Alternative A – Question 1: Springs & Oscillation

Experimental Procedure and Sample Results

Apparatus: helical spring, set of masses, retort stand, clamp and boss, metre rule, stopwatch.

Procedure

• Fix the spring securely to the clamp on the retort stand so that it hangs vertically. Suspend the mass hanger and add masses as required.
• Set the mass m to 150 g (0.150 kg). Displace the mass slightly downward and release it to oscillate vertically.
• Using the stopwatch, measure the time for 20 complete oscillations (to reduce timing error). Record the time t.
• Calculate the period T of oscillation: T = t / 20.
• Repeat the procedure for masses 200 g, 250 g, 300 g and 350 g.

Sample Table of Results

m (g)	Time for 20 oscillations, t (s)	Period, T = t/20 (s)	T2 (s2)
150	15.6	0.78	0.61
200	17.9	0.90	0.81
250	20.0	1.00	1.00
300	21.9	1.10	1.21
350	23.6	1.18	1.39

Graph of m against T² and Slope

On a suitable graph sheet:

• Choose a convenient scale so that the points spread well on the graph paper.
• Plot T² (s²) on the horizontal (x-) axis.
• Plot m (g) on the vertical (y-) axis.
• Mark each pair of values (T², m) as a point and draw the best-fit straight line through the points.

1(a)(xvii) – Slope s₁

1(a)(xvii) — Slope s₁:

s₁ = Change in m / Change in T²

s₁ = (m₂ − m₁) / (T₂² − T₁²)

1(a)(xix) – Slope s₂

1(a)(xix) — Slope s₂:

s₂ = Change in m / Change in T₀²

s₂ = (m₂ − m₁) / (T₀₂² − T₀₁²)

1(a)(xx) – y

1(a)(xx) — y:

y = s₂/s₁

Substitute calculated values of s₁ and s₂ from the graph and evaluate.

Using two approximate points from the sample table for illustration, say (T² = 0.61, m = 150 g) and (T² = 1.39, m = 350 g):

• Slope S = (350 - 150) / (1.39 - 0.61) = 200 / 0.78 ≈ 256 g s-2.

Precautions – 1(a)(xxi) (Pick Any Two)

• Ensure the oscillations are small and strictly vertical.
• Start and stop the stopwatch accurately as the mass passes the same reference point.
• Avoid parallax error when taking readings.
• Take readings in a draught-free environment.
• Repeat readings and obtain the average value.

Question 1(b): Answers

1(b)(i): As the number of springs increases, the frequency of oscillation increases.

1(b)(ii):

Given:

• k = 125 Nm⁻¹
• F = 8.0 N

Using Hooke's Law:

F = ke

e = F/k

e = 8.0/125

e = 0.064 m

Answer: e = 0.064 m

Alternative A – Question 2: Specific Heat Capacity

Experimental Procedure and Sample Results

Apparatus: two beakers A and B, thermometers, measuring cylinder, pendulum bob (copper), heater or Bunsen burner, string, stirrer.

Procedure

• Measure 100 cm³ of water into beaker A and 300 cm³ of water into beaker B (1 cm³ of water ≈ 1 g).
• Heat beaker A strongly and record the steady temperature of the hot water as θ₁.
• Record the initial temperature of water in beaker B as θ₂.
• Quickly pour the hot water from A into B, stir well and record the final steady temperature θ₃.
• Repeat the procedure for other combinations of hot/cold water volumes: 150/300, 200/300, 250/300 and 300/300 cm³.

Sample Table of Results

Volume in A (cm³)	Volume in B (cm³)	θ₁ (°C)	θ₂ (°C)	θ₃ (°C)
100	300	80	28	41
150	300	82	28	47
200	300	85	28	52
250	300	88	28	57
300	300	90	28	61

Question 2(a): Graph, Slope and Precautions

2(a)(xvii) — Slope s

2(a)(xvii) — Slope s:

s = Change in P / Change in Q

s = (P₂ − P₁) / (Q₂ − Q₁)

2(a)(xviii) — Precautions

2(a)(xviii) — Precautions (Pick Any Two):

• Stir the water continuously to ensure uniform temperature.
• Take thermometer readings at eye level to avoid parallax error.
• Transfer the heated pendulum quickly into beaker B to minimize heat loss.
• Ensure the thermometer does not touch the sides or bottom of the beaker.
• Read the temperature immediately after it becomes steady.

