A solid block of wood of density 0.6 gcm⁻³ weighs 3.06 N in air. Determine 

(a) the volume of the block 

(b) the volume of the block immersed when placed freely in a liquid of density 0.9 gcm⁻³? 

Ans. (510 cm³, 340 cm³)


Given:  

Density of the wooden block  = 𝜌 = 0.6 gcm⁻³ =  0.6 ｘ1o³ Kgm⁻³
Weight of the wooden block in air = w = 3.06 N
Gravitational Acceleration  = g = 10 m s⁻²

Mass of the Wooden Block = m = `\frac {w}{g}` = `\frac {3.06 N}{10 m s⁻²}` = 0.306 kg

Density of liquid =  𝞺ₗ  = 0.9 gcm⁻³ =  0.9 ｘ1o³ kg m⁻³


To Find:

The volume of the wooden solid block in air = V₁ = ? 
The volume of the wooden solid block when immersed in liquid = V₂ = ?

Solution:   

As Density is defined as mass per unit volume. So,

Density = `\frac {Mass}{Volume}`

or

Volume = `\frac {Mass}{Density}`

V₁ = `\frac {m}{⍴}`

 by putting values

V₁ = `\frac {0.306 kg}{0.6 ｘ1o³ kg m⁻³}`

V₁ = 0.51 ｘ1o⁻³ m³

V₁ = 510 ｘ1o⁻³ｘ1o⁻³ m³

V₁ = 510 ｘ1o⁻⁶ m³ 

V₁ = 510 cm³     ( 1o⁻⁶ m³ = 1 cm³ )

Thus, the volume of the wooden block in the air  is 510 cm³

Now as we have the formula for upward thrust on the block immersed in liquid, that is 

w = 𝞺ₗ V₂ g     (F = w )

or 

V₂ = `\frac {w}{⍴ₗ g}`

by putting values 

V₂ = `\frac {3.06 N}{0.9 ｘ1o³ kg m⁻³ｘ 10 m s⁻²}`

V₂ = 0.00034 m³ 

V₂ = 340 ｘ1o⁻⁶ m³  

V₁ = 340 cm³     ( 1o⁻⁶ m³ = 1 cm³ )

Thus, the volume of the wooden block when immersed in water  is 340 cm³

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