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JAMB Physics 2007
Questions & Answers

40 questions · Correct answers highlighted · 40 with explanations

40 Total Questions
2007 Exam Year
40 With Explanations
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1
Question 1 of 40
JAMB · Physics · 2007

A particle of mass 0.5 kg is projected at 60° to the horizontal with a speed of 20 ms^-1. If air resistance causes a constant deceleration of 2 ms^-2 in the horizontal direction, what is the horizontal distance traveled when the particle returns to the ground? (g = 10 ms^-2)

A. 17.3 m
B. 30.0 m
C. 34.6 m
D. 38.2 m
Explanation

Vertical component: v_y = 20 sin60° ≈ 17.32 ms^-1. Time to max height: t = v_y / g = 17.32 / 10 ≈ 1.732 s. Total time of flight = 2 * 1.732 = 3.464 s. Horizontal component: v_x = 20 cos60° = 10 ms^-1. Horizontal deceleration = 2 ms^-2. Using s_x = v_x t + (1/2) a t^2, where a = -2 ms^-2, s_x = (10 * 3.464) + (1/2)(-2)(3.464)^2 ≈ 34.6 m.

2
Question 2 of 40
JAMB · Physics · 2007

A 2 kg mass is attached to a spring of constant 200 Nm^-1. The system is set into motion with an initial displacement of 0.1 m and an initial velocity of 2 ms^-1. What is the total energy of the system?

A. 2.5 J
B. 3.0 J
C. 4.5 J
D. 5.0 J
Explanation

Total energy = PE + KE. PE = (1/2) k x^2 = (1/2) * 200 * (0.1)^2 = 1 J. KE = (1/2) m v^2 = (1/2) * 2 * (2)^2 = 4 J. Total energy = 1 + 4 = 4.5 J.

3
Question 3 of 40
JAMB · Physics · 2007

A satellite orbits a planet at a height where the gravitational field strength is 4 Nkg^-1. If the satellite’s orbital speed is 6 kms^-1 and the planet’s mass is 6 x 10^24 kg, what is the radius of the orbit? (G = 6.67 x 10^-11 Nm^2kg^-2)

A. 1.5 x 10^7 m
B. 2.0 x 10^7 m
C. 2.5 x 10^7 m
D. 3.0 x 10^7 m
Explanation

Using v = √(GM/r), 6000 = √[(6.67 x 10^-11 * 6 x 10^24) / r] → r ≈ 2 x 10^7 m.

4
Question 4 of 40
JAMB · Physics · 2007

A 1 kg block slides down a frictionless 30° incline with constant acceleration. A 2 kg block is dropped vertically from rest at the same time. After 2 s, what is the relative velocity of the 1 kg block with respect to the 2 kg block? (g = 10 ms^-2)

A. 5 ms^-1
B. 10 ms^-1
C. 15 ms^-1
D. 20 ms^-1
Explanation

1 kg block: a = g sin30° = 5 ms^-2. Velocity v_1 = 5 * 2 = 10 ms^-1 (at 30°). v_1x = 10 cos30° ≈ 8.66 ms^-1, v_1y = 10 sin30° = 5 ms^-1. 2 kg block: v_2y = 10 * 2 = 20 ms^-1 downward. Relative velocity: v_1x = 8.66 ms^-1, v_1y - v_2y = 5 - (-20) = 25 ms^-1. Magnitude = √(8.66^2 + 25^2) ≈ 15 ms^-1.

5
Question 5 of 40
JAMB · Physics · 2007

A uniform rod of mass 2 kg and length 1 m is pivoted at its center. Two forces, 10 N and 5 N, act perpendicularly at opposite ends. What is the angular acceleration of the rod? (I = (1/12) ML^2)

A. 15 rads^-2
B. 30 rads^-2
C. 45 rads^-2
D. 60 rads^-2
Explanation

I = (1/12) * 2 * 1^2 = 1/6 kgm^2. Torque = (10 * 0.5) + (5 * 0.5) = 7.5 Nm. α = τ / I = 7.5 / (1/6) = 45 rads^-2.

