Question

The maximum excess pressure inside a thin-walled steel tube of radius 'r' and thickness $\Delta$r $(<<r)$ so that the tube won't rupture would be (Breaking stress of steel is ${\sigma }_{max}$).

• A. ${\sigma }_{max}\times \frac{r}{\Delta r}$
• B. ${\sigma }_{max}\times \frac{\Delta r}{r}$
• C. ${\sigma }_{max}\times \frac{\Delta r}{2r}$
• D. ${\sigma }_{max}\times \frac{2\Delta r}{r}$

Answer 1

The correct option is B ${\sigma }_{max}\times \frac{\Delta r}{r}$

$\text{excess pressure}=2T\sin \theta$

$\begin{array}{cc}2T\sin \theta & =\Delta P\times A\\ \theta \text{is}\text{small}\Rightarrow & \sin \theta \approx \theta \\ 2T\theta & =△P\times A\end{array}$

$\begin{array}{cc}T& =\Delta P\times lr\\ {\sigma }_{max}& =\frac{\Delta P\times lr}{l\times \Delta r}=\frac{\Delta Pr}{\Delta r}\end{array}$

$\text{excess pressure}\left(\Delta P\right)=\frac{\Delta Pr}{\delta }\times {\sigma }_{max}$

$={\sigma }_{max}\times \frac{\Delta r}{r}$