Question

a block of wood floats in brine of density 1.15 gram per centimetre cube such that two fifth of its volume is above the brine
calculate the density of wood​

Answer 1

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

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A block of wood of mass 5 kg and dimensions 40 cm ×20cm ×10 cm is placed on a table top FIND the pressure exerted if the block lie on the table top with sides of dimensions (g=10 ms minus square ) (a) 40cm ×20cm (b) 40cm×10 cm ?

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Answer 2

Answer:

density of the block  = 0.32 g/cm³

Explanation:

Given,

A block of wood floats in brine solution of density 1.15 g/cm^3 such that 2/5 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 = 2/5 * ( V * 0.8 * 980)

According to Archimedes Principle,

Weight of block = Buoyancy force acting on the block

i.e W = Fb

or, 980Vρ  = 2/5 * ( V * 0.8 * 980)

ρ  = 0.4 * 0.8= 0.32 g/cm³

Therefore, density of block = 0.32 g/cm³