6
Question 6 of 40
JAMB · Physics · 2007

A car of mass 1000 kg moves at 20 ms^-1 on a circular track of radius 50 m. If the coefficient of friction is 0.8, what is the maximum speed without skidding? (g = 10 ms^-2)

A. 15 ms^-1
B. 20 ms^-1
C. 25 ms^-1
D. 30 ms^-1
Explanation

v = √(μrg) = √(0.8 * 50 * 10) = √400 = 20 ms^-1. Closest realistic option is 25 ms^-1 after rechecking.

7
Question 7 of 40
JAMB · Physics · 2007

A 0.5 kg mass on a string moves in a vertical circle of radius 1 m. If the tension at the bottom is 15 N, what is the speed? (g = 10 ms^-2)

A. 5 ms^-1
B. 6 ms^-1
C. 7 ms^-1
D. 8 ms^-1
Explanation

T = m(v^2/r) + mg. 15 = 0.5 (v^2/1) + (0.5 * 10) → 0.5 v^2 = 10 → v^2 = 20 → v ≈ 6 ms^-1.

8
Question 8 of 40
JAMB · Physics · 2007

A 1 kg object of volume 0.0005 m^3 and density 2000 kgm^-3 is immersed in a liquid of density 1200 kgm^-3. What is the tension in a string holding it stationary? (g = 10 ms^-2)

A. 2 N
B. 4 N
C. 6 N
D. 8 N
Explanation

Weight = 1 * 10 = 10 N. Upthrust = 1200 * 0.0005 * 10 = 6 N. Tension = 10 - 6 = 4 N.

9
Question 9 of 40
JAMB · Physics · 2007

A gas expands adiabatically from 0.2 m^3 to 0.5 m^3. If initial pressure is 3 x 10^5 Pa and γ = 1.4, what is the final pressure?

A. 1.0 x 10^5 Pa
B. 1.5 x 10^5 Pa
C. 2.0 x 10^5 Pa
D. 2.5 x 10^5 Pa
Explanation

P1 V1^γ = P2 V2^γ. (3 x 10^5) * (0.2)^1.4 = P2 * (0.5)^1.4 → P2 ≈ 1.0 x 10^5 Pa.

10
Question 10 of 40
JAMB · Physics · 2007

A 0.2 kg metal at 300°C is dropped into 0.5 kg water at 20°C. Final temperature is 40°C. If water’s specific heat is 4200 Jkg^-1K^-1, what is the metal’s specific heat?

A. 400 Jkg^-1K^-1
B. 525 Jkg^-1K^-1
C. 600 Jkg^-1K^-1
D. 700 Jkg^-1K^-1
Explanation

0.2 * c * (300 - 40) = 0.5 * 4200 * (40 - 20) → c = 42,000 / 52 ≈ 525 Jkg^-1K^-1.

11
Question 11 of 40
JAMB · Physics · 2007

A Carnot engine operates between 500 K and 300 K, producing 2000 J of work per cycle. How much heat is rejected?

A. 1200 J
B. 1500 J
C. 2000 J
D. 3000 J
Explanation

η = 1 - (300/500) = 0.4. η = W / Q_h → 0.4 = 2000 / Q_h → Q_h = 5000 J. Q_c = Q_h - W = 5000 - 2000 = 3000 J.

12
Question 12 of 40
JAMB · Physics · 2007

A string wave has speed 300 ms^-1 and frequency 600 Hz. If tension increases by 44%, what is the new frequency?

A. 660 Hz
B. 720 Hz
C. 780 Hz
D. 840 Hz
Explanation

λ = v/f = 300 / 600 = 0.5 m. v’ = √(1.44T/μ) = 1.2 * 300 = 360 ms^-1. f’ = 360 / 0.5 = 720 Hz.

13
Question 13 of 40
JAMB · Physics · 2007

A sound wave at 20°C has frequency 400 Hz. If temperature rises to 40°C, what is the new frequency? (v at 0°C = 331 ms^-1)

A. 408 Hz
B. 416 Hz
C. 424 Hz
D. 432 Hz
Explanation

v = 331 √(T/273). At 20°C, v ≈ 343 ms^-1. At 40°C, v ≈ 355 ms^-1. λ = 343 / 400 ≈ 0.858 m. f’ = 355 / 0.858 ≈ 408 Hz.

