MCAT Physics Question 47: Answer and Explanation
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Question: 47
2. An anchor made of iron weighs 833 N on the deck of a ship. If the anchor is now suspended in seawater by a massless chain, what is the tension in the chain? (Note: The density of iron is
and the density of seawater is
)
- A. 100 N
- B. 724 N
- C. 833 N
- D. 957 N
Correct Answer: B
Explanation:
The tension in the chain is the difference between the anchor's weight and the buoyant force because the object is in translational equilibrium: T = Fg – Fbuoy. The object's weight is 833 N, and the buoyant force can be found using Archimedes' principle. The magnitude of the buoyant force is equal to the weight of the seawater that the anchor displaces:
Fbuoy = ρwVwg
Because the anchor is submerged entirely, the volume of the water displaced is equal to the volume of the anchor, which is equal to its mass divided by its density
We are not given the anchor's mass, but its value must be the magnitude of the weight of the anchor divided by g. Putting all of this together, we can obtain the buoyant force:
Lastly, we can obtain the tension from the initial equation T = Fg – Fbuoy:
T = 833 N ? 109 N = 724 N