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

Options

A. `sigma_("max")xxr/(/_\r)`
B. `sigma_("max")xx(/_\r)/r`
C. `sigma_("max")`
D. `sigma_("max")xx(/_\2r)/r`

Answer

B

Explanation

Consider a small element of the tube.
`2Tsintheta=/_\pxxA`
Where `/_\p=p_(i)-p_(0)` and `A` is the area of element. As `theta` is very small `sintheta~~theta` so, `2T xxtheta=/_\pxxlxx(2rtheta)`

`implies/_\p=T(lr)sigma` (stress developed in tube) `=(/_\t)/(/_\rxxl)`
where `/_\rxxi` is the cross sectional area
`sigma=(/_\pxxlr)/(/_\rxxl)=/-\pxxr/(/_\r)`
so, `/_\pxxr/(/_\r)lesigma_("max")`
`/_\p("max,value")=sigma_("max")xx(/_\t)/r`