Question 2(b): Theory and Calculations

2(b)(i) — Specific Heat Capacity Definition:

Specific heat capacity is the amount of heat energy required to raise the temperature of unit mass (1 kg) of a substance by 1 K (or 1°C).

2(b)(ii) — Calculation:

Given:

• Mass of copper ball, m₁ = 50 g = 0.05 kg
• Specific heat capacity of copper, c₁ = 400 J kg⁻¹ K⁻¹
• Initial temperature of copper = 100°C
• Initial temperature of water = 23°C
• Final temperature of mixture = 62°C
• Specific heat capacity of water, c₂ = 4200 J kg⁻¹ K⁻¹

Heat lost by copper = Heat gained by water

m₁c₁(100 − 62) = m₂c₂(62 − 23)

0.05 × 400 × 38 = m₂ × 4200 × 39

760 = 163800m₂

m₂ = 760/163800

m₂ = 0.00464 kg

Mass of water = 0.00464 kg or 4.64 g

Alternative A – Question 3: Resistance of Wires

Experimental Procedure for Wires A and B

Apparatus: resistance wires A and B, rheostat, voltmeter, ammeter, connecting wires, micrometer screw gauge, metre rule, switch, battery.

Procedure for Wire A

• Connect the circuit with the battery, switch, rheostat and ammeter in series, and connect the voltmeter in parallel across wire A.
• Close the switch and adjust the rheostat to obtain a small current. Record the corresponding readings of potential difference V and current I.
• Vary the rheostat to obtain at least five different pairs of (V, I) readings, ensuring the wire does not overheat.
• Record all readings in a table.

3(a)(iv) — Table of results (E = 3.00 V)

S/N	R (Ω)	I (A)	1/I
1	2.00	0.60	5.0000
2	4.00	0.42	7.1429
3	6.00	0.32	9.3750
4	8.00	0.28	10.7143
5	10.00	0.22	13.6364

Graph of V against I and Slope (R_A)

On graph paper:

• Plot current I (A) on the horizontal (x-) axis.
• Plot potential difference V (V) on the vertical (y-) axis.
• Mark the points (I, V) and draw the best-fit straight line through them.

The slope of this line gives the resistance R_A:

R_A = slope = ΔV / ΔI

Using two points e.g. (I = 0.10 A, V = 1.0 V) and (I = 0.50 A, V = 5.0 V):

R_A = (5.0 − 1.0) / (0.50 − 0.10) = 4.0 / 0.40 = 10 Ω

Similarly, repeat the entire process for wire B to obtain R_B.

3(a)(vi) — From the graph

From the graph, r = 2.6 Ω.

3(a)(vii) — Precautions (Pick Any Two)

• Neat and tight connections
• Avoid error due to parallax on the ammeter

Diameter Measurements and Resistivity

Use the micrometer screw gauge to measure the diameter of each wire at three different positions, then find the mean.

Sample readings for Wire A

Position	Diameter (mm)
1	0.50
2	0.52
3	0.51

Mean diameter:

d = (0.50 + 0.52 + 0.51) / 3 = 1.53 / 3 = 0.51 mm

Convert to metres:

d = 0.51 mm = 0.51 × 10⁻³ m

Cross-sectional area A of the wire:

A = (π d²) / 4

d² = (0.51 × 10⁻³)² = 0.2601 × 10⁻⁶ m²

A = π × 0.2601 × 10⁻⁶ / 4 ≈ 3.1416 × 0.0650 × 10⁻⁶ ≈ 0.204 × 10⁻⁶ m² = 2.04 × 10⁻⁷ m² (approx.)

