NECO Physics Practical Questions and Answers 2026
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*NECO 2026 PHYSICS PRACTICAL (PAPER I)* *COMPLETE ANSWERS – QUESTIONS 1, 2 & 3* =========================================================== QUESTION 1: SPRING / MASS OSCILLATION =========================================================== (a) Experimental Procedure and Sample Data Mass hanger = 40 g Additional masses = 20 g, 40 g, 60 g, 80 g Total mass M (g) = 40, 60, 80, 100, 120 Time for 20 oscillations (t) – SAMPLE DATA: M = 40 g → t = 10.0 s M = 60 g → t = 12.2 s M = 80 g → t = 14.1 s M = 100 g → t = 15.8 s M = 120 g → t = 17.3 s Period T = t/20 T² = T × T Sample calculation for M = 40 g: T = 10.0/20 = 0.50 s T² = 0.50 × 0.50 = 0.2500 s² Tabulated Results (Sample) M (g) t (s) T (s) T² (s²) 40 10.0 0.50 0.2500 60 12.2 0.61 0.3721 80 14.1 0.705 0.4970 100 15.8 0.79 0.6241 120 17.3 0.865 0.7482 (x) Plot a graph of M on vertical axis and T² on horizontal axis In your answer booklet, plot M (g) on the y-axis and T² (s²) on the x-axis. Plot points and draw the best-fit straight line through the origin. (xi) Determine the slope S of the graph Using two points on the best-fit line (SAMPLE): Point A: T² = 0.25 s², M = 40 g Point B: T² = 0.75 s², M = 120 g Slope S = (120 - 40) / (0.75 - 0.25) = 80 / 0.50 = 160 g/s² Answer: S = 160 g/s² (xii) Evaluate K = 4π²S π = 3.142 4π² = 4 × (3.142)² = 4 × 9.872 = 39.48 K = 39.48 × 160 = 6,316.8 g/s² = 6.32 kg/s² (since 1000g = 1kg) Answer: K = 6.32 kg/s² (or N/m) (xiii) State two precautions taken to ensure accurate results 1. I ensured that the spring oscillated vertically without touching the retort stand to prevent energy loss. 2. I started the stopwatch at the start of an oscillation and counted from zero to reduce timing error. (b)(i) State: I. The two characteristics of oscillatory motion (1 mark) 1. It is periodic – the motion repeats itself after a fixed time interval. 2. It has a restoring force – a force that tries to bring the body back to its equilibrium position. II. Two examples of bodies exhibiting oscillatory motion (1 mark) 1. A simple pendulum swinging 2. A mass suspended on a spiral spring vibrating (b)(ii) A body of mass 0.5 kg is suspended vertically on a spiral spring. If the extension produced is 0.2 m, calculate the force constant of the spring. [g = 10 ms⁻²] (2 marks) Force F = mg = 0.5 × 10 = 5 N Using Hooke's Law: F = k e k = F/e = 5/0.2 = 25 N/m Answer: 25 N/m =========================================================== https://whatsapp.com/channel/0029Vb8TtA8EVccNvTRmUI2u https://t.me/necovipanswer Neco vip 👆👆 *REACT, SHARE AND FORWARD TO EVERYONE ❤️🔥* FOR NECO EXPO 24HRS BEFORE EXAM, MESSAGE ME ON TELEGRAM👇👇: https://t.me/Officialkingjay1 JOIN OUR VIP CHANNEL 👇🏻👇🏻 https://t.me/allexamexporunz ALL STATES(TYPES) QUESTIONS 👇👇 https://t.me/freewassceditto JOIN OUR VIP CHANNEL 👇🏻👇🏻 https://t.me/examgroupng QUESTION 2: RECTANGULAR GLASS PRISM (REFRACTION) =========================================================== (a) Experimental Procedure and Sample Data Angles of incidence (i): 20°, 30°, 40°, 50°, 60° φ = 90° - i Z = l cos φ y = measured displacement from normal Sample Data i (°) φ (°) y (cm) l (cm) Z = l cos φ (cm) 20 70 1.50 6.50 6.50 × 0.342 = 2.22 30 60 2.20 6.80 6.80 × 0.500 = 3.40 40 50 2.80 7.00 7.00 × 0.643 = 4.50 50 40 3.30 7.20 7.20 × 0.766 = 5.52 60 30 3.70 7.40 7.40 × 0.866 = 6.41 (xiv) Plot a graph of y on vertical axis and Z on horizontal axis Plot y (cm) on the y-axis and Z (cm) on the x-axis. Draw the best-fit straight line. (xv) Determine the slope S of the graph Using two points on the line (SAMPLE): Point A: Z = 2.22 cm, y = 1.50 cm Point B: Z = 6.41 cm, y = 3.70 cm Slope S = (3.70 - 1.50) / (6.41 - 2.22) = 2.20 / 4.19 = 0.525 Answer: S = 0.525 (xvi) Evaluate K = 1/S K = 1/0.525 = 1.90 Answer: K = 1.90 (xvii) State two precautions taken to ensure accurate results 1. I ensured that the pins were vertical and fixed firmly on the paper. 