2021-PA-Physics 2A Practical A SOLUTION - MWALA-LEARN

Objectives: 2021-PA-Physics 2A Practical A SOLUTION - MWALA-LEARN

Physics 2A — Q1 Solution (Compound Pendulum)

QUESTION 1 — Compound Pendulum (Worked solution and method)

Below is a clear step-by-step solution for Question 1. The solution uses the standard physical-pendulum relation and shows how to obtain the slope S, intercept b, and the physical meaning of b. For clarity we include a worked example using a set of sample (typical) measured times — treat these as an example only. Replace measured t values from your experiment to get your final answers.

1. Theoretical background (English)

For a physical (compound) pendulum the period T of small oscillations is given by

T = 2π √( I / (m g d) )

where I is the moment of inertia about the pivot, m the mass, g the acceleration due to gravity and d the distance from pivot to the centre of mass. With suitable algebra for the particular pendulum used this relation may be written in the experimental (linear) form

T2 = S (a + b)

Here a is the measured distance (from point N to the suspension), S is the slope of the graph of T2 versus a and b is a constant that depends on the geometry of the pendulum (it is an equivalent-length correction). The relation above is the linear form used in the question.

Important derived relations (how we use the graph)
  • From the plotted straight line: T2 = S·a + (S·b). So the straight-line slope equals S, and the y-intercept equals S·b.
  • Therefore b = (y-intercept) / S.
  • The slope S is related to g (when a is expressed in metres) by the known factor for small oscillations. If the theoretical derivation gives S = 4π² / g, then you can find g as g = 4π² / S.
2. Step-by-step experimental procedure to get S and b (what to write in the answer booklet)
  1. Record the measured times t for 20 oscillations at each value of a (20, 40, 60, 80, 100 cm).
  2. Compute the period for each: T = t / 20.
  3. Compute T2 for each trial.
  4. Plot a graph of T2 (vertical axis) against a (horizontal axis). Draw the best-fit straight line through the points.
    • Find S as the gradient: S = Δ(T2) / Δa (use two well-separated points on the best-fit line or do a linear regression).
    • Find the y-intercept (value of T2 at a = 0) — call it c.
    • Compute b = c / S (ensure units: if a used in cm then b will be in cm; convert to metres if you need g using metres).
3. Worked numerical example (example measured times)

Note: these are example times to show the numerical process. Replace with your own measured t values.

Triala (cm)t for 20 osc (s) — exampleT = t/20 (s)T2 (s2)
12018.000.9000.8100
24025.201.2601.5876
36031.001.5502.4025
48035.801.7903.2041
510040.202.0104.0401
Linear fit (example)

Plotting the example values and performing a least-squares linear fit (T2 = S·a + c) gives:

  • S (slope) = 0.0403835 s2/cm
  • y-intercept c = -0.01415 s2

Because a was taken in centimetres for the plot, convert slope to s2/m before using the formula for g (1 m = 100 cm):

Sm = Scm × 100 = 0.0403835 × 100 = 4.03835 s2/m

Now use the theoretical relation (for small oscillations) Sm = 4π² / g to estimate g:

g = 4π² / Sm = 4π² / 4.03835 ≈ 9.776 m/s²

Compute b (example)

From T2 = S(a + b) we have the y-intercept = S·b = c, so

b = c / S.

Using example numbers (both in cm units):

b = (-0.01415) / 0.0403835 ≈ -0.3504 cm

A small negative b indicates experimental scatter and measurement uncertainties (in a real experiment you expect a small positive correction related to geometry). If your measured t values are accurate the intercept should be small; an exactly zero or small value is acceptable given experimental errors.

4. Final answers and how to state them in the exam
  1. Plot: Draw T2 (y) vs a (x) and draw best-fit line — show points and the line clearly (use a ruler). Label axes with units (s2 and cm or m).
  2. Slope S: From the graph S = gradient = Δ(T2)/Δa. (In the example S = 0.04038 s2/cm = 4.03835 s2/m.)
  3. Intercept parameter b: From the graph y-intercept = S·b so b = (y-intercept)/S. (In the example b ≈ -0.35 cm.)
  4. Physical meaning of b (Swahili for precision):

    Kwa Kiswahili — b ni marekebisho ya urefu unaoonyesha mbali kidogo kati ya nukta ya wasimamizi (pivot/suspension) na urefu sawa wa pendulum ya msongamano (centre of oscillation). Kwa maneno rahisi, b ni kiasi cha kurekebisha ambacho kinafanya period ya pendulum ya mwili (compound pendulum) ifanane na period ya pendulum rahisi ya urefu L = a + b. Hii b inategemea umbo (geometry) na mgawanyiko wa masa (moment of inertia) ya pendulum.

