Let's solve each question one by one: --- ## **Question 1: Metacentric Height Calculation** **Given:** - Length of tank, \( l = 6 \) m - Width of tank, \( b = 4 \) m - Height of tank, \( h = 2 \) m - Depth of immersion, \( d = 1.6 \) m - Center of gravity above bottom = 1 m **Metacentric Height, \( GM \):** \[ GM = BM - BG \] Where: - \( BM = \frac{I}{V} \) - \( BG = \) Distance between center of gravity and center of buoyancy **1. Moment of inertia about longitudinal axis:** \[ I = \frac{b l^3}{12} \text{ (about width)} \Rightarrow \frac{l b^3}{12} \text{ (about length)} \] But for breadth: \[ I = \frac{l b^3}{12} \] So, \[ I = \frac{6 \times 4^3}{12} = \frac{6 \times 64}{12} = \frac{384}{12} = 32 \text{ m}^4 \] **2. Volume of displaced liquid:** \[ V = l \times b \times d = 6 \times 4 \times 1.6 = 38.4 \text{ m}^3 \] **3. \( BM \):** \[ BM = \frac{I}{V} = \frac{32}{38.4} = 0.833 \text{ m} \] **4. \( BG \):** - Center of gravity above bottom = 1 m - Center of buoyancy (center of immersed part) = \( \frac{d}{2} = 0.8 \) m above bottom So, \[ BG = 1 - 0.8 = 0.2 \text{ m} \] **5. \( GM \):** \[ GM = BM - BG = 0.833 - 0.2 = 0.633 \text{ m} \] **Nearest option:** **0.63 m** --- ## **Question 2: Hole-shaft Fit** - **Where the hole is larger than the shaft, it is called:** - **Clearance fit** --- ## **Question 3: Gun Barrel Production** - **Gun barrels are produced by:** - **Tube drawing** --- ### **Summary Table** | Q# | Correct Option | |----|------------------------| | 1 | 0.63 m | | 2 | Clearance | | 3 | Tube drawing | Let me know if you need detailed explanations for any question!