A cylindrical body of diameter 2 m and height 1.5 m having a mass of 2 tonnes and floats in sea water with its axis vertical. A lighting equipment is placed on the body such that the water level is at least 0.5 m below the top surface. What is the maximum mass of the electrical equipment that can be placed on the body. (The specific gravity of sea water may be taken as 1.02 and π = 3.14) 

Correct option is C

​$$\begin{aligned}&{ \ \textbf{Given:}} \\&\text{Diameter } D = 2\,\text{m} \Rightarrow \text{Radius } R = 1\,\text{m} \\&\text{Height of cylinder} = 1.5\,\text{m} \\&\text{Maximum depth of immersion } h = 1.0\,\text{m} \quad (\text{since } 0.5\,\text{m must remain above water}) \\&\text{Mass of cylinder } m_1 = 2\,\text{tonnes} = 2000\,\text{kg} \\&\text{Density of seawater } \rho_{sw} = 1.02 \times 1000 = 1020\,\text{kg/m}^3 \\&\pi = 3.14 \\[1.5em]&{\textbf{Total Buoyant Force}} \\&\text{Buoyant force = weight of displaced water when submerged to } 1.0\,\text{m} \\&V = \pi R^2 h = 3.14 \cdot 1^2 \cdot 1 = 3.14\,\text{m}^3 \\&\text{Buoyant Force} = \rho_{sw} \cdot V \cdot g = 1020 \cdot 3.14 \cdot 9.81 \approx 31412.27\,\text{N} \\[1.5em]&{\textbf{Weight of Cylinder}} \\&W_1 = m_1 \cdot g = 2000 \cdot 9.81 = 19620\,\text{N} \\[1.5em]&{\textbf{Remaining Capacity for Equipment}} \\&\text{Remaining force} = 31412.27 - 19620 = 11792.27\,\text{N} \\&\Rightarrow m_{\text{equipment}} = \frac{11792.27}{9.81} \approx \boxed{1202\,\text{kg}}\end{aligned}$$​​