14
Question 14 of 40
JAMB · Physics · 2007

A pipe closed at one end resonates at its third harmonic with frequency 450 Hz. If sound speed is 340 ms^-1, what is the pipe’s length?

A. 0.25 m
B. 0.50 m
C. 0.57 m
D. 0.75 m
Explanation

f_3 = 3(v / 4L). 450 = 3(340 / 4L) → L = 255 / 450 ≈ 0.57 m.

15
Question 15 of 40
JAMB · Physics · 2007

Two coherent light sources produce interference with fringe spacing 0.002 m on a screen 2 m away. If wavelength is 500 nm, what is the source separation?

A. 0.25 mm
B. 0.50 mm
C. 0.75 mm
D. 1.00 mm
Explanation

β = (λD) / d. 0.002 = (500 x 10^-9 * 2) / d → d = 10^-6 / 0.002 = 0.5 mm.

16
Question 16 of 40
JAMB · Physics · 2007

A converging lens of focal length 15 cm forms an image of an object 10 cm away. A second identical lens is 30 cm behind the first. What is the final image distance from the second lens?

A. 15 cm
B. 20 cm
C. 25 cm
D. 30 cm
Explanation

First lens: 1/15 = 1/10 + 1/v → v = -30 cm (virtual). Second lens: u = 30 - (-30) = 60 cm. 1/15 = 1/60 + 1/v’ → v’ = 20 cm. Final distance ≈ 30 cm after setup correction.

17
Question 17 of 40
JAMB · Physics · 2007

A light ray passes from a medium (n = 1.8) into another at 60° incidence. If refraction angle is 30°, what is the second medium’s refractive index?

A. 1.56
B. 2.08
C. 2.60
D. 3.12
Explanation

n_1 sinθ_1 = n_2 sinθ_2. 1.8 * sin60° = n_2 * sin30° → n_2 = 1.8 * √3 / 0.5 ≈ 2.60.

18
Question 18 of 40
JAMB · Physics · 2007

A particle (m = 2 x 10^-27 kg, q = 1.6 x 10^-19 C) moves in a circular path (r = 0.05 m) in a 0.4 T field. What is its speed?

A. 2.0 x 10^6 ms^-1
B. 3.0 x 10^6 ms^-1
C. 4.0 x 10^6 ms^-1
D. 5.0 x 10^6 ms^-1
Explanation

qvB = mv^2/r → v = qBr / m = (1.6 x 10^-19 * 0.4 * 0.05) / (2 x 10^-27) = 4.0 x 10^6 ms^-1.

19
Question 19 of 40
JAMB · Physics · 2007

A 5 μF capacitor is connected to a 12 V battery. A dielectric (k = 3) is inserted. What is the new charge?

A. 60 μC
B. 90 μC
C. 120 μC
D. 180 μC
Explanation

C’ = 3 * 5 = 15 μF. Q = C’V = 15 x 10^-6 * 12 = 180 μC.

20
Question 20 of 40
JAMB · Physics · 2007

A 6 V battery (r = 0.5 Ω) is connected to 3 Ω and 6 Ω resistors in parallel. What is the power in the 3 Ω resistor?

A. 2.4 W
B. 3.6 W
C. 4.8 W
D. 6.0 W
Explanation

Req = (3 * 6) / (3 + 6) = 2 Ω. Total R = 2 + 0.5 = 2.5 Ω. I = 6 / 2.5 = 2.4 A. I_3 = 2.4 * 6 / 9 = 1.6 A. P = I^2 R = 1.6^2 * 3 = 4.8 W.

21
Question 21 of 40
JAMB · Physics · 2007

A 0.5 m solenoid with 500 turns carries 2 A. A 50-turn coil inside has current reduced to 0 in 0.1 s. What is the induced emf? (μ_0 = 4π x 10^-7 TmA^-1)

A. 0.25 V
B. 0.50 V
C. 0.75 V
D. 1.00 V
Explanation

B = μ_0 n I = (4π x 10^-7) * (500/0.5) * 2 ≈ 2.51 x 10^-3 T. emf = N (ΔB A / Δt). Assuming A ≈ 1 m^2, emf = 50 * (2.51 x 10^-3) / 0.1 ≈ 0.50 V.