5. Practical notes & sources of error (short)
  • Ensure displacements are small (small-angle approximation) so the theory applies.
  • Measure t for many oscillations (20 is good) to reduce reaction-time error.
  • Minimise friction at the pivot and ensure the thread does not stretch.
  • Random scatter in timing and parallax in reading marks can produce small negative/positive intercepts; report uncertainties if required.
6. SVG: Example graph showing points and best-fit line (from example data)
0 20 40 60 80 100 0.0 0.8 1.6 2.4 3.2 4.0 Example graph: T² vs a (points & best-fit line) Vertical: T² (s²) Horizontal: a (cm)
Physics 2A — Q1 Option 2 (Simple)

QUESTION 1 — Option 2 (Simple forging / two-point method)

We use two well-separated measured points on the T² vs a plot and compute slope S, intercept c (y-intercept), parameter b, and estimate g. Example measured data (from earlier worked example):

  • a₁ = 20 cm, T₁² = 0.8100 s²
  • a₂ = 100 cm, T₂² = 4.0401 s²
Step 1 — Calculate slope S (simple Δ/Δ)

ΔT² = T₂² − T₁² = 4.0401 − 0.8100 = 3.2301 (s²)
Δa = a₂ − a₁ = 100 − 20 = 80 (cm)

S = ΔT² / Δa = 3.2301 / 80 = 0.04037625 s² / cm

Step 2 — Find y-intercept c (use point 1)

c = T₁² − S·a₁ = 0.8100 − (0.04037625 × 20)

c = 0.8100 − 0.807525 = 0.0024750 s²

Step 3 — Compute parameter b

From T² = S(a + b) ⇒ y-intercept = S·b = c ⇒ b = c / S

b = 0.0024750 / 0.04037625 ≈ 0.0612984 cm

(Since a was in cm, b is also in cm.)

Step 4 — (Optional) Estimate g using S in SI units

Convert slope to s²/m: Sm = Scm × 100 = 0.04037625 × 100 = 4.037625 s² / m

Using theoretical relation Sm = 4π² / g ⇒ g = 4π² / Sm

g = 4π² / 4.037625 ≈ 9.7776 m/s²

Results (Option 2 — simple forging)
  • S (slope) = 0.04037625 s² / cm (→ 4.037625 s² / m)
  • y-intercept c = 0.0024750 s²
  • b = 0.0612984 cm
  • Estimated g = 9.7776 m/s² (from Sm)
Meaning of b — short Swahili explanation

Kiswahili: b ni marekebisho ya urefu yanayotokana na umbo na mgawanyiko wa masa wa pendulum (inayofanya pendulum ya mwili ifanane na pendulum rahisi ya urefu L = a + b). Ni thamani ndogo inayoonyesha utofauti wa 'centre of oscillation' na nukta N iliyopimwa.

Physics 2A — Q2 Solution (Electrical Measurements)

QUESTION 2 — Electrical Measurements (Two solution options)

Below are two clear solutions for Question 2. Option A (Normal) uses correct experimental logic and honest data. Option B (Forged) shows a manipulated (forged) dataset — we solve it the same way and then highlight how the forgery can be detected. Both options include full steps, worked numbers, vivid real-life examples and SVG graphs.

Theoretical derivation (common to both options)

From the setup: the unknown resistor R is in parallel with a 4 Ω resistor. The voltmeter reads V across the parallel combination, and the ammeter reads the total current I delivered by the cells to the parallel network. Let the equivalent resistance of the parallel block be Req so:

Req = V / I

Also: Req = (R × 4) / (R + 4)

Equate and rearrange:
V / I = 4R / (R + 4).

If we take reciprocals and rearrange to get a simple straight-line relation between 1/I (vertical axis) and 1/V (horizontal axis) we find that 1/I is proportional to 1/V — i.e. the plot passes through the origin with slope m equal to:

m = Δ(1/I) / Δ(1/V) = 4R / (R + 4)

Thus solve for R in terms of the measured slope m:

R = 4m / (4 - m)

Note: m must be < 4 for R to be positive finite. If m ≥ 4 that indicates inconsistent/invalid data or a calculation error.

Option A — Normal (Correct data)
1) Chosen true value for demonstration

For this worked example we assume the true unknown resistance is R = 6.0 Ω. Then the theoretical slope is:

m = 4R / (R + 4) = (4 × 6) / (6 + 4) = 24 / 10 = 2.4 Ω

Therefore the expected relation between measured V and I is (from 1/I = m × 1/V) V = m × I (i.e. V is proportional to I). We'll generate experimental voltmeter readings following V = 2.4 I and then show the analysis you should do in the exam.

2) Example measured table (honest readings)
TrialI (A)V (V) = 2.4·I (example)1/I (A-1)1/V (V-1)
10.100.24010.0004.1667
20.300.7203.33331.3889
30.501.2002.00000.8333
40.701.6801.42860.5952
50.902.1601.11110.46296
3) Plot and slope

Plot 1/I (vertical) against 1/V (horizontal). Because V = m·I the relation reduces to 1/I = m × 1/V so the straight line passes through the origin and the slope equals m.

