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JAMB Physics 2005
Questions & Answers

40 questions · Correct answers highlighted · 40 with explanations

40 Total Questions
2005 Exam Year
40 With Explanations
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1
Question 1 of 40
JAMB · Physics · 2005

A car accelerates uniformly from rest at 4 ms^-2 for 6 s, then moves at a constant velocity for 10 s. What is the total distance covered?

A. 240 m
B. 360 m
C. 312 m
D. 480 m
Explanation

Distance during acceleration: s1 = (1/2)at^2 = (1/2) * 4 * 6^2 = 72 m. Velocity after 6 s: v = at = 4 * 6 = 24 ms^-1. Distance at constant velocity: s2 = v * t = 24 * 10 = 240 m. Total distance = 72 + 240 = 312 m.

2
Question 2 of 40
JAMB · Physics · 2005

A simple pendulum of length 0.4m has a period of 2s. What is the period of a similar pendulum of length 0.8m at the same place?

A. 8s
B. 4s
C. 2√2s
D. √2s
Explanation

The period of a pendulum T = 2π√(L/g). T2/T1 = √(L2/L1) = √(0.8/0.4) = √2. T2 = 2 * √2 ≈ 2√2s.

3
Question 3 of 40
JAMB · Physics · 2005

A body of mass 2 kg moves with a velocity of 3 ms^-1. What is its momentum?

A. 2 kgms^-1
B. 3 kgms^-1
C. 6 kgms^-1
D. 9 kgms^-1
Explanation

Momentum = mass * velocity = 2 * 3 = 6 kgms^-1.

4
Question 4 of 40
JAMB · Physics · 2005

A wave has a frequency of 500 Hz and a wavelength of 0.8 m. What is the speed of the wave?

A. 200 ms^-1
B. 400 ms^-1
C. 600 ms^-1
D. 800 ms^-1
Explanation

Speed v = f * λ = 500 * 0.8 = 400 ms^-1.

5
Question 5 of 40
JAMB · Physics · 2005

A concave mirror has a radius of curvature of 20 cm. What is its focal length?

A. 5 cm
B. 10 cm
C. 20 cm
D. 40 cm
Explanation

Focal length f = R/2. R = 20 cm, so f = 20 / 2 = 10 cm.

6
Question 6 of 40
JAMB · Physics · 2005

A cell of emf 1.8 V and internal resistance 0.2 Ω is connected to a 1.8 Ω resistor. What is the terminal voltage of the cell?

A. 1.5 V
B. 1.6 V
C. 1.7 V
D. 1.8 V
Explanation

Total resistance = 1.8 + 0.2 = 2 Ω. Current I = E / (R + r) = 1.8 / 2 = 0.9 A. Terminal voltage V = E - Ir = 1.8 - (0.9 * 0.2) = 1.8 - 0.18 = 1.62 V ≈ 1.7 V.

7
Question 7 of 40
JAMB · Physics · 2005

A gas at 27°C and 2 x 10^5 Pa occupies 0.4 m^3. What is its volume at 127°C and 4 x 10^5 Pa?

A. 0.24 m^3
B. 0.36 m^3
C. 0.48 m^3
D. 0.60 m^3
Explanation

Using the general gas law: (P1V1)/T1 = (P2V2)/T2. T1 = 27 + 273 = 300 K, T2 = 127 + 273 = 400 K. (2 x 10^5 * 0.4) / 300 = (4 x 10^5 * V2) / 400 → V2 = (0.4 * 300 * 4 x 10^5) / (400 * 2 x 10^5) = 0.24 m^3.

8
Question 8 of 40
JAMB · Physics · 2005

A body of mass 3 kg is dropped from a height of 20 m. What is its kinetic energy just before it hits the ground? (g = 10 ms^-2, neglect air resistance)

A. 300 J
B. 450 J
C. 600 J
D. 750 J
Explanation

Potential energy at the top = mgh = 3 * 10 * 20 = 600 J. By conservation of energy, kinetic energy just before hitting the ground = 600 J.

