Question

Calculate the work done in raising a stone of mass 6 kg of specific gravity 2, immersed in water from a depth of 4m to 1m below the surface of water $(g=10m{s}^{-2})$.

Answer 1

relative density is the ratio of density of object to the density of medium. Here medium is water.

$RD=\frac{\text{density of stone}}{\text{density of water}}=2$

Stone is Two times denser than water.

According to Archimedes principle of floatation, it will sink at the bottom.

But its Apparent weight inside water will be = Actual weight - Buoyant force

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Actual weight = volume of stone x density of stone x g = mg

Buoyant force = Weight of water displaced by the stone = volume of stone x density of water x g

density of water $=\frac{\text{density of stone}}{2}$

So buoyant force = volume of stone x ($\frac{1}{2}$ density of stone) x g $=\frac{1}{2}\times$ actual weight of stone $=\frac{mg}{2}$

Therefore apparent weight of stone in water $=mg-\frac{mg}{2}=\frac{mg}{2}$

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Work done to raise it throught height h = apparent weight x h

$W=\frac{mgh}{2}=\frac{6\times 10\times (4-1)}{2}=90J$