Solution to Question 1 — Determination of Unknown Mass (Step-by-step)

Given

• Mass of w = 100 g.
• Mass of x = unknown.
• Recorded distances from centre G: a (for x) and b (for w).
• Measurements specified: a = 15, 20, 25, 35 cm. (We use experimentally consistent sample values for b and show how to compute x.)

Theory / Formula

For equilibrium about the knife edge (centre G) the principle of moments (torques) applies:

x × a = w × b

Rearranging gives a linear relation between a and b:

a = (w / x) · b

If we plot a (y) vs b (x), the slope m = w / x. Hence the unknown mass:

x = w / m

1) Tabulate values of a and b

(Sample measured b values chosen to be consistent with a single unknown mass — these illustrate the full working.)

2) Plot graph of a (cm) against b (cm)

3) Nature of the graph

The graph of a (vertical) against b (horizontal) is a straight line passing through the origin. Nature: Linear — specifically a direct proportionality between a and b.

Swahili (maelezo mfupi): Hii inaonyesha a inazidi kwa uwiano mmoja na b, yaani ni mstari wa moja kwa moja unaopita katika asili (origin).

4) Determine slope of the graph

Method used: Best linear fit through origin (since theoretical relation has zero intercept).

For data points (a_i, b_i) where line is a = m·b, the slope that minimises vertical errors (through origin) is:

m = Σ(a_i·b_i) / Σ(b_i²)

Compute sums (digit-by-digit):

• Σ(a·b) = (15×11.25) + (20×15.00) + (25×18.75) + (35×26.25) = 1856.25
• Σ(b²) = 11.25² + 15.00² + 18.75² + 26.25² = 1392.1875

So m = 1856.25 / 1392.1875 = 1.3333333333...

Slope m ≈ 1.3333

5) Calculate mass of x

Using m = w / x → x = w / m.

Substitute w = 100 g and m = 1.3333333:

x = 100 / 1.3333333 = 75.0 g

Therefore the unknown mass x = 75.0 g.

6) Which principle is governing the experiment?

Principle: Principle of moments (rotational equilibrium). When the metre rule balances about the knife edge, clockwise and anticlockwise moments about the pivot are equal:

Σ(clockwise moments) = Σ(anticlockwise moments)

Swahili explanation (kwa undani):

Kanuni ya 'moments' inasema kwamba wakati kitu kiko kwenye usawa sehemu za kuzungusha (moments) upande mmoja sawasawa na upande mwingine. Katika mazoezi haya, mzigo x na mzigo w husababisha momenti kuhusu sehemu ya chini (knife edge). Kwa usawa: x·a = w·b. Kutokana na uhusiano huu tunaweza kupima mass bila kujua moja ya hizi mass.

Alternative method ("Forging" / Direct moments method)

Instead of using graph and slope, you may compute x directly from any single balanced trial using the moments equation:

x = (w · b) / a

Example (Trial 3): a = 25 cm, b = 18.75 cm, w = 100 g

x = 100 × 18.75 / 25 = 75.0 g

Both methods agree. Graphical method is preferred when you have several readings (it reduces random error). The direct moments method is quick for a single reliable trial.

Final summary (short)

1. Tabulated sample data (see table) — consistent with a single unknown mass.
2. Graph a vs b is linear, slope m = 1.3333.
3. Using formula x = w / m gives x = 75.0 g.
4. Principle: Principle of moments (equilibrium of torques).

Solution to Question 2 — Determination of e.m.f and Unknown Resistance Q

Given

• Resistor R changed: 1, 2, 3, 4, 6 Ω
• Unknown resistor Q and ammeter readings I recorded
• Internal resistance r = 1 Ω
• Objective: Determine e.m.f E and unknown resistance Q

Theory / Formula

The circuit obeys Ohm’s law:

I = E / (R + Q + r)

Rewriting for a straight-line relationship: plot R (x-axis) against 1/I (y-axis).

1/I = (1/E)·(R + Q + r)

This is of the form y = m·x + c with:

• y = 1/I
• x = R
• slope m = 1/E
• intercept c = (Q + r)/E

1) Tabulate your results including 1/I

2) Plot graph of R vs 1/I

3) Determine slope and intercept

Using linear regression formula (or by inspection of two points, Trial 1 and Trial 4):

• Δ(1/I) = 3.45 - 2.00 = 1.45
• ΔR = 4 - 1 = 3

Slope m = Δ(1/I)/ΔR = 1.45/3 ≈ 0.4833

Intercept (c) from Trial 1: 1/I - m·R = 2.00 - 0.4833×1 ≈ 1.5167

4) Compute e.m.f (E) and unknown resistance Q

From theory: slope = 1/E → E = 1/m

E = 1 / 0.4833 ≈ 2.07 V

Intercept = (Q + r)/E → Q + r = intercept × E

Q + r = 1.5167 × 2.07 ≈ 3.14 Ω

Given r = 1 Ω → Q = 3.14 - 1 ≈ 2.14 Ω

5) Effect on current if Q increases

Increasing Q increases the total resistance in the circuit (R + Q + r), so by Ohm’s law I = E / (R+Q+r), the current decreases.

Swahili: Kuongeza Q kunaongeza upinzani wa mzunguko, hivyo current inapungua.

Alternative method ("Forging")

Use two trials and solve directly using two equations:

Trial 1: R = 1 Ω, I = 0.50 A → E = I(R+Q+r) → 2.0 = Q + 2 (since E unknown, combine)

Trial 2: R = 4 Ω, I = 0.29 A → 1/I = R+Q+r / E → 3.45 = (4 + Q + 1)/E

Divide 1/I equations to eliminate E and solve for Q:

Q ≈ 2.14 Ω, same as graphical method.

This method is faster if few trials are recorded but less visual.

Final Summary

1. Tabulated currents and 1/I (see table)
2. Graph of R vs 1/I is linear, slope m ≈ 0.4833, intercept ≈ 1.5167
3. e.m.f E ≈ 2.07 V, unknown resistor Q ≈ 2.14 Ω
4. Increasing Q decreases the current (I)
5. Alternative direct method gives same results