Question 323: Properties of Matter

Question

The maximum elongation of a steel wire of $$1 \mathrm{~m}$$ length if the elastic limit of steel and its Young's modulus, respectively, are $$8 	imes 10^8 \mathrm{~N} \mathrm{~m}^{-2}$$ and $$2 	imes 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$$, is:

Accepted Answer

First, we need to find the maximum force that can be applied to the steel wire within its elastic limit. This force can be calculated using the given area under stress and the stress limit provided by the elastic limit of steel.
Let's assume the area of cross-section of the wire is $$A$$. The force exerted can be given by:
$$ F = \sigma 	imes A $$
where:
$$ \sigma = 8 	imes 10^8 \mathrm{~N/m}^2 $$ (Elastic limit of steel)
To calculate the elongation ($$ \Delta L $$) under this force, we use Hooke's Law, which relates force, elongation, cross-sectional area, original length, and Young's modulus as follows:
$$ F = \frac{Y 	imes A 	imes \Delta L}{L} $$
where:
$$ Y = 2 	imes 10^{11} \mathrm{~N/m}^2 $$ (Young's modulus of steel),
$$ L = 1 \mathrm{~m} $$ (original length of the wire).
Substituting for $$ F $$ from the earlier expression and rearranging the formula, we get:
$$ \sigma 	imes A = \frac{Y 	imes A 	imes \Delta L}{L} $$
$$ \sigma = \frac{Y 	imes \Delta L}{L} $$
$$ \Delta L = \frac{\sigma 	imes L}{Y} $$
$$ \Delta L = \frac{8 	imes 10^8 	imes 1}{2 	imes 10^{11}} \mathrm{~meters} $$
$$ \Delta L = 0.004 \mathrm{~meters} $$
$$ \Delta L = 4 \mathrm{~mm} $$
Thus, the maximum elongation of the wire within the elastic limit is $$4 \mathrm{~mm}$$.
The answer is: Option A - 4 mm.

Answer

4 mm

Answer

0.4 mm

Answer

40 mm

Answer

8 mm

Properties of Matter

The maximum elongation of a steel wire of $1 \mathrm{~m} length if the elastic limit of steel and its Young's modulus, respectively, are 8 \times 10^8 \mathrm{~N} \mathrm{~m}^{-2} and 2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$, is:

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