The breaking stress for a metal is $7.8 \times 10^{9} N M^{- 2}$ . Calculate the maximum length of the wire made of this metal which may be suspended without breaking. The density of the metal $= 7.8 \times 10^{3} k g m^{- 3}$ . Take $g = 10 N k g^{- 1}$ .

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Solution

We know that stress $= \frac{F o r c e}{A r e a} = \frac{W e i g h t}{A r e a} = \frac{m a s s \times a c c . d u e t o g t a v i t y}{A r e a}$
$S t e s s = \frac{d e n s i t y \times V o l u m e \times g}{A r e a}$
$= \frac{e \times A \times L g}{A}$
$S t r e s s (m a x) = e \times l e n g t h (m a x) \times g$
or $l_{m a x} = \frac{s t r e s s (m a x)}{e \times g} = \frac{7.8 \times 10^{9} k g / m s^{2}}{7.8 \times 10^{3} k g / m^{3} \times 9.8 m / s^{2}}$
$= 1.02 \times 10^{5} m$

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