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Question

A beaker containing water is placed on the platform of a spring balance. The balance reads 1.5 kg. A stone of mass 0.5 kg and density 500 kg/m3 is immersed in water without touching the walls of the beaker. What will be the reading of spring balance now?
  1. 2 kg
  2. 2.5 kg
  3. 1 kg
  4. 3 kg


Solution

The correct option is B 2.5 kg
Volume of stone =mρ=0.5500=103 m3
Force of Buoyancy =ρVg=1000×103×10=10 N

There will be a reaction force to the buoyant force, by the stone on the water.
Therefore, net force on spring balance is sum of force of buoyancy and weight of water and beaker =FB+mg=10+1.5g=10+15=25 N
Net reading of spring balance=2.5 kg

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