Maximum excess pressure inside a thin-walled steel tube of radius r and thickness `/_\r (ltlt r)`, so that the tube would not rupture would be (breaking stress of steel is `sigma_("max")`

Options

A. `sigma_("max") xx (r)/(Delta r)`
B. `sigma_("max") xx (Delta r)/(r)`
C. `sigma_("max")`
D. `sigma_("max") xx (2Delta r)/(r)`

Answer

B

Explanation

Consider a small linerar element fo the tube, which subtends an angle `2theta` at centre as shwon in figure.
`2T sin theta = Delta P xx A` where `Delta P = P_(1) - P_(0)` and `A` is the area of element.
As `theta` is very small, `sin theta approx theta`
So, `2T xx theta = Delta P xx l xx (2r theta)`
`rArr Delta P = (T)/(lr)`
`sigma` (Stress develop in tube) `= (T)/(Delta r xx l)` where `Delta r xx` is `CSA`
`sigma = (Delta P xx l r)/(Delta r xx L) = (Delta P xx r)/(Delta r) L`
For no rupturing, `sigma = sigma_("maximum")`
SO `Delta P xx (r)/(Delta r) le sigma_("maximum")`
`Delta P|` (maximum value)` ( = sigma_("max imum") xx (Delta r)/(r))`