A cube of iron of edge `5 cm` floats on the surface of mercury, contained in a tank. Water is then poured on top, so that the cube just gets immersed. Find the depth of water layer. (Specific gravities of iron-and mercury are `7.8` and `13.6`, respectively.)

Answer

`2cm`

Explanation

Let the depth of water layer be `h_(1) cm` So, the depth to which the cube is immersed in mercury will be as follows:

Weight of water displaced `=(5^(2)h_(1)g)` dyn and weight of mercury displaced `=5^(2)(5-h_(1))13.6gdyn.`
Therefore, upthrust offered by water and mercury together is given by `5^(2)h_(1)g+5^(2)(5-h_(1))13.6g`
`=5^(2)(68-12.6h_(1))g dyn`
Weight of cube `=5^(3)xx8.56g dyn`
From the principle of floatation we have
Upthrust `=` weight of the floating body
`:. 5^(2)(68-12.6h_(1))g=5^(3)xx8.56g`
`implies 68-12.6h_(1)=41.8`
`implies h_(1)=25.2/(12.6)=2cm`