Question

This is the 4 th time I'm asking this question..Please answer part b..Please... Please......

Answer 1

Given values

$F_{1}=4\times10^5N$

$F_{2}=?$

$d_{1}=4 mm$

$d_{2}=2 mm$

We know that :

$\frac{F_{1}}{F_{2}}=\frac{d_{1}^2}{d_{2}^2}$

$\implies \frac{4\times10^5N}{F_{2}}=\frac{(4 mm)^2}{(2 mm)^2}$

$\implies F_{2}=\frac{4 mm^2\times4\times10^5N}{16 mm^2}$

$\implies F_{2}=10^5N$

The breaking stress is    $\boxed{10^5N}$

Hope it helps you :-)

Happy new year ,merry Christmas in advance.

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