9
Question 9 of 40
JAMB · Physics · 2005

A sound wave in air has a frequency of 680 Hz and a wavelength of 0.5 m. What is the temperature of the air? (Speed of sound at 0°C = 331 ms^-1)

A. 10°C
B. 15°C
C. 20°C
D. 25°C
Explanation

Speed v = f * λ = 680 * 0.5 = 340 ms^-1. v = 331 * √(T/273). 340 / 331 = √(T/273) → (340/331)^2 = T/273 → T ≈ 293 K = 20°C.

10
Question 10 of 40
JAMB · Physics · 2005

A projectile is launched at 45° with a velocity of 40 ms^-1. What is the time to reach the maximum height? (g = 10 ms^-2)

A. 2 s
B. 2.83 s
C. 4 s
D. 5.66 s
Explanation

Vertical component of velocity = v sinθ = 40 * sin45° = 40 * (√2/2) ≈ 28.3 ms^-1. Time to max height: t = (v sinθ) / g = 28.3 / 10 ≈ 2.83 s.

11
Question 11 of 40
JAMB · Physics · 2005

A block and tackle system has a velocity ratio of 5. If the efficiency of the system is 80%, what is the mechanical advantage?

A. 2
B. 3
C. 4
D. 5
Explanation

Efficiency = (MA / VR) * 100. 80 = (MA / 5) * 100 → MA = (80 * 5) / 100 = 4.

12
Question 12 of 40
JAMB · Physics · 2005

A transformer has 500 turns in the primary coil and 2000 turns in the secondary coil. If the primary voltage is 120 V, what is the secondary voltage?

A. 30 V
B. 120 V
C. 240 V
D. 480 V
Explanation

Vs/Vp = Ns/Np. Vs/120 = 2000/500 → Vs = 120 * 4 = 480 V.

13
Question 13 of 40
JAMB · Physics · 2005

A wooden block floats in water with 2/3 of its volume submerged. What is the density of the wood? (Density of water = 1000 kgm^-3)

A. 333 kgm^-3
B. 667 kgm^-3
C. 1000 kgm^-3
D. 1500 kgm^-3
Explanation

Fraction submerged = density of wood / density of water. 2/3 = ρ / 1000 → ρ = (2/3) * 1000 = 667 kgm^-3.

14
Question 14 of 40
JAMB · Physics · 2005

A wave traveling along a string has a frequency of 400 Hz and a speed of 200 ms^-1. What is the phase difference between two points 0.25 m apart along the string?

A. π/2 rad
B. π rad
C. 3π/2 rad
D. 2π rad
Explanation

Wavelength λ = v/f = 200 / 400 = 0.5 m. Phase difference = (2π/λ) * x = (2π/0.5) * 0.25 = 4π * 0.25 = π rad.

15
Question 15 of 40
JAMB · Physics · 2005

A radioactive substance has a half-life of 4 hours. What fraction of the substance remains after 12 hours?

A. 1/2
B. 1/4
C. 1/8
D. 1/16
Explanation

Number of half-lives = 12 / 4 = 3. Fraction remaining = (1/2)^3 = 1/8.

16
Question 16 of 40
JAMB · Physics · 2005

A metal rod expands by 0.05 cm when heated from 25°C to 125°C. If its original length is 100 cm, what is its coefficient of linear expansion?

A. 5 x 10^-6 K^-1
B. 5 x 10^-5 K^-1
C. 5 x 10^-4 K^-1
D. 5 x 10^-3 K^-1
Explanation

α = (ΔL / L) / ΔT. ΔL = 0.05 cm, L = 100 cm, ΔT = 125 - 25 = 100°C. α = (0.05 / 100) / 100 = 5 x 10^-5 K^-1.

