Question

a cylinder of wood floats vertically in water with one fourth of it's length out of water.find the density of wood
a)0.25gm/cm
b)0.5gm/cm
c)0.75gm/cm
d)1gm/cm​

Answer 1

Answer:

• Density of Cylinder of wood (σ) = 0.75 g/cm³.

Given:

• One fourth of it's length is out of water.

$\bold{Assumptions}\begin{cases}\rho \: \text{be the Density of Water} \\ \sigma \: \text{be the Density of Wood} \\ \text{V be the Volume of the body}\end{cases}$

Explanation:

$\rule{300}{1.5}$

Firstly we need to find the Immersed Volume of the body.

⇒ Total Volume = V.

⇒ Volume of the body outside the Water = V/4.

⇒ Immersed Volume = Total Volume - Volume outside

⇒ V₂ = V - V₁

• V₂ = Immersed Volume.
• V₁ = Volume outside.

Substituting the values,

⇒ V₂ = V - V/4

⇒ V₂ = 4V - V/4

⇒ V₂ = 3V/4.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

As this is a case where,

Density of Wood < Density of Water.

We need to Apply,

$\large{\boxed{\bold{\dfrac{\sigma}{\rho} = \dfrac{\text{Immersed Volume}}{\text{Total Volume}}}}}$

Where,

• ρ = Density of water = 1g/cm³.
• σ = Density of Wood (Cylinder).

$\large{\tt \hookrightarrow \dfrac{\sigma}{\rho} = \dfrac{V_2}{V}}$

Substituting the values,

$\large{\tt \hookrightarrow \dfrac{\sigma}{\rho} = \dfrac{\dfrac{3V}{4}}{V}}$

$\large{\tt \hookrightarrow \dfrac{\sigma}{\rho} = \dfrac{3V}{4V}}$

$\large{\tt \hookrightarrow \dfrac{\sigma}{\rho} = \dfrac{3\cancel{V}}{4\cancel{V}}}$

$\large{\tt \hookrightarrow \dfrac{\sigma}{\rho} = \dfrac{3}{4}}$

• ρ = 1 g/cm³

$\large{\tt \hookrightarrow \dfrac{\sigma}{1} = \dfrac{3}{4}}$

$\large{\tt \hookrightarrow \sigma = \dfrac{3}{4} \times 1}$

$\large{\tt \hookrightarrow \sigma = \dfrac{3}{4}}$

$\huge{\boxed{\boxed{\tt \sigma = 0.75 \: g/cm^3}}}$

∴ Density of Wood (Cylinder) is 0.75 g/cm³.

$\rule{300}{1.5}$

Answer 2

Answer:

Given:

A cylinder of wood is vertically floating in water with one fourth of it's length outside water.

To find:

Density of wood

Concept:

Since ¼ of the length of cylinder is outside , we can also say that ¼ of the volume of the cylinder is outside water.

Hence ¾ part is inside water.

Calculation:

Let density of water be denoted be ρ

and that of wood be σ

The weight of the wood is being balanced by the Buoyant force of water.

∴ Weight = Buoyant force

=> mg = V(inside water) × ρ × g

=>V(total) × σ × g =V(inside water) ×ρ ×g

=> σ/ρ = V(inside water)/V(total)

=> σ/1 = (¾)/1

=> σ = ¾

=> σ = 0.75 g/cm³

So final answer is

Density of wood is 0.75 gm/cm³