# A block of wood floats in a liquid of density 0.8g/cm sq. with one fourth of its volume submerged. In oil the block floats with 60% of its volume submerged. Find the density of (a) wood? (b) oil?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

22
dk_ch Share
Nov 3, 2016

Let the volume of the block of wood be $V c {m}^{3}$ and its density be ${d}_{w} g c {m}^{-} 3$

So the weight of the block $= V {d}_{w} g$ dyne, where g is the acceleration due to gravity $= 980 c m {s}^{-} 2$

The block floats in liquid of density $0.8 g c {m}^{-} 3$ with $\frac{1}{4} t h$ of its volume submerged.So the upward buoyant force acting on the block is the weight of displaced liquid$= \frac{1}{4} V \times 0.8 \times g$ dyne.

Hence by cindition of floatation

$V \times {d}_{w} \times g = \frac{1}{4} \times V \times 0.8 \times g$

$\implies {d}_{w} = 0.2 g c {m}^{\text{-3}}$,

Now let the density of oil be ${d}_{o} g c {m}^{\text{-3}}$

The block floats in oil with 60% of its volume submerged.So the buoyant force balancing the weight of the block is the weight of displaced oil = 60%xxVxxd_o xxg dyne

Now applying the condition of floatation we get

60%xxVxxd_o xxg=Vxxd_wxxg

$\implies \frac{60}{100} \times \cancel{V} \times {d}_{o} \times \cancel{g} = \cancel{V} \times 0.2 \times \cancel{g}$

$\implies {d}_{o} = 0.2 \times \frac{10}{6} = \frac{1}{3} = 0.33 g c {m}^{-} 3$

• 4 minutes ago
• 8 minutes ago
• 11 minutes ago
• 12 minutes ago
• A minute ago
• A minute ago
• A minute ago
• A minute ago
• 2 minutes ago
• 2 minutes ago
• 4 minutes ago
• 8 minutes ago
• 11 minutes ago
• 12 minutes ago