A thin-walled spherical gas balloon of radius R and wall thickness t (t ≪ R) is subjected to an internal (gauge) pressure p. The maximum tensile and shear stresses in the balloon wall are, respectively:

The correct option is pR/2t and pR/4t.

To understand why this option is correct, let us analyze the stresses developed in a thin-walled spherical pressure vessel subjected to an internal gauge pressure. Let $R$ be the radius of the sphere, $t$ be the wall thickness, and $p$ be the internal gauge pressure.

1. Maximum Tensile Stress (Normal Stress)
For a thin-walled spherical shell under internal pressure, the normal stress is equal in all tangential directions due to symmetry. This stress is commonly referred to as the hoop stress or spherical stress (${\sigma }_s$).
We can find this stress by considering a force balance on a hemispherical section of the balloon:
The bursting force due to internal pressure acting on the projected area is:
$F_{\text{burst}}=p\cdot \pi R^2$
The resisting force developed by the tensile stress in the wall material acting along the circumference is:
$F_{\text{resisting}}={\sigma }_s\cdot (2\pi R\cdot t)$
Equating these forces for equilibrium:
$p\cdot \pi R^2={\sigma }_s\cdot 2\pi Rt$
Solving for ${\sigma }_s$ gives the normal (tensile) stress:
${\sigma }_s=\frac{pR}{2t}$
Because the vessel is thin-walled, the radial stress is negligible compared to the in-plane stresses. Thus, the principal stresses in the plane of the wall are:
${\sigma }_1={\sigma }_2=\frac{pR}{2t}$
Therefore, the maximum tensile stress in the balloon wall is:
${\sigma }_{\text{max}}=\frac{pR}{2t}$

2. Maximum Shear Stress
The maximum out-of-plane shear stress (${\tau }_{\text{max}}$) is determined using Mohr's circle, considering the three-dimensional state of stress. The three principal stresses are:
${\sigma }_1=\frac{pR}{2t}$
${\sigma }_2=\frac{pR}{2t}$
${\sigma }_3\approx 0$ (since the radial stress on the outer surface is zero, and $t\ll R$)
The maximum shear stress is given by the formula:
${\tau }_{\text{max}}=\frac{{\sigma }_{\text{max}}-{\sigma }_{\text{min}}}{2}=\frac{{\sigma }_1-{\sigma }_3}{2}$
Substituting the principal stress values:
${\tau }_{\text{max}}=\frac{\frac{pR}{2t}-0}{2}=\frac{pR}{4t}$

Thus, the maximum tensile stress is pR/2t and the maximum shear stress is pR/4t.