Suppose a spherical object is kept in water and nine tenths of the object are inside the water, while the remaining tenth floats. Find the weight of solid inside liquid.

By Archimedes' principle, if an object is immersed in a fluid, it experiences an apparent loss of weight which is equal to the weight of fluid displaced.

That is, if an object floats completely or partially then the weight of the fluid displaced (water in this case) must be equal to the weight of the object. Right?

My teacher told me that the weight in water will be $\frac{9}{10}\times \text{Weight in air}$

But the weight in water should be zero. Isn't it? How is that 9/10 of weight in air? Please explain.

  • $\begingroup$ The weight is $m\times g$, the mass times $g$. It does not depend on the surrounding fluid. $\endgroup$
    – anderstood
    Nov 25 '14 at 22:48
  • $\begingroup$ @anderstood The weight here means the apparent weight or the net vertical force. I was not talking about true weight/ $\endgroup$ Nov 26 '14 at 12:40

The density of water is 1 g/ml. If the volume of the object is 100 ml and the object weighs 90 g, it is less dense than the water (0.9 g/ml) and hence only 90 ml of the object will be submerged (depending in the shape), leaving 10 ml of the object exposed above the water level.


Let $W_a$ and $W_w$ be the weight of the object in air and water.

V be the volume of the object.

As mentioned 9/10 of the object is immersed in water.

$\rho$ be the density of water, Principle of Archimedes states below expression $$ W_a-W_w = 9/10 * V* \rho. $$


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