Question

a block of wood of length 40cm and area of cross section 15cm3, floats in water with 3/8of its length above water.

Answer 1

Solution:

A block of wood floats in water with 3/8 of its volume above water

Now,it floats in water with its 5/8 part submerged,

Let,the density of the block be D

Now,as it floats in water with its 5/8 part submerged,it means, its weight is being balanced by the buoyant force of water on the /8rd of its total volume.

So,from equation of force,we can write,

v⋅d⋅g=5/8 ×v×1×g (as,density of water is 1g/cm3)

So, density D  =5/8  g/ cm^3 =  0.625 g/ cm^3  ANS

Answer 2

Answer:

Density = 0.625 g/cm³

Step-by-step explanation:

Length of block submerged under water = 1 - 3/8 = 5/8

So, weight of by 5/8 part of wood is equal to upthrust, so

Water Displaced = Volume of Wood * Density of water * g

=> (40*15) * 1 * g

= 600 * g = 600 cm³

Mass of wood = 5/8 * Volume of wood (40*15)

=> 5/8 * 600 = 5 * 75 = 375 gram

Density = Mass/Volume

=> 375/600 = 0.625 g/cm³

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