Question

To hide their loot, thieves put the 1.20 kg gold that they had robbed into a small box and threw it in a lake. The mass of the empty box was 400 grams, and the dimension 12cm3 . When the box is closed with the gold inside it, it is well seated so that water dont enter it. To their surprise, the box floated on the water. Calculate the distance from the bottom of the box to the water level (i.e. how many cm is it inside the water?)

Density: water = 1 g/cm3 = 100 kg/m3 ----------- Gold = 19.3 g/cm3 19,300 kg/m3

Answer 1

11.1 cm

Step-by-step explanation:

Let d be the depth upto which box dipped into water.

Volume of water displaced

.12\times .12\times d m^3\\

=.0144d

Weight of water displaced = upthrust

=.0144d\times 10^3\times g\\ [ density of water is 1000kg /m³]

=14.4dg This will act upwards.

Total downward force = 1.2 +.4 = 1.6 g

For equilibrium

1.6g = 14.4 gd

d = .111 m

=11.1 cm.