Question 5.C-A.53: An ice cube, having a stone of 500 g placed on its top, is f...

(i) By the law of floatation,

mass of cube (ice) + mass of stone]g

= [mass of displaced water]g                      (1)

Then, find the mass of the ice cube from (1).

When a stone weighing 500g placed on its top, slips into water, the ice cube rises by \frac{1}{10} th of its length.

Because of this, the mass of the water displaced changes by 500 g.

The change in the volume of the ice cube is equal to \left(a \frac{1}{10}\right)

Then, this value is equal to the change in the volume of the is equal water displaced.

Then \left(a \frac{1}{10} a l\right)=500 cm ^{-3}                  (2)

Then find the volume of the cube (V_{i} ) from (2)

Here, mass of the cube = 4500 g

This means,  V_{ i } \cdot d_{ i }=4500                  (3)

Substitute ‘V_{i} ’  from (2) in (3).

Find the value of d_{i} .

(ii)  0.9 g cm ^{-3}