Let's break down the problem: **Given:** - Outer radius of spherical vessel, \( r_o = 1 \) m - Internal pressure, \( p = 100 \) kPa = \( 100,000 \) Pa - Maximum allowable normal (hoop) stress, \( \sigma_{max} = 24 \) MPa = \( 24,000,000 \) Pa **Formula for hoop stress in a thin-walled spherical pressure vessel:** \[ \sigma = \frac{p r}{2t} \] where: - \( \sigma \) = hoop stress - \( p \) = internal pressure - \( r \) = inner radius (for thin wall, approximately outer radius) - \( t \) = wall thickness Rearrange to solve for \( t \): \[ t = \frac{p r}{2 \sigma} \] **Substitute the values:** - \( p = 100,000 \) Pa - \( r = 1 \) m = 1000 mm - \( \sigma = 24,000,000 \) Pa \[ t = \frac{100,000 \times 1}{2 \times 24,000,000} \] \[ t = \frac{100,000}{48,000,000} \] \[ t = 0.00208 \text{ m} = 2.08 \text{ mm} \] **Closest answer:** - **2.079 mm** So, the correct answer is: > **2.079 mm**