Question

a block of wood floats in water with 2/5th of its volume above the surface ,calculate the density of wood

Answer 1

Solution:

A block of wood floats in water with 2/5 of its volume submerged it means 3/5th of it's volume is above the water.

Let v be the volume of the object. And p be its density. So, weight of the object, w = vpg

Volume of water displaced = v -3/5v = 2/5 v

Density of water = 1 g/cc

Weight of water displaced = (2/5 v) × 1 × g

By the law of floatation,

Weight of the object = weight of water displaced

= > vpg = 2/5vg

= > p = 2/5 g/cc

Thus, the relative density of the object is = (2/5)/1 = 2/5

Answer 2

Explanation:

we know that , density of water =

${10}^{3} kg/ {m}^{3}$

volume of Dipped wood portion =

$\frac{2}{5}$

of wood volume

To find density of wood :-

Let volume of wood =

$v0$

Volume (v) of dipped wood =

$\frac{2}{5} v0$

$density = \frac{mass}{volume}$

$mg = pvg$

$v0g \: = \frac{2}{5} v0g$

$= \frac{2}{5} p = \frac{2}{5} \times 1000 = 400$

density of wood =

$400kg/ {m}^{3}$