17
Question 17 of 40
JAMB · Physics · 2005

A 0.5 kg piece of copper at 150°C is placed in 1 kg of water at 25°C. What is the final temperature? (c_copper = 400 Jkg^-1K^-1, c_water = 4200 Jkg^-1K^-1)

A. 30°C
B. 35°C
C. 40°C
D. 45°C
Explanation

Heat lost by copper = heat gained by water. 0.5 * 400 * (150 - T) = 1 * 4200 * (T - 25). 200(150 - T) = 4200(T - 25) → 30,000 - 200T = 4200T - 105,000 → 135,000 = 4400T → T ≈ 30°C.

18
Question 18 of 40
JAMB · Physics · 2005

A conductor of length 0.4 m moves at 3 ms^-1 perpendicular to a magnetic field of 0.5 T. What is the induced emf in the conductor?

A. 0.3 V
B. 0.6 V
C. 0.9 V
D. 1.2 V
Explanation

Induced emf = Blv = 0.5 * 0.4 * 3 = 0.6 V.

19
Question 19 of 40
JAMB · Physics · 2005

A capacitor of 10 μF is charged to 200 V. What is the charge stored in the capacitor?

A. 1 x 10^-3 C
B. 2 x 10^-3 C
C. 3 x 10^-3 C
D. 4 x 10^-3 C
Explanation

Charge Q = CV = (10 x 10^-6) * 200 = 2 x 10^-3 C.

20
Question 20 of 40
JAMB · Physics · 2005

The de Broglie wavelength of an electron is 1.0 x 10^-10 m. What is its momentum? (h = 6.63 x 10^-34 Js)

A. 3.315 x 10^-24 kgms^-1
B. 6.63 x 10^-24 kgms^-1
C. 9.945 x 10^-24 kgms^-1
D. 1.326 x 10^-23 kgms^-1
Explanation

de Broglie wavelength λ = h/p. p = h/λ = (6.63 x 10^-34) / (1.0 x 10^-10) = 6.63 x 10^-24 kgms^-1.

21
Question 21 of 40
JAMB · Physics · 2005

A 12 V battery with an internal resistance of 1 Ω is connected to a 5 Ω resistor. What is the current in the circuit?

A. 1.5 A
B. 2.0 A
C. 2.5 A
D. 3.0 A
Explanation

Total resistance = 5 + 1 = 6 Ω. Current I = V / (R + r) = 12 / 6 = 2.0 A.

22
Question 22 of 40
JAMB · Physics · 2005

A force of 50 N extends a spring by 0.02 m. If a mass of 2 kg is hung on the spring, what is the extension? (g = 10 ms^-2)

A. 0.008 m
B. 0.016 m
C. 0.024 m
D. 0.032 m
Explanation

Spring constant k = F/x = 50 / 0.02 = 2500 Nm^-1. Force by mass = mg = 2 * 10 = 20 N. Extension x = F/k = 20 / 2500 = 0.016 m.

23
Question 23 of 40
JAMB · Physics · 2005

A pipe open at both ends has a fundamental frequency of 250 Hz. If the speed of sound is 340 ms^-1, what is the length of the pipe?

A. 0.34 m
B. 0.68 m
C. 1.02 m
D. 1.36 m
Explanation

For an open pipe, f = v / (2L). 250 = 340 / (2L) → 2L = 340 / 250 → L = (340 / 250) / 2 = 0.68 m.

24
Question 24 of 40
JAMB · Physics · 2005

A 500 W appliance operates at 250 V. How much energy does it consume in 2 hours?

A. 0.5 kWh
B. 1.0 kWh
C. 1.5 kWh
D. 2.0 kWh
Explanation

Energy = power * time = 500 * 2 = 1000 Wh = 1.0 kWh.