2. I placed the pins at least 4 cm apart to reduce parallax error in aligning them. (b)(i) State: I. The two types of reflections (1 mark) 1. Regular reflection (specular reflection) – reflection from a smooth surface. 2. Diffuse reflection – reflection from a rough surface. II. Why light ray changes direction when it moves from air to glass (1 mark) Light ray changes direction because it changes speed. Glass is denser than air, so light slows down and bends towards the normal. This is due to the change in refractive index. (b)(ii) Calculate the speed of light in water, if the speed of light in air is 3.0 × 10⁸ ms⁻¹. [Refractive index of water = 1.33] (2 marks) n = speed in air / speed in medium 1.33 = (3.0 × 10⁸) / v v = (3.0 × 10⁸) / 1.33 = 2.26 × 10⁸ ms⁻¹ Answer: 2.26 × 10⁸ ms⁻¹ =========================================================== QUESTION 3: ELECTRIC CIRCUIT (OHM'S LAW) =========================================================== (a) Experimental Procedure and Sample Data E = Open circuit voltage (with key open) = 2.0 V (sample) R = 1 Ω, 2 Ω, 3 Ω, 4 Ω, 5 Ω Sample Data R (Ω) V (V) I (A) 1 1.20 1.20 2 1.40 0.70 3 1.55 0.52 4 1.64 0.41 5 1.70 0.34 (viii) Plot a graph of V on vertical axis and I on horizontal axis Plot V (V) on the y-axis and I (A) on the x-axis. Draw the best-fit straight line. (ix) Determine the slope S of the graph Using two points on the line (SAMPLE): Point A: I = 0.34 A, V = 1.70 V Point B: I = 1.20 A, V = 1.20 V Slope S = (1.70 - 1.20) / (0.34 - 1.20) = 0.50 / (-0.86) = -0.581 The negative slope represents the internal resistance of the cell. Answer: S = -0.581 Ω (x) Determine the intercept C on the vertical axis From the graph, the intercept on the vertical axis (where I = 0) is the open circuit voltage E. Answer: C = 2.0 V (xi) Evaluate K = E/C K = 2.0 / 2.0 = 1.0 Answer: K = 1.0 (xii) State two precautions taken to ensure accurate results 1. I ensured all connecting wires were tight and free from corrosion to avoid additional resistance. 2. I closed the key only when taking readings to prevent unnecessary heating of the wires. (b)(i) Differentiate between a resistor and a capacitor (2 marks) A resistor is a component that opposes the flow of electric current and dissipates electrical energy as heat. Its unit is the ohm (Ω). A capacitor is a component that stores electrical energy in an electric field. It consists of two conductors separated by an insulator. Its unit is the farad (F). (b)(ii) Two identical cells each of e.m.f. E and internal resistance r are connected in parallel to each other. Derive an expression for the current that will pass through a resistor of resistance R connected across the cells (2 marks) When cells are connected in parallel: Total e.m.f. = E (same as one cell) Total internal resistance = r/2 (since two resistances in parallel) Effective resistance = r/2 + R Current I = Total e.m.f. / Total resistance Answer: I = E / (R + r/2) = 2E / (2R + r) =========================================================== END OF ANSWERS – QUESTIONS 1, 2 & 3 =========================================================== INSTRUCTIONS FOR THE EXAM Answer ALL questions in this paper. Draw all graphs neatly on graph paper with appropriate scales and labels. Show all working in your answer booklet. Record all observations and readings accurately. State precautions in your own words (not copied exactly). Time: As specified in your paper. ==============================================
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