε)) × q²/r² = (1/(4π×10ε₀)) × q²/r²
For same force, q² must be 10 times larger
q = √10 × 1.67 μC ≈ 5.28 μC
(c) Earth's surface charge
Electric field E = 150 N/C downward
For conducting sphere: E = σ/ε₀ at surface
σ = Eε₀ = 150 × 8.854×10⁻¹² ≈ 1.33 × 10⁻⁹ C/m²
Earth's radius R ≈ 6.4 × 10⁶ m
Surface area A = 4πR² ≈ 5.15 × 10¹⁴ m²
Total charge Q = σA ≈ 1.33×10⁻⁹ × 5.15×10¹⁴ ≈ 6.85 × 10⁵ C
(d) Final charge on capacitors
Capacitors 4μF and 2μF in series with 100V:
Equivalent capacitance C_eq = (4×2)/(4+2) = 8/6 = 1.33 μF
Total charge Q = C_eqV = 1.33×10⁻⁶ × 100 = 1.33 × 10⁻⁴ C
When disconnected and like terminals joined, charge redistributes:
Common potential V' = Total charge/Total capacitance = 1.33×10⁻⁴/(6×10⁻⁶) ≈ 22.2 V
Final charges: Q₁ = C₁V' = 4×10⁻⁶ × 22.2 ≈ 8.88×10⁻⁵ C
Q₂ = C₂V' = 2×10⁻⁶ × 22.2 ≈ 4.44×10⁻⁵ C
Question 5 (20 Marks)
(a) Charged particle in magnetic field
(i) Parallel to field: Force F = q(v×B) = 0 since sinθ=0, so no change in kinetic energy.
(ii) Perpendicular to field: Force is perpendicular to velocity, so it changes direction but not magnitude of velocity, hence kinetic energy remains constant.
(b) (i) Magnetic force between parallel wires
Distance r = 25 cm = 0.25 m
Current I₁ = I₂ = 50 A
Force per unit length F/L = (μ₀I₁I₂)/(2πr) = (4π×10⁻⁷×50×50)/(2π×0.25) = 2×10⁻³ N/m
(b) (ii) Nature of force
For currents in same direction, the force is attractive.
(c) Magnetic field of circular coil
N = 200 turns, I = 5 A, R = 5 cm = 0.05 m
(i) At center: B = μ₀NI/(2R) = (4π×10⁻⁷×200×5)/(2×0.05) = 1.26×10⁻² T
(ii) At distance x = 10 cm = 0.1 m along axis:
B = μ₀NIR²/(2(R²+x²)^(3/2)) = (4π×10⁻⁷×200×5×0.05²)/(2×(0.05²+0.1²)^(3/2)) ≈ 4.48×10⁻³ T
(d) (i) Mutual vs self induction
Mutual induction: EMF induced in one coil due to current change in nearby coil.
Self induction: EMF induced in a coil due to current change in the same coil.
(d) (ii) Mutual inductance calculation
Short coil: N₂ = 20 turns
Solenoid: n₁ = 1000 turns/m, A = 1.2×10³ m²
Mutual inductance M = μ₀n₁N₂A = 4π×10⁻⁷×1000×20×1.2×10³ ≈ 30.2 H
Question 6 (20 Marks)
(a) de Broglie wavelength of nitrogen molecule
T = 300 K
Mass of N₂ molecule = 28 × 1.674×10⁻²⁷ ≈ 4.69×10⁻²⁶ kg
v_rms = √(3kT/m) = √(3×1.38×10⁻²³×300/4.69×10⁻²⁶) ≈ 516 m/s
de Broglie wavelength λ = h/(mv) = 6.63×10⁻³⁴/(4.69×10⁻²⁶×516) ≈ 2.74×10⁻¹¹ m
Assumptions: Nitrogen behaves as ideal gas, molecular speed is RMS speed.
(b) X-ray spectra graph
(c) X-ray unit calculations
V = 100 kV = 10⁵ V, I = 20 mA = 0.02 A
(i) Electron speed:
Kinetic energy = eV = ½mv²
v = √(2eV/m) = √(2×1.6×10⁻¹⁹×10⁵/9.11×10⁻³¹) ≈ 1.87×10⁸ m/s
(ii) Minimum wavelength:
λ_min = hc/(eV) = (6.63×10⁻³⁴×3×10⁸)/(1.6×10⁻¹⁹×10⁵) ≈ 1.24×10⁻¹¹ m
(d) (i) Meaning of half-life
Half-life of 6 hours means that half of the radioactive atoms will decay in 6 hours.
(d) (ii) Radioactive decay after 20 seconds
Initial atoms N₀ = 10⁸
Half-life T = 40 s
Time t = 20 s = T/2
Fraction remaining = (1/2)^(t/T) = (1/2)^(1/2) ≈ 0.707
Atoms remaining = 0.707 × 10⁸ ≈ 7.07×10⁷
Atoms decayed = 10⁸ - 7.07×10⁷ ≈ 2.93×10⁷
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