A beaker when partly filled with water has total mass 20g. If a piece of metal with density 3g/cm3and volume 1cm3is suspended by a thin string,so that it is submerged in the water but does not rest on the bottom of the beaker,how much does the beaker then appear to weigh if it is resting on a scale?

Dear Sohan,

From Archimedes' principle the upward buoyant force on the metal is 1g and the metal is in equilibrium in the middle of the fluid. Therefore this force,i.e., 1g+20g acts on the bottom of the beaker. Thus, the beaker will appear to weigh 21g. In addition the tension in the string is 3g-1g=2g

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