Question
Two wires as shown in the figure below, made of steel and have breaking stress of $12 \times 10^{8}\text{ }N/m^{2}$. Area of cross-section of upper wire is $0.008{\text{ }cm}^{2}$ and of lower wire is $0.004{\text{ }cm}^{2}$. The maximum mass that can be added to pan without breaking any wire is $\_\_\_\_$ $kg.\left( \right.\ $ take $\left. \ g = 10\text{ }m/s^{2} \right)$
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Solution
Sol. [Key Concept: Stress $= \frac{load}{area}$; the upper wire (carrying $30 + 10 + m$) reaches breaking stress first — set its stress equal to the limit.]
$${T_{1} = mg }{T_{2} = (10 + m)g }{T_{3} = (30 + 10 + m)g = g(40 + m)g }$$Stress in upper wire
$${= \frac{(40 + m)g}{0.008{\text{ }cm}^{2}} }{= \frac{(40 + m) \times 10}{8 \times 10^{- 7}} }$$Stress in lower wire $=$
$${= \frac{(10 + m)g}{0.004{\text{ }cm}^{2}} }{= \frac{(10 + m) \times 10}{4 \times 10^{- 7}} }$$Upper wire breakes :
$$\frac{(40 + m) \times 10}{8 \times 10^{- 7}} = 12 \times 10^{8} $$So, maximum mass, without breaking any wire is 38 kg
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