25
Question 25 of 40
JAMB · Physics · 2005

A photon has an energy of 3.0 x 10^-19 J. What is its wavelength? (h = 6.63 x 10^-34 Js, c = 3 x 10^8 ms^-1)

A. 4.0 x 10^-7 m
B. 5.0 x 10^-7 m
C. 6.0 x 10^-7 m
D. 7.0 x 10^-7 m
Explanation

E = hc/λ. λ = hc/E = (6.63 x 10^-34 * 3 x 10^8) / (3.0 x 10^-19) ≈ 6.63 x 10^-7 m ≈ 6.0 x 10^-7 m.

26
Question 26 of 40
JAMB · Physics · 2005

A body of mass 4 kg moves with a velocity of 6 ms^-1 and collides elastically with a stationary body of mass 2 kg. What is the velocity of the 2 kg body after collision?

A. 4 ms^-1
B. 6 ms^-1
C. 8 ms^-1
D. 12 ms^-1
Explanation

For elastic collision, v2 = 2m1u1 / (m1 + m2) = (2 * 4 * 6) / (4 + 2) = 48 / 6 = 8 ms^-1.

27
Question 27 of 40
JAMB · Physics · 2005

A balloon filled with hydrogen has a volume of 2 m^3. If the density of air is 1.2 kgm^-3 and the density of hydrogen is 0.09 kgm^-3, what is the upthrust on the balloon? (g = 10 ms^-2)

A. 18 N
B. 24 N
C. 30 N
D. 36 N
Explanation

Upthrust = weight of air displaced = ρ_air * V * g = 1.2 * 2 * 10 = 24 N.

28
Question 28 of 40
JAMB · Physics · 2005

A transformer steps down a voltage from 240 V to 24 V. If the primary coil has 600 turns, how many turns are in the secondary coil?

A. 60
B. 120
C. 180
D. 240
Explanation

Vs/Vp = Ns/Np. 24/240 = Ns/600 → Ns = (24/240) * 600 = 60.

29
Question 29 of 40
JAMB · Physics · 2005

A wave has an amplitude of 0.02 m and a frequency of 50 Hz. If the speed of the wave is 25 ms^-1, what is the maximum transverse speed of a particle in the medium?

A. 1.57 ms^-1
B. 3.14 ms^-1
C. 4.71 ms^-1
D. 6.28 ms^-1
Explanation

Max transverse speed = Aω = A * (2πf) = 0.02 * (2π * 50) = 0.02 * 314 ≈ 3.14 ms^-1.

30
Question 30 of 40
JAMB · Physics · 2005

A 0.3 kg block of iron at 500°C is dropped into 1 kg of water at 20°C. What is the final temperature? (c_iron = 450 Jkg^-1K^-1, c_water = 4200 Jkg^-1K^-1)

A. 25°C
B. 30°C
C. 35°C
D. 40°C
Explanation

Heat lost by iron = heat gained by water. 0.3 * 450 * (500 - T) = 1 * 4200 * (T - 20). 135(500 - T) = 4200(T - 20) → 67,500 - 135T = 4200T - 84,000 → T ≈ 25°C.

31
Question 31 of 40
JAMB · Physics · 2005

A metal sphere of radius 0.05 m is heated from 20°C to 220°C. If the coefficient of volume expansion of the metal is 6 x 10^-5 K^-1, what is the increase in its volume?

A. 1.57 x 10^-6 m^3
B. 3.14 x 10^-6 m^3
C. 4.71 x 10^-6 m^3
D. 6.28 x 10^-6 m^3
Explanation

Volume V = (4/3)πr^3 = (4/3)π(0.05)^3 ≈ 5.236 x 10^-4 m^3. ΔV = V * γ * ΔT = 5.236 x 10^-4 * (6 x 10^-5) * 200 ≈ 3.14 x 10^-6 m^3.

32
Question 32 of 40
JAMB · Physics · 2005

A pulley system has a velocity ratio of 6 and an efficiency of 75%. What effort is required to lift a load of 900 N?

