A) 2
B) \[\frac{1}{2}\]
C) 4
D) \[\frac{1}{4}\]
Correct Answer: A
Solution :
\[{{({{\sigma }_{\theta }})}_{cyl}}=\frac{pd}{2t}\] \[{{({{\sigma }_{\theta }})}_{cyl}}=\frac{pd}{4t}\] \[\frac{{{({{\sigma }_{\theta }})}_{cyl}}}{{{({{\sigma }_{\theta }})}_{sph}}}=2\]
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