Education

# Physics Past Questions 2010

Physics Past Questions 2010. PHYSICS it is one of the most important science subjects. Due to the complicated nature of the subject matter, many people are unable to do well in this subject during the scholarship examination. If you are one of the students in this category, then you do not need to bother about it anymore because we have the solution to that problem, and the solution is in having physics Past Questions. However, you need to understand that before going for any examination, maximum preparation is required to get the necessary success. therefore, we bring you information on physics Past Questions, what it is all about, how it is patterned to suit your demand, and the easiest way to download or get it.

Physics Past Questions is the compilation of all the questions that was given to candidates to answer over the years and it is offered in pdf format. We have made it very simple such that every student can get it with their smartphone. All those who have our past questions have been giving testimonies of how it helped them. We are very sure it is going to help you as well.

## How is physics Past Questions Patterned?

We have made it simple for you. we bring all the questions which is usually in objective, theory and practical format. since it is a pure science course. We have put them together but we indicate the specific years for every question and the correct answers in order to save your time. All you need to do is to devote quality time to study the Past Questions.

### The Benefit of Having physics past question

It will expose you to the questions that have been asked years back as most of those questions and answers get repeated every year while some will just be rephrased. So if you don’t have a copy of this past question, then you are losing out. It also provides useful information about the types of questions to expect. You will find out the number of questions you will see in the examination.

### Sample questions from physics past questions

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.

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.

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

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