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Solution
Answer: B) Option 2
The moment of inertia for a hollow circular section can be calculated based on the difference of moment of inertia of two circular sections:
Inner Radius:
\[R_i=6 in\]
Outer Radius:
\[R_o=6\ in+6.5 \ in=6.5 \ in\]
Outer Section Moment of Inertia:
\[I_o=\frac{πR_o^4}{4}= \frac{π×(6.5 in)^4}4=1401.98 \ in^4\]
Inner Section Moment of Inertia:
\[I_i= \frac{πR_i^4}{4}= \frac{π×(6 in)^4}4=1017.88 \ in^4\]
Moment of Inertia of Hollow Section:
\[I_{circular}=I_o-I_i=384.10 \ in^4\]
The moment of inertia for a box section can be calculated based on the difference of moment of inertia of two rectangular sections:
Outer Section Moment of Inertia:
\[I_o = \frac{b_o h_o^3}{12} = \frac{(13 \ in)^4}{12}=2380 \ in^4\]
Inner Section Moment of Inertia:
\[I_i= \frac{b_o h_o^3}{12}= \frac{(12 in)^4}{12}=1728 \ in^4\]
Moment of Inertia of Tube Section:
\[I_{box}=I_o-I_i=652 \ in^4\]
\[I_{box}>I_{circular}\]
Therefore, the box section will minimize deflection.
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