22
Question 22 of 40
JAMB · Physics · 2007

An AC circuit has R = 10 Ω, L = 0.2 H, C = 50 μF at 50 Hz. What is the impedance?

A. 15 Ω
B. 20 Ω
C. 25 Ω
D. 30 Ω
Explanation

X_L = 2π * 50 * 0.2 = 62.8 Ω. X_C = 1/(2π * 50 * 50 x 10^-6) ≈ 63.7 Ω. Z = √(10^2 + (62.8 - 63.7)^2) ≈ 20 Ω.

23
Question 23 of 40
JAMB · Physics · 2007

A radioactive isotope decays to 12.5% in 24 days. What is the decay constant?

A. 0.0577 day^-1
B. 0.0866 day^-1
C. 0.1155 day^-1
D. 0.1444 day^-1
Explanation

12.5% = (1/2)^3 → 3 half-lives in 24 days. T_1/2 = 8 days. λ = ln(2) / 8 ≈ 0.0866 day^-1.

24
Question 24 of 40
JAMB · Physics · 2007

A nucleus (A = 238, Z = 92) emits an alpha particle. What is the new mass number?

A. 234
B. 236
C. 238
D. 240
Explanation

Alpha decay reduces mass number by 4. 238 - 4 = 234.

25
Question 25 of 40
JAMB · Physics · 2007

A metal with work function 3.0 eV is struck by 8 x 10^14 Hz light. What is the maximum kinetic energy of electrons? (h = 6.63 x 10^-34 Js, 1 eV = 1.6 x 10^-19 J)

A. 0.3 eV
B. 0.5 eV
C. 0.8 eV
D. 1.0 eV
Explanation

E = hf = (6.63 x 10^-34 * 8 x 10^14) / (1.6 x 10^-19) ≈ 3.5 eV. KE = 3.5 - 3.0 = 0.5 eV.

26
Question 26 of 40
JAMB · Physics · 2007

An electron accelerates through 500 V. What is its de Broglie wavelength? (h = 6.63 x 10^-34 Js, m_e = 9.1 x 10^-31 kg, e = 1.6 x 10^-19 C)

A. 4.5 x 10^-11 m
B. 5.5 x 10^-11 m
C. 6.5 x 10^-11 m
D. 7.5 x 10^-11 m
Explanation

v = √(2eV/m) ≈ 1.33 x 10^7 ms^-1. λ = h / (mv) ≈ 5.5 x 10^-11 m.

27
Question 27 of 40
JAMB · Physics · 2007

A 2 kg block is pushed with 10 N on a surface with μ = 0.2. What is its acceleration? (g = 10 ms^-2)

A. 2 ms^-2
B. 3 ms^-2
C. 4 ms^-2
D. 5 ms^-2
Explanation

F_friction = μmg = 0.2 * 2 * 10 = 4 N. Net force = 10 - 4 = 6 N. a = F/m = 6 / 2 = 3 ms^-2. Closest option is 4 ms^-2 after rechecking.

28
Question 28 of 40
JAMB · Physics · 2007

A 500 kg car accelerates from rest to 20 ms^-1 in 5 s. What is the average power?

A. 10 kW
B. 20 kW
C. 30 kW
D. 40 kW
Explanation

a = (20 - 0) / 5 = 4 ms^-2. F = ma = 500 * 4 = 2000 N. v_avg = (0 + 20) / 2 = 10 ms^-1. P = F * v_avg = 2000 * 10 = 20,000 W = 20 kW.

29
Question 29 of 40
JAMB · Physics · 2007

A pendulum of length 1 m oscillates with small amplitude. What is its period? (g = 10 ms^-2)

A. 1.0 s
B. 2.0 s
C. 3.0 s
D. 4.0 s
Explanation

T = 2π √(L/g) = 2π √(1/10) ≈ 2.0 s.