Let the length of the wire, L, be 1.0 m (measured with the metre rule). From the graph, we obtained R_A ≈ 10 Ω.

The resistivity ρ of the material of wire A is given by:

ρ = R A / L

Substitute:

ρ = 10 × (2.04 × 10⁻⁷) / 1.0 = 2.04 × 10⁻⁶ Ω m

Resistivity of wire A ≈ 2.0 × 10⁻⁶ Ω m (sample value).

Question 3(b): Short Answers

(i) Function of a rheostat in the circuit

A rheostat is used to vary or control the current in the circuit by changing the resistance, thereby adjusting the potential difference across the component under test.

3(b)(i) — Factors affecting the resistance of a wire:

• Temperature
• Nature of wire (resistivity)
• Cross-sectional area
• Length of wire

3(b)(ii) — Calculation:

Given: lost voltage = Ir = 3 V (note: this should be the total e.m.f. E = 3 V) and I = 2 A.

2r = 3

r = 3 / 2 = 1.5 Ω

Internal resistance, r = 1.5 Ω.

Alternative B – Question 2: Heat Capacity and Heating of Water

(i) Definitions

Heat capacity of a body is the quantity of heat required to raise the temperature of the whole body by 1 K (or 1°C).

Specific heat capacity of a substance is the quantity of heat required to raise the temperature of 1 kg of the substance by 1 K (or 1°C).

(ii) Difference Between Boiling and Evaporation

• Boiling occurs at a fixed temperature (boiling point) for a given pressure and takes place throughout the liquid. Bubbles form in the liquid and rise to the surface.
• Evaporation occurs at all temperatures below the boiling point and takes place only at the surface of the liquid. It is usually a slower process and does not involve bubble formation in the liquid.

(iii) Energy Required to Heat Water

Given:

• Mass of water, m = 2 kg
• Initial temperature, θ₁ = 20°C
• Final temperature, θ₂ = 100°C
• Specific heat capacity of water, c = 4200 J kg⁻¹ K⁻¹

Temperature change:

Δθ = θ₂ − θ₁ = 100 − 20 = 80 K

Heat energy required:

Q = m c Δθ

Q = 2 × 4200 × 80

Q = 2 × 336,000 = 672,000 J

Energy required = 6.72 × 10⁵ J.

Alternative B – Question 3: Ohm's Law and Resistances

(i) Statement of Ohm's Law

Ohm's Law states that the current flowing through a metallic conductor is directly proportional to the potential difference across its ends, provided the temperature and other physical conditions remain constant.

(ii) Resistance of a Wire

Given:

• Length of wire, L = 2 m
• Diameter, d = 0.5 mm = 0.5 × 10⁻³ m
• Resistivity, ρ = 1.7 × 10⁻⁸ Ω m

First, find the cross-sectional area A of the wire:

A = (π d²) / 4

d² = (0.5 × 10⁻³)² = 0.25 × 10⁻⁶ m²

A = π × 0.25 × 10⁻⁶ / 4 = π × 0.0625 × 10⁻⁶ ≈ 3.1416 × 6.25 × 10⁻⁸ ≈ 1.96 × 10⁻⁷ m²

Resistance is given by:

R = ρ L / A

Substitute:

R = (1.7 × 10⁻⁸ × 2) / (1.96 × 10⁻⁷)

R = 3.4 × 10⁻⁸ / 1.96 × 10⁻⁷

Divide the numbers and powers of ten:

R ≈ (3.4 / 1.96) × 10⁻¹ ≈ 1.73 × 0.1 ≈ 0.173 Ω

Resistance of the wire ≈ 0.17 Ω.

(iii) Combined Resistance of Two Parallel Resistors

Given two resistors of 6 Ω and 3 Ω connected in parallel.

For parallel connection:

1 / R_eq = 1 / R₁ + 1 / R₂

1 / R_eq = 1 / 6 + 1 / 3

1 / R_eq = 1 / 6 + 2 / 6 = 3 / 6 = 1 / 2

Therefore:

R_eq = 2 Ω

Combined resistance = 2 Ω.