Free Jamb Physics Past Questions with Solutions

Tip: To pass JAMB Physics, focus on thoroughly understanding the JAMB syllabus for physics, practice extensively with past questions, master key physics formulas, prioritize understanding concepts rather than cramming, manage your time effectively during the exam, and consider seeking additional help from tutors or online resources if needed.

Practice online with our comprehensive JAMB Physics Past Questions with solutions here:

JAMB: PHYSICS

Here are free JAMB physics past questions for your practice.

Question 1:

A particle slows down from a velocity of 10 ms^-1 at 2 ms^-2 to come to rest after 4 seconds. The displacement of the particle in the last second of motion is

A. 3 m
B. 9 m
C. 21 m
D. 24 m

Solution:

u = 10 ms^-1
a = –2 ms^-2
t = 4 seconds.
S = ?

Using S = \( \scriptsize ut \: + \: \normalsize \frac{1}{2} \scriptsize at^2\)

At 4s ………. S₄ = \( \scriptsize 10 \: \times \: 4 \: + \: \normalsize \frac{1}{2} \scriptsize \: \times \: – 2\: \times \: 4^2 \\ = \scriptsize 40 \: – \: 16 \\ = \scriptsize 24 \: m\)

At 3s ………. S₃ = \( \scriptsize 10 \: \times \: 3 \: + \: \normalsize \frac{1}{2} \scriptsize \: \times \: – 2\: \times \: 3^2 \\ = \scriptsize 30 \: – \: 9\\ = \scriptsize 21 \: m\)

Displacement in the last 1 second; S₄ – S₃ = 24 – 21 = 3 m

Answer = 3 m

Question 2:

An object weighs 30 N in air and 21 N in water. The weight of the object when completely immersed in a liquid of relative density 1.4 is

A. 25.2 N
B. 17.4 N
C. 12.6 N
D. 9.0 N

Solution:

Weight in air = 30 N
Weight in water = 21 N
Weight in liquid = ?
Relative density of liquid = 1.4

R.D = \( \frac{upthrust \: in \: liquid}{upthrust\:in\:water}\)

1.4 = \( \frac{upthrust \: in \: liquid}{30\:-\:21}\)

1.4 = \( \frac{upthrust \: in \: liquid}{9}\)

upthrust in liquid = 1.4 × 9 = 12.6 N

Weight in liquid = 30 – 12.6 = 17.4 N

Answer = 17.4 N

Question 3:

A uniform meter rule QR is balanced on a knife edge, which is 55 cm from R. When a mass of 10 g is hung at P as shown above, the mass of the meter rule is

A. 550 g
B. 350 g
C. 70 g
D. 35 g

Solution:

Distance of mass of the ruler ‘m’ from the pivot = 55 – 50 = 5 cm

Distance of 10 g mass from pivot = 35 cm (as shown in the diagram)

anticlockwise moment = clockwise moment

10 × 35 = m × 5

5m = 350

m = \( \frac{350}{5} \\ = \scriptsize 70\:g\)

Answer = 70 g

Question 4:

A simple pendulum, 0.6 m long, has a period of 1.5 s. What is the period of a similar pendulum 0.4 m long in the same location?

A. \( \scriptsize 1.5 \normalsize \sqrt{\frac{2}{3}} \scriptsize\:s \)
B. \( \scriptsize 1.5 \normalsize \sqrt{\frac{3}{2}} \scriptsize\:s\)
C. \( \scriptsize 2.25\: s\)
D. \( \scriptsize 1.00\: s\)

Solution:

T₁ = 1.5 s
L₁ = 0.6 s
T₂ = ?
L₂ = 0.4 m

From \( \frac{T_1}{T_2} = \frac{\sqrt{L_1}}{\sqrt{L_2}} \)

⇒ \( \frac{1.5}{T_2} = \frac{\sqrt{0.6}}{\sqrt{0.4}} \)

cross multiply

⇒ \(\scriptsize T_2 \: \times \: \sqrt{0.6} = 1.5\: \times \: \sqrt{0.4}\)

⇒ \(\scriptsize T_2 = 1.5\: \times \: \normalsize \frac{\sqrt{0.4}}{\sqrt{0.6}}\)

⇒ \(\scriptsize T_2 = 1.5\: \times \: \normalsize \frac{\sqrt{4}}{\sqrt{10}} \: \div \: \frac{\sqrt{6}}{\sqrt{10}}\)

