Question

A block of a plastic material floats in water with 35.6% of its volume under water. What is the density of the block in kg/m3

Answer 1

The density is
`\rho_(block) =  356 \ kg/m^3`

Step-by-step explanation:

From the question we are told that

The proportion of the plastic under water is p = 0.356

Generally at equilibrium the buoyant force is equal to the weight of the block i.e

`B =  mg`

Generally the Buoyant force is mathematically represented as

`B  =  \rho  *  V *  g`

Here
`\rho` density of the water with value

`\rho =  1000 kg/m^3` ,

V is the volume of water displaced by the block which is
`p * V_{block` ,

So

`1000 *  0.356 V_(block)  *  g  =  m *  g`

Here m is the mass of the block which is mathematically represented as

`m  =  \rho_(block)  *  V_(block)`

So

`1000 *  0.356 V_(block)  *  g  =   \rho_(block)  *  V_(block)*  g`

`1000 *  0.356 V_(block)  =   \rho_(block)  *  V_(block)*  g`

=>
`\rho_(block) =  1000 *  0.356`

=>
`\rho_(block) =  356 \ kg/m^3`