Let's solve each part step-by-step: --- ## **Given Data** - **Density of freshwater,** \(\rho = 100 \, \text{kg/m}^3\) - **Length of barge,** \(L = 12 \, \text{m}\) - **Breadth of barge,** \(B = 5 \, \text{m}\) - **Depth of barge,** \(d = 3 \, \text{m}\) - **Draft (submerged depth),** \(h = 1.2 \, \text{m}\) - **Centre of gravity above base,** \(G = 2.2 \, \text{m}\) --- ## **(i) Metacentric Radius (BM) when Axis of Rotation is Parallel to Length** ### **Formula:** \[ BM = \frac{I}{V} \] where - \(I\) = second moment of area (moment of inertia) of waterplane area about the axis - \(V\) = submerged volume #### **Step 1: Submerged Volume** \[ V = \text{Length} \times \text{Breadth} \times \text{Draft} = L \times B \times h = 12 \times 5 \times 1.2 = 72 \, \text{m}^3 \] #### **Step 2: Waterplane Area Moment of Inertia (Axis parallel to length)** - Axis parallel to length means rotation about breadth (transverse axis) \[ I_{breadth} = \frac{L B^3}{12} = \frac{12 \times (5)^3}{12} = \frac{12 \times 125}{12} = 125 \, \text{m}^4 \] #### **Step 3: Metacentric Radius (BM)** \[ BM_{breadth} = \frac{I_{breadth}}{V} = \frac{125}{72} = 1.736 \, \text{m} \] --- ## **(ii) Metacentric Radius (BM) when Axis of Rotation is Parallel to Breadth** - Axis parallel to breadth means rotation about length (longitudinal axis) \[ I_{length} = \frac{B L^3}{12} = \frac{5 \times (12)^3}{12} = \frac{5 \times 1728}{12} = \frac{864}{12} = 720 \, \text{m}^4 \] \[ BM_{length} = \frac{I_{length}}{V} = \frac{720}{72} = 10. \, \text{m} \] --- ## **(iii) Metacentric Height (GM) for Both Axes** \[ GM = BM - BG \] Where \(BG\) = distance from center of buoyancy to center of gravity #### **Find Center of Buoyancy (B):** - Center of Buoyancy is at half the draft from the base \[ \text{Depth of Center of Buoyancy from base} = \frac{h}{2} = \frac{1.2}{2} = .6 \, \text{m} \] - Distance \(BG = G - B = 2.2 - .6 = 1.6 \, \text{m}\) #### **Metacentric Height (GM):** - **Transverse (Breadth):** \[ GM_{breadth} = BM_{breadth} - BG = 1.736 - 1.6 = .136 \, \text{m} \] - **Longitudinal (Length):** \[ GM_{length} = BM_{length} - BG = 10. - 1.6 = 8.4 \, \text{m} \] --- ## **(iv) Stability of the Barge** - The barge is **stable** if \(GM > \). - Both \(GM_{breadth}\) and \(GM_{length}\) are positive (**.136 m** and **8.4 m**), so the barge is **stable in its current configuration**. --- # **Summary Table** | Quantity | Value | |-----------------------------|------------------------| | BM (axis parallel to length) | \(1.736 \, \text{m}\) | | BM (axis parallel to breadth)| \(10. \, \text{m}\) | | GM (breadthwise axis) | \(.136 \, \text{m}\) | | GM (lengthwise axis) | \(8.4 \, \text{m}\) | | Stability | **Stable** | --- ### **Final Answers** **(i)** Metacentric radius (axis parallel to length): **1.736 m** **(ii)** Metacentric radius (axis parallel to breadth): **10. m** **(iii)** Metacentric height (breadthwise axis): **.136 m** **(iii)** Metacentric height (lengthwise axis): **8.4 m** **(iv)** The barge is **stable** in its current configuration.