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JAMB Physics 2023 Paper

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Question : 16

Total: 65

A wire of radius 0.2 mm is extended by 0.5% of its length when supported by a load of 1.5 kg. Determine the Young's modulus for the material of the wire.

[Take g = 10 ms

^{-2}

]

$2.4 \times 10^{10} (\mathrm{Nm}^{-2})$
$1.5 \times 10^{10} (\mathrm{Nm}^{-2})$
$2.4 \times 10^{9} (\mathrm{Nm}^{-2})$
$1.3 \times 10^{10} (\mathrm{Nm}^{-2})$

Solution:

Concept:

Young's modulus is the ratio of tensile stress to tensile strain.

Explanation:

Radius:

r = 0.2\ \text{mm} = 2 \times 10^{-4}\ \text{m}

.

Extension is

0.5\%

of the original length

L

, so

e = 0.005L

.

Load force:

F = mg = 1.5 \times 10 = 15\ \text{N}

.

Cross-sectional area:

A = \pi r^2 = \frac{22}{7} \times (2 \times 10^{-4})^2 = 1.26 \times 10^{-7}\ \text{m}^2

.

Young's modulus:

Y = \frac{FL}{Ae} = \frac{15L}{A(0.005L)}

.

The

L

cancels:

Y = \frac{15}{A \times 0.005} = \frac{15}{1.26 \times 10^{-7} \times 0.005}

.

This gives

Y = \frac{15}{6.3 \times 10^{-10}} = 2.4 \times 10^{10}\ \text{Nm}^{-2}

.

Answer:

A.

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

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