A spherical object of 1·45 m diameter is completely immersed in a water reservoir and chained to the bottom. If the chain has a tension of 5·2 kN, the weight of the object when it is taken out of the reservoir into the air will be nearly
- A15·5 kN
- B12·5 kN
- C10·5 kNCorrect
- D7·5 kN
Solution
Concept
Equilibrium of a submerged chained body
Formula
T + W = FB
Step-by-step solution
1. Volume = (π/6) × 1.45³ = 1.596 m³. 2. Buoyant force = 9.81 × 1.596 = 15.66 kN. 3. W = FB − T = 15.66 − 5.2 = 10.46 kN ≈ 10.5 kN.
Explanation
The object is held down by its weight and the chain tension against the buoyancy. Its weight in air is the buoyant force minus the tension.
Why the other options are wrong
15.5 kN is nearly the buoyancy, and the others do not follow from FB − T.
Shortcut
W = ρ g V − T.
Common mistake
Adding the tension to the buoyancy.
Correct answer: C) 10·5 kN