Question

A block of wood floats in brine solution of density 1.15 g/cm3 such that 3/5th of its volume immersed into the brine. Calculate the density of wood.

Answer 1

Answer:

Explanation:

The density of the block is d  = 0.32 g/cm³

Explanation:

Let the volume of wood = V cm³

Density of wood =  d g/cm³

Weight of block of wood = Vdg = V × d × 980 cm/s²]

The block floats in a liquid of density 1.15 g/cm³ with 2/5th of its volume submerged.

Therefore, upward buoyancy force acting on the block is the weight displaced of liquid = 2/5 ( V × density of liquid × g)

= 2/5 ( V × 0.8 × 980)

Now apply the condition:

weight of block = buoyancy force acting on the block

V × d × 980 = 2/5 × ( V × 0.8 × 980)

d  = 0.4 x 0.8

d  = 0.32 g/cm³

Thus the density of the block is d  = 0.32 g/cm³

Answer 2

Answer:

density of the block  = 0.48 g/cm³

Explanation:

Given,

A block of wood floats in brine solution of density 1.15 g/cm^3 such that 3/5 th of its volume immersed into the brine.

To find: the density of wood.

Solution:

Let density of wood be ρ g/cm³ and volume be V cm³.

Weight of block of wood = Vρg = 980 Vρ cm/s²

Buoyancy force acting on the block = 3/5 * ( V * 0.8 * 980)

According to Archimedes Principle,

Weight of block = Buoyancy force acting on the block

i.e W = Fb

or, 980Vρ  = 3/5 * ( V * 0.8 * 980)

ρ  = 0.6 * 0.8= 0.48 g/cm³

Therefore, density of block = 0.48 g/cm³