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WAEC Physics 2013 Paper

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Question : 12

Total: 46

A body of mass 11kg is suspended from a ceiling fan by an aluminum wire of length 2m and diameter 2mm. Calculate the elastic energy stored in the wire [Young's modulus of aluminum is 7.0 x

10^{10}\mathrm{Nm}^{-2}

\mathrm{s}^{-2}

\pi

= 3.142]`

1.1 x 10^-1J
5.5 x 10^-2J
1.1 x 10^-4J
5.5 x 10^-8J

Solution:

Concept:

The elastic energy stored in a stretched wire equals the work done in stretching it by its extension due to the hanging weight.

Explanation:

The force stretching the wire is the weight of the body:

F = mg = 11 \times 10 = 110 \, \mathrm{N}

.

Given: length

L = 2 \, \mathrm{m}

, diameter

d = 2 \, \mathrm{mm} = 2 \times 10^{-3} \, \mathrm{m}

.

So

d^2 = (2 \times 10^{-3})^2 = 4 \times 10^{-6} \, \mathrm{m}^2

.

Using the extension formula:

e = \frac{4FL}{Y\pi d^2}

.

e = \frac{4 \times 110 \times 2}{7.0 \times 10^{10} \times 3.142 \times 4 \times 10^{-6}}

.

This gives

e \approx 1.0 \times 10^{-3} \, \mathrm{m}

.

Elastic energy stored is:

E = \frac{1}{2} F e = \frac{1}{2} \times 110 \times 1.0 \times 10^{-3}

.

E = 5.5 \times 10^{-2} \, \mathrm{J}

.

Answer:

The correct option is B. $5.5 \times 10^{-2} \, \mathrm{J}$ .

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