⇒ \(\scriptsize T_2 = 1.5\: \times \: \normalsize \frac{\sqrt{4}}{\sqrt{10}} \: \times \: \frac{\sqrt{10}}{\sqrt{6}}\)

⇒ \(\scriptsize T_2 = 1.5\: \times \: \normalsize \frac{\sqrt{4}}{\sqrt{6}}\)

⇒ \(\scriptsize T_2 = 1.5 \normalsize \frac{\sqrt{2}}{\sqrt{3}}\)

⇒ \(\scriptsize T_2 = 1.5 \normalsize \sqrt{\frac{2}{3}}\scriptsize \:s\)

Question 5:

When vibration occurs in an air column, the distance between a node and an antinode is equal to

A. one-quarter of the wavelength
B. one-half of the wavelength
C. the wavelength
D. twice the wavelength

Solution:

The distance between two consecutive nodes N N is \(\frac{\lambda}{2}\)

while the distance between a node and an antinode NA is \( \frac{\lambda}{4}\)

Question 6:

40 m³of liquid P is mixed with 60 m³ of another liquid Q. If the densities of P and Q are 1.00 kg/m³and 1.6 kg/m³, respectively, what is the density of the mixture?

A. 0.05 kg/m³
B. 1.25 kg/m³
C. 1.30 kg/m³
D. 1.36 kg/m³

Solution:

The formula for density is mass/volume. The volume and density of two liquids, P and Q, are given in the mixture. The first step is to calculate their respective masses. Then, you add the masses of P and Q to get the total mass of the mixture. Next, you add the volumes of P and Q to get the total volume of the mixture. Finally, calculate the density of the mixture by dividing the mass of the mixture by the volume of the mixture.

First liquid: P

Density = \(\frac{mass}{volume} \)

1 = \(\frac{m_P}{40}\)

∴ m_P = 1 × 40 = 40 kg

Second liquid: Q

Density = \( \frac{mass}{volume} \)

1.6 = \( \frac{m_Q}{60}\)

∴ m_Q = 1.6 × 60 = 96 kg

Total mass of mixture = m_P + m_Q = 40 + 96 = 136 kg

Total volume of mixture = V_P + V_Q = 40 + 60 = 100 m³

∴ Density of mixture = \( \frac{Mass\:of\:mixture}{Volume\:of\:mixture} \\ = \frac{136}{100} \\ = \scriptsize 1.36\: kg/m^3 \)

Density of mixture = \(\scriptsize 1.36\: kg/m^3 \)

Question 7:

Which of the following is in a neutral equilibrium?

A. A heavy weight suspended on a string
B. Cone resting on its slant edge
C. A heavy-based table lamp
D. The beam of a balance in use

Explanation

A cone resting on its slant edge is considered to be in neutral equilibrium because if slightly disturbed, it will simply roll to a new position along the slant edge, maintaining the same center of gravity level without returning to its original position or tipping over further; essentially, it can settle in a new balanced state at any point on the side of the cone.

Question 8:

When two objects, P and Q, are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. If the masses of P and Q are the same, the ratio of the specific heat capacity of P to Q is

A. 1:4
B. 4:1
C. 1:1
D. 1:2

Explanation

P = \( \scriptsize mc_P \Delta \theta_P \)

Q = \( \scriptsize mc_Q \Delta \theta_Q \)

Heat supplied to P = Heat supplied to Q

∴ \( \scriptsize mc_P \Delta \theta_P = mc_Q \Delta \theta_Q \)

We are given that Δθ_P = 2Δθ_Q. Therefore, substituting this into the above equation, we get:

⇒ \( \scriptsize mc_P \Delta 2\theta_Q = mc_Q \Delta \theta_Q \)

⇒ \( \frac{\not{m} c_P}{\not{m}c_Q} = \frac{\Delta \theta_Q}{\Delta 2\theta_Q} \)

⇒ \( \frac{c_P}{c_Q} = \frac{1}{2} \)

∴ c_P : c_Q = 1 : 2

Question 9:

In a ray diagram for a thin converging lens, a ray that is not parallel to the optic axis but passes through the optic centre will

A. pass through undeviated
B. pass through the centre of curvature after refraction
C. emerge parallel to the principal axis
D. pass through the principal focus after refraction

Explanation

Any ray of light passing through the optical centre of a convex lens goes straight through without deviation.

Construction of Ray Diagrams for a Convex Lens