Question

Maximum excess pressure inside a thin-walled steel tube of radius $r$ and thickness $\Delta r(<<r)$, so that the tube would not 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. ${\sigma }_{\max }$
• D. ${\sigma }_{\max }\times \frac{\Delta 2r}{r}$

Answer 1

Answer: (B)

Consider a small element of the tube.
$2T\sin \theta =\Delta p\times A$
where $\Delta p=p_i-p_0$ and $A$ is the area of element.
As $\theta$ is very small, $\sin \theta \approx \theta$
so, $2T\times \theta =\Delta p\times l\times (2r\theta )$

$\Rightarrow \Delta p=\frac{T}{lr}\sigma$ (stress developed in tube)
$=\frac{\Delta T}{\Delta r\times l}$ where $\Delta r\times i$ is the cross-sectional area.
$\sigma =\frac{\Delta p\times lr}{\Delta r\times l}=\Delta p\times \frac{r}{\Delta r}$
For no rupturing, $\sigma \le {\sigma }_{\max }$
So, $\Delta p\times \frac{r}{\Delta r}\le {\sigma }_{\max }$
$\Delta p(\max ,\text{ value })={\sigma }_{\max }\times \frac{\Delta r}{r}$