1. **Problem Statement:** Calculate the polar moment of inertia $J$ for a hollow circular tube with outer diameter $D_o = 2$ in, inner diameter $D_i = 1.5$ in. 2. **Formula:** The polar moment of inertia for a hollow circular shaft is given by: $$J = \frac{\pi}{32} \left(D_o^4 - D_i^4\right)$$ 3. **Explanation:** - $D_o$ is the outer diameter. - $D_i$ is the inner diameter. - The formula subtracts the inner hollow part's moment from the outer solid cylinder's moment. 4. **Calculation:** Calculate $D_o^4$ and $D_i^4$: $$D_o^4 = 2^4 = 16$$ $$D_i^4 = 1.5^4 = (1.5)^4 = 5.0625$$ 5. Substitute values into the formula: $$J = \frac{\pi}{32} (16 - 5.0625) = \frac{\pi}{32} \times 10.9375$$ 6. Simplify: $$J = 0.09817477 \times 10.9375 = 1.074 \text{ in}^4$$ 7. **Answer:** The polar moment of inertia $J$ is approximately $1.074$ in$^4$. This corresponds to option B.