Question

What is the depth to which the block is submerged?

A rectangular block of base dimensions `20cm*20cm` and weighs 50 Newtons, floats in water. Calculate the depth to which it is submerged given that the density of water is `1000kgm^-3`.

Answer 1

Let the depth to which the block is submerged be `x` cm.

The base area of the block `=20xx20cm^2`

So volume of the block submerged will be `V=20xx20xx xcm^3`

`=400x xx10^-6m^3`

So weight of water displaced by the block
`W=Vm^3xx1000kgm^-3xxg` N, where acceleration due to gravity `g=10m"/"s^2`

Given that the weight of the block is 50 N ,

By condition of floatation we get

`Vxx1000xxg=50`

`=>400x xx10^-6xx1000xx10=50`

`=>x=50/4=12.5cm`