0.0 0.5 1.0 1.5 2.0 2.5 3.0 0 2 4 6 8 10 Option A: 1/I vs 1/V (normal) Slope m = 2.4 Ω
4) Compute slope & determine R (from table)

Because our chosen data satisfy exactly V = m·I, a least-squares fit (or using two widely separated points) gives m = 2.4 Ω

R = 4m / (4 - m) = (4 × 2.4) / (4 - 2.4) = 9.6 / 1.6 = 6.0 Ω

5) How to present in the exam (steps)
  1. Tabulate I and V, compute 1/I and 1/V.
  2. Plot 1/I (y) vs 1/V (x). Draw best-fit straight line through points; it should pass near the origin.
  3. Measure slope m = Δ(1/I) / Δ(1/V) (use two distant points on best-fit line).
  4. Compute R by R = 4m / (4 - m). Show units and round sensibly.
6) Vivid real-life example

Suppose you're a technician testing two batteries powering two load branches (one heater R and one lamp 4 Ω). You measure total current with an ammeter and voltage across the loads with a voltmeter at different rheostat settings. If the plot of 1/I vs 1/V is a straight line through the origin, then the method is valid and you can calculate the unknown heater resistance directly. This mirrors how electronics labs characterise components quickly with load sweeps.

Option B — Forged / Manipulated data (demonstration)
1) What we mean by "forging"

Here "forging" means someone altered voltmeter readings (intentionally or accidentally) so that the computed result for R will be wrong. We show a plausible forgery: add a small systematic offset (+0.12 V) to each voltmeter reading and also introduce one bad outlier measurement. We then solve and show how to detect the forgery.

2) Forged dataset (example)
TrialI (A)V (V) — forged1/I (A-1)1/V (V-1)
10.100.44010.0002.2727
20.300.9003.33331.1111
30.501.3502.00000.7407
40.701.8201.42860.5495
50.902.3001.11110.4348
3) Fit line and determine slope

If you plot 1/I vs 1/V for these forged points and determine the slope (use two distant points or least-squares), you'll obtain an apparent slope mf (we compute below):

Take point 1 and point 5 as well-separated: slope mf ≈ (1/I1 - 1/I5) / (1/V1 - 1/V5) = (10 - 1.1111) / (2.2727 - 0.4348) ≈ 8.8889 / 1.8379 ≈ 4.839 Ω

Compute forged R':

R' = 4 mf / (4 - mf) = (4 × 4.839) / (4 - 4.839) = 19.356 / (-0.839) ≈ -23.06 Ω

A negative value (or very large positive) indicates inconsistent/invalid data — here the forged slope mf > 4 which makes the formula invalid. That is a red flag: the dataset cannot represent a physical positive resistance with the assumed arrangement.

4) How to spot the forgery (practical tips)
  • Check linearity through origin: For correct data, 1/I vs 1/V should be linear and pass near the origin. Forged data often shifts points so the line does not pass through origin or produces impossible slopes (m ≥ 4).
  • Compute R from each trial: Use R = 4V / (4I - V) derived directly from the single-reading formula. For honest data these values should cluster around one number. Large scatter or negative/huge values indicate problems.
  • Plot residuals: Subtract the predicted 1/I from measured 1/I and inspect residuals — a forged dataset will show systematic bias, not random noise.
  • Repeat measurements: Real experiments include repeat measurements; consistent repeats reduce chance of forgery being undetected.
5) Example: compute R from each forged reading individually (to show inconsistency)

Using R = 4V / (4I - V) for each row:

TrialIVR = 4V / (4I - V)
10.100.4404×0.44/(0.4 - 0.44) = 1.76 / (-0.04) = -44.0 Ω
20.300.9003.6 / (1.2 - 0.9) = 3.6 / 0.3 = 12.0 Ω
30.501.3505.4 / (2.0 - 1.35) = 5.4 / 0.65 = 8.3077 Ω
40.701.8207.28 / (2.8 - 1.82) = 7.28 / 0.98 = 7.4286 Ω
50.902.3009.2 / (3.6 - 2.3) = 9.2 / 1.3 = 7.0769 Ω

These single-trial R values show wild scatter (from -44 Ω to 12 Ω), which is impossible if R is a single fixed resistor; this strongly suggests data tampering or measurement error. In contrast Option A produced consistent R ≈ 6 Ω for all trials.

6) Real-life example of detect-forgery scenario

Imagine a student wants to claim a particular resistor has value 7 Ω to match a report. They might (dishonestly) adjust voltmeter readings upward. However, when the lab instructor computes R from individual readings, results scatter widely and physical checks (e.g., using an ohmmeter or measuring V vs I proportionality) catch the forgery. In industry, technicians flag sensors that give inconsistent load-sweep curves — those sensors are replaced or retested.

7) What to do if you suspect forged/incorrect data (exam answer phrase)

Write: "Data shows inconsistency: slope m ≥ 4 or large scatter in R computed from single readings. Repeat measurements, check meter zero, ensure correct connections and check internal resistances of cells."

Reference Book: N/A

Author name: SIR H.A.Mwala Work email: biasharaboraofficials@gmail.com
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