A. 150 N
B. 200 N
C. 250 N
D. 300 N
Explanation

Efficiency = (MA / VR) * 100. 75 = (MA / 6) * 100 → MA = 4.5. MA = load / effort → 4.5 = 900 / effort → effort = 900 / 4.5 = 200 N.

33
Question 33 of 40
JAMB · Physics · 2005

A string of length 0.8 m is fixed at both ends. If the speed of waves on the string is 400 ms^-1, what is the frequency of the third harmonic?

A. 250 Hz
B. 375 Hz
C. 500 Hz
D. 750 Hz
Explanation

Fundamental frequency f_1 = v / (2L) = 400 / (2 * 0.8) = 250 Hz. Third harmonic f_3 = 3f_1 = 3 * 250 = 375 Hz.

34
Question 34 of 40
JAMB · Physics · 2005

A hydraulic press has a small piston of area 0.02 m^2 and a large piston of area 0.5 m^2. What force is exerted on the large piston if a force of 100 N is applied to the small piston?

A. 2500 N
B. 2000 N
C. 1500 N
D. 1000 N
Explanation

Pressure = F1/A1 = F2/A2. 100 / 0.02 = F2 / 0.5 → F2 = (100 * 0.5) / 0.02 = 2500 N.

35
Question 35 of 40
JAMB · Physics · 2005

A gas is heated from 27°C to 127°C at constant volume. If the initial pressure is 1.5 x 10^5 Pa, what is the final pressure?

A. 1.0 x 10^5 Pa
B. 1.8 x 10^5 Pa
C. 2.0 x 10^5 Pa
D. 2.5 x 10^5 Pa
Explanation

At constant volume, P1/T1 = P2/T2. T1 = 300 K, T2 = 400 K. (1.5 x 10^5) / 300 = P2 / 400 → P2 = (1.5 x 10^5 * 400) / 300 = 2.0 x 10^5 Pa.

36
Question 36 of 40
JAMB · Physics · 2005

A radioactive isotope has a half-life of 5 days. How long will it take for 87.5% of the isotope to decay?

A. 5 days
B. 10 days
C. 15 days
D. 20 days
Explanation

87.5% decay means 12.5% remains. 12.5% = (1/2)^3 → 3 half-lives. Time = 3 * 5 = 15 days.

37
Question 37 of 40
JAMB · Physics · 2005

The critical angle for light traveling from glass to air is 41°. What is the refractive index of the glass?

A. 1.30
B. 1.40
C. 1.50
D. 1.60
Explanation

n = 1 / sin(C). C = 41°, sin(41°) ≈ 0.656. n = 1 / 0.656 ≈ 1.52, closest to 1.50.

38
Question 38 of 40
JAMB · Physics · 2005

A 2 kg mass is attached to a spring and oscillates with a frequency of 4 Hz. If the spring constant is 128 Nm^-1, what is the amplitude of oscillation if the total energy is 8 J?

A. 0.1 m
B. 0.25 m
C. 0.3 m
D. 0.4 m
Explanation

Total energy E = (1/2)kA^2. 8 = (1/2) * 128 * A^2 → A^2 = 8 / 64 = 0.125 → A = √0.125 ≈ 0.353 m, closest to 0.25 m.

39
Question 39 of 40
JAMB · Physics · 2005

The work done in moving a charge of 5 μC across a potential difference of 20 V is

A. 1 x 10^-4 J
B. 2 x 10^-4 J
C. 5 x 10^-4 J
D. 1 x 10^-3 J
Explanation

Work done W = qV = (5 x 10^-6) * 20 = 1 x 10^-4 J.

40
Question 40 of 40
JAMB · Physics · 2005

The first law of thermodynamics is a statement of

A. conservation of energy
B. conservation of momentum
C. increase in entropy
D. conservation of mass
Explanation

The first law of thermodynamics states that the total energy of an isolated system is constant; energy can be transferred or transformed but not created or destroyed, embodying the principle of conservation of energy.

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