Question

A tank of base area 4 m² is initially filled with water up to height 2 m. An object of uniform cross-section 2 m² and height 1 m is now suspended by wire into the tank, keeping distance between base of tank and that of object 1 m. Density of the object is 2000kg/m³. Take atmospheric pressure 1 × 10⁵N/m²; g = 10 m/s²

The buoyant force on the object is :

• A. (a) 0.1 × 10⁵N
• A. (b) 0.2 × 10⁵N
• A. (c) 0.3 × 10⁵N
• A. (d) 0.4 × 10⁵N

Answer 1

(b)

F_b = V.ρ _wg

= 2 × 1000 × 10 =0.2 × 10⁵N

It is also equal to net contact force by the liquid= P₂A - P.,A

= 0.2 × 10⁵N

Note: Net contact force and byoyant force are same