Question 5.C-A.65: An ice cube has a stone of 500 g placed on its top, is float...

Let volume of the ice cube be  =\ell^{3} cm ^{3}

Let its density be  =d_{i}  g  cm ^{-3}

Weight of the floating body = weight of the displaced liquid

⇒ (500 + m)g = 5000 g

m = 5000 – 500

m = 4500 g                        (1)

Now, when stone slips and falls into water.

⇒       500 g =\ell^{2} \cdot \frac{1}{10} \ell 1 . g

\ell^{3}=5000 cm ^{3}                           (2)

∴  The density of the ice cube