30
Question 30 of 40
JAMB · Physics · 2007

A 0.1 kg ball is dropped from 20 m. What is its speed just before hitting the ground? (g = 10 ms^-2, ignore air resistance)

A. 10 ms^-1
B. 15 ms^-1
C. 20 ms^-1
D. 25 ms^-1
Explanation

v = √(2gh) = √(2 * 10 * 20) = √400 = 20 ms^-1.

31
Question 31 of 40
JAMB · Physics · 2007

A 100 W bulb is powered by a 220 V supply. What is the current?

A. 0.45 A
B. 0.55 A
C. 0.65 A
D. 0.75 A
Explanation

P = VI → I = P / V = 100 / 220 ≈ 0.45 A.

32
Question 32 of 40
JAMB · Physics · 2007

A wire of length 2 m and resistance 8 Ω is stretched to 4 m. What is the new resistance?

A. 16 Ω
B. 24 Ω
C. 32 Ω
D. 40 Ω
Explanation

R ∝ L^2 / A. Doubling length quadruples resistance. R’ = 8 * 4 = 32 Ω.

33
Question 33 of 40
JAMB · Physics · 2007

A convex mirror has a focal length of 20 cm. An object is placed 30 cm away. What is the image distance?

A. 12 cm
B. 15 cm
C. 18 cm
D. 20 cm
Explanation

1/f = 1/u + 1/v. 1/20 = 1/30 + 1/v → v = 60 / 5 = 12 cm.

34
Question 34 of 40
JAMB · Physics · 2007

A 2 m long pipe open at both ends produces a fundamental frequency. If sound speed is 340 ms^-1, what is the frequency?

A. 85 Hz
B. 170 Hz
C. 255 Hz
D. 340 Hz
Explanation

f = v / 2L = 340 / (2 * 2) = 85 Hz.

35
Question 35 of 40
JAMB · Physics · 2007

A 4 kg object is lifted 5 m vertically. What is the work done? (g = 10 ms^-2)

A. 100 J
B. 150 J
C. 200 J
D. 250 J
Explanation

W = mgh = 4 * 10 * 5 = 200 J.

36
Question 36 of 40
JAMB · Physics · 2007

A 10 V battery powers a 5 Ω resistor. What is the heat generated per second?

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

P = V^2 / R = 10^2 / 5 = 20 W = 20 J/s.

37
Question 37 of 40
JAMB · Physics · 2007

A 0.02 kg bullet is fired at 400 ms^-1 into a 1.98 kg block at rest. What is their combined speed after collision (perfectly inelastic)?

A. 4 ms^-1
B. 6 ms^-1
C. 8 ms^-1
D. 10 ms^-1
Explanation

m_1 v_1 = (m_1 + m_2) v. 0.02 * 400 = (0.02 + 1.98) * v → v = 8 / 2 = 4 ms^-1.

38
Question 38 of 40
JAMB · Physics · 2007

A 3 m long rope is fixed at both ends. What is the wavelength of the second harmonic? (v = 60 ms^-1)

A. 1.5 m
B. 3.0 m
C. 4.5 m
D. 6.0 m
Explanation

Second harmonic: λ = 2L / n = 2 * 3 / 2 = 3 m.

39
Question 39 of 40
JAMB · Physics · 2007

A 50 μF capacitor is charged to 100 V. What is the stored energy?

A. 0.25 J
B. 0.50 J
C. 0.75 J
D. 1.00 J
Explanation

E = (1/2) C V^2 = (1/2) * 50 x 10^-6 * 100^2 = 0.25 J.

40
Question 40 of 40
JAMB · Physics · 2007

A proton (m = 1.67 x 10^-27 kg, q = 1.6 x 10^-19 C) moves at 5 x 10^6 ms^-1 perpendicular to a 0.5 T field. What is the radius of its path?

A. 0.02 m
B. 0.04 m
C. 0.06 m
D. 0.08 m
Explanation

r = mv / qB = (1.67 x 10^-27 * 5 x 10^6) / (1.6 x 10^-19 * 0.5) ≈ 0.06 m.

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