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# JAMB Physics sample past questions solutions Here are comprehensive Physics questions and their solutions based on JAMB Physics past questions and answers and UTME syllabus. These are based on JAMB syllabus and past questions from JAMB exams. The goal is to help the student understand every topic and learn how to solve any question during her JAMB CBT exam.

In your JAMB exam, you don’t really have to solve the questions step-by-step as done in examples below because that will waste a lot of your time. The purpose of step-by-step solutions is to help the student have a deeper understanding of the solutions.

### From JAMB Physics past question 2017:

Question: Which of the following statements is correct?

A. The pressure exerted is higher with a narrow shoe than a flat heel shoe.
B. No pressure is exerted in both narrow and flat heel shoes.
C. The pressure exerted is lower with narrow heel shoes than a flat heel shoe.
D. The pressure exerted is the same in both narrow and flat heel shoes.

Solution:
Recall that the formula for pressure is:

P = F/A,

Where F = applied force and A = surface area.

Hence, assuming the force is the same (and should be for this comparison), e have that:

P = c/A, where c = constant (constant force).

Hence, P is inversely proportional to A such that as A increases, P decreases and vice-versa.

Since a narrow heel shoe has a smaller surface area (smaller A) than a flat heel shoe, it means that P is greater for a narrow heel shoe.

Hence, at constant force (F), a narrow heel shoe will exert more pressure than a flat heel shoe.

### From JAMB Physics past question 2013:

Question: When a brick is taken from the earth’s
surface to the moon, its mass

A. Becomes zero
B. Remains constant
C. Reduces
D. increases

Explanation:

Unlike weight which varies with the acceleration due to gravity for a given mass, the mass of an object does not change with location or place.

Recall that in the formula for weight,

w = mg, where m = constant.

The value of will change from one location to another within the earth and, in the moon, the value of g drops to about 16m/s2.

These changes in the value of g lead to changes in the weight of the object. However, the changes, which affects the weight, do not affect the mass of the given object.

### From JAMB Physics past question 2013:

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

A. √2s
B. 8s
C. 4s
D. 2√2s

Solution:

Given:

l1 = 0.4m

T1 = 2s

l2 = 0.8m

Find: T2

The pendulum in the question performs Simple Harmonic Motion (SHM). The formula for finding the period (T) of such motion is:

T = 2π√(l/g)

Where:

T = period of oscillation,

l = length of the pendulum.

g = acceleration due to gravity = 10m/s(assumed approximate value).

The next step is to factor out the constants (2, π and g)  from the equation:

» Square both sides: T2 = (2π)2(l/g)

» T2 = ((2π)2)/g) * l

Since 2, π and g are constants, we can bundle them together and call them one constant, say k. Hence,

» T2 = kl, where k = (2π)2)/g

T2  = kl

T2/l = k

Let’s use subscript 1 to denote the first pendulum and subscript 2 to denote the second, hence,

T12/l1 = k, also

T22/l2 = k

Hence, T12/l1 = T22/l2 = k

» T22/T12 = l2/l1

» T22/(2)2 = (0.8)/(0.4)

» T22/4 = 2

» T22/4 = 2*4 = 8

» T2 = √8 = √(42) = √(4)√(2) = 2√(2)

Hence, T2 = 2√2s

### From JAMB Physics past question 2013:

A train with an initial velocity of 20ms-1 is subjected to a uniform deceleration of
2ms-2. The time required to bring the train to a complete halt is

A. 40s
B. 5s
C. 10s
D. 20s

Solution

Given:

u = 20ms-1

-a = 2ms-1 (Note: -a because it is deceleration)

Find: v

From Newton’s law of linear motions, v = u + at

Where t = time, u = initial velocity, and v = final velocity.

By the time the train is brought to a complete halt, v = 0

Therefore,

0 = u + at

at = -u

t = -(u/a)

But -a = 2 implies that a = -2

Hence, t = -(20/-2) = 20/2 = 10

t = 10s.